Solution

Solutions to Problems in Fundamentals of Structural Mechanics An instructor’s manual to accompany the book Fundamentals of Structural Mechanics

Keith D. Hjelmstad

University of Illinois at Urbana-Champaign

Solutions to Problems in Fundamentals of Structural Mechanics An instructor’s manual to accompany the book Fundamentals of Structural Mechanics by Keith D. Hjelmstad

Keith D. Hjelmstad

University of Illinois at Urbana-Champaign

With assistance from Ghadir Haikal, Dan Turner, and Cara (Liverant) Phillips

Solutions to Problems in Fundamentals of Structural Mechanics An instructor’s manual to accompany the book Fundamentals of Structural Mechanics by Keith D. Hjelmstad by Keith D. Hjelmstad University of Illinois at Urbana-Champaign With assistance from Ghadir Haikal, Dan Turner, and Cara (Liverant) Phillips

E

2007 by Keith D. Hjelmstad All rights reserved. No part of this document may be translated or reproduced in any form without the written permission of the author.

Contents

Preface

vii

1

Vectors and Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

The Geometry of Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39

3

The Transmission of Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

4

Linear Elastic Constitutive Theory . . . . . . . . . . . . . . . . . . . . . . . . .

95

5

Boundary Value Problems in Elasticity . . . . . . . . . . . . . . . . . . . . . 129

6

The Ritz Method of Approximation . . . . . . . . . . . . . . . . . . . . . . . . 141

7

The Linear Theory of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

8

The Linear Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

9

Energy Principles and Static Stability . . . . . . . . . . . . . . . . . . . . . . . 263

10 Fundamental Concepts in Static Stability . . . . . . . . . . . . . . . . . . . . 285 11 The Planar Buckling of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 12 Numerical Computation for Nonlinear Problems . . . . . . . . . . . . . . 359 Appendix A. Fortran Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

Preface

A significant part of the book Fundamentals of Structural Mechanics lives in the problems at the back of each chapter. No student can hope to master the material covered in the book without solving at least some of these problems. An instructor who uses the book as a text for a course on structural mechanics may want to use these problems as a source of examples to illuminate points in the text or as a source of homework problems for the student to work out. This manual should relieve some of the burden of preparation of course materials. This manual provides detailed solutions to almost all of the 302 problems in Fundamentals of Structural Mechanics 2e. The style in which the solutions are presented is similar to the examples presented in the book, with narrative to help get from one step to the next as well as some discussion of the results. The notation agrees with that used in the book. In solving the problems it became quite apparent that a program like MATHEMATICAT is indispensable in carrying out some of the more tedious computations. For those problems that we used MATHEMATICAT we have provided a listing of the script. Most of these are fairly short, but might be useful to someone who is not very familiar with MATHEMATICAT.

The problems in Chapter 12, in almost every case, involve modifying the programs presented in the book. When I revised the book for the second edition I converted all fo the programs from FORTRAN to MATLAB. However, I have not found time to change the problem solutions. Therefore, those are still in FORTRAN. I have included the FORTRAN versions from the first edition of the book at the back so that you can follow the solutions presented. You can get the FORTRAN source, for the programs in the text as well as the programs in the solutions manual, from me. Make your request by e-mail to: [email protected]. I will send you the files.

viii

Preface

We have tried very hard to get the answers right and to get them into this book with the highest possible fidelity. Alas, I am sure some errors remain. Please forgive those errors. I would appreciate learning of the errors that you discover so that I can eliminate them in future editions. Putting this manual together was hard work, the burden of which I gladly shared with Ghadir Haikal, Cara (Liverant) Phillips, and Dan Turner. Ghadir managed to locate and organize the solutions to quiz problems that I had collected over the years. She then completed Chapters 1 through 5 while I did 6 through 12. The problems from the first edition were available to us as were the problems that Cara had typset from old hand-written examination solutions a few years ago. Dan proofread a large portion of the document. I would also like to (re)acknowledge the contributors to the solutions manual for the first edition: Ertugrul Taciroglu, Jiwon Kim, Eric Williamson, and Quihai (Ken) Zuo. The problems they worked out are still correct! I am pleased to acknowledge the contributions of Ghadir, Dan, Cara, Ertugrul, Jiwon, Eric and Ken to the preparation of this manual. Keith D. Hjelmstad

Chapter 1 Vectors and Tensors

1. Compute the values of the following expressions (a) δii (b) δij δij (c) Cij δik δ jk (d) δab δbc δ cd ... δxy δ yz (enough terms to exhaust the whole alphabet)

(a) δii. Compute the value by summing on the repeated index i as δii = δ11+δ 22+δ33 = 3. (b) δij δij. The two indices are repeated. Contract first on index j, then on index i. δij δij = δii = 3 . (c) Cij δik δ jk. The three indices are repeated. Contract first on index i, then on index j, and finally on index k. Cij δik δ jk = Ckj δjk = C kk = C 11+C22+C 33. (d) δab δbc δ cd ... δxy δ yz (enough terms to exhaust the whole alphabet). All of the indices are repeated except for a and z. Contract first on index b, then on index c, etc. until all indices have been summed δab δbc δ cd ... δxy δ yz = δac δ cd ... δxy δ yz = δ ad ... δxy δ yz = ... = δ az

2

Fundamentals of Structural Mechanics

2. Let two vectors, u and v, have components relative to some basis as u = (5, −2, 1) and v = (1, 1, 1). Compute the lengths of the vectors and the angle between them. Find the area of the parallelogram defined by u and v.

The length of a vector can be computed as ‖ u ‖= u ⋅ u = ui ui. Thus, the lengths of the vectors can be compute as ‖ u ‖ = 25+4+1 = 30

‖ v ‖ = 1+1+1 = 3

The dot product of the vectors is u ⋅ v = 5−2+1 = 4 cos θ(u, v) = u ⋅ v∕ ‖ u ‖ ‖ v ‖ = 4∕ 30 3 ,

θ = 1.136 rad

The area of the parallelogram can be computed either of two ways: A = ‖ (u × v) ‖ = ‖ (−3, −4, 7) ‖ = 9+16+49 = 8.6023 A = ‖ (u × v) ‖ = ‖ u ‖ ‖ v ‖ sin θ = 30 3 sin(1.136 rad) = 8.6023

3. The vertices of a triangle are given by the position vectors a, b, and c. The components of these vectors in a particular basis are a = (0, 0, 0), b = (1, 4, 3), and c = (2, 3, 1). Using a vector approach, compute the area of the triangle. Find the area of the triangle projected onto the plane with normal n = (0, 0, 1). Find the unit normal vector to the triangle.

(a) Using a vector approach, compute the area of the triangle. The area of the original triangle can be computed as A1 =

1 2

‖ (c × b) ‖ =

1 2

‖ (5, −5, 5) ‖ =

1 25+25+25 2

= 4.330

(b) Find the area of the triangle projected onto the plane with normal n = (0, 0, 1). The projected vectors can be found as follows c′ = c − (c ⋅ n) n = (2, 3, 1) − 1(0, 0, 1) = (2, 3, 0) b′ = b − (b ⋅ n) n = (1, 4, 3) − 3(0, 0, 1) = (1, 4, 0) The area of the projected triangle can be computed as A2 =

1 2

‖ (c′ × b′) ‖ =

1 2

‖ (0, 0, 5) ‖ =

1 0+0+25 2

= 2.5

Chapter 1

3

Vectors and Tensors

x3

b A1

n

c

a

x2

b′

A2

c′ x1

(c) Find the unit normal vector to the triangle. n =

ac × ab ‖ ac × ab ‖

where ab is the vector pointing from a to b. Noting that a is located at the origin of the coordinate system, we find ab = b = (1, 4, 3)

ac = c = (2, 3, 1)

ac × ab =  2, 3, 1  ×  1, 4, 3  =  5, −5, 5  ;

‖ ac × ab ‖= 5 3

⇒ n = 1 (1, − 1, 1) 3

4. Let the coordinates of four points a, b, c and d be given by the following position vectors a=(1, 1, 1), b=(2, 1, 1), c=(1, 2, 2), and d=(1, 1, 3) in the coordinate system shown. Find vectors normal to planes abc and bcd. Find the angle between those vectors. Find the area of the triangle abc. Find the volume of the tetrahedron abcd.

x3

d x2

a

x1

(a) Find vectors normal to planes abc and bcd. nabc = ab × ac =  b − a ×  c − a = (1, 0, 0) × (0, 1, 1) = (0, −1, 1) nbcd = bc × bd =  c − b ×  d − b = (−1, 1, 1) × (− 1, 0, 2) = (2, 1, 1) (b) Find the angle between those vectors. cos θ =

n abc ⋅ nbcd ‖ n abc ‖‖ nbcd ‖

nabc ⋅ n bcd = 0 ⇒ θ = π 2

c b

4


(c) Find the area of the triangle abc. Aabc = ‖ nabc ‖ = 2 (d) Find the volume of the tetrahedron abcd. Vabcd =

1 ab × ac 6

⋅ ad =

1 (0, −1, 1) ⋅ 6

(0, 0, 2) = 13

5. Demonstrate that (u × v) ⋅ w = u i vj wk ε ijk from basic operations on the base vectors.

Equality can be demonstrated by simply substituting the component forms of the vectors into the triple scalar product, collecting the scalars to the front and recognizing the definition of the components of the permutation tensor. To wit, (u × v) ⋅ w = (u i ei × vjej) ⋅ wk e k = ui v j w k (ei × e j) ⋅ ek = u i v j wk Á ijk

6. Show that the triple scalar product is skew-symmetric with respect to changing the order in which the vectors appear in the product. For example, show that (u × v) ⋅ w = − (v × u) ⋅ w To generalize this notion, any cyclic permutation (e.g., u, v, w → w, u, v) of the order of the vectors leaves the algebraic sign of the product unchanged, while any acyclic permutation (e.g., u, v, w → v, u, w) of the order of the vectors changes the sign. How does this observation relate to swapping rows of a matrix in the computation of the determinant of that matrix?

For the case given the demonstration is straightforward as, by definition of the cross product, the vector u × v = −(v × u), i.e., a vector of the same magnitude, pointing in the opposite direction. The negative sign persists in the triple product. To see the more general case, consider the component form of the triple scalar product (u × v) ⋅ w = Á ijk u i vj w k To swap an entry, for example v and w we would have (u × w) ⋅ v = Á ijk u i wj v k = Á ikj ui v j w k Clearly, then, swapping entries in the triple scalar product is the same as swapping indices on the permutation tensor. A cyclic permutation of indices leaves the sign of

Chapter 1

Vectors and Tensors

5

the product unchanged while an acyclic permutation changes the sign. Two successive acyclic permutations change the sign and then change it back and hence must be equivalent to a single cyclic permutation. The determinant of a matrix and the triple scalar product are equivalent because  u × v ) ⋅ w = det

u v w

where the components of u, v, and w comprise the rows of the matrix. The effects of permutation of rows of the matrix on the determinant are clearly the same as swapping elements of the triple scalar product.

7. Use the observation that ‖ u−v ‖ 2 =  u−v  ⋅  u−v  along with the distributive law for the dot product to show that u⋅v ≡

1 ‖ 2

u ‖ 2 + ‖ v ‖ 2 − ‖ v−u ‖ 2 

The solution depends upon the computation of the length of u−v ‖ u − v ‖ 2 =  u−v  ⋅  u−v  = u ⋅ u − 2u ⋅ v + v ⋅ v = ‖ u ‖ 2 − 2 u ⋅ v + ‖ v ‖2 Rearrange to give desired result. Note that ‖ u−v ‖ = ‖ v−u ‖.

8. Prove the Schwarz inequality, | u ⋅ v | ≤ ‖ u ‖ ‖ v ‖. Try to prove this inequality without using the formula u ⋅ v =‖ u ‖ ‖ v ‖ cos θ(u, v). Consider any two vectors u ≠ 0 and v ≠ 0 (equality holds if either one of the vectors is zero). The length of any vector is greater than or equal to zero. 0 ≤ ‖ α u+β v ‖ 2 =  α u+β v  ⋅  α u+β v  = α 2 ‖ u ‖ 2 + 2α β u ⋅ v + β 2 ‖ v ‖ 2 Let α =‖ v ‖ 2 and β = −u ⋅ v. Making this specialization gives 2 2 0 ≤ ‖ α u+β v ‖ 2 = ‖ v ‖ 2 ‖ v ‖ 2 ‖ u ‖ 2 − 2 ‖ v ‖ 2  u ⋅ v  +  u ⋅ v  ‖ v ‖ 2



= ‖ v ‖2 ‖ v ‖2 ‖ u ‖2 −  u ⋅ v 

2



6


Since ‖ v ‖ 2 ≥ 0 we can conclude that 0 ≤ ‖ v ‖2 ‖ u ‖2 −  u ⋅ v  2

⇒

‖ v ‖2 ‖ u ‖2 ≥  u ⋅ v  2

9. Show that [u  v] T = v  u using the definition of the transpose of a tensor and by demonstrating that the two tensors give the same result when acting on arbitrary vectors a and b.

A tensor A has transpose AT defined by a ⋅ Ab = b ⋅ A Ta for any vectors a and b. Let A ≡ u  v and proceed as follows: a ⋅ u  v b = a ⋅  b ⋅ v  u

By definition tensor product of vectors

=  a ⋅ u  b ⋅ v 

Dot product is a scalar

=  b ⋅ v  a ⋅ u 

Scalars are commutative

= b ⋅ a ⋅ uv

Dot product is a scalar

= b ⋅ v  u  a

By definition tensor product of vectors

Therefore AT = v  u and hence [u  v] T = v  u.

10. Show that the components of a tensor T and its transpose TT satisfy [TT] ij = [T] ji.

a ⋅ Tb = a ⋅ T jk  e j  ek  b

Component form of T

= Tjk a ⋅  ej  e k  b

Gather scalars to front

= Tjk  a ⋅ e j   e k ⋅ b 

By definition of tensor product

= Tjk  e k ⋅ b  a ⋅ e j 

Scalars are commutative

= Tjk  b ⋅ e k   ej ⋅ a 

Dot product is commutative

= Tjk b ⋅ e k  ej a

Definition of tensor product

= b ⋅ Tjk e k  ej a

Scatter scalars

= b ⋅ Tkj e j  ek a

Rename dummy indices

= b ⋅ TT a

Definition of tensor transpose

Chapter 1

Vectors and Tensors

7

By identification with the penultimate line of the derivation we can observe that the components satisfy [TT] kj = [T] jk.

11. Show that the tensor TT T is symmetric.

A tensor A is symmetric if u ⋅ Av = v ⋅ Au for any vectors u and v. Let A ≡ T T T and proceed as follows: u ⋅ TT T v = u ⋅ T T Tv

By definition of composition of tensors

= Tu ⋅ Tv

By definition of transpose of tensor

= TT Tu) ⋅ v

By definition of transpose of tensor

= TT T u ⋅ v

By definition composition of tensors

= v ⋅ TT T u

By commutativity of dot product of vectors

12. Consider any two tensors S and T. Prove the following: (a) det  T T  = det  T  (b) det  ST  = det  S  det  T  (c) [ ST] T = T T ST (d) [ ST] −1 = T−1 S −1

(a) det  T T  = det  T . The determinant of a tensor T can be computed from its components as det  T  = 16 T im Tjn T kp Á ijk Á mnp Noting that the components of TT are [ T T ] ij = T ji det  T T  = 16 Tmi T nj Tpk Á ijk Á mnp = 16 Tmi T nj Tpk Á mnp Á ijk Changing the dummy index i to m, j to n, k to p, and vice--versa ⇒ det  TT  = 16 T im Tjn T kp Á ijk Á mnp = det  T 

8


(b) det  ST  = det  S  det  T . Recall that  u × v ) ⋅ w = det

u v w

Therefore, we can compute the determinant of a tensor S as det ( S ) = det

S1 S2 S2

=  S1 × S 2 ) ⋅ S3

where the components of a vector Si are  S i ] j = S ij. Thus, we can compute the determinant of S as det  S  =  S1 i  S 2 j  S3  k Á ijk = S1i S 2j S3k Á ijk det  T  =  T1 i  T 2 j  T3  k Á ijk = T1i T 2j T3k Á ijk det  S  det  T  = S1i S 2j S3k Á ijk det  T  Note that Á ijk det  T  = Á ijk T 1p T2m T 3n Á pmn = T ip Tjm T kn Á pmn ⇒ det  S  det  T  = S 1i S2j S 3k Tip T jm Tkn Á pmn = ( S1i T ip ) ( S 2j Tjm )( S 3k Tkn )Á pmn = [ ST] 1p [ ST] 2m [ ST ] 3nÁ pmn = det  ST  (c) [ ST] T = T T ST. From the definition of the product of two tensors ST = S ik Tkj [ e i  ej ] [ ST] T = S ik Tkj [ e j  ei ] = Sjk T ki [ ei  e j ] Similarly, TTS T = [ T T ] ik [ ST ] kj [ ei  e j ] = Tki S jk [ ei  e j ] = Sjk T ki [ ei  e j ] = [ ST] T (d) [ ST] −1 = T−1 S −1. From the definition of the inverse of a tensor, for this relationship to hold, the following has to be true

Chapter 1

Vectors and Tensors

9

[ ST] −1[ ST] = [ T−1 S−1 ][ ST] = I But, [ T −1 S−1 ][ ST ] = T−1 [ S −1 S ] T = T −1 [ I ] T = T −1 T = I ⇒ [ ST] −1 = T−1 S −1

13. Consider two Cartesian coordinate systems, one with basis {e1 , e2 , e3 } and the other with basis {g1, g2, g3}. Let Qij ≡ gi ⋅ e j be the cosine of the angle between gi and ej. (a) Show that gi = Qij e j and ej = Qij g i relate the two sets of base vectors. (b) We can define a rotation tensor Q such that ei = Qgi. Show that this tensor can be expressed as Q ≡ Q ij [gi  g j], that is, Qij are the components of Q with respect to the basis [ gi  g j]. Show that the tensor can also be expressed in the form Q = [e i  gi]. (c) We can define a rotation tensor Q T, such that gi = Q Te i (the reverse rotation from part (b). Show that this tensor can be expressed as Q T ≡ Qij [e j  ei], that is, Qij are the components of Q T with respect to the basis [ ej  e i]. Show that the tensor can also be expressed in the form Q T = [gi  e i]. (d) Show that Q TQ = I, which implies that the tensor Q is orthogonal.

(a) Show that gi = Qij e j and ej = Qij g i relate the two sets of base vectors. Qik = gi ⋅ e k = Qij ej ⋅ ek = Qij δ jk = Qik Qkj = gk ⋅ e j = gk ⋅ Q ij gi = Qij δ ki = Qkj (b) We can define a rotation tensor Q such that ei = Qgi. Show that this tensor can be expressed as Q ≡ Q ij [gi  g j], that is, Qij are the components of Q with respect to the basis [ gi  g j]. Show that the tensor can also be expressed in the form Q = [e i  gi]. Qgk = Qijg i  gj g k = Qij gj ⋅ g k  gi = Qij δ jk gi = Qik g i = ek Qgk = ei  g i gk =  gi ⋅ g k  ei = δik e i = ek (c) We can define a rotation tensor Q T, such that gi = Q Te i (the reverse rotation from part (b). Show that this tensor can be expressed as Q T ≡ Qij [e j  ei], that is, Qij are the components of Q T with respect to the basis [ ej  e i]. Show that the tensor can also be expressed in the form Q T = [gi  e i]. Q Te k = Q ijej  e i ek = Q ij e i ⋅ ek  e j = Q ij δik e j = Q kj ej = g k Q Te k = g i  ei e k =  e i ⋅ ek  g i = δ ik ei = e k (d) Show that Q TQ = I, which implies that the tensor Q is orthogonal.

10

Fundamentals of Structural Mechanics Q TQ = gi  e i [ej  g j] =  e i ⋅ ej [g i  gj] = δ ij[gi  g j] = [g i  gi] = I

14. The components of tensors T and S and the components of vectors u and v are T~

1 2 0

2 0 1

0 1 2

0 2 --1

S~

--2 0 1

1 --1 0

1 1 1

v~

1 u~ 1 2

Compute the components of the vector Su. Find the cosine of the angle between u and Su. Compute the determinants of T, S, and TS. Compute Tij Tij and ui Tik S kj vj. (a) Compute the components of the vector Su. 0 Su ~ 2 --1

--2 0 1

1 1 2

1 --1 0

=

0 0 0

(b) Find the cosine of the angle between u and Su. Since Su has zero length, the angle between u and Su cannot be determined. (c) Compute the determinants of T, S, and TS. det (T) = --9;

det (S) = 0;

det(TS) = det(T)det(S) = 0

(d) Compute Tij Tij and ui Tik S kj vj. Note that Tij Tij = tr T T T = tr T 2 since T is symmetric 1 T2 ~ 2 0

2 0 1

1 2 0

0 1 2

2 0 1

0 1 2

5 2 2

=

2 5 2

2 2 5

⇒ Tij T ij = tr T 2 = 15 Similarly ui Tik S kj vj = u ⋅ TSv TSv ~

1 2 0

2 0 1

0 1 2

0 2 --1

--2 0 1

1 --1 0

1 1 1

=

1 --2 1

ui Tik S kj vj = u ⋅ TSv = 1

15. Verify that, for the particular case given here, the components of the tensor T and the components of its inverse tensor T−1 are

Chapter 1

T ~

2 --1 0

--1 2 --1

0 --1 2

11

Vectors and Tensors

T−1 ~

1 4

3 2 1

2 4 2

1 2 3

That this is actually the inverse can be verified by multiplying the components of the tensor and its inverse together to get the identity 1 4

3

2

1

2

--1

0

2

4

2

--1

2

--1

1

2

3

0

--1

2

=

1

0

0

0

1

0

0

0

1

16. Consider two bases:  e 1, e 2, e 3  and  g 1, g 2, g 3 . The basis  g 1, g 2, g 3  is given in terms of the base vectors  e 1, e 2, e 3  as g 2 = 1  2e1−e 2−e3 ,

g1 = 1  e 1+e2+e 3 ,

g3 = 1  e2−e 3 

6

3

2

The components of the tensor T and vector v, relative to the basis  e 1, e 2, e 3  are 0 1 --1

T~

--1 0 1

1 --1 0

v~

1 2 3

Compute the components of the vector Tv in both bases. Compute the nine values of Tij Tjk T kl (i.e., for i, l = 1, 2 ,3). Find the components of the tensor [T+TT]. Compute Tii.

(a) Compute the components of the vector Tv in both bases. In the basis  e 1, e 2, e 3  Tv =

0 1 --1

--1 0 1

1 --1 0

1 2 3

1 = --2 1

Let bi =  Tvi be the i--th component of Tv relative to the basis  g 1, g 2, g 3  bi = Tv ⋅ gi = (e 1 --2 e 2+e3) ⋅ gi

⇒

b1 = 1 -- 2 + 2 = 0 3 3 3 b2 = 2 + 2 -- 1 = 3 6 6 6 6 2 1 3 b3 = -- -= -2 2 2

12


(b) Compute the nine values of Tij Tjk T kl (i.e., for i, l = 1, 2 ,3). 0 1 --1

Tij T jk =

--1 0 1

--2 1 1

 T ij Tjk  T kl =

0 1 --1

1 --1 0

1 --2 1

--1 0 1

0 1 --1

1 1 --2

1 --1 0

--1 0 1

--2 = 1 1

1 --2 1

0 = --3 3

1 --1 0

1 1 --2 3 0 --3

--3 3 0

(c) Find the components of the tensor [T+TT].

T + TT =

0 1 --1

--1 0 1

1 --1 0

0 + --1 1

1 0 --1

--1 1 0

=

0 0 0

0 0 0

0 0 0

(d) Compute Tii. Tii = T11 + T 22 + T33 = 0

17. Consider two bases:  e 1, e 2, e 3  and  g 1, g 2, g 3 , where g1 = e1+e 2+e3,

g2 = e 2+e3,

g3 = e 2−e3

Compute Qij for the given bases. Compute the value of Qik Qjk. Explain why the identity Qik Qjk = δ ij does not hold in this case. Now consider a vector v = e 1+2e2+3e 3 and a tensor T given as T =  e 2  e1−e 1  e2  +  e3  e 1−e1  e 3  +  e3  e 2−e2  e 3  Compute the components of the vector Tv in both bases, i.e., find vi and v^ i so that the following relationship holds Tv = v i ei = v^ i g i. Find the cosine of the angle between the vector v and the vector Tv. Find the length of the vector Tv.

(a) Compute Qij for the given bases. Recall that Qij = gi ⋅ e j or the ij--th component of Q is the j--th component of gi in the basis  e 1, e 2, e 3  ⇒ Qij =

1 0 0

1 1 1

1 1 --1

(b) Compute the value of Qik Qjk. Explain why the identity Qik Qjk = δ ij does not hold in this case. Qij Q kj =

1 0 0

1 1 1

1 1 --1

1 1 1

0 1 1

0 1 --1

3 = 2 0

2 2 0

0 0 2

Chapter 1

Vectors and Tensors

13

For the idendity Qik Qjk = δ ij to hold, the base vectors  g 1, g 2, g 3  have to be orthogonal and of unit length. It can be observed that g1 and g2 are not orthogonal, which leads to the non--zero off diagonal terms in Q. The diagonal entries in Q are not equal to 1 because the base vectors are not of unit length. (c) Now consider a vector v = e 1+2e2+3e 3 and a tensor T given as T =  e 2  e1−e 1  e2  +  e3  e 1−e1  e 3  +  e3  e 2−e2  e 3  Compute the components of the vector Tv in both bases, i.e., find vi and v^ i so that the following relationship holds Tv = v i ei = v^ i g i. In the basis  e 1, e 2, e 3  Tv =

0 1 1

--1 0 1

--1 --1 0

1 2 3

--5 = --2 3

Let v^ i =  Tvi be the i--th component of Tv relative to the basis  g 1, g 2, g 3  v^ i = Tv ⋅ gi = (--5 e1 --2 e 2+3 e3) ⋅ g i

⇒

v^ 1 = − 5 − 2 + 3 = − 4 v^ 2 = − 2 + 3 = 1 v^ 3 = − 2 − 3 = − 5

(d) Find the cosine of the angle between the vector v and the vector Tv. cos θ(Tv, v) =

Tv ⋅ v = − 5(1) − 2(2) + 3(3) = 0 ‖ Tv ‖ ‖ v ‖ ‖ Tv ‖ ‖ v ‖

(e) Find the length of the vector Tv. ‖ Tv ‖=  Tv ⋅ Tv = 38

18. A general nth-order tensor invariant can be defined as follows f n(T) ≡ Ti

1 i2

Ti

2 i3

⋅⋅⋅ Tin i

1

where {i 1, i 2, . . ., i n} are the n indices. For example, when n = 2 we can use {i, j} to give f 2(T) = Tij T ji; when n = 3 we can use {i, j, k} to give f 3(T) = Tij T jk Tki. Prove that f n(T) is invariant with respect to coordinate transformation.

Let the sets of indices { j 1, j 2, . . ., j n} and {k1 , k2 , . . ., kn } be identical to the set {i 1, i 2, . . ., i n}. Use the formula for change of basis to transform the components of

14

Fundamentals of Structural Mechanics ^

the tensor T from one coordinate system to another as Tij = QimQ jn Tmn . Observe that the Qij terms can be rearranged and the identity QikQ ij = δ ij can be used pairwise to eliminate the Qij . To wit, f n(T) ≡ Ti

1 i2

Ti

2 i3

⋅⋅⋅ Tin i

1

= Qi

1 j1

Qi

^ T Q Q T ⋅⋅⋅ 2 k2 j1 k2 i2 j2 i3 k3 j2 k3

Qi n j nQi

= Qi

1 j1

Qi

^ ^ T Q Q T ⋅⋅⋅ 1 k1 j1 k2 i2 k2 i2 j2 j2 k3

Qi n k nQi n j nT j n k

= δj

^

^

^

T δ T ⋅⋅⋅ 1 k1 j1 k2 j2 k2 j2 k3

^

= Tk

^

1 k2

Tk

^ T 1 k1 jn k1 ^

1

^

δ j n k nT j n k

1

^

2 k3

⋅⋅⋅ Tkn k

1

Since the final expression does not depend upon the basis it is invariant.

19. Use the Cayley-Hamilton theorem to prove that for n ≥ 4 all of the invariants f n(T), defined in Problem 18, can be computed from f 1(T), f 2(T), and f 3(T).

First let n = 4, f 4(T) = tr(T4). According to the Cayley--Hamilton theorem T3 − I TT2 + II TT − III TT = 0 Operating on both sides with T T4 − I TT3 + II TT2 − III TT = 0 ⇒ T4 = I TT3 − II TT2 + III TT Taking the trace of both sides leads to f 4T) = trT4) = 2 I T III T − (II T) 2 Following a similar procedure we can find that T5 = I TT 4 − II TT 3 + III TT 2 f 5T) = I I f 4(T) = 2 I 2T III T − II 2T I T By extension, it can be seen that for any n ≥ 4 all of the invariants f n(T) can be computed from f 1(T), f 2(T), and f 3(T)

Chapter 1

15

Vectors and Tensors

20. From any tensor T one can compute an associated deviator tensor Tdev which has the property that the deviator tensor has no trace, i.e., tr  T dev  = 0. Such a tensor can be obtained from the original tensor T simply by subtracting α ≡ 13 tr  T  times the identity from the original tensor, i.e., Tdev = T−α I. Show that tr  T dev  = 0. Show that the principal directions of Tdev and T are identical, but that the principal values of Tdev are reduced by an amount α from those of the tensor T.

(a) Show that tr  T dev  = 0. tr T′ = tr  T−p I  = tr T − 3p = tr T − tr T = 0 (b) Show that the principal directions of Tdev and T are identical, but that the principal values of Tdev are reduced by an amount α from those of the tensor T. Let m and n be a principal value and a principal direction of the stress tensor S. By definition, then Sn = m n. Now S = S′+p I. Therefore, the eigenvalue problem can be recast as  S′+p I  n = m n,

S′ n =  m−p  n

⇒

By definition of the eigenvalue problem we can observe that n is, indeed, a principal direction of the deviatoric stress tensor S′ corresponding to principal value  m−p  .

21. Consider a tensor T that has all repeated eigenvalues m 1 = m 2 = m 3 ≡ m. Show that the tensor T must have the form T = m I.

Let n1 , n2 , and n3 be eigenvectors of T. Further assume that these vectors are orthogonal (remember, if they are not already orthogonal they can always be orthogonalized by the Gram-Schmidt process). By the spectral decomposition theorem 3

T =

3

 m  n  n  = m  n  n  i

i

i

i

i=1

i

= mI

i=1

Since the sum of outer products of any orthogonal set of vectors is equal to the identity tensor.

22. Prove that the product of a tensor with itself n times can be represented as 3

Tn =

m  i

i=1

n

n i  ni

16


Hint: Observe that [ni  n i][nj  n j] = δ ij [ni  n j] (no summation implied).

Let us assume that we have proven the result for Tn−1. Compute Tn as the product of Tn−1 and T: 3

Tn = Tn−1 T =

m

3

n−1 i

ni  n i

i=1

3

=

 m n n j

j

j

j=1

3

 m

n−1 i

m j  n i  ni   nj  n j 

n−1 i

m j  n i  nj  δij

i=1 j=1 3

=

3

 m i=1 j=1 3

=

m

n i

 n i  ni 

i=1

Since the spectral decomposition clearly holds for n=1, the proof is complete by induction.

23. Show that the determinant of the tensor T can be expressed as follows det  T  =

1  3 tr T − 12 I T tr  T2  + 16  I T  3 3

where I T = tr  T  = T ii is the first invariant of T. Use the Cayley-Hamilton theorem.

The Cayley-Hamilton theorem says that T3 − I T T 2 + II T T − III T I = 0 where I T = tr T, II T, and III T = det T are invariants of T. Taking the trace of the above expression gives tr(T3) − I T tr(T 2) + II T tr(T) − 3 det T = 0 Note that ( tr T) 2−tr(T2) = 2II T (one can show this identity with a component computation). Substituting for II T in the above expression gives tr(T3) − I T tr(T 2) + 12  I 2T−tr(T2)  I T − 3 det T = 0 Finally, rearranging the expression we get det T =

1 tr(T 3) − 12 I T tr(T2) 3

+ 16 I 3T

Chapter 1

Vectors and Tensors

17

As an alternate approach we can observe that if the identity is proven in one coordinate system then it is proven in all coordinate systems. Let us select the principal coordinates. Then tr(T3) = m 31+m 32+m 33 and tr(T2) = m 21+m 22+m 23 . Now we can compute det T =

1 tr(T 3) − 12 I T tr(T2) 3

=

+ 16 I 3T

1 3 m 1+m 32+m 33  − 12 m 1+m 2+m 3 m 21+m 22+m 23 + 16  m 1+m 2+m 3  3 3

= m 1 m 2 m 3 = det T

24. A certain state of deformation at a point in a body is described by the tensor T, having the components relative to a certain basis of 3 --1 0

T~

--1 5 1

0 1 2

Find the eigenvalues and eigenvectors of T. Show that the invariants of the tensor T are the same in the given basis and in the basis defined by the eigenvectors for the present case.

(a) Find the eigenvalues and eigenvectors of T.The coefficients of the characteristic equation are the invariants of the tensor: I T = 10, II T = 29 and III T = 25 . Thus, the characteristic equation is −m 3 + 10m 2 − 29m + 25 = 0 The three roots of the characteristic equation are m 1 =1.6228, m 2 =2.7261 and m 3 = 5.6511. The corresponding principal directions are n1 = 0.2482, 0.3419, −0.9064 n2 = 0.9064, 0.2482, 0.3419 n3 = −0.3419, 0.9064, 0.2482 (b) Show that the invariants of the tensor T are the same in the given basis and in the basis defined by the eigenvectors for the present case. The tensor T in the basis defined by eigenvectors is a diagonal matrix and its components are the eigenvalues themselves.

T~

1.6228 0

0 2.7261

0 0

0

0

5.6511

Invariants of this tensor can be obtained by Eqn. (63) in the text.

18

Fundamentals of Structural Mechanics I T = m 1 + m 2 + m 3 = 10 II T = m 1 m 2 + m 1 m 3 + m 2 m 3 = 29 III T = m 1 m 2 m 3 = 25

They are the same as the invariants computed above.

25. Find the tensor T that has eigenvalues m 1 =1, m 2 =2, and m 3 =3 with two of the associated eigenvectors given by n1 = 1  e 1+e2, 2

n2 = 13 −2e 1+2e2+e 3 

Is the tensor unique (i.e., is there another one with these same eigenproperties)?

Since the eigenvalues are distinct we can find n3 by taking the cross product of the other two eigenvectors n3 = n1 × n 2 =

1

 18

e 1−e 2+4e 3

Based on the spectral decomposition theorem in the text, T can be computed as 3

T =

m

i

ni  n i

i=1

In components we have T ~

=

1 2

1 1 0

1 1 0

1 18

28 --10 --10 28 4 --4

0 0 0

+ 29

4 --4 --2

--4 4 2

--2 2 1

3 + 18

1 --1 4

--1 1 --4

4 --4 16

4 --4 52

The tensor T is unique by the spectral decomposition theorem.

26. Find the tensor T that has eigenvalues m 1 =1, m 2 =3, and m 3 =3, with two of the associated eigenvectors given by n1 = 1  e 1+e2+e 3, 3

Are the eigenvectors unique?

n 2 = 1 −e2+e 3  2

Chapter 1

19

Vectors and Tensors

The solution is similar to Problem 25. 1

n3 = n1 × n 2 =

6

 2e 1−e 2−e 3

3

T =

m

i

ni  n i

i=1

T ~ 1 3

= 1 6

1 1 1

1 1 1

1 1 1

14 --4 --4

--4 14 --4

--4 --4 14

+3 2

0 0 0

0 1 --1

0 --1 1

+3 6

4 --2 --2

--2 1 1

--2 1 1

The eigenvectors n2 and n3 are not unique because the eigenvalues are repeated. Any linear combination of n2 and n3 is an eigenvector. The tensor T, however, is unique.

27. A certain state of deformation at a point in a body is described by the tensor T, having the components relative to a certain basis of

T ~ 10 −2

14 2 14 2 --1 --16 14 --16 5

Let the principal values and principal directions be designated as m and n. Show that n1 = (--1, 2, 2) is a principal direction and find m 1 . The second principal value is m 2 = 9 × 10−2 , find n2. Find m 3 and n3 with as little computation as possible.

(a) Let the principal values and principal directions be designated as m and n. Show that n1 = (--1, 2, 2) is a principal direction and find m 1 . Assume that n1 is a principal direction of T, with eigenvalue m 1 × 10 −2 . Then 14−m1

10

−2

2 14

2 14 −1−m1 − 16 5−m1 − 16

−1 2 = 2

0 0 0

From the first equation we have −14 − m 1 + 4 + 28 = 0, or m 1 = −18. To verify, substitute into the second and third equations − 2 + 2 −1 + 18 − 32 = 0 −14 − 2(16) + 25+18 = 0

20


Therefore n1 is a principal direction. (b) The second principal value is m 2 = 9 × 10−2 , find n2. 10

−2

(14−9) 2 14

2 14 (−1−9) − 16 (5−9) − 16

1 a b

=

0 0 0

From the first and second rows 2a + 14b = −5 10a + 16b = 2 The corresponding eigenvector is n2 =

a = 1

⇒ 1 2, 3

b = −0.5

2−1  .

(c) Find m 3 and n3 with as little computation as possible. The eigenvalue m 3 can be found from the first invariant of Lagrangian strain tensor and n3 can be found from the cross product of the two other principal directions. I T = T11 + T 22 + T33 = 18 = m 1 + m 2 + m 3 = −18 + 9 + m 3 The third eigenvalue is m 3 = 27 × 10−2 n3 = n1 × n 2 =

1 −2, 3

1, −2 

28. The equation for balance of angular momentum can be expressed in terms of a tensor T and the base vectors ei as ei ×  Tei  = 0 (sum on repeated index implied). What specific conditions must the components of the tensor T satisfy in order for this equation to be satisfied?

Carry out the computation for the tensor S as follows ek × S ij [e i  ej]e k = Sij e k × (ej ⋅ e k) e i = Sij δ jk e k × ei ∴ ek × Se k = S ikek × e i = 0 If k = i then ek × e i = 0 ⇒ S 11, S 22, S 33 can be anything. For the rest of the double summation we have S12(e 2 × e1) + S 13(e3 × e 1) + S21(e 1 × e2) + S 23(e3 × e 2) + S31(e 1 × e3) + S 32(e2 × e 3) = 0 S12(−e 3) + S13(e 2) + S21(e 3) + S23(−e 1) + S31(−e 2) + S32(e 1) = 0 [S32 --S 23] e 1 + [S13 --S 31] e2 + [S 21 --S12] e 3 = 0

Chapter 1

21

Vectors and Tensors

Since e1, e2, e3 ≠ 0, we must have S32 = S23

S 13 = S31

S 21 = S12

29. The tensor R that operates on vectors and reflects them (as in a mirror) with unit normal n is given by

n

v

R ≡ I−2 n  n Compute the vector that results from [RR]v. Compute the length of the vector Rv in terms of the length of v. What is the inverse of the tensor R? Compute the eigenvalues and eigenvectors of R.

(a) Compute the vector that results from [RR]v. RRv = I − 2n  n I − 2n  n v = I − 2n  n v − 2(n ⋅ v)n = v − 2(n ⋅ v)n − 2(n ⋅ v)n + 4[n  n]n(n ⋅ v) = v − 4(n ⋅ v)n + 4(n ⋅ n)n(n ⋅ v) = v (b) Compute the length of the vector Rv in terms of the length of v.  Rv  2 = Rv ⋅ Rv = (v − 2(n ⋅ v) n) ⋅ (v − 2(n ⋅ v) n) 2 v ⋅ v − 4(n ⋅ v) 2 + 4(n ⋅ v) = v ⋅ v =  v 

∴  Rv  =  v  (c) What is the inverse of the tensor R? RRv = I v = v ⇒ R --1 = R (d) Compute the eigenvalues and eigenvectors of R. R = I − 2 n  n =  I − n  n − n  n From the spectral decomposition theorem, we have one distinct eigenvalue and a pair of repeated eigenvalues m 1 = 1; m 2 = m 3 = − 1;

n1 = n

n 2, n 3 are any two vectors in the plane ⊥ n

22


30. Let v(x) and u(x) be two vector fields, and T(x) be a tensor field. Compute the following expressions in terms of the components ( vi, ui, and Tij) of these fields relative to the basis  e 1, e 2, e 3 : div  Tv , ∇  u ⋅ Tv , ∇  Tv , u  Tv

(a) div  Tv  Tv = T ij [ ei  e j ] vk e k = T ij vk [ e i  ej ] e k = T ij vk δ jk ei = T ij vj e i ∂Tij ∂v j div(Tv) = ∂ ( T ij vj ) = v + T ij ∂x i ∂x i j ∂x i (b) ∇  u ⋅ Tv  ∇(u ⋅ Tv) = ∂ ( u i Tij v j ) ek = ∂x k

∂x∂u T v + u ∂T∂x v + u T ∂x∂v  e i

k

ij

ij

j

i

j

j

k

i

ij

k

k

(c) ∇  Tv  ∇(Tv) = ∂  T ij vj e i  ek = ∂x k

∂T∂x v + T ∂x∂v  e  e  ij k

j

j

ij

k

i

k

(d) u  Tv. u  Tv = u k ek  T ij vj e i =

 u k Tij v j   ek  e i 

31. Evaluate the following expressions: (a) div  div  x  x   (d) div  x  div  x  x  

(b) div  x div  x div x   (e) ∇  x div x 

(c) ∇  ‖  ∇ ‖ x ‖ 2  ‖ 2  (f) ∇  x ⋅ ∇ x ⋅ x  

where x = x 1 e1+x 2 e2+x 3 e 3 is the position vector in space and all derivatives are with respect to the coordinates xi.

(a) div  div  x  x   div(x  x) = ( x i xj) ,j ei = [ xi,j x j + xi x j,j ] ei = [δ ij xj + 3 xj ] e i = 4x div  div[x  x]  = div(4 x) = 4(xi ei) ,j ⋅ ej = 4 x i,j ei ⋅ e j = 4 δij δ ij = 12 (b) div  x div  x div x   div x = ( x i ei ) ,j ⋅ ej = 3

Chapter 1

Vectors and Tensors

23

div( x div x ) = div( 3x ) = 3 div x = 9 div  x div  x div x   = div( 9x ) = 9 div x = 27 (c) ∇  ‖  ∇ ‖ x ‖ 2  ‖ 2   x 2 = x x i i

∇ x 

2

= ( xi x i) ,j e j = [ x i,j xi + x i xi,j] e j = 2 δij x i ej = 2 x  2x  2 = 4  x  2 2 2 ∇ 2x  = 4 ∇  x  = 8x

(d) div  x  div  x  x   [x  x] ij = xi x j div[x  x] = ( x i xj ) ,j ei = ( xi,j x j + xi x j,j ) ei = (δij x j + xi δ jj) ei = 4 x i ei = 4 x div  x  div  x  x   = div[x  4 x] = 4 div[x  x] = 16 x (e) ∇  x div x  div x = x i,i = δii = 3 ∇( x div x ) = ∇(3x) = 3 ∇x = 3 I (f)

∇  x ⋅ ∇ x ⋅ x   ∇(x ⋅ x) = ( x i xi) ,j ej = ( xi,j x i + xi x i,j ) ej = 2 xj e j = 2 x ∇( x ⋅ 2x ) = 2 ∇( x ⋅ x ) = 4 x

32. Let v(x) =  x 2−x3  e1 +  x 3−x1  e2 +  x 1−x2  e3 . Evaluate the following expressions: ∇v, ∇ x ⋅ v  , div  x  v , and ∇  x × v  , where x = x i ei is the position vector. Evaluate the expressions at the point x = e 1+2 e2+e 3.

(a) ∇v ∇v =

0 --1 1

1 0 --1

--1 1 0

24


(b) ∇ x ⋅ v  x ⋅ v = ( x 1x2− x1x3 ) + ( x2x 3−x2x 1 ) + ( x 3x1−x 3x2 ) = 0 ∴∇(x ⋅ v) = 0 (c) div  x  v  xv ~

x 1 x 2 - x 3

x 1 x 3 - x 1

x 1 x 1 - x 2

x 2 x 2 - x 3

x 2 x 3 - x 1

x 2 x 1 - x 2

x 3 x 2 - x 3

x 3 x 3 - x 1

x 3 x 1 - x 2

div  x  v  ~

 x 2 - x 3  x 3 - x 1  x 1 - x 2

(d) ∇  x × v  x 2 x 1 - x 2 − x 3 x 3 - x 1

x×v ~

∇ x × v ~ (e)

x 3 x 2 - x 3 − x 1 x 1 - x 2 x 1 x 3 - x 1 − x 2 x 2 - x 3

 x 2+x 3

- 2 x2

- 2 x1

 x 1+x 3

- 2 x3

- 2 x1

- 2 x2

 x 1+x 2

- 2 x3

Evaluate the expressions at the point x = e 1+2 e2+e 3. div  x  v  ~

1 0 −1

3   -2 ∇ x×v ~ --2

33. Let v(x) be given by the following explicit function v(x) =  x 21+x2 x 3  e 1 +  x22+x 1 x 3  e2 +  x23+x 1 x2  e3

--4

--2

2 --4

--2 3

x3 x2

where x is the position vector of any point and has components B {x1, x2, x3} relative to the Cartesian coordinate system shown. x1 The vector field is defined on the spherical region B of unit ran dius as shown in the sketch. Give an explicit expression for the unit normal vector field n(x) to the surface of the sphere. Compute the gradient of the vector field v(x). Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. Compute the divergence of the vector field v(x). Compute the integral of div v over the volume of the sphere. Compute the integral of v ⋅ n over the surface of the sphere.

Chapter 1

25

Vectors and Tensors

(a) Give an explicit expression for the unit normal vector field n(x) to the surface of the sphere. n = x = xi ei (b) Compute the gradient of the vector field v(x). ∇v =

∂v i [ e  ej ] ~ ∂x j i

x3 2 x2 x1

2 x1 x3 x2

x2 x1 2 x3

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. x3 2x 2 x1

2x 1 x3 x2

[∇v] n ~

x1 x2 = x3

x2 x1 2x 3

2 x 12 + 2 x 2x 3 2 x 22 + 2 x 1x 3 2 x 32 + 2 x 1x 2

(d) Compute the divergence of the vector field v(x). div v = tr (∇v) = 2 (x1+x 2+x3) (e) Compute the integral of div v over the volume of the sphere. Given the geometry of the problem, it would be best to compute the integral in spherical coordinates using the following transformations (see figure below) x1 = r cos φ cos θ

x2 = r cos φ sin θ

x3 = r sin φ

dV = ( r dφ ) ( r cos φ dθ ) dr = r 2 cos φ dφ dθ dr x2

x3 r* d θ dr

dθ θ

r dφ dφ

r * = r cos φ x1

z

r

φ

π∕2 2π 1

 div v dV = 2    r cos φ cos θ + r cos φ sin θ + r sin φ  r cos φ dr dθ dφ = 0 2

B

−π∕2 0

0

(f) Compute the integral of v ⋅ n over the surface of the sphere. v ⋅ n =  x 21+x2 x3  x 1 +  x22+x 1 x 3  x2 +  x23+x 1 x2  x3 = x 31 + x32 + x 33 + 3 x1 x 2 x 3

26


At the surface we have x1 = cos φ cos θ

x2 = cos φ sin θ

x3 = sin φ

dA = ( 1 dφ ) ( 1 cos φ dθ ) = cos φ dφ dθ π∕2 2π

 v ⋅ n dA =   [ cos φ cos θ + cos φ sin θ + sin φ 3

Ω

−π∕2 0

3

3

3

3

+ 3 cos 2 φ sin φ cos θ sin θ ] cos φ dθ dφ = 0

34. Let v(x) be a vector field given by the following explicit function v(x) =  x 1 e1+x 2 e 2  ln x21+x 22 

x2

x3 h

B

x2 R

x1

where ln(⋅) indicates the natural logarithm of (⋅). x1 The vector field is defined on the cylindrical region B of height h and radius R as shown in the sketch. Give an expression for the unit normal vector field n(x) to the for the cylinder (including the ends). Compute the divergence of the vector field v(x) and the integral of div v over the volume of the cylinder.

(a) Give an expression for the unit normal vector field n(x) to the for the cylinder (including the ends). nt = e3 for the top surface nb = − e 3 for the bottom surface ns = 1 ( x 1e1 + x 2e2 ) for the side surface R (b) Compute the divergence of the vector field v(x) and the integral of div v over the volume of the cylinder. Let r 2 ≡ x21 + x 22. Since v has no components in the direction of, we will assume that the index range is 1 to 2 instead of 1 to 3 for this problem. Thus, ∂r 2 = 2x i ∂xi Note that vi = xi ln r 2 ⇒ div v =

2 ∂vi x x = δ ii ln r 2 + x i 12 ∂r = δii ln r 2 + 2 i 2 i = 2 ln r 2 + 2 ∂xi r ∂x i r

Chapter 1

27

Vectors and Tensors

where a repeated index implies summation from 1 to 2 . Computing the integral of div v over the volume would be most conveniently done in cylindrical coordinates. However, we can apply the divergence theorem to find the result more easily. Note that v ⋅ n = 0 on the top and bottom surfaces. On the side surface r 2 = R2 = const. v ⋅ n = ln R2 ( x 1 e1 + x 2e2 ) ⋅ 1 ( x 1 e1 + x 2e2 ) = R ln R 2 R ⇒

 div v dV =  v ⋅ n dA =  B

Ω

v ⋅ ns dA = R ln R 2 As = 2 π R 2 h ln R2

Ωs

35. Consider the scalar field g(x) =  x ⋅ x  2 . Compute div ∇  div  ∇g( x )   .

g(x) = (x i xi) 2 ∇g = 2(x i xi) ∂ (xk x k) ej = 2(xi x i)[δkj x k + xk δ kj] e j = 4 xi xi x j ej ∂xj div(∇g) = ∂ (4 x i xi x j) = 4[δij x i xj + x i δij x j + xi x i δjj] = 20 xi x i ∂x j ∇(div(∇g)) = ∂ (20 x i xi) e j = 20[δij x i + xi δ ij] ej = 40 xj e j ∂x j div(∇(div(∇g))) = ∂ (40 x j) = 40 δjj = 120 ∂x j

36. Let v(x) be given by the following explicit function v(x) =  x 2+x3  e1 +  x 1+x3  e2 +  x 1+x2  e3

g(x) = 0

x3 x2

B where x is the position vector of any point and has components  x 1, x 2, x 3  relative to the Cartesian coordinate system as n x1 shown. The vector field is defined on the ellipsoidal region B whose surface is described by the equation g(x) = 2 x 21+x22+ 2 x23−4 = 0. Give an expression for the unit normal vector field n(x) to the ellipsoid. Compute the gradient of the vector field v(x). Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. Compute the divergence of the vector field v(x).

28


(a) Give an expression for the unit normal vector field n(x) to the ellipsoid. n=

∇g ∇g 

∇g ~

4 x1 2 x2 4 x3

 ∇g  = (4 x ) 2+(2 x ) 2+(4 x ) 2 1 2 3

(b) Compute the gradient of the vector field v(x). 0 1 1

∇v ~

1 0 1

1 1 0

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. [∇v]n =  1  ∇g

0 1 1

1 0 1

1 1 0

4x1 2x2 4x3

= 1  ∇g

2 x2 + 4 x 3 4 x1 + 4 x 3 4 x1 + 2 x 2

(d) Compute the divergence of the vector field v(x). div v = tr(∇v) = 0

37. Evaluate the expression div ∇x ⋅ Ax  , where A is a constant tensor (i.e., it does not depend upon x), and the vector x has components x = x i ei. The derivatives are to be taken with respect to the independent variables xi. Express the results in terms of the components of A and x.

x ⋅ Ax = x i Aij x j

∇g =

∂g e ∂xk k

div v =

∂v k ∂x k

∇(x ⋅ Ax) = ∂ (x i Aij x j) ek = (δ ij Aij x j + xi A ij δjk) ek = (A kj xj + x i Aik) e k ∂x k div(∇(x ⋅ Ax)) = ∂ ( A kj xj + x i Aik ) = Akj δ jk + δik A ik = Akk + A kk = 2 A kk ∂x k

2

38. Let g(x) = e −‖ x ‖ be a scalar field in three-dimensional space, where ‖ x ‖ is the distance from the origin to the point x. Qualitatively describe the behavior of the function (a one- or two-dimensional analogy might be helpful). Compute the gradient ∇g of the field. Where does the gradient of the function go to zero?

Chapter 1

29

Vectors and Tensors

Let r 2 ≡‖ x ‖ 2 = x i xi be the radial distance from the origin. Note that

r2, j = 2r r, j = xi, j xi + x i xi, j = 2δij xi = 2x j ∇g(x) =

x r, i = ri

⇒

x ∂g ∂r 2 2 e = e −r (−2r) ri e i = −2e−r x i e i = −2g(x) x ∂r ∂xi i 2

Therefore, the gradient of the given function is ∇g(x) = −2 x e−‖ x ‖ . The function has value g(0) = 1 at the origin and decays exponentially to zero with distance from the origin. The function has a zero gradient at the origin.

39. Consider a tensor field T defined on a tetrahedral region bounded by the coordinate planes x1 = 0, x2 = 0, x3 = 0, and the oblique plane 6x1+3x 2+2x3 = 6, as shown in the sketch. The tensor field has the particular expression T = b  x, where b is a constant vector and x is the position vector x = x i ei. Compute the integral of div  T  over the volume and the integral of Tn over the surface of the tetrahedron (and thereby show that they give the same result, as promised by the divergence theorem). Note that the volume of the tetrahedron of the given dimensions is one.

x2 2 1 3 x3

(a) Compute the integral of div  T  over the volume. Tij = bi x j ⇒ div T = bi

∂x i e = bi δ jj ei = 3 b ∂x j i

 div T dV =  3 b dV = 3 b V = 3 b B

B

(b) Compute the integral of Tn over the surface of the tetrahedron. Face x1 = 0 ; n = −e 1 Tn =  b  x (−e1) = −( x ⋅ e 1 ) b = −x 1 b = 0 Face x2 = 0 ; n = −e 2 Tn =  b  x (−e2) = −( x ⋅ e 2 ) b = −x 2 b = 0 Face x3 = 0 ; n = −e 3 Tn =  b  x (−e3) = −( x ⋅ e 3 ) b = −x 3 b = 0

x1

30


Face 6x1+3x 2+2x3 = 6 ; The normal to the oblique face can be found as n = ∇  6x1+3x2+2x 3 − 6 = 6 e 1 + 3 e2 + 2 e3 Normalizing to unit length n = 1  6 e1 + 3 e2 + 2 e 3  7 Tn = ( x ⋅ n ) b = 1  6x 1+3x2+2x 3  b = 6 b 7 7

 Tn dA =  67 b dA = 67 Ab = 67 ( 72 ) b = 3 b Ω

Ω

40. Let v(x) = x on a spherical region of radius R, centered at the origin. Compute the integral of div  v  over the volume of the sphere and compute the integral of the flux v ⋅ n, where n is the unit normal to the sphere, over the surface of the sphere. Give the result in terms of the radius R.

div  v  =

⇒

∂vi ∂x = i = δ ii = 3 ∂xi ∂x i

 div v dV = 3  dV = 3 V = 3 4 π3R = 4 π R 3

B

3

B

The normal to the sphere surface is n = x∕ R ⇒

 v ⋅ n dA = R1  x ⋅ x dA = R1  R dA = R A = 4 π R 2

Ω

Ω

3

Ω

What does this calculation tell you about the ratio of surface area to volume of a sphere? From the divergence theorem

 div v dV =  v ⋅ n dA B

Ω

Substituting with the results obtained above ⇒3V=RA

V=R 3 A

Chapter 1

Vectors and Tensors

31

41. The Laplacian of a scalar field is a scalar measure of the second derivative of the field, defined as ∇2g(x) ≡ div  ∇g(x) . Write the component (index) form of the Laplacian of g in Cartesian coordinates. Compute the Laplacian of the scalar field of Problem 38.

In components, ∇2g(x) ≡ div (∇g(x)) = g, ii. From Problem 38 note that g, i = −2g(r) xi

where

r 2 ≡‖ x ‖ 2 = x i xi.

Thus, g, ii = −2  g, i x i + g xi, i  = −2 −2g x i x i + 3g  =  4r 2−6  g(x) 2 Therefore, the Laplacian is ∇2g(x) =  4r 2−6  e−‖ x ‖

42. Compute div  T  , where T(x) =  x ⋅ x  I−2 x  x is a tensor field.

Tij = ( xk x k ) δ ij − 2 xi x j div T =

∂T ij e =  2 x k δjk δij − 2 δ ij xj − 2 x i δjj e i ∂xj i

=  2 xj δij − 2 x j δij − 6 x i ei = − 6xi e i = − 6 x

43. Let u(x), v(x), and w(x) be vector fields and let T(x) be a tensor field. Compute the component forms of the following derivatives of products of vectors (a) ∇(u ⋅ v) = (u i vi), j e j =  u i, j vi + u i vi, j e j (b) div (u × v) =  ui v j Á ijk , k =  u i, k vj + u i v j, k Á ijk (c) ∇(u × v) =  u i v j Á ijm , n  em  e n  =  u i, n vj + u i vj, n Á ijm  em  e n  (d) div (Tv) =  Tij v j , i = Tij, i v j + Tij v j, i (e) ∇(u ⋅ Tv) =  u i T ij vj , k e k =  u i, k Tij v j + ui Tij, k v j + ui T ij vj, k  e k (f)

∇(Tv) =  T ij vj , k  ei  e k  =  T ij, k vj + Tij v j, k   ei  e k 

(g) div (u  v) =  ui v j , j ei =  ui, j v j + ui v j, j  ei (h) div  [u  v] w =  ui vj w j , i =  ui, i v j wj + u i v j, i wj + u i vj w j, i 

32 (i)

Fundamentals of Structural Mechanics ∇ (u × v) ⋅ w =  ui v j wk Á ijk , m e m =  u i, m vj w k + ui v j, m w k + ui v j wk, m  Á ijk em

44. Use the same reasoning that was used to derive the three-dimensional version of the divergence theorem to develop (a) a one-dimensional version

f (x)

xi−1 xi

a n(xi−1)

b

x

n(xi) ∆xi

Noting that in one dimension n = 1 for the normal on the right side and n = −1 for the normal on the left side of the segment div1 f (x) ≡ lim

∆x i→0

1  f (x ) − f (x )  ≡ df i i−1 dx ∆x i

Divide the region [a,b] into segments with endpoints defined by the x values  x o, x 1, x 2, . . ., x n−1, x n 

where xo ≡ a and xn ≡ b. Let us do the following simple computation n

 f (x ) − f (x i

i−1)

 = f (x 1) − f (a) + f (x 2) − f (x 1) + . . .

i=1

+ f (xn−1) − f (xn−2) + f (b) − f(x n−1) = f (b) − f (a)

Each term appears once with a plus sign and once with a minus sign except for the two end points. Define the “almost divergence” as follows D i ≡ 1  f (xi) − f (x i−1)  ∆xi Now we can write n

 D ∆x i

i=1

n

i

=

 f (x ) − f (x i

i=1

i−1)

 = f (b) − f (a)

(*)

Chapter 1

33

Vectors and Tensors

In the limit as ∆xi → 0, D i → div1 f (x), n → ∞, and the above sum becomes an integral (this is the definition of the Riemann integral, the limit of a sum!) as follows lim

i

∆x i→0

 dxdf  dx b

n

 D ∆x

i

=

i=1

a

From (*) we can see that the result is independent of the limiting process and we get the one-dimensional version of the divergence theorem

 dxdf  dx = f (b) − f (a) b

a

a very familiar result from the integral calculus. (b) The two-dimensional version version is identical to the three-dimensional version if the domain B is interpreted as a planar region, the boundary Ω a closed curve in the plane. Also, let v(B) be the area of the domain, the integral over B an area integral, and the integral over Ω a line integral. With these reinterpretations of notation, the argument proceeds exactly as the three-dimensional case.

45. Consider a vector field v(x) on a region B with surface Ω having unit normal field n. The “curl” of the vector field can be defined as curl  v  ≡

lim

V(B)→0

1 V(B)

 v × n dA Ω

Show (using the cuboid for B, as in the text) that the expression for curl  v  is curl  v  = ∂v xi × ei =

∂v∂x − ∂v∂x  e + ∂v∂x − ∂x∂v  e + ∂v∂x − ∂v∂x  e 2

3

3

2

1

3

1

1

3

2

1

2

2

1

3

Note that many authors define the curl to be the negative of the definition given here, which is easily achieved by using the flux n × v instead. The form presented here seems to be more consistent with our other definitions of derivatives of vector fields.

Let Ωi be the rectangular region with area Ai over which xi is constant. The integral over the surface is then the sum of the integrals over the six faces.

 v × n dA =    v( x + ∆x e ) × e + v( x ) × ( −e )  dA

x3

3

i i

i=1

Ω

i

i

  v( x + ∆x e ) − v( x )  × e dA i i

i=1

Ωi

i

− e1 e2

− e2

3

=

e3

i

Ωi

i

e1 x1

− e3

x2

34


The ratio of flux to volume is 1 V(B)

 v × n dA =  A1   v( x + ∆x∆xe ) − v( x )  × e dA 3

i i

i

i=1

Ω

i

i

Ωi

i

The curl can be computed as 3

curl  v  =

 lim i=1

=

A i→0

1 Ai



Ωi

lim

 v( x+∆x i e i ) − v( x ) 

∆xi

∆x i→0

3

× e i dAi =

 ∂x∂v × e i

i=1

 ∂v∂x e + ∂v∂x e + ∂x∂v e  × e +  ∂v∂x e + ∂v∂x e + ∂v∂x e  × e ∂v ∂v ∂v + e + e + e ×e ∂x ∂x ∂x 1 1

2

1

3

2

1

1

3

1

1

3

1

1

2

2

3

2

3

3

2

1

3

3

2

2

2

3

i

2

3

Therefore, curl  v  =

∂v∂x − ∂v∂x  e + ∂v∂x − ∂x∂v  e + ∂v∂x − ∂v∂x  e 2

3

3

2

1

3

1

1

3

2

1

2

2

1

3

46. Consider variously a scalar field g(x), a vector field v(x), and a tensor field T(x) on a region B with surface Ω with unit normal vector field n. Prove the following theorems

 ∇g dV =  g n dA, B

(a) Prove that

 ∇v dV =  v  n dA,

Ω

 div T dV =  Tn dA

Ω

B

B

Ω

 ∇g dV =  g n dA Ω

B

Chop the region B into N small subvolumes B i with surfaces Ωi. Let each subvolume be centered at location xi and have volume v(B i). The surface of each subvolume can be characterized by its outward normal field ni. Now the surface of each subvolume either will be adjacent to another subvolume, or will lie on the original boundary Ω. Since the normal vector points outward from each subvolume, the normals of two adjacent subvolumes point in exactly the opposite direction. Consequently, if we sum the fluxes over all of the subvolumes, we get simply

 N

i=1

Ωi

g ni dA i =

 g n dA

(*)

Ω

because all of the interior surfaces cancel each other out. Let us define the “almost gradient” of the scalar field to be

Chapter 1

35

Vectors and Tensors

Gi[g] ≡

1 v(B i)

 g n dA i

i

Ωi

The only difference between G[g] and ∇g is that ∇g is the limit of G[g] as the region becomes vanishingly small. From (*) we can see that

 g n dA

N

 G [g] v(B )

=

i

i

i=1

Ω

Upon taking the limit as the number of subvolumes becomes large, the discrete elements pass to their infinitesimal limits, that is, Gi[g] → ∇g and v(B i) → dV, the infinitesimal volume element. The limit of the sum is the integral over the volume N

 G [g] v(B )

lim

i

N→∞

i

=

i=1

 ∇g dV B

thereby completing the proof. (b) Prove that

 ∇v dV =  v  n dA Ω

B

Chop the region B into N small subvolumes B i with surfaces Ωi. Let each subvolume be centered at location xi and have volume v(B i). The surface of each subvolume can be characterized by its outward normal field ni. Now the surface of each subvolume either will be adjacent to another subvolume, or will lie on the original boundary Ω. Since the normal vector points outward from each subvolume, the normals of two adjacent subvolumes point in exactly the opposite direction. Consequently, if we sum the fluxes over all of the subvolumes, we get simply

 N

i=1

v  ni dA i =

Ωi

 v  n dA (*) Ω

because all of the interior surfaces cancel each other out. Let us define the “almost gradient” of the vector field to be Gi[v] ≡

1 v(B i)

 v  n dA i

i

Ωi

The only difference between G[v] and ∇v is that ∇v is the limit of G[v] as the region becomes vanishingly small. From (*) we can see that

 v  n dA Ω

N

=

 G [v] v(B ) i

i

i=1

Upon taking the limit as the number of subvolumes becomes large, the discrete elements pass to their infinitesimal limits, that is, Gi[v] → ∇v and v(B i) → dV, the infinitesimal volume element. The limit of the sum is the integral over the volume

36

Fundamentals of Structural Mechanics N

lim

N→∞

 G [v] v(B ) i

i

=

i=1

 ∇v dV B

thereby completing the proof. (c) Prove that

 div T dV =  Tn dA Ω

B

T

First observe that div (T h) = h ⋅ divT and that (TT h) ⋅ n = h ⋅ Tn. Since the vector h is constant it can be taken outside the integral. Therefore, from the original statement of the divergence theorem we have h⋅

 divT dV

= h⋅

 Tn dA Ω

B

Since this result must hold for any h the theorem is proved.

47. Use the divergence theorem for a vector field to show the following identities (a) Green’s first identity for scalar functions u(x) and v(x), (Hint: Let v(x) = u ∇ v)

  u∇ v + ∇u ⋅ ∇v dV =  n ⋅  u∇v dA 2

Ω

B

(b) Green’s second identity for scalar functions u(x) and v(x), (Hint: Let v(x) = u ∇ v−v ∇ u)

  u∇ v−v∇ u dV =  n ⋅  u∇v−v∇u dA 2

2

Ω

B

(a) Green’s first identity for scalar functions u(x) and v(x)

  u∇ v + ∇u ⋅ ∇v dV =  n ⋅  u∇v dA 2

Ω

B

Let v = u∇v. In components div v = (u v, i ), i = u, i v, i + u v, ii. Observe that, by definition ∇2v = v, ii . Thus, div v = ∇u ⋅ ∇v + u∇2v. (b) Green’s second identity for scalar functions u(x) and v(x)

  u∇ v−v∇ u dV =  n ⋅  u∇v−v∇u dA 2

B

2

Ω

Let v = u∇v − v∇u. In components div v = (u v, i −v u, i ), i = u v, ii − v u, ii.

Chapter 1

37

Vectors and Tensors x3

48. Many problems are more conveniently formulated and solved in cylindrical coordinates ( r, θ, z). In cylindrical coordinates, the components of a vector v can be expressed as

ez e θ (θ)

v(r, θ, z) = v r er + v θ eθ + v z ez

e r (θ)

z

where the components vr, vθ , and vz are each functions of the x2 r coordinates ( r, θ, z). However, now the base vectors er (θ) θ and eθ (θ) depend upon the coordinate θ. We must account x1 for this dependence of the base vectors on the coordinates when computing derivatives of the vector field. Using the coordinate-free definition of the divergence of a vector field, Eqn.(78), show that the divergence of v in cylindrical coordinates is given by ∂v ∂v div v(r, θ, z) = 1r ∂  r vr  + 1r θ + z ∂r ∂z ∂θ (Hint: Observe from the figure that n1 = eθ (θ+∆θ) and n2 = −eθ (θ) and are constant over the faces 1 and 2, respectively. The normal vectors n3 = er (ξ) and n4 = −er (ξ), with ξ ∈ [θ, θ+∆θ], vary over faces 3 and 4. Finally, note that n5 = ez and n6 = −ez are constant over faces 5 and 6.) n5

n1 n3 ∆r

n4

∆θ

θ

n2

r

n4

∆z

r∆θ

n2 ∆r

r

n6 z

z

Note that the volume of the wedge is V(B) = r ∆θ ∆r ∆z plus terms of higher order that vanish more quickly in the limit as V(B) → 0. The contribution of the two surfaces with normal vectors n1 and n2 give lim

∆ θ, ∆ r, ∆ z→0

1 r∆ θ∆ r∆ z



v(x+∆ θ n1) ⋅ n 1 dr dz +

F1



v(x) ⋅ n 2 dr dz

F2



Noting that the region of integration is the same for both integrals, F1 = F2 with area ∆ r ∆ z, and substituting the expressions for the normal vector fields we get lim

∆ θ, ∆ r, ∆ z→0

1 r∆ r∆ z



F1

v(x+∆ θ e θ) ⋅ eθ(θ+∆ θ) − v(x) ⋅ e θ(θ) dr dz ∆θ

Taking the limit, and noting that v ⋅ eθ = vθ we find that the above expression reduces to 1 ∂(v ⋅ e θ) = 1 ∂vθ r ∂θ r ∂θ

38


Similarly, the contribution of the faces with normal vectors n3 and n4 give

lim

∆ θ, ∆ r, ∆ z→0

1 r∆ θ∆ r∆ z



v(x+∆ r n3) ⋅ n 3 (r+∆r) dθ dz +

F3



v(x) ⋅ n 4 r dθ dz

F4



In this case, the regions of integration are not the same, but F3 = F4 since the variables of integration are θ and z, and at every point in the region, n3 = −n4 = e r(θ). Thus, we have lim

∆ θ, ∆ r, ∆ z→0

1 r ∆ θ∆ z



F4

(r+∆r) v(x+∆ r er) ⋅ e r(θ) − r v(x) ⋅ er(θ) dθ dz ∆r

Taking the limit, and noting that v ⋅ er = vr we find that the above expression reduces to 1 ∂(r v ⋅ er) = 1 ∂(r vr) = vr + ∂v r r ∂r r r ∂r ∂r Finally, the contribution of the faces with normal vectors n5 and n6 give

lim

∆ θ, ∆ r, ∆ z→0

1 r∆ θ∆ r∆ z



v(x+∆ z n5) ⋅ n 5 r dθ dr +

F5



v(x) ⋅ n 6 r dθ dr

F6



In this case, the regions of integration are the same, but F5 = F6 , and at every point in the region, n5 = −n6 = e z. Thus, we have lim

∆ θ, ∆ r, ∆ z→0

1 ∆ θ∆ r



F5

v(x+∆ z ez) ⋅ e z − v(x) ⋅ ez dθ dr ∆z

Taking the limit, and noting that v ⋅ ez = vz we find that the above expression reduces to ∂vz ∂(v ⋅ ez) = ∂z ∂z Adding the contributions together gives the desired result.

Chapter 2 The Geometry of Deformation

Note: Unless otherwise indicated we shall assume that the base vectors in the deformed and undeformed configurations coincide, i.e., that  e i  =  g i . 49. Consider a unit cube in the positive octant with a vertex positioned at the origin of coordinates subjected to the following deformation map φ(z) =  z 1+Á z 2 z 3  e 1 +  z 2+Á z 1 z 3  e 2 +  z 3+Á z 1 z 2  e 3 where Á is a constant. Compute the deformation gradient F, the Green deformation tensor C, and the Lagrangian strain tensor E for the given deformation. Using graph paper, plot the deformed position of a square in the x1−x 2 plane by locating the positions of a grid of points. (Select a value of Á to execute the plot.)

(a) Compute the deformation gradient F, the Green deformation tensor C, and the Lagrangian strain tensor E for the given deformation. The components of the deformation gradient are given by Fij = ∂φi(z)∕∂z j. The components of F can, therefore, be computed as

F ~

1 Áz3 Áz2

Áz3 1 Áz1

Áz2 Áz1 1

The Green deformation tensor has components Cij = Fki F kj, which can be computed in matrix format as follows

40

Fundamentals of Structural Mechanics 1 Áz3 Áz2

T

C=F F ~

=

Áz3 1 Áz1

Áz2 Áz1 1

1 Áz3 Áz2

Áz3 1 Áz1

Áz2 Áz1 1

1+Á2z 2 2+Á 2z3 2 2Áz3+Á 2z1z 2

2Áz3+Á 2z1z 2

2Áz2+Á 2z1z 3

1+Á2z 1 2+Á 2z3 2

2Áz1+Á 2z2z 3

2Áz2+Á 2z1z 3

2Áz1+Á 2z2z 3

1+Á2z 1 2+Á 2z2 2

The components of the Lagrangian strain tensor can be computed in matrix format as follows:

E =

1 C−I  2

~

1 2

Á2z 2 2+Á 2z3 2 2Áz3+Á 2z1z 2 2

2Áz3+Á 2z1z 2

2Áz2+Á 2z1z 3

Á2z 1 2+Á 2z3 2

2Áz1+Á 2z2z 3

2

2Áz2+Á z1z 3

2Áz1+Á z2z 3

Á2z 1 2+Á 2z2 2

(b) Using graph paper, plot the deformed position of a square in the x1−x 2 plane by locating the positions of a grid of points. (Select a value of Á to execute the plot.) z2, x2

z3 = 1 0.25

Á = 0.25

1

x1 = z 1+0.25z2 x2 = z 2+0.25z1 0.25

0

1

z1, x1

50. The deformation gradient that results from deforming the body shown through a deformation map φ(z) has the following components relative to the standard basis at the point P F(P) ~

1.1 0.3 0.1 0.1 1.2 0.2 0.2 0.3 1.3

P

z3

n

z2 z1

Find the stretch of a line oriented in the direction of the vector n = (1,1,0) at the point P. What is the value of the Lagrangian strain of that same line at that same point? Calculate the tensors C and E.

(a) Find the stretch of a line oriented in the direction of the vector n = (1,1,0) at the point P. First, the vector n must be normalized to unit length. The length of n is 2 .

Chapter 2

41

The Geometry of Deformation

Define the unit vector m ≡ n∕ ‖ n ‖. The tangent vector m in the deformed configuration is F(P) m =

1 (1.4, 1.3, 0.5) 2

λ2(m) = Fm ⋅ Fm = 1.95 λ(m) = 1.396 (b) What is the value of the Lagrangian strain of that same line at that same point? Calculate the tensors C and E. E =

1 2 λ (m)−1  2

= 0.475

The components of the Green deformation tensor are C = F TF ~

1.26 0.51 0.39 0.51 1.62 0.66 0.39 0.66 1.74

The components of the Lagrangian strain tensor are

E =

1 C−I 2

51. Consider a square piece of material of unit thickness with a round hole in it of radius 1. The material is subjected to a deformation described by the map shown in the diagram. The deformation map shown has the following explicit expression

0.13 0.255 0.195 0.255 0.31 0.33 0.195 0.33 0.37

~

z2

φ(z)

x2

4

0

4

z1

x1

φ(z) = z 1  1+βz 2  e1 + z 2  1+3βz 1  e2 + z 3 e 3 Compute the the volume of the hole in the undeformed and deformed configurations. Compute the perimeter area of the square in the undeformed and deformed configurations. Compute the perimeter area of the circle in the undeformed and deformed configurations.

(a) Compute the the volume of the hole in the undeformed and deformed configurations. The volume of the undeformed hole can be obtained easily as V = π. The volume of the deformed hole can be obtained as

42

Fundamentals of Structural Mechanics v =



dv =

 det F dV Ω

φ(Ω)

The components of the deformation gradient are 1+βz2 3βz2 0

F ~

βz1 1+3βz1 0

0 0 1

det F = 1+β z 2 1+3β z 1 − 3β 2 z 1 z 2 = β 3 z 1+z 2 + 1 v =

 β 3 z +z  + 1 dz dz dz 1

2

1

2

3

Let us carry out the volume integral in cylindrical coordinates r and θ, centered at z 1 = 2, z 2 = 2. The relationship can be described as z 1 = 2+r cos θ

z 2 = 2+r sin θ

Substituting into the equation for deformed volume we have

    β 8+3r cos θ+r sin θ + 1 r dr dθ dz 1

v =

0

2π

0

1

3

= π (1+8β)

0

(b) Compute the perimeter area of the square in the undeformed and deformed configurations. The perimeter area of the undeformed square is A = 16. To obtain the deformed perimeter area we must integrate over each of the four faces. Based on Eqn. (156) in the text, the deformed area can be obtained from a =



da =

–T

N  dA

Ω

φ(Ω)

J F −T ~

 JF

1+3βz1 −βz1 0

−3βz2 1+βz2 0

0 0 J

where J = det F is the Jacobian of F. The faces with normals −e1 and −e2 are positioned at z 1 = 0 and z 2 = 0, respectively. Because we have JF−Te 1 = e 1 along z 1 = 0 and JF−Te 2 = e 2 along z 2 = 0 these faces are not deformed. Therefore, we only need to integrate the two faces with normals e1 and e2 . At the face with normal e1 with z 1 = 4 J  F −Te1  =

1+12β 2 + 16β 2

and the area of the face can be computed as

Chapter 2

  1+12β + 16β 1

1

2

ae1 =

0

43

The Geometry of Deformation 2

dz 2 dz 3 =

1+24β+160β 2

0

Similarly, at the face with normal e2 with z 2 = 4

  12β + 1+4β 1

1

2

ae2 =

0

2

dz 2 dz 3 =

1+8β+160β 2

0

Thus, total deformed perimeter area of the square is a = 8 + 1+24β+160β 2 + 1+8β+160β 2 (c) Compute the perimeter area of the circle in the undeformed and deformed configurations. The undeformed perimeter area of the circle is simply A = 2π. To obtain the deformed perimeter area, we follow the same procedure as part (b) of this problem. The normal to the undeformed circle can be parameterized in terms of the angle θ and represented with the vector N = cos θe1+ sin θe2. Thus J F −TN =

1+3βz1 cos θ−3βz 2 sin θ 2+−βz 1 cos θ+1+βz2 sin θ 2

We can change to cylindrical coordinates to carry out the integration. Noting that r = 1 is constant along the undeformed circle we have J F –TN =

c21 + c 22

where c1 ≡ 1+6β cos θ−6β sin θ+3β cos 2θ c2 ≡ 1+2β sin θ−2β cos θ−β cos 2θ The deformed area of the cylinder can be computed numerically for specific values of the deformation parameter β as

 1

a=

0

β

a

2π

0

J  F−T N  dθ dz 3

0.0 0.5 1.0 2.0

6.2831 22.5961 40.1483 75.6558

52. Prove that (unit) eigenvectors n1 and n2 , of the tensor C, associated with distinct eigenvalues m 1 and m 2, respectively, point in the direction of extreme stretch by computing the stretch for a unit vector m = sin θ n1+ cos θ n 2, where θ is a parameter. Plot the stretch in the direction m as a function of θ.

44


Let us compute the stretch in the direction of the vector m m ⋅ Cm =  sin θ n1+ cos θ n 2  ⋅ C  sin θ n1+ cos θ n 2  =  sin θ n 1+ cos θ n2  ⋅  sin θ Cn 1+ cos θ Cn2  =  sin θ n 1+ cos θ n2  ⋅  m 1 sin θ n 1+m 2 cos θ n 2  = m 1 sin 2 θ  n1 ⋅ n 1  + m 2 cos 2 θ  n 2 ⋅ n2  +  m 1 + m 2  sin θ cos θ  n1 ⋅ n 2  Since the eigenvectors corresponding to distinct eigenvalues are orthogonal, and, by definition, of unit length, we have n1 ⋅ n 2 = 0,

n1 ⋅ n 1 = n2 ⋅ n 2 = 1

⇒ λ2(m) = m ⋅ Cm = m 1 sin 2 θ + m 2 cos 2 θ It can be seen from this equation that, for θ = 0,

λ 2(m) = m 2

for θ = π 2,

λ 2(m) = m 1

for any 0 ≤ θ ≤ 2π,

minm 1,m 2 ≤ λ 2(m) ≤ maxm 1,m 2

Therefore m 1 and m 2 are the extreme values of stretch (squared). This result can be further verified by plotting the stretch in the direction m as a function of θ as shown in the sketch below. λ2(m)

λ2(m)

m2

m1

m1

m2

0

π 2

3π 2 m1 < m2 π

2π

θ

53. Consider a square piece of material of unit thickness. The material is subjected to a deformation described by the map shown in the diagram. The deformation map shown has the following explicit expression

π 2

0

3π 2 m1 > m2 π

z2

φ(z) = z 1  1+z 2  e1 + z 2  1+3z 1  e2 + z 3 e 3

x2 φ(z) 1

1

0

θ

2π

1

z1

0

1

x1

Chapter 2

45


Compute the components (with respect to the standard basis e1 , e2 , and e3 ) of the Green deformation tensor C and the Lagrangian strain tensor E at the point z = (1,1,0). Find the principal stretches and principal directions of C at z = (1,1,0). Find the eigenvalues and eigenvectors of E at z = (1,1,0).

(a) Compute the components (with respect to the standard basis e1 , e2 , and e3 ) of the Green deformation tensor C and the Lagrangian strain tensor E at the point z = (1,1,0). Obtain the components of the deformation gradient and evaluate it at the position z = (1, 1, 0) F ~

1+z2

z1

3z2 0

1+3z1 0

0 0 1

2 3 0

=

1 4 0

0 0 1

The components of the Green deformation tensor are C = F TF ~

13 14 0

14 17 0

0 0 1


E =

1 C−I 2

~

6 7 0

7 8 0

0 0 0

(b) Find the principal stretches and principal directions of C at z = (1,1,0). The principal stretches are the solutions to the characteristic Eqn. (49), with coefficients I C = 31, II C = 55 and III C = 25. The three solutions of the characteristic equation are m 1 = 0.8579, m 2 = 1 and m 3 = 29.1421. The corresponding principal directions are n1 = 0.7555, − 0.6552, 0 n2 = 0, 0, 1 n3 = 0.6552, 0.7555, 0 Note that, from Gershgorin’s theorem one can deduce, by inspection, that one of the stretches is 1. One can use this information to reduce the cubic to a quadratic by synthetic division. (c) Find the eigenvalues and eigenvectors of E at z = (1,1,0). The eigenvalues and eigenvectors of E can be found easily by Eqn. (136) in the text. The eigenvalues are γ1 = − 0.07105, γ2 = 0 and γ3 = 14.0711 and eigenvectors are the same as the principal directions in (b).

46


54. Prove that it is impossible to deform the vertex of a solid cube into a flat face (e.g., the deformation map shown in the sketch deforms the cube into a tetrahedron with the vertex at a deformed onto the flat plane). Hint: You do not need to find an explicit expression for the map to do this problem. Consider a neighborhood of the point a.

φ(z) a

a

Let N1, N2 and N3 be unit vectors normal to the three faces at point a in the undeformed configuration. These vectors are orthogonal by virtue of the initial geometry. These vectors deform to vectors n1, n2 and n3 according to ni = FNi If the corner is deformed to a flat face at a then n1 = n2 = n 3 ≡ n. Observe that (n × n) ⋅ n = 0. Therefore  n 1 × n2  ⋅ n 3 =  FN 1 × FN2  ⋅ FN3 = 0

Since  N 1 × N2  ⋅ N 3 = 1 we have  FN 1 × FN2  ⋅ FN = det F  N 1 × N2  ⋅ N3 = det F

Thus, det F = 0, implying that a finite volume of material in the vicinity of the point a must be deformed to zero volume, which is impossible.

55. A semi-infinite half-space (i.e., the body occupies every point in space that satisfies z 3 > 0) has a deformation map z2 given by the following explicit expression φ(z) = 1+β e −R  z + γe −R e3

B

z1

z3

where β and γ are constants and R is the distance from the origin to any point with position vector z, that is, R2 ≡ z ⋅ z. Plot the variation of displacement along the coordinate axes. Compute the displacement of the point that was originally located at z = (0, 0, ln2). Compute the deformation gradient F(z) of the motion in general and evaluate it at z = (0, 0, ln2). Compute the Green deformation tensor C(z). Find the value of the stretch of a line in the neighborhood of z=(0, 0, ln2) and initially oriented in the direction e1 .

(a) Plot the variation of displacement along the coordinate axes. Compute the displacement of the point that was originally located at z = (0, 0, ln2). The displacement map can be obtained from deformation map as u(z) = φ(z) − z =  β e −R  z + γe −R e3

Chapter 2


47

The displacement at the point z = (0, 0, ln2) is u(0, 0, ln2) =  β ln2 + γ  e −ln 2 e3 =

1 β ln2 + γ  e3 2

(b) Compute the deformation gradient F(z) of the motion in general and evaluate it at z = (0, 0, ln2). Compute the Green deformation tensor C(z). The deformation gradient can be computed as follows F = ∇φ = 1+β e–R  I + β ∇e–R  z + γ∇e –R  e3 Noting that ∇R = z∕R we have ∇e–R = −e–R z∕R. Thus, F = 1+β e –R  I − e β z  z + γ z  e3 R –R

Evaluating F at the position z = (0, 0, ln2) we get F(0, 0, ln2) = 1+ 12 β  I − 12  β ln2+γ  e3  e 3 To obtain Green’s deformation tensor, first compute the transpose of the deformation gradient FT = 1+β e–R  I − e  β z  z + γ e 3  z  R –R

Noting that z  z z  z = R 2 z  z and e3  z z  e3 = R 2 e3  e 3, we can represent Green’s deformation tensor as C = F TF = a I + b z  z + c  z  e 3 + e 3  z + d e 3  e 3 where a =  1+β e−R  2 c =

b =

γe −2R  β R−e R−β  R

βe −2R  β R−e R−β  R

d = γ 2 e−2R

(c) Find the value of the stretch of a line in the neighborhood of z=(0, 0, ln2) and initially oriented in the direction e1 . λ2(e 1) = Fe 1 ⋅ Fe1 = 1+ 12 β 2 λ(e1) = 1+ 12 β

48


z2, x2 56. A beam theory is characterized by a specific deformation map that is parameterized φ(z) θ(z1) by a set of deformation variables that depend only on the axial coordinate z 1. The depenw(z1) dence of the map on z 2 and z 3 is explicit. Let z1, x1 u(z1) u(z 1) represent the displacement of the cen0 troid of the beam in the z 1 direction, w(z 1) the displacement of the centroid of the beam ℓ in the z 2 direction, and θ(z 1) the rotation of a vector normal to the deformed cross section relative to the horizontal. The deformation map for finite planar motion of the beam then takes the form shown in the diagram. The deformation map has the following mathematical expression

φ(z) =  z 1+u(z 1)−z 2 sin θ(z 1)  e1 +  w(z 1)+z 2 cos θ(z 1)  e2 + z 3 e 3 Compute the deformation gradient F of the given deformation map. Compute the Green deformation tensor C, and the Lagrangian strain E. Linearize the deformation map by assuming that cos θ ≈ 1 and sin θ ≈ θ, and compute F, C, and E for the linearized kinematic description. Is the strain linear in the displacement variables u(z 1), w(z 1), and θ(z 1)? Linearize E by neglecting all squares and products of the generalized variables u, w, and θ. What are the consequences of neglecting the higher-order terms?

(a) Compute the deformation gradient F of the given deformation map. Compute the Green deformation tensor C, and the Lagrangian strain E. The components of the deformation gradient are given by Fij = ∂φi(z)∕∂z j. In matrix format we have − sin θ cos θ 0

1+u′−z2 θ′ cos θ w′+z2 θ′ sin θ 0

F ~

0 0 1

The components of the Green deformation tensor are

C = F TF ~

C 11

C 12

0

C 21

1

0

0

0

1

where

C 12

C 11 = 1+u′−z 2 θ′ cos θ  2 +  w′+z2 θ′ sin θ  2 = C 21 = − sin θ (1+u′) + w′ cos θ + 2z 2 θ′ sin θ cos θ

The components of the Lagrangian strain tensor are E =

1 C−I  2

~

1 2

C 11−1 C 21 0

C 12

0

0 0

0 0

Chapter 2

49


(b) Linearize the deformation map by assuming that cos θ ≈ 1 and sin θ ≈ θ, and compute F, C, and E for the linearized kinematic description. Is the strain linear in the displacement variables u(z 1), w(z 1), and θ(z 1)? The linearized deformation map can be written as φ(z) =  z 1 + u(z 1) − z 2 θ(z 1) e1 +  w(z 1) + z 2 e2 + z 3 e3 The components of the deformation gradient of the linearized deformation map are, in matrix form 1+u′−z2 θ′ w′ 0

F ~

0 0 1

−θ 1 0

The components of the Green deformation tensor are C 11

C = F TF ~

0

C 12 2

C 21

θ +1

0

0

0

1


E =

1 C−I 2

~

1 2

C 11−1

C 12

C 21

θ

0

0

2

0 0 0

where C 11 = 1+u′−z 2 θ′ 2 +  w′  2 C 12 = C 21 = −θ (1+u′) + w′ + z2 θ θ′

The strain is not linear in the displacement variables because strain tensor contains squares and products of the displacements and derivatives. (c) Linearize E by neglecting all squares and products of the generalized variables u, w, and θ. What are the consequences of neglecting the higher-order terms? EL =

1 C L−I 2

~

1 2

2u′−2z2 θ′ w′−θ 0

w′−θ 0 0

0 0 0

The consequences of neglecting higher-order terms is that the accuracy of the strains decrease as the deformations increase.

57. Does the linearized strain tensor ever have the same eigenvectors and eigenvalues as the Lagrangian strain tensor? If so, provide an explicit example.

50


The Lagrangian strain tensor cannot have the same eigenvalues as the linearized strain tensor. However, for a rigid body motion they can have the same eigenvectors. Suppose Q is an orthogonal tensor, then a rigid body motion can be expressed as φ(z) = z + α Qz The deformation gradient is F = I+α Q and the Lagrangian strain tensor is

=

1 T F F − I 2 1 α Q+α Q T+α 2 Q TQ  2

=

1 α Q + α Q T + α2 I  2

E =

= Elinear + 12 α 2 I If the difference of two tensors is a multiple of the identity tensor, then the two tensors have identical eigenvectors. Suppose that the identical eigenvector is ni, then En i = Elinear + 12 α 2I n i The relationship between eigenvalues of Lagrangian and linearized strain tensors are γi = (γlinear) i + 12 α 2

x2 58. Find the mathematical expresz2 sion for the map that takes a strip of φ(z) length 2π and deforms it into a semicircular arc without changing the depth of the strip. The deformation 0 0 z1 map is illustrated in the sketch. Com2π 2π pute the deformation gradient F, the Green deformation tensor C, and the Lagrangian strain tensor E for the map.

x1

This problem is similar to the example in the text. The difference is that this beam is bent to a half-circle instead of a complete circle. Since the fiber along the initial z 1 axis does not change length we can determine the radius of the arc. The arc length is πR = 2π, implying that the radius is R = 2. The position of a point, say P in the sketch, is easily described in polar coordinates (r, θ). The position can be described as x1 = r cos θ− π2 ,

x 2 = R + r sin θ− π2 

Since the arc length from the origin to the point P is z 1 and the radius of the arc is R = 2 we can compute the angle θ = z 1∕2. Furthermore, the radial distance to the

Chapter 2

51


x2

P

r θ

R

z1

0

x1

z2 2π

point P is r = 2−z 2. Substituting these into the above equations, noting that cos(θ−π∕2) = sin θ and sin(θ−π∕2) = − cos θ, gives the resulting deformation map as φ(z) = (2−z 2) sin(z 1∕2) e1 + 2−(2−z 2) cos(z 1∕2) e2 + z 3 e3 To verify the solution, check out the location of the point originally located at position z=(2π, 0, 0) which gives φ(2π, 0, 0) = 4 e 2.

59. Consider the rectangular piece of material with the triangular cutout. The body is subjected to the deformation map

z2 4 3 a 2 1 c b 0 0 1 2 3

φ(z) =  z 1+βz 2  e1 + z 2 e2 + z 3 e3 Find the angle of the triangle at the vertex at a before and after deformation. Find the equation describing the inclined line a-b before and after deformation. Find the area of the triangle abc before and after deformation.

z1

(a) Find the angle of the triangle at the vertex at a before and after deformation. Let N1 = (0, −1, 0) and N2 = 1 (1, −2, 0) be the unit vectors before deformation 5

cos θ = N1 ⋅ N 2 =

2 5

θ = 26.57

⇒

Let n1 = FN1 and n2 = FN2 be the deformed vectors. The angle between them can be computed as n ⋅n FN 1 ⋅ FN2 cos ψ =  1   2  =  n n FN   FN  1

F ~

1 β

0

0 0

0 1

1 0

FN 1 =

2

1

−β −1 0

2

FN 2 =

1 5

1−2β −2 0

52

Fundamentals of Structural Mechanics cos ψ =

2–β (1−2β )

1+β 2 4 + (1−2β )2

(b) Find the equation describing the inclined line a-b before and after deformation. The coordinates of the points a and b are (1, 3, 0) and (2, 1, 0), respectively. The equation of the line connecting them before deformation is z 2 = 5−2 z 1 . From the deformation map we have x1 = z 1+β z 2

⇒

x2 = z2

z 1 = x1−β x 2 z2 = x2

The equation after deformation can be obtained by substitution x2 = −2(x1−β x 2) + 5 =

1 (5−2x ) 1 1−2β

(c) Find the area of the triangle abc before and after deformation. The area before deformation is simply 1. The area after deformation is a =



da =

 det F F

–T

N dA

Ω

φ(Ω)

The normal vector of plane is e3 F−Te 3 =

1

0

0

−β

1 0

0 1

0

0 0 1

=

0 0 1

Therefore, det F ‖ F–TN ‖= 1 and the deformation preserves the area at 1.

60. Consider a square piece of material of unit thickness. The material is subjected to a deformation described by the following explicit expression φ(z) =  αz 1+βz 2  e1 +  γz 1+δz 2  e2 + z 3e3 where α, β, γ, and δ are constants. For what values of the constants is the given deformation map physically impossible to realize? Assume that we have scribed a line on the body before deforming it according to the above map. The equation of that line in the undeformed configuration was z 2 = 1−3z 1. What is the equation of the line after deformation? Will the given map ever deform straight lines into curved lines? Why or why not?

(a) For what values of the constants is the given deformation map physically impossible to realize? The determinant of the deformation gradient must always be positive if the deformation is physically realizable. The components of the deformation map given above are

Chapter 2

53


F ~

α γ 0

β δ 0

0 0 1

The condition for a physically reasonable deformation is det F = αδ−β γ > 0. Another way to look at the deformation is that the stretch in any direction should not be less than zero. Let n have components n1, n2, and n3. The square of the length of the vector Fn is  αn 1+βn2  2 +  γn 1+δn2  2 + n23 > 0,

∀ n 1, n 2, n 3

subject to the constraint that n21+n 22+n23 = 1. (b) Assume that we have scribed a line on the body before deforming it according to the above map. The equation of that line in the undeformed configuration was z 2 = 1−3z 1. What is the equation of the line after deformation? The relationship can be obtained by inverting the deformation map. x1 = α z1 + β z2

⇒

x2 = γz 1 + δ z 2

z1 =

δ x1−β x 2 αδ−βγ

z2 =

α x2−γx 1 αδ−βγ

Substituting these into the equation of the original line gives x2 =

αδ−β γ γ−3δ + x α−3β α−3β 1

(c) Will the given map ever deform straight lines into curved lines? Why or why not? Straight lines always deform to straight lines because the deformation map is linear in the undeformed coordinates z 1, z 2 and z 3.

z1

61. The deformation map for the pure twist of a circular shaft of length ℓ and radius r can be expressed in terms of the rate of twist β (a constant) as follows φ(z) =  z 1 cos( βz 3 )−z 2 sin( βz 3 )  e1 +  z 1 sin( βz 3 )+z 2 cos( βz 3 )  e 2 + z 3 e 3 Compute the deformation gradient F(z). Find the displacement of the point initially located at the position z=(r, 0, ℓ) in the undeformed configuration. Find the volume of the deformed shaft in terms of the angle of twist β. A horizontal line is etched on the surface of the undeformed shaft, parallel to the z 3 axis as shown. Find the length of the line in the deformed configuration.

r z3

z2 ℓ x1

x2

φ(z)

x3

54


(a) Compute the deformation gradient F(z). Find the displacement of the point initially located at the position z=(r, 0, ℓ) in the undeformed configuration. F ~

cos βz 3

− sin βz3

sin βz 3 0

cos βz 3 0

−βz1 sin βz3+z 2 cos βz 3 βz1 cos βz3−z 2 sin βz 3 1

u(z) = φ(z) − z u(r, 0, ℓ) = rcos(β ℓ) − 1 e 1 + r sin(β ℓ) e2 (b) Find the volume of the deformed shaft in terms of the angle of twist β.



v =

dv =

φ(B)

 det F dV B

det F = cos 2 βz 3 + sin2 βz 3 = 1 The map is, therefore, volume preserving, and the deformed volume is πr 2ℓ. (c) A horizontal line is etched on the surface of the undeformed shaft, parallel to the z 3 axis as shown. Find the length of the line in the deformed configuration.

−β z 1 sin βz 3+z2 cos βz3 β z 1 cos βz 3−z2 sin βz3 1

Fe 3 =

To obtain the length of the deformed line, we must integrate the stretch along the original line. The normal vector of the original line is n = e 3. The line is etched on the surface so that z 21 + z 22 = R 2 . λ2(e 3) = Fe 3 ⋅ Fe3 = β 2 z 21 + z 22 + 1 = β 2 R 2+1 The deformed length can be obtained by integration

 λ(e ) dz ℓ

L=

3

3

= ℓ β 2 R 2+1

0

z3

62. Consider the unit cube shown. Let the cube be subjected to the deformation map given by

1

φ(z) =  z 1+z 1 z 2  e 1 +  z 2+z 1 z 2  e2 + z 3 e 3 z1

1

e1

1

z2

Chapter 2

55


Compute the volume of the cube in the deformed configuration. Find the area in the deformed configuration of the face with normal e1 in the undeformed configuration.

(a) Compute the volume of the cube in the deformed configuration. F ~

1+z2

z1

z2 0

1+z1 0

0 0 1

det F = 1+z 1+z 2

   1+z +z  dz dz dz 1

v =

1

1

1

0

0

2

1

2

3

= 2

0

(b) Find the area in the deformed configuration of the face with normal e1 in the undeformed configuration. The unit normal of shaded face is N = e 1 with z 1 = 1 on that face

F−T N =

1 2+z 2

2

−z2 1+z2 0

−1 0

0 0 2+z2

1 0 0

=

1 2+z 2

2 --1 0

Therefore, det F F–TN = 5 on this face. The deformed area is then

  det F F 1

a =

0

1

–T

N dz 2 dz 3 = 5

0

63. A thin flexible wire of initial length ℓ, originally oriented along the z 3 axis, is wrapped around a hub (with negligible friction between the wire and the hub). The deformation map that accomplishes the motion is given by

T

φ(z) = sin αz 3 e1+ cos αz 3 e2+βz 3 e3

x3

T where α and β are known constants. What is the radius of the hub? How many times does the wire wrap around it? What is the spacing between adjacent passes of the wire? What is the length of the wire after it is wrapped?

(a) What is the radius of the hub? How many times does the wire wrap around it? What is the spacing between adjacent passes of the wire? The magnitude of

56


sin αz 3 e1+ cos αz 3 e2 is 1, therefore the radius of hub is 1. The wrapping goes according to a periodic function. The original distance required to make one cycle is z 3 = 2π∕α. The total number of wraps is the initial length ℓ divided by the distance per wrap. Therefore, the number of wraps N is given by N = ℓα∕2π. When ∆ z 3 = 2π∕α the wire makes a complete wrap. In this distance, the x3 coordinate advances by β ∆ z 3 = 2πβ∕α, which is the spacing between adjacent wires. (b) What is the length of the wire after it is wrapped? The deformed length of the wire can be obtained in a manner similar to Problem 61 0

F ~

α cos αz3

0

0 0

C ~

0 −α sin αz3 0 β

0

0

0

0 0

0 0 0 α2+β 2

λ2(e 3) = e 3 ⋅ Ce3 = α 2+β 2

 λ(e ) dz ℓ

L=

3

3

= ℓ α 2+β 2

0

64. The displacement map of a certain solid body can be expressed as follows: u(z) = αz 2 z 3 e 1 + αz 1 z 3 e 2 + αz 1 z 2 e 3 where α is a constant. Compute the deformation gradient of the motion. Find the principal stretches at the point originally located at z = {0, 0, 1}, in terms of α. Is n ~ (1, 1, 0) a principal direction for the specified motion? Find the principal (Lagrangian) strains at z = {0, 0, 1} in terms of α.

(a) Compute the deformation gradient of the motion. 0 αz 3 αz2

∇u ~ α

αz3 0 αz 1

αz 2 αz 1 0

F = I + ∇u ~

1 αz3 αz2

αz3 1 αz 1

αz 2 αz 1 1

(b) Find the principal stretches at the point originally located at z = {0, 0, 1}, in terms of α.

T

C=F F=

1 α 0

α 1 0

0 0 1

Solving the characteristic equation of C

1 α 0

α 1 0

0 0 1

=

1+α2 2α 0 2α 1+α2 0 0 0 1

Chapter 2



57




2

det C − mI =  1 + α 2 − m − 4α 2  1 − m = 0

⇒



 1 − m = 0 ⇒ m 1 = 1

1 + α 2 − m = 2α ⇒ m 2, m 3 = α 2  2α + 1 = (α  1) 2

(c) Is n ~ (1, 1, 0) a principal direction for the specified motion?

1+α2 2α 0 2α 1+α2 0 0 0 1

1 1 0

=

1+α2+2α 2α+1+α2 0

=  α 2 + 2α + 1

1 1 0

Therefore n is an eigenvector and the corresponding eigenvalue is α 2 + 2α + 1. (d) Find the principal (Lagrangian) strains at z = {0, 0, 1} in terms of α. The principal values of strain are the eigenvalues of E and can be computed from the eigenvalues of C as follows γi = 1 (m i − 1) 2 Thus, γ1 = 0 γ2, γ3 = 1 α 2  2α + 1 − 1  = 1 (α 2  2α) 2 2

65. The expansion of a hollow sphere can be described by the deformation map

z2

φ(z)

φ(z) = λ(r) z

x2

x z where z is the position vector of a point in the z1 x1 undeformed configuration and λ(r) is a given  function of the radial distance r(z) ≡ z ⋅ z. Compute the deformation gradient F for the map. Compute the stretch through the thickness of the sphere in terms of λ, r, and dλ∕dr.

(a) Compute the deformation gradient F for the map. F = ∇φ = λ′( r) ∇r  x + λ(r) ∇x  ∇x = ij

∂xi = δij ⇒ ∇x = I ∂xj

58

Fundamentals of Structural Mechanics ∴F=

λ′(r) r x  x + λ(r) I

(b) Compute the stretch through the thickness of the sphere in terms of λ, r, and dλ∕dr.





λ′(r) λ′(r) λ λ Fn = r  x  x xr + λI xr = 2  x ⋅ x x + r x = λ′ + r x r where λ′(r) ≡ dλ∕ dr . Let us denote the stretch by m (since λ is already used).





2





2

m 2 =  Fn 2 = Fn ⋅ Fn = λ′ + λr x ⋅ x = λ′ + λr r 2 ⇒ m = λ′r + λ

66. The Green deformation tensor that results from deforming the body shown through a deformation map φ(z) has the following components relative to the standard basis at the point P: C(P) ~

P

z3 z2

1.0 0.2 0.5 0.2 3.0 0.2 0.5 0.2 2.0

m2 m1

z1

Undeformed Configuration

Find the stretch of a line oriented in the direction of the vector m 1 = (1,1,1) at the point P. Find the angle, after deformation, between two lines with tangent vectors m 1 = (1,1,1) and m 2 = (0,1,1) in the undeformed configuration at the point P. Is the vector m 1 = (1,1,1) an eigenvector of the tensor C at the point P?

(a) Find the stretch of a line oriented in the direction of the vector m 1 = (1,1,1) at the point P. First we need to normalize m 1 to unit length m 1 = 1 (1, 1, 1) 3 λ2 m 1 = m 1 ⋅ Cm 1 = 1 (1, 1, 1) ⋅ (1.7, 3.4, 2.7) = 7.8 3 3 ⇒ λ m 1 = 1.612 (b) Find the angle, after deformation, between two lines with tangent vectors m 1 = (1,1,1) and m 2 = (0,1,1) in the undeformed configuration at the point P. cos θ(Fm 1,Fm 2) =

m 2 ⋅ Cm 1 λ(m 1) λ(m 2)

Chapter 2

59


m 2 ⋅ Cm 1 = 1 1 (0, 1, 1) ⋅ (1.7, 3.4, 2.7) = 2.4903 2 3 λ2 m 2 = m 2 ⋅ Cm 2 = 1 (0, 1, 1) ⋅ (0.7, 3.2, 2.2) = 2.7 ⇒ λ m 2 = 1.6432 2 cos θ(Fm 1, Fm 2) =

2.4903 = 0.9399 , (1.612) (1.6432)

θ = 0.348 rad

(c) Is the vector m 1 = (1,1,1) an eigenvector of the tensor C at the point P? Since Cm 1 = (1.7, 3.4, 2.7)∕3 ≠ m (1, 1, 1) where m is a scalar, then m 1 is not an eigenvector of C.

67. A 4 by 3 by 1 in. block of material is scribed with a straight line from corner to corner on one of its broad faces as shown. The block is then subjected to a deformation described by the following map:

1 in. z1

line

3 in. z2

φ(z) =  z 1+0.2z 2  e1 +  z 2+0.3z 3  e2 +  z 3+0.1z 1  e3

4 in.

Compute the length of the line in the deformed configuration. Compute the Lagrangian strain of the line due to the motion. Compute the Lagrangian strain tensor E associated with the motion. Compute the volume of the block in the deformed configuration.

(a) Compute the length of the line in the deformed configuration. F = ∇φ ~

1.0 0.2 0.0 0.0 1.0 0.3 0.1 0.0 1.0

3 n ~ 1 --4 5 0

Fn ~ 1 5

2.2 --4 0.8

λ2(n) = Fn ⋅ Fn = 1  (2.2) 2 + (4) 2 + (0.8) 2 = 0.8375 ⇒ λ(n) = 0.914 25 Since the deformation is homogeneous l = λl o = 4.575 in. (b) Compute the Lagrangian strain of the line due to the motion. E( n) = 1  λ2(n) − 1  = 1 0.8372 − 1 = − 0.0816 2 2 (c) Compute the Lagrangian strain tensor E associated with the motion. E = 1 FTF − I ~ 1 2 2

0.01 0.20 0.10

0.20 0.04 0.30

0.10 0.30 0.09

60


(d) Compute the volume of the block in the deformed configuration. v=

 det FdV = det F  dV = det F * V = 1.006(12) = 12.072 in B

3

B

68. The components of the deformation tensor C at a certain point in a solid body, relative to the basis  e 1, e 2, e3 , are given as C~ 1

10

11 --1 0

--1 11 0

0 0 10

Compute the eigenvalues and eigenvectors of C. What is the direction in which the stretch of the body is greatest at the given point? What is the magnitude of that stretch? What is the ratio of deformed volume to undeformed volume in the neighborhood of the point?

(a) Compute the eigenvalues and eigenvectors of C. What is the direction in which the stretch of the body is greatest at the given point? What is the magnitude of that stretch?

detC − mI = det

⇒ (1 − m)



11 − m 10



2

11∕10--m --1∕10 0 --1∕10 11∕10--m 0 = 0 0 0 1−m



− 1 = 1 − m 1  121 − 220 m + 100 m 2 − 1 = 0 100 100

⇒ 1 (1 − m)10m 2 − 22m + 12 = 0 10 ∴ m 1 = 1, m 2 = 1, m 3 = 6 5 Stretch is greatest in direction n3 where λ2(n 3) = 6∕5 ⇒ λ max = λ(n 3) = 1.095. (b) What is the ratio of deformed volume to undeformed volume in the neighborhood of the point? dv = det F dV The determinant of F can be found from the determinant of C by noting that 2 C = F TF ⇒ det C = det FTF = det F T det F =  det F

Chapter 2

61

The Geometry of Deformation ∴ det F = det C

From the spectral decomposition of C we can find det C = m 1 m 2 m 3 = 6∕5 ⇒ dv = det F = dV

65 = 1.095 = const

⇒ v = 1.095 V

z2

69. A right tetrahedral block of material, with edges of length 1, 2, and 3 along the coordinate axes, is subjected to a deformation described by the following map:

2

φ(z) = 6z 1 e1 + 3z 2 e2 + 2z 3 e 3

3

Find the volumes of undeformed and deformed bodies. Find the areas of the four faces in the deformed and undeformed configurations. Compute the principal stretches and principal directions. Compute the volume of the block in the deformed configuration.

1

z3

(a) Find the volumes of undeformed and deformed bodies. The volume of the undeformed body is V = 1 (1)(2)(3) = 1 6 The deformed volume can be computed using dv = det F dV F~

6 0 0

0 3 0

0 0 2

det F = (6)(3)(2) = 36

v = (det F) V = 36

(b) Find the areas of the four faces in the deformed and undeformed configurations. First compute undeformed areas. Let Ai be the undeformed area of the surface with normal −ei. A1 = 1 (2)(3) = 3 2

A 2 = 1 (1)(3) = 3 2 2

A3 = 1 (1)(2) = 1 2

The undeformed area of the oblique face can be computed from the cross product of any two of the three vectors defining the edges to that face

AN = (0, − 2, 3) × (1, − 2, 0) = (6, 3, 2) ⇒ A N = 1  AN  = 7 2 2 And the normal to that face is N ~ 17  6 e 1+3 e2+2 e 3 .

z1

62


Now compute the deformed areas. aN =

  JF

N dA N

−T

Ω

JF

−T

1/6 0 0 0 1/3 0 ~ 36 0 0 1/2

6 0 0

=

--6 0 0

0 12 0

0 0 18

⇒ a1 = 6 A 1 = 18

For N = −e 1,

JF −TN ~

For N = −e 2,

0 JF −TN ~ 12 0

⇒ a2 = 12 A 2 = 18

For N = −e 1,

JF −TN ~

--6 0 0

⇒ a3 = 18 A 3 = 18

6/7 For N = 3/7 2/7

JF−TN ~

36/7 36/7 36/7

3 A = 10 3 ⇒ aN = 36 N 7

(c) Compute the principal stretches and principal directions. C = F TF ~

36 0 0

0 9 0

0 0 4

By Gershgorin’s theorem m 1 = 36 ⇒ λ1 = 6;

n1 = e1

m 2 = 9 ⇒ λ2 = 3;

n2 = e2

m 3 = 4 ⇒ λ3 = 2;

n3 = e3

(d) Compute the volume of the block in the deformed configuration. Note, this was computed in the first part of the problem.

z3

70. A thin square plate of dimension π (the number 3.14...) and thickness t is subjected to the deformation φ(z) =  z 1−β z 3 cos z 1  e1 + z 2 e 2 +  z 3+β sin z 1  e3

0

z2

π π

z1

Chapter 2

63


where ei is the ith base vector in the deformed configuration and β≪1 (very small compared to 1) is a constant that describes the motion. Compute the strain tensor associated with the map (you can neglect all terms of order β 2 and higher). Where is the strain the greatest? Sketch the deformed shape of the plate.

(a) Compute the strain tensor associated with the map (you can neglect all terms of order β 2 and higher). Where is the strain the greatest? Since β << 1 the linearized strain is sufficient E = 12 ∇u + ∇uT u = φ(z) − z = − β z 3 cos z 1 e1 + β sin z 1e3

∇u ~

β z3 sin z1 0 --β cos z1 0 0 0 β cos z1 0 0

⇒ E~

β z3 sin z1 0 0 0 0 0

0 0 0

The strain is greatest at z 3 = t and sin z 1 = 1 ⇒ z 1 = π∕2 giving E11 = β t. (b) Sketch the deformed shape of the plate. x3

0

β

x1 π

71. The unit cube shown is subjected to a homogeneous deformation (i.e., the deformation gradient is constant). The deformation tensor C is given by C ≡ γI−nn

n

a

b

where γ is a constant that characterizes the deformation and n is a unit vector normal to one of the faces of the cube, as shown on the sketch. Find the principal stretches associated with this state of deformation. Find the stretch λ and the (scalar) Lagrangian strain E of the line ab. What is the smallest possible value of the constant γ for which the deformation is physically reasonable? Explain why smaller values are not possible. If det C = 1 then the volume of the deformed cube is the same as the volume of the undeformed cube. For what value of γ is the volume unchanged?

(a) Find the principal stretches associated with this state of deformation.

64

Fundamentals of Structural Mechanics C = γI − n  n = γ  I − n  n +  γ − 1 n  n

From the spectral decomposition theorem, we have one distinct eigenvalue and a pair of repeated eigenvalues m 1 = γ − 1 ⇒ λ1 = γ − 1 ; n 1 = n. The other two are given by m 2 = m 3 = γ ⇒ λ2 = λ 3 = γ ; n2, n3 are any two vectors in the plane ⊥ n. Also, Cn = γ n −  n ⋅ n n =  γ−1 n Cm = γ m −  m ⋅ n n = γ m if m ⊥ n (b) Find the stretch λ and the (scalar) Lagrangian strain E of the line ab. First we define a unit vector along the line ab in the undeformed configuration nab = 1  m+n , where m ⊥ n and m is a unit vector 2 Note that nab ⋅ n = nab ⋅ m = 1∕2 due to the orthonormality of m and n. Thus, Cnab = γ nab −  n ab ⋅ n n = γ n ab − n 2

λ2(n ab) = n ab ⋅ Cnab = γ n ab ⋅ nab −  n ab ⋅ n = γ − 1∕2 λ(nab) =  γ−1∕2 ,

E(nab) = 12  λ2−1  = 14  2 γ−3 

(c) What is the smallest possible value of the constant γ for which the deformation is physically reasonable? Explain why smaller values are not possible. γ > 1 because if γ ≤ 1 lines along n deform from finite length to zero length. (d) If det C = 1 then the volume of the deformed cube is the same as the volume of the undeformed cube. For what value of γ is the volume unchanged? det C = γ2  γ−1  = 1 ⇒ γ3 − γ 2 − 1 = 0 Solving this equation by Newton’s method leads to γ = 1.4656

72. A 2 by 2 by 2 unit solid cube is subjected to the deformation described by the map (the center of the block is at the origin of coordinates): φ(z) = z 1  1+α z 1  e1 + z 2  1+α z 1  e 2 + β z 3 e3

z3

b

a

2 z1

z2

2 Compute the values of the constants α and β that are consistent 2 with the observation that the total volume of the block is unchanged by the deformation. Compute the length of the line ab in the deformed configuration. Compute the Lagrangian strain tensor E associated with the motion. Compute the deformed area of the side with original normal e1 .

Chapter 2

65


(a) Compute the values of the constants α and β that are consistent with the observation that the total volume of the block is unchanged by the deformation. 1+z1 0 1+z 1 z2

F = ∇φ ~

0

0 0

0

β

det F =  1+z 1  2 β The volume of the undeformed block is V = 8. After deformation, the volume of the block is

 det F dV =    1

v=

V

1

−1

= 4β

−1



1

1

β  1+z 1  2 dz 1 dz 2 dz 3

−1 1

 1+2z 1+z21 dz 1 = 4βz 1+z21+ 1 z31 = 32 β 3 --1

−1

3

If v = V ⇒ β = 4∕3. (b) Compute the length of the line ab in the deformed configuration. Let n =  −e 1+e2∕ 2 be a unit vector pointing in the direction of ab in the undeformed configuration −(1+z1) Fn ~ 1 −z2+(1+z 1) 2 0

−(1+z1) 1+2 z1 0

= 1 2

given that z 2 = −z 1 along ab.





2 2 λ2(n) = Fn ⋅ Fn = 1  1+z 1 + 1+2 z 1 = 1  2+6 z 1+5 z 21 2 2

Therefore the length after deformation is

 λ(z )dz =   2+6 z +5 z  dz = 2.3144 1

l=

1

1

1

--1

1 2

2 1

1

1

--1

(c) Compute the Lagrangian strain tensor E associated with the motion.

C = F TF ~

 1+z 1  2+z 22

z2  1+z1 

z2  1+z1 

 1+z1 2

0

⇒ E~1 2

z1  2+z1 +z 22 z2  1+z1  0

0 0 β2

0 z2  1+z1 

0

z 1  2+z 1  0

0 2

β −1

66


(d) Compute the deformed area of the side with original normal e1 . 1+z1 −z 2

1 F−T ~  1+z 1 2

0

1+z1

0

0

0 0  1+z 1  2 β

J = det F = β  1+z 1  2 Given that z 1 = 1 for the side with original normal e1 ⇒  JF−Te 1 = β 1+z 1 = 2β Therefore, the area after the deformation is a = 2β A = 8.

z3

73. A circular cylinder with initial inside radius of 1 and outside radius of 2 is subjected to a deformation with displacement map h

u(z) =  z 1 e1+z 2 e 2  ln z 21+z 22  where ln(⋅) indicates the natural logarithm of (⋅). Find the deformation gradient F for the given motion. Compute the stretch of the cylinder in the radial direction. Compute the Lagrangian strain of a line in the radial direction. What are the height, inside radius, and outside radius of the cylinder after the deformation?

z2

B z2

z1 1 2

(a) Find the deformation gradient F for the given motion. Let r 2 ≡ z 21+z 22 so that d ln r 2 = 1 dr 2 = 2 z i dz i r2 r 2 dz i ∇u ~

ln r 2+ 2z1z 2 r2 0

2z21

2z1z 2 r2 2z 2 ln r 2+ 22 r 0

r2

0 0 0

We can define the unit vector n = 1r  z 1e 1+z 2e 2 so that ∇u = ln r 2  I − e 3  e3 + n  n F = I + ∇u = I + ln r 2  I − e3  e 3  + n  n (b) Compute the stretch of the cylinder in the radial direction.

z1

Chapter 2


67

Fn = n + ln r 2  n −  n ⋅ e3 e 3 +  n ⋅ n n =  3+ ln r 2  n λ2(n) = Fn ⋅ Fn =  3+ ln r 2  2 ⇒ λ(n) =  3+ ln r 2  (c) Compute the Lagrangian strain of a line in the radial direction. E(n) = 1  λ2−1  = 1  3+ ln r 2  2 − 1 2 2 (d) What are the height, inside radius, and outside radius of the cylinder after the deformation? Fe3 = e3 λ 2(e3) = e 3 ⋅ e3 = 1 the height does not change

 λ(n) dr =   3+ ln r  dr = 1 1

r inside =

1

2

0

0

Thus, the inside radius does not change

 λ(n) dr =   3+ ln r  dr = 4.7726 2

r outside =

2

2

0

0

74. Consider a deformation map φ(z) given by the explicit expression φ(z) =  1+ε z ⋅ z  z Compute the deformation gradient F of the given motion. Compute the stretch in the radial direction (i.e., in the direction z). Compute the Lagrangian strain tensor E for the given motion. Is the direction z an eigenvector of E or not?

(a) Compute the deformation gradient F of the given motion. Let r 2 ≡ z ⋅ z = z iz i so that 2 z ∂z ∇r 2 = ∂r ej = 2 z i i e j = 2 2i e i = 22 z. ∂z j ∂z j r r

Also, note that ∇z = I so that F = ∇φ =  1+ε r 2  I + 2ε z  z (b) Compute the stretch in the radial direction (i.e., in the direction z). The unit vector in the direction z is n = z∕r . Thus,





1  2 Fn = 1r  1+ε r 2  z + 2ε r r z = r +3ε r z





λ(n) =  Fn  = 1r +3ε r  z  = 1+3ε r 2

68


(c) Compute the Lagrangian strain tensor E for the given motion. C = F TF =   1+ε r 2  I + 2ε z  z  1+ε r 2  I + 2ε z  z 2

=  1+ε r 2  I + 4ε  1+ε r 2   z  z+4ε 2 z  z z  z 2

=  1+ε r 2  I +  4ε+8ε 2 r 2   z  z



2 E = 1  C − I = 1  1+ε r 2  I +  4ε+8ε 2 r 2   z  z − I 2 2







= ε r 2+ 1 ε 2 r 4 I +  2ε+4ε 2 r 2   z  z 2 (d) Is the direction z an eigenvector of E or not?









2 Ez = ε r 2+ ε r 4 z +  2ε+4ε 2 r 2   z  z z = 3ε r 2+ 9 ε 2r 4 z 2 2

Therefore z is an eigenvector of E and the corresponding eigenvalue is 3ε r 2+ 9 ε 2r 4 2

75. A spherical shell in the undeformed configuration has an inside radius of R and an outside radius of 2R. The shell is subjected to a deformation described by the following map: φ(z) =  1+α 4R2−z ⋅ z   z where α is a given constant of the motion and z is the position vector of a point in the undeformed configuration. Find the displacement of the point originally located at z=(0, 0, R)? Compute the deformation gradient F of the motion. What is the change in thickness of the shell? How much does the inside surface of the shell stretch? (Note: the stretch is the same in all directions because of the spherical symmetry).

(a) Find the displacement of the point originally located at z=(0, 0, R)? u = φ − z = α 4R2−z ⋅ z  z At z=(0, 0, R) we have z ⋅ z = R 2 ⇒ u(0, 0, R) = α 4R2−R 2  R e3 = 3αR3 e 3 (b) Compute the deformation gradient F of the motion. Let r 2 ≡ z ⋅ z = z iz i so that 2 ∂z ∇r 2 = ∂r ej = 2 z i i e j = 2 z i e i = 2 z. ∂z j ∂z j

⇒ F = ∇φ =  4R 2−z ⋅ z  I − 2α z  z

Chapter 2

69


(c) What is the change in thickness of the shell? u(0, 0, 2R) = α(4R 2 − 4R2) 2R e3 = 0 ⇒ ∆t =  u(0, 0, 2R) − u(0, 0, 2) ⋅ e3 = − 3αR3 (d) How much does the inside surface of the shell stretch? (Note: the stretch is the same in all directions because of the spherical symmetry). On the inside surface z ⋅ z = R 2 and z points in the radial direction. Therefore any vector m⊥n is tangent to that surface. Let  m  = 1 Fm =  1 + 3α R 2 m λ(m) =  Fm  =  1 + 3α R2

76. A sphere (exploded view shown in sketch) with initial inside radius of 1 and outside radius of 2 is subjected to a deformation with a radially symmetric displacement map given by u(z) = β z ln z ⋅ z  where z is the position vector and ln(⋅) indicates the natural logarithm of (⋅). Find the deformation gradient F for the given motion. Compute the stretch of the sphere in the radial direction. What is the inside radius and the outside radius of the sphere after the deformation? Compute the stretch of the sphere in any direction perpendicular to the radial direction and evaluate that stretch at the surface.

z3

z2

B

z1

z2 1 2

(a) Find the deformation gradient F for the given motion. Let r 2 ≡ z ⋅ z so that d ln r 2 = 1 dr 2 = 2 z i dz i r 2 dz i r2 ⇒ F = I+∇u =  1 + β ln r 2 I + 2

β zz r2

=  1 + β ln r 2 I + 2β n  n where n = z∕ r is a unit vector pointing in the radial direction (b) Compute the stretch of the sphere in the radial direction. Fn =  1 + β ln r 2 n + 2β n ⋅ n n =  1 + β ln r 2 + 2 n

z1

70

Fundamentals of Structural Mechanics λ(n) =  Fn  = 1 + β ln r 2 + 2

(c) What is the inside radius and the outside radius of the sphere after the deformation?

 λ(n)dr =   1 + β ln r + 2dr = 1 1

r inside =

1

2

0

0

Therefore, the inside radius does not change. The outside radius is

 λ(n)dr =   1 + β ln r + 2dr = 2 + 2.7726 β 2

r outside =

2

2

0

0

(d) Compute the stretch of the sphere in any direction perpendicular to the radial direction and evaluate that stretch at the surface. For any m⊥n Fm =  1 + β ln r 2 n + 2β m ⋅ n n =  1 + β ln r 2 n λ(m) =  Fm  = 1 + β ln r 2 At the surface r = 2, λ(m) =  Fm  = 1 + β ln 4

z2

77. A circle of unit radius is etched on a plate. The plate is then subjected to a homogeneous deformation that stretches according to the following map:

θ

φ(z) = 2 z 1 e1 + z 2 e 2 + z 3 e 3

z1

Find the expression for the stretch of the line under the deformation map (as a function of θ). Find the length of the etched line in the deformed configuration.

(a) Find the expression for the stretch of the line under the deformation map (as a function of θ). F=

2 0 0

0 1 0

0 0 1

Let n = − sin θ e 1+ cos θ e 2 be a unit vector tangent to the line λ2(θ) = n(θ) ⋅ Cn(θ) =  − sin θ cos θ 0 

4 0 0

0 1 0

0 0 1

− sin θ cos θ 0

Chapter 2

71


λ2(θ) = 4 sin 2 θ + cos 2 θ = 1 + 3 sin2 θ ⇒ λ(θ) = 1 + 3 sin 2 θ (b) Find the length of the etched line in the deformed configuration.



l=

2π

λ(θ) Rdθ =

0



2π

1 + 3 sin 2 θ dθ = 9.688

0

z1

78. A circular cylinder of length ℓ and radius R experiences the deformation characterized by the following map:

R z2

φ(z) = α z 1 e1 + β z 2 e 2 + γz 3 e3 where α, β, and γ are constants of the motion. Find the volume of the deformed cylinder. Find the total surface area of the deformed cylinder. Find the principal stretches of the motion. What are the limits on the constants α, β, and γ?

φ(z)

z3

ℓ

(a) Find the volume of the deformed cylinder.

F=

α

0

0

0

β

0

0

0

γ

So that J = det F = α β γ. Since J is constant, v = J V = αβγ πR2ℓ.

(b) Find the total surface area of the deformed cylinder. The total area of the deformed cylinder is the sum of the areas of the top, bottom and side surfaces of the cylinder after deformation. For each of these surfaces, da =  JF −TN  dA, where

F

−T

1∕α

0

0

1∕β

0

0

=

0 0

βγ

JF

−T

= 0 0

1∕γ

0

0

αγ

0

0

αβ

For the top and bottom surfaces N = e 3 ⇒  JF−TN  = αβ ; ae3 = αβ Ae3 = αβ πR 2 For the side surface Ns = cos θ e 1+ sin θ e 2 ⇒ JF −TN = βγ cos θ e1+αγ sin θ e2;

  (βα) cos θ + (αγ) sin θ R dθ dz + 2αβ πR 2π

atotal =

l

2

0

0

2

2

2

3

2

72


(c) Find the principal stretches of the motion.

T

C=F F=

α2

0

0

2

β

0

0

0 0 γ

2

By the spectral decomposition theorem, the principal stretches are α, β, and γ. (d) What are the limits on the constants α, β, and γ? α>0

β>0

γ>0

z2

79. Consider a thin (i.e., it has essentially no thickness in the z3 direction) circular membrane of radius R initially lying in the z1 --z2 plane as shown in the sketch. Under pressure the membrane deforms into a bubble according to the following deformation map



R r

z1 z3



φ(z) = z 1e1+z 2e2 + β cos π z 21+z 22 ∕2R e3

where β is a known constant and R is the radius of the circle. Compute the deformation gradient of the given map. Comz2 R pute the stretch in the initial radial direction (i.e., the direcz1 tion of the vector r = z 1 e1+z 2 e2). Also compute the stretch x3 in the direction that is in the initial plane of the membrane but is orthogonal to r (i.e., tangent to a circle centered at the origin). Are these two directions principal directions? Why x2 or why not? What is the deformed length of the line that was R x1 the radial line from the origin to the edge of the circle along the z 1 direction in the undeformed configuration? What is the slope of the membrane at the edge after deformation? Note: The stretch through the thickness of the membrane is zero, but that is acceptable because we are assuming that the thickness is very small compared to the diameter of the membrane.

(a) Compute the deformation gradient of the given map.

F = ∇φ ~



1 0 0 1 αz 1∕ r αz 2∕ r

0 0 0



where α = α(r) is defined as follows. Let r = z 21 + z 22 and compute the partial derivatives

Chapter 2


∂r = z 1 , r ∂z 1



∂ β cos πr ∂z i 2R



= −

73

∂r = z 2 r ∂z 2

πβ z z sin πr ri = α(r) ri 2R 2R

for i = 1, 2

where α(r) = −

πβ sin πr 2R 2R

(b) Compute the stretch in the initial radial direction (i.e., the direction of the vector r = z 1 e1+z 2 e2). Also compute the stretch in the direction that is in the initial plane of the membrane but is orthogonal to r (i.e., tangent to a circle centered at the origin). Are these two directions principal directions? Why or why not? C = F TF = I + α 2 n  n where n = 1r  z 1 e 1+z 2 e2. Note that n is the unit vector in the direction r. Cn =  I+α 2n  n  n = n+α 2 n ⋅ n  n =  1+α 2 n ⇒ n ⋅ Cn = 1 + α 2,

λ(n) = 1 + α 2

Let m = 1r  −z 2 e 1+z 1 e2 ⇒ m ⋅ n = 0 Cm =  I+α 2n  n  m = m+α 2 n ⋅ m  n = m ⇒ m ⋅ Cm = 1,

λ(m) = 1

Therefore m and n are the eigenvectors of C. (c) What is the deformed length of the line that was the radial line from the origin to the edge of the circle along the z 1 direction in the undeformed configuration? l =  λ(n) ds =  1 + α (r) dr R

R

2

0

0

(d) What is the slope of the membrane at the edge after deformation? The slope is the angle between n and Fn 1 cos θ n, Fn  = n ⋅ Fn = 1 = 1+α 2 λ(n) λ(n)

θ

n Fn

74

Fundamentals of Structural Mechanics z3

80. Consider the deformation map defined on a sphere of unit radius φ(z) = (1+ε) z − β (n ⋅ z) n

n

z2

where ε and β are known (small) constants of the motion and n is a B z1 known constant direction. Compute the ratio of the volume of the Ω sphere after deformation to the volume of the sphere before deformation. Compute the surface area of the sphere after deformation. What makes this calculation complicated? Is the deformed area larger or smaller than the original area? What is the stretch of the sphere in the radial direction? What is the radius of the sphere after the deformation?

(a) Compute the ratio of the volume of the sphere after deformation to the volume of the sphere before deformation. To compute the ratio of the volumes we need the deteminant of the deformation gradient F φi = ( 1 + ε ) z i − ε nk zk n i ∂φi = ( 1 + ε ) δ ij − ε nk ni δ kj = ( 1 + ε ) δ ij − ε nk nj δ kj ∂z j ⇒ F = (1 + ε)I − εn  n The eigenvalues of F can be easily determined. Note that Fn = γn. First consider m = n as a possible eigenvector Fn = ( 1 + ε ) n − ε  n ⋅ n  n = 1 n ⇒ γ1 = 1 Now consider m⊥n as a possible eigenvector Fm = ( 1 + ε ) m − ε  m ⋅ n  n = ( 1 + ε ) n ⇒ γ 2 = γ3 = ( 1 + ε ) Therefore det ( F) = (1 + ε ) 2

v=

 det ( F) dV = det ( F) dV B

v = (1 + ε ) 2 V

(b) Compute the surface area of the sphere after deformation. What makes this calculation complicated? Is the deformed area larger or smaller than the original area? To compute the deformed area we must compute a=

 det ( F) ‖ F

−T

m ‖ dA

Ω

The normal vector in the undeformed configuration is simply m = z. Computing the inverse transpose of F is not particularly difficult. Executing the integral over the surface would be done in spherical coordinates, which may be difficult. The deformed area would be larger than the undeformed area because all of the principal stretches are larger than or equal to 1.

Chapter 2


75

(c) What is the stretch of the sphere in the radial direction? What is the radius of the sphere after the deformation? Let m = z∕‖ z ‖= z∕ r be a unit vector in the radial direction Fm = ( 1 + ε ) m − ε  m ⋅ n  n = ( 1 + ε ) n ⇒ γ 2 = γ3 = ( 1 + ε ) λ2 = Fm ⋅ Fm =  ( 1 + ε ) m − ε  m ⋅ n  n  ⋅  ( 1 + ε ) m − ε  m ⋅ n  n  ( 1 + ε ) 2 m ⋅ m − 2 ε  1 + ε  n ⋅ m + ε2 n ⋅ n ( 1 + ε ) 2 + ε2 − 2 ε  1 + ε  n ⋅ m λ( z ) = ( 1 + ε ) 2 + ε 2 − 2 ε  1 + ε  n ⋅ m ( z ) The radius of the sphere after deformation is

 λ( z ) dr =  ( 1 + ε ) + ε − 2rε  1 + ε  n ⋅ z dr 1

Rdef =

1

2

0

0

where r =‖ z ‖= z ⋅ z

2

76


Chapter 3 The Transmission of Force

81. The stress tensor S at a certain point in a body has components with respect to a set of coordinate axes {x1 , x2 , x3 } of S~

5 3 --8

3 0 --3

--8 --3 11

On a plane whose normal n makes equal acute angles with the coordinate axes, find the traction vector t n, the component of the traction vector that is normal to the plane, and the shearing component of the traction vector.

The cosine of the angles between the normal vector and the base vectors must all be the same. Hence, cos θ = n ⋅ ei = n i, where ni is the i--th component of the vector n. Hence unit normal vector has components n = 1 (1, 1, 1). The traction can be com3 puted as

tn = S n =

1 3

5

3

–8

1

3

0

–3

1

–8

–3

11

1

indicating that the surface is traction free. σ = tn ⋅ n = 0 τ =

 t n  2 − σ 2

= 0

0

=

0 0


78

82. Resolve Problem 81 with S22 changed to 10 3 .

–8

1

3

3 10 3

–3

1

–8

–3

11

1

5

tn = S n =

1 3

σ = tn ⋅ n = τ =

 t n  2 − σ 2

0

=

10 0

10 3

= 10

23

83. Find the principal values and principal directions of the two stress tensors having components with respect to the standard basis of S~

3 1 2

1 --6 0

2 0 15

S~

20 --5 --5 --10 0 0

0 0 0

(a) The first stress tensor. The principal values can be obtained by solving the characteristic equation with coefficients I S = 12, II S = −68, and III S = −261. Thus, −m 3 + 12m 2 + 68m − 261 = 0 The roots are m 1 = −6.1121, m 2 = 2.7863, and m 3 = 15.3258. The corresponding principal directions are n1 = 0.1114, −0.9937, −0.0105 n2 = 0.9807, 0.1116, −0.1606 n3 = 0.1608, 0.0075, 0.9870 (b) The second stress tensor. The principal values can be obtained by solving the characteristic equation with coefficients I S = 10, II S = −225, and III S = 0 as above. Note that, by Gershgorin’s theorem one of the eigenvalues must be zero because the off-diagonals corresponding to the zero diagonal are all zeros. Thus, the characteristic equation can be factored and reduced to a quadratic −m  m 2−10m−225  = 0 The roots are m 1 = −10.8114, m 2 = 0, and m 3 = 20.8114. The corresponding principal directions are

79

Chapter 3 The Transmission of Force n1 = 0, 0, 1 n2 = 0.1602, 0.9871, 0 n3 = −0.9871, 0.1602, 0

84. The condition called plane stress is characterized by the stress state S33 = S23 = S13 = 0. Show that if the remaining stress components are given by S11 =

∂2ψ(x 1, x 2) , ∂x 22

∂ 2ψ(x1, x 2) , ∂x21

S 22 =

S12 = −

∂2ψ(x 1, x 2) ∂x 1∂x2

and the body force b = 0, then the equations of equilibrium are satisfied for any sufficiently smooth function ψ(x1, x2). How smooth must the function be? The components of the stress tensor should satisfy the equilibrium equations ∂3ψ(x 1, x 2) ∂3ψ(x 1, x 2) − = 0 ∂x 1∂x22 ∂x 1∂x22

S11, 1 + S12, 2 = 0 ⇒

∂3ψ(x 1, x 2) ∂3ψ(x 1, x 2) − = 0 ∂x 21∂x2 ∂x 21∂x2

S21, 1 + S22, 2 = 0

Thus, the equations are automatically satisfied when the stress components are defined in this way. The function ψ(x1, x2) must be sufficiently differentiable to satisfy equilibrium.

85. The state of stress at a point is characterized by the stress tensor S, given below S~

4 --4 0

--4 4 0

0 0 8

Consider the vectors n and m given by n =

1  e −e 2−e3 , 3 1

1  e +e 2  2 1

m =

Are the two given vectors n and m eigenvectors of S? Find the principal stresses for the given stress tensor S. (a) Are the two given vectors n and m eigenvectors of S? If n and m are eigenvectors of S, then Sn and Sm should be proportional to n and m, respectively, i.e., S n = σ 1 n and S m = σ 2 m where σ1 and σ2 are eigenvalues. 4 1 3

-- 4 0

-- 4 4 0

0

1

0

--1

8

--1

1

=

8 3

--1 --1


80

Proportionality holds, so n is an eigenvector. The eigenvalue is σ1 = 8. 1 3

4

-- 4

0

1

-- 4

4

0

1

0

0

8

0

1

=0

1 0

Proportionality holds, so m is an eigenvector. The eigenvalue is σ2 = 0. (b) Find the principal stresses for the given stress tensor S. From part (a), two of the principal stresses are already found. They are 0 and 8. One last principal stress can be obtained from the first invariant. I S = S11 + S 22 + S33 = σ 1 + σ2 + σ 3 = 16 Therefore last one is σ3 = 8.

x3

86. Consider the tetrahedron shown in the figure, with edges along the coordinate axes of length 4, 2, and 1, respectively. The state of stress in the tetrahedron is given by the expression S(x) = S o  x  x  where So is a constant and x is the position vector. The equation of the oblique plane is x1+2 x2+4 x 3 = 4. Compute the body force b required for the tetrahedron to be in equilibrium. Compute the tractions on the four faces of the tetrahedron required for equilibrium.

1

x1

n

4

2

x2

(a) Compute the body force b required for the tetrahedron to be in equilibrium. b = − div S = −Sij, j e i = − So x i x j , j ei = −So δ ij xj+x i δ jj ei = −4So x i ei = −4So x

(b) Compute the tractions on the four faces of the tetrahedron required for equilibrium. First note that the traction on any face with normal ei is given by t ei = S ei = S o  x  x  ei = S o  x ⋅ e i  x = S o xi x On the face with normal −ei the coordinate xi = 0. Therefore, by the above formula, the traction on those faces is identically zero. The vector normal to the oblique face is n = 1 (1, 2, 4). The traction on that face is 21

81

Chapter 3 The Transmission of Force tn = S n = S o  x  x  n = S o  x ⋅ n  x =

4S o So  x +2x2+4x 3 x = x 21 1 21

where the position vector x is restricted to lie in the plane.

87. Find an expression for the following derivatives of the principal invariants with respect to tensor components ∂II S , ∂Smn

∂I S , ∂Smn

and

∂III S ∂Smn

The derivatives of the first and second invariant can be easily computed in components. ∂S ii ∂I S = = δ im δin = δmn ∂Smn ∂Smn The second invariant has derivative ∂II S = ∂Smn

1 2

S

1 2

=

ii

∂Sjj ∂Sij ∂S ii + Sjj − 2 S ij ∂S mn ∂Smn ∂S mn



Sii δmn + S jj δmn − 2 S ij δ im δjn

= Sii δ mn − Smn We can compute the derivative of det S = III S formally as follows ∂Sil S jp S kq ∂Smn ∂Sjp ∂Skq ∂S il 1 = 6 Á ijk Á lpq S S + Sil S + Sil S jp ∂Smn jp kq ∂S mn kq ∂S mn

∂III S = ∂Smn

1 Á Á 6 ijk lpq





 S jp S kq δim δ ln + Sil Skq δ jm δpn + S il S jp δkm δ qn 

=

1 Á Á 6 ijk lpq

=

1 Á Á S S 6 mjk npq jp kq

+ 16 Á imk Á lnq S il Skq + 16 Á ijm Á lpn S il Sjp

Alternatively, from Problem 44 (and the Cayley-Hamilton theorem) we have the relationship det S =

1 tr(S 3) − 12 I Str(S 2) + 16 I 3S 3

Noting that ∂(tr(S3))∕∂S = 3S2 and ∂(tr(S2))∕∂S = 2S we can compute the derivative of detS as follows:

82


∂  det S  = S2 − I S − 1 tr(S 2) I + 1 I 2 I S 2 2 S ∂S = S2 − I S S + II S I where we have noted that 2 II S = I 2S−tr(S 2). One can see that the component form is the same as the direct form by multiplying the direct form by S as follows S ∂  det S  = S3 − I S S 2 + II S S = III S I ∂S where we have used the Cayley-Hamilton theorem. Multiplying through by the inverse of S we get ∂  det S  = III S −1 S ∂S Note the similarity with the component form of the derivation.

88. Consider the sphere of radius R shown in the figure. The state of stress in the sphere is given by the stress field

x3

S(x) = S o  m  x + x  m

m

d

x2

c

a

m b

x1 where Sois a constant, x is the position vector of the point in question, and m is a constant unit vector field. What is the body force b(x) required for equilibrium? Compute the tractions acting on the surface of sphere. Sketch the traction vectors at points a, b, c, and d shown on the figure (line segment ca points in the direction of m).

(a) What is the body force b(x) required for equilibrium? b = − div S = −Sij, j e i = − So m i xj + x i mj , j e i = −So m i δjj + δ ij m j  ei = −4 So m i ei = −4So m (b) Compute the tractions acting on the surface of sphere. A vector normal to the surface of a sphere is a radial vector parallel to the position vector x. Hence, n = x∕ x  = x∕R ≡ r. The tractions on the sphere are tn = S n = S o  m  x + x  m  r = So   x ⋅ r  m +  m ⋅ r  x  = So R  m +  m ⋅ r  r 

83

Chapter 3 The Transmission of Force

(c) Sketch the traction vectors at points a, b, c, and d shown on the figure (line segment ca points in the direction of m). At point A, r = m and r ⋅ m = 1. At points B and D, r⊥m and r ⋅ m = 0. At point C, r = −m and r ⋅ m = −1. The traction vectors are shown on the sketch. x3 So Rm

m

A

2 S o Rm

D

m

C

x2 x1

B

So Rm

2 S o Rm

n3

89. Consider a state of stress S that has principal values {σ1 , σ2 , σ3 } with corresponding (orthogonal) principal directions {n1 , n2 , n3 }. Let us consider one of the eight (octahedral) planes whose normal vector m i makes equal angles with the principal directions (one of the eight vectors is shown in the sketch). Show that the normal component of the traction on any of the eight octahedral planes is given by σ =  σ 1+σ2+σ 3 ∕3. Show that the shearing component of the traction on any of the eight octahedral planes is

m1 n2 n1

m 1 = 1  n 1+n2+n 3 3

τ2 = 19  (σ 1−σ2) 2 + (σ2−σ 3) 2 + (σ 3−σ1) 2  ≡ τ 2oct Express τoct in terms of the principal invariants of the stress tensor I S and II S. (a) Show that the normal component of the traction on any of the eight octahedral planes is given by σ =  σ 1+σ2+σ 3 ∕3. The traction on the face with normal vector m i can be computed as 3

σ(i) = m i ⋅ t m i = m i ⋅

 σ n  n  ⋅ m j

j

j

3

i

=

j=1

 σ n ⋅ m  j

j

i

2

j=1

The only difference among the eight normal vectors is sign in front of each vector component m i = 1 γ ik nk 3

(sum implied on k), i.e., γik = 1. There are eight choices of the three pairings of plus and minus values. Therefore, noting the identity nj ⋅ m i = 1 γij 3

3

σ(i) =

 σ n ⋅ m  j

j=1

j

i

3

2

=

 13 σ γ j

j=1

2 ij


84

Thus, for all i (i.e., for any of the eight faces) we have σ(i) =

1 σ +σ2+σ 3  3 1

(b) Show that the shearing component of the traction on any of the eight octahedral planes is τ2 = 19  (σ 1−σ2) 2 + (σ2−σ 3) 2 + (σ 3−σ1) 2  ≡ τ 2oct We can compute the normal traction as 3

 σ n ⋅ m  n

tmi =

j

j

i

3

j

=

j=1

 13 σ γ

j ij

nj

j=1

We can compute the length of this vector as follows ‖ tmi ‖ 2 =

1 3

3

3

  σ γ

j ij

n j  σk γik n k  =

j=1 k=1

1 3

3

σ γ

2 2 j ij

=

j=1

1 3

3

σ

2 j

j=1

Finally, we can compute the square of the shearing component on the face as τ2 =  t mi  2 − σ 2 =

1 2 σ + σ22 3 1

+ σ 23  − 19  σ 1+σ2+σ 3  2

=

1 2 σ 21 + 2 σ22 9

=

1 (σ 1−σ2) 2 + (σ2−σ 3) 2 9

+ 2 σ23 − 2 σ 1 σ2 − 2 σ2 σ 3 − 2 σ1 σ3  + (σ 3−σ1) 2  ≡ τ2oct

(c) Express τoct in terms of the principal invariants of the stress tensor I S and II S. First, compute the values of the invariants I S2 =  σ1+σ 2+σ3 2 = σ 21 + σ22 + σ 23 + 2 σ1 σ2 + 2 σ 1 σ3 + 2 σ2 σ 3 II S = σ1 σ 2 + σ1 σ 3 + σ2 σ3 From part (b) we found that τ2oct = =

2 9

 σ 21 + σ 22 + σ23 − σ 1 σ 2 − σ2 σ 3 − σ 1 σ3 

2 2 I − 3 II S  9 S

Therefore, τoct =

2 3

I S2 − 3 II S

85

Chapter 3 The Transmission of Force 90. A thick-walled sphere of inside radius 1 and outside radius 2 is subjected to an internal pressure of magnitude p. The principal directions of stress are the radial and tangential directions. The principal values of stress are given by the expressions σ1 = σ2 =





p 1 + 43 , 7 r

σ3 =



p 1 − 83 7 r

x2

σ3 r

σ1 or σ 2 x1



where r is the radial distance to an arbitrary point (with position vector x) from the center of the sphere, i.e., r 2 = x21+x 22+x23 . Find the expression for the stress tensor S in the cartesian coordinate system  x 1, x 2, x 3 . Prove that the outside surface of the sphere is traction free. Find the body force b that must be present to maintain equilibrium.

(a) Find the expression for the stress tensor S in the cartesian coordinate system  x 1, x 2, x 3 . Using the spectral decomposition theorem in the case of a repeated eigenvalue S = σ 1  I − n 3  n3  + σ 3  n3  n 3  where n3 = x∕r is the unit radial vector associated with the distinct eigenvalue σ3. Simplifying S = σ 1 I + ( σ 3 − σ 1 ) n 3  n 3  = σ1 I + 12 ( σ 3 − σ1 ) x  x  r (b) Prove that the outside surface of the sphere is traction free. The normal to the outside surface is n3 with r = 2 Sn3 = σ1 n 3 + ( σ3 − σ 1 ) n 3 ⋅ n3  n3 = σ3 n 3 =





p 1 − 83 n3 = 0 7 2

(c) Find the body force b that must be present to maintain equilibrium. b = − div S = −

∂Sij e ∂x j i

Sij = σ1 δ ij + 12 ( σ3 − σ 1 ) x i xj r





∂Sij = σ′ 1 ∂r δij + − 23 ( σ 3 − σ1 ) + 12 ( σ′ 3 − σ′ 1 ) ∂r x i xj ∂xj ∂x j ∂x j r r + 12 ( σ3 − σ 1 ) δij x j + 12 ( σ3 − σ 1 ) x i δjj r r


86 where

σ′ 1 =

dσ1 12 p =− , dr 7 r4

σ′ 3 =

dσ3 24 p = , dr 7 r4

∂r = ∂ ( x x ) 1∕2 = 1 ( x x ) −1∕2 ( 2 x δ ) = 1 x k kj r j ∂xj k k ∂xj 2 k k





∂Sij = 1r σ′ 1 xi + − 23 ( σ3−σ 1 ) + 12 ( σ′ 3−σ′ 1 ) ( xj x j ) x i + 42 ( σ3−σ 1 ) x i = 0 ∂x j r r r b = −div S = 0

91. Consider a state of stress S that has principal values {σ1 , σ2 , σ3 } with corresponding (orthogonal) principal directions {n1 , n2 , n3 }. Let us consider a plane parallel to n3 described by the normal vector

n3 n2 m

n2 θ

m ≡ cos θ n1+ sin θ n 2

m

θ n3

n1

n1

parameterized by the angle θ as shown in the figure. Show, using the spectral decomposition theorem, that the traction vector and its normal component, acting on this plane, are given by t m(θ) = σ1 cos θ n 1 + σ2 sin θ n 2,

σ(θ) = σ 1 cos 2 θ + σ 2 sin2 θ

Show that the shear and normal components of the traction vector satisfy the relationship τ2+σ 2 = σ 21 cos 2 θ+σ22 sin2 θ Now let σ ≡ 12  σ1+σ 2 ,

and ρ ≡ 12  σ 1−σ2 

Show that the shear and normal components of the traction vector satisfy the relationship τ2 +  σ−σ  2 = ρ 2 Note that this problem proves that the shear and normal components of the traction vector on a plane with any value of θ lies on a circle of radius ρ, centered at ( σ,0) in the τ−σ plane. This result, discovered by Otto Mohr, is usually called Mohr’s circle and is shown in the figure above. Clearly, the same results hold for all three pairings of the principal directions.

(a) Show, using the spectral decomposition theorem, that the traction vector and its normal component, acting on this plane, are given by t m(θ) = σ1 cos θ n 1 + σ2 sin θ n 2,

σ(θ) = σ 1 cos 2 θ + σ 2 sin2 θ

87

Chapter 3 The Transmission of Force From the spectral decomposition theorem, the stress tensor can be constructed as 3

 σ n  n 

S =

i

i

i

i=1

The traction on a face with normal m can be computed as 3

t m(θ) = S m =

 σ  n  n  cos θ n + sin θ n  i

i

i

1

2

i=1

3

=

 σ  cos θ δ i

i1 n i+ sin θ δ i2 n i



i=1

= σ1 cos θ n 1 + σ2 sin θ n 2 The normal component of the traction vector is σ(θ) = t m(θ) ⋅ m =  σ1 cos θ n 1 + σ2 sin θ n2 ⋅  cos θ n1 + sin θ n2 = σ1 cos 2 θ + σ 2 sin 2 θ (b) Show that the shear and normal components of the traction vector satisfy the relationship τ2+σ 2 = σ 21 cos 2 θ+σ22 sin2 θ. The shear and normal components satisfy τ2+σ 2 =  t n  2. Thus, from above, τ2 + σ 2 =  t n  2 = σ21 cos 2 θ + σ22 sin2 θ (c) Now let σ ≡  σ 1+σ2 ∕2 and ρ ≡  σ 1−σ2 ∕2. Show that the shear and normal components of the traction vector satisfy the relationship τ2 +  σ−σ  2 = ρ 2. The identity can be shown by the following computation τ2 + (σ − σ) 2 = τ 2 + σ2 − 2 σ σ + σ2 = σ21 cos 2 θ + σ 22 sin 2 θ − 2  σ1 cos 2 θ + σ 2 sin 2 θ 12 σ 1+σ2 + σ 2 = −σ 1 σ 2 cos 2 θ + σ1 σ2 sin 2 θ  + 14 σ1+σ 2 =

1 4

2

 σ 1−σ2  2 = ρ 2

92. Prove the identity S ⋅ ∇(h × x) = h ⋅ [ e j × Sej ], where h is constant.

First compute the derivative ∇(h × x) as ∇( h × x) = ∂ ( h × x)  e k = ( h × e k )  ek = [ h ×] (e k  ek) = [ h ×] ∂x k

88


where we have noted that (ek  e k) = I and that  h × is the skew-symmetric tensor associated with the vector h. Since A ⋅ B = tr A T B  = e i ⋅  A T B  e i we can write S ⋅ ∇(h × x) = S ⋅ [h ×] = ei ⋅ S T  h × ei = Sei ⋅  h × e i  = h ⋅  ei × Se i 

93. A block of material is subjected to a homogeneous state of stress described by the constant stress tensor with x1 3ℓ S = 10 [ e1  e 1 ]−2 [ e1  e 2+e2  e 1 ]+5 [ e 2  e2 ]. 4ℓ The triangular wedge shown is cut out of the block as a x2 freebody. Compute the tractions that must act on each side of the freebody diagram. Demonstrate that the freebody is in overall equilibrium. Assume that the block has unit width.

(a) Compute the tractions that must act on each side of the freebody diagram. n = 45 e1 + 35 e 2 −e1 e3 −e2

The normals to the wedge surfaces are shown in the figure. The corresponding traction vectors are, 10 t −e1 = −Se1 = − --2 0

t −e2 = −Se2 = −

8 -- 6/5 t n = Sn = --8/5 + 3 0

=

--2 5 0

34/5 7/5 0

There are no tractions on the faces with normal e3 . (b) Demonstrate that the freebody is in overall equilibrium. Assume that the block has unit width. Since S is constant and in the absence of body forces (b = 0), the equilibrium equation div S + b = 0 is satisfied. Also, S = S T by inspection, therefore equilibrium of angular momentum is satisfied. We further need verify that the total tractions acting on the wedge surfaces are in equilibrium. 4 t −e1 + 3 t −e2 + 5 t n =

6 --40 8 + --15 0 0

+

34 7 0

=0

89

Chapter 3 The Transmission of Force Therefore the freebody is in overall equilibrium.

x2

94. A triangular prism of material (with base b, height h, and unit thickness) has an internal stress given by the stress field

ρ  x1 h x −h   bx2+hx 1−bh   e2  e 2  bh 1 1 x3 where ρ is the (constant) unit weight of the material and ei is the b unit base vector in the direction of the coordinate axis xi . Find the body force b required for equilibrium. Find the tractions of all of the faces of the prism. Sketch the normal (σ) and tangential (τ) components of traction on the three faces whose normals are orthogonal to the e3 direction. S(x) =

(a) Find the body force b required for equilibrium. For the body to be in equilibrium, the body force vector has to satisfy the equation div S + b = 0 or in component form Sij,j + b i = 0. S11,1 + S 12,2 + S13,3 + b 1 = 0

⇒

S21,1 + S 22,2 + S23,3 + b 2 = 0 S31,1 + S 32,2 + S33,3 + b 3 = 0

⇒ ⇒

b1 = 0 ρ b 2 = −  x 1−h h b3 = 0

Therefore the body force vector required for equilibrium is ρ b = −  x 1−h e2 h (b) Find the tractions of all of the faces of the prism. Face 1 ( x3 = 1): t 1 = Se3 = 0

Face 3

Face 4 ( b x 2+h x1−b h = 0): t 4 = Sn4 = 0 because S = 0 everywhere on that face.

Face 2 Face 4

Face 2 ( x3 = 0): t 2 = −Se3 = 0 Face 3 ( x1 = 0): t 3 = −Se1 = 0

x2

Face 1

x1 x3

Face 5

Face 5 ( x2 = 0): The traction is ρ t 5 = −Se2 = −  x1−h   x1−b  e 2 b (c) Sketch the normal (σ) and tangential (τ) components of traction on the three faces whose normals are orthogonal to the e3 direction.


90

x2

x2

0

0

0

0

σ

τ x1

x1 0

−ρh

95. A spherical shell has an inside radius of R and an outside radius of 2R. In the center of the sphere there is a magnetic core that sets up a stress field in the shell. The state of stress in the shell is S(x) =



ρ 1− 3 3r r

8R 3

x  x

x1

x3 t (traction)

where r is a magnetic constant of the material, x is the position vector, and r is the radial distance to the point x defined as r ≡ x ⋅ x

B

x2

(Note that ∂r∕∂xj = x j∕r)

x1

x2

Find the body force vector field b in the shell. What is the pressure at the inside surface of the shell? Take a freebody of the shell by slicing it along the plane x3 = 0. What are the tractions t on the shell that must act at the slice?

x3

(a) Find the body force vector field b in the shell. Let f (r) =



3 ρ 1 − 8R3 3r r



The components of the stress tensors are given by Sij = f (r) xi x j. Then we can compute  div S  = Sij,j = f′(r) ∂r x i x j + f (r)x i,j x j + x i x j,j i ∂x j xj = f′(r) r xi x j + f (r)δij x j + xi δ jj

x = f′(r) x ⋅ x ri + 4f (r) xi Thus, div S =  r f′(r)+4f (r) x, in which r f′(r) + 4f (r) =



 

3 3 ρ r -- 12 + 32R + 4 1r − 8R4 3 r4 r r

Therefore, the body force vector required for equilibrium is

 = f

r

91

Chapter 3 The Transmission of Force b = − div S = − f r x

(b) What is the pressure at the inside surface of the shell? The normal to the inside surface of the shell is n = −x∕r , with r = R. The traction acting on that surface is Sn =





ρ (−7)[x  x] − x = 73 ρx R 3R

which acts along the normal to the surface, therefore the pressure is  Sn  = 7 ρR 3

(c) Take a freebody of the shell by slicing it along the plane x3 = 0. What are the tractions t on the shell that must act at the slice? The normal to the surface along the plane x3 = 0 is n = − e3 . The traction acting on that surface is Se3 = f (r) x  x e 3 = f (r) x ⋅ e3 x Also, on the slice x3 = 0, x = x 1 e1+x 2 e 2 ⇒ x ⋅ e3 = 0 and t = Se 3 = 0

96. The stress tensor S can be expressed in cylindrical coordinates ( r, θ, z) as S(r, θ, z) = S rr  er  e r  + Srθ  er  e θ  + Srz  er  e z  + Sθr  e θ  er  + S θθ  e θ  eθ  + S θz  e θ  ez  + Szr  e z  er  + S zθ  e z  eθ  + S zz  e z  ez  where the components (e.g., Srz) are each functions of the coordinates ( r, θ, z). However, now the base vectors er (θ) and eθ (θ) depend upon the coordinate θ. x3 ≡ z ez eθ (θ) z r x1

θ

n5

n1

er (θ)

∆θ

x2

n3 ∆r

n4 θ

r

n2

n4

r∆θ ∆z

∆r n6

r

z

n2

z

Using the coordinate-free definition of the divergence of a tensor field, Eqn. (87), show that the divergence of S in cylindrical coordinates is given by div S(r, θ, z) = 1r ∂  r Ser  + 1r ∂  Se θ + ∂  Se z ∂z ∂r ∂θ


92

Observe from the figure that n1 = eθ (θ+∆θ) and n2 = −eθ (θ), and are constant over the faces 1 and 2, respectively. The normal vectors n3 = er (ξ) and n4 = −er (ξ) with ξ ∈ [θ, θ+∆θ] varying over faces 3 and 4. Finally, note that n5 = ez and n6 = −ez are constant over faces 5 and 6. The volume of the wedge is ∆V = r ∆θ ∆r ∆z plus terms of higher order that vanish more quickly in the limit as ∆V → 0. To compute the component expression for the divergence of the stress tensor, we must expand the vectors Ser, Seθ, and Sez. Show that Ser = Srr e r + Sθr e θ + Szr ez Seθ = Srθ e r + Sθθ e θ + Szθ ez Sez = Srz e r + Sθz e θ + Szz ez Before we take derivatives, we must observe that in terms of the standard constant basis {e1 , e2 , e3 }, the radial and angular base vectors have the form er (θ) = cos θ e 1 + sin θ e 2 eθ (θ) = − sin θ e1 + cos θ e2 and, therefore, ∂er∕∂θ = eθ and ∂eθ∕∂θ = −er. Show that the component expression of the divergence of S is

 ∂S∂r + 1r ∂S∂θ + ∂S∂z + 1r  S −S  e ∂S ∂S ∂S +  ∂r + 1r + + 1r  S +S  e ∂z ∂θ

div S =

+

rr

rθ

rz

θr

θθ

θz

rr

rθ

θr

 ∂S∂r + 1r ∂S∂θ + ∂S∂z + 1r  S  e zr

zθ

zz

zr

r

θθ

θ

z

Begin with Eqn. (87) in text.

div S = lim

v(B)→0

1 v(B)



+

S( r+∆ r, θ, z ) er(θ) (r+∆r) dθ dz −

Ω3



S( r, θ+∆ θ, z ) e θ( θ+∆ θ) dr dz −



S( r, θ, z+∆ z ) e z r dθ dr −

Ω1

+

Ω5

Substituting v(B) = r ∆ r ∆ θ ∆ z we get



Ω6





S( r, θ, z ) er (θ) r dθ dz

Ω4

S( r, θ, z ) e θ ( θ) dr dz

Ω2

S( r, θ, z ) ez r dθ dr



Chapter 3 The Transmission of Force div S(x) =

lim

∆θ ∆z→0

1 r ∆θ ∆z

+ lim

∆r ∆z→0

+ lim

∆θ ∆r→0



r ∆θ ∆z

1 r ∆r ∆z 1 ∆θ ∆r

( r+∆ r) S( r+∆ r, θ, z ) − r S( r, θ, z ) er(θ) dθ dz ∆r



∆r ∆z



r ∆θ ∆r

S(r, θ+∆θ, z) e θ(θ+∆θ) − S( r, θ, z ) e θ(θ) dr dz ∆θ S( r, θ, z+∆ z ) − S( r, θ, z ) e z dθ dr ∆z

Taking the limits we find that div S = 1r ∂  r Ser  + 1r ∂  Seθ + ∂  Sez ∂z ∂r ∂θ Substituting the components expression for the tractions we get div S = 1r ∂  r  S rr er+S θr eθ+S zr ez   + 1r ∂  Srθ er+S θθ eθ+S zθ ez ∂r ∂θ + ∂  S rz er+S θz eθ+S zz e z  ∂z Carrying out the explicit differentiations we get ∂S ∂S ∂S div S = 1r  Srr er + S θr eθ + S zr ez  + rr er + θr e θ + zr ez ∂r ∂r ∂r ∂S ∂S ∂S + 1r  S rθ eθ − S θθ er  + 1r rθ e r + 1r θθ eθ + 1r zθ ez ∂θ ∂θ ∂θ +

93

∂S ∂Srz ∂S e + θz e θ + zz ez ∂z r ∂z ∂z

Finally, rearranging the terms we arrive at the desired result.

94


Chapter 4 Linear Elastic Constitutive Theory

97. The constitutive equations for a three-dimensional isotropic, linearly elastic material can be expressed in the form Sij = λEkk δ ij+2m E ij where the subscripts i, j, and k range over the values 1, 2, and 3. Find equivalent expressions for the constitutive equations that already reflect the plane stress condition S33 = S23 = S 13 = 0, that is, find new material constants λ* and m * such that the two-dimensional relationship can be written as Sαβ = λ* E γγ δαβ + 2m * E αβ where the Greek indices range only over the values 1 and 2. Express the new constants ( λ* , m *) in terms of the constants ( λ, m) of the three-dimensional theory.

To find equivalent constitutive equations for plane stress condition, use S33 = 0 to get E33 which is not zero. S33 = λ  E11+E 22  +  λ+2m  E33 = 0 E33 = −

λ(E 11+E22) λ+2m

Substitute E33 into S11 to get S11 = λ(E22+E 33) + (λ+2m) E11 = λE22 − = Similarly, for S22

λ 2 (E11+E 22) + (λ+2m) E 11 λ+2m

2 λm  E +E22 + 2m E 11 λ+2m 11


96

S22 = λ(E11+E 33) + (λ+2m) E22 = λE11 − =

λ 2 (E11+E 22) + (λ+2m) E 22 λ+2m

2 λm  E +E22 + 2m E 22 λ+2m 11

Finally for S12 we have S12 = 2m E12. Combine these three equations and make a single constitutive equation with indices α, β and γ, which have the range (1, 2). Sαβ =

2 λm E γγ δαβ + 2m Eαβ λ+2m

which are in the proposed form if we let λ =

2 λm , λ+2m

m = m

98. Demonstrate that the following relationships between the elastic constants λ, m, C, K, and ν hold for an isotropic, linearly elastic material λ =

m(C−2m) 2mν Cν = = = 3Kν (1+ν)(1−2ν) 1+ν 3m−C 1−2ν

K = λ + 23 m =

mC λ(1+ν) C = = 3ν 3(1−2ν) 3(3m−C)

C = 2m(1+ν) =

m(3λ+2m) 9Km λ(1+ν)(1−2ν) = = ν 3K+m λ+m 3K(1−2ν) λ(1−2ν) = 2(1+ν) 2ν

m =

C = 2(1+ν)

ν =

3K−2m λ = C −1 = = 3K−C 2m 6K 2( λ+m ) 2( 3K−m )

3 (K − λ) 2

=

We can observe that, in each case, one of the elastic constants is expressed in terms of two of the others from the set of five constants. There are some natural limits to the values that the constitutive parameters can take. Assume that under compressive hydrostatic pressure it is impossible for the volume to increase, and that in uniaxial tension it is impossible for a bar to get shorter. What do these hypotheses imply about the other moduli?

Start from Eqn. (240), (241), (242) and what we have already proved in Chapter 4:

Chapter 4

Linear Elastic Constitutive Theory

λ =

Cν (1+ν)(1−2ν)

(a)

m =

C 2(1+ν)

(b)

ν =

λ 2 (λ+m)

(c)

C =

m(3λ+2m) λ+m

(d)

K = λ + 23 m

(e)

Noting that C = 2 m(1+ν) we can eliminate C from (a) to give λ =

2 m(1+ν) ν 2 mν = (1+ν)(1−2ν) 1−2ν

Noting that ν = (C−2 m)∕2 m we can eliminate ν from (a) to give λ =

C



C−2 m 2m

1−2

C−2 m 1+ 2 m

=



C−2 m 2m

Cm (C−2 m) m(C−2m) = 3m−C (2 m−C−2 m) (m−C+2 m)

Noting that C = m(3λ+2m)∕(λ+m) we can eliminate C from (a) to give m(3λ+2m)

λ+m ν ν C λ = = = ν (3λ+2m) = 3Kν (1+ν) (1−2ν) (1+ν) 1− λ  1+ν 1+ν λ+m

Noting that λ = m(C−2m)∕(3m−C) we can eliminate λ from (e) to give K = λ + 23 m =

m(C−2m) 2 3m(C−2m) + 2m(3m−C) mC + 3m = = 3m−C 3(3m−C) 3(3m−C)

Noting that 2m = λ(1−2ν)∕ν from (b), we can eliminate m from (e) to give K = λ + 23 m = λ +

λ(1−2ν) λ(1+ν) = 3ν 3ν

Substitute (a) and (b) into (e) to give K = λ + 23 m =

3 Cν + C(1−2ν) Cν C + C = = (1+ν)(1−2ν) 3(1+ν) 3(1−2ν) 3(1+ν)(1−2ν)

From (b), we can directly manipulate C C = 2m(1+ν) From (a), we can directly manipulate C C =

λ(1+ν)(1−2ν) ν

97


98

Noting that λ = K− 23 m we can eliminate λ from (d) to give C =

m ( 3K − 2m+2m) 3 mK 9Km m(3λ+2m) = = = 2 1 3K+m λ+m K − 3 m+m K+ 3m

From (e), we can directly manipulate m m =

3 (K − λ) 2

Noting that C = 3K(1−2ν) we can eliminate C from (b) to give m =

3K(1−2ν) C = 2(1+ν) 2(1+ν)

Noting that C = λ(1+ν)(1−2ν)∕ν we can eliminate C from (b) to give m =

λ(1+ν)(1−2ν) λ(1−2ν) C = = 2ν(1+ν) 2ν 2(1+ν)

99. Show that the isotropic elasticity tensor with components Cijkl = λδij δ kl + m  δ ik δjl+δ il δ jk is invariant with respect to coordinate transformation since the components of the tensor in the two coordinate systems are related by cabcd = Cijkl Q ai Qbj Qck Q dl,, where, as usual, Qij = gi ⋅ e j are the components of the orthogonal change-of-basis tensor. (Hint: another way to view change of basis is gi = Qij e j.) Substitute (1) into (2) cabcd = λδij δ kl + m δ ik δjl+δ il δjk Qai Q bj Qck Qdl = λδij δ kl Q ai Qbj Q ck Q dl + m δik δ jl Q ai Qbj Q ck Qdl + m δil δ jk Qai Qbj Q ck Qdl = λQai Q bi Q ck Qdk + m Q ai Q bj Qci Q dj + m Q ai Qbj Q cj Qdi = λδab δ cd + m δ ac δbd+δ ad δbc Therefore the isotropic elasticity tensor is invariant with respect to coordinate transformation.

100. Consider a linearly elastic, isotropic material with Lamé parameters λ and m, subjected to the following displacement map u(x) = β  x21 e 1+x22 e 2 . Assume that the linearized strain tensor is adequate to characterize the strain field, and compute the body forces required to satisfy equilibrium.

Chapter 4


99

The linearized strain tensor can be given by Eqn. (129). E =

1 ∇u(x)+∇uT(x) 2

= 2βx 1 e1  e 1 + 2βx2 e2  e 2

The stress tensor can be obtained from the strain tensor by Eqn. (236) as S = λ( tr E) I + 2m E Noting that tr E = 2 β ( x1+x 2 ) we have S = 2β  λ( x1+x 2 ) + 2mx1 e 1  e1 + 2β  λ( x1+x 2 ) + 2mx 2 e2  e 2 The body force required for equilibrium is b = −div S = −2β  λ+2m  e 1+e2 

101. Arrange the six independent stress and strain components in column matrices as follows: S = ( S 11, S 22, S 33, S 12, S23, S13 ) T and E = ( E 11, E 22, E 33, 2E 12, 2E23, 2E13 ) T. Assume that the constitutive equations of linear elasticity hold. Show that the constitutive equations can be expressed in matrix form as S = DE, where D is a six by six matrix. Expand constitutive relationship equation as a component form. S11 = ( λ+2m ) E 11 + λE22 + λE 33 S22 = λE11 + ( λ+2m ) E22 + λE 33 S33 = λE11 + λ E22 + ( λ+2m ) E 33 S12 = 2m E12 S13 = 2m E13 S23 = 2m E23 The normal stress terms and shear stress terms are independent of each other, therefore the upper right and lower left blocks of D are zero

D =

λ+2m

λ

λ

0

0

0

λ

λ+2m

λ

0

0

0

λ

λ

λ+2m

0

0

0

0

0

0

m

0

0 0 m

0

0

0

0

m

0

0

0

0

0

100


102. Consider the thin rectangular sheet with Young’s 15 psi modulus C=1000 ksi and Poisson’s ratio ν=0.45. The sheet is subjected to a uniform state of stress through the 25 psi tractions given in the sketch. The thickness of the sheet before the tractions were applied was 0.1 in. What is the thickness of the sheet after the tractions are applied?

25 psi 15 psi

The stress tensor can be determined from the given tractions. We can note that the tractions Se1 = 25e1, Se2 = 15e2, and Se3 = 0. Thus, the components of the stress tensor must be 25 0 0

S =

0 15 0

0 0 0

The change in thickness is t E 33, where t is undeformed thickness. From Eqn. (244) in the text E33 = − ν ( tr S) + 1+ν S33 = − 0.45 ( 25+15) = −0.018 C C 1000 Thus, the thickness change is -- 0.0018.

15 psi 103. Consider the thin rectangular sheet with Young’s modulus C=1000 ksi and Poisson’s ratio ν=0.2. The sheet is fixed between two immovable frictionless plates 35 psi 35 psi and is subjected to a uniform state of stress through the tractions around the edges as shown in the sketch. The 15 psi thickness of the sheet before the tractions were applied was 0.3 in. What is the state of stress in the sheet after 35 psi 35 psi the tractions are applied? What are the reacting tractions provided by the plates? Find the ratio of the change in volume to the original volume of the sheet.

(a) What is the state of stress in the sheet after the tractions are applied?

S~

--35 0 0 0 --15 0 0 0 S 33

The unknown component in S can be computed as follows. E = 1 + ν S − ν tr( S) C C

Chapter 4


101

where tr( S) = − 50 + S33 Since the plates cannot move in the vertical direction E33 = 0

⇒

1 + ν S − ν ( − 50 + S ) = 0 33 33 C C

S33 = − ν(50) = − 10 ksi (b) What are the reacting tractions provided by the plates? t = Sn On the top face n = e 3 ⇒ t = −10 e 3 On the bottom face n = −e 3 ⇒ t = 10 e 3 10 psi

(c) Find the ratio of the change in volume to the original volume of the sheet. The change in volume is ∇V ≈ tr( E) V The strain components are E11 = 1 + ν (−35) − ν (−60) = --42+12 = − 0.030 C C 1000 E22 = 1 + ν (−15) − ν (−60) = --18+12 = − 0.006 C C 1000 E33 = 0 Therefore, ∇V ≈ tr( E) = − 0.030 − 0.006 = − 0.036 V

t t 104. A disk made of isotropp ic, linearly elastic material is p p subjected to a known uniform Top Side p p pressure p around its perimet ter. The faces of the disk are t clamped between immovable, frictionless plates so that the strain through the thickness is zero. Assume that the stress state is homogeneous throughout the disk and that the Lamé constants are known. Find the tractions t acting on the faces.


102

We know that E11 = 0 because the top and bottom of the disk are immovable. Therefore, the traction t = S 11 is an unknown reaction force. The applied tractions around the cylinder suggest that S22 = S33 = p . To find p, use Eqn. (237) in the text E11 = −

λ tr S + 1 S11 2m 2m ( 3 λ+2m)

Substituting tr S = t+2p gives E11 = −

λ ( t+2p) + 1 t = 0 2m 2m ( 3 λ+2m)

Solving for t we find t =

λ p λ+m

σ1 105. In a triaxial test, a cylindrical specimen is subjected to a uniform pressure σ1 on the σ1 = − 50 psi ends of the cylinder and a uniform pressure σ2 σ2 = − 20 psi on the sides. The change in height ∆h and the σ2 σ2 ε 1 = − 0.00367 change in diameter ∆d are measured. Let ε 2 = + 0.00133 ε 1 ≡ ∆h∕h and ε 2 ≡ ∆d∕d, where h is the σ1 original height and d the original diameter. The values measured in a test are given in the diagram. Assume that the material is linear, isotropic, and elastic. What is the volume of the deformed cylinder? Compute the value of the bulk and shear moduli (K and m) for this sample.

(a) What is the volume of the deformed cylinder? The deformed volume can be obtained (approximately) as V = V o ( 1+tr E) = Vo ( 1+Á 1+2 Á 2 ) = 0.99899 Vo (b) Compute the value of the bulk and shear moduli (K and m) for this sample. The bulk and shear moduli can be determined from the parameters λ and m. S ~

σ1 0 0 σ2 0 0

0 0 σ2

E ~

Á1 0 0

0 Á2 0

0 0 Á2

From the constitutive relationship Sij = λEkk δ ij + 2m E ij The trace of the stress tensor is Sii = ( 3 λ+2m) E ii. The bulk modulus can be found from the pressure and volume change as follows:

Chapter 4


103

S σ +2 σ 2 = 29703 psi K = λ+ 23 m = 1 ii = 1 1 3 Eii 3 Á 1+2 Á 2 To determine the shear modulus, use the constitutive relationship S11 = λEkk + 2m E11 = ( λ+2m ) E11 + λ( E22+E33) Substitute λ = K − 2 m∕ 3, then σ1 = K+ 43 m  Á 1 + 2  K− 23 m  Á 2 Solving for m and observing that σ1+2σ 2 = K(Á 1+2Á 2) we find that m =

3 4

Á +2 Á ) σ +2 σ )  σ −K(Á −Á  =  3 σ −(Á −Á  = 3000 psi 1

1

1

1 4

2

2

1

1

1

2

2

106. Prove the identity det[ I+2E] = 1+2 I E+4 II E+8 III E . (Hint: Use the component expression for the determinant of a tensor. The Á−δ identity from Chapter 1 may also be useful.)

det[ I+2E] =

1 Á Á ( δ +E il ) ( δ jm+Ejm ) ( δkn+E kn ) 6 ijk lmn il

=

1 Á Á 6 ijk lmn

 δil δ jm δkn

+ 2 ( Eil δ jm δ kn+Ejm δ il δkn+E kn δ il δjm )

+ 4 ( Eil E jm δ kn+Eil E kn δjm+E jm E kn δil ) + 8 Eil E jm Ekn  Noting that III E = det( E) = 16 Á ijk Á lmn Eil Ejm E kn and det( I) = 16 Á ijk Á lmn δ il δjm δkn = 1 we have det[ I+2E] = 1 + 8 III E +

1 6

 2  E ilÁijk Á ljk+Ejm Áijk Á imk+Ekn Á ijk Á ijn 

+ 4  Eil E jm Á ijk Á lmk+E il Ekn Á ijk Á ljn+Ejm E kn Á ijk Á imn   From the ε—δ identity proved in Chapter 1 Á ijk Á imn = δ jm δkn − δ jn δ km Substitute j = m into the ε—δ identity, then Á ijk Á ijn = δ jj δkn−δ jn δ kj = 3 δkn−δ kn = 2 δ kn Substitute them into the original expression to get det[ I+2E] = 1 + 8 III E +

1 6

 4 ( E il δil+E jm δ jm+Ekn δ kn )

+ 4  Eil E jm ( δ il δjm−δ im δ jl ) + Eil E kn ( δ il δkn−δ in δ kl ) + Ejm E kn ( δjm δkn−δ jn δkm ) 

104


Contracting the terms involving the Kronecker deltas we get det[ I+2E] = 1 + 8 III E +

1 6

 4  E ii+Ejj+E kk 

+ 4  Eii E jj−Eml Elm+E ii Ekk −E nl E ln +Ejj E kk −E nm Emn   We know that I E = Ekk and II E = 12  E ii Ejj−E ij E ij . Substituting these into the above expression we obtain det[ I+2E] = 1 + 2 I E + 4 II E + 8 III E

107. Consider a beam of length ℓ with its axis oriented along the z 3 direction. The cross section of the beam lies in the z 1−z 2 plane, and its second moment of the area is equal to I. The beam is subjected to equal and opposite end moments of magnitude M, bending it about the axis with second moment of the area I. The beam is made of elastic material with moduli C and ν. The displacement field in the beam is given by the expression u(z) = M  12  z 23+νz 21−ν z 22  e 1 + νz 1 z 2 e 2 − z 1 z 3 e 3  CI Assume that the applied moment is small enough relative to CI that the displacements are quite small. Compute the components of the strain tensor. Compute the components of the stress tensor from the strain tensor and the linear elastic constitutive equations. Verify that the stress field satisfies the equations of equilibrium.

(a) Compute the components of the strain tensor. Use linearized theory for the strain tensor. ∇u ~ M CI

E =

1 2

ν z 1 --ν z 2 z3 ν z2 ν z1 0 --z3 0 --z1

[∇u+∇uT] ~ M CI

ν z1 0 0 0 ν z1 0 0 0 −z1

(b) Compute the components of the stress tensor from the strain tensor and the linear elastic constitutive equations. From the linear elastic constitutive relationship in Eqn. (243) in text S =

Cν ( tr E) I + C E (1+ν) (1+ν) (1−2 ν)

The trace of the strain tensor given above is tr E = M (2 ν−1) z 1 CI

Chapter 4

105


Substituting into the constitutive relationship we get



Cνz 1 I + C E S = M − 1+ν CI 1+ν

 = − MI z  e 1

3

 e3 

(c) Verify that the stress field satisfies the equations of equilibrium. Suppose b=0, and from stress equation of equilibrium since S33 is the only nonzero component of the stress tensor b = −div S = − M z 1, 3 e 3 = 0 I

z3

108. Consider the displacement map u(z) for a sphere of unit radius, given by the explicit expression u(z) = ε  z ⋅ z  z

z2 B

z1 where ε is a (very small) constant of the motion. Assume that the maΩ terial is isotropic and linearly elastic with material constants λ and m (i.e., the Lamé parameters). Compute the body force b required to maintain equilibrium. Compute the traction forces t that must be acting on the surface of the sphere. Determine the principal stress field associated with the given motion.

(a) Compute the body force b required to maintain equilibrium. ∇u = ∂  ε  z k z k z i  e i  ej = ε 2 z k δ kj z i + z k z k δij  ei  e j ∂z j = ε 2 z j z i + z k z k δ ij  e i  ej = ε 2 z  z +  z ⋅ z  I  The linearized strain tensor can be given by Eqn. (129). E =

1 ∇u+∇uT  2

= ε 2 z  z +  z ⋅ z  I 

tr ( E) = ε 2 z i z i + z k z k δ ii  = 5 ε z ⋅ z  The stress tensor can be obtained from the strain tensor as S = λ( tr E) I + 2m E = 5 λε z ⋅ z  I + 2 m ε 2 z  z +  z ⋅ z  I  = ε  5 λ + 2 m  z ⋅ z  I + 4 m ε z  z  The body force required for equilibrium is b = −div S = − ∂  ε  5 λ + 2 m  z k z k  δ ij + 4 m εz i z j e i ∂z j

106


= − 2 ε  5 λ + 2 m  z k δ kj δ ij + 4 m εδ ij z j + 4 m εδjj z i ei = −ε  10 λ + 20 m  z (b) Compute the traction forces t that must be acting on the surface of the sphere. Note that the radial vector z is normal to the surface of the sphere and is of unit length (since the radius of the sphere is 1). Sz = ε  5 λ + 2 m  z ⋅ z  z + 4 m ε z  z  z = ε  5 λ + 6 m  z ⋅ z  z (c) Determine the principal stress field associated with the given motion. First we can rewrite the stress tensor as S = α I + β  z  z  = α  I − z  z +  α+β  z  z where α = ε  5 λ + 2 m  z ⋅ z  , β = 4 m ε. From the spectral decomposition theorem, we have one distinct eigenvalue and a pair of repeated eigenvalues m 1 = α; n 1 = z (radial direction) and m 2 = m 3 = α + β; n2, n3 are any two vectors in the plane ⊥ z (tangent to the sphere surface).

109. Let the elasticity tensor be given by Cijkl = λδij δ kl + m δ ik δjl+δ il δjk. Show that the expression Sij = Cijkl E kl reduces to Eqn. (236).

Substitute the elasticity tensor into the constitutive equation. Sij = λδij δ kl + m δ ik δjl+δ il δjk Ekl = λδij δ kl E kl + m δik δ jl E kl + m δil δ jk E kl = λδij E kk + m E ij + m Eji The strain tensor is symmetric, therefore Sij = λEkk δ ij + 2 m E ij

110. A cube of isotropic elastic material, having Lamé constants λ = 1000 psi and m = 1000 psi is in a homogeneous (i.e., does not vary with position) state of stress given by a stress tensor with components S~

10 2 1

2 5 1

1 1 S33

Chapter 4

107


Find the stress component S33 that is consistent with the observation that the cube decreases in volume (from the stress-free state) by 5%. Now compute the components of the deviatoric stress tensor and the strain components E13 and E33. (a) Find the stress component S33 that is consistent with the observation that the cube decreases in volume (from the stress-free state) by 5%. S = λ tr(E)I + 2mE tr(E) = --0.05

⇒ tr(S) = (3λ + 2m) tr(E)

⇒ 15 + S 33 = (5000)(--0.05) = − 250 ∴ S33 = − 265

(b) Now compute the components of the deviatoric stress tensor ∴ S33 = − 265 10+ 250 3 2 S′ = S − 1 ( tr S ) I ~ 3 1

S′ ~

280 3 2

2

1

5+ 250 3 1

1

2

1

265 3 1

1

−265+ 250 3

1 543 3

(c) and the strain components E13 and E33. S13 = 2 m E13 S33 = λ( trE) + 2 m E33 E33 =

⇒

E13 = 1

2000

⇒

= 0.0005

E 33 = 1 [S 33 − λtr( E) ] 2m

1 (− 365 − 1000( −0.05) ) = − 0.1075 2000

111. A block of elastic material, having Frictionless Ao Lamé constants λ = 1000 psi and σ σ h m = 1000 psi is subjected to a lateral ho compressive pressure of σ = 80 psi and clamped between two frictionless rigid plates that reduce the height of the block to 99% of its original height. Compute the total force required on the plates to accomplish the motion. Compute the volume of the block after deformation. Compute the change in the area of the block on the faces in contact with the plates.


108

(a) Compute the total force required on the plates to accomplish the motion. To compute the total force acting on the plates, we need to find the stress component normal to the plates S11. Given that S22 = S33 = − 80 psi Let E22 = E33 = Á. From the change of height we can deduce that E11 = − 0.01, tr( E) = 2 Á − 0.01 Substituting in the constitutive equation ⇒ S22 = λ(2Á − 0.01) + 2mÁ = − 80 2  λ+m  Á = − 80 + 0.001 λ = − 80 + 10 = − 70 ⇒ Á = − 70 = − 7 4000



S11 = λ − 7 − 2 200

200

400

 − 0.01  2m = − 45 − 20 = − 65 psi

And the total force = − 65 A o lbs. (b) Compute the volume of the block after deformation. ∆V ≈ tr( E ) = E + E + E = − 7 − 2 = − 9 11 22 33 V 200 200 200 v = V − 9 V = 191 V 200 200 (c) Compute the change in the area of the block on the faces in contact with the plates. V = h o A; v = h a = 0.99 ho a Using the deformed volume of the block calculated in the previous part v = 191 V = 191 Aho 200

200

Substituting, 0.99 a h o = 191 Aho ⇒ a = 191 A 200

198

a−A =− 7 A 198

112. Consider a body B subjected to the following displacement map:

Chapter 4

109


u(z) = β  z 21−2z 2 z 3  e 1 + β  z 22+2z 1 z 3  e2 + β  z 23−2z 1 z 2  e3 where β is a (very small) constant. Find the stress tensor associated with this motion, assuming that the material is linear, isotropic, and elastic with moduli λ and m and that the stress is zero when the displacement is zero. Find the body force field required to maintain equilibrium for the given motion. Find the principal values of the (linearized) strain tensor.

(a) Find the stress tensor associated with this motion, assuming that the material is linear, isotropic, and elastic with moduli λ and m and that the stress is zero when the displacement is zero. The gradient of displacement is

∇ u ~ 2β

z1

−z 3

−z2

z3

z2

z1

−z2 −z 1

z3

Therefore, the strain is

E = 1 (∇u + ∇uT) ~ 2β 2

z1

0

−z 2

0

z2

0

−z2

0

z3

with a trace of tr( E) = 2β( z 1 + z 2 + z 3) The stress can be computed from teh constitutive equation S = λ tr(E) I + 2m E 0

λ(z1+z 2+z3)+2m z1

S ~ 2β

0

−2 m z 2 0

λ(z 1+z2+z 3)+2 m z2

−2 m z 2

0

λ(z 1+z2+z 3)+2 m z3

(b) Find the body force field required to maintain equilibrium for the given motion. div S + b = 0 div S =

∂Se i ~ 2β  λ+2m  ∂z i

b = − div S

1 1 1

b = −2β  λ+2m 

(c) Find the principal values of the (linearized) strain tensor. det  E−γI  = 0

1 1 1


110

Since E is proportional to 2 β it is sufficient to find the eigenvalues of E∕2 β then multiply the result by 2 β

det

z1 − γ

0

−z 2

0

z2 − γ

0

−z2

0

z3 − γ

= ( z 2 − γ)( z 1 − γ)( z 3 − γ) − z 22 = 0

One root of this equation is γ1 = z 2 . The other two are the roots of the quadratic γ2 − ( z 1 + z 3 ) γ + ( z 1 z 3 − z 22) = 0 which has solutions ⇒ γ2,3 =

( z 1 + z 3)  ( z 1 + z 3 ) 2 − 4( z 1 z 3 − z 22) 2

The principal values of strain are 2β γi

i = 1, 2, 3

113. The state of the deformation at a certain point in a solid body is such that it has the following principal strains Á 1 = ε and Á 2 = Á 3 = 3ε, where ε is a known value. The principal directions associated with these principal strains are known to be n1 , n2 , and n3 . Assuming linear, isotropic response, find the principal values and principal directions of the stress tensor S. The material constants are λ = 1000 psi and m = 1000 psi. What is the average pressure p at the point in question? What is the change in volume in the neighborhood of the point in question?

(a) Assuming linear, isotropic response, find the principal values and principal directions of the stress tensor S. The material constants are λ = 1000 psi and m = 1000 psi. S = λtr ( E) I + 2m E Sn = σ n = λtr ( E) n + 2 m En =  λtr ( E) + 2 m Á n ∴ σi =  λ  Á 1+Á 2+Á 3  + 2m Á i  The principal directions of S and E are the same. tr ( E) = Á + 3Á + 3Á = 7Á ⇒ σ1 = 1000 ( 7 Á ) + 2000( Á ) = 9000 Á σ2 = σ3 = 1000 ( 7 Á ) + 2000( 3 Á ) = 13000 Á

Chapter 4


111

(b) What is the average pressure p at the point in question? p = 1 tr ( S ) = 1 ( σ1 + σ 2 + σ3 ) = 35000 Á 3 3 3 (c) What is the change in volume in the neighborhood of the point in question? ∆V ≈ tr (E) = 7Á ⇒ ∆V = 7 Á V V

114. The strain energy function of a nonlinear hyperelastic material is given by W(E) =

1 a Eii E jj + 12 b E ij E ij + 13 c E ij Ejk Eki 2

where a, b, and c are material constants and E = E ij [ e i  ej ] is the strain tensor. Find the stress tensor S as a function of the strain E implied by the strain energy function.

Smn =

∂W ( E) = 12 a ( δ im δin Ejj + E ii δjm δ jn ) + 12 b ( δim δjn E ij + Eij δ im δ jn ) + ∂ Emn 1 3

c ( δim δjn E jk Eki + E ij δ jm δkn E ki + Eij Ejk δ kmδin )

Smn = 12 a ( δmn E jj + δmn E ii ) + 12 b ( Emn + E mn ) + 13 c ( Enk E km + Eim E ni + Emj E jm )

115. The strain energy function of a nonlinear hyperelastic material is given by W(E) = a E ii ln1+E jj  + 32 b Eij E ij where ln(.) indicates the natural logarithm of (.), a and b are known material constants, and E = E ij [ e i  ej ] is the strain tensor. Find the stress tensor S as a function of the strain E implied by the strain energy function. Consider a hydrostatic state of stress with pressure p in which the stress tensor is given by S = pI. Set up a relationship between the change in volume and the pressure p. What is the pressure required to decrease the volume to 95% of the original volume (assume that the linearized strain tensor is adequate)?

(a) Find the stress tensor S as a function of the strain E implied by the strain energy function. Skℓ =

∂W ( E) = a ∂  Eii ln ( 1+E jj )  + 3 b ∂ ( Eij Eiℓ ) ∂Ekℓ 2 ∂Ekℓ ∂ E kℓ


112



= a ln( 1 + Ejj ) +



Eii δ + 3 b Ekℓ 1 + Ejj kℓ

Let e ≡ tr ( E) ≈ ∆V∕V. Then, S = a ( ln(1 + e) +

e ) I + 3bE 1+e

(b) Consider a hydrostatic state of stress with pressure p in which the stress tensor is given by S = pI. Set up a relationship between the change in volume and the pressure p. What is the pressure required to decrease the volume to 95% of the original volume (assume that the linearized strain tensor is adequate)?

S = pI

⇒

tr ( S ) = 3 p

Thus, taking the trace of S from above tr ( S ) = 3 p = 3 a ( ln(1 + e) + p = a ( ln(1 + e) +

e ) + 3be 1+e

e ) + 3be 1+e

For a 5% volume increase e = − 0.05 ⇒ p = − 0.1039 a − 0.05 b

116. The strain energy function of a nonlinear hyperelastic material is given by the (component) expression W(E) = a 0 Eii E jj E kk+a1 E ij Ejk Eki , where a0 and a1 are known material constants, and Eij is the ijth component of the strain tensor E. Find the stress tensor S as a function of the strain E implied by the strain energy function. Is the material isotropic? Explain your answer. Is the material linear? Explain. Consider a uniform state of shearing in which the strain tensor has components E12 = E21 = γ and all other components equal zero, where γ is a given constant. Find the principal values of the stress tensor S for the given constitutive model under the given state of strain.

(a) Find the stress tensor S as a function of the strain E implied by the strain energy function. Smn =

∂W ( E) = a 0 ∂  E ii Ejj Ekk  + a 1 ∂  E ij Ejk Eki  ∂E mn ∂E mn ∂ Emn

= a0 δ im δin Ejj E kk + Eii δ jm δ jn Ekk + E ii Ejj δkm δ kn  + a1 δ im δjn E jk E ki + Eij δ jm δkn Eki + E ij Ejk δ km δ in 

Chapter 4

113


Smn = a0  E jj E kk + Eii E kk + Eii E jj  δ mn + a1  Enk E km + Eni E im + Enj Ejm  (b) Is the material isotropic? Explain your answer. Is the material linear? Explain. The strain energy function is written in terms of the invariants of the strain tensor, therefore the material is isotropic. The stress components are quadratic in strain, therefore the material is not linear. (c) Consider a uniform state of shearing in which the strain tensor has components E12 = E21 = γ and all other components equal zero, where γ is a given constant. Find the principal values of the stress tensor S for the given constitutive model under the given state of strain. Eii = 0 ⇒ Smn = a1  E nk E km + Eni E im + Enj E jm  which can be alternatively written as (given the symmetry of E) S = 3 a 1( EE) T = 3 a 1ETET = 3 a1EE

⇒ S ~ 3 a1

0

γ

0

0

γ

0

γ

0

0

γ

0

0

0

0

0

0

0

0

= 3 a1

γ2

0

0

0

γ2

0

0

0

0

Therefore the principal values of stress are m 1 = m 2 = γ 2, m 3 = 0

117. The strain energy function of a nonlinear hyperelastic material is given by W(E) = ln1+α E ij Eij  + β  e Eii − Eii  where ln(.) indicates the natural logarithm of (.) and e(.) indicates the exponential of (.), α and β are known material constants, and Eij is the ijth component of the strain tensor E. Find the stress tensor S as a function of the strain E implied by the strain energy function. How do the constants α and β relate to the Lamé parameters of linear isotropic elasticity? Consider a uniform state of dilatation in which the strain tensor is given by E = ε I, where ε is a constant. Find the principal values of the stress tensor as a function of ε.

(a) Find the stress tensor S as a function of the strain E implied by the strain energy function. Smn =

∂W ( E) = ∂ ln1+α E ij Eij  + β ∂  eEii − E ii  ∂E mn ∂Emn ∂ Emn =

2α E δ δ + β  eEkk − 1  δim δ in 1+α Ekℓ E kℓ ij im jn


114

=

2α E + β  eEkk − 1  δ mn 1+α Ekℓ E kℓ mn

In direct notation the stress has the form S=

2α

E + β  e tr E − 1  I ( )

2 1+α trE

(b) How do the constants α and β relate to the Lamé parameters of linear isotropic elasticity? When strains are small, the quadratic term Ekℓ Ekℓ becomes negligible. Furthermore, expanding eEkk in a Taylor series around Ekk = 0 gives etr(E) ≈ 1 + trE + 1 trE +  2

2

Dropping the terms of order 2 and higher, and substituting in the stress tensor obtained above ⇒ S = 2 α E + β tr  E  I from which it can be observed that a → m, b → λ for small strains. (c) Consider a uniform state of dilatation in which the strain tensor is given by E = ε I, where ε is a constant. Find the principal values of the stress tensor as a function of ε. Let tr( E) = 3 ε so that S=





2α ε 2α ε I + β  e3 ε − 1  I = + β e 3 ε − 1  I 1+9 αε 2 1+9 αε 2

The principal values of the stress tensor are m1 = m2 = m 3 =

2αε + β e 3 ε − 1  1+9 αε 2

118. The strain energy function of a nonlinear hyperelastic material is given by the expression W(e, γ) = a0 e 2+a1 γ+a 2 e γ, where a0 , a1 , and a2 are known material constants, and the scalar invariant strain measures e and γ, which are functions of the strain tensor E, are defined as e ≡ tr  E  and γ ≡ tr  E′ E′ , where E′ ≡ E−e I∕3 is the deviator strain. Observe that ∂e∕∂E = I and ∂γ∕∂E = 2E′. Find the stress tensor S as a function of the strain E implied by the strain energy function. Consider a state of hydrostatic pressure S = p I, where p is a given pressure. Find the relationship among p, e and γ. Next consider a sample of the material subjected to a state of pure shear strain described by E = g  n  m + m  n , where g is a given constant describing the motion and m and n are given orthogonal unit vectors. Will there be a change in volume of the sample? Do you expect that you would need a confining pressure to execute this motion? Why?

Chapter 4

115


(a) Find the stress tensor S as a function of the strain E implied by the strain energy function. The stress can be computed as ∂γ S = ∂W = ∂W ∂e + ∂W ∂e ∂E ∂γ ∂E ∂E = 2a0 eI + 2a 1 E + a2 γ I + 2e E  =  2a0 e+a 2 γ  I +  2a1+2 a 2 e  E (b) Consider a state of hydrostatic pressure S = p I, where p is a given pressure. Find the relationship among p, e and γ. Compute the trace of the stress tensor tr S  = p tr(I) =  2a 0 e+a2 γ  tr(I) +  2a1+2 a2 e  tr(E) 3p = 3  2a0 e+a 2 γ  p = 2a 0 e+a2 γ (c) Next consider a sample of the material subjected to a state of pure shear strain described by E = g  n  m + m  n , where g is a given constant describing the motion and m and n are given orthogonal unit vectors. Will there be a change in volume of the sample? Do you expect that you would need a confining pressure to execute this motion? Why? The two strain measures can be computed for the present case as follows: e = tr(E) = g  tr[n  m]+tr[m  n]  = g  n ⋅ m+m ⋅ n  = 0 e = ∆V∕V = 0

(no volume change)

also

E=E

γ = tr(EE) = g 2 tr [n  m+m  n] [n  m+m  n]  = g2 tr [n  m][n  m] + [n  m][m  n] + [m  n][n  m] + [m  n][m  n]  = g2 tr [n ⋅ m][n  m] + [m ⋅ m][n  n] + [n ⋅ n][m  m] + [n ⋅ m][m  n]  = g2 tr [n  n] + [m  m]  = 2g2 Therefore, the stress is S = a 2 γI + 2a 1 E,

tr(S) = 6a 2 g2

The confining pressure is proportional to tr S. Therefore, this deformation will require confining pressure to execute.


116

119. A cube of elastic material, having Lamé constants λ = 1000 2σ psi and m = 1000 psi is subjected to purely normal tractions on its faces as shown in the sketch. Compute the value of σ required −σ to change the volume of the block by 2% of its original volume. 3σ

S~

--σ

0

0

0

2σ

0

0

0

3σ

3σ −σ 2σ

tr ( S ) = 5 σ

S = λtr ( E) I + 2 m E ⇒ tr ( S ) = ( 3λ+2m ) tr ( E) tr ( E) ≈

∆V = 3 V 100

∴ 5 σ = 5000 ( 3 ) ⇒ σ = 300 psi 100

120. Three unit cubes (1×1×1) are uniformly compressed between two rigid plates with an aggregate force of P. The change in height is the same for all three cubes. The two outer cubes are made of material A, while the inner cube is made of material B. Both of the materials are linearly elastic with Lamé constants λA = 1000 psi,

m A = 1000 psi

λB = 500 psi,

m B = 2000 psi

P

A

B

A P

Compute the force P required to change the volume of the middle block by 3% of its original volume. What is the final area of the compressed face of the outer cubes?

(a) Compute the force P required to change the volume of the middle block by 3% of its original volume. The plates are compressed, therefore the volume of block B must be decreased by 3 percent ⇒ tr ( EB ) = −0.03. The stress tensors have the form,

SA ~

α

0

0

0

0

0

0

0

0

SB ~

β

0

0

0

0

0

0

0

0

⇒ β ≡ tr (SB) = 3 ( λB + 2 m B ) tr ( EB ) = 3[500 + 2 (2000)](−0.03) = −165 psi 3

3

Chapter 4

117


Also, SB11 = λB tr ( EB ) δ 11 + 2 m B EB11 − 165 = 500(−0.03) + 2(1000) EB11

⇒

EB11 = − 0.0375

Since the vertical displacements (and strains) are the same for all the blocks. EA11 = EB11 = −0.0375 EB22 = EB33 = − EA11 νA = 0.0375 × 0.25 = 0.009375 where νA = ⇒

λA 1000 = = 0.25 2(1000 + 1000) 2(λA + m A)

tr ( EA ) = − 0.0375 + 2 × 0.009375 = − 0.01875

Thus,





α = tr ( SA ) = 3 λ A + 2 m A tr (EA) = 3(1000 + 2 1000)(− 0.01875) = − 93.75 psi 3

3

P = − 165 psi × 1 in2 − 2 × 93.75 psi × 1 in 2 = 352.5 lbs assuming that the change in area is relatively small. (b) What is the final area of the compressed face of the outer cubes? A = (1 + E A22)(1 + EA33) = (1.009375) 2 = 1.0188 in 2

121. Two cubes with dimensions 2×2×2 are uniformly compressed between two rigid plates with an aggregate force of P. Assume that there is no friction between any of the contacting surfaces. The top cube is made of material A, while the bottom cube is made of material B. Both of the materials are linearly elastic with Lamé constants λA = 1000 psi,

m A = 1000 psi

λB = 500 psi,

m B = 2000 psi

P

A B P

Compute the force P required to change the total volume of the two cubes by 5% of the original volume. What are the final dimensions of the two cubes?

(a) Compute the force P required to change the total volume of the two cubes by 5% of the original volume. Let σA , σ B be stresses in blocks a, b along force axis and Á A , Á B be the corresponding strains.


118

0 0 0

σi 0 0

Si ~

0 0 0

Ái 0 0 0 E i22 0 0 0 E i22

Ei ~

i = A, B

where Ei22 = − ν i Á i is the lateral strain. The material constants are C= ⇒ CA = 2500

m(3λ + 2m) λ+m

ν=

C B = 4400

λ 2(λ + m)

νA = 0.25

νB = 0.10

From equilibrium σ A = CA Á A = σ B = CB Á B

⇒

ÁA =

CB Á CA B

Since the total change in volume to be 5% of the original volume ∆VA + ∆V B = − 0.05 VA + V B Noting that ∆Vi = ei V i where ei ≡ tr ( Ei ) ⇒

eA V A + eB V B = − 0.05 V A + VB ⇒ eA + e B = − 0.1

Substituting ei ≡ ( 1 − 2 ν i ) Á i leads to

( 1 − 2 ν ) CC + ( 1 − 2 ν )Á A

B A

B

B

= − 0.10

⇒

Á B = − 0.0595

Á A = − 0.1048 P = 4 σ aA = 4 C A Á A = − 1048 lb assuming that the change in area is relatively small. (b) What are the final dimensions of the two cubes? Each block decreases in height by Á i and increases in width by −νi Á i times the original dimensions of 2. Block A is V=(1.7905)(2.0524)(2.0525), thus Vol. = 7.542 Block B is V=(1.8810)(2.0120)(2.0120), thus Vol. = 7.61 Note: ∆V∕ V| total = − 0.0527 ≠ − 0.05 because of the small strain assumption

122. The strain at a point in a body is given by E ~ 10 −3

2 3 4

3 5 1

4 1 1

Chapter 4

119


Find the components of the stress tensor assuming linear, isotropic, elastic material behavior, with λ = 16,000 ksi and m = 11,000 ksi.

Sij = λ Ekk δ ij + 2m Eij Ekk = 10−3(2 + 5 + 1) = 8 × 10 −3 ⇒ Sij = (16 × 10 3)(8 × 10 --3) δ ij + (22 × 103) E ij 1 ⇒ S ~ 128 0 0

0 1 0

0 0 1

+ 22

2 3 4

3 5 1

4 1 1

172 66 88 66 238 22 88 22 150

=

z3

123. A 2 by 2 by 2 unit solid cube, centered at the origin of coordinates, is subjected to the deformation described by the map: φ(z) =  z 1+ 13 a z 31  e1 +  z 2+ 13 a z 32  e2 + b z 3 e3

z2

2

2 Compute the values of the constants a and b that are consistent with z 1 2 the observations that the total volume of the block is unchanged by the deformation and the total area of the side with original normal e1 decreases by 5% due to the deformation. Assuming that the cube is made of a linear, elastic, isotropic material with Lamé parameters λ and m, find the body forces and surface tractions required for equilibrium. (You may assume that the linearized strain tensor is adequate to describe the strains for this problem).

(a) Compute the values of the constants a and b that are consistent with the observations that the total volume of the block is unchanged by the deformation and the total area of the side with original normal e1 decreases by 5% due to the deformation. The original volume and area are V = 8 and a = 4. The values after deformation are v=

 det FdV = 8

a=

B

 det F ‖ F

−T

e1 ‖ dA = 0.95(4)

B

The following quantities can be computed from the deformation 1+a z 21

F~

0 0

0

1+ a z 22 0

0 0 b

F

−T

~

1 1+a z 21 0 0

0 1 1+ a z22 0

0 0 1 b

F−T e1 ~

1 0 0


120

det F = b ( 1 + a z 21 ) ( 1 + a z 22 )

det F ‖ F−T e 1 ‖= b ( 1 + a z 22 )

Now the volume can be computed as

   1

v=

1

−1

−1

1

b  1+a z 21  1+a z 22  dz 1 dz 2 dz 3

−1

1 1 = 2 b ( z 1 + a z 31 ) | ( z 2 + a z 32 ) | = 8 b ( 1 + a ) 2 3 −1 3 −1 3

The area can be computed as

 1

area =

−1

1

−1



b  1+a z 22  dz 2 dz 3 = 2b z 2+ a z 32 3



1 −1



= 4 b 1+ a 3



Substituting in the given constraints on the deformation we get ∴ b ( 1 + a )2 = 1 3 ⇒ a = 0.158

b ( 1 + a ) = 0.95 3 b = 0.903

(b) Assuming that the cube is made of a linear, elastic, isotropic material with Lamé parameters λ and m, find the body forces and surface tractions required for equilibrium. (You may assume that the linearized strain tensor is adequate to describe the strains for this problem). u(z) = φ(z) − z = 13 a z 31 e 1 + 13 a z 32 e 2 + b z 3 e 3

∇u ~

a z 21

0

0

0

a z 22

0

0

0

b

E ≈ 1 [ ∇ u + ∇ uT ] = ∇ u 2

The stress is given by S = λtr ( E) I + 2 m E. In components we have

S11 = λ( a z 21 + a z 22 + b ) + 2 m a z 21 2 2 2 ⇒ S22 = λ( a z 1 + a z 2 + b ) + 2 m a z 2 2 2 S33 = λ( a z 1 + a z 2 + b ) + 2 m b all other Sij = 0. Therefore, the body force required to maintain equilibrium is

b = − div S ~

2 a (λ + 2 m ) z1 2 a (λ + 2 m ) z2 0

The surface tractions are as follows

Chapter 4


121

Face z 1 = 1, n = e 1, t e1

= S11( z 1 = 1 ) e 1 = [ λ( a + a z 22 + b ) + 2 m a ] e1

Face z 1 = −1, n = −e 1, t −e1 = −S11( z 1 = −1 ) e 1 = −[ λ( a + a z 22 + b ) + 2 m a ] e1 Face z 2 = 1, n = e 2, t e2

= S22( z 2 = 1 ) e 2 = [ λ( a z 21 + a + b ) + 2 m a ] e2

Face z 2 = −1, n = −e 2, t −e2 = −S22( z 2 = −1 ) e 2 = −[ λ( a z 21 + a + b ) + 2 m a ] e2 Face z 3 = 1, n = e 3, t e3

= S3( z 3 = 1 ) e 3 = [ λ( a z 21 + a z 22 + b ) + 2 m b ] e3

Face z 3 = −1, n = −e 3, t −e3 = −S33( z 3 = −1 ) e 3 = −[ λ( a z 21 + a z 22 + b ) + 2 m b ] e3

124. Consider a displacement map u(z) given by the explicit expression u(z) = ε A z, where A is a given constant tensor and ε is a given scalar (which is very small compared to 1). The vector z is the position vector of a point in the undeformed configuration. Compute the strain tensor E of the given motion. Compute the stress tensor S assuming that the material is linear and elastic and has Lamé parameters λ and m. Compute the body force b required to maintain equilibrium with the stress.

(a) Compute the strain tensor E of the given motion. ∇u = ∂  ε A ik z k  ei  e j = ε A ik δkj e i  ej = ε A ∂z j E =

1 ∇u+∇uT  2

= εA

(b) Compute the stress tensor S assuming that the material is linear and elastic and has Lamé parameters λ and m. S = λ( tr E) I + 2m E = λε ( tr A) I + 2m ε A constant (c) Compute the body force b required to maintain equilibrium with the stress. b = −div S = 0


122

z3

125. Consider the unit cube with vertex at the origin of coordinates as shown in the sketch. The cube is subjected to the following deformation map:

1

φ(z) =  z 1+z 2 sin γ  e 1 + z 2 cos γ e 2 + z 3e 3

1

1

z2

z1 Note that γ is a constant. Compute the tractions and body forces required to achieve the given deformation for the specific shearing angle of γ=0.2 rad assuming that the material is linear, elastic, and isotropic with Young’s modulus of 1000 psi and Poisson’s ratio of 0.499. Does it make any difference if you use the linearized strain tensor as opposed to the Lagrangian strain tensor in the constitutive equation for this problem? Explain.

(a) Compute the tractions and body forces required to achieve the given deformation for the specific shearing angle of γ=0.2 rad assuming that the material is linear, elastic, and isotropic with Young’s modulus of 1000 psi and Poisson’s ratio of 0.499.

1

F = ∇φ ~

sin γ

0

0

cos γ

0

0

0

1

E = 1 [ FT F − I ] ~ 1 2 2

1 T

F F~

0

sin γ

0

sin γ

0

0

0

0

0

sin γ

0

sin γ

1

0

0

0

1

;

tr ( E) = 0

The stress can be computed from S = λ( tr E) I + 2m E (constant), where

m=

C = 1000 = 333 3 2 ( 1 + ν)

λ=

⇒ S12 = S 21 = m sin γ;

2mν m = = 166, 6667 0.002 2 ( 1 − 2 ν) all other S ij = 0

The body force vector required for equilibrium is, b = −div S = 0 The surface tractions are as follows Face z 1 = 1, n = e 1, t e1 = S12 e 2 = m sin γ e 2 Face z 2 = 1, n = e 2, t e2 = S21 e 1 = m sin γ e 1

(b) Does it make any difference if you use the linearized strain tensor as opposed to the Lagrangian strain tensor in the constitutive equation for this problem? Explain.

Chapter 4

0

∇u = F − I ~

sin γ 0

0 1 ∇u+∇u T  2

0

0 cos γ−1 0 0

Elinear =

123


~1 2

0

sin γ

sin γ 2 cos γ−2 0

0

0 0 0

tr ( E linear ) = cos γ − 1 ⇒ S12 = S 21 = m sin γ S22 = ( λ + 2 m ) ( cos γ − 1 ); S11 = S33 = λ ( cos γ − 1 ) Comparing these results with those obtained using Lagrangian strain we find that: ⇒ S12 = S 21 = 67 psi in both cases whereas S11 = S33 ≈ − 166, 667( 0.02) = − 3322 psi S22 ≈ − ( 166, 667 + 667 )( 0.02) = − 3333 psi as opposed to S11 = S22 = S 33 = 0 when the Lagrangian strain tensor was used. Therefore, even though the term cos γ − 1 ≈ −γ2∕2 is a second--order term, it gives rise to large stresses because ν = 0.499 is a nearly incompressible material with λ ≫ m.

126. A bar of length ℓ has an elliptical cross section. The equation of the ellipse is b2 z 21+a 2 z 22 = a 2 b 2, where a and b are the major and minor semi-axis dimensions. The bar experiences the following displacement map: u(z) = −β z 2 z 3 e1 + β z 1 z 3 e2 − β c z 1 z 2 e3

z2

z1 z3 z2

b a where β and c are constants. Find the stress tensor associated with this motion, assuming that the material is linz1 ear, isotropic, and elastic with moduli λ and m. Find the body force required for equilibrium. What value must the constant c have in order that the lateral surface of the bar be traction-free?

124


(a) Find the stress tensor associated with this motion, assuming that the material is linear, isotropic, and elastic with moduli λ and m.

∇u ~ β

0

−z3

−z 2

z3

0

z1

−cz2 −cz 1

0

0

0

−x2( 1+c)

0

0

x1( 1−c)

E = 1 ∇u + ∇u ~ β 2

−x2( 1+c) x1(1−c )

0

We thus have tr ( E) = 0. The stress is given by S = λtr ( E)I + 2 mE = 2 m E. Thus,

S ~ 2m β

0

0

−x2( 1+c)

0

0

x1( 1−c)

−x2( 1+c) x1(1−c )

0

(b) Find the body force required for equilibrium. b = −div S = 0 (c) What value must the constant c have in order that the lateral surface of the bar be traction-free? The normal to the lateral surface is 2 b 2 z1

nΓ = ∇ b2 z 21+a2 z 22 − a 2 b2  ~ γ

2 a 2 z2 0

0

t Γ = SnΓ ~ 2 γ

0 − b z1 z2 1+c)+a 2 z1 z2 (1−c ) 2

For the lateral surface to be traction free − b2 z 1 z 2 1+c )+a2 z 1 z 2 ( 1−c ) = 0 ⇒ c = a2 − b 2 a +b 2

2

Chapter 4


125

127. Consider a displacement map u(z) =  z  z  a, where a is a given constant vector (which has a magnitude very small compared to 1). The vector z is the position vector of a point in the undeformed configuration. Compute the linearized strain tensor E of the given motion. Compute the stress tensor S assuming that the material is linear and elastic and has Lamé parameters λ and m. Compute the body force b required to maintain equilibrium for the given motion. (a) Compute the linearized strain tensor E of the given motion. E = 12  ∇u+∇uT ,

∇u =

∂u i [ e  ej ] ∂z j i

∂ui = ak  z k, j z i+z k z i, j  = a k δ kj z i+a k z k δij = z i aj+a k z k δij ∂z j

u i = a k z k z i,

= ak  z k, j z i+z k z i, j  = a k δ kj z i+a k z k δij = z i aj+a k z k δij In direct notation, ∇u = z  a + (a ⋅ z) I ∇uT = a  z + (a ⋅ z) I E = (a ⋅ z) I + 12  z  a + a  z  (b) Compute the stress tensor S assuming that the material is linear and elastic and has Lamé parameters λ and m. The stress is given by S = λtr(E) I + 2m E. Thus, tr(E) = 3(z ⋅ a) + (z ⋅ a) = 4(z ⋅ a) S = 4λ(a ⋅ z) I + 2m (a ⋅ z) I + m  z  a + a  z  =  4λ+2m (a ⋅ z) I + m  z  a + a  z  (c) Compute the body force b required to maintain equilibrium for the given motion. div S + b = 0, div S =

b = −div S

∂S ij e = S ij, j ei ∂z j i

Sij =  4λ+2m  a k z k δij + m  z i aj + a i z j  The divergence of the stress tensor is Sij, j =  4λ+2m  ak z k, j δ ij + m  z i, j a j + ai z j, j  =  4λ+2m  a k δkj δij + m  δ ij aj + a i δjj  =  4λ+2m  a i + m  a i + 3ai  =  4λ + 6m  a i

126


Therefore, the body force required for equilibrium is b = −  4λ+6m  a

z1

128. A linearly elastic solid body is subjected to forces that give rise to the following displacement map: u(z) = γ 12  z 23+β z 21−β z 22  e1 + γβ z 1 z 2 e2 − γz 1 z 3 e3

z3

z2

where γ ≪ 1 (i.e., very small) and β are constants describing the motion. Assume that the elastic response is adequately characterized by Hooke’s law with known material constants λ and m (the Láme parameters). Find β in terms of the constants λ and m such that S11 = 0 (S is the stress tensor). Find S22 and S33 for the conditions given previously. Find the traction on the surface with normal−e3 at z 3 = 0.

(a) Find β in terms of the constants λ and m such that S11 = 0 (S is the stress tensor). Since the displacements are very small, the linearized strain tensor is appropriate. E = 12  ∇u+∇uT , β z1 ∇ u ~ γ β z2 − z3

−β z2 z3 β z1 0 0 − z1

∇u =

∂u i [ e  ej ] ∂z j i

E~γ

β z1 0 0

0 β z1 0

z3 0 − z1

The stress can be determined from Hooke’s Law S = λtr(E) I + 2m E,

tr(E) = −z 1  1−2β  γ

S11 =  − 1−2β  λ + 2mβ  z 1 γ = 0

⇒ β=

λ 2(λ+m)

(b) Find S22 and S33 for the conditions given previously. S22 =  − 1−2β  λ + 2mβ  z 1 γ = 0 S33 =  − 1−2β  λ−2m  z 1 γ = − 2mβ+2m  z 1 γ

⇒ S22 = 0 ⇒ S 33 = −2 m 1+β  z 1 γ

(c) Find the traction on the surface with normal−e3 at z 3 = 0. The traction on the surface can be computed with Cauchy’s relationship t n = Sn with n = −e 3 S = S 33  e3  e3  S(−e3) = S33  e 3  e3 (−e3) = −S 33 (e3 ⋅ e 3) e 3 = −S33 e 3

Chapter 4


127

t n = 2m 1+β  z 1 γe3

129. The state of stress S as a function of position x in a certain solid body is given by the expression S(x) = x  Bx, where B is a given constant tensor. Find the body force (as a function of position x) required to maintain equilibrium of the body. Express the result in both index and direct (vector) notation. What are the restrictions, if any, on the constant tensor B in order for the stress field S to be an admissible stress state? (Please describe any restrictions explicitly in terms of the components of B, not in terms of x and S.)

(a) Find the body force (as a function of position x) required to maintain equilibrium of the body. Express the result in both index and direct (vector) notation. The components of the stress tensor are S = S ij  ei  ej  = x i ei  Bjk  e j  ek  xl e l = xi B jk xl ei   e j  ek  el = xi B jk xl δkl e i  ej  = xi B jk xk ei  e j  ⇒ Sij = x i Bjk xk The body force vector is, therefore ∂Sij b = −div S = − x ei j ∂Sij ∂x ∂x = i Bjk x k + xi Bjk k = δ ij Bjk xk + x i Bjk δ kj ∂x j ∂xj ∂xj ∂Sij = Bik x k + xi Bkk ∂x j bi = − Bik xk + x i Bkk  b = − Bx + tr(B) x  = − B+tr(B) I  x (b) What are the restrictions, if any, on the constant tensor B in order for the stress field S to be an admissible stress state? (Please describe any restrictions explicitly in terms of the components of B, not in terms of x and S.) The stress state must also satisfy balance of angular momentum. Hence, the stress tensor must be symmetric. Sij = xi B jk xk

128


S12 = x1  B 21 x 1 + B22 x 2 + B23 x 3 

B21 = B23 = B 21 = B 13 = 0

S21 = x2  B 11 x 1 + B12 x 2 + B13 x 3 

B11 = B22

S13 = x1  B 31 x 1 + B32 x 2 + B33 x 3 

B31 = B32 = B 12 = B 13 = 0

S31 = x3  B 11 x 1 + B12 x 2 + B13 x 3 

B11 = B33

S23 = x2  B 31 x 1 + B32 x 2 + B33 x 3 

B31 = B32 = B 21 = B 23 = 0

S32 = x3  B 21 x 1 + B22 x 2 + B23 x 3 

B22 = B33

Therefore, the tensor B must be proportional to the identity tensor, i.e., B = BI.

The solution above was given in component form. Here is a component-free derivation of the conditions on B implied by symmetry of the stress tensor. Thanks to Xiang Ding for pointing out this elegant solution. The stress state must satisfy balance of angular momentum. Hence, the stress tensor must be symmetric. S is symmetric if

u ⋅ Sv = v ⋅ Su

for any vectors u, v

u ⋅  x  Bx  v = v ⋅  x  Bx  u u ⋅  v ⋅ Bx  x = v ⋅  u ⋅ Bx  x  u ⋅ x   v ⋅ Bx  =  v ⋅ x 

This equality can be rewritten as  v ⋅ Bx   u ⋅ Bx  = = b v ⋅ x u ⋅ x

where b is a constant (because that is the only way a function of v and a function of u can be equal to each other). Therefore, (for either u or v) we must then have  u ⋅ Bx  = b u ⋅ x ,

[or

 v ⋅ Bx  = b v ⋅ x ]

u ⋅  Bx − b x  = 0 u ⋅  B−b I  x = 0 Since the above equation must hold for an arbitrary vector u and must also be true for any position vector x then we must have B = bI In other words, the tensor B must be proportional to the identity.

Chapter 5 Boundary Value Problems in Elasticity

130. Consider the uniaxial rod shown below, fixed at x = 0, free at x = ℓ, and subjected to the linearly varying body force indicated. The rod is made from a composite material with a variable elastic modulus C(x) = C o  2−x∕ℓ  , making it twice as stiff at x = 0 as it is at x = ℓ. The governing differential equation for a rod with variable modulus is

b(x) = b o 1−x∕ℓ  x ℓ

 C(x) u′ ′ + b(x) = 0

where a prime indicates differentiation with respect to x. Find the exact (classical) solution to the problem by directly integrating the governing equations.

Let ξ ≡ x∕ℓ. The original differential equation is given by

 C o 2−ξ  u′  ′ = −bo  1−ξ  Integrating once we get Co 2−ξ  u′ = −b oℓ ξ− 12 ξ2 + a 1 = σ(ξ) The constant a1 can be determined from the boundary condition σ(1) = 0 to have the value a1 = boℓ∕2 . Dividing through by Co  2−ξ  we have u′ =





b oℓ 1 −ξ 2C o 2−ξ

Integrating this equation gives u(ξ) =

b oℓ 2  − ln(2−ξ) − 12 ξ2 + a 2 2C o

130


The constant a2 can be determined from the boundary condition u(0) = 0 to have the value a2 = boℓ 2 ln 2∕2C o . Thus, the displacement map has the explicit form u(ξ) =

b oℓ 2  ln 2− ln(2−ξ) − 12 ξ2 2C o

131. Consider the rod of unit length and modulus C(x) that varies as shown in the sketch. The rod is fixed at the left end, is free at the right end, and is subjected to a linearly varying body force b(x) as shown. Consider the following displacement map: u(x) = a  x3+2x2−3x  where a is some constant. Is the displacement map a solution to the given problem? Why or why not?

b(x) = 15x x

C(x) = 2−x 2 1

For the given map to be a solution to the problem, it has to satisfy the governing differential equation and boundary condition. ( Cu′ )′ + b = 0 ∀ x,

u ( 0 ) = 0,

Cu′ ( 1 ) = 0

Substituting, u′(x) = a  3 x2+4x−3 ,

u′′(x) = a  6 x+4 

( Cu′ )′ + b = Cu′′ + C′ u′ + b = a( 2 − x2 )  6 x + 4  + a( − 2 x )  3 x2 + 4x − 3  + 15 x = a( − 12 x 3 − 12 x 2 + 18 x + 8 ) + 15 x ≠ 0 u ( 0 ) = 0,

Cu′ ( 1 ) = 4 a ≠ 0

Therefore the given map is not a solution to the problem.

132. Prove that S ⋅ E = S ⋅ ∇u when the virtual strain is defined to be the strain that would occur if the virtual displacement actually took place, i.e., E = 12 ∇u +∇uT. Upon what property of the stress tensor S does this identity rely?

Chapter 5

Boundary Value Problems in Elasticity

131

S ⋅ ∇u = S ij u i, j = Sij  12  ui, j + uj, i  + 12  ui, j − uj, i  = Sij E ij + Sij 12  ui, j − uj, i  = Sij E ij + 12  S ij ui, j − Sij uj, i  = Sij E ij + 12  S ij ui, j − Sji uj, i  = Sij E ij = S ⋅ E The penultimate line of the proof relies on the symmetry of the stress tensor.

133. Show that ∇u ⋅ C∇u = λ  div u   div u +m  ∇u +∇u T  ⋅ ∇u for an isotropic, linear, elastic material. Express this equation in component form.

Since the components of C are Cijkl = λδij δkl + m  δ ik δ jl+δil δ jk  we have ui, j Cijkl u k, l = u i, j  λδ ij δ kl + m  δik δjl+δ il δjk   uk, l = λui, i u k, k + m  ui, j u k, l δik δ jl+ui, j u k, l δil δ jk  = λui, i u k, k + m  uk, j u k, j + u i, j uj, i  = λui, i u k, k + m  uk, j u k, j + u k, j uj, k  = λui, i u k, k + m  uk, j + u j, k  uk, j = λdiv u div u + m  ∇u+∇uT  ⋅ ∇u

134. Carry out the derivation of the principle of virtual work for the case in which the real displacements are known and a system of virtual forces are applied to the body, and thereby deduce the principle of virtual forces. Specifically, apply virtual body forces b and virtual surface tractions t, and define the complementary external virtual work as ^

WE ≡

 u ⋅ b dV +  u ⋅ t dA Ω

B

where u is the real displacement of the body. Perform a derivation similar to the one for the principle of virtual displacements to demonstrate that an appropriate definition of complementary internal virtual work is ^

WI ≡

 S ⋅ E dV B

132


where S is the virtual stress associated with the applied virtual force system and E is the strain tensor associated with the real displacements. Prove the principle of virtual forces, ^ ^ which states that if W E = W I for all virtual stresses S in equilibrium with the applied virtual forces b and t, then E = 12  ∇u+∇uT . State precisely the conditions that must hold in order for the principle to be valid.

The external complementary virtual work has the following equivalent expressions ^

 t ⋅ u dA +  b ⋅ u dV =  t−Sn ⋅ u dA +  Sn ⋅ u dA +  b ⋅ u dV

WE =

Ω

B

Ω

=

Ω

B

 t−Sn ⋅ u dA +   div  Su  + b ⋅ u  dV Ω

B

 t−Sn ⋅ u dA +   div S +b  ⋅ u dV +  S ⋅ ∇u dV

=

Ω

B

B

 t−Sn ⋅ u dA +   div S +b  ⋅ u dV +  S ⋅ E dV −  S ⋅  E − ∇u + ∇u  dV

=

Ω

B

B

1 2

T

B

The third term is the complementary internal virtual work. The difference between complementary external and internal virtual work will be defined as the complementary virtual work functional ^

^

^

G(S, u) ≡ W E − W I ^

G(S, u) =

  t−Sn  ⋅ u dA +   div S +b  ⋅ u dV −  S ⋅  E − ∇u + ∇u  dV Ω

B

1 2

T

B

From the fundamental theorem of the calculus of variations, we obtain the principle of ^ virtual work for a three-dimensional solid: If G(S, u) = 0 for all virtual stress fields S that are in equilibrium with the applied virtual forces, i.e., divS + b = 0 in B and t = Sn on Ω, then the strains are compatible with the displacements, that is, they satisfy E = 12 ∇u + ∇uT .

Chapter 5

135. The virtual-work functional for the little boundary value problem is given by

b(x) to

  σ u′−b u dx − t u(0) − t u(ℓ) o

tℓ

x

ℓ

G(σ, u) =

133


ℓ

ℓ

0

The body force b(x) and the traction at the free end t ℓ are known while the stress σ(x) and the reaction t o are unknown. Using the principle of virtual work, select a virtual displacement field that allows you to compute the reaction force in terms of only the known applied forces. Give the expression for t o in terms of b(x) and t ℓ.

Note that G(σ, u) = 0 represents a statement of equilibrium for any choice of u. If we chose u(x) = u o (a constant) we get an interesting result because u′ = 0. Recognizing that, for this choice, u(0) = u(ℓ) = u o we get

 b u dx − t u − t u ℓ

−

o

o o

ℓ

o

= 0

0

Since uo factors out of every term we get the result

 b dx + t ℓ

o

+ tℓ = 0

0

which represents overall horizontal equilibrium of the applied forces and the reactions.

136. In the little boundary value problem, we saw that the virtual-work functional could be stated as a weighted residual functional. A weighted residual functional for a three-dimensional solid body B with boundary Ω (with normal vector field n) can be defined as G(S, w) ≡ −

  div S+b  ⋅ w dV −   t−Sn  ⋅ w dA B

Ω

where div S+b is the equilibrium residual in the domain B, t−Sn is the equilibrium residual on the boundary Ω, and w is an arbitrary weighting function. Show that the weighted residual functional is identical to the virtual-work functional given in Eqn. (295), and, therefore, that the arbitrary weighting function is identical to the virtual displacement, i.e., w = u.

The correspondence can be shown by applying the divergence theorem

134


G(S, w) ≡ −

 t−Sn ⋅ w dA −   div S +b  ⋅ w dV Ω

= −

B

 t−Sn ⋅ w dA −   div  Sw  + b ⋅ w  dV +  S ⋅ ∇w dV Ω

= −

B

B

 t ⋅ w dA −  b ⋅ w dV +  S ⋅ ∇w dV Ω

B

B

If we interpret w as a virtual displacement then we have G(S, w) = W I − W E.

137. Consider the pile of length ℓ, constant modulus C P (w/ unit area), embedded in an elastic medium with modulus k (force per unit displacement per unit length), and subjected to a load P at x = 0. The pile is elastically rex k (elastic strained at the end x = ℓ giving an end force of the medium) ℓ amount F = 2kℓu(ℓ) as shown. The governing differential equation for the system is Cu′′(x)−k u(x) = 0. What C (pile) must be the value of the constant α for the solution to have the form u(x) = Ae αx + Be −αx? F = 2 kℓ u(ℓ) What are the values of the constants A and B that satisfy the problem shown in the figure? Does this function u(x) represent a classical solution to the given problem? Why or why not? Are there any other solutions to this specific problem?

(a) What must be the value of the constant α for the solution to have the form u(x) = Ae αx + Be −αx? The given displacement has to satisfy the differential equation u′ = α( Ae αx − Be --αx ) u′′ = α 2( Aeαx + Be --αx ) = α 2x

∴ Cα 2 u − k u = 0 [ Cα 2 − k ] u(x) = 0

⇒

α=

Ck

(b) What are the values of the constants A and B that satisfy the problem shown in the figure? Does this function u(x) represent a classical solution to the given problem? Why or why not? Are there any other solutions to this specific problem? u′ = α( Ae αx − Be --αx )

∴ Cα 2 u − k u = 0

The constants A and B have to satisfy the boundary conditions at x = ℓ

at x = 0 P

σ(0)

σ(ℓ)

F = 2kℓ u(ℓ)

Chapter 5


135

σ(ℓ) = − 2kℓ u(ℓ)

σ(0) = − P Cu′(0) = − P (1)

Cu′(ℓ) = − 2kℓ u(ℓ) (2)

Let m ≡ 2αℓ and P ≡ P∕ αC (1) → Cα ( A − B) = − P ⇒ A = B − P (2) → α Ae αℓ − Be --αℓ  + 2α 2ℓ Ae αℓ + Be --αℓ  = 0 e2αℓ 1 + 2αℓ A − 1 − 2αℓ B = 0 em(1 + m) A = (1 − m) B From (1) A = P − B. Substituting in (2) em(1 + m) (P − B) = (1 − m) B ⇒B=P

em 1−m e m − 1+m

A=P

1−m  1+m 1−m em − 1+m

Since u satisfies the classical differential equation and boundary conditions, it is an exact solution to the problem, and this solution is unique.

138. Consider the rod of length ℓ=1 and constant modub(x) = −(1−x 2) lus C=1. The rod is restrained by an elastic spring of modulus k=1 at each end and rests on an elastic foundation, x also with modulus k=1. The rod is subjected to a quadratically varying body force as shown. The displacement 1 u(x), positive in the x direction, is governed by the following differential equation u′′−u = 1−x 2. What are the boundary conditions for this problem? Is the following displacement function a classical solution to this problem? u(x) = 1 + x 2 − 2ex − 12 e−x Why or why not? If it is not then modify it so that it is. (a) What are the boundary conditions for this problem? at x = 0 k u(0)

at x = 1 σ(0)

σ(ℓ)

k u(ℓ)

Cu′(0) − k u(0) = 0

Cu′(1) + k u(1) = 0

u′(0) − u(0) = 0

Cu′(1) + k u(1) = 0

136


(b) Is the given displacement function a classical solution to this problem? Why or why not? If it is not then modify it so that it is. The solution has to satisfy the differential equation and boundary conditions u′(x) = 2 x − 2e x + 12 e −x

u′′(x) = 2 − 2e x − 12 e −x

Substituting in the differential equation u′′(x) − u(x) = ( 2 + 2e x − 12 e −x ) − ( 1 + x 2 − 2ex − 12 e−x ) = 1 − x2 Therefore the solution satisfies the differential equation for all values of x. Checking the boundary conditions: u′(0) − u(0) = ( − 2 + 1 ) − ( 1 − 2 − 1 ) = 0 OK 2 2 u′(1) + u(1) = ( 2 − 2 e + 1 ) − ( 2 − 2 e − 1 ) = 4 (1 − e) ≠ 0 2e 2e The boundary condition at x = 1 is not satisfied. Thus the given function is not a solution to the problem. We can modify the function to u(x) = 1 + x 2 − 2e e x − 12 e −x u′(x) = 2 x − 2e ex + 12 e−x

u′′(x) = 2 − 2e ex − 12 e−x

u′′(x) − u(x) = ( 2 + 2e ex − 12 e−x ) − ( 1 + x2 − 2e e x − 12 e −x ) = 1 − x 2 u′(0) − u(0) = ( − 2e + 1 ) − ( 1 − 2e − 1 ) = 0 2 2 u′(1) + u(1) = ( 2 − 2 + 1 ) − ( 2 − 2 − 1 ) = 0 2e 2e The modified function is the exact solution to the problem.

139. The principle of virtual work for a certain boundary value problem can be stated as

  A(x) u′′u′′ + B(x) u u − b(x) u  dx = 0 ℓ

G(u, u) ≡

for all u(x) ∈ F(0, ℓ)

0

where A, B, and b are known functions of x, u(x) is the unknown field, and a prime denotes derivative with respect to x. What is the classical differential equation that is equivalent to this variational statement?

Chapter 5

137


To apply the fundamental theorem of the calculus of variations, we need to have a functional in the form

  ⋅  u dx + Boundary terms ℓ

G(u, u) =

0

Hence, need to work on Au′′u′′ term using integration by parts. Note that,

  Au′′′ u′ =  Au′′′′ u +  Au′′′ u′   Au′′ u′′ =  Au′′′ u′ +  Au′′ u′′ Hence,

Au′′ ′ u′ − Au′′  u′′ = Au′′ ′′ u − Au′′ u′′ or Au′′ u′′ =  Au′′′′ u +  Au′′ u′′ −   Au′′′ u′

 Au′′ u′′ dx =   Au′′′′ u dx +  Au′′ u′ −  Au′′ ′ u l

⇒

l

0

ℓ

ℓ

0

0

0

Thus,

   Au′′′′ + Bu − b  u dx + Boundary terms l

G(u, u) =

0

By the fundamental theorem of the calculus of variations:  Au′′ ′′ + Bu − b = 0

for x ∈ (0, l)

is the classical differential equation

140. The classical (4th order) differential equation and boundary conditions for a certain boundary value problem are Au′′′′ + Bu′′ + Cu = b u(0) = 0,

u(ℓ) = 0,

for all x ∈ [0, ℓ]

Au′′(0) = 0,

Au′′(ℓ) = 0

where A, B, C, and b are known constants, u(x) is the unknown field, and a prime denotes derivative with respect to x. Find an expression for the virtual-work functional associated with the classical differential equation. In other words, find the functional G that has the property that the statement “G(u, u) = 0 for all u ∈ F e ” is equivalent to the classical differential equation and the highest derivative that appears in G is second order. Describe any restrictions that must be placed on Fe .

138

Fundamentals of Structural Mechanics Because Au′′′′ + Bu′′ + C u − b = 0 we can define

 Au′′′′ + Bu′′ + C u − bu dx = 0 l

G(u, u) ≡

0

By the fundamental theorem of the calculus of variations, G(u, u) = 0 for all u implies Au′′′′ + Bu′′ + C u − b = 0. To balance derivatives integrate by parts

 A u′′′′ u dx = −  A u′′′ u′ dx +  A u′′′ u  ℓ

ℓ

0

ℓ 0

0

 A u′′ u′′ dx −  A u′′ u′′  +  A u′′′u  ℓ

=

ℓ

ℓ

0

0

0

=  A u′′ u′′ dx +  A u′′′u  ℓ

ℓ 0

0

 B u′′ u dx = −  B u′ u′ dx +  B u′ u  ℓ

0

ℓ

ℓ 0

0

Therefore,

  A u′′ u′′−B u′ u′+C u u−bu  dx +   A u′′′+B u′ u  ℓ

G(u, u) =

ℓ 0

0

We can eliminate the boundary term by restricting u to satisfy essential B.C.’s u(0) = 0 and u(ℓ) = 0.

141. Consider the solid spherical region B with surface Ω having a unit normal vector field n, as shown in the sketch. Assume that there exists a scalar field w(x), of the position vector x, for which we can define the functional G(w, v) ≡

  ∇v ⋅ ∇w − v  dV −  t v dA B

x3 B n Ω

x2 x1

Ω

that has the property that if G(u, v) = 0 for all (virtual) scalar functions v(x) then the classical differential equations governing the real field w(x) are satisfied (i.e., G(w,v) is a “virtual-work” functional). Note that the scalar field t(x) is defined on the surface of the solid region. Find the classical governing differential equation for w(x) that is implied by the variational statement “ G(u, v) = 0 for all v”. Determine what must be the relationship between t(x) and w(x) on the surface of the sphere.

Chapter 5

139


First note that





2 div( ∇w v ) = ∂ ∂w v = ∂ w2 v + ∂w ∂v = ∆w v + ∇v ⋅ ∇w ∂xi ∂xi ∂xi ∂xi ∂xi

  ∇v ⋅ ∇w  dV =  B

div( ∇w v ) dV −

B



∆wv dV

B

Applying the divergence theorem

  ∇v ⋅ ∇w  dV = 

∇w ⋅ n v dV −

Ω

B



∆wv dV

B

where n is the normal to the surface. Substituting back into the virtual work functional G(w, v) = −

  ∆w + 1  v dV +  ( ∇w ⋅ n − t ) v dA = 0 B

Ω

Applying the fundamental theorem of the calculus of variations ∆w + 1 = 0 in B (governing differential equation) t = ∇w ⋅ n on Ω (natural boundary condition)

∀v

140


Chapter 6 The Ritz Method of Approximation

142. Consider the uniaxial rod shown in the sketch, fixed at bo x = 0, free at x = ℓ, and subjected to the linearly varying body force indicated. The rod has a variable elastic modulus x C(x) = C o  2−x∕ℓ  , making it twice as stiff at x = 0 as it is ℓ at x = ℓ. Using the principle of virtual work, find the expression for the displacement u(x) and stress σ(x) for the given body force and variable modulus, approximating the real and virtual displacements with polynomials.

Consider first a two-term polynomial approximation u(x) = a 1 x + a 2 x 2 , ℓ ℓ

u(x) = a1 x + a 2 x 2 ℓ ℓ

2

2

Let ξ ≡ x∕ℓ. The derivative of u is u′ = (a 1+2a2 ξ)∕ℓ, similarly for u. Substituting into the virtual-work functional, noting that b(x) = b o(1−ξ), we get

 Cℓ (2−ξ) a +2a ξ  a +2a ξ  − b ℓ (1−ξ) a ξ+a ξ  dξ 1

G(u, u) ≡

o

1

2

1

2

o

1

2

2

0

In matrix form, then the coefficients a1, and a2 must satisfy the equation Co ℓ

3 2 4 3

4 3 5 3

a1 a2

= b oℓ

1 6 1 12

⇒

Thus, the displacement field is given by u(ξ) =

b oℓ 2  12ξ−7ξ 2  52 C o

a1 a2

b ℓ2 = o Co

3 13 7 − 52

142


The stress can be computed from σ = Cu′. Thus, we have σ(ξ) =

b oℓ  52

2−ξ   12−14ξ 

Note that the approximation of the stress at the free end is σ(1) = −b oℓ∕26, about a 4% error. We can examine the accuracy of the solution by using a cubic approximation. Let ξ ≡ x∕ℓ and let u(ξ) = a 1 ξ+a2 ξ 2+a3 ξ3 and u(ξ) = a 1 ξ+a2 ξ 2+a3 ξ3 be the approximation for the real and virtual displacement fields. The derivative of u can be computed as u′ = (a 1+2a2ξ+3a 3ξ2)∕ℓ, similarly for u. Substituting into the virtualwork functional, we get

 Cℓ (2−ξ) a +2a ξ+3a ξ  a +2a ξ+3a ξ  dξ −  b ℓ (1−ξ) a ξ+a ξ +a ξ  dξ 1

G(u, u) ≡

o

1

0

2

3

2

1

2

3

2

1

o

1

2

2

3

3

0

In matrix form, then, the coefficients a1, a2, and a3 must satisfy the equations

Co ℓ

3 2 4 3 5 4

4 3 5 3 9 5

5 4 9 5 21 10

a1 a2

= b oℓ

a3

1 6 1 12 1 20

a1 ⇒

a2 a3

=

b oℓ 2 Co

16 63 53 − 252 10 189

Thus, u(ξ) =

b oℓ 2  192ξ−159ξ 2+40ξ3  756 C o

The stress can be computed from σ = Cu′. Thus, σ(ξ) =

b oℓ 

756

2−ξ  192−318ξ+120ξ 2 

The approximation of the stress at the free end is σ(1) = −b oℓ∕126 which is less than 1% error.

143. Reconsider the nonprismatic rod of problem 142 subjected to the linearly varying body force. However, consider the condition in which the rod is fixed at both ends with prescribed end displacements of u(0) = u o and u(ℓ) = u 1. Solve the problem with the Ritz method using polynomial base functions.

Let ξ ≡ x∕ℓ and let u(ξ) = a 0+a1 ξ+a 2 ξ2 and u(ξ) = a 0+a1 ξ+a 2 ξ2 be the approximation for the real and virtual displacement fields. We must first satisfy the essential boundary conditions

Chapter 6

143

The Ritz Method of Approximation

u(0) = a 0 = uo u(1) = a 0 + a1 + a 2 = u1 Thus, we have a1 = u1−u o−a2. Substituting these expressions into the original approximations gives u(ξ) =  u o +  u 1−uo  ξ  + a 2  ξ2 − ξ  The virtual displacement field must satisfy homogeneous essential boundary conditions, u(0) = 0 and u(1) = 0. Therefore, u(ξ) = a 2  ξ2−ξ . From these results we can identify uo(ξ) = uo +  u 1−uo  ξ and h2(ξ) = ξ2−ξ. The virtual-work functional is

 Cℓ (2−ξ) u −u +a (2ξ−1)  a (2ξ−1)  dξ −  b ℓ (1−ξ) a (ξ −ξ)  dξ = 0 1

G(u, u) ≡

o

o

1

0

2

2

1

o

2

2

0

Carrying out the integrals we find that the coefficient a2 must satisfy the equation −

Co  C bℓ u −uo  + o a2 + o = 0 12 6ℓ 1 2ℓ

a2 =

⇒

bℓ 1 u −u o  − o 3 1 6 Co

2

Thus, the displacement field is given by bℓ u(ξ) =  u o + 13  u1−u o   2ξ+ξ 2   − o  ξ 2−ξ  6C o 2

144. Consider using a basis for the virtual displacement different from the basis used for the real displacement. What would be the ramifications of using a different number of terms in the expansions for real and virtual displacements? That is N

u(x) =

 a h (x), n

n=1

n

M

u(x) =

 a h (x) n

n

n=1

where N ≠ M. What happens if N > M? What happens if N < M? Perform some computations on the little boundary value problem to investigate this issue.

The number of terms in the approximation of u determines the degree of refinement of the approximation while the number of terms in the expansion of u determines the degree to which we satisfy the weak form of equilibrium. If N > M then the system of equations resulting from the Ritz method is underdetermined. The equations are all valid equilibrium statements, but are insufficient to determine all of the coefficients of the displacement approximation. As such we get a family of possible displacement approximations that satisfy equilibrium in a limited sense. If N < M then the system

144


of equations is overdetermined. Since all of the equations are valid equilibrium equations they should be consistent and therefore solvable. For example, the little boundary value problem with a uniform load with N = 3 and M = 4 gives rise to the following self-consistent overdetermined equations.

Co ℓ

1

1

1

a0

1

4 3 6 4 8 5

6 4 9 4 12 6

a1

1 1

1 2 1 3 1 4 1 5

= b oℓ

a2

a0 ⇒

1

a1

= b oℓ

−1

a2

2

0

These equations can be solved by a least-squares projection. For an overdetermined system of equations Ax = b, the least squares solution can be obtained as the solution to AT Ax = AT b (least--squares projection).

145. Consider using a basis for the virtual displacement different from the basis used for the real displacement. What would be the ramifications of using different base functions for the real and virtual displacements? That is N

u(x) =

 a h (x), n

n

N

u(x) =

n=1

a

n

gn(x)

n=1

where gn(x) ≠ hn(x). Perform some computations on the little boundary value problem to investigate this issue using, for example, polynomials for the real displacements and trigonometric functions for the virtual displacements.

Let ξ ≡ x∕ℓ and let u(ξ) = a 1 ξ+a2 ξ 2 and u(ξ) = a 1 sin πξ+a2 sin(πξ∕2) be the approximation for the real and virtual displacement fields for the little boundary value problem, fixed at the left end free at the right, subjected to a uniform load. The virtualwork functional is

 Cℓ  a + a ξ  πa cos πξ+ π a cos πξ  dξ πξ −  b ℓ  a sin πξ+a sin  dξ = 0 1

G(u, u) ≡

o

1

2

2

0

1

2

2

2

2

2

1

o

1

0

Integrating we get the following system of equations Co ℓ

0 1

4 −π

a1

2− π4

a2

= b oℓ

2 π 2 π

⇒

a1 a2

=

b oℓ 2 Co

1 −1 2

Which is the exact solution. Clearly the coefficient matrix and right side are different. The coefficient matrix is not symmetric. If the displacement approximation is not ex-

Chapter 6

145


act, then the choice of functions for the virtual displacements will bias the satisfaction of weak equilibrium.

146. The uniaxial rod shown has unit area, length ℓ, and elastic modulus C. The body force is characterized by b(x) =

0

0 < x ≤ ℓ∕2

bo

ℓ∕2 < x ≤ ℓ

bo x ℓ∕2

ℓ∕2

Assume that the real and virtual displacements can be approximated by the expressions u(ξ) = a 0+a1 ξ+a 2 ξ2

u(x) = a 0+a1 ξ+a 2 ξ2

where ξ ≡ x∕ℓ. Using the principle of virtual work, compute the displacement field u(x) and the tractions at the two ends σ(ℓ) and σ(0).

First, the real and virtual approximations must satisfy the essential boundary condition. Setting u(0) = 0 and u(0) = 0 gives a0 = 0 and a0 = 0. The virtual-work functional is

 Cℓ  a +2a ξ  a +2a ξ  dξ −  b ℓ  a ξ+a ξ  dξ = 0 1

G(u, u) ≡

1

1

2

1

2

1 2

0

o

1

2

2

resulting in the equations C ℓ

1

1

a1

1

4 3

a2

= b oℓ

3 8 7 24

a1 ⇒

a2

b ℓ2 = o C

5 2 −1 4

Therefore, the approximate displacement is u(ξ) =

b oℓ 2  5ξ − 2ξ 2  8C

The approximate end tractions are σ(0) = 0.625 b oℓ and σ(ℓ) = 0.125 b oℓ.

147. The uniaxial rod shown has unit area and length ℓ. It is fixed at the left end, is free at the right end, and is subjected to a constant body force field b(x) = b o along its length. The elastic modulus C(x) is characterized by C(x) =

2C

0 < x ≤ ℓ∕2

C ℓ∕2 < x ≤ ℓ

bo x ℓ∕2

ℓ∕2

146


Find the classical solution to the governing differential equation. Using the principle of virtual work, compute a stress field σ(x) assuming a Ritz approximation as follows u(ξ) = a 0+a1 ξ+a 2 ξ2

u(x) = a 0+a1 ξ+a 2 ξ2

where ξ ≡ x∕ℓ. Because the modulus changes abruptly at x = ℓ∕2, the stresses and strains are discontinuous at that point. Why is this discontinuity a problem for the polynomial base functions suggested? What happens if you increase the order of the approximation?

(a) Within each segment the modulus is constant. Therefore, the solution in each segment satisfies Cu 2′′ + bo = 0

2Cu1′′ + b o = 0

Where u1 is the solution in the region 0 < x < ℓ∕2 and u2 is the solution in the region ℓ∕2 < x < ℓ. Both equations can be integrated by direct quadrature. Let ξ ≡ x∕ℓ, the two general solutions are u1 =

b oℓ 2  1 2 − 2 ξ + a1 ξ + a2  2C

u2 =

b oℓ 2  1 2 − 2 ξ + b1 ξ + b2  C

The four constants can be found by the two boundary conditions (1) u1(0) = 0 and (2) Cu2′(1) = 0 and the two continuity conditions (3) u1(1∕2) = u2(1∕2) (continuity of displacement) and equilibrium of force (4) 2Cu1′(1∕2) = Cu2′(1∕2), as shown in the sketch. σ2(ℓ∕2)

σ1(ℓ∕2) The boundary conditions can be implemented as (1) u1(0) = 0

a2 = 0

(2) Cu2′(1) = 0

boℓ 2 −1+b 1  = 0

(3) u1(1∕2) = u2(1∕2)

b ℓ2 b oℓ 2  1 1 − 8 + 2 a 1+a2  = o  − 18 + 12 b1+b 2  2C C

(4) 2Cu1′(1∕2) = Cu2′(1∕2)

boℓ 2 − 12 +a1  = boℓ 2 − 12 +b1 

These four equations can be solved to give a1 = 1, a2 = 0, b1 = 1, b2 = −3∕16 . Thus, the displacement map is given by

u(ξ) =

b oℓ 2 C

− 14 ξ2+ 12 ξ

0 ≤ ξ ≤ 12

3 − 12 ξ2+ ξ− 16

1 2

And the stresses are given by σ(ξ) = C(ξ) u′(ξ) to be

≤ξ≤1

Chapter 6

147


σ(ξ) = b oℓ

1−ξ

0 ≤ ξ ≤ 12

1−ξ

1 2

≤ξ≤1

(b) The approximations must satisfy the essential boundary condition. Setting u(0) = 0 and u(0) = 0 gives a0 = 0 and a0 = 0. The virtual-work functional is G(u, u) ≡



1 2

2C  a +2a ξ  a +2a ξ  dξ + 1 2 1 2

0

ℓ

1

1 2

2

1

2

1

2

2

 b ℓ  a ξ+a ξ  dξ = 0 1

−

 Cℓ  a + a ξ  a + a ξ  dξ

o

1

2

2

0

Carrying out the integrations results and invoking the Fundamental Theorem of the Calculus of Variations results in the equations C ℓ

3 2 5 4

5 4 3 2

1 2 1 3

a1 a2

= b oℓ

a1

⇒

a2

b ℓ2 = o C

16 33 − 2 11


b oℓ 2  16ξ − 6ξ2  33C

The approximate stress is given by

σ(ξ) =

b oℓ 33

32 − 24ξ

0 ≤ ξ ≤ 12

16 − 12ξ

1 2

≤ξ≤1

Exact

32 20

33σ

b oℓ

10

4

Although the approximate displacement is reasonably accurate, the stress field is not. The base functions are smooth and thus the approximate strain is smooth, forcing the stress to jump at the midpoint. In the exact solution, the stress is smooth and the strain jumps at the midpoint.

148. Using a piecewise linear finite element basis, resolve Problem 147. Does the finite element basis suffer from the same problem as the polynomial basis? Why? What general conclusions can we make about the smoothness of the approximation?

The piecewise linear finite element base functions have the explicit expressions

h1(ξ) =

2ξ

0 ≤ ξ ≤ 12

2 − 2ξ

1 2

≤ξ≤1

1

148


h2(ξ) =

0

0 ≤ ξ ≤ 12

2ξ − 1

1 2

1

≤ξ≤1

The virtual-work functional can be expressed as

G(u, u) ≡



1 2

0

2C  2a  2a  dξ + 1 1

ℓ

 Cℓ  − a + a  − a + a  dξ 1

2

1 2

1 2

1

2

2

2

1

2

2

−  b ℓ  a ξ  dξ −  b ℓ  a ( − ξ)+a ( ξ− )  dξ = 0 1

o

2

1

1 2

0

o

1

2

2

2

1

=

b oℓ 2 C

2

Resulting in the equations 6 −2 C ℓ −2 2

a1 a2

= b oℓ

1 2 1 4

a1

⇒

a2

3 16 5 16


b oℓ 2  3 5 h (ξ) + 16 h 2(ξ)  C 16 1

which is exact at the free end. The approximate stress is given by

σ(ξ) =

b oℓ 32

12h1′ + 20h 2′

0 ≤ ξ ≤ 12

6h1′ + 10h 2′

1 2

≤ξ≤1

Exact 24

32 24

32σ

b oℓ

8

8

The stress again takes a jump at the midpoint, but this time it jumps because the linear approximation of displacement within each element implies a constant variation of stress in each element. The approximation is trying to capture the linear variation of stress of the exact solution within these constraints.

149. Consider the rod of length 2π and constant modulus b(x) = sin x C=1, free at both ends and subjected to the sinusoidal body force, as shown. The general classical solution for the given x loading is u(x) = a 0+a1 x+ sin x. Show that the given solution satisfies the governing differential equation for the bar, 2π and state the essential and natural boundary conditions. Use the boundary conditions to find the integration constants a0 and a1. Explain any peculiar features of the solution to this problem. Use a polynomial Ritz basis to find a two-term approximate solution for the displacement field, using the principle of virtual work. Explain any peculiar features of the Ritz approximate solution.

Chapter 6

149


By direct differentiation of the solution: u′ = a 1+ cos x and u′′ = − sin x, which is clearly equivalent to the governing equation for C=1. The boundary conditions are both natural: u′(0) = 0 and u′(2π) = 0. Substituting we get a1+1 = 0 and a1+1 = 0, respectively. These two equation give only the constant a1 = −1, a0 is arbitrary, implying u(x) = a 0−x+ sin x. It is not possible to establish the position of the bar, only the strains. The loading is called a self-equilibrating load. Let u = a 0+a1 x be the approximate displacement ( u′ = a 1) and u = a 0+a1 x be the virtual displacement ( u′ = a 1). The virtual-work functional then gives

G(u, u) ≡



2π

 a 1 a 1 − (a0+a 1 x) sin x  dx = 2π a 1+1  a 1

0

The approximate configuration is found by setting G = 0 for all values of the virtual displacement. Hence we get a1 = −1, as in the classical solution. Note that, again, the constant a0 is undetermined.

150. Consider the rod of length 3, constant unit modulus C = 1 (and unit area), fixed at x = 0. The rod is subjected to a certain (unspecified) distribution of body force b, as shown. Three piecewise linear finite element basis functions are shown in the sketch. The functional expressions for the basis function hi (x) is

hi (x) =

x − i + 1,

i -- 1 ≤ x ≤ i

i + 1 − x, 0,

i ≤ x ≤ i+1 elsewhere

b(x) x 1

1

1

h1 h2 h3

An approximate displacement field can be constructed from the base functions as u(x) = a 1 h1 (x)+a 2 h 2 (x)+a3 h 3 (x). Find the stiffness matrix K consistent with this approximation and the principle of virtual work. Assume that the values of the coefficients are a1 = 1, a2 = 2, and a3 = 4. Plot the stress field associated with the approximation. Find the equivalent force vector f.

(a) Find the stiffness matrix K consistent with this approximation and the principle of virtual work. The first derivatives of the base functions are (graphically)

h 1′

1

h 2′ --1

1

h 3′ --1

1

150


 h ′ h ′ dx 3

K ij =

i

(note: C = 1)

j

0

 h ′ h ′ dx +  h ′ h ′ dx +  h ′ h ′ dx 1

=

2

i

j

3

i

0

j

i

1

j

2

Carrying out the integrations we get 2 --1 0

K=

--1 2 --1

0 --1 1

(b) Assume that the values of the coefficients are a1=1, a2=2, and a3=4. Plot the stress field associated with the approximation and find the equivalent force vector f. The stress can be computed from the base functions as

σ = Cu′ =

 a h ′(x) = i

0≤x≤1

a1

3

i

i=1

a 2−a 1

1≤x≤2

a 3−a 2

2≤x≤3

Substituting the values of the coefficients given the stress is

σ 2 1 0

1

2

3

(c) The force vector can be computed as 2

f = K a = --1 0

--1 2 --1

0 --1 1

1 2 4

151. Consider the rod of length 3ℓ, constant modulus C = 2kℓ, and unit area, fixed at x = 0 and spring supported at x = 3ℓ, with spring constant k. The rod is subjected to a point load F = 3b oℓ at midspan and a uniform body force bo, as shown. Three piecewise linear finite element basis functions are shown in the sketch. The functional expressions for the basis function hi (x) is hi (x) =

x∕ℓ − i + 1,

i -- 1 ≤ x∕ℓ ≤ i

i + 1 − x∕ℓ,

i ≤ x∕ℓ ≤ i + 1

0

= --1 2

b(x) = b o x ℓ h1 h2 h3

F ℓ

ℓ

k

Chapter 6


151

Set up the equilibrium equations implied by the principle of virtual work using the Ritz method (i.e., find K and f). Express your answer in terms of k, bo , and ℓ (not F and C). The displacement interpolation is

u(x) = a 1 h1(x)+a 2 h 2(x)+a3 h3(x) and the virtual work functional has the form

G(u, u) =



3ℓ

 Cu′u′ − bu  dx − F u3ℓ∕2 + k u(3ℓ)u(3ℓ)

0

Therefore, the stiffness matrix and force vector

K ij =



3ℓ

C h i′hj′ dx + k hi (3ℓ) hj (3ℓ)

fi =

0



3ℓ

b h i dx + F h i 3ℓ∕2

0

The components of K can be computed as

 C1ℓ1ℓ dx +  ℓ

K 11 =

0

K 12

2ℓ

0

=

2ℓ

0

  

C − 1 − 1 dx = 2C = 4k = K 22 ℓ ℓ ℓ

  

C − 1 1 dx = − C = −2k = K 21 = K23 = K 32 ℓ ℓ ℓ K 13 = K31 = 0

K 33 =



3ℓ

2ℓ

  

C 1 1 dx + k(1)(1) = C + k = 3k ℓ ℓ ℓ

The components of f can be computed as

f1 =



2ℓ

b 0 h1(x) dx + F h13ℓ∕2 = b 0ℓ + 12 F = 52 b0ℓ = f 2

0

f3 =



3ℓ

b 0 h3(x) dx = 12 b 0ℓ

2ℓ

Summarizing, we have 4

K = k --2 0

--2 4 --2

0 --2 3

f=

b 0ℓ 2

5 5 1

152


152. Solve the problem of the rod subjected to a triangular load shown in the sketch using the quadratic Lagrangian finite element base functions. Use at least two quadratic elements (i.e., five nodes with four segments of length ℓ∕4) to carry out the solution.

bo x ℓ

The virtual-work functional can be put into the form 4

G(u, u) ≡

4

a K i

4

ij a j −

i=1 j=1

a f

i i

i=1

Where the coefficients are computed as Kij = C ℓ

 h ′(ξ) h ′(ξ) dξ

 (1−ξ) h (ξ) dξ

1

i

1

f i = b oℓ

j

i

0

0

For example, K24 = C ℓ

f 2 = b oℓ

1 2

 1

1 2

2C 3ℓ

16ξ−14 (16ξ−10) dξ =

 (1−ξ)( ξ − ξ) dξ + b ℓ  (1−ξ)( ξ − 1

8

2

2

o

0

2

8

1 2

1 14ξ+6) dξ = 12 b oℓ

The remainder of the values are summarized in the following equations

C 3ℓ

32

−16

0

0

a1

−16

28

−16

2

a2

0

−16

32

−16

a3

0

2

−16

14

a4

= b oℓ

1 4 1 12 1 12

0

a1

37

a2 ⇒

a3

=

a4

The approximate displacement is then computed as u(ξ) =

b oℓ  37h 1(ξ) + 56h2(ξ) + 63h 3(ξ) + 64h4(ξ) 

384 C

Note that the displacement at the free end is boℓ∕6 C, which is exact.

b oℓ 2 384 C

56 63 64

Chapter 6

153


153. Consider the rod of length 3ℓ and constant modulus C, fixed at both ends, and subjected to point loads of magnitude F and 2F at the third points, as shown. Use a piecewise linear finite element approximation with nodes at the ends and at the third points. Write the expressions for the base functions hi (x). Compute K and f associated with the discrete virtual-work function. Compute the coefficients a from Ka = f. Sketch the approximate displacement field. Compute the approximate stress field.

2F

F x ℓ

ℓ

ℓ

(a) Let ξ ≡ x∕ℓ. The base functions are h1(ξ) =

h2(ξ) =

ξ

0≤ξ≤1

2−ξ

1≤ξ≤2

0

2≤ξ≤3

0

0≤ξ≤1

ξ−1

1≤ξ≤2

3−ξ

2≤ξ≤3

h1

1

h2

1

(b) The coefficients are computed as Kij = C ℓ

 h ′(ξ) h ′(ξ) dξ 3

i

j

f i = Fhi(1) + 2Fh i(2)

0

The resulting equations are 2 −1 C ℓ −1 2

a1 a2

a1

1

=F

2

⇒

= Fℓ C

a2

Therefore, the approximate displacement is 3Cu Fℓ

u(ξ) = Fℓ  43 h 1(ξ) + 53 h 2(ξ)  C

4

5

(c) The approximate stress is σ = Cu′. Therefore, from the interpolation,

σ(ξ) = F  43 h 1′(ξ) + 53 h 2′(ξ) 

3σ F

4

1 −5

4 3 5 3

154


154. Consider the rod of length ℓ and constant modulus bo C. The rod is fixed at the left end and restrained by an elasx k C tic spring of modulus k at the right end. The spring accrues force equal to the product of spring constant and stretch of ℓ the spring, i.e., f s = k u(ℓ). The rod is subjected to a constant body force b(x) = b o , as shown. What are the essential and natural boundary conditions for this problem? (Hint: Take a freebody diagram of the right end of the rod to get the mixed boundary condition at that end). Find the classical solution to the boundary value problem. At what point is the strain in the rod the greatest? Consider the two limiting cases (1) k → ∞, and (2) k → 0. What are the boundary conditions in these two limiting cases? What is the solution in these two cases? Find an approximate solution with the Ritz method and a polynomial approximation.

(a) What are the boundary conditions for this problem? The boundary condition on the left end of the rod is u(0) = 0. To get the condition at the right end consider a freebody of the right end of the rod.

u(ℓ) σ(ℓ)

k u(ℓ)

From equilibirum we have

σ(ℓ) + k u(ℓ) = 0

⇒

Cu′(ℓ) + k u(ℓ) = 0

(b) Find the classical solution to the boundary value problem. The classical equation governing the response is Cu′′ = −b o. The general solution (homogeous plus particular) of this equation is

Cu = − 12 b o x 2 + a1 x + a 2 Substituting the condition u(0) = 0 gives a 2 = 0. Substituting the boundary condition Cu′(ℓ) + k u(ℓ) = 0 gives

−b o ℓ + a 1 + k − 12 b o ℓ 2 + a 1 ℓ = 0 C Upon simplification we find that







a 1 1+ kℓ = b oℓ 1+ kℓ C C



⇒

a 1 = 12 boℓ

2+β  1+β

where β ≡ kℓ∕C is a nondimensional measure of the spring stiffness relative to the modulus of the rod. Substituting this constant back into the general solution we find the expression for the displacement as

u(x) =

   

2 b oℓ2 2x 2+β − x2 2C ℓ 1+β ℓ

Chapter 6

155


(c) At what point is the strain in the rod the greatest? Compute the strain as

Á = u′ =

b oℓ C

   2+β −x ℓ 1+β

which is a linear function. Therefore, the maxima are at the ends of the rod.

Á(0) =

 

b oℓ 2+β C 1+β

Á(ℓ) =

 

b oℓ 1 C 1+β

Thus, the strain is greatest at x = 0. (d) Consider the two limiting cases (1) k→∞, and (2) k→0. What are the boundary conditions in these two limiting cases? What is the solution in these two cases?

k→∞ k→0

u(ℓ) = − 1 σ(ℓ) → 0 k σ(ℓ) = −k u(ℓ) → 0





b oℓ 2 2x x 2 − 2 ℓ 2C ℓ



u(x) = u(x) =

2 b oℓ 2 x −x 2C ℓ ℓ 2



The first represents a fixed end, the second is a free end.

155. Consider the rod of length ℓ and constant modulus C, b(x) = b o fixed at both ends, and subjected to a uniform body force bo as shown. The left end moves to the right by an amount 3uo uo 3uo and the right end moves to the left by an amount uo. x What are the essential and natural boundary conditions. ℓ Compute the displacement field u(x) using the principle of virtual work and the Ritz method with a quadratic approximation. Sketch the approximate displacement field. Compute the approximate stress field.

(a) What are the essential and natural boundary conditions. Essential B.C.’s: u(0) = 3u o and u(ℓ) = −u o Natural B.C.’s: none. (b) Compute the displacement field u(x) using the principle of virtual work and the Ritz method with a quadratic approximation. Take a solution of the form u(x) = a 0+a1 x+a 2 x 2. Substituting the essential boundary conditions we get

u(0) = a 0 = 3uo u(ℓ) = a 0+a1 ℓ+a 2 ℓ 2 = −uo

⇒

a1 = −4

uo − a2 ℓ ℓ

Therefore, the displacment approximation can be expressed in the form

156






u(x) = u o(x) + a 2 h 2(x) = uo 3−4 x + a 2  x 2−x ℓ  ℓ Now, the virtual displacement can be taken to satisfy the homogeous boundary conditions u(0) = 0 and u(ℓ) = 0. Hence, the virtual displacement is

u(x) = a 2  x 2−x ℓ  The virtual work functional for this problem is

  Cu′u′ − b u dx = 0 ℓ

G(u, u) =

∀u

o

0

Substituting the interpolation we get

 C −4uℓ  2x−ℓ  +  2x−ℓ  a  − b  x −x ℓ  dx = 0 ℓ

a2

0

2

o

2

2

0

Carrying out the integrals we get

a 2 0 + 13 Ca 2ℓ3 + 16 b oℓ3 = 0

⇒

a2 = −

bo 2C

Thus, the approximation solution is





b u(x) = u o 3−4 x − o  x 2−x ℓ  2C ℓ (c) Compute the approximate stress field.



σ(x) = Cu′(x) = C − =−



4u 0 b − o  2x−ℓ  2C ℓ

4Cuo bo  − 2x−ℓ  2 ℓ

P 156. Consider the rod of unit length ℓ = 1, constant unit modulus C = 1 (w/ unit area), embedded in an elastic medium that gets stiffer with depth. The elastic medium has variable modulus given by k(x) = 12(1+x), and the resistance to motion is linearly proporx k tional to the displacement. The rod is subjected to a unit load at ℓ x = 0, i.e., P = 1, and is traction free at the end x = 1. The classiC cal governing differential equation for the displacement field u(x) of the rod is u′′−12 (1+x) u = 0. Calculate an approximate value of the displacement at the point of load using a linear approximation of the displacement field. Use the Ritz method to carry out the calculations. Is the approximate solution a good one? Why or why not? Does the accuracy of the approximation depend upon the relative flexibilities of the rod and the elastic medium? How?

Chapter 6

157


(a) Calculate an approximate value of the displacement at the point of load using a linear approximation of the displacement field. The virtual work functional for this problem is

  u′u′ + 12 1+x  uu  dx − u(0) 1

G(u, u) =

0

Take the displacement approximation in the form

u(x) = a 0 + a 1 x

u(x) = a 0 + a 1 x u′ = a 1

u′ = a 1

Note that there are no essential boundary conditions. Substituting the approximation into the virtual work functional we get

  a a + 12(1+x)a +a x a +a x   dx − a = 0 1

1

1

0

1

0

1

0

∀ a 0, a 1

0


a 0  18a 0+10a 1 − 1  + a1  10a0+8 a1  = 0 or, invoking the Fundamental Theorem of the Calculus of Variations,

18a 0+10a 1 = 1 10a 0+ 8a 1 = 0 which can be solve to give a 0 = 8∕44 and a 1 = −10∕44. Thus, the approximation displacement is 1  u(x) = 22 4−5x 

2 u(0) = 11

(b) Is the approximate solution a good one? Why or why not? Does the accuracy of the approximation depend upon the relative flexibilies of the rod and the elastic medium? How? Compute the error as the residual of the classical differential equation

 1+x  4−5x  e ≡ u′′ − 12(1+x)u = − 12 22 ~1

e 0

1

x

~2

Also, u′(0) = −5∕22 ≠ − 1 (the value of the applied load at the end) and u′(1) = −5∕22 ≠ 0 (traction free at the other end). Therefore, this is not a very accurate approximation.

158


157. Consider the rod of Problem 156, now with length ℓ, constant modulus C (w/ unit area), subjected to a load P at x = 0. The rod is embedded in an elastic medium such that the force developed is linearly proportional to the displacement at each point. The modulus k is constant. The elastic constants are related by kℓ2∕C = 1. The rod is pointed so that the traction at the end x = ℓ is zero. Find the virtual-work functional for the given problem. What are the essential and natural boundary conditions? Find an approximation to the displacement field using the Ritz method with a two-term approximation with the following base functions h1(x) = eαx, h2(x) = e −αx, where α ≡ k∕C. The virtual work functional for the given problem is

  Cu′u′ + k uu  dx − P u(0) ℓ

G(u, u) =

0

There are no essential boundary conditions. The natural boundary conditions are σ(0) = −P and σ(ℓ) = 0. Approximate the displacement field using the Ritz method with a two-term approximation with the following base functions h 1(x) = e αx , h 2(x) = e −αx , with α ≡ k∕C. The components of K and f are

  Ch ′h ′+k h h  dx ℓ

K mn =

m

n

m

n

f m = P h m(0)

0

Substituting the base functions we get

  α C+k  e ℓ

K 11 =

2

2α x

0

2 k  e 2α ℓ−1  dx = α C+k  e 2α ℓ−1  = α 2α

 − α C−k  dx =  k−α C  ℓ = 0 = K ℓ

K 12 =

2

2

21

0

=   α C+k  e ℓ

K 22

2

−2α x

dx =

0

α 2 C+k  1−e −2αℓ  = αk  1−e −2αℓ  2α

The force vector has components f 1 = f 2 = P. Solving Ka = f gives

a1 =

αP k  e 2αℓ−1 

a2 =

αP k  1−e −2αℓ 

Note that, from the problem description αℓ = kℓ 2∕C = 1. Hence,

u(x) =

P  e x∕ℓ + e (2−x∕ℓ)  kℓ  e 2−1 

Chapter 6

159


158. Consider the rod of length ℓ, constant modulus C (w/ unit area), subjected to a load P at x = 0. The rod is embedded in an elastic medium that provides a resisting force proportional to the displacement at each point with a modulus k(x) that increases linearly with depth, so that the force is f (x) = k o x u(x)∕ℓ. The end resistance can be modeled as a spring of modulus ko ℓ. The elastic constants are related by koℓ 2∕C = 1. Find the virtual-work functional for the given problem. What are the essential and natural boundary conditions? Find the displacement field using the Ritz method with a two-term polynomial approximation.

P

x ℓ

f (x)

k oℓ

The virtual work functional for the given problem.

  Cu′u′ + k ℓx u u  − Pu(0) + k ℓu(ℓ) u(ℓ) = 0 ℓ

G(u, u) =

o

o

0

There are no essential boundary conditions. The natural boundary conditions can be found by taking a freebody diagram of each end as enforcing the equilibirum equations. They are σ(0) = −P and σ(ℓ) = −k oℓu(ℓ). Approximate the displacement field using a two-term polynomial approximation. Let ξ ≡ x∕ℓ and

u = a0 + a1 ξ

u = a0 + a1 ξ

u′ = a 1∕ℓ

u′ = a 1∕ℓ

Substituting the interpolation into the virtual work functional gives

  ℓC a a +k ξ  a +a ξ  a +a ξ   dξ 1

G(u, u) =

2

0

1

1

o

0

1

0

1

− P a 0 + k oℓ  a 0+a 1  a 0+a1 

Carrying out the integrals, noting C = k oℓ2 , gives

a 0  12 k oℓa0 + 13 k oℓ a1 − P + k oℓa 0 + k oℓa1 + a 1  k oℓa 1 + 13 k oℓa0 + 14 k oℓa1 + k oℓ a0 + k oℓ a1  = 0 Invoking the Fundamental Theorem of the Calculus of Variations we obtain 3 k ℓa 2 o 0 4 k ℓa 3 o 0

These equations can be solved to give

+ 43 k oℓ a1 = P + 94 k oℓ a1 = 0

160


a 0 = 1.409 P , k oℓ

a 1 = −0.835 P k oℓ

Therefore, the approximate displacement field is



u(x) = P 1.409−0.835 x k oℓ ℓ



159. Consider the rod of length ℓ and constant modulus C. The rod is fixed at the left end and restrained by a linear elastic spring of modulus k at the right end. The elastic constants are related by kℓ/C=2. The rod is subjected to a constant body force b(x) = bo and a prescribed displacement at the left end of uo , as shown. Set up, the discrete equations Ka=f that result from applying the Ritz method to the principle of virtual work using the base functions shown in the sketch. Express the answer in terms of k, ℓ, bo , and uo . Do you expect the solution using the Ritz method to be the exact classical solution to the boundary value problem? What base functions would you need to add to make the Ritz approximation exact?

uo

bo x

C

k

ℓ uo(x) uo h1

1

h2

1

h3

1

(a) Set up the discrete equations Ka=f that result from applying the Ritz method to the principle of virtual work using the base functions shown in the sketch. The virtual work functional has the form

  Cu′u′ − b u  dx + k u(ℓ)u(ℓ) ℓ

G(u, u) =

o

0

Take a displacement approximation in the form

u(x) = u 0(x) + h(x) ⋅ a,

u(x) = h(x) ⋅ a

where h(x) = {h 1, h2, h 3 }, a = {a 1, a2, a 3 }, and a = {a 1, a2, a 3 }. We can take u o(x) = uo (constant), so that u o′(x) = 0. With these definitions the virtual work functional takes the form G(a, a) = a T  K a−f , where

 C h′  h′ dx + k h(ℓ)  h(ℓ) ℓ

K=

0

 b h dx − k u (ℓ)h(ℓ) ℓ

f=

o

0

o

Chapter 6

161


Note that

 C h′  h′ dx = 3Cℓ I ℓ

h(ℓ) = {1, 1, 1 }

0

Carrying out the remaining integrals and making the substitutions we get

K = 3C ℓ

1 0 0

f = b0

0 1 0

0 0 1

1 1 1

+k

5ℓ∕6

1

ℓ∕2

− k u0 1

ℓ∕6

1

1 1 1

1 1 1

bℓ = o 6

5 2 2

=k 2

5 3

2 5 2

2 2 5

1

− k u0

1

1 1

(b) Do you expect the solution using the Ritz method to be the exact classical solution to the boundary value problem? No. The exact solution is quadratic for this problem. Thus, we need quadratic base functions to represent it exactly.

160. Consider the rod of length ℓ = 1 and a constant modulus C = 10. The rod is restrained by an elastic spring of modulus k = 2 at each end and is subjected to a constant body force b(x) = 1, as shown. The virtual-work functional is

b(x) = 1 k

x

  Cu′u′−bu  dx + ku(0)u(0) + ku(ℓ)u(ℓ)

C

k

1

ℓ

G(u, u) =

0

What are the boundary conditions for this problem? Are they essential or natural boundary conditions? Explain. Set up, the system of equations that result from using a quadratic approximation of the displacement field with the Ritz method. Consider solving this problem using the trigonometric approximation u(x) = a 1 sin πx + a2 sin 2πx + a3 sin 3πx Is this approximation likely to give a good solution to the problem or not?

(a) The boundary conditions for this problem are both natural because they represent equations of equilibrium. They can be obtained by taking a freebody diagram of each end of the beam.

σ(0) − ku(0) = 0

⇒

Cu′(0) − ku(0) = 0

σ(ℓ) − ku(ℓ) = 0

⇒

Cu′(ℓ) − ku(ℓ) = 0

162


(b) Set up the system of equations that result from using a quadratic approximation of the displacement field with the Ritz method. Take an approximation of the form

u = a0 + a1 x + a2 x2

u = a0 + a 1 x + a 2 x 2

u′ = a 1 + 2a 2 x

u′ = a 1 + 2a 2 x

Substituting the approximations into the virtual work functional gives

K=C



1

0

K=

0

0

0

0

1

2x

0

2x 4x2

1

4

2

2

2

12

12

2

12

46 3

f=b

0

1

0



0

1

1

x

1

0

1

1

1

dx = 1

x

0

1

+k 1 1 1

dx + k 0 0 0 0

1

2

1 2 1 3

The governing equations are then K a = f . (c) Consider solving this problem using the trigonometric approximation

u(x) = a 1 sin πx + a 2 sin 2πx + a 3 sin 3πx Is this approximation likely to give a good solution to the problem or not? Why or Why not? This basis is not good for this problem because each base function satisfies h(0) = 0 h(1) = 0. Hence, the springs cannot be mobilized by the approximation.

161. Consider the solid cubical region B shown in the sketch having unit dimensions. Let the scalar field w(x) characterize the response of the system. The field w is a function of the position vector x. If we define the functional G(w, v) ≡

  ∇w ⋅ ∇v − 3 v  dV

x3 1

1

x2

B 1

x1

B

then G = 0 (for all v) is a “virtual-work” statement of the equations governing w. The essential boundary conditions are such that w = 0 on the coordinate faces. Use the Ritz method with a single term approximation of the form w(x) = a 0 x1 x 2 x3 to determine the unknown field w. Describe how you would improve the approximation.

Chapter 6

163


Let the virtual displacement field be v = a 0 x 1 x 2 x 3. Computing the gradients

x 2 x3 x 1 x3 x 2 x1

∇w = a 0

x 2 x3 x 1 x3 x 2 x1

∇v = a 0

∇w ⋅ ∇v = a0 a 0  x 22 x 23 + x 21 x 23 + x 22 x 21  Substituting into the virtual work functional, invoking the Fundamental Theorem of the Calculus of Variations, we get

     x x +x x +x x  a − 3x x x  dx dx dx = 0 1

1

1

2 2

0

0

2 3

2 2 1 3

2 2

2 1

0

1

2

3

1

2

3

0

1 3

a 0 − 38 = 0

⇒

a 0 = 98

Therefore, the approximate displacement is

w(x) = 98 x 1 x 2 x 3 To improve the approximation add more terms, w(x) = a ⋅ h(x), where each h i(x) satisfies the essential B.C. (i.e., equal zero on the coordinate faces).

162. Reconsider the 2ℓ × 2ℓ × h block shown in Fig. 84. The block is fixed at the base, (i.e., u(x1, x2, 0) = 0) and subjected to a body force (self-weight) of b = −ρ o e3 throughout the volume. Let the material have independent elastic constants λ and m. Solve the problem by the Ritz method using the following assumption about the displacement field u(x) = a 1 x3 ℓ e3 + a 2  x1 x3 e 1 + x2 x 3 e 2 

The base functions are h1(x) = x3ℓ e 3 and h2(x) = x1 x 3 e1 + x 2 x 3 e2. These base functions are identical to the example problem in the book. The necessary derivatives of the functions are div h 1(x) = ℓ

∇h 1(x) = ℓ e 3  e 3

div h 2(x) = 2x 3

∇h 2(x) = x 3 e1  e1 + x3 e2  e2 + x1 e 1  e3 + x2 e 2  e 3

 ∇h 1 + ∇h T1  ⋅ ∇h 1 = 2ℓ 2  ∇h 2 + ∇h T2  ⋅ ∇h 1 = 0

 ∇h 1 + ∇h T1  ⋅ ∇h 2 = 0  ∇h 2 + ∇h T2  ⋅ ∇h 2 = x21 + x22 + 4 x23

The coefficient matrix Kij and right side f i can be computed from the equations

  h

Kij =

0

ℓ

ℓ

−ℓ

−ℓ

 λdiv h i div h j + m ∇h i + ∇hTi ⋅ ∇h j dx1 dx 2 dx3

164


  h

fi =

0

ℓ

ℓ

−ℓ

−ℓ

−ρ o e 3 ⋅ hi(x 1, x 2, x 3) dx1 dx 2

Carrying out the indicated integrations gives the following system of equations for the unknown coefficients a1 and a2 4 hℓ 4 3

3 λ+2m 

3β λ

a1

3β λ

2+4β2 λ+m 

a2

= − 2ρ ohℓ4β

1 0

where β ≡ h∕ℓ. These equations can be solved to give the constants a1 = − 12 ρ o β a2 =

2m+4β 2(λ+m) 2m(λ+2m)+β 2(λ 2+12λm+8m 2)

λβ 3 ρ β 2 o 2m(λ+2m)+β 2(λ2+12λm+8m 2)

It is interesting to note that when λ = 0, 2m = C, and there is no Poisson’s effect. In this case we get a1 = −ρ o β∕2C and a2 = 0 so that u3 = −ρ o x 3 h∕2C.

163. Resolve Problem 162 with the following assumed displacement field u(x) = a 1 x3 ℓ e3 + a 2 x23 e 3 + a3  x1 x 3 e 1 + x2 x 3 e2  What is the contribution of the term  x 1 x3 e 1 + x2 x3 e 2  (i.e., the lateral displacement) to the response in parts (a) and (b)? Compute the stress tensor S implied by the displacement fields of parts (a) and (b). What body forces and surface tractions are implied by these maps? What terms would you add to the solution to improve the Ritz approximation?

The base functions are h1(x) = x3ℓ e 3, h2(x) = x23 e 3 , h3(x) = x1 x 3 e1 + x 2 x 3 e2. The divergence and gradient for each base function can be computed as follows div h 1(x) = ℓ

∇h 1(x) = ℓ e 3  e 3

div h 2(x) = 2x 3

∇h 2(x) = 2x 3 e3  e3

div h 3(x) = 2x 3

∇h 3(x) = x 3 e1  e1 + x3 e2  e2 + x1 e 1  e3 + x2 e 2  e 3

∇S h 1 ⋅ ∇h 1 = 2ℓ 2

∇S h 1 ⋅ ∇h 2 = 4x 3ℓ

∇S h 1 ⋅ ∇h 3 = 0

∇S h 2 ⋅ ∇h 1 = 4x 3ℓ

∇S h 2 ⋅ ∇h 2 = 8x23

∇S h 2 ⋅ ∇h 3 = 0

∇S h 3 ⋅ ∇h 1 = 0

∇S h 3 ⋅ ∇h 2 = 0

∇S h 3 ⋅ ∇h 3 = x 21+x22+4 x 23

where ∇S hi ≡ ∇h i+∇hTi is the symmetric gradient. Carrying out the indicated integrations as in part (a) gives the following system of equations for the unknown coefficients a1, a2 and a3

Chapter 6

4 hℓ 4 3

165


3 λ+2m 

3  λ+2m  β

3λβ

a1

3  λ+2m  β

4 λ+2m  β 2

4λβ2

a2

3β λ

4λβ2

2m+4β2 λ+m 

a3

3 β 2

= − 43 ρ ohℓ4

β2 0

where β ≡ h∕ℓ. These equations can be solved to give the constants a1 = −ρ o a2 = ρo a3 = ρo

β λ+2m

m(λ+2m) + β 2(4m 2+6λm−λ2)

2m(λ+2m)λ+2m+2β 2(3λ+2m) 

λβ 2  2m λ+2m + 2β 2(3λ+2m) 

It is interesting to note that when λ = 0, 2m = C, and there is no Poisson’s effect. In this case we get a1 = −ρ o β∕C, a2 = ρ o∕2C, and a3 = 0 which is the classical result for a truss bar, with u3 = ρ o x 23−2 x3 h ∕2C. The contribution of the term  x 1 x3 e 1 + x2 x3 e 2  is driven by the Poisson effect. The stress can be computed as S = λdiv u + m ∇Su =

 a  λdiv h + m a ∇ h  i

i

i

S i

i

For example, in part (a) the stress tensor has the explicit form (with divergence and tractions as indicated):

S=

λa 1ℓ+2ma 2 x 3

0

a 2 mx1

0

λa 1ℓ+2ma 2 x 3

a 2 mx2

a 2 mx1

a 2 mx2

a 1ℓ(λ+2m)+λa 2 x 3

0 a 2 mx1

Se2 =

λa 1ℓ+2ma 2 x 3 a 2 mx2

0

div S =

0

λa 1ℓ+2ma 2 x 3

Se1 =

0

a 2 λ+2m 

a 2 mx1

Se3 =

a 2 mx2 a 1ℓ(λ+2m)+λa 2 x 3

Observe that the given displacement field does not do a very good job of representing the stresses for this problem (except when Poisson’s effect goes to zero). It is clearly a difficult problem trying to select base functions that improve the representation of stresses. Perhaps a good choice for the next term in the expansion would have the specific form  x 1 x23 e 1 + x2 x23 e 2 .

166


164. Reconsider the 2ℓ × 2ℓ × h block shown in Fig. 84. As in Problem 162, the block is fixed at the base, (i.e., u(x1, x2, 0) = 0) and subjected to a body force b = −ρ o e3 throughout the volume. Let the material have independent elastic constants λ and m. Let ξ1 = 12  x 1∕ℓ+1 , ξ2 = 12  x 2∕ℓ+1  and ξ3 = x3∕h be a change of variables that maps the block onto the unit cube with one vertex at the origin of the coordinate system ( ξ1, ξ2, ξ3). Define the following functions φ1(ξ) = ξ1 ξ 2 ξ3

φ5(ξ) = ξ1 ξ 2 (1−ξ3)

φ2(ξ) = (1−ξ1) ξ 2 ξ3

φ6(ξ) = (1−ξ1) ξ 2 (1−ξ3)

φ3(ξ) = ξ1 (1−ξ 2) ξ3

φ7(ξ) = ξ1(1−ξ 2) (1−ξ3)

φ4(ξ) = (1−ξ1) (1−ξ 2) ξ3

φ8(ξ) = (1−ξ1) (1−ξ 2) (1−ξ3)

These functions have the property that, at each of the eight vertices, one of the functions has unit value while the others are zero. They are, in fact, the finite element base functions for a hexahedron element. Let the displacement be approximated as 8

 a φ (ξ)

u(ξ) =

i

i

i=1

where ai is a vector constant with component expression ai = aij e j (no sum on j). What are the base functions hi(ξ) associated with this expansion? What is the physical significance of the coefficient vector ai? What does the essential boundary condition u(ξ1, ξ2, 0) = 0 imply about the values of the coefficients in the expansion? Solve the block problem using the base functions identified, as restricted by the essential boundary condition.

It is convenient to write the expansion of the displacement in terms of scalar coefficients as 8

u(ξ) =

3

a

ij

φi (ξ) ej

i=1 j=1

Then the base functions can be represented as hij = φi (ξ) e j. Let the nodes be numbered as shown in the figure. Then the function φi (ξ) has unit value at node i and is zero at all of the other nodes. The coefficient aij is the nodal displacement at node i in the direction j. ξ3 2

4

1

3

ξ2

6

8

5

7

ξ1

Chapter 6

167


Since the base functions φ5, φ6, φ7, and φ8 are not zero at ξ3 = 0 the coefficients must be zero to enforce the essential boundary conditions. Therefore, aij = 0 for the values i=5, 6, 7, and 8 and j=1, 2, and 3. Since the base functions φ1, φ2, φ3, and φ4 are zero at ξ3 = 0 the remaining coefficients are not restricted by the essential boundary conditions. where differentiation of φi is with respect to x which can be done by the chain rule. ∇ h ij = (φie j ), k  e k = φ i, k e j  ek ∇S hij ⋅ ∇ h kl = φi, α φ k, α δjl + φ i, l φk, j The derivatives of the shape functions are φ 1, 1 = a ξ 2 ξ 3

φ 1, 2 = a ξ 1 ξ 3

φ 1, 3 = b ξ 1 ξ 2

φ2, 1= −a ξ 2 ξ 3

φ2, 2 = a (1−ξ 1)ξ 3

φ2, 3 = b (1−ξ 1)ξ 2

φ3, 1 = a (1−ξ 2)ξ 3

φ3, 2 = −a ξ 1 ξ 3

φ3, 3 = b ξ 1 (1−ξ 2)

φ4, 1 = −a (1−ξ 2)ξ 3

φ4, 2 = −a (1−ξ 1)ξ 3

φ4, 3 = b (1−ξ 1)(1−ξ 2)

where a = 1∕2ℓ and b = 1∕h. The components of the coefficient and right side matrices can be computed as

    λφ , φ , + m  φ , φ , δ +φ , φ ,   J dξ dξ dξ 1

K(ij)(kl) =

1

1

i j

0

0

i α

k l

k α

jl

i l

k j

1

2

3

0

   −ρ φ δ 1

f (kl) =

1

1

o

0

0

k 3l

J dξ1 dξ 2 dξ3

0

where J = 4ℓ 2h is the determinant of the Jacobian. The double indices can be put into standard matrix form by ordering the components in a particular way. For example, [11, 21, 31, 41, 12, 22, 32, 42, 13, 23, 33, 43], which simply runs through the first index (associated with node number) for each value of the second index (the coordinate direction). With this ordering we can put the results into matrix form by defining the vectors φ1, φ2, and φ3 to have components [φj] i = φ i, j. In other words φj contains the derivatives of the base functions with respect to xj. In the present case each of these vectors has length 4 since there are four base functions. Let us further define the matrices φ 11 φ 12 φ 13 A≡

φ 21 φ 22 φ 23

φ 11 φ 21 φ 31 B≡

φ 31 φ 32 φ 33

C≡

φ 12 φ 22 φ 32 φ 13 φ 23 φ 33

where 3

φij = φi  φ j

and

Φ≡

φ  φ i

i=1

i

Φ

0

0

0

Φ

0

0

0

Φ

168


Note that A, B, and C are 12×12. The coefficient matrices of the system of equations can now be written

    λA + m  B+C   J dξ dξ dξ 1

K =

1

1

1

0

0

   −ρ  φ0  J dξ dξ dξ 1

f =

2

1

0

1

o

0

3

0

0

1

2

3

0

where φ ≡{φ1, φ2, φ3, φ4}. The problem can be solved as Ka = f to determine the constants however there is a great deal of symmetry in this problem that one can use to reduce the system of equations to a 2×2. Let us select a^ ≡{a11 , a13 } as our vector of primary unknowns. By symmetry the remaining unknowns can be related to the present ones as a = Γa^ where Γ is given by ΓT =

1 −1

1 −1

1

1 −1 −1

0

0

0

0

0

0

0

0

1

1

1

1

0

0

0

0

The problem can now be solved as  Γ T K Γ  a^ = Γ Tf. The computation can be simplified by defining Γ T ≡  Γ T1 Γ T2 Γ T3  along the partition lines described above. Further, define cij ≡ Γ Ti φ j. Now we can observe that ^ A ≡ Γ T AΓ =  c 11+c22+c 33    c 11+c22+c 33 

and that 3

^

B ≡ Γ T BΓ =

3

c

3

^

C ≡ Γ T CΓ =

 cji

ij

i=1 j=1

 Γ ΦΓ T i

i

i=1

^

^

Finally, the system of equations can be put into the form Ka = f, where

    λA + m  B+C  J dξ dξ dξ 1

^

K =

1

1

^

^

1

0

0

f =

2

3

0

   −ρ Γ φ J dξ dξ dξ 1

^

^

1

1

o

0

0

T 3

1

2

3

0

The MATHEMATICAt commands needed to form and solve the problem are given below: p = { x y z , (1-x) y z , x (1-y) z ,(1-x) (1-y) z} p1 = D[p,x]/(2 a) p2 = D[p,y]/(2 a) p3 = D[p,z]/b gam1 = {{1,0},{-1,0},{ 1,0},{-1,0}} gam2 = {{1,0},{ 1,0},{-1,0},{-1,0}}

Chapter 6

169


gam3 = {{0,1},{ 0,1},{ 0,1},{ 0,1}} c11 = Transpose[gam1].p1 c12 = Transpose[gam1].p2 c13 = Transpose[gam1].p3 c21 = Transpose[gam2].p1 c22 = Transpose[gam2].p2 c23 = Transpose[gam2].p3 c31 = Transpose[gam3].p1 c32 = Transpose[gam3].p2 c33 = Transpose[gam3].p3 AA = c11 + c22 + c33 AAA = Outer[Times,AA,AA] BB = Outer[Times,c11,c11] + Outer[Times,c12,c21] + Outer[Times,c13,c31] + Outer[Times,c21,c12] + Outer[Times,c22,c22] + Outer[Times,c23,c32] + Outer[Times,c31,c13] + Outer[Times,c32,c23] + Outer[Times,c33,c33] PHI = Outer[Times,p1,p1] + Outer[Times,p2,p2] + Outer[Times,p3,p3] CC = Transpose[gam1].PHI.gam1 + Transpose[gam2].PHI.gam2 + Transpose[gam3].PHI.gam3 KK = ( lam AAA + mu ( BB + CC )) (4 a^2 b) K = Integrate[KK, {x,0,1},{y,0,1},{z,0,1}] FF = -(4 a^2 b)(Transpose[gam3].p) F = Integrate[FF, {x,0,1},{y,0,1},{z,0,1}] DD = Simplify[Inverse[K].F]

165. Consider a cube of dimension 2 × 2 × 2 fixed at the base and subjected to a body force b = −ρ o e3. Describe a method for refining the finite element approximation by establishing a local coordinate system for each element that allows the creation of the finite element base functions from the eight basic element functions φi (ξ) described in Problem 164. Notice that each element is associated with eight nodes while the entire block is associated with 27 nodes. Continuity of displacements can be assured by associating the element base functions with element nodal displacements (i.e., finite element functions) and by associating elements nodal displacements with a common global displacement parameter where elements share a common node. x3

x3

2 x2

2

1

x1

1 1

1

2

1

x2

1 1

B

ξ3

x3

1

1

ξ2

1 1

ξ1

1 x1

x1

Increase the order of interpolation. To refine the approximation a node can be placed at the center of each edge and face and one at the center of the cube. Then, the quadratic Lagrangian functions can be used to create the functions. There are 27 nodes and hence 27 functions φ i that take unit value at node i and are zero at the other nodes. At each node i there exist two other nodes that have the same ( ξ 2 , ξ 3 ) coordinates. Call the ξ 1 coordinates of those nodes ξ a1 and ξ b1 . There are two other nodes that have

x2

170


the same ( ξ 1 , ξ 3 ) coordinates. Call the ξ 2 coordinates of those nodes ξ c2 and ξ d2 . There are two other nodes that have the same ( ξ 1 , ξ 2 ) coordinates. Call the ξ 3 coordinates of those nodes ξ e3 and ξ f3 . Define Lj1(ξ 1) =

(ξ 1−ξ a1)(ξ 1−ξ b1) (ξ j1−ξ a1)(ξ j1−ξ b1)

Lj2(ξ 2) =

(ξ 2−ξ c2)(ξ 2−ξ d2) (ξ j2−ξ c2)(ξ j2−ξ d2)

Lj3(ξ 3) =

(ξ 3−ξ e3)(ξ 3−ξ f3) (ξ j3−ξ e3)(ξ j3−ξ f3)

The base function for node j can be constructed by multiplying these three functions together

φ j(ξ1, ξ 2, ξ3) = L j1(ξ 1)L j2(ξ 2)L j3(ξ 3) For example, consider the case shown in the figure with the node j taken at the corner with ξ 1 = ξ2 = ξ 3 = 1. The functions are

L j1(ξ1) = ξ 1 (2ξ 1−1),

L j2(ξ 2) = ξ 2 (2ξ2−1),

L j3(ξ 3) = ξ3 (2ξ 3−1)

Therefore, the base function for this node is

φ j(ξ1, ξ 2, ξ3) = ξ 1 (2ξ 1−1)ξ 2 (2ξ2−1) ξ3 (2ξ 3−1) Observe that this function has unit value at node j and is zero at all other nodes.

ξ3 b a d

c

j e

ξ2

f

ξ1 There are many other possible higher-order finite element base functions with different definitions of the nodal positions. Refine the mesh. An alternative to increasing the order of interpolation is to use the same base functions, but to have them describe the solution over a smaller portion of the domain. For example, we can subdivide the original problem into eight regions of half the dimension on each side. For each region we can define a change of coordinates to map that region onto the unit cube. Now, for each region (for example, the shaded region in the sketch), the base functions given for the eight-node cube provide a satisfactory building block for the finite element base functions. Since the original base functions have the property that they have unit value at one node and are zero at the other nodes, they can be pieced together to get base functions that are continuous between adjacent regions. Let us assume that all of the nodes have the local numbering show in the sketch for the single isolated element. The base functions can be constructed by piecing to-

Chapter 6

171


gether the local base functions φ i(ξ) defined in the problem statement. For example, the global base function that has unit value at node k and is zero at all other nodes can be defined as follows:

h kj(x) =

φ 3(ξ A)e j

x∈A

φ 4(ξ B)e j 0

x∈B x ∉ {A or B}

The change of variable is different for each region. For example

x 1∕ℓ x 2∕ℓ+1 2x 3∕h−1

ξA =

x 1∕ℓ+1 x 2∕ℓ+1 2x 3∕h−1

ξB =

These functions can be constructed systematically by establishing a global node numbering (each node in the global mesh has a distinct number) and establishing the correspondence between the local and global node numbers. The details of these transformations are covered in books on finite elements.

x3

ξ3

ξA

2

4

h∕2

B

h∕2

k

8

A ℓ

ℓ

1

7

5

ξ2

ξ1

x2 ℓ

3

ℓ x1

166. What is the appropriate definition of f in Eqn. (366) when there is a nonzero boundary displacement term u^ (x) in the Ritz approximation?

Substituting Eqn. (352) into the virtual-work functional G(u, u) =

  λ div u  div u  + m  ∇ u ⋅ ∇u   dV −  b ⋅ u dV − 

^

t ⋅ u dA

S

B

Ωt

B

we get, for the jth component of f fj =

 t ⋅ h dA +  b ⋅ h − λ div u  div h  + m  ∇ u ⋅ ∇h   dV ^

j

Ωt

j

B

o

j

S

o

j

172


where ∇S u ≡ ∇u+∇uT is the symmetric gradient of u.

Chapter 7 The Linear Theory of Beams

x2 167. The beam shown below has a rectangular cross x1 section of depth 4 and width 2, and has a length of 50 15 length units. It has a uniform mass density that gives rise to a constant body force of b(x) = −2 e 2 (force x units per length units cubed), and is subjected to a surface traction on its top surface that is bilinear with 50 x2 respect to x2 and x3 ≡ x reaching a maximum value of 15 (force units per length units squared), as shown. 4 x1 Find an expression for the applied tractions t Γ(x). Find the resultant applied loads q(x) and m(x) equivalent to the surface tractions and body force. Find the 2 distribution of resultant force Q(x) and resultant moment M(x) along the beam. Find the displacements w(x) and the rotations θ(x) along the beam.

(a) Find an expression for the applied tractions t Γ(x). The distribution of the tractions on the top surface is bilinear, i.e., at any fixed value of x1 the variation with x3 is linear and at any fixed value of x3 the variation with x1 is linear, as shown in the sketch. x2 x2 c1(x 1) x 3 c2(x 3)  1+x 1 

x3

x1

Since for any x1 the value of the function is zero at x3 = 0 it must have the form c1(x 1) x 3. Similarly, since for any x3 the value of the function is zero at x1 = −1 it must also have the form c2(x 3)  1+x 1 . Combining these two observations we conclude

174


that the distribution must have the form c x 3  1+x1 . At the point x1 = 1, x3 = 50 the function takes the value 15. Therefore, we must have c = 3∕20. The tractions act in the direction −e2. Therefore, the applied tractions are given by 3  − 20 1+x1  x3 e 2

t Γ(x) =

on Ω1 elsewhere

0

where Ω1 is the top surface of the beam ( x2 = 2). (b) Find the resultant applied loads q(x) and m(x) equivalent to the surface tractions and body force. The applied loading per unit length is q x  =

 b dA +  t ds Γ

Ω

Γ

 2

=

−2

1

−2 e 2 dx1 dx2 +

−1



1

3  − 20 x 1+x 1  e 2 dx1

−1

3  = −16 + 10 x e2

The applied moment per unit length is m x  =

 p × b dA +  p × t ds Γ

Ω

Γ

The position vector px1 , x 2 on the top surface is given by p = x 1 e1+2 e 2. Thus, m x  =

 x e +x e  × −2 e  dA +  1 1

2 2

2

Ω

 x 1 e 1+2 e 2  × t Γ dx 1

−1

  2

=

1

−2

1

−2 x 1 e 3 dx1 dx2 −

−1



1

3  2 x x 1+x1  e 3 20

dx 1

−1

1 = − 10 x e3

(c) Find the distribution of resultant force Q(x) and resultant moment M(x) along the beam. The resultant force Q(x) is related to the applied loading q(x) by the equation Q′ x+q x  = 0. Integrating with respect to x, we get 3 2 Q x = 16 x+ 20 x e2 + c

Since Q50 = 0 , then c = −1175 e 2. Thus, 3 2 x −1175 e2 Q x = 16 x + 20

Chapter 7

175

The Linear Theory of Beams

The resultant moment M(x) is related to the applied moment m(x) by M′ x + e 3 × Q x + m x  = 0 By plugging in the value of the resultant force Q(x) into the equation above and integrating it with respect to x, we get, 1 3 1 2 x e 1 + 20 x e3 + d M x = −1175 x−8 x2− 20

Since M 50 = 0 , then d = 32500 e 1 −125 e 3. Thus, 1 3 1 2 x −32500  e 1 +  20 x −125  e 3 M x = −1175 x−8 x2− 20

≡ M1 x  e1 + M 3 x  e3 (d) Find the displacements w(x) and the rotations θ(x) along the beam. The tensor S is a zero tensor since the origin is located at the centroid of the cross section. Therefore, the equations for the shear stress and moment resultants are uncoupled. To wit, Q = A Á o ⇒ Á o = A−1 Q M = I κ o ⇒ κ o = I −1 M The cross-sectional constants can be evaluated as A=

 dA = 8

I 11 =

 x d A = 323

I 12 =

Ω

I 22 =

 x d A = 83 2 1

Ω

2 2

Ω

 x x dA = 0 1 2

Ω

So, 1

Áo =

κo =

8m

0

0

0

0

1 8m

0

Q 2 (x )

0

0

8E

0

3 32 E

0

0

M 1 (x )

0

3 8E

0

0

0

0

3 40 m

M 3 (x )

1

3 2 = 1 −1175+16 x+ 20 x e2 8m

= 3 M 1 x e 1 + 3 M3 x  e3 32 E 40 m

It follows that, since θ′ x = κo  x , the rotation vector can be computed as









1 x 3 − 125x e θ(x) = 3 32500x − 1175 x 2 + 8 x 3 + 1 x4 e1 + 3 3 3 80 40m 60 2 32E

176


where the vector constant of integration vanishes because θ 0 = 0. By using the result for the rotation above, we can integrate the following equation w′ x = Áo  x  − e3 × θ x  =

81m Q x−θ x e 2

1

2

to get the displacement vector





w(x) = 1 −1175x + 8x2 + 1 x3 e 2 8m 20





− 3 16250x 2 − 1175 x 3 + 2 x 4 + 1 x5 e2 3 200 6 32E where the vector constant of integration vanishes because w 0 = 0.

168. Consider the beam with square cross section, of di- x 1 τo mension h by h and length ℓ. The beam has Young’s modulus C and shear modulus m. The beam is subjected to horix3 zontal tractions on its top face, as shown. The body forces acting on the beam are negligible. The coordinate axes ℓ τo shown are principal and centroidal. Find expressions for the applied force and moment per x3 unit length of beam, q(x) and m(x), where x = x3 is the axial coordinate. Find the displacement and rotation field caused by the loading by integrating the governing beam equations (that is, find the classical solution).

(a) Find expressions for the applied force and moment per unit length of beam, q(x) and m(x), where x = x3 is the axial coordinate. The traction vector can be written as, t Γ = τo e 3 which act only at the top surface. Therefore, the applied load is q=



h 2 h 2

τ o e 3 dx2 = τ o h e 3

−

and the applied moment is

m=



h 2

h − 2

 12 h e 1 +x2 e2  ×  τ o e 3  dx2 = − 12 τ o h 2 e2

(b) Find the displacement and rotation field caused by the loading by integrating the governing beam equations (that is, find the classical solution).

Chapter 7

177


x3 A = h2 h

Sα = 0

x2

1 4 I 11 = I 22 = 12 h

h The resultant force Q(x) is related to the applied loading q(x) by the differential equation Q′ x = −q  x. Integrating with respect to x, we get Q x = −τo h x e3 + c 1 Since Qℓ = 0 , we get c1 = τo h ℓ e3 . Thus, Q x = τo h ℓ−x e3 Since, e3 × Qx  = 0, the resultant moment M(x) is related to the applied moment m(x) by M′ x+m x  = 0. Integrating this with respect to x, we get M x = 12 τo h2 x e2 + c 2 Since Mℓ = 0 , we get c2 = − 12 τo h 2 ℓ e 2. Thus, M x = − 12 τo h 2  ℓ−x  e2 It follows that, since θ′(x) = I –1M(x), we can integrate to obtain the rotation field 3 τo θ x = − 2  2ℓ x−x2  e2 + c 3 Eh

Since θ0 = 0 , we get c3 = 0. Also since Áo = A−1 Q the displacement can be obtained by integrating the equation w′ x = A−1 Q − e 3 × θ x. Thus,





τ τ w x = − o 2 3ℓ x2−x 3 e1 + o ℓ x− 1 x 2 e3 + c4 2 Eh Eh Since w0 = 0, we have c4 = 0.

169. Consider a beam of length ℓ with square 2×2 cross section. The beam is subjected to the applied traction field over the cross section at the end of the beam t ℓ = τ x 1  1+x 22  e3

x2 1

1 1 1 x1

tℓ

x3

where τ is the known magnitude of the loading. Find the resultant force and moment acting on the end of the beam. Assume that u(x1, x2, 0), i.e., the beam is fixed at x3 = 0. Find the

178


resultant force field Q(x) and the resultant moment field M(x) that equilibrate the applied forces. Compute the displacement and rotation fields that result from the applied loads.

The resultant force at the right end can be computed as

 t dA =   1

Q(ℓ) =

1

τ x1  1+x 22  e 3 dx1 dx 2

ℓ

Ω

−1

=



−1

1

τ  1+x 22  e3 dx 2

−1

  x21 2

1

=0 −1

The resultant moment at the right end, noting that p = x 1 e1+x 2 e2 M(ℓ) =

 p × t dA ℓ

Ω

=  1

1

−1

−1



1

1

=

−1

 −τ x 21  1+x 22  e 2 + τ x1 x 2  1+x 22  e 1  dx1 dx 2 −τ x 21  1+x 22  e2 dx 1 dx2

−1

 

x3 =τ 1 3

1

−1

x3 x 2+ 2 3



1

−1

e 2 = − 16 τ e 2 9

b 170. The hollow box beam shown has a square cross x1 section of dimension b, thickness t ≪ b, and unit weight ρb . It is submerged in a fluid of unit weight ρo . t Recall that the pressure at any point in a static fluid is x2 proportional to the depth h according to the relationship p = hρo . The unit weight of the air inside the beam can ℓ x3 be taken as zero. The end is capped so that fluid cannot get inside. Plot the typical traction field tΓ acting on the lateral surface of the beam. Compute the resultant apx2 plied load q(x) and the resultant applied moment m(x) that would be appropriate in order to treat the problem using beam theory.

(a) Plot the typical traction field tΓ acting on the lateral surface of the beam.

Chapter 7

179


ρo b

ρo b

Sketch traction field here. (b) Compute the resultant applied load q(x) and the resultant applied moment m(x) that would be appropriate in order to treat the problem using beam theory. The traction, pressure, and body force fields are ds (4) (1)

(3)

−ρ o x2 e 1 (1) dx2 −ρ o b e2 (2) dx 1 ρ o x2 e 1 (3) dx2 0 (4) dx1

tΓ =

(2)

b e 1 + x2 e 2 (1) x1 e1 + b e 2 (2) x2e 2 (3) x1 e1 (4)

p=

b = ρ b t e2 all segments. The loading q is defined as q(x) =

 t ds +  ρ e t ds b 2

Γ

Γ

Ω

Using the quantities defined previously we have

 −ρ x e dx +  − ρ b e dx +  ρ x e dx + 4 b t ρ e b

q=

b

o 2 1

b

o

2

0

2

o 2 1

1

0

2

b 2

0

=  4 bt ρ b−ρ o b 2  e 2 The loading m is defined as m(x) =

 p × t ds +  Γ

Γ

p ×  ρ b e2 ) t ds

Ω

Using the quantities defined previously we have

  b e +x e ) ×  −ρ x e +t ρ e ) dx b

m=

1

o 2 1

2 2

b 2

2

0

  x e +b e  ×  −ρ b e +t ρ e  dx b

+

1 1

2

o

2

b 2

1

0

+



b

0

  x e  ×  t ρ e  dx b

 x 2 e 2  ×  −ρ o x 2 e 1+t ρ b b e 2  dx 2 +

1 1

0

b 2

1

180





 

3 3 2 m = ρ o b e 3 + ρ b t b2 e 3 + −ρ o b e 3 + ρ b t b e 3 3 3 2



 

3 2 + −ρ o b e3 + 0 + ρ b t b e3 3 2









= 2 ρb t b2 − 1 ρo b3 e 3 2

171. The Saint-Venant torsion problem is restricted to problems with constant rate of twist and traction-free lateral surfaces. One feature of this solution is that, at the fixed end, the rotation is restrained but the out-of-plane warping is not. Physically, such a boundary condition would be very difficult to realize. One solution to this problem is to create a model in which the amplitude of warping is independent from the rate of twist of the beam. Consider the deformation map u(x) = θ(x 3)  e3 × p  + φ(x3) ψ(x 1, x 2) e3 where θ(x3) is the angle of twist, φ(x3) is the amplitude of warping, ψ(x1, x2) is the warping function, and p = x 1e1+x 2 e2. Further, assume that the warping function ψ(x1, x2) is the Saint-Venant warping function derived in the text. Compute the strain tensor and use the linear elastic constitutive equations to show that the tractions on a cross section are Se3 = m φ∇ψ + θ′e3 × p  + Eφ′ψe3 where E ≡ λ+2m, and a prime denotes differentiation with respect to x3 . In addition to the polar moment of inertia J, define the cross-sectional properties J1 ≡

  p × ∇ψ ⋅ e dA, 3

J2 ≡

Ω

  ∇ψ ⋅ ∇ψ dA,

J3 ≡

Ω

 ψ dA 2

Ω

which can be computed once ψ(x1, x2) is known. Define the stress resultants T ≡ e3 ⋅

 p × Se dA, 3

W ≡ e3 ⋅

Ω

 ψ Se dA, 3

B≡

Ω

 ∇ψ ⋅ Se dA 3

Ω

where T is the usual torque. The stress resultants W and B are often called the bi-moment and the bi-shear. Substitute the expression for Se3 to show that T = mJ 1 φ + mJθ′,

W = EJ 3 φ′,

B = mJ2 φ + mJ 1 θ′

Show that if φ(x3) = θ′(x3) = α, then the above results are consistent with the Saint-Venant problem discussed in the text. Using the definitions of the stress resultants, compute T′ and W′ and show that T′ + t = 0,

W′ − B + w = 0

What are the appropriate definitions of the applied loads t(x) and w(x)? Substitute the resultant constitutive equations into the resultant equilibrium equations, and show that the equations

Chapter 7

181


mJθ′′ + mJ1φ′ + t = 0,

EJ 3 φ′′ − mJ2 φ − mJ 1 θ′ + w = 0

govern the spatial variation of rotation θ and warping φ. These equations constitute a pair of second-order ordinary differential equations, and, therefore, we can expect four constants of integration that must be found from boundary conditions. What are the boundary conditions for a free end and a fixed end?

(a) Compute the strain tensor and use the linear elastic constitutive equations to show that the tractions on a cross section are given by Se3 = m φ∇ψ + θ′e3 × p  + Eφ′ψe3 where E ≡ λ+2m, and a prime denotes differentiation with respect to x3 . The gradient of the displacement field is given by ∇u = u, i  ei where u, 1 = θ x3 e3 × e 2 + φ ψ, 1 e3 = θ x3 e2 + φ ψ, 1 e 3 u, 2 = −θ x3 e1 + φ ψ, 2 e3 u, 3 = θ′ x3 e3 × p + φ′ ψ e3 Thus, the strain tensor can be expressed as E = 12 ∇u+∇uT = 12 φ ψ, 1 e 1  e3+e 3  e1+φ ψ, 2 e 2  e3+e 3  e2 + θ′ e3 × p  e 3+e3  e 3 × p+2 φ′ ψ e 3  e3  Noting that, from the equation above, trE = φ′ ψ, we have (see text) Se3 = λ φ′ψ e 3 + 2m Ee3 where the term Ee 3 can be computed as follows Ee3 =

1 2

 φ ψ,1 e1+ψ, 2 e 2 + θ′ e3 × p + θ′ e3 × p ⋅ e3 e 3 + 2 φ′ ψ e 3 

=

1 2

 φ ∇ψ + θ′ e3 × p + 2 φ′ ψ e3 

since e3 × p ⋅ e3 = 0. By plugging this result into the relationship above, we get Se3 =  λ+2m  φ′ψ e3 + m φ ∇ψ + θ′ e3 × p = Eφ′ψ e3 + m φ ∇ψ + θ′ e3 × p (b) In addition to the polar moment of inertia J, define the cross-sectional properties J1 ≡

  p × ∇ψ ⋅ e dA 3

Ω

J2 ≡

  ∇ψ ⋅ ∇ψ dA Ω

J3 ≡

 ψ dA 2

Ω

182


which can be computed once ψ(x1, x2) is known. Define the stress resultants T ≡ e3 ⋅

 p × Se dA

W ≡ e3 ⋅

3

Ω

 ψ Se dA 3

B≡

Ω

 ∇ψ ⋅ Se dA 3

Ω

where T is the usual torque. The stress resultants W and B are often called the bi-moment and the bi-shear. Substitute the expression for Se3 to show that W = EJ 3 φ′

T = mJ 1 φ + mJθ′

B = mJ2 φ + mJ1 θ′

Show that if φ(x3) = θ′(x3) = α, then the above results are consistent with the SaintVenant problem discussed in the text. Using the result obtained in (a), we get T ≡ e3 ⋅

 p × Se dA 3

Ω

= e3 ⋅

  Eφ′ψ p × e  + m φ p × ∇ψ + m θ′ p × e × p  dA 3

3

Ω

= m φ J 1 + m θ′ J where we have made use of the following e 3 ⋅ p × e 3  = 0

J = e3 ⋅

 p × e × p dA 3

Ω

Similarly, since e3 ⋅ p × e3 = 0 and e3 ⋅ e3 = 1, W ≡ e3 ⋅

 ψ Se dA 3

Ω

= e3 ⋅

  Eφ′ψ e + m φ ψ ∇ψ + m ψ θ′ e × p  dA 2

3

3

Ω

= EJ3 φ′ And finally, since e3 ⋅ ∇ψ = 0, we have B≡

 ∇ψ ⋅ Se dA 3

Ω

=

  Eφ′ψ ∇ψ ⋅ e + m φ ∇ψ ⋅ ∇ψ + m ψ θ′ ∇ψ ⋅ e × p  dA 3

Ω

= m J2 φ + m J1 θ′

3

Chapter 7

183


If φ(x3) = θ′(x3) = α, then θ = α x 3 + c = α x 3 , since θ(0) = 0. If we plug this result into the given expression for the displacement field we get u(x) = α x 3  e 3 × p  + α ψ(x1, x2) e3 which gives the same results as the Saint-Venant problem discussed in the text. (c) Using the definition of the stress resultants, compute T′ and W′ and show that T′ + t = 0 and W′ − B + w = 0. What are the appropriate definitions of the applied loads t(x) and w(x)? Using the definition of the stress resultant T, we can obtain its derivative with respect to the x3 coordinate as follows,

 p × ∂x∂S e dA = e ⋅  p × div S− ∂x∂S e  dA

T′ = e 3 ⋅

3

Ω

3

3

α

Ω

α

where α = 1, 2. Then, T′ = e 3 ⋅

 p × div S dA − e ⋅  p × t ds − e ⋅ e × Q 3

= e3 ⋅

3

Γ

Ω

3

Γ

 p × div S dA − e ⋅  p × t ds 3

Γ

Ω

Γ

since e3 ⋅ e 3 × Q = 0. Let us define t (applied torque per unit length) as t ≡ e3 ⋅ m = e3 ⋅

 p × b dA + e ⋅  p × t ds 3

Ω

Γ

Γ

Adding this term to T′, we get the equation T′ + t = 0 Similarly, by using the definition of the stress resultant W, we can obtain its derivative with respect to the x3 coordinate as follows W′ = e 3 ⋅

 ψ ∂x∂S e dA = e ⋅  ψ div S− ∂x∂S e  dA 3

3

Ω

3

α

Ω

α

where α = 1, 2. Then, from equilibrium (i.e., div S+b = 0), we have W′ = −e 3 ⋅

 ψ  ∂x∂S e + b  dA α

Ω

= −e3 ⋅

 ∂x∂ ψ S  e + ψ b − ∂x∂ψ S e  dA Ω

Observe that

α

α

α

α

α

184

Fundamentals of Structural Mechanics e3 ⋅

 ∂x∂ψ S e dA = e ⋅  ∇ψ ⋅ S dA = B α

α

Ω

3

Ω

Thus, we get W′ = −e 3 ⋅

 ψ b dA − e ⋅  ∂x∂ ψ S  e dA + B 3

Ω

= −e3 ⋅

 ψ b dA − e ⋅  ψ S n ds + B 3

Ω

= −e3 ⋅

α

α

Ω

Γ

Γ

 ψ b dA − e ⋅  ψ t ds + B 3

Ω

Γ

Γ

At this point, let us define w (warping force per unit length) as

 ψ b dA + e ⋅  ψ t ds

w ≡ e3 ⋅

3

Ω

Γ

Γ

Adding this term to W′, we get the equation W′ − B + w = 0 (d) Substitute the resultant constitutive equations into the resultant equilibrium equations, and show that the equations mJθ′′ + mJ1φ′ + t = 0 EJ3 φ′′ − mJ2 φ − mJ 1 θ′ + w = 0 govern the spatial variation of rotation θ and warping φ. These equations constitute a pair of second-order ordinary differential equations, and, therefore, we can expect four constants of integration that must be found from boundary conditions. What are the boundary conditions for a free end and a fixed end? Using the results we have obtained in (b), we have the constitutive equations T′ =  m φ ′ J 1 +  m θ′ ′ J = m φ′ J 1 + m θ′′ J W′ = EJ 3 φ′′ Plugging these results into the equations T′+t = 0 and W′−B+w = 0, obtained in part (c), we get the desired result. At a free end the tractions are prescribed. At a fixed end displacements are prescribed. Let us illustrate the boundary conditions for a free end and a fixed end on the beam in the figure. The essential boundary conditions (i.e., no twist and no warping at the fixed end) are given by,

x ℓ

Chapter 7

185


θ(0) = 0

φ(0) = 0

The boundary conditions at the free end require some care. To be able to tell what has to be prescribed at a free end, we shall resort to the principle of virtual work. The internal virtual work is WI =

 S ⋅ ∇u dV B

where, in the present case ∇u = θ′ e 3 × p  e3 + φ′ ψ e3  e 3 + θ e3 × p, α  eα + φ ψ, α e3  e α Then, since S ⋅ ∇u = tr S ∇u = ei ⋅ S T ∇u e i = ei ⋅ S ∇u ei (since S is symmetric), it follows that WI =

  θ′ e ⋅ Se × p + φ′ ψ e ⋅ Se 3

3

3

3

B

+ θ e α ⋅ S e 3 × p, α + φ ψ, α e α ⋅ Se3  dV Note that ψ, α e α = ∇ψ and e3 ⋅ Se 3 × p = e 3 ⋅ p × Se3. From balance of angular momentum we know that ei × Se i = 0. Also, a ⋅ a × b = 0 for any pair of vectors a and b. Therefore, since p, α = eα we have eα ⋅ Se 3 × p, α = eα ⋅ Se 3 × eα = e3 ⋅ e α × Seα = e3 ⋅ −e 3 × Se3 = 0 Using the definitions of stress resultants T, W, and B, and integrating by parts we get

 θ′ T + φ′ W + φ B  dx ℓ

WI =

0

 −θ T′ + φ W′− B   dx + T θ  ℓ

=

0

 θt + φw dx + T θ  ℓ

=

0

ℓ 0

+ Wφ



ℓ

+ Wφ

0



ℓ 0

ℓ 0

where we have recognized that T′ = −t and W′− B = −w. The end resultant force resultants can be defined as to ≡ e3 ⋅



t o dA



ψ t o dA

tℓ ≡ e3 ⋅

Ωo

wo ≡ e 3 ⋅

Ωo



t ℓ dA

Ω

wℓ ≡ e 3 ⋅



Ω

ℓ

ψ t ℓ dA ℓ

186


Thus, the external work is given by

 θt + φw dx + t θ (ℓ) + t θ (0) + w φ (ℓ) + w φ (0) ℓ

WE =

ℓ

o

ℓ

o

0

Finally, the virtual work functional G(θ, φ, θ , φ ) = W I − W E can be stated as

 θ T′ + t  + φ W′− B + w  dx ℓ

G(θ, φ, θ , φ ) =

0

+  T(ℓ)−t ℓ  θ (ℓ) −  T(0)+t o  θ (0) +  W(ℓ)−wℓ  φ (ℓ) −  W(0)+wo  φ (0) We can observe from the above variational statement that, in the sense of virtual work, T is conjugate to θ and W is conjugate to φ. Thus, the appropriate boundary conditions at a free end are W(ℓ) = EJ 3 φ′(ℓ) = 0 T(ℓ) = m φ(ℓ) J1 + m θ′(ℓ) J = 0

172. The method of initial parameters integrates the governing equations and substitutes the values at x = 0 to give the general form of the displacement function. For the Bernoulli-Euler beam, the transverse deflection can be computed as M o x 2 Qo x 3 − + 1 wBE(x) = wo + θ o x + EI 2EI 6EI

  (x−ξ) q(ξ) − x

1 6

3

1 (x−ξ) 2m(ξ) 2

 dξ

0

where wo = w(0), θ o = w′(0), M o = M(0), Qo = Q(0) are the initial parameters. Verify that the expression satisfies the governing differential equations of Bernoulli-Euler beam theory. This equation is particularly useful for those cases where M o and Qo can be determined from overall equilibrium. Use the method of initial parameters to solve the problem of the cantilever beam under uniform load given as an example in the text.

(a) Verify that the expression satisfies the governing differential equations of Bernoulli-Euler beam theory. The Leibnitz Rule of the calculus affords us a method for differentiating an integral when the limits of integration are functions of the variable of differentiation. In general, we have d dx



b(x)

a(x)

f  ξ, x  dξ =



b(x)

a(x)

df dξ + f  b(x), x  db − f  a(x), x  da dx dx dx

Chapter 7

187


If we specialize this to the case where a(x) = 0 and b(x) = x we get, d dx Let f (ξ, x) =

 f ξ, x  dξ =  dxdf dξ + f x, x  x

x

0

0

1 (x−ξ) 3q(ξ) − 12 (x−ξ) 2 m(ξ). 6

M o x Qo x 2 dwBE = θo + − + 1 EI EI dx 2EI because f (x, x) = 0. Now let f (ξ, x) = the Leibnitz rule we get EI

Applying the Leibnitz rule we get

  (x−ξ) q(ξ)−(x−ξ) m(ξ) dξ x

1 2

2

0

1 (x−ξ) 2q(ξ) − (x−ξ) m(ξ). 2

d2w BE = Mo − Qo x + dx 2

Again, applying

 (x−ξ) q(ξ) − m(ξ) dξ x

0

where, again f (x, x) = 0. Let f (ξ, x) = (x−ξ) q(ξ) − m(ξ). By the Leibnitz rule we get



2 d EI d wBE dx dx2

 = −Q

 q(ξ) dξ − m(x) x

o

+

0

And finally, letting f (ξ, x) = q(ξ) we get



d 2 EI d2w BE dx2 dx 2

 = q(x) − m′(x)

which is the differential equation for Bernoulli-Euler beam theory. (b) This equation is particularly useful for those cases where M o and Qo can be determined from overall equilibrium. Use the method of initial parameters to solve the problem of the cantilever beam under uniform load given as an example in the text. For the cantilever beam with length ℓ, loaded with a uniform downward acting load q, we have, from static equilibrium of the entire beam: M o = − 12 qℓ2 and Qo = −qℓ. The boundary conditions give wo = 0 and θ o = 0. Plugging these values into the expression for the displacement, we get, EI w (x) = − 1 ℓ 2x2 + 1 ℓ x3 − q BE 4 6



x

1 (x−ξ) 3 dξ 6

0

=

1 −4

1 6

1 4 ℓ2x 2 + ℓx 3 − 24 x

Letting ξ ≡ x∕ℓ, we get wBE(x) =

qℓ 4  −ξ 4+4 ξ3−6 ξ 2  24EI

188


173. The method of initial parameters can be applied to the Timoshenko beam to give Qo x − 1 wT(x) = wBE(x) + GA GA

 (x−ξ) q(ξ) dξ x

0

where wT(x) is the deflection according to Timoshenko beam theory and wBE(x) is the deflection according to Bernoulli-Euler beam theory (as given in Problem 172). Verify that the expression satisfies the governing differential equations of Timoshenko beam theory. Use the method of initial parameters to solve the problem of the cantilever beam under uniform load given as an example in the text.

(a) Verify that the expression satisfies the governing differential equations of Timoshenko beam theory. Differentiating the second of Eqns. (407) twice allows us to solve for θ′′′ in terms of w as θ′′′ = w iv +

q′′ GA

Substituting into the first of Eqns. (408) gives the alternative form of the Timoshenko beam equations EI w iv + EI q′′ − q + m′ = 0 GA The displacement wT(x) must satisfy this fourth-order differential equation in order to be consistent with the Timoshenko beam theory. Using Leibnitz rule with f (ξ, x) = (x−ξ) q(ξ) we get d wT d wBE Qo = + – 1 dx dx GA GA

 q(ξ) dξ x

0

Similarly, with f (ξ, x) = q(ξ), we can differentiate again to give d 2 wBE q(x) d 2 wT = – 2 GA dx dx 2 Differentiating this result twice and multiplying the result by EI we get EI w ivT = EI wivBE − EI q′′ GA Substituting the expression for the wBE we get the desired result EI w ivT + EI q′′ − q + m′ = 0 GA (b) Use the method of initial parameters to solve the problem of the cantilever beam under uniform load given as an example in the text. For the cantilever beam with length ℓ, loaded with a uniform downward acting load q, we have, from static equilibrium of the entire beam: M o = − 12 qℓ2 and Qo = −qℓ. The boundary conditions give

Chapter 7

189


wo = 0 and θ o = 0. Plugging these values into the expression for the displacement, we get EI w (x) = − 1 ℓ 2x2 + 1 ℓx3 − 1 x4 − q T 4 6 24

β ℓ  (x−ξ) dξ x

1 12

3

βℓ x +

1 12

2

0

where β ≡ 12 EI∕GAℓ 2. Let ξ ≡ x∕ℓ and multiply through by q∕EI to get wT(x) =

qℓ 4  −ξ 4 + 4 ξ3 − 6−β ξ2 − 2 β ξ  24EI

174. The three-dimensional rotation tensor Λ can be expressed in terms of three parameters e1 , e2 , and e3 as e 20 + e 21 − e 22 − e 23

Λ(e1, e2, e3) =

2(e 1 e 2 − e 0 e 3) 2(e 1 e 3 + e 0 e 2)

2(e 1 e 2 + e 0 e 3) e 20

− e 21

+ e 22

− e 23

2(e 2 e 3 − e 0 e 1)

2(e 1 e 3 − e 0 e 2) 2(e 2 e 3 + e 0 e 1) e 20 − e 21 − e 22 + e 23

where the parameter e0 has been introduced for convenience. This fourth parameter does not represent an independent parameter, but rather satisfies the constraint equation e20+e 21+e22+e 23 = 1. These parameters are called the Euler parameters. Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT. Show that for small values of the parameters e1 , e2 , and e3 the tensor can be expressed in the form Λ ≈ I+W, where I is the identity and W is a skew-symmetric tensor. Show, therefore, that when the three parameters are small, they can be viewed as the components of the rotation vector θ, with θ i = 2ei, and that W = θ ×. (a) Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT. If Λ−1 = ΛT, then ΛT Λ = I must hold. The tensor ΛT is given by

Λ(e1, e2, e3) ~

e20 + e 21 − e 22 − e 23

2(e1 e2 − e0 e 3)

2(e1 e2 + e0 e 3)

e20 − e 21 + e 22 − e 23

2(e1 e3 − e0 e 2)

2(e2 e3 + e0 e 1)

2(e1 e3 + e0 e 2) 2(e2 e3 − e0 e 1) e20

− e 21 − e 22 + e 23

T

Since ΛT Λ = Λ T Λ, then ΛT Λ is a symmetric tensor. Therefore, we need only compute the upper (or lower) triangle and diagonal entries of this tensor. It follows that,

ΛT Λ = (e 20+e21−e 22−e23 ) 2 + 4(e 1 e2−e 0 e3) 2 + 4(e1 e3+e 0 e2) 2 11 = e40 + e 41 + e42 + e 43 + 2 (e 20 e21 − e 20 e 22 − e20 e 23 − e21 e 22 − e21 e23 + e 22 e23 ) + 4 (e21 e 22+e20 e 23 ) − 8 e 0 e1 e2 e 3 + 4 (e21 e22+e 20 e22 ) + 8 e0 e 1 e2 e3 = (e20+e 21+e22+e 23 ) 2 = 1

190


Likewise, ΛT Λ22 = Λ T Λ 33 = 1. The (1,2) term of ΛT Λ can be computed as

ΛT Λ = 2 (e20+e 21−e22−e 23 ) (e1 e2+e 0 e3 ) 12 + 2 (e20−e 21+e22−e 23 ) (e 1 e2−e 0 e 3 ) + 4 (e1 e3+e 0 e2 ) (e2 e 3−e0 e 1 ) = 4 e 20 e1 e 2+4 e 21 e0 e 3−4 e 22 e0 e 3−4 e23 e1 e 2 + 4 e 23 e1 e 2 − 4 e 20 e1 e 2 − 4 e 21 e0 e 3 + 4 e 22 e0 e 3 = 0 Likewise, ΛT Λ13 = Λ T Λ 23 = 0. Therefore, ΛT Λ = I holds. (b) Show that for small values of the parameters e1 , e2 , and e3 the tensor can be expressed in the form Λ ≈ I + W, where I is the identity and W is a skew-symmetric tensor. Show, therefore, that when the three parameters are small, they can be viewed as the components of the rotation vector θ, with θ i = 2ei, and that W = θ ×. ^

Let us define W ≡ Λ − I. Then −2 (e 22 + e 23 ) ^

W~

2(e1 e2 − e0 e 3) 2(e1 e3 + e0 e 2)

2(e1 e2 + e0 e 3) −2 (e 21

2(e1 e3 − e0 e 2)

e 23 )

2(e2 e3 + e0 e 1)

2(e2 e3 − e0 e 1)

−2 (e 21 + e 22 )

+

For small values of the parameters e1 , e2 and e3 , the squared terms above are very small. Since e20 + e 21 + e22 + e 23 = 1 then e20 ≈ 1 or e0 ≈  1. Furthermore the terms e1 e2 , e1 e3 and e2 e3 are very small. If we neglect the higher order terms and take e0 < 0, we get ^

W≈W~

0 2 e3 −2 e 2

2 e2 −2 e 1

−2 e 3 0 2 e1

0

Then, with θ i ≡ 2ei we get ^

W≈W~

0 θ3 − θ2

θ2 −θ1

−θ3 0 θ1

0

Thus, W = θ ×. Note that, if we had assumed e0 > 0 then, W = −θ ×.

175. Reconsider the cantilever beam of length ℓ, fixed at x = 0, with bending modulus EI and shear modulus GA solved as an example in the text (Fig. 96). Examine the results of the Ritz method as you increase the number of basis functions taken from the sets





2 3 N hi (x) ∈ x, x , x 2 , . . ., xN−1 , ℓ ℓ ℓ



2 3 M gi (x) ∈ x , x 2 , x 3 , . . ., x M ℓ ℓ ℓ ℓ



Chapter 7

191


In particular, find general expressions for the ijth components of K aa , K ab , K ba, K bb , and the ith components of f a and f b when hi (x) = xi∕ℓ i−1 and gi (x) = xi∕ℓ i. Solve the problem for (N, M) = (2, 2), (3, 2), (3, 3), (4, 3), (4, 4). What do you expect to happen for higher-order approximations? Comment on the differences in the solutions obtained when M = N−1 versus those obtained with M = N.

(a) Find general expressions for the ijth components of K aa, K ab, K ba, K bb, and the ith components of f a and f b when hi(x) = xi∕ℓ i−1 and gi(x) = xi∕ℓ i. The essential boundary conditions w(0) = 0 and θ(0) = 0 are satisfied if the displacement and rotation are approximated as N

w(x) =

M

a h , i

θ(x) =

i

i=1

b g i

i

i=1

Let ξ ≡ x∕ℓ. The base functions and their derivatives can be expressed as hi(ξ) = ξiℓ

gi(x) = ξi

hi′(ξ) = i ξi−1

gi′(ξ) = i ξi−1∕ℓ

It follows that ij  GAℓ h ′ h ′ dξ = GAℓ i+j−1 1

 K aa  = ij

i

j

0

 −GAℓ h ′ g dξ = −GAℓ i+ji =  K 1

 K ab  = ij

i

j

ba



ji

0

  EIℓ g ′ g ′+GAℓ g g  dx = GAℓ  1

 K bb  = ij

i

j

i

j

0

12

ijβ 12 + i+j+1 i+j−1



where β ≡ 12 EI ∕ GAℓ2. The right side vectors have components

 q(ξ) h ℓ dξ = −qℓ i+11 1

f a = i

i

2

0

and f b = 0. (b) Solve the problem for (N, M) = (2, 2), (3, 2), (3, 3), (4, 3), (4, 4). Solving the system of equations K aa a + K ab b = f a K ba a + K bb b = f b where the coefficient matrices are computed from part (a) yields, the results presented in the following table,

192


(N, M) →

( 2, 2 )

( 3, 2 )

( 3, 3 )

( 4, 3 )

13 + 70 β 6 + 70 β

−1

15 + 210 β 3 β + 35 β2

–6 β

15 + 165 β + 35 β 2 6 + 70 β2

β−6 2β

a1

−

7 + 10 β 2 + 10 β

−1

b1

−

2 + 25 β β + 5 β2

−5 β

a2

β−2 2β

β−5 2β

b2

15 1 + 5β

3 β

9 + 210 β 3 β + 35 β2

6 β

a3

h

1 β

1 β

2 β

b3

h

h

a4

h

h

− − −

−

70 3 + 35 β

−2 β

h

− 1 2β

where all the coefficients in the table are normalized by qℓ ∕ GA. The results for the case (4, 4) are the same as the case (4, 3) with b4 = 0. These results were obtained using MATHEMATICAT and the commands needed to solve the problem are given below n = 4 m = 4 hi = xî / L^(i - 1) hj = x^j / L^(j - 1) gi = xî / Lî gj = x^j / L^j kaa = G*A*D[hi, x]*D[hj, x] kab = -G*A*gj*D[hi, x] kbb = G*A*(Beta*L^2/12*D[gi, x]*D[gj, x] + gi*gj) fa = (-q)*hi fb = 0*i Kaa = Table[Integrate[kaa, {x, 0, L}], {i, 1, n}, {j, 1, n}] Kab = Table[Integrate[kab, {x, 0, L}], {i, 1, n}, {j, 1, m}] Kbb = Table[Integrate[kbb, {x, 0, L}], {i, 1, m}, {j, 1, m}] Kba = Transpose[Kab] fa = Table[Integrate[fa, {x, 0, L}], {i, 1, n}] fb = Table[Integrate[fb, {x, 0, L}], {i, 1, m}] a = Inverse[Kaa - Kab . Inverse[Kbb] . Kba] a = a . (fa - Kab . Inverse[Kbb] . fb) b = Inverse[Kbb] . fb - Kba . a FullSimplify[a]//MatrixForm FullSimplify[b]//MatrixForm hvec = Table[xî/L^(i-1), {i,1,n}] gvec = Table[x^j/L^j , {j,1,m}]

Chapter 7


193

w = Simplify[a . hvec] theta = Simplify[b . gvec]

(c) What do you expect to happen for higher-order approximations? Since the solution for the case (4, 3) corresponds to the exact solution, higher-order approximations will yield the same results as the case (4, 3). (d) Comment on the differences in the solutions obtained when M = N−1 versus those obtained with M = N. The cases with the rotation interpolated one polynomial order less than the displacement give results with the rotation independent of the shear stiffness of the beam, in accord with the exact solution, even for low-order approximations.

176. Prove that S ⋅ W = 0, or SijW ij = 0, when S is a symmetric tensor and W is a skewsymmetric tensor. Note that S ⋅ W = e i ⋅ SWei. Since S is a symmetric tensor, ST = S, or Sij = Sji . Also, since W is a skew-symmetric tensor, WT = −W, or W ij = −W ji . It follows that Sij W ij = S ji W ij = −Sji W ji = −S ij W ij where the first step uses the symmetry of S, the second step uses the skew-symmetry of W, and the third step is just a change of labels. As a consequence we have Sij W ij +S ij W ij = 2 Sij W ij = 0

⇒

Sij W ij = 0

An alternative proof is given below. For any tensor A, the difference between the tensor and its transpose A−AT is always skew-symmetric. The trace of a skew-symmetric tensor is always zero. Therefore, noting that S ⋅ W = tr SW , we have trSW− SW T  = 0 Observing that  AB  T = B T AT, that tr A+B  = tr A+tr B, and noting that W is skew-symmetric, we find that tr SW  + tr WS  = 0 Since tr  AB  = tr  BA  for and tensors A and B, we have, finally, 2 trSW = 0, thereby proving the result.

177. The prismatic beam shown has a cross section that is symmetric with respect to the plane of the page. The cross section has axial modulus EA, shear modulus GA, ℓ and flexural modulus EI. The beam is subjected to a unix form transverse load q(x) = −q o. Find the displacements and rotations for the beam by directly integrating the governing equations.

qo

194


For the Timoshenko beam, EI θ′′′ = q(x) − m′ (x). Since q(x) = −q o and there are no applied moments ( m(x) = 0) we can integrate the governing equation by direct quadrature qo EI qo θ′′ = − x + a 0 EI qo 2 θ′ = − x + a0 x + a1 2EI θ′′′ = −

Since M(x) = EI θ′ we can enforce the boundary conditions on the simple ends to determine the two constants of integration obtained thus far. To wit, M(0) = 0

a1 = 0

⇒

M(ℓ) = 0

⇒

a0 =

qo ℓ 2EI

Substituting these values and integrating again we get θ =

qo − 16 x3 + 14 ℓx2 + a2 EI

Also, for the Timoshenko beam, GAw′ = GAθ − EI θ′′ − m . Substituting the results we have obtained above for the rotation θ and integrating again we get w=

qo − 241 x4 + 121 ℓx3 + a2 x + qo  x2−ℓ x  + a3 EI 2GA

The displacement boundary conditions at the two ends give w(0) = 0

⇒

a3 = 0

w(ℓ) = 0

⇒

a2 = −

qo ℓ3 24 EI

Letting ξ ≡ x∕ℓ we get, after some simplification, w= θ =

qo ℓ 4  4 qo  2 −ξ +2 ξ3−1  + ξ −ξ  24EI 2GA qo ℓ 3  −4 ξ 3+6 ξ2−1  24EI

178. Resolve Problem 177 after making the Bernoulli-Euler assumption that w′ = θ (i.e., there is no shear deformation). What is the difference between the two solutions?

For the Bernoulli--Euler beam,



 = q(x) − m′(x)

d 2 EI d2w dx2 dx2

Chapter 7


195

With q(x) = −q o and m(x) = 0, we can integrate the differential twice to get M(x) = EI w′′ = − 12 q o x 2 + a0 x + a 1 The force boundary conditions at the simple ends give M(0) = 0

⇒

a1 = 0

M(ℓ) = 0

⇒

a 0 = 12 q o ℓ

Substituting these values and integrating twice more gives w=

qo − 241 x4 + 121 ℓx3 + a2 x + a 3 EI

The displacement boundary conditions at the ends give w(0) = 0

⇒

a3 = 0

w(ℓ) = 0

⇒

a2 = −

qo ℓ3 24 EI

Letting ξ ≡ x∕ℓ we get, with some rearrangement w=

qo ℓ 4 −ξ4+2 ξ3−1 24EI

w′ = θ =

qo ℓ3  −4 ξ 3+6 ξ2−1 24EI

Note that, the results we have here are identical to the ones we got for Problem 177 for the extreme case when GA → ∞. In other words, for a beam with a very large shear stiffness, the response of a Timoshenko beam and the Bernoulli-Euler beam are nearly identical.

179. Use a polynomial basis to find an approximate solution to Problem 177 using the principle of virtual work for the beam including shear deformation. Which terms should you include? What order approximation is adequate?

The virtual work functional is G(w, θ, w, θ) =   EIθ′θ′ + GA(w′−θ)(w′−θ) − qw  dx = 0 ℓ

0

The essential boundary conditions w(0) = w(ℓ) = 0 are satisfied by the approximate displacement field

196


w(x) =

M

a h , i

θ(x) =

i

i=1

b g i

i

i=1

when the trial functions are given by hi(ξ) = ξℓ 1−ξ hi′(ξ) = 1−ξ

i

i−1

gi(ξ) = ξi−1

1−1+i ξ

gi′(ξ) = ( i−1 ) ξ i−2∕ℓ

where ξ ≡ x∕ℓ. The coefficient matrices have the components

 GAℓ h ′ h ′ dξ

 K ab  = − ij

  EIℓ g ′ g ′+GAℓ g g  dξ

f a = i

1

 K aa  = ij

i

j

0

i

i

j

0

 q(ξ) h ℓ dξ

1

 K bb  = ij

 GAℓ h ′ g dξ 1

j

i

j

0

1

i

0

Carrying out the indicated integrations yields 1 3

K aa = GA ℓ

1 6

1 10

2 15

1 10

symm.

K ab = −GA ℓ

3 25

1 2

1

β+4 12

K bb = GA ℓ symm.

f

= −q ℓ

2

1 12 1 20 1 30

1 6

1 6

3 20

0

1 12

1 15

1 20

0

1 20

1 30

3 140

1 3

1 4

β+3 12

5 β+12 60

5 β+9 45

3 β+4 24 21 β+20 140

1 6

a

0

0 0

f b =

0 0

where the portion of the matrices for N = 1, M = 2 are shaded. Note that, by symmetry, we have K ba = K ab T and the parameter β ≡ 12 EI∕GA ℓ2. The results for various cases are presented in the table below.

Chapter 7 (N, M) → a1

( 1, 2 )

−

1+β 2β

197

The Linear Theory of Beams ( 2, 3 )

−

1+β 2β

( 3, 3 )

−

1+β 2β

( 3, 4 )

−

1+β 2β

a2

h

0

0

− 1 2β

a3

h

h

0

1 2β

b1

− 1 2β

− 1 2β

− 1 2β

−1 β

b2

1 β

1 β

1 β

0

b3

h

0

0

3 β

b4

h

h

h

−2 β

where all the coefficients in the table are normalized by qoℓ ∕ GA. The result for the case (3, 4) corresponds to the exact solution which is given by q oℓ 4  4 −ξ + 2 ξ 3 − β ξ2 + 1+β ξ  24EI q ℓ3 θ(x) = o −4 ξ3 + 6 ξ 2 − 1  24EI

w(x) =

where ξ ≡ x∕ℓ. These results were obtained using MATHEMATICAT with the following commands n = 3 m = 4 hi = x*(1 - x/Lî) hj = x*(1 - x/L^j) gi = x/L^(i - 1) gj = x/L^(j - 1) kaa = G*A*D[hi, x]*D[hj, x] kab = -G*A*gj*D[hi, x] kbb = G*A*(Beta*L^2/12*D[gi, x]*D[gj, x] + gi*gj) fa = -q*hi fb = 0*i Kaa = Table[Integrate[kaa, {x, 0, L}], {i, 1, n}, {j, 1, n}]) Kab = Table[Integrate[kab, {x, 0, L}], {i, 1, n}, {j, 1, m}]) Kbb = Table[Integrate[kbb, {x, 0, L}], {i, 1, m}, {j, 1, m}]) Kba = Transpose[Kab] fa = Table[Integrate[fa, {x, 0, L}], {i, 1, n}] fb = Table[Integrate[fb, {x, 0, L}], {i, 1, m}] a = Inverse[Kaa - Kab . Inverse[Kbb] . Kba] a = a . (fa - Kab . Inverse[Kbb] . fb b = Inverse[Kbb] . (fb - Kba . a) FullSimplify[a]//MatrixForm FullSimplify[b]//MatrixForm

198

Fundamentals of Structural Mechanics hvec = Table[x*(1-x/L)î, {i,1,n}] gvec = Table[(x/L)^(j-1), {j,1,m}] w = FullSimplify[a . hvec] theta = FullSimplify[b . gvec]

180. Repeat the virtual-work computation in Problem 179 for the Bernoulli-Euler beam.

The virtual work functional is

  EIw′′w′′ − qw  dx = 0 ℓ

G(w, w) =

0

The essential boundary conditions, w(0) = w(ℓ) = 0, are satisfied for the approximate displacement field N

w(x) =

a h i

i

i=1

when the base functions are given by i hi(ξ) = ξℓ 1−ξ i−1 hi′(ξ) = 1−ξ 1−1+i ξ



i−2 i−1 hi′′(ξ) = 1  i−1  1−ξ 1−1+i ξ − (1+i )1−ξ ℓ



where ξ ≡ x∕ℓ. The stiffness matrix and the force vector are given by

 EIℓ h ′′ h ′′ dξ , 1

K = ij

i

j

0

 qℓ h dξ 1

f = − i

i

0

Thus, we get the system of equations (the shading is to indicate the appropriate parts for N=1 (darkest), N=2, and N=3 (lightest)) 4

K = EI ℓ symm.

2

2

4

4 24 5

1 6

f = −q ℓ 2

1 12 1 20

with which we solve for the vector of unknown coefficients a from the system of equations Ka=f. For the case N=1, we get

Chapter 7 a1 = −

199

The Linear Theory of Beams q o ℓ3 qo ℓ 4 2 ⇒ w= ( ξ −ξ) 24 EI 24 EI

For the case N=2, we get the same result as the case N=1 (i.e., a2 = 0). For the case N=3, the coefficients are a1 = −

q o ℓ3 q o ℓ3 qo ℓ 3 , a2 = − , a3 = 24 EI 24 EI 24 EI

and the displacement is given by w=

q o ℓ4  −ξ 4 + 2 ξ3 − ξ  24 EI

which corresponds to the exact solution. These results were obtained using MATHEMATICAT and the commands needed to solve the problem are given below n=3 hi = x*(1 - x/L)î hj = x*(1 - x/L)^j kx = EI*D[hi, {x,2}]*D[hj, {x,2}] fx = -q*hi K = Table[Integrate[kx, {x, 0,L}], {i,1,n}, {j,1,n}] f = Table[Integrate[fx, {x,0,L}], {i,1,n}] a = Inverse[K].f FullSimplify[a]//MatrixForm hvec = Table[x*(1-x/L)î, {i,1,n}] w = FullSimplify[a.hvec]

181. Carry out the integrations in Eqn. (385) to show that the tensors given in Eqn. (387) result.

Throughout the computations we shall assume that the material constants λ and m are constant throughout the cross section. We shall also use the following definitions A≡

 dA

Sα ≡

Ω

 x dA

I αβ ≡

α

Ω

 x x dA α β

Ω

The tensor A is defined as

A≡

  (λ+m) e  e 3

3

+ m I  dA

Ω

Carrying out the integration we have A =  (λ+m) e3  e 3 + m I  A. In component form this tensor can be expressed as

200

Fundamentals of Structural Mechanics mA

0

0

mA

0

0

0

EA

A~

0

The tensor S is defined as S≡

  (λ+m) e  (p × e ) − m (p × )  dA 3

3

Ω

The components of the tensors e3  ( p × e3 ) and p × are given as follows e3  (p × e3 ) ~

0

0

x2

0

0

0

0 −x1

0

0

0

x2

0

0

− x1

−x2

x1

0

(p × ) ~

Therefore, we get

S ~ (λ+m)

0

0

S2

0

0

0

0 −S 1

0

0

0

−m S 2

0

0

m S1

E S2

E S1

0

=

−m

0

0

0

0 − S1

S2 0

S1

−S 2

The tensor I is defined as

I≡

  (λ+m) (p × e )  (p × e ) − m [p × p × ]  dA 3

3

Ω

The components of the tensors ( p × e 3 )  ( p × e3 ) are given as

(p × e3 )  (p × e3 ) ~

x22

−x1 x2

0

−x1 x2

x21

0

0

0

0

and the components of p × p × are given as (p × p × ) ~

0

0

x2

0

0

0

0 − x1

0

0 − x1

−x2

=

x1

0

−x22

x1 x2

x1 x2

−x21

0

0

−x2 0 0 −x21−x 22

x1

x2

0

Chapter 7

201


Therefore, we get I22

−I12

0

I ~ (λ+m) −I12

I11

0

0

0

0

E I 22 −E I 12

=

−E I 12 0

−m

−I22

I12

0

I12

−I11

0

0

0

−J

0

E I 11

0

0

mJ

182. Consider the beam in Problem 177. Find an expression for the transverse displacement w(x) using the principle of virtual work, using a quartic polynomial basis. Note that the problem has two essential boundary conditions and two natural boundary conditions.

See problem 179. The quartic terms add no new information and therefore the solution does not improve.

183. The principle of virtual work does not require that the assumed displacement functions satisfy the natural boundary conditions a priori. Is there an advantage to satisfying the natural boundary conditions, too? What happens in Problem 182 if we do enforce the natural boundary conditions?

Enforcing the natural boundary conditions improves the approximation’s ability to satisfy the exact solution at the end points at the expense of using the additional terms to satisfy the equations in a weak sense throughout the domain. In general, it is difficult to justify enforcing equilibrium exactly at certain points in the domain.

q 184. A continuous beam is one that has one or more intermediate supports. The extra boundary conditions are in excess of the four end conditions. ℓ ℓ Describe an approach to solving the following x1 x2 problem that exactly satisfies the differential equations everywhere in the domain, as well as the boundary and intermediate conditions. Find the classical solution to the given problem by integrating the governing differential equations. (Hint: It is useful to describe the solution independently in each segment and to enforce continuity by equating state variables at the place where the two segments join.)

202


Let us divide the beam at the middle support into two segments and refer to the variables associated with each beam with indices 1 and 2. ^

θ

q

M 2 (0)

M 1 (ℓ)

x1

ℓ

x2

w1 (0) = 0

w1 (ℓ) = 0

M 1 (0) = 0

θ 1 (ℓ) = θ

w2 (0) = 0

^

^

θ 2 (0) = θ

For the first beam, EI w iv1 = −q EI w 1′′′ = −q x1+a 1 EI w 1′′ = − 12 qx21 + a 1 x1 + b 1 EI w 1′ = − 16 qx31 + 12 a1 x 21 + b1 x1 + c 1 1 EI w 1 = − 24 qx41 + 16 a1 x 31 + 12 b 1 x 21 + c1 x 1 + d1

Since w1(0) = 0

⇒ d1 = 0

M 1(0) = EI w1 ′′(0) w1(ℓ) = 0

⇒ b1 = 0 2 qℓ3 ⇒ c1 = − a1 ℓ 24 6

Similarly, for the second beam, since EI w iv2 = 0 we have EI w 2′′′ = a2 EI w 2′′ = a2 x2 + b 2 EI w 2′ = 12 a 2 x 22 + b2 x 2 + c2 EI w 2 = 16 a 2 x 32 + 12 b 2 x22 + c 2 x 2 + d2 Evaluating the boundary conditions w2(0) = 0

⇒ d2 = 0

M 2(ℓ) = EI w2 ′′(ℓ) = 0 w2(ℓ) = 0

⇒ b2 = −a2 ℓ

⇒ c2 = 13 a2 ℓ2

ℓ

w2 (ℓ) = 0 M 2 (ℓ) = 0

Chapter 7


203

Then, since M 1 (ℓ) = M 2 (0) a1 + a 2 = 12 qℓ

(1)

and since θ 1 (ℓ) = θ 2 (0) or w1′ (ℓ) = w2′ (0) a1 − a 2 = 38 qℓ

(2)

we can use equations (1) and (2) to get 7 a1 = 16 qℓ

1 a2 = 16 qℓ

Letting ξα ≡ xα∕ℓ we get w1 =

q ℓ4 −ξ 41 + 74 ξ 31 − 34 ξ 1 24 EI

w2 =

q ℓ4 24 EI

14 ξ2 3 − 34 ξ22 + 12 ξ 2

q 185. The following prismatic beam has a cross seckℓ 2 tion that is symmetric with respect to the plane of the page. The cross section has flexural modulus EI. Axial k and shear deformations can be neglected (i.e., use Berx noulli-Euler beam theory). The beam is subjected to a ℓ uniform transverse load q acting downward. The beam has deformable spring supports at the ends. At the left end, the support prevents translation in the vertical and horizontal directions and the spring elastically restrains rotations. The moment developed by the spring is related to the rotation at that point by M s = kℓ2θ s, where θ s = θ(0) is the rotation experienced by the spring. The right end of the beam is free to translate horizontally and to rotate, but the spring elastically restrains vertical motion. The force developed by the spring is related to the deflection at that point by Fs = kws , where ws = w(ℓ) is the deflection experienced by the spring. What are the appropriate boundary conditions for this problem? Solve the problem by integrating the differential equations and using the boundary conditions to find the constants of integration. Revise the principle of virtual work to account for the work done by the springs. Estimate the deflection of the beam using a two-term polynomial expansion for the transverse deflection. That is, assume the real and virtual transverse deflections to be of the form 2 2 w(x) = a 1 x + a2 x , w(x) = a 1 x + a2 x ℓ ℓ What constraints do the assumed displacement field add to the problem?

(a) What are the appropriate boundary conditions for this problem?

204

Fundamentals of Structural Mechanics Q(ℓ)

M(0)

kℓ2 w′(0) k

k w(ℓ)

There is one essential boundary condition: w(0) = 0 The natural boundary conditions can be established by taking freebody diagrams of each end (see sketch). At the left end we have M(0) = kℓ 2w′(0). At the right end we have M(ℓ) = 0 and Q(ℓ)+kw(ℓ) = 0. Moment and shear in the beam are related to the transverse displacement of the beam as M = EI w′′ and Q = −EI w′′′. Therefore, the natural boundary conditions can be expressed as EIw′′(ℓ) = 0 EI w′′(0) − k ℓ 2 w′(0) = 0 EI w′′′(ℓ) − k w(ℓ) = 0 (b) Solve the problem by integrating the differential equations and using the boundary conditions to find the constants of integration. 1 q x 4 + d x3ℓ + c x2ℓ 2 + b xℓ3 + aℓ 4 EIw = − 24

EI w′ = − 16 qx 3 + 3d x2ℓ + 2c xℓ2 + bℓ 3 EI w′′ = − 12 qx 2 + 6dxℓ + 2cℓ2 EI w′′′ = −qx + 6dℓ Let us define the parameter η ≡ k ℓ 3∕EI to be the ratio of spring stiffness to bending stiffness. Then, w(0) = 0

⇒ a=0

EIw′′(ℓ) = 0

⇒ − 12 qℓ 2 + 6d ℓ 2 + 2cℓ2 = 0

η EI w′′(0)− EIw′(ℓ) = 0 ℓ η EIw′′′(ℓ)− 3 EI w(ℓ) = 0 ℓ

⇒ 2cℓ2−ηbℓ 2 = 0



⇒ −q ℓ + 6dℓ−η −

qℓ 24



+bℓ+cℓ+dℓ = 0

Solving these equations for b, c, and d we get b=−

q 8η

+η 126+η ,

c=−

q 16

+η 126+η ,

d=

q 48

366++η5η

Chapter 7

205


Letting ξ ≡ x∕ℓ, we get w=

qℓ 4 24 EI



−ξ 4 +

3612++52ηη ξ − 3612++32ηη ξ − 366η++η3η ξ 3

2

2

Note that as η → ∞, we get the solution for a propped cantilever beam q

w=

qℓ 4  −ξ 4 + 52 ξ 3 − 32 ξ 2 24 EI

ℓ

x

(c) Revise the principle of virtual work to account for the work done by the springs. Recall that the virtual work expression without the springs is

  EIw′′w′′ − qw  dx = 0 ℓ

G(w, w) =

0

The work associated with the springs is internal work (see text for a discussion). Thus,

  EIw′′w′′ − qw  dx + k w(ℓ) w(ℓ) + kℓ w′(0) w′(0) ℓ

G(w, w) =

2

0

(d) Estimate the deflection of the beam using a two-term polynomial expansion for the transverse deflection. That is, assume the real and virtual transverse deflections to be of the form w(x) = a 1 x + a2 x , ℓ 2

w(x) = a 1 x + a2 x ℓ

2

What constraints does the assumed displacement field add to the problem? Let us define the vector of base functions and unknown coefficients as h=

ξℓ

a=

2

ξℓ

a1 a2

a=

a1 a2

where ξ ≡ x∕ℓ. It follows that, w = a T h and w′′ = a T h′′. By using this notation, the virtual work functional becomes

  EIℓ  h′′  h′′  1

G(w, w) = a

T

T

a + qℓ  h   dξ

0

+ k a T h(1)h(1) T+ℓ 2 h′(0)h′(0) T  a

206


Thus, the stiffness matrix K and the force vector f are given by

 EIℓ  h′′  h′′  1

K≡

T

dξ + k h(1)h(1) T+ℓ 2 h′(0)h′(0) T 

0

 −qℓ  h  dξ 1

f≡

0

Letting η ≡ k ℓ 3∕EI, we find that K ~ EI ℓ

2η

η

f ~ −qℓ 2

4+η

η

1 2 1 3

Solving the system of equations Ka = f we get a1 = −





qℓ 3 12+η , 6 EI 8η+η 2

a2 = −

qℓ 3 1 6 EI 8+η

Thus, w = a 1 h1 + a2 h 2 = −

qℓ 4  η ξ2 + 12+η ξ 48+6η η EI

The approximate displacement is quadratic. The second derivative is constant. Therefore, the approximation imposes a constant bending moment field, which is clearly at odds with the exact solution.

P P P 186. In the derivations of beam theory, both in a classical sense and in a variational sense, no mention was made of concentrated forces. Describe a way to account for concentrated forces in solving the classical ℓ x differential equations (for example, for a BernoulliEuler beam). Describe how the concentrated forces should be implemented into the principle of virtual work.

(a) Describe a way to account for concentrated forces in solving the classical differential equations (for example, for a Bernoulli-Euler beam). Similar to the previous problem, we can divide the beam into four segments (see the figure below) and refer to the variables associated with each beam with indices i = 1,. . ., 4.

Chapter 7 P

M1 (ℓ∕4)

Q 1 (ℓ∕4)

P

M2 (0)

M2 (ℓ∕4)

Q 2 (0)

207


M3 (0)

Q 2 (ℓ∕4)

x1

P

Q 3 (0)

M3 (ℓ∕4)

M4 (0)

Q 3 (ℓ∕4)

x2

x3

Q 4 (0)

x4

For each beam beam EI w ivi = 0 EI w i′′′ = −Qi = ai EI w i′′ = M i = ai xi + bi EI w i′ = 12 a i x 2i + bi x i + ci EI w i = 1 a i x 3i + 1 b i x2i + c i x i + di 6

2

where, i = 1, , 4 and xi denotes the local coordinate for the ith beam ( 0 < x i < ℓ∕4). So we have 16 (4 per each beam) unknown coefficients to be determined. We get the 16 equations required as follows Boundary conditions at the ends w1(0) = 0

w4(ℓ∕4) = 0

M 1(0) = 0

M 4(ℓ∕4) = EI w4′′(ℓ∕4) = 0

continuity of deflection and slope at the nodes wi(ℓ∕4) = wi+1(0) wi′(ℓ∕4) = wi+1′(0) and finally equilibrium at the nodes M i(ℓ∕4) = M i+1(0) Qi(ℓ∕4) − Q i+1(0) = −P where ( i = 1, 2, 3 ). (b) Describe how the concentrated forces should be implemented into the principle of virtual work. Concentrated forces must be added to the external work. So, for this beam the virtual work functional is given by

208


  EIw′′w′′  dx − −P w(ℓ∕4)+w(2ℓ∕4)+w(3ℓ∕4)  = 0 ℓ

G(w, w) =

0

(c) Solve the problem pictured above, assuming that the forces are applied at the quarter points of the span, using the principle of virtual work for the Bernoulli-Euler beam theory and the approximation basis functions h1 = 1−3ξ2+2ξ 3,

h 2 = ξ2(3−2ξ)

h3 = ξ(1−2ξ+ξ2)ℓ, h4 = ξ 2(ξ−1)ℓ where ξ ≡ x∕ℓ, so that the transverse displacement can be approximated by 4

w(x) =

a h , i

i

4

w(x) =

i=1

a h i

i

i=1

Can you ascribe a physical interpretation to the coefficients ai? The first derivatives of the base functions are h1′ = 6ξ2−6ξ

h2′ = −6ξ2+6ξ

h3′ = (1–4ξ+3ξ2) ℓ

h4′ = (3ξ2−2ξ)ℓ

We can observe that the coefficients ai are related to the end displacement and end rotations by w(0) = a 1

w(ℓ) = a 2

w′(0) = θ(0) = a 3 ℓ

w′(ℓ) = θ(ℓ) = a 4 ℓ

For this problem, since the essential boundary conditions are w(0) = 0 and w(ℓ) = 0 so that a1 = 0 and a2 = 0. Let us define a matrix of shape functions hT ≡ [ h3, h4 ] and a matrix of unknown coefficients aT ≡ [ a3, a4 ]. Thus, w = a T h and w′′ = a T h′′. By using this notation, the virtual work functional becomes G(w, w) = a  EIℓ  h′′  h′′  1

T

T

a dξ + P a T  h(1∕4)+h(1∕2)+h(3∕4) 

0


 EIℓ  h′′  h′′  1

K≡

T

f ≡ −P h(1∕4)+h(1∕2)+h(3∕4)

dξ

0

Carrying out the indicated integrations we get K = EI ℓ 2

4

2

2

4

5 f = 16 P

1 −1

Chapter 7

209


Then solving Ka = f yields P , a3 = − 5 32 EIℓ

a4 = 5 P 32 EIℓ

Thus, 3 3 w = a 3 h3+a 4 h4 = 5 Pℓ −ξ(1−2ξ+ξ 2) + ξ2(ξ−1) = 5 Pℓ  ξ2−ξ  32 EI 32 EI

For a better solution, we can divide the beam into segments, approximate the displacement in each beam segment and enforce continuity of state variables among neighboring segments.

187. A beam on a Winkler elastic foundation accrues force in the foundation in proportion to the deflection of the beam according to f (x) = kw(x), where k is the modulus of the foundation. q(x)

k x

kw(x) ℓ

Show that a simply supported Bernoulli-Euler beam on an elastic foundation is governed by the following differential equation and boundary conditions EIwiv + kw = q(x) w(0) = 0,

w′′(0) = 0,

w(ℓ) = 0,

w′′(ℓ) = 0

Verify that w(x) = e βx  a1 cos βx+a 2 sin βx  + e−βx  a3 cos βx+a4 sin βx  is the displacement field that satisfies the homogeneous differential equation, if 4β 4 ≡ k∕EI.

(a) Show that a simply supported Bernoulli-Euler beam on an elastic foundation is governed by the following differential equation and boundary conditions EIwiv + kw = q(x) w(0) = 0,

w′′(0) = 0,

w(ℓ) = 0,

w′′(ℓ) = 0

The reaction force due to the elastic foundation can be treated as an applied distributed load which acts always in the direction opposite the displacement. The loading shown on the sketch above is downward and hence negative according to our sign convention. Let us take q(x) in the positive sense as shown below

210

Fundamentals of Structural Mechanics q(x)

kw(x)

Then let us replace the load term in Eqn. (417) in the text with the effective loading q^ (x) ≡ q(x) − kw(x) So, for the Bernoulli-Euler beam we have EIwiv = q^ (x) = q(x) − kw(x). Rearranging terms gives the governing equation for the beam on an elastic foundation (with no applied moment) EIwiv + kw = q The boundary conditions are same as those of a simply supported Bernoulli-Euler beam. (b) Verify that the displacement w(x) = e βx a 1 cos βx + a2 sin βx  + e−βx a3 cos βx + a 4 sin βx  where 4β 4 ≡ k∕EI, satisfies the homogeneous differential equation. w′ = β e βx  a1 cos βx − sin βx + a2 cos βx + sin βx  −β e −βx  a3 cos βx + sin βx − a 4 cos βx − sin βx  w′′ = −2β 2 eβx  a 1 sin βx − a2 cos βx  + 2β 2 e −βx  a3 sin βx − a 4 cos βx  w′′′ = −2β 3 eβx  a 1 cos βx + sin βx + a2 cos βx − sin βx  +2β 3 e−βx  a 3 cos βx − sin βx + a4 cos βx + sin βx  wiv = −4β 4 e βx  a1 cos βx + a 2 sin βx  − 4β 2 e−βx  a 3 cos βx − a4 sin βx  = −4β 4w Substituting these results into the governing differential equation we get EIwiv + kw = EI  wiv + 4β 4w  = EI −4β 4w + 4β 4w = 0 Thus, the given displacement satisfies the homogeneous differential equation.

188. Consider the beam on a Winkler elastic foundation of Problem 187, subjected to a uniform load q. Show that the principle of virtual work, accounting for the work done by the elastic foundation, is

Chapter 7

211


  EI w′′w′′ + k ww − q w  dx = 0 ℓ

G(w, w) =

0

Find the displacement map of the system using the Ritz method, assuming that the real and virtual displacements are approximated as N

w(x) =

a

n

n=1

sin nπx , ℓ

N

w(x) =

a

n sin

n=1

nπx ℓ

(a) Show that the principle of virtual work, accounting for the work done by the elastic foundation, is

  EI w′′w′′ + k ww − q w  dx = 0 ℓ

G(w, w) =

0

Similar to the discussion in Problem 187, the work done by the Winkler foundation can be treated as work done by an external (applied) load. So, the contribution of the Winkler foundation to the external work is, = −  k ww dx ℓ

W WINKLER E

0

Then G(w, w) ≡ W I − W E yields

  EI w′′w′′ + k ww − q w  dx ℓ

G(w, w) =

0

(b) Find the displacement map of the system using the Ritz method, assuming that the real and virtual displacements are approximated as N

w(x) =

a

n=1

n

sin nπx , ℓ

N

w(x) =

a

m sin

m=1

mπx ℓ

The derivatives of the approximate displacement are N

w′ =

 a nπℓ  cos nπx , ℓ n

N

w′′ = −

n=1

2

 a nπℓ  sin nπx ℓ n

n=1

The derivatives of the virtual displacement are N

w′ =

 a mπℓ  cos mπx , ℓ m

m=1

Then, the virtual work statement yields

N

w′′ = −

2

 a mπ  sin mπx ℓ ℓ m

m=1

212


m

m=1

a 

−

n

n=1

N

2

2

4

4

4

0

ℓ

m

m=1

 n mℓ π + 4β  sin nπxℓ sin mπxℓ dx  ℓ

N

 a a

G(w, w) =

0

−q sin mπx dx ℓ



G(w, w) = 0 for all choices of am is satisfied only if N

 a n mℓ π 2

n

2

4

i=n

4

+ 4β  

ℓ

4

0

sin nπx sin mπx dx = −q ℓ ℓ

 sin mπxℓ dx ℓ

0

The integrals in the equation above are easily evaluated to be

 sin nπxℓ sin mπxℓ dx = ℓ

0

1 ℓ 2

n=m

0

n≠m

 sin mπxℓ dx = mπℓ 1− cos mπ = mπℓ 1−−1  ℓ

m

0

Substituting these results into the virtual work equation we get





n4 π 4 ℓ 1 − −1 n + 4 β 4 = −q nπ an ℓ 2 ℓ4 which can be immediately solved for an to give









n −1 n−1 2q 2qℓ4 −1 −1 an = nπ 4 4 4 = n π ∕ℓ + 4β 4 π4 n 4+γ nπ

where γ ≡ kℓ 4∕EI is a dimensionless constant. The displacement is given by N

w(x) =

a

n=1

nπx = n sin ℓ

N

 2qℓ nπ

n=1

4





−1 n−1

π 4 n4+γ

sin nπx ℓ

P 189. A semi-infinite beam on a Winkler elastic EI foundation extends to infinity in one direction. Find the classical solution to the problem of a beam of modulus k x EI on a foundation with modulus k subjected to a concentrated force P at x = 0. There is no boundary at the right end of the beam, but you can argue that a1 = a2 = 0 to have finite displacements. Plot the deflected shape of the beam.

Chapter 7

213


Using the displacement given in Problem 187 for the homogeneous solution and noting that a1 = a2 = 0 for finite displacements, we have w(x) = e –βx a 3 cos βx + a4 sin βx  w(x) = e βx a 1 cos βx + a2 sin βx  The boundary conditions are M(0) = EI w′′(0) = 0 Q(0) − P = −EI w′′′(0) − P = 0 These conditions yield a3 = − P , a 4 = 0 EI So the displacement is w(x) = − P e –βx cos βx EI w(x) 0

2

4

βx

−P EI

P 190. A beam of length ℓ and modulus EI rests on two EI linearly elastic springs, each of modulus k. The springs accrue force in proportion to the amount by which they k k stretch. The beam is pinned at the left end and is subx jected to a point load P at the right end. Axial and shear deformations of the beam can be neglected. What is the ℓ∕3 ℓ∕3 ℓ∕3 virtual-work form of the equilibrium equations? What are the essential and natural boundary conditions? Use the Ritz method to find an approximation of the displacement field using the two-term polynomial w(x) = a 1 x+a2 x 2∕ℓ.

(a) What is the virtual work form of the equilibrium equations?

  EIw′′w′′  dx ℓ

G(w, w) =

0

+ k w(ℓ∕3) w(ℓ∕3) + k w(2ℓ∕3) w(2ℓ∕3) + Pw(ℓ)

214


(b) What are the essential and natural boundary conditions? The essential boundary condition is w(0) = 0 Natural boundary conditions are M(0) = M(ℓ) = 0 and Q(ℓ)+P = 0. These boundary conditions can be expressed in terms of w as w′′(0) = 0 w′′(ℓ) = 0 EI w′′′(ℓ) − P = 0 (c) Use the Ritz method to find an approximation of the displacement field using the two-term polynomial w(x) = a 1 x + a2 x2∕ℓ. Let ξ ≡ x∕ℓ and let us define the following matrices ξℓ

h=

a1

a=

2

ξℓ

a1

a=

a2

a2

It follows that w = a T h and w′′ = a T h′′. By using this notation, the virtual work functional becomes

  EIℓ  h′′  h′′  1

G(w, w) = a

T

0

T

a + qℓ  h   dξ + P aT h(1) 

+ k a T h(1∕3)h(1∕3) T+ h(2∕3)h(2∕3) T  a


 EIℓ  h′′  h′′  1

K≡

dξ + k h(1∕3)h(1∕3) T+ h(2∕3)h(2∕3) T 

T

0

f ≡ −P  h(1)  Carrying out the indicated integrations we get

K = kℓ

5 9

2

1 3

1 3 4η+

f = −Pℓ

17 81

1 1

where, η ≡ EI∕k ℓ 3 . Solving the system of equations Ka=f yields a1 = −





9 P 162η−5 , kℓ 2+810η

a2 = −



Therefore, the approximate displacement map is w(x) = −



81 P 1 kℓ 2+810η

9P  162η−5 ξ + 9 ξ2  k(2+810η)

Chapter 7

215


191. Consider the beam of modulus EI, pinned at one k P EI end, free at the other, and restrained by a rotational spring as shown. The beam is subjected to a tip load P at the free end. Shear and axial deformations can be neglected. Estiℓ mate the deflection of the beam at the point where load is applied. Discuss the accuracy of your estimate and discuss one other possible method for making the estimate.

The simplest method to solve this problem is to use the classical solution to the fixed cantilever beam and add the effect of the discrete rotation at the spring (which adds a linear motion). The classical solution for the fixed cantilever gives a deflections of Pℓ3∕3EI. The moment felt by the spring is Pℓ so the rotation is Pℓ∕k. The deflection at the end caused by that rotation can be computed to give Pℓ2∕k. Thus, the end deflection is w(ℓ) = Pℓ + Pℓ 3EI k 3

2

One can more formally solve the classical equation by noting the homogeneous solution (and its derivatives) is w′′ = 12  2c+6d ξ  ℓ

w = a +b ξ+c ξ2+d ξ 3 w′ = 1  b+2c ξ+3d ξ 2  ℓ

w′′′ = 13  6d  ℓ

The boundary conditions are w(0) = 0,

M(0) = k w′(0),

M(ℓ) = 0,

Q(ℓ) = −P

Substituting the solution into these conditions give w(0) = 0,

a=0

EIw′′(0) = k w′(0), EIw′′(ℓ) = 0, EIw′′′(ℓ) = P,

EI 2c = k b, ℓ ℓ2

b = 2c EI kℓ

EI  2c+6d  = 0, ℓ2 EI 6d = P, ℓ3

c = −3d

d = Pℓ 6EI 3

Thus, a = 0,

b=−P, kℓ

c = − Pℓ , 2EI 3

so that the displacement field is w = Pℓ  ξ 3−3ξ2  − Pℓ ξ 6EI k 3

2

d = Pℓ 6EI

3

216


which yields the results mentioned earlier. This solution is exact. An alternate approach to this problem is to solve it using the Ritz method and the Principle of Virtual Work.

q 192. A flexible beam of length ℓ and moduRigid EI lus EI is welded to a rigid beam of length ℓ, and rests on an elastic foundation of modulus k x k=60EI/ℓ4. The beam is simply supported and is subjected to a transverse force q over ℓ ℓ the rigid part of the span. The elastic foundation accrues a transverse force in proportion to the transverse displacement w. Shear and axial deformations in the beam are negligible. Write the virtual-work functional G for the system. What are the essential and natural boundary conditions for the flexible beam? Find an approximate solution for the displacement w(x) using a two-term polynomial Ritz basis.

(a) Write the virtual work functional G for the system. q θ

Q(ℓ)

RA

M(ℓ)

∆

Q(ℓ)

w(x) x

x

ξ

k ∆+θ ξ  ξ

ℓ

ℓ

Let us define ∆ ≡ w(ℓ), θ ≡ w′(ℓ), and a coordinate axis ξ, which starts from the left end of the rigid part of the beam. From the geometric constraint that the beam must reach the right support we must have ∆ + ℓθ = 0 Thus, the effective force on the rigid part of the beam is q^ ≡ −q + kθ  ℓ−ξ , taken positive upward. The “external work done by these forces in a virtual displacement given by w(ξ) = ∆+θ ξ = −θ  ℓ−ξ  is

 w(ξ) q(ξ) dξ =   qθ ℓ−ξ  − kθθ  ℓ−ξ   dξ ℓ

WE =

ℓ

^

0

=

1 qℓ2 θ − 13 kℓ 3 θ θ 2

2

0

=

1 qℓ2 w′(ℓ) 2

− 13 kℓ3 w′(ℓ) w′(ℓ)

The virtual work functional G(w, w) = W I−W E is, therefore,

Chapter 7


  EIw′′w′′ + kww  dx + ℓ

G(w, w) =

1 3

217

kℓ 3 w′(ℓ) w′(ℓ) − 12 qℓ2 w′(ℓ)

0

or, noting the constraint ∆+ℓ θ = 0, equivalently

  EIw′′w′′ + kww  dx + ℓ

G(w, w) =

1 3

kℓ w(ℓ) w(ℓ) + 12 qℓ w(ℓ)

0

(b) What are the essential and natural boundary conditions for the flexible beam? Since the right portion of the beam is rigid (see figure above), it follows that the essential boundary conditions on the beam are w(0) = 0,

w(ℓ) + ℓ w′(ℓ) = 0

Also, taking the rigid part of the beam as a freebody diagram, the sum of the moments about the right support is zero. To wit,

 q  ℓ−ξ  dξ = 0 ℓ

M(ℓ) − ℓ Q(ℓ) +

^

0

^

Substituting the expression for q from above, and carrying out the integrations, we get the expression M(ℓ) − ℓ Q(ℓ) − 12 qℓ2 + 13 kℓ3θ = 0 Finally, substituting the constitutive relationships M = EIw′′ and Q = −EIw′′′ we find that the natural boundary conditions are w′′(0) = 0 EI  w′′(ℓ)+ℓ w′′′(ℓ)  + 13 kℓ 3 w′(ℓ) − 12 qℓ 2 = 0 (c) Find an approximate solution for the displacement w(x) using a two-term polynomial Ritz basis. Let the approximate displacement be given by the following polynomial expansion w(x) = a 0 ℓ3 + a 1 ℓ2 x + a2 ℓ x 2 + a3 x3 w′(x) = a 1 ℓ2 + 2 a2 ℓ x + 3 a 3 x 2 The essential boundary conditions must be satisfied a priori. Therefore, w(0) = 0

⇒ a0 = 0

w(ℓ) + ℓ w′(ℓ) = 0

⇒ a1 = − 3 a 2 − 2 a3 2

Substituting these relationships into the original expansion we get w(x) = a 2  ℓ x 2− 2 ℓ2 x  + a 3  x 3−2 ℓ2 x  3

w′(x) = a 2 ( 2 ℓ x − 3 ℓ 2 ) + a3 ( 3 x2 − 2 ℓ 2 ) 2 w′′(x) = a 2 ( 2 ℓ ) + a3 ( 6 x )

218


Let ξ ≡ x∕ℓ and define the matrices ℓ 3 ξ 2− 2 ξ  3

h=

a2

a=

ℓ3 ξ3−2ξ 

a=

a3

a2 a3

It follows that w = a T h and w′′ = a T h′′. Using this notation, the virtual work functional becomes G(w, w) =   EIw′′w′′ + kww  dx + ℓ

1 3

kℓ 3 w′(ℓ) w′(ℓ) − 12 qℓ2 w′(ℓ)

0

  EIℓ  h′′  h′′  + kℓ  h  h   a dξ 1

= aT

T

T

0

+ a T  13 kℓ3 h′(1)h′(1) T  a − a T  12 qℓ 2 h′(1)  Thus, the stiffness matrix K and the force vector f are given by K =

  EIℓ  h′′  h′′  + kℓ  h  h   dξ

f =

1 qℓ2 h′(1) 2

1

T

T

+

1 3 kℓ h′(1)h′(1) T 3

0

Carrying out the indicated integrations, noting that k=60EI/ℓ4, we get K = EI ℓ 3

21

38

38

508 7

f = − q ℓ4

1 4 1 2

Solving the system of equations Ka=f yields a2 =

3 qℓ , 280 EI

a2 = −

qℓ 80 EI

The approximate displacement is given by w(ξ) =

qℓ4  −7ξ3+6ξ 2+5ξ 

560EI

193. Consider the beam of modulus EI, fixed at one wℓ end, pinned at the other. The beam is subjected to a EI prescribed displacement of wℓ at the right end. Shear and axial deformations can be neglected. Find the ℓ expression for the displacement w(x) that satisfies the governing equations exactly. Approximately solve the problem using the principle of virtual work, assuming a cubic polynomial deflection field.

Chapter 7


219

Using the results from Problem 191 without the spring we have w = Pℓ  ξ 3−3ξ2  6EI 3

where P is an unknown reaction force required to obtain the prescribed motion. Use the boundary condition w(ℓ) = w ℓ to get w(1) = − Pℓ = w ℓ, 3EI 3

P=−

3wℓ EI ℓ3

Thus, w=

wℓ 2

 3ξ 2−ξ3 

To solve the problem using the Ritz method we have a virtual work functional

 EI w′′w ′′ dx = 0 ℓ

G(w, w) =

0

Take a Ritz approximation of w = w ℓ ξ2 + a  ξ3−ξ 2 ,

w = a  ξ3−ξ 2 

w′′ =  2 wℓ + a ( 6 ξ−2 ) ∕ℓ2 ,

w′′ = a  6 ξ−2 ∕ℓ 2

Note that the fixed function is quadratic so that the displacement function satisfies both essential boundary conditions at the left end. The base functions are chosen to satisfy the homogeneous essential boundary conditions. Since there are three conditions a cubic function only has one degree of freedom left. Therefore, the Ritz approximation is a one--term approximation. Substituting the approximation into the virtual work functional gives G = EI4 ℓ

  2 w + a  6 ξ−2   a  6 ξ−2  ℓdξ = 0 1

ℓ

0

Invoking the fundamental theorem of the calculus of variations we get

  6 ξ−2  dξ + a  36 ξ −24 ξ+4  dξ = 0 1

2 wℓ

1

2

0

0

2 w ℓ  3−2  + a 12−12+4  = 0,

⇒

a = −w ℓ∕2

Thus, the Ritz approximation gives w = w ℓ ξ2 − which is the exact solution.

wℓ 3  ξ −ξ 2  = wℓ  3 ξ 2−ξ3  2 2

220


M 194. A beam of unit length and variable modulus is fixed at the right end and is subjected to a moment M at the left end. The bending and x shear stiffnesses of the beam are variable with EI(x) = EI 0  1+x  and GA(x) = GA 0  1+x  , where EI 0 and GA0 are known constants. 1 Recall that the governing equations for the Timoshenko beam are given by Eqns. (406). Find the deflection and rotation at the left end of the beam (i.e., at x = 0) by finding the classical solution to the governing differential equations.

The governing equations for the Timoshenko beam are Q′ + q = 0,

Q = GA(w′−θ)

M′ + Q + m = 0,

M = EIθ′

Note the following differentiation relationships

 dxx = ln x,

d ln x = 1 x dx

The boundary conditions are (letting the applied moment at the left be called M 0 to distinguish it from the moment field) Q(0) = 0,

M(0) = M 0,

w(1) = 0,

θ(1) = 0

Note that, for this problem, we have q = 0 and m = 0. Therefore, we can integrate the equilibrium equations and use the boundary conditions at the left end to find two of the constants of integration. To wit, Q′ = 0

Q = a0

⇒

Q(0) = a 0 = 0

⇒

Q(x) = 0

⇒

M(x) = M 0

Similarly for moment we have M′ = 0

M = a1

⇒

M(0) = a 1 = M 0

Now, substituting this result into the constitutive equation for moment we can find the rotation by integration EI 0  1+x  θ′ = M 0 θ′ =

M0 1 EI 0 1+x

⇒

θ(x) =

M0  ln 1+x  + a 2 EI 0

Substituting the boundary condition on rotation at the right end we get θ(1) = 0 =

M0 ln(2) + a 2 = 0 EI 0

⇒

a2 = −

M0 ln 2 EI 0

Chapter 7


221

Therefore, the rotation field is θ(x) =



M0 ln 1+x EI 0 2



Now we can do the same for the constitutive equation for shear. To wit, Q = GA(w′ − θ) = 0

w′ = θ

⇒

Integrating this equation we get w′(x) = w(x) =



M0 ln 1+x EI 0 2



 





M0  1+x  ln 1+x −  1+x  + a 3 EI 0 2

Substituting the boundary condition for displacement at the right end we get w(1) = 0 = −

2M 0 + a3 EI 0

⇒

a3 =

2M 0 EI 0

Thus, the displacement field is w(x) =







M0  1+x  ln 1+x +  1−x  EI 0 2



Finally, we can evaluate the rotation and displacement at the left end to get θ(0) = −

M0 ln 2, EI 0

w(0) =

M0  1− ln 2  EI 0

EI M 195. A beam of length ℓ and modulus EI rests on two linearly elastic springs, each of modulus k. The springs k k accrue force in proportion to the amount by which they stretch. The beam is pinned at both ends and is subjected ℓ∕3 ℓ∕3 ℓ∕3 to a concentrated moment M at the right end. Axial and shear deformations of the beam can be neglected. What is the virtual-work form of the equilibrium equations? What are the essential and natural boundary conditions? Use the Ritz method to find a polynomial approximation of the displacement field.


 EI w′′w ′′ dx + k  wℓ∕3 wℓ∕3+w2ℓ∕3 w2ℓ∕3  − Mw′(ℓ) ℓ

G(w, w) =

0

222


(b) What are the essential and natural boundary conditions? Essential B.C.’s:

w(0) = 0,

w(ℓ) = 0

Natural B.C.’s:

w′′(0) = 0,

EI w′′(ℓ) = M

(c) Use the Ritz method to find an approximation of the displacement field using a quadratic approximation. w(x) = a 0 + a1 x + a 2 x2 The approximation must satisfy the essential boundary conditions. Thus, w(0) = 0

⇒

a0 = 0

w(ℓ) = 0

⇒

a 1 = −a 2ℓ

Letting a2 ≡ a, the displacement field, and its derivatives are w(x) = a  x 2−x ℓ 

w′ = a  2x−ℓ )

w′′ = 2a

Thus, the discrete form of the virtual work functional is

 EI 2a 2a dx + k a a    ℓ +   ℓ  − M a ℓ = 0 ℓ

G(a, a) =

2 9

2

2 9

2

2

2

0

Carrying out the integration we get 8 a   4EI ℓ + 81 k ℓ 4  a − M ℓ = 0

This equation must hold for all a therefore we get a=

81 M 324 EI+8kℓ3

Thus, the approximate displacement field is w(x) =

81 M 324 EI+8k   x −x ℓ  ℓ 3

2

196. Consider a beam of length ℓ, elastic moduli E and G and rectangular cross section of width 2a and depth 2b. Let c be a (very small) constant and let ξ ≡ x 3∕ℓ be the normalized axial coordinate. The beam has the following displacement and rotation fields θ 1 = c  6ξ−6ξ2 ,

θ 2 = c 3ξ2−2ξ ,

w1 = c ℓ  ξ 3−ξ2 ,

w2 = c ℓ 2ξ3−3ξ 2 ,

θ3 = 0 w3 = 0

x2 x1

2b

2a

Chapter 7

223


Find the resultant moment and resultant force at ξ = 1, i.e. M(ℓ) and Q(ℓ). Find the total displacement u of the point located at the position x = (a, b, ℓ).

(a) Find the resultant moment and resultant force at x = ℓ, i.e. M(ℓ) and Q(ℓ). w 1′ − θ 2 w′ + e3 × θ = w ′ + θ 2 1

=c

3ξ 2 − 2ξ − 3ξ 2 + 2ξ 6ξ 2 − 6ξ + 6ξ − 6ξ 2

w 3′

=0

0

Therefore, Q = 0. The curvature can be computed from the given field as 6∕ℓ − 12∕ℓ −6∕ℓ θ′(ℓ) = 6∕ℓ − 2∕ℓ c = 4∕ℓ c 0

0

Noting that I 11 = 16a3b∕12 and I 22 = 16b3a∕12, the moment is M=

4 ab 3E 3

0 0

0 4 3 a bE 3 0

0 0 GJ

− 24 ab 3c 3 −6∕ℓ 16 3 4∕ℓ c = a bc 3 0 0

(b) Find the total displacement u of the point located at the position x = (a, b, ℓ). w 1 − x 2 θ3

u =w−p×θ =

w 2 + x 1 θ3 w 3 − x 1 θ 2 + x2 θ 1

0 u(a, b, ℓ) = −cℓ −ca

197. A beam of length ℓ is fixed at the left end, free at the right end, and is subjected to a concentrated transverse load P at the right end. The bending and shear stiffnesses of the beam are EI and GA, respectively. What is the virtual-work functional for the system? What are the essential and natural boundary conditions? Let ξ ≡ x∕ℓ. Find the deflection and rotation fields for the given loading by the Ritz method using the following approximation w(ξ) = a 2 ℓ ξ+ 12 a0 ℓ ξ2+ 13 a1 ℓ ξ3,

θ(ξ) = a0 ξ+a1 ξ 2

P EI, GA x ℓ

224


(a) What is the virtual work functional for the system?

  EIθ′θ′ + GA  w′−θ   w′−θ ′   dx + Pw(ℓ) ℓ

G(w, θ, w, θ) =

0


w(0) = 0,

θ(0) = 0

Natural B.C.’s:

M(ℓ) = 0,

Q(ℓ)+P = 0

(c) Find the deflection and rotation fields for the given loading by the Ritz method using the following approximation w(ξ) = a 2 ℓ x + 12 a 0ℓ ξ 2 + 13 a1ℓ ξ3

θ(ξ) = a 0 ξ + a1 ξ2 θ′ = 1  a0+2a 1 ξ  ℓ

w′ = a 2 ξ + a0 ξ + a1 ξ 2

Note that w′− θ = a 2 ξ. Take w, θ using the same interpolation as w, θ. Then

 EIℓ   a +2a ξ  a +2a ξ + 1γ a a  dξ+Pℓ  a2 + a3 +a  = 0 1

G=

0

1

0

1

0

2 2

1

2

0

where γ ≡ EI∕GAℓ 2 . Carrying out the integrals, and invoking the fundamental theorem of the calculus of variations, we get

EI ℓ

1

1

0

1

4∕3

0

a0 a1

0

0

1∕γ

a2

1∕2 + Pℓ

=0

1∕3 1

Solving these equations gives a0 = − Pℓ EI

2

a1 = Pℓ 2EI

2

2 a2 = − Pℓ γ = − P EI GA

The displacement and rotation fields are w(ξ) = Pℓ  −3ξ 2 + ξ3 − 6 γξ , 6EI 3

θ(ξ) = Pℓ  −2ξ+ξ2  2EI 2

P P 198. A beam of length ℓ and modulus EI rests on two linEI early elastic springs, each of modulus k. The beam is subk k x jected to point loads P at the ends. Axial and shear deformations of the beam can be neglected. What is the virtual-work form of the equilibrium equations? What are the essential ℓ∕3 ℓ∕3 ℓ∕3 and natural boundary conditions? Solve the discrete virtualwork equations Ka = f for this system using a three-term polynomial Ritz approximation.

Chapter 7

225



 EIw′′w′′ dx + k  wℓ3wℓ3+w2ℓ3 w2ℓ3  +Pw(0)+Pw(ℓ) ℓ

G(w, w) =

0


None

Natural B.C.’s:

M(0) = 0

⇒

w′′(0) = 0

M(ℓ) = 0

⇒

w′′(ℓ) = 0

P−Q(0) = 0

⇒

P+EIw′′′(0) = 0

P+Q(ℓ) = 0

⇒

P − EIw′′′(ℓ) = 0

(c) Set up, but do not solve, the discrete virtual work equations Ka = f for this system using a three-term polynomial Ritz approximation. Let ξ = x∕ℓ and take an approximation of the form w(ξ) = a 0 + a1 ξ + a 2ξ2

w(ξ) = a0 + a1 ξ + a2ξ 2

The base functions are then h =  1, ξ, ξ 2  and h′′ =  0, 0, 2  ∕ℓ2 . The stiffness matrix is then

 EI h′′  h′′ ℓ dξ + k  h1∕3  h1∕3 + h2∕3  h2∕3  1

K=

0

and the load vector is f = P  h0+h1 Carrying out the integrals we get 0 K = EI3 0 ℓ 0

0 0 0

0 0 4

+k

2 1 5∕9

1 5∕9 1∕3

5∕9 1∕3 17∕81

,

f = −P

2 1 1

P 199. A semi-infinite (i.e., x extends to infinity) beam EI of modulus EI rests on an elastic foundation of modulus k and is subjected to a concentrated force P at x = 0. The k x classical differential equation for the beam is given by iv iv EIw +kw = 0, where w means fourth derivative. The shear in the beam is given by Q(x) = −EIw′′′. Use the Ritz method with the principle of virtual work to find an approximation of the displacement field using the single term approximation w(x) = a e –βx, where β is a given constant and a is the unknown Ritz displacement parameter. Find an expression for the error in equilibrium at each point x in the beam. For what value of β is the shear boundary condition at x=0 satisfied exactly?

226


(a) Use the Ritz method with the principle of virtual work to find an approximation of the displacement field using the single term approximation w(x) = a e –βx where β is a given constant and a is the unknown Ritz displacement parameter. The Ritz function and its derivatives are given as w = a e --βx

w′ = − β a e --βx

w′′ = β 2 a e --βx

w = a e --βx

w′ = − β a e --βx

w′′ = β 2 a e --βx

The virtual work functional is thus G(w, w) =



∞

 EI w′′w′′+k ww  dx + Pw(0)

0



= a  EI β 4+k  a





∞

e−2 β x dx + P

0

=0

If G = 0 for all a, then the value of a can be determined as (and therefore the approximate displacement as) a=

−2 β P EI β 4+k

w(x) =

 EI−2β β+kP  e 4

−β x

(b) Find an expression for the error in equilibrium at each point x in the beam. The error is the amount by which the approximation fails to satisfy the classical differential equation. Substituting into the equations we get error = EI β 4 a e −β x + k a e −β x =  EI β 4+k 

EI−β2β+kP  e 4

−β x

= −2β P e−β x

(c) For what value of β is the shear boundary condition at x=0 satisfied exactly? The boundary conditions at the left end is P−Q(0) = 0. Substituting the approximation we get P − Q(0) = P + EI w′′′(0) = P +

2EI β 4 P = 0 EI β 4+k

Thus, P  3EI β 4 + k  = 0 EI β 4+k

⇒

β 4 = − 3EI k

Since β is a positive constant, there is no value that will satisfy the boundary condition exactly.

Chapter 7

227


M 200. A flexible beam of length ℓ and modulus EI is conEI nected to a rigid beam of length ℓ at a point that rests on a k k roller support. The left end of the rigid part of the beam is x restrained by a linear elastic spring of modulus k as is the right end of the flexible part of the beam. The beam is subℓ ℓ jected to an end moment M. Axial and shear deformations of the flexible beam can be neglected. What is the virtual-work form of the equilibrium equations? What are the essential and natural boundary conditions for the flexible segment of the beam? Solve the discrete virtual-work equations Kaf for this system using a three-term polynomial Ritz approximation.

(a) What is the virtual work form of the equilibrium equations? w(x) w(ℓ) ℓ w′(0)

Q(0)

w′(0)

R M

M(0)

M(ℓ)

Q(ℓ)

k w(ℓ)

k ℓ w′(0)

The kinematics and freebody diagrams are given in the sketch. The virtual work form of the differential equations

 EI w′′w′′ dx + k ℓ w′(0) w′(0) + k w(ℓ) w(ℓ) − M w′(0) ℓ

G(w, w) =

2

0

The principle of virtual work suggests that if G(w, w) = 0 for all w then equilibrium holds. (b) What are the essential and natural boundary conditions for the flexible segment of the beam? Essential B.C.’s : w(0) = 0 Natural B.C.’s: From the freebody diagram above. M(0)+M−k ℓ 2 w′(0) = 0;

M(0) = EI w′′(0)

M(ℓ) = EI w′′(ℓ) = 0 Q(ℓ)+k w(ℓ) = 0;

Q = −EI w′′′

(c) Set up, but do not solve, the discrete virtual work equations Ka = f for this system using a three-term polynomial Ritz approximation. Let the Ritz approximation be w = a T h(ξ) and w = a T h(ξ) where the base functions (and derivatives) are h(ξ) =  ξ, ξ2, ξ3 , h′(ξ) = 1  1, 2ξ, 3ξ 2 , h′′(ξ) = 12  0, 2, 6ξ  ℓ ℓ The stiffness and force vectors then have the form

228


 EI h′′  h′′ ℓ dξ + k ℓ h′(0)  h′(0) + k h(1)  h(1) 1

K=

2

0

f = M h′(0) Carrying out the indicated integrations K = EI3 ℓ

0 0 0

0 4 6

0 6 12

+k

2 1 1

1 1 1

1 1 1

f=M ℓ

q(x) = 2x 201. Consider the simply supported beam of length 1 and constant modulus EI1, subjected to a linearly varying force q(x)2x, as shown. The beam is supported by a spring at x midspan that has modulus k64. Shear and axial deformations can be neglected. What are the natural boundary condi1∕2 tions? What are the essential boundary conditions? Find an approximate solution with the Ritz method. Use a cubic polynomial.

1 0 0

k 1∕2

(a) What are the natural boundary conditions? What are the essential boundary conditions? Natural B.C.’s : w′′(0) = 0

w′′(1) = 0

Essential B.C.’s: w(0) = 0

w(1) = 0

Note: The spring is not a boundary! (b) Find an approximate solution with the Ritz method. Use a polynomial approximation of cubic order. The essential boundary conditions can be used to establish a suitable set of base functions. Take a cubic approximation as follows w(x) = a 0+a1 x+a 2 x2+a 3 x 3 w(0) = 0

⇒

a0 = 0

w(1) = 0

⇒

a 1 + a2 + a 3 = 0

w(x) = a 2  x2−x  + a3  x 3−x  Let the Ritz approximation be w = a T h(ξ) and w = a T h(ξ) where the base functions (and derivatives) are h(x) =  x 2−x, x 3−x , h′(x) =  2x−1, 3x 2−1 , h′′(x) =  2, 6x  The virtual work functional is

 w′′w′′ dx + 64 w(1∕2)w(1∕2) −  1

G(w, w) =

0

1

0

2x w dx

Chapter 7

229


Therefore, the stiffness and force vectors then have the form

 h′′  h′′ dx + 64 h(1∕2)  h(1∕2),

 2x h dx

1

K=

1

f=

0

0

Carrying out the indicated integrations we get K=

8

12

12

21

− 0.1667

f=

− 0.10

Thus, the coefficients satisfy the equations

a2 a3

8

12

a2

12

21

a3

21 −12 = 1 24 −12 8

− 0.1667

=

− 0.10

− 0.1667

=

− 0.10

− 0.0958 0.050

Therefore, the approximation solution for the displacement field is w(x) = −0.0958  x2−x  + 0.05  x3−x 

qo

202. A beam of length ℓ is fixed at the left end, pinned at the right end, and is subjected to a uniform load, as shown. The shear and bending moduli are related as EI/ GAℓ2=1. Find the displacement and rotation fields for the beam by solving the classical governing equations. Find the reaction forces at the supports.

EI, GA x ℓ

(a) Find the displacement and rotation fields for the beam by solving the classical governing equations. Start by integrating the equation for rotation EI θ′′′ = −q 0

⇒

θ(x) = −



2 q 0ℓ 3 x 3 + a1 x 2 + a 2 x + a 3 6EI ℓ 3 ℓ ℓ

θ′(x) = −

q 0ℓ 2 6EI

 3 ℓx

2 2

+ 2a 1 x + a 2 ℓ

Substitute the boundary conditions that involve the rotation field:





230

Fundamentals of Structural Mechanics (1)

θ(0) = 0

(2)

θ′(ℓ) = 0

a3 = 0

⇒ ⇒

2a 1 + a2 + 3 = 0

Now integrate the remaining governing equation w′ = θ−ℓ 2 θ′′ = −

q 0ℓ 3 6EI

 ℓx

2 + a 1 x 2 + a2 x − 6 x − 2a 1 ℓ ℓ ℓ

3 3



to give



2 q 0ℓ 4 1 x 4 1 x 3 1 x 2 + a + a − 2 x 2 − 2a 1 x + a 4 6EI 4 ℓ 4 3 1 ℓ 3 2 2 ℓ 2 ℓ ℓ

w(x) = −



Now substitute the boundary conditions that involve the displacement field: (3) w(0) = 0

⇒

a4 = 0

(4) w(l) = 0

⇒

− 20a1 + 6a 2 − 33 = 0

From (2) and (4): a1 = − 51∕32, a2 = 6∕32. Therefore, The displacement and rotation fields have the final form w(x) = −



4 3 2 q 0ℓ 4 8 x 4 − 17 x3 − 93 x 2 + 102 x 192EI ℓ ℓ l ℓ

θ(x) = −



3 2 q 0ℓ 3 32 x3 − 51 x 2 + 6 x 192EI ℓ l ℓ





(b) Find the reaction forces at the supports. The reactions can be found from: M(0) = EIθ′(0) = − q0ℓ 2∕32. To wit,

Q(0) = −EIθ′′(0) = −

51q0ℓ 96

Q(ℓ) = −EIθ′′(ℓ) =

45q 0ℓ 96

qo q ℓ2 − 0 32

ℓ 51 q ℓ 96 0

45 q ℓ 96 0

M 203. A beam of length ℓ and modulus EI rests on two linEI early elastic springs, each of modulus k. The springs ack k crue force in proportion to the amount by which they ℓ stretch. The beam is subjected to a concentrated moment M at the right end. Axial and shear deformations of the beam can be neglected. Assume that motion along the axis of the beam is restrained. What is the virtual-work form of the equilibrium equations? What are the essential and natural boundary conditions? Solve the discrete equations of equilibrium Ka=f using the Ritz method with a quadratic approximation of the displacement field. Describe the error in approximation.

Chapter 7

231



 EIw′′w′′ dx + k  w(0) w(0) + w(ℓ) w(ℓ)  − M w′(ℓ) ℓ

G(w, w) =

0

(b) What are the essential and natural boundary conditions? There are no essential B.C.’s (remember mixed B.C.’s are natural). The natural boundary conditions can be obtained from freebody diagrams of the ends of the beam. (1) EI w′′′(0) + k w(0) = 0

Q(0)

M

M(ℓ)

(2) EI w′′(0) = 0

M(0)

(3) −EI w′′′(ℓ) + k w(ℓ) = 0

Q(ℓ)

(4) M − EI w′′(ℓ) = 0

k w(0)

k w(ℓ)

(c) Solve the discrete equations of equilibrium Ka=f using the Ritz method with a quadratic approximation of the displacement field. Take a quadratic approximation to the displacement field as w(x) = a 0 + a1 x + a2 x 2 ℓ ℓ

w(x) = a0 + a 1 x + a2 x 2 ℓ ℓ

2

2

Let the Ritz approximation be w = a T h(ξ) and w = a T h(ξ) where the base functions (and derivatives) are



2 h(x) = 1, x , x 2 ℓ ℓ





h′(x) = 0, 1 , 2x ℓ ℓ



h′′(x) =

 0, 0, ℓ2  2

Substituting these into the virtual work functional we get the stiffness and force vectors

 EI h′′  h′′ dx + k  h(0)  h(0) + h(ℓ)  h(ℓ) , ℓ

K=

f = M h′(ℓ)

0

Carrying out the integrals we find K = EI3 ℓ

0 0 0

0 0 0

0 0 4

+k

2 1 1

1 1 1

1 1 1

f=M ℓ

0 1 2

(d) Describe the error in approximation. Shear is constant, moment in linear for the exact solution. The quadratic approximation gives constant moment, no shear.

232


M 204. A flexible beam of length ℓ and modulus EI is conrigid EI nected to a rigid beam of length ℓ at a point that rests on k k a roller support. The left end of the rigid beam is restrained x by a linear elastic spring of modulus k. The beam is subjected to an end moment M. Axial and shear deformations ℓ ℓ of the flexible beam can be neglected. What is the virtualwork form of the equilibrium equations? What are the essential and natural boundary conditions for the flexible segment of the beam? Solve the discrete equations of equilibrium Ka=f using the Ritz method with a quadratic approximation of the displacement field. Describe the error in approximation.


w(x) w(ℓ) ℓ w′(0)

Q(0)

w′(0)

R

M(0) M(ℓ)

Q(ℓ)

M

k w(ℓ)

k ℓ w′(0)

The kinematics and freebody diagrams are given in the sketch. The virtual work form of the differential equations

 EI w′′w′′ dx + k ℓ w′(0) w′(0) + k w(ℓ) w(ℓ) − M w′(ℓ) ℓ

G(w, w) =

2

0

The principle of virtual work suggests that if G(w, w) = 0 for all w then equilibrium holds. (b) What are the essential and natural boundary conditions for the flexible segment of the beam? Essential B.C.: w(0) = 0 Natural B.C.’s : From the freebody diagram above. M(0)−k ℓ 2 w′(0) = 0;

M(0) = EI w′′(0)

M(ℓ) = EI w′′(ℓ) = M Q(ℓ)+k w(ℓ) = 0;

Q = −EIw′′′

(c) Solve, the discrete virtual work equations Ka = f for this system using a three-term polynomial Ritz approximation. Let the Ritz approximation be w = a T h(ξ) and w = a T h(ξ) where the base functions (and derivatives) are h(ξ) =  ξ, ξ2, ξ3 , h′(ξ) = 1  1, 2ξ, 3ξ 2 , h′′(ξ) = 12  0, 2, 6ξ  ℓ ℓ The stiffness and force vectors then have the form

Chapter 7

233


 EI h′′  h′′ ℓ dξ + k ℓ h′(0)  h′(0) + k h(1)  h(1) 1

K=

2

0

f = M h′(1) Carrying out the indicated integrations K = EI3 ℓ

0 0 0

0 4 6

0 6 12

+k

2 1 1

1 1 1

1 1 1

1 2 3

f=M ℓ

P 205. A semi-infinite beam (i.e., the beam extends to inEI finity in the positive x direction) of modulus EI rests on an elastic foundation of modulus k. The beam is supk ported at x=0 and is subjected to a concentrated force P x at a distance ℓ from the support. Discuss how you would ℓ solve this problem. Include in your discussion comments on both classical and variational approaches. Assume that Bernoulli-Euler beam theory is adequate to describe the response of this system. The classical equations of a Bernoulli-Euler beam on an elastic foundation are

EIwiv + kw = 0

M(x) = EIw′′

Q(x) = − EIw′′′

Classical Solution: The homogenous solution is given in the form of an exponential w(x) = e λx. Substituting this expression into the governing equation gives the characteristic equation to establish the value of λ

 λ 4 EI+k  eλx = 0

⇒

λ4 = k EI

Let β ≡ 4k∕EI be the positive fourth root of k∕EI. Then the characteristic equation has four roots λ1 = β,

λ 2 = −β,

λ 3 = i β,

λ 4 = −i β

where i is the imaginary unit. Then the homogeneous solution has the form w(x) = a 0 e β x + a1 e −β x + a2 e i β x + a3 e −i β x = b0 sin β x + b 1 cos β x + b2 sinh β x + b3 cosh β x Break the domain into two segments: P

(1)

M 1(0) Q 1 Q 2 Q 1(0)

M 2(0) Q 2(0)

R

(2)

∞

234


There are 4 constants per segment, which gives 8 constants of integration to establish from boundary and continuity conditions.: 2 B.C.’s at x = −ℓ (1) M 1(−ℓ) = 0 (2) Q1(−ℓ) + P = 0 4 B.C.’s at x = 0 (1) w1 = 0 (2) w2 = 0 (3) w1′(0) = w2′(0) (4) M 1(0) = M 2(0) Note that as x → ∞, w2 → 0, w2′ → 0. These two conditions provide the remaining two boundary conditions. These eight conditions would be used to establish the eight constants. The two displacement fields are w1(x) = b10 sin β x + b 11 cos β x + b12 sinh β x + b13 cosh β x w2(x) = b20 sin β x + b 21 cos β x + b22 sinh β x + b23 cosh β x It is possible that a different parameterization of distance in the two domains would simplify the algebra of solving the equations that result from the boundary and continuity conditions (i.e., measure x from the left end of segment 1 and use ξ in segment 2 with origin at the left of segment 2). Variational Approach: The virtual work functional can be written as

G(w, w) =



∞

 EI w′′w′′ + k ww  dx + Pw(−ℓ)

−ℓ

We can use a Ritz approximation in the following form N

w(x) =

 a h (x) i

i=1

i

N

w(x) =

 a h (x) i

i

i=1

where hi(x) must satisfy hi(0) = 0. The moment will have a kink and the shear will have a jump at x = 0, therefore, the FE functions should be used. Also, the solution must approach zero as x → ∞. Use the ordinary cubic Hermitian functions in segment (1) then use decaying functions, e.g., e−βx in the second segment matching value and derivative at x = 0.

Chapter 7

h1


235

1 1

h2 h3

1 1

h4

206. Consider a beam of unit length and circular cross section, fixed at x = 0 and free at x = 1. The axis of the beam ( x3 = x) points along the e3 direction. The origin of the cross sectional coordinates x1 and x2 are at ℓ = 1 the center of the centroid of the section. The internal resultant force is given by the explicit expression Q(x) =  x 2−1  e 1+ x3−1  e2 . Find the applied force q(x) that must be present. Assume that the applied moment m(x) = 0. Find the internal moment field M(x). Find the rotation field θ(x) (a) Find the applied force q(x) that must be present. The equations of equilibrium of a beam assert that Q′ + q = 0 Therefore, the applied load that is implied by the net internal force is q = −Q′ = −2 x e1 − 3 x 2 e 2 (b) Assume that the applied moment m(x)=0. Find the internal moment field M(x). The equilibrium equations governing moment equilibrium for the beam are given by (for m = 0) is M′ + e3 × Q = 0. First we can compute the second term because Q is given e3 × Q =  x2−1  e3 × e 1 +  x3−1  e3 × e 2 =  x2−1  e2 −  x 3−1  e 1 That establishes the rate of change of the moment. M′ = −e 3 × Q =  x 3−1  e1 −  x2−1  e 2 The moment can be computed from integration M =  14 x4−x  e1 −  13 x3−x  e 2 + c

x

236


The boundary condition at x = 1 can be used to establish the constant of integration M(1) =  14 −1  e 1 −  13 −1  e2 + c = 0 c =  4  e1 −  23  e2 3

Thus, the moment field is M = 14  x4−4x+3  e1 − 13  x3−3x+2  e2 (c) Find the rotation field θ(x). The rotation field can be found by integrating the constitutive equation for moment--rotation. Note that I 11 = I 22 ≡ I and J = 2 I, where I is the second moment of the cross--sectional area. The constitutive equation for moment is M = I θ′ where 0 EI 0

EI 0 0

I=

0 0 2GI

Consider the equations by components. Integrating we get EIθ 1′ = 14  x 4−4x + 3 

⇒

5 2 EIθ 1 = x − x + 3x + a 1 4 20 2

EIθ 2′ = 13 −x 3 + 3x−2 

⇒

4 2 EIθ 2 = − x + x − 2x + a2 3 12 2

2 GI θ 3′ = 0

2 GI θ 3 = a3

⇒

Using the boundary condition θ(0) = 0 gives a1 = a2 = a 3 = 0. Therefore, the rotation field is θ(x) = 1 EI

  20x − x2 + 3x4  e +  − 12x + x2 − 2x3  e  5

2

4

1

2

2

P 207. A flexible beam of length 2ℓ and modulus EI EI rests on an elastic foundation of modulus k. The properties have values such that the dimensionless ratio k x kℓ4∕EI = 15. The beam is subjected to a load P at its midpoint. Axial and shear deformations of the flexℓ ℓ ible beam can be neglected. Find the deflection at the middle and ends of the beam using virtual work and the Ritz method with a polynomial approximation. (Note: due to symmetry odd functions—i.e., linear, cubic, etc.—need not be included.)

Chapter 7

237


The virtual work form of the equilibrium equations? G(w, w) =



ℓ

 EIw′′w′′+k ww  dx − Pw(0)

−ℓ

There are no essential B.C.’s. Let the Ritz approximation be w = a T h(ξ) and w = a T h(ξ) where the base functions (and derivatives) are h′(ξ) = 1  0, 2ξ, 4ξ 3  ℓ

h(ξ) =  1, ξ2, ξ4 

h′′(ξ) = 12  0, 2, 12ξ2  ℓ

Substituting these into the virtual work functional we get the stiffness and force vectors K = 2   EI h′′  h′′ + k h  h  ℓ dξ, 1

f = −P h(0)

0

Carrying out the integrals we find 0

K = 2EI ℓ3

0 0

0

1

1 3

1 5

4

8

1 3

1 5

1 7

8

144 5

1 5

1 7

1 9

0

+ 2kℓ

1

f=P

0 0

Noting that kℓ4∕EI = 15 we have

K = 2EI ℓ3

K = EI3 ℓ

0

0

0

0

4

8

0

8

144 5

0

0

0

0

8

16

0

16

288 5

K = EI3 ℓ

+ 30EI ℓ3

+ EI3 ℓ

1

1 3

1 5

1 3

1 5

1 7

1 5

1 7

1 9

30

10

6

10

6

30 7

6

30 7

30 9

30

10

6

10

14

142 7

6

142 7

2742 45

Note that for a two--term solutions we can use the upper two by two corner of K and the upper two terms of f. The two--term solution is 7 Pℓ a1 = − 160 , EI 3

5 Pℓ a2 = 160 EI

3

238


So the approximate displacement is



3 2 7−5 x 2 w(x) = − Pℓ 160EI ℓ



We can evaluate the displacement at the middle and ends to be 3 w(0) = − 7Pℓ , 160EI

3 w(ℓ) = − 2Pℓ 160EI

Chapter 8 The Linear Theory of Plates

208. The three-dimensional rotation tensor Λ without drilling rotation can be obtained from two successive rotations, first ψ about the x1 axis, and then φ about the new x2 axis Λ(ψ, φ) =

cos φ 0

0 1

sin φ 0

1 0

− sin φ

0

cos φ

0 − sin ψ

0 cos ψ

0 sin ψ cos ψ

Compute the product of the two tensors to find Λ. Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT. Show that for small values of the parameters ψ and φ, the tensor can be expressed in the form Λ ≈ I+W, where I is the identity and W is a skewsymmetric tensor. Show, therefore, that when the parameters are small, they can be viewed as the components of the rotation vector θ with ψ = θ 1, φ = − θ 2, and θ 3 = 0 such that W = θ ×. (a) Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT. If Λ−1 = ΛT, then ΛT Λ = I must hold. Let us denote Λ = Q 1 (φ) Q 2 (ψ), where

Q 1 (φ) ≡

cos φ

0

sin φ

0

1

0

0

cos φ

-- sin φ

0 cos ψ

sin ψ

0 -- sin ψ

cos ψ

1 0

Q 2 (ψ) ≡

0

Note that, Q 1 (φ) and Q 2 (ψ) are orthogonal tensors. To wit, cos φ

Q T1 (φ) Q 1 (φ)

=

0 − sin φ

0

1

sin φ

0

0 cos φ

cos φ

0

sin φ

0

1

0

0

cos φ

− sin φ

=

1

0

0

0

1

0

0

0

1

240

Fundamentals of Structural Mechanics 1

Q T2 (ψ) Q 2 (ψ)

=

0

0 0 cos ψ − sin ψ

0

sin ψ

1

0 cos ψ

0

cos ψ

0 sin ψ

=

0 − sin ψ cos ψ

1

0

0

0

1

0

0

0

1

where we have used the trigonometric identity sin2 α+ cos 2 α = 1. It follows that ΛT Λ = (Q 2 Q 1 ) T (Q 2 Q 1 ) = Q T1 Q T2 Q 2 Q 1 = Q T1 I Q 1 = Q T1 Q 1 = I (b) Show that for small values of the parameters ψ and φ, the tensor can be expressed in the form Λ ≈ I + W, where I is the identity and W is a skew-symmetric tensor. Show, therefore, that when the parameters are small, they can be viewed as the components of the rotation vector θ with ψ = θ 1, φ = -- θ 2, and θ 3 = 0 such that W = θ ×. Since sin α ≈ α and cos α ≈ 1, for small values of the angle α 1

0

φ

1

0

1

0

0

0 1

ψ

-- φ

0

1

0

-- ψ

1

Λ(ψ, θ) ≈

1 -- φψ

φ

0

1

ψ

-- φ

-- ψ

1

1

0

φ

0

1

ψ

-- φ

-- ψ

1

=

≈

0

where we have noted that −φψ ≈ 0, since ψ and φ are very small. It follows that 0

0

φ

0

0

ψ

-- φ

-- ψ

0

Λ≈ I+

= I+W

Thus, by letting ψ = θ 1, φ = -- θ 2, we have W=

0

0

-- θ 2

0

0

θ1

θ2 -- θ 1

So, W = θ × with θ 3 = 0.

0

Chapter 8

The Linear Theory of Plates

209. Write out the explicit constitutive expressions for the stress resultants Nαβ = h  λw γ, γ δαβ + m  w β, α +wα, β   Qα = hm  w3, α +Á αβ θ β  M αβ = h12  λÁ αβ Á γη θ η, γ + m Á γβ Á αη θ η, γ +θ β, α   3

For the normal force resultants we get N11 = h  λwγ, γ + m (w1, 1 + w1, 1 )  = h (λ+2m) w1, 1 + λw2, 2 = h Ew1, 1 + λw2, 2 N12 = hm w1, 2 + w2, 1 where E = λ+2 m. Similarly N22 = h Ew2, 2 + λw 1, 1 N21 = hm w1, 2 + w2, 1 = N12 For the shear force resultants we get Q1 = hm w3, 1 + Á 11 θ 1 + Á 12 θ 2 = hm w3, 1 + θ 2 Q2 = hm w3, 2 − θ 1 For the moment resultants we get 3 M 11 = h12 m Á γ1 Á 1η θ η, γ + θ 1, 1 3 = h12 m θ 1, 1 − θ 2, 2 3 M 12 = h12 λÁ γη θ η, γ + m Á γ2 Á 1η θ η, γ + θ 2, 1 3 = h12 λ θ 2, 1 − θ 1, 2 + m  θ 2, 1 + θ 2, 1 3 = h12 E θ 2, 1 − λθ 1, 2

And similarly 3 M 22 = h12 m θ 2, 2 − θ 1, 1 3 M 21 = h12 Eθ 1, 2 − λθ 2, 1

241

242


210. Write out the explicit equilibrium expressions for the Mindlin plate equations h 3  λÁ Á θ , + m  Á Á θ , +θ ,   + hm Á  w , +Á θ  + m = 0 αγ γ 3 α αβ γν ν γα γβ αν γ να β αα αβ β 12 hm  w 3, αα +Á αβθ β, α  + q 3 = 0

Note that β is a free index in this equation. The following simplifications can be made λÁ αβ Á γν θ ν, γα = λÁ αβ ( θ 2, 1α − θ 1, 2α ) m Á γβ Á αν θ γ, να + θ β, αα = m Á γβ ( θ γ, 12 − θ γ, 21 ) + θ β, 11 + θ β, 22 = m  θ β, 11 + θ β, 22 Thus, for β = 1 h3  λÁ ( θ , − θ , ) + m ( θ , + θ , )  + hm Á w , + Á θ  + m = 0 αγ γ α1 2 1α 1 2α 1 11 1 22 α1 3 α 1 12 h3 −λ( θ , − θ , ) + m ( θ , + θ , ) − hm w , −θ  + m = 0 2 12 1 22 1 11 1 22 3 2 1 1 12 And for β = 2 h3  λÁ ( θ , − θ , ) + m ( θ , + θ , )  + hm Á w , + Á θ  + m = 0 αγ γ α2 2 1α 1 2α 2 11 2 22 α2 3 α 2 12 h3 −λ( θ , − θ , ) + m ( θ , + θ , ) − hm w , −θ  + m = 0 2 11 1 21 1 11 2 22 3 1 2 2 12 The equation hm w 3, αα +Á αβ θ β, α + q 3 = 0 yields hm w 3, 11 + w 3, 22 + θ 2, 1 − θ 1, 2 + q3 = 0

211. For a smooth boundary, the expression for the external virtual work for the Kirchhoff-Love plate is WE =

 q w dA +   Q w−M w, n  ds ^

^

s

Ω

s

γ

γ

Γ

Modify the equation to account for point loads in the domain and corners on the boundary. What terms need to be added for a plate that has the shape of a regular polygon with n sides?

Chapter 8

243

The Linear Theory of Plates Γ i+1 ti

Pm

ℓi

ni Γi

x2

si ym Γ i−1

x1

The external virtual work is given by (Eqn. 484 in the text) WE =

 q wdA +   m ⋅ θ + n Q w  ds α

Γ

Ω

α

Γ

Let us divide the boundary Γ into subdomains Γ i which represent each edge of the plate where i = 1,. . ., n. Thus, the external virtual work becomes

 q wdA +     m n

WE =

i=1

Ω

Γi

Γi

 

⋅ θ + niα Qα w ds i

For the Kirchhoff-Love plate, we have from Eqn. (488) in the text

 m Γi

Γi



⋅ θ ds i =

 m Γi

Γi



⋅ ni  ∇w ⋅ t i  ds i −

 m Γi

Γi



⋅ t i  ∇w ⋅ ni  ds i

and by the Lemma on page 260 of the text we have

 v ∇w ⋅ t ds =  i

i

i

Γi

∇ v i w  ⋅ t i ds i −

Γi

 w  ∇v ⋅ t  ds i

i

i

Γi

ℓ

= vi w 0 i −

 w  ∇v ⋅ t  ds i

i

i

Γi

where we have denoted vi ≡ m Γ ⋅ n i . Thus, by defining i

^

Ms ≡ mΓ ⋅ t ^

Qs ≡ nα Q α − ∇( m Γ ⋅ n ) ⋅ t and carrying out the summation, we get WE =

 q w dA +  Q w−M w, n  ds +  w m ^

Ω

Γ

n

^

s

s

γ

γ

i=1

Γi

⋅ ni



ℓi 0

In addition to this, for the concentrated loads Pm , applied at the locations denoted by the position vectors ym in the domain, external virtual work functional can be augmented to

244


WE =

 q w dA +  Q w−M w, n  ds +  w m ^

Ω

n

^

s

s

γ

γ

i=1

Γ

Γi

⋅ ni



ℓi 0

M

−

P

m

w(y m)

j=1

where M denotes the number of concentrated loads.

212. Consider the simply supported square plate of depth h, sides of length π, moduli λ* and m, subjected to a downward uniform load of intensity −qo (shown in Fig. 110). Assume an approximate transverse displacement of the form w(x1, x2) =

∞

∞

 a

mn

sin mx1 sin nx 2

m=1 n=1

with a similar approximation for the virtual displacement. Assume that shear deformations are negligible and compute the coefficients amn using the principle of virtual work. Is it possible to consider an infinite number of terms in the displacement function? How should the solution be modified to solve the problem of a rectangular plate of dimensions ℓ1 × ℓ 2?

The virtual displacement field can be written as w(x1, x2) =

∞

∞

a

kl

sin kx1 sin lx 2

k=1 l=1

The derivatives of the approximate displacement are given by w, 11 = −

∞

∞

 a

mn

m2 sin mx 1 sin nx2

mn

n2 sin mx 1 sin nx2

mn

mn cos mx1 cos nx 2

m=1 n=1

w, 22 = −

∞

∞

 a

m=1 n=1

w, 12 = −

∞

∞

 a

m=1 n=1

with w, 21 = w, 12. The derivatives of the virtual displacements can be computed in the same way. For convenience, let s ij ≡ sin ix 1 sin jx 2 and cij ≡ cos ix 1 cos jx 2. Thus, since

w, αα w, ββ =

∞

∞

∞

kl

k=1 l=1

w, αβ w, αβ =

∞

a  a m=1 n=1

a   a ∞

∞

∞

∞

kl

k=1 l=1

mn ( m

m=1 n=1

mn

2

+n 2 ) ( k2+l 2 ) s mn s kl

 

 (k 2 m2 + l 2n 2 ) s mn s kl + 2 klmn c mn c kl 

Chapter 8

245


The virtual work equation G(w, w) = 0 for all w, where



h3 12

G(w, w) ≡

Ω

λw, αα w, ββ + 2m w,αβ w,αβ − qw dA

yields the following system of equations for the coefficients amn

  a ∞

Ω

∞

mn

m=1 n=1

=−

  λ( m 2+n 2 ) ( k2+l 2 )+2m (k 2 m2+l 2n 2 )  smn s kl +4m klmn c mn ckl   dA

 12hq s o

3

kl

dA

Ω

The integrals can be evaluated with the help of the following identities cos α+β  + cos α−β  = −2 cos α cos β cos α+β  − cos α−β  = 2 sin α sin β To wit,

s

   cos(m+n)x − cos(m−n)x   cos(k+l)x − cos(k−l)x  dx dx π

mn s kl

dA = 14

Ω

1

=

c

dA = 14



0 π2∕4

2

2

1

π

1



1

2

2

1

0

if if

0 π2∕4

m ≠ k OR n ≠ l m = k AND n = l

π

1

0

2

m ≠ k OR n ≠ l m = k AND n = l

  sin kx sin lx dx dx = kl1  (−1) −1 (−1) −1 π

kl dA =

if if

1

   cos(m+n)x + cos(m−n)x   cos(k+l)x + cos(k−l)x  dx dx

=

Ω

0

0

Ω

s

0

π

mn c kl

π

2

1

2

k

l

0

By substituting these results into the virtual work equation above, we get amn  λ( m 2+n 2 ) 2+2m ( m 4+n 4 ) + 4m m 2n 2  = – Thus, we can find the coefficients explicitly as

12 q o 2 2  (−1)m−1  (−1)n−1 mnh 3 π



2

246

Fundamentals of Structural Mechanics 2



  (−1) −1 (−1) −1





amn = –

qo 2 mnD π m 2 + n 2

amn = –

qo 4 mnD π m 2 + n 2

m

n

2

for all m, n that are odd integers.

where D ≡  λ+2m  h 3∕12. (a) Is it possible to consider an infinite number of terms in the displacement function? We can, indeed, consider an infinite number of terms because of the orthogonality properties of the base functions. The equations for the displacement coefficients amn are uncoupled. (b) How should the solution be modified to solve the problem of a rectangular plate of dimensions ℓ1 × ℓ 2? The displacement approximation can be written as w(x1, x2) =

∞

∞

 a

mn

sin

m=1 n=1

mπℓ x  sinnπℓ x  1

2

1

2

with a similar expression for the virtual displacement. Then we follow the same procedure as in part (a) to obtain the coefficients amn .

213. Reconsider the plate in Problem 212. We can compute an approximate solution using the Ritz method with a polynomial basis. Note that, in order to satisfy the boundary conditions, the polynomial must have the form w(x1, x2) = x1 x 2 (x1−π)(x 2−π) a 00+a10 x 1+a01 x 2+a11 x 1 x 2+⋅⋅⋅ (a) Compute the approximate displacement considering only the first term a00 . (b) Unlike the beam, where we can add one term at a time with good results, the next term we might want to add to improve the solution is more complicated for the plate. Since the displacement is a function of two variables, it is possible to introduce an asymmetry if we are not careful in the introduction of new terms. One strategy is to select all terms with the same exponents, as shown below by the dashed lines. 1 x1 x 21 x 31 x 41

x2

x1 x 2 x 21 x 2

x 31 x 2

x 22

x 1 x 22

x 21 x 22

x 32

x 1 x 32

x 42

Resolve the problem using a four-term and a nine-term approximation. (c) A mixed strategy for selecting the basis functions might also be fruitfully employed. Solve the problem with the four-term approximation

Chapter 8

247


w(x1, x2) = sin x1 sin x 2  a 00 + a10 x 1 + a01 x 2 + a11 x1 x 2 

Let g( x 1 , x 2 ) ≡ x 1 x2(x 1−π)(x2−π), and let us define a matrix containing the base functions as hT = g(x1, x2)  1, x 1, x 2, x1 x 2,   and a matrix containing the unknown coefficients as aT =  a00, a 10, a 01, a11,   so that we can express the displacement as w(x1, x2) = aT h(x 1, x 2) Similarly, we can write w = a T h. The derivatives of the displacement field can be computed as w, 11 = aT h, 11 and w, 22 = aT h, 22 and w, 12 = aT h, 12 . It follows that w, αα w, ββ = aT ( h, 11 +h, 22 )( h, 11 +h, 22 ) T a w, αβ w, αβ = aT  h, 11 hT, 11 + h, 22 h T, 22 +2 h, 12 hT, 12  a We can substitute these results into the virtual work expression to get the standard linear equations Ka = f. After some manipulation, the coefficient matrix K is given by K=D

  h,

11 +h, 22

  h, 11 +h, 22  T− (1−ν)  h, 11 hT, 22 + h, 22 h T, 11 −2 h, 12 hT, 12   dA

Ω

1  λ+2 m  h 3, and the right-side matrix f is given by where D ≡ 12

f≡−

 q h dA o

Ω

Carrying out the indicated integrals we find the explicit matrix components as 22 π 6 45

11 π 7 45

11 π 7 45

11 π 8 90

6π 8 35

6 π8 35

3 π9 35

3 π9 35

61 π 10 1050

82 π 8 315

11 π 8 90

41 π 9 315

47 π 9 210

3 π9 35

47 π 10 420

44 π 10 525

7 π 11 100

82 π 8 315

41 π 9 315

3 π9 35

47 π 9 210

44 π 10 525

47 π 10 420

7 π 11 100

176 π 10 1575

47 π 10 420

47 π 10 420

31 π 11 350

31 π 11 350

71 π 12 1050

367 π 10 1575

61 π 10 1050

367 π 11 3150

7 π 11 100

223 π 12 3150

367 π 10 1575

7 π 11 100

367 π 11 3150

223 π 12 3150

19 π 12 225

71 π 12 1050

97 π 13 1575

19 π 12 225

97 π 13 1575

K=D

symm.

194 π 14 3675

248


and π6 36

f T = −qo

π7 72

π7 72

π8 144

π8 120

π8 120

π9 240

π9 240

π 10 400

The different shading represents a different level of approximation. The darkest shading is the one-term approximation, the lighter shading is the four-term approximation, and the unshaded is the nine-term approximation. Clearly, all of the matrix components computed for the lower-order approximations are apropos to the higher-order approximation since the same base functions are reused. The equations can be solved for the coefficients a. The MATHEMATICAT commands needed to form and solve the problem are given below for the nine-term approximation. To get the other approximations simply change the value of nterms. g = x*y*(x - Pi)*(y - Pi) nterms = 4 binom = List [ 1, x, y, x y, x^2, y^2, x^2 y, x y^2, x^2 y^2 ] h = g* Take[binom, nterms] h11 = D[h, {x,2}] h22 = D[h, {y,2}] h12 = D[h, {x,1}, {y,1}] k1 = Outer[Times, h11 + h22, h11 + h22] k2 = Outer[Times, h11, h22] + Outer[Times, h22, h11] k3 = 2 Outer[Times, h12, h12] ki = k1 - (1-n)(k2 - k3) K = Integrate[ki, {x,0,Pi},{y,0,Pi}] f = Integrate[h , {x,0,Pi},{y,0,Pi}] a = Simplify[ Inverse[K].f ]

The results are presented in the following table N

a 00

1

− 5 88

4

− 5 88

9

a 10

a 01

a 11

0

0

0

− 3661 − 4515 77948 155896

− 4515 155896

− 5145 77948π 2

a 20

a 02

4515 155896π 2

4515 155896π 2

a 21

5145 77948π 3

a 12

a 22

5145 77948π 3

− 5145 77948π 4

where all the coefficients amn are normalized by qo ∕D. Note that the coefficient amn is the multiplier of the base function xm1 xn2. As the table indicates, the four-term approximation yields the same result as the one-term approximation. The reason that the new coefficients come out to be zero is that these are the values that best represent the symmetry of the loading. The nine-term approximation allows the development of additional symmetric displacement.

Chapter 8

249


(c) A mixed strategy for selecting the basis functions might also be fruitfully employed. Solve the problem with the four-term approximation w(x1, x2) = sin x1 sin x 2  a 00 + a10 x 1 + a01 x 2 + a11 x1 x 2  We follow the same procedure as in part (a) with g( x 1 , x 2 ) ≡ sin x 1 sin x2. The first line of the MATHEMATICAT script needs to be changed to the line g = Sin[x]*Sin[y]

also, the number of terms in the approximation must be specified as nterms=1 or as nterms=4 as appropriate. Carrying out the indicated integrals gives the components of the coefficient matrix K π3 2

π3 2

π4 4

π2 (9+2π2) 6

π4 4

π3 (9+2π2) 12

π2 (9+2π2) 6

π3 (9+2π2) 12

π2

K=D

symm.

π6 − 5 π2+π 4+ 9 4

and the right-side matrix as 4

f T = −qo

2π

π2

2π

The results are presented in the following table N

a 00

1

4 π2

4

4 π2

a 10

a 01

a 11

0

0

0

where, again, all the coefficients aij are normalized by qo ∕D. As the table indicates, the four-term approximation again yields the same result as the one approximation.

214. Consider a square plate of depth h, sides of length ℓ, moduli λ* and m, subjected to a downward uniform load of intensity −qo . Assume that shear deformations are negligible (i.e., Kirchhoff-Love plate theory is applicable). Find an approximate displacement for the following boundary conditions

−qo

x2

x1

250


(a) x2 ℓ

(b) x2

ss

ℓ

fx

ss

ss

(c)

ss

x2 ℓ

fr

fx

fx

fr

x1 0 x1 0 x1 ℓ fx ℓ ss fx ℓ where the designation ss indicates a simple support, fx indicates a fixed edge, and fr indicates a free edge. 0

The virtual work functional is given by the expression G(w, w) ≡



h3 12

Ω

λw, αα w, ββ + 2m w,αβ w,αβ − qw dA

Observing that λ = νC∕(1−ν 2) and 2m = (1−ν) C∕(1−ν 2), the virtual work functional can be put into the form G(w, w) ≡ D

  w,

αα w, ββ +

(1−ν)  w, αβ w, αβ − w, αα w, ββ   dA −

Ω

 qw dA Ω

where D = Ch 3∕12(1−ν2). This form will prove convenient for organizing the computations for this problem. For the purposes of this problem we shall use a one-term expansion. Thus, let us define the displacement as w = a g( x1 , x2 ) where g( x 1 , x 2 ) = h 1 (x1 ) h2 (x 2 ) is the product of one-dimensional functions. The derivatives can be computed as w, 11 = a g, 11

w, 22 = a g, 22

w, 12 = a g, 12

It follows that w, αα w, ββ = a a  g, 11 + g, 22  2 w, αβ w, αβ − w, αα w, ββ = 2 a a  g, 12 g, 12 − g, 11 g, 22  By substituting these results into the virtual work expression we get the standard linear equations Ka = f, where the coefficient K is K=D

   g,

11

+ g, 22 2 − 2(1−ν)  g, 11 g, 22 −  g, 12  2  dA

Ω

and the right-side f is f≡−

 q g dA Ω

Chapter 8

251


For part (a), the essential boundary conditions are w(0, x2 ) = w(π, x 2 ) = 0 w(x 1 , 0 ) = w( π, x 2 ) = 0 w′(x 1 , 0 ) = w′( 0, x2 ) = 0 Thus, an appropriate choice of base functions is h1 = ξ31 − ξ 21

h2 = ξ32 − ξ 22

where ξi ≡ xi∕π. The MATHEMATICAT commands needed to form and solve the problem are given below h1 = (x1/Pi)^3 - (x1/Pi)^2 h2 = (x2/Pi)^3 - (x2/Pi)^2 g = h1*h2 g11 = D[g,{x1,2}] g22 = D[g,{x2,2}] g12 = D[g,{x1,1},{x2,1}] q = - qo ki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2)) K = Integrate [ ki, {x1,0,Pi},{x2,0,Pi}] f = Integrate [q*g, {x1,0,Pi},{x2,0,Pi}] a = Simplify[ f/K ]

It follows that K=

176 D 1575 π 2

f =−

q o π2 144

⇒

a =−

175 q o π4 2816 D

For part (b), the essential boundary conditions are w(0, x2 ) = 0 w(x 1 , 0 ) = w( π, x 2 ) = 0 Thus, an appropriate choice of base functions is h1 = x1∕π and h2 = sin x 2. The first two lines of the MATHEMATICAT script need to be changed to h1 = x1/Pi h2 = Sin[x2]

It follows that

6

2 K = D π +1−ν



f = −q o π

⇒

a = −

For part (c), the essential boundary conditions are w(0, x2 ) = 0 w(x 1 , 0 ) = w( x1 , π ) = 0 w′(x 1 , 0 ) = w′(x1 , π ) = w′( 0, x 2 ) = 0

6q o π D ( π2+6−6ν)

252


Thus, an appropriate choice of base functions is h1 =  x1∕π  2 and h2 = 1− cos 2 x 2. The first two lines of the MATHEMATICAT script need to be changed to h1 = (x1/Pi)^2 h2 = 1 - Cos[x2]^2

It follows that



3 2π 2 K = D 4 − 2ν + 2 + 5 3 2π



f =−

qo π 2 6

The displacement coefficient is, therefore a = −

5q o π 4  D 45+40π2−60π 2ν+12π4 

215. Solve any of the above variations on the square-plate problem with any combination of the following differences (a) Consider a rectangular plate of dimension ℓ1 × ℓ 2. (b) Consider a load form different from a uniform load.

(a) Consider a rectangular plate of dimension ℓ1 × ℓ 2. Let us consider the plate shown in the figure below to be subjected to a uniform downward load of magnitude qo x2 ss

ℓ2

ss

fx

fx

0

ℓ1

x1

where the designation ss indicates a simple support and fx indicates a fixed edge. For the purposes of this problem we shall use a one-term expansion. Thus, let us define the displacement as w = a g( x1 , x2 ) where the function g( x 1 , x 2 ) = h 1 (x1 ) h2 (x 2 ) is the product of one-dimensional base functions. Now, the suitable base functions for this case are 3

h1 =

ℓx  − ℓx  1

1

1

1

2

3

h2 =

ℓx  − ℓx  2

2

2

2

2

Performing the integrals associated with K and f as in Problem 214 we get K=

 

4D 15 γ+ 1γ 1575ℓ 1ℓ2

2

 −16

f =−

q o ℓ 1ℓ 2 144

Chapter 8

253


where γ ≡ ℓ 2∕ℓ1 is the aspect ratio of the plate. Solving Ka = f for a we get a =−

175 qo ℓ21ℓ 22 64 D

2

15 γ+ 1γ  −16

−1

Note the symmetry in the lengths ℓ1 and ℓ2. Note also that, when ℓ1 = ℓ2 = π, we get the same result as we did in Problem 214(a). The MATHEMATICAT commands needed to form and solve the problem are given below h1 = (x1/L1)^3 - (x1/L1)^2 h2 = (x2/L2)^3 - (x2/L2)^2 g = h1*h2 g11 = D[g,{x1,2}] g22 = D[g,{x2,2}] g12 = D[g,{x1,1},{x2,1}] q = - qo ki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2)) K = Integrate [ ki, {x1,0,L1},{x2,0,L2}] f = Integrate [q*g, {x1,0,L1},{x2,0,L2}] a = Simplify [f/K]

(b) Consider a load form different from a uniform load. x2

-- q min

-- q max

x1

Let us again consider the same plate, but define the loading as 2

  1− xπ 1− xπ 

4 q(x) = −q max + qmax−q min x1 x2 π

1

2

which is a parabolic loading that takes on its minimum value qmin at the center of the plate. Thus, f≡−

 q(x) g(x) dA = −π q144 + q 2

Ω

max

225

Since K does not change, the solution is given by a=

7 π4 f  9 qmax+16 qmin  =− K 2816 D

Note that, if qmax = qmin = qo , then a=−



min−q max

7 π4 175 qo π 4  9 qo+16 q o  = − 2816 D 2816 D

254


which is what we got in Problem 214(a). The MATHEMATICAT commands needed to form and solve this problem are as follows: h1 = (x1/Pi)^3 - (x1/Pi)^2 h2 = (x2/Pi)^3 - (x2/Pi)^2 g = h1*h2 g11 = D[g,{x1,2}] g22 = D[g,{x2,2}] g12 = D[g,{x1,1},{x2,1}] q = -qmax + qmin*(4/Pi)^2*x1*x2*(1-x1/Pi)*(1-x2/Pi) ki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2)) K = Integrate [ ki, {x1,0,Pi},{x2,0,Pi}] f = Integrate [q*g, {x1,0,Pi},{x2,0,Pi}] a = Simplify[ f/K ]

216. Solve the square ( π × π), simply supported plate under uniform load (shown in Fig. 110) considering shearing deformations. Note: You can still neglect the in-plane problem because it is uncoupled from the bending problem. Use the Ritz method. Note that w(x1, x2) must vanish on the boundary, but that θ 1(x 1, x 2) and θ 2(x 1, x 2) are not restricted. Let w(x1, x2) = a1 sin x 1 sin x 2. Select the functions for θ 1 and θ 2 so that the shearing deformation is linear, that is θ 1 = w, 2 + b 1 2x2−π ,

θ 2 = −w, 1 + b2 2x1−π 

What is the result of making such an assumption? Noting that the tangential moment must vanish on the boundary, is there a means of finding a better displacement function? Can we use statical considerations to improve the approximation of the rotation field?

The virtual work functional is given by G(w, w, θ, θ) ≡

 Q ⋅ ε α

α

+ Mα ⋅ κ α − q ⋅ w  dA

Ω

Neglecting the in-plane actions, the first term is Q α ⋅ ε α = Q α  w, α +ε αη θ η  = hm w, α +ε αβ θ β   w, α +ε αη θ η  However, for the present problem, we are given θ β = ε βν w, ν +η β where the functions η β are given as η 1 = b1  2x 2−π  and η 2 = b2  2x 1−π  . Thus, since w, α +ε αβ θ β = w, α +ε αβ ε βν w, ν +ε αβη β = w, α −δαν w, ν +ε αβη β = ε αβη β and noting that ε αβ ε αν η β η ν = δβν η β η ν = η β η β, we get Q α ⋅ ε α = hm η β η β

Chapter 8

255


Similarly, neglecting the in-plane actions, the second term of the virtual work functional can be written as Mα ⋅ κ α = M αβ θ β, α. The components of the bending moment field are given by 3 M αβ = h12 λÁ αβ Á γη θ η, γ + m Á γβ Á αη θ η, γ + θ β, α 3 = h12  λ( θ β, α −θ α, β ) + m (2 θ β, α −δαβ θ ν, ν ) 

where we have used the identity Á αβ Á γη = δ αγ δβη−δ αη δ βγ. Thus, the internal work associated with bending is given by M αβ θ β, α = D  θ β, α θ β, α − ν θ α, β θ β, α − (1−ν) θ α, α θ β, β  Let x1 ≡  2x2−π  e2 , x2 ≡  2x1−π  e3 , and aT =  a1, b1, b2  so that η β = aTx β. Further more let w ≡ sin x1 sin x2 e1 and let y1 ≡ w, 2 +x1 and y2 ≡ −w, 1 +x2 so that the rotation field can be computed as θ α = aTy α. The derivatives of the rotation fields can be computed as θ α , β = a Tyα , β. The principle of virtual work yields the standard linear system of equations Ka = f where the matrix K can be constructed from a bending component and a shear component as K = K b+K s where Kb = D

  y , y , − ν y , y , − ( −ν) y , y ,  dA β α

T β α

α β

T β α

1

α α

T β β

Ω

K s = hm

 x x dA T α α

Ω

The load vector can be computed as f=−

 q wdA o

Ω

Evaluating the integrals, we get the system matrices π2

−8 (1+ν ) π2  12+γ π 2  3

K=D

8 (1+ν )

4

−4π2ν

f = −q o

π2  12+γ π 2  3

symm.

0 0

where γ ≡ mh∕D = 6(1−ν)∕h 2 is the ratio of shear to bending modulus. Solving the linear system of equations yields

  96 q 1+ν   = −b b =− D 384(1+ν) + 12π (1+ν)–γπ

a1 = −

1

4π 2 q o 12(1−ν) + γ π 2 D 384(1+ν)2 + 12π 2(1+ν)–γπ 6 o

2

2

6

2

256


Note that in the thin plate limit, as γ → ∞, we get lim a 1 = −

γ→∞

4q o π 2D

and b1 = b2 = 0, which is the exact solution for a Kirchhoff-Love plate. The MATHEMATICAT commands needed to form and solve this problem are as follows: z1 = { 0, 2 x2 - Pi , 0} z2 = { 0, 0, 2 x1 - Pi } w = { Sin[x1]Sin[x2], 0, 0 } y1 = D[w, x2] + z1 y2 = -D[w, x1] + z2 y11 = D[y1,x1] y12 = D[y1,x2] y21 = D[y2,x1] y22 = D[y2,x2] Ks = M1 ( Outer[Times, z1, z1] + Outer[Times, z2, z2] ) B11 = Outer[Times,y11,y11] B12 = Outer[Times,y12,y12] B21 = Outer[Times,y21,y21] B22 = Outer[Times,y22,y22] C12 = Outer[Times,y12,y21] C21 = Outer[Times,y21,y12] K1 = B11 + B12 + B21 + B22 K2 = B11 + C12 + C21 + B22 K3 = Outer[Times,y11 + y22, y11 + y22] Kb = M2 ( K1 - n K2 - (1-n) K3 ) K = Integrate[Ks + Kb, {x1,0,Pi},{x2,0,Pi}] f = Integrate[w, {x1,0,Pi},{x2,0,Pi}] a = Simplify[Inverse[K].f]

Alternatively, we can try the displacement given by w(x1, x2) = a1 sin x 1 sin x 2 θ 1 = w, 2 + b 1 sin x1 cos x2 θ 2 = –w, 1 + b 2 cos x 1 sin x2 Which leads to the matrices π2 2

−

π2 2

π2  1+γ  4

−

π2 4

π2

K=D

4

f = −q o

π2  1+γ  4

symm.

0

Solving these we get a1 = −



4q o 1+ 2γ π 2D



0

b1 = −

8q o = −b2 π 2mh

Chapter 8

257


Note that, for both solutions, deflection at the center of the plate is w(π∕2, π∕2) = a1 . In the limit as the thickness of the plate goes to zero γ → ∞ and lim a 1 = −

γ→∞

4q o π 2D

and b1 = b2 = 0, which is the same as the center deflection of the exact solution of a simply supported Kirchhoff-Love Plate under uniform loading. The MATHEMATICAT commands needed to form and solve this problem are the same as above except for the first two lines which should be replace by z1 = { 0, Sin[x1]Cos[x2] , 0} z2 = { 0, 0, Cos[x1]Sin[x2] }

217. Consider a square ( π × π) plate subjected to a uniform transverse load q. The plate is simply supported along the edges x2 = 0 and x2 = π with any boundary conditions along the other two edges. Assume that shear deformations are negligible. According to the method of Kantorovich let us assume that the real and virtual displacement fields can be represented as w(x1, x2) = W(x1) sin x2,

x2

ss

π any 0

w(x1, x2) = W(x 1) sin x2

any ss

π

x1

where W(x1) and W(x1) are unknown functions. (a) Substitute these functions into the virtual-work functional G(w, w) ≡

  D  ν w,

αα w, ββ +

 1−ν  w, αβ w, αβ  − qw  dA

Ω

and show that W(x1) must satisfy the ordinary differential equation 4q πD (b) Verify that the following function satisfies the above equation W IV − 2W′′ + W =

4q +  a1+a 2 x1 cosh x 1 +  a 3+a4 x 1 sinh x1 πD (c) What are the possible boundary conditions for W(x1)? Find the constants of integration for the case where the plate is simply supported on all four edges. W(x1) =

(a) The second derivatives of the displacement function are given by w, 11 = W′′ sin x2

w, 22 = −W sin x2

w, 12 = W′ cos x2

with w, 21 = w, 12. Thus, we have w, αα w, ββ =  W′′−W   W′′−W  sin 2 x 2 w, αβ w, αβ =  W′′ W′′+W W  sin2 x2 + 2 W′ W′ cos 2 x 2

258


Substituting these into the virtual work functional yields h3 12

 λ  W′′−W   W′′−W  + 2m  W′′ W′′+W W   sin x  4m W′ W′ cos x dA =  qW sin x dA 2

2

dA

Ω

h3 12

2

2

2

Ω

Ω

The integrals above are

 sin x dx =  π

2

2

π

2

0

0

 sin x dx = 2 π

cos x 2 dx 2 = π 2 2

2

2

0

Thus, upon evaluating the integrals with respect to x2 we get h3 12

 λ  W′′−W  W′′−W  + 2m  W′′ W′′+W W +2 W′ W′  dx = π4  qW dx π

π

1

0

1

0

We can integrate by parts to shed the derivatives from the W terms as follows



π

π

 W′′−W W′′ dx 1 =  W′′−W W′− W′′′−W′ W  + 0

0



π

 W IV−W′′  W dx1

0

 W′′ W′′ dx =  W′′ W′ − W′′′ W  +  π

π

π

1

0

0

W IV W dx1

0

 W′ W′ dx = W′ W  –  π

π

π

1

W′′ W dx 1

0

0

0

Using these results, noting that λ∕(λ+2m) = ν and 2m∕(λ+2m) = 1−ν, we get

 W  W −2W′′+W  dx + Dν   W′′−W  W′− W′′′−W′  W  4  qW dx D  1−ν   W′′ W′−W′′′ W+2 W′ W  = π π

D

IV

1

π 0

0

π

π

1

0

0

where D =  λ+2m  h3∕12. Rearranging the boundary terms, we get

 W W −2W′′+W  dx + ( W′′−νW) W′–( W′′′+ ν−2  W′ ) W  = 4  qW dx πD π

IV

0

1

π 0

π

1

0

Invoking the fundamental theorem of the calculus of variations, we can conclude that the function W must satisfy the differential equation

Chapter 8

259


W IV − 2W′′ + W =

4q

πD

We shall see in detail in part (c) what are the implications of the boundary terms. (b) Verify that the following function satisfies the above equation W(x1) =

4q +  a1+a 2 x1 cosh x 1 +  a 3+a4 x 1 sinh x1 πD

The derivatives of the displacement field are W′ = a 2 cosh x1 + a 4 sinh x1 +  a 1+a2 x 1 sinh x1 +  a3+a 4 x1 cosh x 1 W′′ = 2a 2 sinh x1 + a 4 cosh x1 + W −

4q πD

(*)

W′′′ = 2a 2 cosh x1 + a 4 sinh x1 + W′ W IV = 2a2 sinh x 1 + a4 cosh x 1 + W′′

(* *)

Subtracting (*) from (**) we arrive at the result W IV − 2W′′ + W =

4q πD

(c) What are the possible boundary conditions for W(x1)? Find the constants of integration for the case where the plate is simply supported on all four edges. Since the normal vector at the boundaries at x1 = 0 and x1 = π are e1 and −e1 respectively, the four possible boundary conditions are as follows (see page 259 in the text): w = W sin x 2 = 0 w, 1 = W′ sin x2 = 0 w, 11 + ν w, 22 =  W′′ − νW  sin x2 = 0 w, 11 + ν w, 22 =  W′′′ +  ν−2  W  sin x2 = 0

vanishing displacement vanishing tangential rotation vanishing tangential moment vanishing effective shear

Since these equations must hold for all x2 , we have W=0 W′ = 0 W′′ − νW = 0 W′′′ +  ν−2  W = 0

vanishing displacement vanishing tangential rotation vanishing tangential moment vanishing effective shear

Note that, on any edge, either the displacement or the effective shear is prescribed, but not both; either the tangential rotation or the tangential moment is prescribed but not

260


both. Note that, the boundary terms in part (a) vanish from the functional if the the function W is made to satisfy the homogeneous essential boundary conditions. For a plate that is simply supported on all four edges, we have the conditions W(0) = W(π) = 0 W′′(0) − νW(0) = 0 ⇒ W′′(0) = 0 W′′(π) − νW(π) = 0 ⇒ W′′(π) = 0 For simplicity, let us write the displacement function as follows W(x1) =

4q  1 +  a1+a 2 x1 cosh x 1 +  a 3+a4 x 1 sinh x1  πD

W′′(x1) =

4q  2a2 sinh x 1 + 2a4 cosh x 1 − 1  + W(x 1) πD

Thus, W(0) = 0

⇒

1 + a1 = 0

W′′(0) = 0

⇒

2a4 − 1 = 0

W′′(π) = 0

⇒

2a2 sinh π + 2a4 cosh π − 1 = 0

W(π) = 0

⇒

1 + −1+a2π  cosh π +  a 3− π2 sinh π = 0

These equations can be solved to give the constants a1 = −1

a2 = − 12 tanh π2

a3 = − 14  π− sinh π  sech 2 π2

a4 = 12

218. There are many possibilities available within the context of the method of Kantorovich. In general, we use an approximation of the form w(x1, x2) = W(x1) φ(x 2), where φ(x2) is a known function. In Problem 217, the choice was φ(x2) = sin x 2. A more general approach is to use the deflected shape of a simply supported beam subjected to the applied loading (assuming that it does not vary with x2 ). Find the general expression for the coefficients a, b, c, and q^ of the differential equation a W IV − b W′′ + c W = q^ that results from applying the method of Kantorovich with the known function φ(x2). Express the coefficients as integrals of φ(x2). Can Kantorovich’s method be applied to boundary conditions other than simple supports along parallel edges? How would the approach change to accommodate more general boundary conditions?

Chapter 8

261


The derivatives of the displacement functions are given by w, 11 = W′′ φ

w, 22 = W φ′′

w, 12 = W′ φ′

Thus, we can compute w, αα w, ββ =  W′′ φ−W φ′′   W′′ φ−W φ′′  2

w, αβ w, αβ = W′′ W′′ φ2 + W W φ′′  + 2 W′ W′  φ′ 

2

Substituting these into the virtual work functional yields

 λ  W′′ W′′ φ +W φ φ′′ + W W′′ φ φ′′ +W φ′′    dA 2

2

Ω

+

 2m  W′′ W′′ φ + W W φ′′  +2 W′ W′φ′  dA =  qW φ dA 2

2

12 h3

2

Ω

Ω

Since the function φ(x2) is specified a priori, the following integrals can be evaluated

 φ dx π

α1 ≡

2

2

0

 φ φ′′ dx

α4 ≡

2

2

2

0

π

α3 ≡

  φ′  dx π

α2 ≡

0



α ≡  φ dx π

π

 φ′′  2 dx 2

5

2

0

0

Noting that λ∕(λ+2m) = ν and 2m∕(λ+2m) = 1−ν, and carrying out the integrals with respect to x2 , we get

 D W′′ α W′′+ν α W  + W′ 2 −ν  α W′ π

1

1

3

2

0

+ W  α 4 W+να 3 W′′  − α 5 q W  dx 1 = 0 Defining a new set of parameters γi ≡ α i ∕α 1 and integrating by parts yields

 W  W +2ν γ +(ν−1)γ  W′′+ γ W dx + B.T. = γD  qW dx π

π

IV

3

2

4

5

1

0

1

0

where B.T. stands for the boundary terms. To wit,



B.T. ≡  W′′−ν γ3 W  W′ +  W′′′ + ν γ 3+2(ν−1)γ 2 W′  W



π 0

By identification, then, the coefficients of the problem statement have the form a = 1,

b = −2 νγ 3 + 2(1−ν)γ 2 ,

c = γ4 ,

q^ =

q γ D 5

262


Observe that, for φ(x2) = sin x 2 a = c = 1,

b = −2 ,

q^ =

4q πD

and the boundary terms are B.T. ≡   W′′−ν W  W′ +  W′′′ +  ν−2 W′  W



π 0

Substituting these results into the equation above, we get the same result as Problem 217. Kantorovich’s method can be applied to other boundary conditions, but the function φ(x2) must be chosen to satisfy the boundary conditions.

Chapter 9 Energy Principles and Static Stability

219. Consider a one-dimensional rod of length ℓ and modulus C subjected to a body force b(x) and a traction τo at the right end. The left end has a prescribed displacement of u(0) = u o. The Hu-Washizu energy functional for the rod is given in terms of the independent variables σ(x) (stress), u(x) (displacement), and Á(x) (strain) as

τo x ℓ

  CÁ −b u −σ  Á−u′  dx − τ u(ℓ) + σ(0)  u(0)−u  ℓ

H(u, σ, Á) ≡

b(x)

uo

1 2

2

o

o

0

Show that by taking the directional derivative of H in the direction of variations of each of the variables, i.e., DH(u, σ, Á) ⋅ (u, σ, Á) = d  H( u+α u, σ+ασ, Á+α Á) α=0 dα and setting the result equal to zero (to find the extremum), all of the classical governing equations for the one-dimensional bar result from applying the fundamental theorem of the calculus of variations σ′ + b = 0 Á − u′ = 0 σ − CÁ = 0

x ∈ (0, ℓ)

σ(ℓ) − τo = 0 u(0) − uo = 0

Note that the fields that appear in the functional are each functions of x themselves.

The directional derivatives of the functional in the directions of a variation in each of the argument functions are

264


  σu′−b u  dx − τ u(ℓ) + σ(0) u(0) ℓ

D1H ⋅ u =

0

 − σ′+b  u dx +  σ(ℓ)−τ  u(ℓ) ℓ

=

0

 −σ Á−u′  dx ℓ

D2H ⋅ σ =

+ σ(0)  u(0)−u o

0

 CÁ−σ  Á dx ℓ

D3H ⋅ Á =

0

where the notation D1H ⋅ u means take the directional derivative by considering u to be the argument holding σ and Á constant (i.e., like the partial derivative). Setting each of these expression equal to zero and invoking the fundamental theorem of the calculus of variations, considering the fields u, σ, and Á as independent gives the desired result.

220. Find the energy functional E(u, w, θ) for a Timoshenko beam.

The energy functional for the Timoshenko beam is

  EA  u′  + ℓ

E(u, w, θ) =

1 2

2

1 EI  θ′  2 + 12 GA  w′−θ  2 − p u − q w 2

− mθ dx

0

plus boundary terms if there are applied loads at the end points. One can demonstrate that this is the correct functional by taking the directional derivative to get the virtual work functional and then applying the Fundamental Theorem of the Calculus of Variations to show that minimization of the energy is equivalent to the classical differential equations.

221. Find a Hu-Washizu energy functional for a simply supported Bernoulli-Euler beam of length ℓ and modulus EI subject to a transverse load q(x). The appropriate field variables are the transverse displacement w(x), the moment M(x), and the curvature κ(x). Show that the extremum of the energy functional with respect to the three field variables gives the classical equations of Bernoulli-Euler beam theory. How must the functional change if the beam is fixed at x = 0 and pinned at x = ℓ?

The functional can be created by analogy with the little boundary value problem with the following identifications u → w, σ → M, and Á → κ.

Chapter 9

  EIκ − q w − Mκ−w′′  dx + M′w  ℓ

H(w, M, κ) =

265

Energy Principles and Static Stability 1 2

0

2

ℓ 0

Differentiating with respect to w we get

  Mw′′ − q w  dx + M′w  ℓ

D1H ⋅ w =

0

ℓ 0

  −M′w′ − q w dx + Mw′  + M′w  ℓ

=

0

  M′′−q  w dx + Mw′  ℓ

=

0

ℓ

ℓ

0

0

ℓ 0

Setting the expression equal to zero to find the extremum we find, from the fundamental theorem of the calculus of variations, that M′′−q = 0 in [0, ℓ] and that M(0) = 0 and M(ℓ) = 0. Differentiating with respect to M we get

 M  κ−w′′  dx + M′w  ℓ

D2H ⋅ M = −

0

ℓ 0

Setting the expression equal to zero to find the extremum we find, from the fundamental theorem of the calculus of variations, κ = w′′ in [0, ℓ] and w(0) = 0 and w(ℓ) = 0. Finally, differentiating with respect to κ we get

 EIκ − M  κ dx ℓ

D3H ⋅ κ =

0

Setting the expression equal to zero to find the extremum we find, from the fundamental theorem of the calculus of variations, M = EIκ in [0, ℓ]. For the beam that is fixed at x = 0 and pinned at x = ℓ we must add a term to account for the additional essential boundary condition at x = 0. To wit,

  EIκ − q w − Mκ−w′′  dx + M′w  − M(0) w′(0) ℓ

H(w, M, κ) =

0

1 2

ℓ

2

0

222. The Hellinger-Reissner energy functional for a three-dimensional hyperelastic solid body B with boundary Ω is given by R(u, S) =

  S ⋅ ∇u−b ⋅ u−U(S)  dV −  τ ⋅ u dA −  ^

B

Ωt

Ωu

Sn ⋅  u−u^  dA

266


where u is the displacement field, S is the stress field, τ^ is the prescribed traction over Ωt (the portion of the boundary where tractions are prescribed), u^ is the prescribed displacement over Ωu (the portion of the boundary where displacements are prescribed), b is the body force, and U(S) in the stress-energy function. What do the necessary conditions for an extremum imply? (Hint: take the directional derivative of the functional in the directions of u and S, and apply the fundamental theorem of the calculus of variations.) The directional derivative of R in the direction of u is D1R ⋅ u =

  S ⋅ ∇u − b ⋅ u dV −  t ⋅ u dA −  ^

Ωt

B

=

Sn ⋅ u dA

Ωu

  div S u − div S ⋅ u − b ⋅ u dV −  t ⋅ u dA −  T

B

^

Ωt

= −

n ⋅ STu dA

Ωu

  div S + b ⋅ u dV +   Sn − t  ⋅ u dA ^

Ωt

B

From the fundamental theorem of the calculus of variations we get the conclusions that ^ div S+b = 0 in B and Sn = t on ∂Ωt. The directional derivative of R in the direction of S is D2R ⋅ S =

  S ⋅ ∇u − ∂U∂S ⋅ S  dV − 

=

Sn ⋅  u − u^  dA

B

Ωu

  ∇u − ∂U∂S  ⋅ S dV − 

Sn ⋅  u − u^  dA

Ωu

B

From the fundamental theorem of the calculus of variations we get the conclusions that u = u^ on ∂Ωu , and that E ≡

1 ∇u+∇uT  2

= ∂U ∂S

Note that because the expression in parentheses (i.e., ∇u−∂U∕∂S) is contracted with S, which is symmetric, the antisymmetric part of it is annihilated. Since S is symmetric one can conclude that it is appropriate to take only the symmetric part of ∇u. This issue can be eliminated if 12  ∇u+∇uT  is used in the original functional R in place of ∇u.

223. Show that the energy functional for the Kirchhoff-Love plate, given in Eqn. (513), can be expressed in the equivalent form

Chapter 9 E(w) =



1 2

267

Energy Principles and Static Stability

D   w, 11 +w, 22  2 − 2  1−ν   w, 11 w, 22 − w, 212   − qw  dA

Ω

where D is the plate modulus and ν is Poisson’s ratio. The term w, 11 w, 22 −w, 212 is an approximation of the Gaussian curvature of the deformed reference surface of the plate.

The expression for the energy in Eqn. (513) is E(w) =

  Dν w, 1 2





2

αα

 − (1−ν) w, αβ w, αβ − qw dA

Ω

Expanding the terms inside the parentheses we get 2

(⋅) ≡ ν w, αα + (1−ν) w, αβ w, αβ = ν w, 211 +2w, 11 w, 22 +w, 222 + (1−ν)w, 211 +2w, 212 +w, 222 = w, 211 +w, 222 + 2νw, 11 w, 22 + 2(1−ν)w, 212 = w, 211 +w, 222 + 2w, 11 w, 22 − 2(1−ν)w, 11 w, 22 + 2(1−ν)w, 212 2 = w, 11 +w, 22 − 2(1−ν)w, 11 w, 22 − w, 212 

which demonstrates the equivalence.

224. Show that the energy functional for a three-dimensional linear elastic solid is E(u) =

  λ div u 1 2

2

+ 12 m  ∇uT+∇u  ⋅ ∇u − b ⋅ u  dV −

 t ⋅ u dA Ωt

B

Show also that the extremum of the energy gives the same equations as the principle of virtual work.

The virtual work functional has the form G(u, u) =

 λdiv u div u + m ∇u +∇u ⋅ ∇u − b ⋅ u dV −  t ⋅ u dA T

Ωt

B

First check symmetry of the virtual work functional DG(u, u) ⋅ u^ =

 λdiv u div u + m ∇u +∇u ⋅ ∇u  dV ^

B

Noting that u^ j, i ui, j − u j, i u^ i, j = 0 we have

^T

^

268


 m ∇u

^T

DG(u, u) ⋅ u^ − DG(u, u^ ) ⋅ u =

⋅ ∇u − ∇u T ⋅ ∇u^  dV = 0

B

Thus, an energy functional exists. Applying Vainberg’s theorem, noting the linearity of differentiation, i.e., div (t u) = t div u and ∇(t u) = t ∇u, we find

 G(t u, u) dt 1

E(u) =

0

Substituting into the virtual work functional we get

   t λ div u + t m ∇u +∇u ⋅ ∇u − b ⋅ u  dV −  t ⋅ u dA dt 1

E(u) =

2

0

T

Ωt

B

Carrying out the indicated integration with respect to the scalar t we obtain E(u) =



1 2

λ div u 2 + 12 m ∇uT+∇u ⋅ ∇u − b ⋅ u dV −



t ⋅ u dA

Ωt

B

225. Show that the energy functional for a Bernoulli-Euler beam on an elastic foundation can be expressed in the form

 ℓ

E(w) =

 dx

1 EI  w′′  2 + 12 kw 2 − qw 2

0

where EI is the flexural modulus of the beam, k is the modulus of the foundation, and q is the transverse load.

The virtual work functional can be obtained by taking the directional derivative of the energy functional in the direction of the function w G(w, w) =   EI w′′ w′′ + k ww − qw  dx =   −EI w′′′ w′ + k ww − qw  dx + EIw′′w′  ℓ

0

ℓ

0

  EI w ℓ

=

0

iv



ℓ

ℓ 0

+ k w − q  w dx + EIw′′w′ − EIw′′′w 0



ℓ 0

Equilibrium is satisfied if G(w, w) = 0 for all w. Applying the fundamental theorem of the calculus of variations we obtain the classical equilibrium equations for the beam on elastic foundation

Chapter 9

269

Energy Principles and Static Stability EI w iv + k w − q = 0

Either EI w′′ = 0 or w′ = 0

at x = 0 and x = ℓ

Either EI w′′′ = 0 or w = 0

at x = 0 and x = ℓ

The classical equations can also be obtained from first principles using Cauchy-Newton balance of momentum.

226. Find the Euler equation and boundary conditions for the functional

 F(w, w′, w′′) dx b

J(w) =

a

Use the Euler equation on the energy functional for a Bernoulli-Euler beam on an elastic foundation to find the classical differential equation governing the beam.

We can find the Euler equations at the extremum of the functional J. Taking the directional derivative in the direction of the function w we get ∂F w′ + ∂F w′′  dx   ∂w∂F w + ∂w′ ∂w′′ b

DJ(w) ⋅ w =

a

∂F w − d ∂F w′  dx   ∂w∂F w − dxd ∂w′ dx ∂w′′ b

=

a





b

 

b

b

+ ∂F w + ∂F w′ ∂w′ a ∂w′′ =

a

∂F + d ∂F  w dx   ∂w∂F − dxd ∂w′ dx ∂w′′ b

2

2

a



+ ∂F − d ∂F w + ∂F w′ ∂w′′ ∂w′ dx ∂w′′ a



b a

The Euler equation for the functional J is, therefore, ∂F − d ∂F + d2 ∂F = 0 ∂w dx ∂w′ dx 2 ∂w′′ With boundary conditions indicated at a and b as Either ∂F − d ∂F = 0 ∂w′ dx ∂w′′ Either

∂F = 0 ∂w′′

or

w = 0

at x = a and x = b

or

w′ = 0

at x = a and x = b

270


For the Bernoulli-Euler beam we can take F(w, w′, w′′) =

1  EI w′′  2 2

+ 12 kw2 − qw

The derivatives with respect to the various arguments are ∂F = 0 ∂w′

∂F = kw − q ∂w

∂F = EIw′′ ∂w′′

Therefore, the Euler equation is d 2 EIw′′ + kw − q = 0 dx2 One can observe that the boundary terms come out the same as the previous problem.

227. Using the fundamental theorem of the calculus of variations, find the classical form of the governing differential equation for w(x) implied by the minimum of the energy functional

  EI w w − qw  dx − ℓ

E(w) =

1 2

0

iv



ℓ

1 EI w′′′ w 2 0

+



ℓ

1 EI w′′w′ 2 0

Taking the directional derivative of the functional in the direction of the function w we get

  EI  w w+w w  − qw  dx ℓ

DE(w) ⋅ w =

1 2

iv

iv

0

−



ℓ

1  EI w′′′w+w′′′w  2 0



ℓ

1  EI w′′w′ +w′′w′  2 0

+

To apply the fundamental theorem of the calculus of variation we must integrate the first term by parts four times. The result is

  EI  ww +w w  − qw  dx ℓ

DE(w) ⋅ w =

1 2

iv

iv

0



ℓ

+

1 EI w′′′w 2 0

−

1  EI w′′′w+w′′′w  2 0

−



ℓ

1 EI w′′w′ 2 0



ℓ

+

+



ℓ

1 EI w′w′′ 2 0





ℓ

1  EI w′′w′ +w′′w′  2 0

Simplifying we finally arrive at the result

  EIw ℓ

DE(w) ⋅ w =

0

iv

− q  w dx − EIw′′′w



ℓ 0

+ EIw′′w′

ℓ

1 EI w w′′′ 2 0

−



ℓ 0

Chapter 9

271


Clearly, setting this functional equal to zero and applying the fundamental theorem of the calculus of variations results in the equations of the Bernoulli-Euler beam.

228. The potential energy of a simply supported, symmetrically loaded circular plate of radius R is

  D  r  w′′ + 1r  w′ R

2

E(w) = π

2





+ 2νw′w′′ − 2rqw dr

0

where the function w(r) is the transverse deflection of the plate, D and ν are constants, and q is a known function of r. Find the variational (virtual work) form of the governing differential equation. Find the classical (strong) form of the governing differential equation. What can you say, if anything, about the boundary conditions for the problem?

(a) Find the variational (virtual work) form of the governing differential equation.

  D  2r w′′w′′ + 2r w′w′ + 2ν  w′w′′+w′w′′  − 2rqw dr a

DE(w) ⋅ w = π

0

Integrating the term w′w′′ + w′w′′ by parts eliminates these terms to give

  D  r w′′w′′ + 1r w′w′  − rqw dr + 2πDνw′ w′  a

G(w, w) = 2π

0

a 0

(b) Find the classical (strong) form of the governing differential equation. Integrating by parts, using the above formulas, until all derivatives are shed from the w terms gives

 Dw a

DE(w) ⋅ w = 2π

0

iv

 

+ 2r w′′′ − 12 w′′ + 13 w′ − q wr dr r r

+ 2πD  rw′′+νw′  w′



a 0



 

+ 2πD 1r w′−w′′−rw′′′ w

a 0

The classical equations can then be stated as q w iv + 2r w′′′ − 12 w′′ + 13 w′ − = 0 D r r (c) What can you say, if anything, about the boundary conditions for the problem? The boundary conditions can be determined from the boundary terms of the last expression for DE ⋅ w. Because the loading is axisymmetric we have the condition w′(0) = 0 as an essential boundary condition. The other boundary condition at r = 0 is the natural condition (shear)

272




lim 1r w′−w′′−rw′′′ r→0



= 0

At the edge r = a a simply supported plate would have the conditions w(a) = 0,

aw′′(a) + νw′(a) = 0

For a fixed edge at r = a the conditions would be w′(a) = 0,

1 w′(a) − w′′(a) − aw′′′(a) = 0 a

229. Find an approximate solution to the problem of a simply supported, circular plate of radius R and modulus D, subjected to a uniform load of q. Use the Ritz method with the energy functional given in Problem 228. Assume that the displacement is of the form

 

w(r) = a 0 cos πr 2R

where a0 is the, as yet, undetermined constant. If you had to pick additional terms in the approximation, what would you choose? Why is the cosine function a good choice?

The virtual work functional is

  D  r w′′w′′ + 1r w′w′  − rqw  dr + Dνw′ w′  a

G(w, w) =

0

a 0

The derivatives of w can be computed from the approximation as

   

w′ = −c o π sin πr , 2a 2a

   

2 w′′ = −c o π 2 cos πr 2a 4a

Let ξ = πr∕2a and β ≡ 4a 2∕π2. The virtual work functional then becomes G(w, w) =

π 2

  Dβ c c  ξ cos ξ + 1ξ sin ξ  − q β c ξ cos ξ  dξ + Dνβ c c 2

o o

2

o

0

Carrying out the indicated integrals G(co, co) = D c o co  1.191+ν  − 0.5708 q β c o β Setting G equal to zero and solving for the constant we get co = −

qa 4 D



0.09376 1.191+ν



o o

Chapter 9


273

The exact solution to this problem is co = −





qa 5+ν 64D 1+ν 4

The approximation compares very favorably. For example, for ν = 0.3 the approximate solution is co = 0.0629 and the exact is co = 0.0637. Additional terms in the approximation. To increase the order of approximation one could continue to chose cosine functions of higher wave number. The cosine function is a good choice because it automatically satisfies the boundary condition w′(0) = 0, as required by symmetry. The term cos(3πr∕2a) would be the next choice as it satisfies w(a) = 0.

230. A beam of length ℓ and modulus EI rests on a q(x) nonlinearly elastic foundation that accrues transEI verse force in proportion to the cube of the transverse displacement, i.e., f (x) = k o w3. The beam is subko x jected to downward transverse loading q(x). Axial ℓ and shear deformations are negligible. Take w(x) as positive when it is upward. Find the virtual-work form of the equilibrium equations. Find the energy functional E(w) for the system.

(a) Find the virtual work form of the equilibrium equations. The virtual work per unit length done by the nonlinear force is simply −f w. Thus, the “external” work per unit length is  q−f ) w. This term can be substituted directly into the expression for virtual work to give

  EI w′′ w′′ + k w w − qw  dx ℓ

G(w, w) =

o

3

0

(b) Find the energy functional E(w) for the system. First check the symmetry of the virtual work functional

  EI w′′ w′′ + 3k w w w  dx ℓ

DG(w, w) ⋅ w^ =

^

o

2 ^

0

The virtual work functional is clearly symmetric. Now the energy can be found from Vainberg’s theorem

274


 G(tw, w) dt 1

E(w) =

0

   EI (t w′′) w′′ + k (t w) w − qw  dx dt 1

=

ℓ

o

0

0

  EI  w′′  + ℓ

=

3

1 2

2

1 k w 4 − qw 4 o

 dx

0

231. Consider a functional that takes scalar functions u(x) as input. The independent scalar variable is defined on the range x∈[0, 1]. The functional has the explicit form: 1  u′(x)  − 2u(x)  dx   1+x 1

E(u) =

2

0

where a prime indicates differentiation with respect to x. The functions are constrained at the boundary to satisfy the conditions u(0) = 0 and u′(1) = 0. Find the classical differential equation implied by stationarity (i.e., max, min, or saddle point) of the functional. Solve the classical differential equation. Compute the second derivative functional associated with the given functional.

DE( u ) ⋅ u =



1

0

1 u′u ′−2 u  dx  1+x 1 u′  ′ + 2  u dx +   1 u′  u   1+x 1+x 1

=−

0

1 0

1 u′  ′ + 2  u dx − u′(0) u(0)   1+x 1

=−

0

The classical differential equation can be deduced from the fundamental theorem of the calculus of variations to be. 1 u′  ′ + 2 = 0 1+x

232. The virtual-work functional for a system is given by the following expression

Chapter 9

275


  a  w′′ w+w′′ w  + bw w  dx ℓ

G(w, w) =

3

0

where a, b, and ℓ are constants and w(x) and w(x) are functions of the independent variable x. Does this functional have an associated energy? Find the energy functional for the system, if it exists. The existence of the energy functional depends upon the symmetry of the directional derivative DG(w, w) ⋅ w^ = DG(w, w^ ) ⋅ w

  a  w ′′ w+w′′ w  + 3bw w w  dx ℓ

DG(w, w) ⋅ w^ =

^

^

2 ^

0

  −2a w ′ w′ + 3bw w w  dx + a  w ′ w+w′ w  ℓ

^

=

2 ^

^

^

ℓ 0

0

This functional has an energy only for certain boundary conditions, but not in general.

233. Resolve Problem 232 with the functional

  −a w′′ w + bww  dx − a w(ℓ) w′(ℓ) + a w(0) w′(0) ℓ

G(w, w) =

0

The existence of the energy functional depends upon the symmetry of the directional derivative DG(w, w) ⋅ w^ = DG(w, w^ ) ⋅ w

  −a w ′′ w + b w w  dx − a w(ℓ) w′(ℓ) + a w(0) w′(0) ℓ

DG(w, w) ⋅ w^ =

^

^

^

^

0

  a w ′ w′ + b w w  dx − a  w ′ w  + a  w w′  ℓ

=

^

^

^

ℓ 0

^

ℓ 0

0

Again, this functional is not symmetric because of the presence of the asymmetric boundary terms. It is worth noting that if the original functional had been

  −a w′′ w + bww  dx − a w′(ℓ) w(ℓ) + a w′(0) w(0) ℓ

G(w, w) =

0

then the functional would be symmetric because the boundary terms from integration by parts would cancel the original boundary terms.

276


234. The deformation state of a particular system is characterized by the scalar function θ(x), where the scalar variable x∈[0, ℓ]. The virtual-work functional for the system is given by the following expression

  a θ′ θ′ + b θ sin θ  dx ℓ

G(θ, θ) =

0

where a, b, and ℓ are constants. Equilibrium of the system holds if G(θ, θ) = 0 for all θ. Does this functional have an associated energy? Find the energy functional for the system, if it exists.

(a) Does this functional have an associated energy? DG( θ, θ ) ⋅ ψ =



ℓ

 a ψ′θ′+b θ cos θ ψ  dx

0

which is symmetric because swapping θ and ψ does not change the value. Therefore an energy functional E(θ) exists. (b) Find the energy functional for the system, if it exists. The energy functional can be found by application of Vainberg’s theorem

 G(t θ, θ) dt =    a t θ′ θ′ + b θ sin θ  dx dt 1

E(θ) =

1

0

ℓ

0

0

  a t ℓ

=

 ℓ

 

 θ′  2 + −b cos t θ dx 2

0

=

2

1 2

1 0

a  θ′  2 + b 1− cos θ dx

0

235. The deformation state of a particular system is characterized by the scalar function u(x), where the scalar variable x∈[0, 1]. The virtual-work functional for the system is given by the following expression G(u, u) =   a u′ u′ + bu′ u + g(u) u  dx 1

0

where a and b are constants and g(u) is a given nonlinear function of the displacement function u(x). Equilibrium of the system holds if G(u, u) = 0 for all u. For what values of the constants a and b does this functional have an associated energy? Find the energy functional for the system, if it exists.

Chapter 9

277


(a) For what values of the constants a and b does this functional have an associated energy? An energy functional exists if DG(u, u) ⋅ u^ = DG(u, u^ ) ⋅ u. Compute the directional derivative as DG(u, u) ⋅ u^ = d dε

   a  u+ε u ′ u′ + b  u+ε u ′ u + g u+ε u  u  dx 1

^

^

^

0

ε=0

=   a u′u′ + b u′u + g′u u  dx 1

^

^

^

0

Note: Symmetry holds only if b = 0, a can have any value. (b) Find the energy functional for the system, if it exists. For b = 0 we can compute the energy by Vainberg’s theorem.

 G(t u, u) dt =    a t(u′) +g(t u) u  dx dt 1

E(u) =

1

1

2

0

0

0

Define the function f (u) to be

 g( t u ) u dt 1

f (u) ≡

0

such that Df (u) ⋅ u = g(u) u. With this definition the energy functional can be written as

 1

E(u) =

1 2

a  u′  2 + f (u)  dx

0

236. A beam of length ℓ rests on an elastic foundation of modulus k (per unit length). It is pinned at the left end and is subjected to a point load P at the right end. The elastic foundation accrues a transverse force in proportion to the transverse displacement w. The energy of the system can be expressed as: E(w) =



ℓ

P

x

k ℓ

1 EI(w′′) 2+kw2  dx − Pw(ℓ) 2

0

Find the virtual-work form of the equilibrium equations. What are the essential boundary conditions? Find the classical form of the equilibrium equations and the boundary conditions. Which of the three functions given below are suitable for approximating the solution with the Ritz method? Explain why or why not in each case.

278

Fundamentals of Structural Mechanics w(x) = x(ℓ−x)(a 1+a2x),

w(x) = x2(a 1+a2x),

w(x) = x(a1+a 2x)

(a) Find the virtual work form of the equilibrium equations. What are the essential boundary conditions? Essential Boundary Condition: w(0) = 0, the rest are natural boundary conditions. The virtual work functional can be found by taking the directional derivative of the energy functional G(w, w) = DE(w) ⋅ w = d  E(w+αw)  α=0 dα Carrying out this operation gives

  EI w′′w′′ + kww  dx − Pw(ℓ) ℓ

G(w, w) =

0

G(w, w) = 0 for all virtual displacements that satisfy the essential boundary condition is the weak form of equilibrium. (b) Find the classical form of the equilibrium equations and the boundary conditions. To put the virtual work functional in a form suitable to apply the fundamental theorem of the calculus of variations, integrate G(w, w) by parts to get rid of w′′ terms:

  EI w′′′′+k w  w dx +  EI w′′w′  −  EIw′′′ w  − Pw(ℓ) ℓ

G(w, w) =

ℓ

ℓ

0

0

0

From the Fundamental Theorem of the Calculus of Variations we can deduce G(w, w) = 0

∀w

⇒

EI w′′′′+k w = 0

for

0
The virtual work functional tells us something about the possible boundary conditions too. In particular, the boundary terms vanish if w(0) = 0

(essential BC)

EI w′′(0) = 0

(natural BC)

EI w′′(ℓ) = 0

(natural BC)

EI w′′′(ℓ) + P = 0

(natural BC)

(c) Which of the three functions shown are suitable for approximating the solution with the Ritz method? Explain why or why not in each case. The function w(x) = x  ℓ−x  a1+a 2 x  is no good because w(ℓ) = 0 so the load P does no work. The function w(x) = x 2  a1+a 2 x  is no good because w′(0) = 0 implying an additional essential boundary condition (fixed end).

Chapter 9

279


The function w(x) = x  a1+a2 x  is fine.

237. Consider a rectangular (rigid) block of height h and width ℓ and weight W. The block is prevented from sliding by a small obstruction at the lower right corner and is pushed by a force P at the upper left corner. Write the potential energy of the system in terms of the angle of rotation θ of the block. Find the force P as a function of h, ℓ, W, and the angle θ needed to have static equilibrium. Find an expression for the angle at which equilibrium goes from being stable to being unstable.

ℓ

P

h W

W θ

(a) Write the potential energy of the system in terms of the angle of rotation θ of the block. P

h sin θ

h cos θ 2

W

ℓ sin θ 2

Datum for W ℓ cos θ 2

Datum for P

Using terms defined in the sketch, we can write the energy of the system as E(θ) = P  ℓ cos θ−h sin θ  + 12 W  ℓ sin θ+h cos θ  (b) Find the force P as a function of h, ℓ, W, and the angle θ needed to have static equilibrium. Equilibrium holds when the energy is stationary, or DE(θ) ⋅ θ = 0

∀ θ

Computing the derivative of the energy, we get E(θ) ⋅ θ = P  −ℓ sin θ−h cos θ  θ + 12 W  ℓ cos θ−h sin θ  θ = 0 This holds for values of the load that satisfy ℓ cos θ − h sin θ P=W 2 ℓ sin θ+h cos θ

280


(c) Find and expression for the angle at which equilibrium goes from being stable to being unstable. To establish the stability we must compute the second derivative A(θ, θ) =

 dεd

2 2

E(θ+εθ)



ε=0

Carrying out this computation for the system at hand we get A(θ, θ) = P  −ℓ cos θ+h sin θ  θ 2 + 12 W  −ℓ sin θ−h cos θ  θ 2 but P must satisfy equilibrium (i.e., it must satisfy the above expression) A(θ, θ) = − W 2

θ−h sin θ    ℓℓ cos ℓ cos θ−h sin θ  θ sin θ+h cos θ

2

− W  ℓ sin θ+h cos θ  θ 2 2 = −W 2

+h  ℓ sinℓθ+h θ cos θ 2

2

2

Thus, we find that 0<θ<π 2

A(θ, θ) ≤ 0, Therefore, equilibrium is always unstable.

x3

238. Consider the solid spherical region shown in the sketch. Assume that there exists a scalar field w(x) for which we can define the functional G(w, v) ≡

  ∇v ⋅ ∇w + v w  dV −  p v dA B

Ω

B n

x2 x1

Ω

that has the property that if G(w, v) = 0 for all (virtual) scalar functions v(x) then the classical differential equations governing the real field w(x) are satisfied (i.e., G(w, v) is a “virtual-work” functional). Note that the scalar field p is defined on the surface of the solid region. Show that an “energy” functional exists for this theory if the function p depends only upon the position vector x and not the function w(x), i.e., p = p(x). Determine the energy functional in terms of the field w.

(a) Show that an “energy” functional exists for this theory if the function p depends only upon the position vector x and not the function w(x), i.e., p=p(x). An energy functional exists if the virtual work functional is symmetric in the sense of DG(w, v) ⋅ u = DG(w, u) ⋅ v. Hence, we must compute the derivative

Chapter 9

281


DG(w, v) ⋅ u =

 dεd G(w+εu, v) 

=

ε=0

  ∇u ⋅ ∇u + v u  dV B

Therefore, an energy functional does exist. (b) Determine the energy functional in terms of the field w. The energy functional can be determined by Vainberg’s theorem E(w) =



1

G( t w, w) dt

0

Carrying out these operations we get

    ∇w ⋅ ∇(t w) + w(t w)  dV −  1

E(w) =

0

= 12

B

p w dA

Ω

  ∇w ⋅ ∇w + w  dV −  2

B



dt

p w dA

Ω

239. Reconsider Problem 238 for the case where the function p depends upon the field w and the position x. Under what conditions would an “energy” functional exist in this case? (Hint: would an energy exist if p depends upon w itself? What if it depends upon derivatives of w, i.e., p = p(x, w(x), ∇w(x), )?

The directional derivative of the function p can be carried out as Dp(w) ⋅ u =

 dεd p(w+εu) 

ε=0

This derivative takes the form Dp(w) ⋅ u =

∂p ∂p u+ ⋅ ∇u +  ∂w ∂∇w

Thus, last term of the given functional can be written as ∂p  ∂w∂p uv + ∂∇w ⋅ ∇u v +  dA Ω

282


Hence, if terms in p involve w alone, then energy exists. Energy may not exist for higher derivatives.

240. A beam of length ℓ rests on an elastic foundation of modulus k (per unit length). It is fixed at the left end, pinned at the right end, and is subjected to a point load P at midspan. The energy of the system can be expressed in terms of the transverse displacement w(x) as: E(w) =



ℓ

1 EI(w′′) 2 + kw2  2

P EI k

x ℓ∕2

ℓ∕2

dx + Pw(ℓ∕2).

0

Find the virtual-work form of the equilibrium equations. What are the essential and natural boundary conditions? Use the Ritz method to find a one-term approximation of the displacement field (use a polynomial approximation).

(a) Find the virtual work form of the equilibrium equations. G(w, w) = DE(w) ⋅ w =

 dεd E(w+εw) 

ε=0

  EI w′′ w′′ + k ww  dx + P w  ℓ

=

ℓ 2

0

(b) What are the essential and natural boundary conditions? Essential: w(0) = 0, Natural: M(ℓ) = 0

w′(0) = 0,

w(ℓ) = 0

(or w′′(ℓ) = 0)

(c) Use the Ritz method to find a one-term approximation of the displacement field (use a polynomial approximation). Start with a quartic polynomial and substitute the three essential boundary conditions to get a one term polynomial approximation. Let ξ ≡ x∕ℓ and w = a + b ξ + c ξ2 + d ξ 3 w′ = 1  b + 2c ξ + 3d ξ 2  ℓ Substituting the boundary conditions we have w(0) = a = 0 w′(0) = 1  b  = 0 ℓ w(ℓ ) = a + b + c + d = 0

⇒ c = −d

Chapter 9

283


Thus, the approximation (and its derivatives) can be written as w′ = c  2 ξ−3 ξ 2  ℓ

w = c  ξ2−ξ3 

w′ = c2  2−6 ξ  ℓ

with similar expressions for the virtual displacement. Substituting this approximation into the virtual work functional we have

  EIℓ  2−6ξ  c c + k  ξ −ξ  c c  ℓ dξ + P c  1

G=

2

4

2

3 2

1 1 − 4 8

=0

∀ c

0

The integrals have the values

  2−6ξ  1

dξ =

0

  ξ −ξ  1

2

0

  4−24ξ+36ξ  dξ = 4−12+12 = 4 1

2

3 2

2

0

dξ =   ξ −2ξ +ξ  dξ = 1

4

5

0

6

1 2 1 − + 5 6 7

=

Thus, kℓ  4EIℓ + 105  c + P8 = 0 3

which allows the determination of the value of the constant c.

42 − 70 + 30 5⋅6⋅7

1 = 105

284


Chapter 10 Fundamental Concepts in Static Stability

241. The frame shown is composed of two rigid members conP nected by rotational springs. The moment developed by a spring k is related to the rotation by M = k φ, where φ is the rotation expe- ℓ rienced by the spring. Both springs have modulus k. The frame is k subjected to a load P acting downward. The motion of the structure can be completely characterized by the rotation of the vertical ℓ ℓ member from its original position. Examine the stability of the system. In particular, find the critical load and plot the bifurcation diagram. Note that the bifurcation diagram is not symmetric. Can you explain, in physical terms, why it is not?

P

ℓ(1− cos θ) θ

θ

β

datum

Let θ be the rotation of the vertical member from its original position (see the figure). From geometry, the vertical drop of the hinge in the beam must be the same measured from the left or from the right. Therefore, we get the relationship between the angles β and θ as sin β = sin θ + 1 − cos θ The potential energy of the frame is E(θ) =

1 k θ2 2

+ 12 k  θ+β  2 + Pℓ cos θ

286


The virtual-work functional can be computed from the energy as G(θ, θ) = DE(θ) ⋅ θ =  k θ + k  θ+β   1+dβ∕dθ  − Pℓ sin θ  θ where, up to the second-order of θ, β ≈ θ + 12 θ 2,

dβ ≈ 1 + θ, dθ

sin θ ≈ θ

The equilibrium condition G(θ, θ) = 0 for all θ yields two solutions: θ = 0,

P(θ) ≈ k  5+3θ  ℓ

The value of P(θ) at θ = 0 is the critical load Pcr = 5k ℓ The bifurcation diagram is asymmetric about θ because G(θ, θ) contains even terms in θ, which in turn causes P(θ) to have odd terms. A physical explanation is that for the same amount of rotation of the vertical bar, the amount of rotation the horizontal bar experiences is larger for θ > 0 than for θ < 0, hence, more energy is stored in the frame in the former case. More energy leads to a higher bifurcation load. The second derivative of the energy is





A(θ, θ) = k 1 +  θ+β  β′′ +  1+β′  2 − Pℓ cos θ θ 2 k





= k 1 +  θ+β  +  2+θ  2 − Pℓ cos θ θ 2 k For the trivial path θ = 0 A(0, θ) = θ 2  5k − Pℓ  therefore, the trivial path is stable for P < P cr and unstable for P > P cr. For the bifurcated path A(θ, θ) = k θ 2 3θ + O(θ 2)  where O(θ 2) denotes second and higher powers of θ. Therefore, the bifurcated path is stable for θ > 0 and unstable for θ < 0. The equilibrium paths are plotted in the attached figure. P(θ)ℓ k 5

3 1

θ

Chapter 10

Fundamental Concepts in Static Stability

287

242. Examine the effect of an imperfection in the system of Problem 241. Let the imperfection be an initial value of the angle of rotation used to describe the motion, and assume that the springs are such that they have no moment at this initial position. Plot the maximum load versus the size of the initial imperfection.

The potential energy of the frame with an initial imperfection θ o is E(θ) =

1  k θ−θ o  2 2

+ 12 k  θ−θ o+β−β o  2 + Pℓ cos θ

Hence, G(θ, θ) = DE(θ) ⋅ θ =  k θ−θ o  + k  θ−θ o+β−β o   1+dβ∕dθ  − Pℓ sin θ  θ From geometry, up to the second-order of θ β ≈ θ + 12 θ 2,

dβ ≈ 1+θ dθ

β o ≈ θ o + 12 θ 2o,

G(θ, θ) = 0 for all θ gives



θ P(θ) ≈ k 5 1− o ℓ θ

 + 2 θ−θ  + (2+θ)2θ θ −θ   2

o

2 o

Setting P′(θ) = 0 gives the equation 3θ 2 + 5θ o = 0 Hence, there exists a limit load for θ o < 0. It occurs at θ cr = − −5θ o∕3. Substituting θ cr in the expression for P(θ), and retaining only the terms up to the first order gives the maximum load as a function of the imperfection



Pmax ≈ Pcr 1 − 2

− 35 θ

o

− 17 θ 6 o



where Pcr = 5k∕ℓ is the critical load for the perfect system. The maximum load as a function of the imperfection is shown in the sketch. P max P cr

−5 o

θo

1

0

288


P 243. Consider the two rigid bars hinged together and subjected to axial load P, as shown. The k k k bars have length ℓ and 3ℓ, and are restrained by three elastic springs, with modulus k, that resist ℓ ℓ 2ℓ vertical motion. Find all equilibrium paths for the system. Find the bifurcation loads of the system. Assess the stability of the straight and bent configurations.

θ θ P datum

The system has one degree of freedom. The potential energy and virtual work functionals of the system are, respectively E(θ) = 3kℓ 2 sin 2 θ + 4Pℓ cos θ G(θ, θ) = ℓ sin θ  6kℓ cos θ − 4P  θ G(θ, θ) = 0 for all θ gives the equilibrium paths for the system as θ = 0,  π,  2π,  P(θ) =

3 kℓ cos θ 2

The bifurcation loads are Pcr = 32 kℓ, corresponding to θ = 0 and θ = π, respectively. The second derivative functional is A(θ, θ) = ℓ θ 2  6kℓ  cos 2 θ− sin 2 θ  − 4P cos θ  For θ = 0, the second derivative functional is A(0, θ) = ℓ θ 2  6kℓ − 4P  Hence, the configuration θ = 0 is stable for P < 32 kℓ, and unstable for P > 32 kℓ. For θ = π, A( π, θ) = ℓ θ 2  6kℓ + 4P  Hence, the straight configuration θ = π is stable for P > − 32 kℓ, and it is unstable for P < − 32 kℓ. For the bent configuration P(θ) = 32 kℓ cos θ, A(θ, θ) = −6kℓ2 θ 2 sin 2 θ < 0 Hence, the bent configuration is always unstable.

Chapter 10


289

244. Consider the three-bar rigid linkage P shown. The bars are hinged together and are rek k strained by elastic springs that resist vertical motion. The springs accrue force in proportion ℓ ℓ ℓ to their extension, with modulus k. The system is subjected to an axial force P. Write an expression for the potential energy of the system. What are the equations of equilibrium governing the response of the system? Find the critical loads and the buckling mode shapes of the system. Feel free to linearize the geometry of deformation as you see fit.

θ1

θ2

β

P

datum

The system has two degrees of freedom. Let θ 1, θ 2, and β(θ) be the rotations of the bars (see sketch). For a linearized bifurcation analysis β ≈ θ 1+θ 2. Let θ={θ 1, θ 2} be the degrees of freedom. The linearized potential energy functional of the system is E(θ) =

1 2 2 kℓ θ 1 +  θ 1+θ 2  2  − 12 Pℓ  θ 21 2

+ θ 22 +  θ 1+θ 2  2 

and the linearized virtual-work functional of the system is G(θ, θ) = kℓ2  θ 1 θ 1 +  θ 1+θ 2   θ 1+θ 2   − Pℓ  θ 1 θ 1 + θ 2 θ 2 +  θ 1+θ 2   θ 1+θ 2   =  2θ 1+θ 2   kℓ −P  ℓ θ 1 +  kℓ θ 1+θ 2  − P θ 1+2θ 2   ℓ θ 2 G(θ, θ) = 0 for all θ gives the equations of equilibrium  kℓ − P   2θ 1+θ 2  = 0

kℓ  θ 1+θ 2  − P  θ 1+2θ 2  = 0 Let m ≡ P∕kℓ. The matrix form of the equations is kℓ

2−2m

1−m

1−m

1−2m

 θ1 θ2



0 = 0

The solution is either θ = 0, or the determinant of the coefficient matrix is equal to zero

290

Fundamentals of Structural Mechanics  2−2m  1−2m  −  1−m  2 =  1−m  1−3m  = 0

which yields the bifurcation loads as P1cr =

1 kℓ, 3

P2cr = kℓ

The corresponding buckling mode shapes are θ1 = ( 1, −2 ), and θ2 = ( 1, 0 ), as shown in the sketch.

θ1 = ( 1, −2 )

θ2 = ( 1, 0 )

P 245. Consider the three-bar rigid linkage shown. The bars are hinged together and are k k restrained by elastic rotational springs. The ℓ ℓ ℓ springs accrue force in proportion to the relative angle of distortion, with modulus k. The system is subjected to an axial force P. Write an expression for the potential energy of the system. What are the equations of equilibrium governing the response of the system? Find the critical loads and the buckling mode shapes of the system. Feel free to linearize the geometry of deformation as you see fit.

θ1

θ2

β+θ2 β

P

datum

The system has two degrees of freedom. Let θ = (θ 1, θ 2) be the degrees of freedom. Again, the linearized system has β ≈ θ 1+θ 2 . The potential energy of the system is E(θ) =

1  k θ 1−θ 2  2+ θ 1+2θ 2  2  − 12 Pℓ  θ 21 2

+ θ 22 +  θ 1+θ 2  2 

The virtual work functional can be found as G(θ, θ) = DE(θ) ⋅ θ. To wit, G(θ, θ) = k   θ 1−θ 2  θ 1−θ 2 + θ 1+2θ 2  θ 1+2θ 2   − Pℓ  θ 1 θ 1+θ 2 θ 2+ θ 1+θ 2  θ 1+θ 2  

Chapter 10


291

Rearranging we get G(θ, θ) =  2k θ 1+k θ 2 − Pℓ θ 1 − Pℓ  θ 1+θ 2   θ 1 +  k θ 1+5k θ 2 − Pℓ θ 2 − Pℓ  θ 1+θ 2   θ 2 The linearized equations of equilibrium are  k − Pℓ   2θ 1 + θ 2  = 0  k − Pℓ  θ 1 +  5k − 2Pℓ  θ 2 = 0

Let m ≡ Pℓ∕k. The matrix form of the equations is k

2−2m

1−m

1−m

5−2m

 θ1 θ2



0 = 0

The solution is either θ = 0, or the determinant of the coefficient matrix is equal to zero 2 1−m  5−2m  −  1−m  2 = 3 1−m  3−m  = 0 The bifurcation loads are P1cr = k ℓ

P2cr = 3k ℓ

The corresponding buckling mode shapes are θ1 = ( 1, 0 ), and θ2 = ( 1, −2 ).

246. Consider the rigid bar subjected to axial load P and transverse load ε P as shown. The bar has length 2ℓ, is restrained against horizontal and vertical motion at the midpoint, and is supported by two elastic springs that resist vertical motion at the ends. The springs accrue force in proportion to their extension, with modulus k. The deformation can be characterized by the rotation of the bar relative to the horizontal position. εP P P ε = 0.1 k k π −π θ ℓ ℓ Find all equilibrium paths P(θ) for the system ( −π < θ < π). Determine the stability of these branches. Find the critical load of the system when ε = 0. Locate the limit point on the bifurcation diagram plotted for ε = 0.1. Is the limit load at this point greater or less than the critical load?

ÁP

θ

P

292


The system has a single degree of freedom. The potential energy, virtual work functional, and second derivative functional of the system are E(θ) = kℓ 2 sin 2 θ + Pℓ  cos θ−Á sin θ  G(θ, θ) =  kℓ sin 2θ − P  sin θ+Á cos θ   ℓ θ A(θ, θ) =  2kℓ cos 2θ − P  cos θ−Á sin θ   ℓ θ 2 G(θ, θ) = 0 for all θ gives the nonlinear equilibrium configurations P(θ) = kℓ

sin 2θ sin θ + Á cos θ

There are two equilibrium paths asymptotic to θ o ≡ − tan –1 Á where P(θ) goes to positive infinity at the left and negative infinity at the right, as shown in the figure. The second derivative of the energy is A(θ, θ) = =

=

 2kℓ cos 2θ − kℓ sin θsin+ 2θÁ cos θ  cos θ−Á sin θ   ℓ θ sin θ  kℓ θ  ÁÁcoscos θθ − + sin θ 3

3

2

2

2

>0

−π < θ < −π+θ cr

(stable)

<0

−π+θ cr < θ < θ o

(unstable)

>0

θ o < θ < θ cr

(stable)

<0

θ cr < θ < π−θ o

(unstable)

>0

π−θ o < θ < π

(stable)

where θ cr ≡ tan–1 Á 1∕3 . Note that P(θ) reaches an extremum at θ cr. When Á = 0, the principle of virtual work, i.e., G(θ, θ) = 0 for all θ, yields the equilibrium configurations θ = 0,  π,  2π,  P(θ) = 2kℓ cos θ The critical load is Pcr = 2kℓ. When Á = 0.1, the limit load is found to be 0.75Pcr, located at θ cr = 0.43. Hence, the limit load for the imperfect system with Á = 0.1 is 25% lower than that for the perfect system ( Á = 0).

Chapter 10

293


247. Consider the frame composed of rigid bars subjected to the load P as shown. The rigid members are hinged at the top right corner, with an elastic spring that resists relative rotation. The rotational spring accrues force in proportion to its relative angle change, with modulus k. Find the buckling load for this system. Express the deformation of the system in terms of the angle of rotation of the vertical member on the right side of the structure. What happens if you use the angle of rotation of the vertical member on the left?

P k ℓ

ℓ

ℓ

P

θ

θ β

The system has a single degree of freedom. We shall take β as the measure of deformation, with the angle θ depending on β (see sketch). Based on the deformed geometry, ℓ cos θ − 2ℓ sin θ = ℓ cos β, which implies that θ ≈ 14 β 2 . The potential energy of the system is E(β) =

1  k θ + β  2 + Pℓ  cos θ− sin θ  2

The virtual work functional is G(β, β ) = k  θ+β  θ′+1  β − Pℓ  sin θ+ cos θ  θ′β ≈ k  14 β 2+β  12 β+1  β − Pℓ  14 β 2+1  12 β β ≈

 k − 12 Pℓ  β β

Therefore, the linearized equation of equilibrium is

 k − 12 Pℓ  β

= 0

The buckling load is then Pcr = 2k ℓ If θ is used instead of β for the primary unknown, then the constraint θ > 0 must be accounted for in the formulation. Consequently, the solution becomes messier.

294

Fundamentals of Structural Mechanics P

248. Consider the rigid bar subjected to axial load P as shown. The bar has length 2ℓ and is supported by elastic springs that resist vertical motion. The springs accrue force in proportion to their extension, with modulus k. Find the critical loads and linearized buckling mode shapes of the system. Note that this system has two degrees of freedom.

k

k

ℓ

k

ℓ

P θ u

The system has two degrees of freedom. Let u (vertical displacement at the center of the bar) and θ be the degrees of freedom of the system, as shown in the sketch. Then, the linearized potential energy of the system is E(u, θ) =

1  k u−ℓθ  2 2

+ u 2 +  u+ℓθ  2  − Pℓ θ 2

The virtual work functional is G(u, θ, u, θ) = k  u−ℓθ  u−ℓθ  + u u +  u+ℓθ  u+ℓθ   − 2Pℓ θ θ = k 3u u + 2ℓ2θθ  − 2Pℓ θ θ Therefore, the linearized equations of equilibrium are 3k u = 0 2 kℓ 2−Pℓ  θ = 0 The trivial position ( θ, u )=(0, 0) is an equilibrium configuration for any value of P. There is one critical load, Pcr = kℓ, with corresponding mode ( θ, u )=(1, 0).

249. Consider the three-bar rigid linkage P k k k shown. The bars are hinged together and are restrained by elastic springs that resist vertical motion. The springs accrue force ℓ ℓ ℓ ℓ in proportion to their extension, with moθ1 dulus k. The system is subjected to an axial force P. Write an expression for the potenθ2 tial energy of the system. What are the linearized equations of equilibrium governing the response of the system? Find the critical loads of the system. A convenient choice of degrees of freedom is shown in the diagram.

Chapter 10


295

Let θ = (θ 1, θ 2). The potential energy of the system is 1 2 kℓ 2 θ 21 2

E(θ) =

+  θ 1+θ 2  2  − 12 Pℓ  3 θ 21+θ 22 

The virtual work functional is G(θ, θ) = kℓ2  2 θ 1θ 1+ θ 1+θ 2  θ 1+θ 2   − Pℓ  3 θ 1 θ 1+θ 2 θ 2  =  kℓ2  3 θ 1+θ 2  − 3Pℓ  θ 1 +  kℓ2  θ 1+θ 2  − Pℓ  θ 2 The linearized equations of equilibrium are kℓ2  3θ 1+θ 2  − 3Pℓ θ 1 = 0 kℓ2  θ 1+θ 2  − Pℓ θ 2 = 0 Let m ≡ P∕kℓ. The matrix form of the equations is kℓ 2

3−3m

1

1

1−m

 θ1 θ2



0 = 0

The solution is either θ = 0, or the determinant of the coefficient matrix is equal to zero 3 1−m  2 − 1 = 3m 2 − 6m + 2 = 0 The bifurcation loads are P1cr =

 1−   kℓ 1 3

P2cr =

 1+   kℓ 1 3

The corresponding buckling mode shapes are θ1 = ( 1, − 3 ), θ2 = ( 1, 3 ).

250. Two rigid bars are hinged together and rest on a linP early elastic foundation. The foundation accrues a force k per unit length proportional to the transverse displaceℓ ℓ ment, i.e., f(x)=kw(x). The system is subjected to an axial load P as shown. Find an expression for the energy functional for the system. Find an expression for the virtual-work functional for the system. Find the buckling loads of the system by solving the linearized buckling eigenvalue problem.

(a) Find an expression for the energy functional for the system. The system has two degrees of freedom. The displacement in the two rigid segments can be expressed as w1(x) = x sin θ 1 and w2(ξ) = ℓ sin θ 1+ξ sin θ 2. These quantities are shown in the sketch.

296

Fundamentals of Structural Mechanics θ2 w 2(ξ)

θ1

w 1(x)

x

E(θ 1, θ 2) =



ℓ

1  k x sin θ 1 2 2

dx +

0

ξ



ℓ

1  k ℓ sin θ 1+ξ sin θ 2  2 2

dξ + Pℓ cos θ 1+ cos θ 2 

0

 



3 3 = 1 k sin 2 θ 1 ℓ + 1 k ℓ3 sin2 θ 1+ℓ 3 sin θ 1 sin θ 2+ ℓ sin 2 θ 2 2 2 3 3



+ Pℓ  cos θ 1+ cos θ 2 





= 1 k ℓ 3 4 sin 2 θ 1 + sin θ 1 sin θ 2 + 1 sin 2 θ 2 + Pℓ cos θ 1+ cos θ 2 2 3 3 (b) Find an expression for the virtual work functional for the system. G(θ, θ) = 12 k ℓ 3  83 sin θ 1 cos θ 1 θ 1 + sin θ 2 cos θ 1 θ 1  + 12 k ℓ 3 sin θ 1 cos θ 2 θ 2 + 23 sin θ 2 cos θ 2 θ 2  − Pℓ  sin θ 1 θ 1 + sin θ 2 θ 2  ≈

12 k ℓ  83 θ 3

1



+ θ 2  − Pℓ θ 1 θ 1 +

 12 k ℓ  θ 3

1



+ 23 θ 2  − Pℓ θ 2 θ 2

If we use a first order approximation of the nonlinear terms. (c) Find the buckling loads of the system by solving the linearized buckling eigenvalue problem. Let λ ≡ 2 P∕k ℓ2. The equations of equilibrium can be written as 8 −λ 3

1

2 3

1

θ1

−λ

θ2

=

0 0

This system of equations as a nontrivial solution only if the determinant of the coefficient matrix is zero. Thus,

 83 − λ  23 − λ − 1 = 0 or

Chapter 10


297

9λ2 − 30λ + 7 = 0 This quadratic equation has the solutions λ1,2 =

30  900 − 4 * 7 * 9 = 0.2525, 3.0809 2*9

Therefore, the critical loads are P1 = 0.126 k ℓ2

P 2 = 1.540 k ℓ 2

251. The vertical rigid bar is subjected to axial load P and P is hinged to the horizontal rigid bar which has length 2ℓ. A rotational spring restrains the change in angle between the ℓ two bars. The horizontal bar is restrained against horizontal kθ = 2kℓ2 and vertical motion at the midpoint, and is supported by two k k elastic springs that resist vertical motion at the ends. The springs accrue force in proportion to their extension, with ℓ ℓ modulus k. Find an expression for the energy E of the system. Find the (nonlinear) equations of equilibrium of the system. Find the critical loads of the system.

P

θ2 θ1

The system has two degrees of freedom. Let θ = (θ 1, θ 2). The potential energy of the system is E(θ) = kℓ 2  sin 2 θ 1+ θ 1+θ 2  2  + Pℓ cos θ 2 The virtual work functional can be found from G(θ, θ) = DE(θ) ⋅ θ to be G(θ, θ) = 2kℓ2  sin θ 1 cos θ 1 θ 1 +  θ 1+θ 2  θ 1+θ 2   − Pℓ sin θ 2 θ 2 =  2kℓ2  sin θ 1 cos θ 1 + θ 1 + θ 2   θ 1 +  2kℓ2  θ 1 +θ 2  − Pℓ sin θ 2  θ 2 The nonlinear equations of equilibrium are 2kℓ2 θ 1+θ 2+ sin θ 1 cos θ 1  = 0 2kℓ2  θ 1+θ 2  − Pℓ sin θ 2 = 0

298


Let m ≡ P∕kℓ. The matrix form of the linearized equations is kℓ 2

4

2

2

2−m

 θ1 θ2



0 = 0

The solution is either θ = 0, or the determinant of the coefficient matrix is equal to zero 4 2−m  − 4 = 4 1−m  = 0 Since the characteristic equation is linear there is only one critical load. The critical load is Pcr = kℓ. The buckling mode is θ = ( 1, −2 ).

252. A rigid bent of height 2ℓ and length 3ℓ rests on three elastic springs, each with modulus k. The springs accrue force in proportion to the amount by which they stretch. The bent is pinned at the corner end and is subjected to a load P at the top and a load of P at the right end. Find the virtual-work form of the equilibrium equations. Find the second-derivative functional A for the system. Find all equilibrium configurations of the system and assess their stability. Sketch the result on a bifurcation diagram.

P ℓ ℓ

k

P k ℓ

k ℓ

ℓ

P

θ θ P

The system has a single degree of freedom. The energy of the system is E(θ) =

1  k (ℓ sin θ) 2+(ℓ sin θ) 2+(2ℓ sin θ) 2  + 2Pℓ cos θ − 3Pℓ cos θ 2

with virtual work functional G(θ, θ) = 6kℓ2 sin θ cos θ θ + Pℓ sin θ θ and second-derivative functional A(θ, θ) =  6kℓ cos 2θ + P cos θ  ℓ θ 2 G(θ, θ) = 0 for all θ gives the equilibrium configurations of the system

Chapter 10

299

Fundamental Concepts in Static Stability θ = 0,  π,  2π,  P(θ) = −6kℓ cos θ

Since A(0, θ) =  6kℓ+P  ℓ θ , the straight configuration θ = 0 is stable for P > −6kℓ, and unstable for P < −6kℓ. At θ = π we have A( π, θ) =  6kℓ−P  ℓθ 2. Thus, the straight configuration θ = π is stable for P < 6kℓ, and unstable for P > 6kℓ. For the bent configuration we have 2

A(θ, θ) =  6kℓ cos 2θ − 6kℓ cos 2 θ  ℓ θ 2 = −6kℓ 2 θ 2 sin 2 θ ≤ 0 Thus, the bent configuration is always unstable.

P 253. A rigid bar of length ℓ is pinned and restrained by a rotational spring of modulus k at the bottom. It is subjected to a force P at the top. The force changes its direction with the motion of the bar. If the bar rotates by an angle θ then ℓ the load rotates an angle αθ in the opposite sense (α is a known constant). Find a suitable virtual work function for k the system? (Hint: start with a classical equilibrium equation from a freebody diagram of the bar). Does an energy function exist? If so, then find it. Estimate the buckling load of the system.

αθ P θ

(a) Find a suitable virtual work function for the system? Consider a freebody diagram with moment M (and reaction forces) acting at the fixed end. Summing moments about the base gives M = kθ = P cos αθ ℓ sin θ + P sin αθ ℓ cos θ = Pℓ cos αθ sin θ+ sin αθ cos θ  = Pℓ sin(1+α) θ Hence, k θ−Pℓ sin(1+α) θ = 0 is the classical equilibrium equation. We can develop a virtual work functional by the method of weighted residuals G(θ, θ) ≡  k θ − Pℓ sin(1+α) θ  θ Then G(θ, θ) = 0 for all θ implies equilibrium by the fundamental theorem of the calculus of variations. This is the principle of virtual work. (b) Does an energy function exist? If so, then find it. Symmetry of the virtual work functional determines the existence of the energy functional. Taking the directional derivative of the virtual work functional gives ^



^

^



DG(θ, θ) ⋅ θ = k θ − (1+α) Pℓ cos (1+α)  θ θ ^

^

=  k − (1+α) Pℓ cos(1+α)θ  θ θ = DG(θ, θ) ⋅ θ

300

Fundamentals of Structural Mechanics ^

The symmetry is evident simply by swapping θ and θ in the expression above. The symmetry proves that an energy functional E(θ) exists. The energy functional can be computed by Vainberg’s theorem.

  k t θ − Pℓ sin (1+α) t θ   θ dt 1

E(θ) =

0

The second term in the integral can be integrated by letting u =  1+α  t θ so that du =  1+α  θ dt. Carrying out the integral we get E(θ) = 12 k θ 2 +

Pℓ Pℓ cos(1+α)θ − 1+α 1+α

Of course, this can be verified by taking the directional derivative of the energy and showing that it gives the virtual work functional. (c) Estimate the buckling load of the system. The governing equation is k θ − Pℓ sin(1+α) θ = 0. It is evident that one solution is θ = 0. For θ ≠ 0 the equation is satisfied only for certain values of the load. In particular, for loads that satisfy θ P=k ℓ sin (1+α)θ  The critical load (buckling load) is the limit of this load as θ → 0. To wit, k 1 lim P = k = ℓ (1+α) cos(1+α)θ ℓ (1+α) θ→0 Hence, there is a bifurcation at Pcr =

k ℓ (1+α)

We can check the stability of the straight configuration with the second derivative test. The second derivative of the energy is A(θ, θ) =  k − (1+α) Pℓ cos(1+α) θ  θ 2 At θ = 0 we have A(0, θ) =  k − (1+α) Pℓ  θ 2 , therefore the system is stable for P<

k ℓ (1+α)

254. Two rigid bars, each of length ℓ are hinged together and attached to two linearly elastic springs of modulus k. The bottom end of the vertical member is on a roller that rolls on a horizontal plane. The right end of the horizontal member is on a roller that rolls on a slope. The column is subjected to a vertical force P. Find an expression for the energy of the system. Find the equilibrium configurations of the system. Find the critical loads of the structure.

P ℓ

k

k

4

ℓ

3

Chapter 10

301


(a) Find an expression for the energy of the system. ∆ + ℓ sin θ x

θ ℓ cos θ

5 x 4

DOF : ∆ , θ ∆

The energy functional can be established from the quantities shown in the sketch. For a linearized motion x ≈ ∆ + ℓ θ. Therefore, E(∆ , θ) = 12 k ∆ 2 + 12 k  54  ∆+ℓ θ   2 − Pℓ cos θ = 0 (b) Find the equilibrium configurations of the system. The virtual work functional can be found by differentiating the energy functional. To wit, G(∆, θ, ∆, θ ) = k ∆ ∆ + 25 k   ∆+ℓ θ  ∆+ℓ θ  − P ℓ θ θ = 0 16 Grouping terms with common virtual displacement measure we have

 41 k ∆ + 25 k ℓ θ ∆ + 25 k ℓ ∆ + 25 k ℓ 2 θ − P ℓ θ θ = 0 16 16 16 16

∀ ∆, θ

Therefore, the equations of equilibrium are 41 k 16

∆ + 25 kℓ θ = 0 16

25 k 16

ℓ ∆ + 25 k ℓ2 θ − P ℓ θ = 0 16

(c) Find the critical loads of the structure. From the first equilibrium equation we can determine ℓθ ∆ = − 25 41 Substituting this result into the second equation gives 25 kℓ 16

− 25 ℓ θ + 25 k ℓ2 θ − Pℓ θ = 0 41 16 16 k ℓ2 41

θ − 16 Pℓ θ = 0 25

One solution for this equation is θ = 0. A nontrivial solution is possible for the load level Pcr = which is the buckling load of the system.

25 kℓ 41

302


255. A ladder of length ℓ=20 ft leans against a wall with the base 4 ft from the wall. Both ends are frictionless and the bottom end is restrained by an elastic spring of modulus k=10 lb/ft. What is the maximum height x (measured along the ladder as shown) that a person of weight W=200 lb can climb? The ladder can be assumed rigid, the rollers are very small relative to the length of the ladder, and the person climbs slowly enough to neglect dynamic effects.

W x k

4 ft

W

x

x cos θ

θ

sin θ 0 = 3 5

x sin θ ℓ sin θ

The energy of the system can be established from the quantities given in the sketch. To wit, E(θ) = 12 k ℓ2(sin θ− sin θ 0) 2 + W x cos θ The virtual work functional can be determined by differentiating the energy as G(θ, θ) = DE(θ) ⋅ θ =  k ℓ 2  sin θ− sin θ 0  cos θ − W x sin θ  θ The condition G(θ, θ) = 0 for all θ gives the equation k ℓ 2  sin θ− sin θ 0  cos θ − W x sin θ = 0 Any ( x, θ) pair that satisfies this equations is an equilibrium configurations. Hence, we can write x(θ) = kℓ W

2

 1− sinsinθθ  cos θ 0

The value of θ that gives the greatest value of x can be determined from



 



dx = kℓ 2 sin θ 0 cos 2 θ − 1− sin θ 0 sin θ = 0 W sin θ dθ sin θ from which we find sin θ 0  cos 2 θ+ sin2 θ  − sin3 θ = 0 From trigonometry we know that

⇒

sin θ cr = sin1∕3 θ 0

Chapter 10

303

Fundamental Concepts in Static Stability cos θ cr =  1− sin2∕3 θ 0  1∕2

Substituting these results into the equation for x(θ) we get 2 xmax = kℓ  1− sin 2∕3 θ 0 3∕2 W

=



2∕3 (10)(20) 2 1 −  0.6  (200)



3∕2

= 3.1 ft < 20 ft Therefore, the maximum occurs inside the range [0,20]. Check stability for all x Á  0, 20  A(θ, θ) =  kℓ2  cos 2 θ− sin2 θ + sin θ sin θ 0  − W x cos θ  θ 2





= kℓ2 cos 2 θ− sin2 θ+ sin θ sin θ 0− 1−

= kℓ2

 sin θsin−θsin θ θ 0

3





sin θ 0 cos 2 θ θ 2 sin θ

2

Because A(θ, θ) > 0 for all θ < θ c r it is unsafe to climb up the ladder past the distance x = 3.1 ft!

256. Four rigid bars are hinged together and subjected to the load P as shown. The two horizontal bars are restrained by a linear, elastic rotational spring of modulus k. Find an expression for the energy of the system. Find an equation describing the equilibrium configurations of the system. Find the bifurcation load.

P k

ℓ

4 ℓ

3

ℓ

ℓ

(a) Find an expression for the energy of the system. The geometry of the system can be characterized by the angles α and θ(α). Note that the system has a single degree of freedom with relations shown P 2α ℓ

α

ℓ

2ℓ cos α + 52 ℓ cos θ = 4ℓ 5 ℓ 4

θ

5 cos θ = 8 − 4 cos α −5 sin θ Dθ ⋅ α = 4 sin α α Dθ ⋅ α = − 4 sin α α 5 sin θ

304


The energy of the system can be established from the quantities shown in the sketch as E(α) =

1 2

k  2α  2 + P  54 ℓ  sin θ(α)

(b) Find an equation describing the equilibrium configurations of the system. The virtual work functional can be found by differentiating the energy as G(α, α) = DE(α) ⋅ α = 4 k α α + 54 Pℓ cos θ  Dθ ⋅ α  The condition G(α, α) = 0 for all α (upon substituting the expression for Dθ ⋅ α) gives the equation 4 k α − P ℓ cos θ sin α = 0 sin θ where, from the kinematic constraint we have cos θ =  8−4 sin α ∕5 and we can get sin θ =  1− cos 2 θ  −1∕2. (c) Find the bifurcation load. One solution for the equilibrium equation is α = 0. A nontrivial solution exists for certain values of the load P. Linearizing the equilibrium equation for α ≪ 1, noting that sin α ≈ α,

cos θ ≈ cos θ 0 = 45 ,

sin θ ≈ sin θ 0 = 35

we find the linearized equation 4  k− 13 Pℓ α = 0 from which the critical buckling load can be found as Pcr =

3k ℓ

kθ 257. Consider the linkage of two rigid bars subjected to P axial load P as shown. The linkage has length 4ℓ, is k k pinned at the left end, and has elastic springs that resist motion. The translational springs accrue force in proporℓ ℓ ℓ ℓ tion to their extension, with modulus k. The rotational spring accrues force in proportion to the its relative angle change, with modulus kθ = kℓ2. Find the critical loads and linearized buckling mode shapes of the system.

θ2 θ1

Chapter 10


305

The energy can be established from the quantities defined in the sketch as E(θ) = 12 k  ℓ sin θ 1  2 + 12 k  2ℓ sin θ 1+ℓ sin θ 2  2 + 12 k ℓ 2  θ 2−θ 1  + 2Pℓ  cos θ 1+ cos θ 2−2  To solve the bifurcation problem we need only a quadratic energy functional. Using the Taylor series expansion we can quadratify the energy to E(θ) = 12 k ℓ2  6θ 21 + 2θ 1θ 2 + 2θ 22  − Pℓ θ 21+θ 22  The virtual work functional can be found by differentiating the energy as G(θ, θ) = DE(θ) ⋅ θ = k ℓ2  6θ 1θ 1+θ 2θ 1+θ 1θ 2+2θ 2θ 2  − 2 Pℓ  θ 1θ 1+θ 2θ 2  The condition G(θ, θ) = 0 for all θ gives the equilibrium equations k ℓ 2  6θ 1+θ 2  − 2 Pℓ  θ 1  = 0 k ℓ 2  θ 1+2θ 2  − 2 Pℓ  θ 2  = 0 Letting m ≡ 2P∕ k ℓ we can write these equations in the form 1

6−m 1

0

θ1

2−m

θ2

=

0

This system of equations has a nontrivial solution only if the determinant of the coefficient matrix is equal to zero. This gives the equation  6−m   2−m  − 1 = 0

This equation can be simplified to m 2−8m+11 = 0, which has roots m 1 = 4− 5 ,

m 2 = 4+ 5

The buckling mode shape can be found by substituting back into the governing equation for the critical values, specifically for m 1 = 4− 5 we have 2+ 5 1

1

−2+ 5

1 a

=

0

a = −2− 5

0

φ1 =  1, −2− 5 

0

a = −2+ 5

and m 2 = 4+ 5 we have 2− 5 1

1

−2− 5

1 a

=

0

φ2 =  1, −2+ 5 

306


258. Consider the frame composed of rigid bars subjected to the load P as shown. The rigid members are hinged at the top left corner, with an elastic spring that resists relative rotation. The rotational spring accrues force in proportion to the its relative angle change, with modulus k. Find the buckling load for this system. Determine the post-buckling response of the system in terms of the rotation of the left column. Does it make a difference if the frame buckles to the left or to the right?

P

k

ℓ ℓ 3ℓ∕2

P 3ℓ∕2

2ℓ  1− cos ψ 

ψ

θ ℓ

2ℓ

ψ

ℓ + ℓ cos θ + 32 ℓ sin ψ = 2 ℓ cos ψ

The energy can be established in terms of the quantities shown in the sketch. The system has one degree of freedom. Hence, the angles θ and ψ are related. In particular, the kinematic constraint has the form 2 + 2 cos θ + 3 sin ψ − 4 cos ψ = 0

We can think of ψ(θ) as a dependent function. Then the energy can be written as E(θ) = 12 k  θ+ψ(θ)  2 + 2 Pℓ cos ψ(θ) The virtual work functional can be found by differentiating the energy G(θ, θ) = k  θ+ψ   θ + Dψ ⋅ θ  − 2Pℓ sin ψ Dψ ⋅ θ From the constraint −2 sin θθ +  3 cos ψ+4 sin ψ  Dψ ⋅ θ = 0. Hence, Dψ ⋅ θ =

2 sin θ θ ≈ 23 θθ 3 cos ψ+4 sin ψ

Also, from the constraint equation we can determine from Taylors theorem, for small values of the angle −θ2 + 3ψ = 0 . Hence, the virtual work functional can be written approximately as G(θ, θ) = k  θ + 13 θ 2  1+ 23 θθ − 2Pℓ 13 θ 2  23 θθ  = 0 The condition G(θ, θ) = 0 for all θ gives the equilibrium equations θ  k  1+ 13 θ  1+ 23 θ  − 49 Pℓθ 2  = 0

Chapter 10

307


The trivial solution θ = 0 satisfies this equation. For a nontrivial solution we must have the load level of  3+θ  1+θ  P= k 2ℓ θ2

θ = 0 is stable for any P. There is no bifurcation load.

259. Two rigid bars, each of length 2ℓ are connected by a linear elastic spring of length ℓ and modulus k. The right vertical bar is subjected to a force P as shown. The left bar is attached to a vertical spring of modulus k that has been stretched into place, giving it an initial tension force of To (i.e., when P=0). Write the energy functional for the system. Find the lowest buckling load Pcr of the system.

P

ℓ ℓ

k

ℓ ℓ

k, T o ℓ

No solution available.

260. Two rigid bars, each of length 2ℓ are connected by a single P rigid bar of length ℓ which is hinged at the ends. A weight of fixed ℓ value W hangs from the left bar while the right vertical bar is subℓ jected to a force P as shown. Note: there are no elastic elements in ℓ this system! How many degrees of freedom does the system have? ℓ W Write the exact energy functional for the system. (Hint: You can describe the deformation in terms of the rotation angles of each member, but you must write equations of constraint relating those angles ℓ to your chosen degrees of freedom). Find the critical value of P at which buckling of the system takes place. Is the post-buckling behavior symmetric or asymmetric? Do you expect the post-buckling behavior to be stable or unstable?

(a) How many degrees of freedom does the system have? System has 1 DOF. Because if we fix the rotation of any bar, we cannot rotate the remaining bars freely. (b) Write the exact energy functional for the system. (Hint: You can describe the deformation in terms of the rotation angles of each member, but you must write equations of contraint relating those angles to your chosen degrees of freedom).

308


ℓ cos β ℓ sin β

2ℓ cos θ β θ

α 2ℓ cos α

2ℓ sin α

The energy can be established from the quantities defined in the sketch. The system has one degree of freedom, θ. It is convenient to express the position in terms of two additional angles α(θ) and β(θ). The constraint equations relating these angles 3ℓ = ℓ cos θ + ℓ sin β + 2ℓ cos α ℓ = ℓ sin θ + ℓ cos β − 2ℓ sin α The energy has the form E(θ) = −2ℓ W cos θ + 2ℓ P cos α(θ) (c) Find the critical value of P at which buckling of the system takes place. The virtual work functional can be determined from the energy by differentiation. To wit, G(θ, θ) = −2ℓ W sin θ θ + 2ℓ P sin α Dα ⋅ θ For small angles we have sin θ ≈ 2 sin α, θ ≈ 2α, and θ ≈ 2α. Hence, the work functional can be linearized to G(α, α) = −2ℓ  4 Wα − Pα  α = 0

∀ α

Therefore, the critical load can be computed as Pcr = 4W (d) Is the post-buckling behavior symmetric or asymmetric? Do you expect the post-buckling behavior to be stable or unstable? Asymmetric. Stable in one direction, unstable in the other.

Chapter 11 The Planar Buckling of Beams

261. Vainberg’s theorem is simply a statement of integrability. This theorem can be applied to the strain variations that we derive through a virtual work argument for a nonlinear planar beam. Let u ≡  u, w, θ  and u ≡  u, w, θ  be the real and virtual displacements and rotation. From the principle of virtual work, we have found that the virtual curvature is given by κ o(u, u) = θ′. Show that the symmetry condition holds for the virtual curvature, i.e., Dκ o(u, u) ⋅ u^ = Dκ o(u, u^ ) ⋅ u and, hence, that it is integrable. Show that the real curvature is given by κ o = θ′. Note that the directional derivative of κ o is Dκ o(u, u) ⋅ u^ = d  κ o(u+ε u^ , u)ε=0 dε and the integral of the virtual curvature can be computed by Vainberg’s formula as

 κ ( t u, u ) dt 1

κo =

o

0

where t u =  tu, tw, tθ . Repeat the calculation for the virtual shear and axial strains β o(u, u) = w′ cos θ − u′ sin θ − [w′ sin θ + (1+u′) cos θ] θ Á o(u, u) = u′ cos θ + w′ sin θ + [w′ cos θ − (1+u′) sin θ] θ to get the real shear and axial strains β o = w′ cos θ − (1+u′) sin θ Á o = w′ sin θ + (1+u′) cos θ − 1 Take the directional derivatives of the real strains to verify that these results are correct.

The virtual curvature is integrable because ^

^

Dθ(Ω, Ω) ⋅ Ω = 0 = Dθ(Ω, Ω) ⋅ Ω

310


The integral of the virtual curvature is simply

 κ (tΩ, Ω) dt =  θ′ dt = θ′ 1

κo =

1

o

0

0

It follows from the definition of the virtual shear strain that ^

^

Dβ o(Ω, Ω) ⋅ Ω = −(w′ sin θ + u′ cos θ) θ



^



− w^ ′ sin θ + u^ ′ cos θ +  w′ cos θ − (1+u′) sin θ  θ θ ^

^

= − w′θ+w^ ′θ  sin θ +  u′θ+u^ ′θ  cos θ ^

−  w′ cos θ − (1+u′) sin θ  θ θ which is clearly symmetric (just swap hats for bars and observe that the result does not change). Hence, ^

^

Dβ o(Ω, Ω) ⋅ Ω = Dβ o(Ω, Ω) ⋅ Ω and the real shear strain is given by Vainberg’s theorem as

 β (tΩ, Ω) dt 1

βo =

o

0

  w′ cos tθ − u′ sin tθ −  tw′ sin tθ + (1+tu′) cos tθ  θ  dt 1

=

0

 w′θ sin tθ + u′θ cos tθ − w′θ  sin tθ−tθ cos tθ  − sin tθ − u′θ  cos tθ−tθ sin tθ  

=

1 0

= w′ cos θ − (1+u′) sin θ

The real axial strain can be obtained in exactly the same way. One can again demonstrate the required symmetry. Therefore, the axial strain can be computed from the virtual axial strain by Vainberg’s theorem as

 Á (tΩ, Ω) dt 1

Áo =

o

0

=   w′ sin tθ + u′ cos tθ +  tw′ cos tθ − (1+tu′) sin tθ  θ  dt 1

0

=

 − w′θ cos tθ + u′θ sin tθ + w′θ  cos tθ+tθ sin tθ  + cos tθ − u′θ  sin tθ−tθ cos tθ  

= w′ sin θ + (1+u′) cos θ − 1

1 0

Chapter 11

The Planar Buckling of Beams

311

Note that the following two integral identities have been used

 x sin x = sin x − x cos x,

 x cos x = cos x + x sin x

The directional derivatives of the real shear and axial strains are Dβ o(Ω) ⋅ Ω = w′ cos θ − w′ sin θ θ − u′ sin θ − (1+u′) cos θ θ = β o(Ω, Ω) DÁ o(Ω) ⋅ Ω = w′ sin θ + w′ cos θ θ + u′ cos θ − (1+u′) sin θ θ = Á o(Ω, Ω)

262. Consider the simply supported column of length ℓ P EI and flexural modulus EI. Assume that shear and axial deformations are negligible, so that the constraints of Euler’s elastica are appropriate. Compute the critical loads for this ℓ column by solving the equation EIwiv+Pw′′ = 0 with the appropriate boundary conditions. Carry out the stability analysis parallel to the analysis done for the cantilever model problem in the text.

The general solution to the the fourth-order differential equation is w(x) = a 0 + a1 x + a 2 sin mx + a3 cos mx where m 2 ≡ P∕EI, and a0, a1, a2, and a3 are integration constants. The boundary conditions at the left end, w(0) = 0 and w′′(0) = 0 immediately yield the constants a0 = a3 = 0. At the right end, w(ℓ) = 0 and w′′(ℓ) = 0 give a2 sin mℓ = 0

a1ℓ + a 2 sin mℓ = 0

The first of these substituted into the second gives a1 = 0. The condition for existence of a non-trivial solution (i.e., w(x) ≠ 0) is, therefore, sin mℓ = 0 This equation is the characteristic equation, and has solutions only for values of the load parameter m n = nπ∕ℓ, for n = 1, 2, 3, . The critical loads of the system are, therefore, Pn = n2π 2EI∕ℓ2. Noting that a2 ≠ 0 only when m = m n we can substitute back into the expression for the displacement to get the buckling modes that correspond to the critical loads. The nth buckling mode is

 

wn(x) = a2 sin nπx ℓ where the amplitude a2 is undetermined.

(b) Carry out the stability analysis parallel to the analysis done for the cantilever model problem in the text. Since the functions wn(x) form a complete basis, the test function w(x) can be written as a linear combination of them. To wit,

312

Fundamentals of Structural Mechanics w(x) =

∞

 a w (x) n

n

n=1

Then the second-derivative of E(w) for the trivial equilibrium is (see page 415 of the text) A(0, w) =

∞



 (w ′) dx ℓ

a 2n ( Pn−P)

n=1

n

2

0

Therefore, the trivial equilibrium is stable for P < P 1 and unstable for P > P 1.

263. Consider the simply supported column of length ℓ q(x) P and flexural modulus EI. Assume that shear and axial deformations are negligible so that constraints of Euler’s EI elastica are appropriate. Can the classical elastica theory ℓ be extended to accommodate the transverse load q(x)? What difficulties do you encounter when you attempt to do so? Can the virtual-work principle for the elastica be modified to account for the transverse load?

For this problem, p = m = 0. The horizontal and vertical forces are readily available from overall equilibrium of the freebody diagram shown in the sketch. To wit,

 q(ξ) dξ ℓ

H = −P,

V(x) =

x

The moment equation (Eqn. 549 in the text) reduces to EIθ ′′ + V(x) cos θ + P sin θ = 0 with the boundary condition θ(0) = 0 and θ′(ℓ) = 0. The differential equation is harder to solve than that for Euler’s elastica because the coefficient of cos θ varies with x. The inextensibility constraints imply that Á o = β o = 0. Furthermore, we have Mθ| ℓ0 = 0 and Vw| ℓ0 = 0, either from the natural boundary conditions by imposing homogeneous essential boundary conditions on w and θ. Consequently, the virtual work functional for the problem reduces to

  EI θ′θ′ − q w  dx + Pu  ℓ

G(Ω, Ω) =

0

ℓ 0

Using the inextensibility constraints w′ = sin θ and u′ = cos θ−1, we get w′ = cos θ θ,

u′ = − sin θ θ

Chapter 11


313

Noting that V′ = −q we can do the following computation

 q(x) wdx = −  V′(x) w dx =  V(x) w′ dx − Vw  ℓ

ℓ

0

0 ℓ

0

ℓ 0

 V(x) cos θ θ dx ℓ

=

0

Also, we can compute

u



 u′dx = −  sin θ θ dx ℓ

ℓ

=

0

ℓ

0

0

Substituting these results into the virtual work functional we get

  EI θ′θ′ −  V(x) cos θ + P sin θ  θ  dx ℓ

G(θ, θ) =

0

which, by the fundamental theorem of the calculus of variations, is equivalent to the moment equilibrium equation. If we had, for example, q(x) = q o then we can integrate the equation for the transverse force as V = q o ℓ−x  and the virtual work functional in this particular case would be

  EI θ′θ′ − q  ℓ−x  cos θ θ − P sin θ θ  dx ℓ

G(θ, θ) =

o

0

264. Consider the bar of length ℓ with bending modulus EI P EI, fixed at the left end, propped at the right end, and subjected to axial load P as shown. Assume that shear and axial ℓ deformations are negligible. Compute the critical loads for this column by solving the classical differential equation. Estimate the critical loads using the principle of virtual work in conjunction with the Ritz method. Use a polynomial basis.

(a) Compute the critical loads for this column by solving the classical differential equation. The linearized equation governing the column is EIwiv + Pw′′ = 0 The general solution to the equation is w(x) = a 0 + a1x + a 2 sin mx + a3 cos mx

314


where m 2 ≡ P∕EI. Substituting the boundary conditions w(0) = 0, w′(0) = 0, w(ℓ) = 0, and w′′(ℓ) = 0 into the general solution yields a0 + a 3 = 0 a1 + ma 2 = 0 a0 + a 1ℓ + a2 sin mℓ + a 3 cos mℓ = 0 a2 sin mℓ + a 3 cos mℓ = 0 The first two equations give a0 = −a3 and a1 = −ma2. These can be substituted into the second two equations to give the system sin mℓ−mℓ

cos mℓ−1

sin mℓ

cos mℓ

0

a2 a3

=

0

Non-trivial solution w(x) ≠ 0 exists only if the determinant of the coefficient matrix vanishes. Thus, sin mℓ − mℓ cos mℓ = 0 The first solution to the transcendental equation is found to be m 1 = 4.4934∕ℓ. The critical load is P1 = 20.16EI∕ℓ2 . The corresponding buckling mode is w1(x) = a2  sin m 1x−m 1x  − a 3  1− cos m 1x  Observing that a2 sin mℓ+a 3 cos mℓ = 0 we can express the buckling mode as w1(x) = a^ 2   sin m 1x−m 1x  cos m 1ℓ +  1− cos m 1x  sin m 1ℓ  where a^ 2 = a2∕ cos m 1ℓ. The buckling mode shape is shown below. P

(b) Estimate the critical loads using the principle of virtual work in conjunction with the Ritz method. Use a polynomial basis. The virtual work functional for the problem takes the form G(w, w) =   EIw′′w′′ − Pw′w′  dx ℓ

0

Let the approximate displacement be 2 3 4 w(x) = b 0ℓ + b1 x + b 2 x + b3 x 2 + b 4 x 3 ℓ ℓ ℓ

The function must satisfy the essential boundary conditions of the beam, i.e., it must satisfy w(0) = w′(0) = w(ℓ) = 0. The first two conditions give b0 = 0 and b1 = 0. The third condition gives b2+b 3+b4 = 0. Substituting these conditions into the original function, letting a1 = b3, a2 = b4, and ξ ≡ x∕ℓ, gives

Chapter 11


315

w(ξ) = a 1  ξ3−ξ 2 ℓ + a 2  ξ4−ξ 2 ℓ Thus, h1(ξ) =  ξ3−ξ 2 ℓ and h2(ξ) =  ξ4−ξ 2 ℓ are the base functions. Let the virtual displacement be w = a 1 h1+a 2 h2. The virtual work functional can be expressed in the form G(w, w) = a T  Ka−λGa  where the components of the coefficient matrices are computed as

 h ′′ h ′′ dξ 1

Kij =

i

j

 h ′h ′ dξ 1

Gij =

0

i

j

0

and λ ≡ Pℓ 2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional. We can perform the indicated integrals to get the equations 4− 2 λ

15 7 8− 30 λ

7 8− 30 λ

a1

84 44 − 105 λ 5

a2

0 =

0

The characteristic equation is obtained by setting the determinant of the coefficient matrix equal to zero 32 1 2 λ − 175 λ + 16 700 5

= 0

which is readily solved to give λ1 = 20.919 and λ2 = 107.1. Therefore, the critical loads are P1 = 20.919EI∕ℓ2 and P2 = 107.1EI∕ℓ2. The fundamental critical load obtained by the Ritz method is about 4% higher than the exact value found by the classical method. The MATHEMATICAT commands needed to solve this problem are given below h = { x^3 - x^2, x^4 - x^2} hp = D[h,x] hpp = D[hp,x] K = Integrate[ Outer[Times,hpp,hpp],{x,0,1}] G = Integrate[ Outer[Times,hp ,hp ],{x,0,1}] F = K - m G d = Det[F] Solve[d==0,m]

316


265. The prismatic beam shown below has a cross seck P EI tion that is symmetric with respect to the plane of the page. The cross section has flexural modulus EI. Axial ℓ and shear deformations can be neglected. The beam has a deformable spring support at the left end that elastically restrains rotations. The moment developed by the spring is related to the rotation at that point by M s = kθ s, where θ s = w′(0) is the rotation experienced by the spring. The right end of the beam is free to translate and to rotate. Solve the linearized buckling problem by the classical method, i.e., by integrating the differential equation. What are the appropriate boundary conditions for this problem? Solve the problem by integrating the differential equations and using the boundary conditions to find the constants of integration. What are the critical loads of the system? What is the smallest critical load as k → 0? What is the smallest critical load as k → ∞? Into what shapes does the beam deform at the critical loads? What are the shapes as k → 0? What are the shapes as k → ∞? (a) What are the appropriate boundary conditions for this problem? M(0)

P V(ℓ)

kw′(0)

The essential boundary condition for the beam is w(0) = 0, the natural boundary conditions are M(0) = kw′(0), M(ℓ) = 0, and V(ℓ) = 0. These conditions can be expressed in terms of w as follows: M(0) = kw′(0)

⇒ w′′(0) − αw′(0)∕ℓ = 0

M(ℓ) = 0

⇒ w′′(ℓ) = 0

V(ℓ) = 0

⇒ w′′′(ℓ) + m 2 w′(ℓ) = 0

where m 2 ≡ P∕EI and α ≡ kℓ∕EI have been introduced. (b) Solve the problem by integrating the differential equations and using the boundary conditions to find the constants of integration. The linearized equation governing the beam is EIwiv + Pw′′ = 0 The general solution to the equation is w(x) = a 0 + a1 x + a 2 sin m x + a3 cos m x Substituting the general solution into the boundary conditions yields a0 + a 3 = 0 a2 sin mℓ + a 3 cos mℓ = 0 m 2 a3 + α  a 1+ma2 ∕ℓ = 0 m  a 3 sin mℓ − a2 cos mℓ  + a 1 + ma2 cos mℓ − ma3 sin mℓ = 0

Chapter 11


317

The first equation gives a0 = −a3 and the last equation gives a1 = 0. The remaining two equations can be expressed as sin mℓ

cos mℓ

a2

m2

a3

αm∕ℓ

0 =

0

A nontrivial solution exists only when the determinant of the coefficient matrix above is equal to zero. Hence, the characteristic equation is mℓ sin mℓ − α cos mℓ = 0 There may be solutions to the characteristic equation for certain values of m (see part (c) below). However, observe that the four equations that result from the boundary conditions must be used to determine five unknowns, the constants a0, a1, a2, a3 and the load parameter m. If there exist solutions to the characteristic equation then we cannot completely determine the constants a0, a1, a2, and a3. The best we can hope to do is find three of them in terms of one that cannot be determined. If there is no solution to the characteristic equation then the solution is a0= a1= a2= a3=0. One way to visualize the solutions of the characteristic equation is to let x ≡ mℓ and rewrite the characteristic equation as tan x =

α x

The functions f (x) = tan x and f (x) = α∕x are easily visualized as shown in the sketch below. The solutions to the equation, xi, are the intersections of these two curves. The first two solutions are shown in the sketch. Since the tangent function is asymptotic to the value π∕2 we can observe that x1 < π∕2 and that π∕2 < x 2 < 3π∕2. The hyperbolic function f (x) = α∕x is asymptotic to zero for large values of x. Since the tangent function crosses the axis at π, 2π, 3π, . . ., we can observe that xn ≈  n−1  π for large values of n. f (x)

tan x

tan x

α x x1

π 2

x2

x

(c) What are the critical loads of the system? What is the smallest critical load as k → 0? What is the smallest critical load as k → ∞? The characteristic equation from part (b) can be solved numerically for specific values of the parameter α. For example,

318


if α = 2 the equation is satisfied for the following values of the load parameter: m 1 = 1.077∕ℓ, m 2 = 3.644∕ℓ, and m 3 = 15.83∕ℓ. The fundamental critical load is, = 1.160EI∕ℓ2. therefore, P(α=2) 1 As k → 0 the parameter α → 0, and the characteristic equation reduces to the equation mℓ sin mℓ = 0. The first solution is m 1 = 0, which yields a critical load of zero. This is expected since, when k = 0, the beam can rotate about the left end with no resistance. A stability analysis will reveal that the straight configuration is unstable for any value of P > 0. The higher critical values are given by m n = (n−1) π∕ℓ for n=2, 3, .... These values are essentially those of a pinned-pinned beam. We can consider the other extreme, i.e., as k → ∞, by dividing the characteristic equation by α and then taking the limit as α → ∞. In the limit, the characteristic equation is cos mℓ = 0. The solution to the characteristic equation is = π2 EI∕4ℓ 2. This result m n = (2n−1) π∕2ℓ. The fundamental critical load is P(α→∞) 1 is again in agreement with our intuition since now the beam is cantilevered at the left end. (d) Into what shapes does the beam deform at the critical loads? What are the shapes as k → 0? What are the shapes as k → ∞? The buckling mode is wn(x) = a^ 3  α  1− cos m n x  + m 2nℓ sin m n x  where a^ 3 = a3∕α. Since the mℓa3+α a2 = 0 we can determine that as k → ∞ we must have a2 = 0 and the buckling mode shape is w(α→∞) = a3  1− cos m nℓ  n The case where k → 0 is more complicated because m → 0 as α → 0. Consider m 1(α) to be a function of α. Because m ≪ 1 we can expand the sine and cosine functions approximately as 1− cos θ ≈ 12 θ 2 and sin θ ≈ θ. Thus, we have w1 = a^ 3  12 α m 21 x2 + m 31ℓ x  The characteristic equation reduces to m 2ℓ 2 − α 1− 12 m 2ℓ 2  = 0 which means that α ∝ m 2 . Consequently, the term αm 21 vanishes faster than the term m 31 as α → 0. In the limit, then, the mode shape for the case k → 0 is simply w(α→0) = a~ x 1 The rigid body motion for the mode shape is the expected result.

266. For Problem 265, the virtual-work functional that accounts for the work done by the springs and by the axial force is given by the expression

Chapter 11

319


  EI w′′w′′ − Pw′w′ dx + k w′(0) w′(0) ℓ

G(w, w) =

0

Estimate the buckling loads of the beam using a two-term polynomial expansion for the transverse deflection. That is, assume the real and virtual transverse deflections to be 2 2 w(x) = a 1 x + a2 x , w(x) = a 1 x + a2 x ℓ ℓ Repeat the calculation with a three-term polynomial. The classical solution gives an infinite number of critical loads. How many did the two-term approximation give? Why? Were the critical loads higher or lower than the exact values? Why? Discuss what is good and bad about the assumed shapes. Could the approximating functions be improved easily? Suggest a better approximation.

The base functions are h1(ξ) = ξℓ and h2(ξ) = ξ2ℓ. The virtual work functional can be expressed as G(w, w) = a T  Ka−λGa  where the components of the coefficient matrices are computed as

 h ′′ h ′′ dξ + α h ′(0) h ′(0) 1

Kij =

i

j

i

 h ′h ′ dξ 1

Gij =

j

0

i

j

0

where α ≡ kℓ∕EI and λ ≡ Pℓ 2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional. We can perform the indicated integrals to get the equation −λ

a1

4− 43 λ

a2

α−λ −λ

0 =

0

The characteristic equation is obtained by setting the determinant of the coefficient matrix equal to zero 1 2 λ 3

−  12+4α  λ + 12α  = 0

which is readily solved to give λ1, 2 = 6+2α  2 9+3α+α 2 Therefore, the critical loads are P1 = λ1(α)EI∕ℓ 2 and P2 = λ2(α)EI∕ℓ 2. In the limit as α → 0 we have λ1 → α 2 → 0 (show this by taking the Taylor expansion of λ1(α) about α = 0). Dividing the characteristic equation by α and taking the limit as α → ∞ gives the equation 4 λ−12 = 0 giving λ1 = 3 (compare with the exact solution of λ1 = 2.47). (b) Repeat the calculation with a three-term polynomial. Let the three-term approximation be 3

w(x) =

 a h (x), i

i=1

i

3

w(x) =

 a h (x) i

i=1

i

320


with h3(ξ) ≡ ξ3ℓ and h1, h2 unchanged. Following the principle of virtual work the equations for a1, a2 and a3 are α−λ

−λ

−λ

a1

−λ

4− 43 λ

6− 2 λ

3

a2

9 12− λ 5

a3

−λ

3

6− 2 λ

0 =

0 0

The characteristic equation is now cubic in the load parameter λ λ3 − 9 α+8  λ2 + 24 13α+30  λ − 720α = 0 The closed-form expressions for the roots of a cubic equation are very cumbersome. However, for a specific value of α one can numerically calculate the values of the roots. For example, for α = 2 the first three critical loads can be determined as λ1 = 1.160, λ2 = 17.36, and λ3 = 71.48. As usual, the corresponding critical loads are given by Pi = λi EI∕ℓ 2. The exact values of the critical loads were computed in Problem 265(c). The first two of were found to be λ1 = 1.160 and λ2 = 13.28. Obviously, the three-term approximation gives very good results for the fundamental critical load. (c) The classical solution gives an infinite number of critical loads. How many did the two-term approximation give? Why? Were the critical loads higher or lower than the exact values? Why? The number of the critical loads one gets from a Ritz approximation is equal to the number of unknown constants a1, a2, , a N in the expansions of the real and virtual displacement fields. The reason is that for an N-term approximation, the matrices K and G have dimension N by N. Therefore, the characteristic equation det K−λG  = 0 is an Nth-order polynomial in λ with N roots. (Because K and G are symmetric, the roots are all real). For a two-term approximation, then, we get two critical loads. The approximate critical loads are always higher than the exact values due to the constraints introduced by the approximating functions. In a qualitative sense one can view the base functions as containing errors in the form of deviations from the eigenfunctions. Differentiation magnifies the errors. Thus, the product aT Ka is augmented more than the product aT Ga because the former has second derivatives and the latter has first derivatives of the base functions. As a consequence, more constraint leads to a stiffer system and hence higher buckling loads since λ ≈ a T Ka∕aT Ga. (d) Discuss what is good and bad about the assumed shapes. Could the approximating functions be improved easily? Suggest a better approximation. The good aspect of the assumed shape w(x) = a 1 x + a2 x2∕ℓ is that it satisfies the essential boundary condition of the problem w(0) = 0. The bad aspect is that it introduces a nonphysical constraint w′′(x) = 2a 2∕ℓ, a constant curvature over the whole length of the beam. We have improved the approximation by simply adding a cubic term to the approximation.

Chapter 11

321


267. The prismatic beam shown below has flexural EI P modulus EI. Axial and shear deformations can be nek glected. The beam has a spring of modulus k located at x the middle of the span. The force developed by the ℓ spring is related to the deflection at that point by f (ℓ∕2) = kw(ℓ∕2). Find a suitable expression for the virtual-work functional that accounts for the virtual work done by the spring. Estimate the critical loads of the column using the principle of virtual work in conjunction with the Ritz method. Use a polynomial basis.

(a) Find a suitable expression for the virtual work functional that accounts for the virtual work done by the spring. The virtual work functional for the beam is simply

  EIw′′w′′−Pw′w′ dx + k w(ℓ∕2) w(ℓ∕2) ℓ

G(w, w) =

0

(b) Estimate the critical loads of the column using the principle of virtual work in conjunction with the Ritz method. Use a polynomial basis. Let ξ ≡ x∕ℓ. The base functions h1 = ξ  1−ξ  ℓ and h2 = ξ2  1−ξ  ℓ satisfy the

essential boundary conditions for the pinned-pinned beam. The virtual work functional can be expressed as G(w, w) = a T  Ka−λGa  where the components of the coefficient matrices are computed as

 h ′′ h ′′ dξ + i

j

 h ′h ′ dξ 1

1

Kij =

1 α hi (1∕2) h j (1∕2) 2

Gij =

i

j

0

0

where α ≡ 2kℓ∕EI and λ ≡ Pℓ 2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional. We can perform the indicated integrals to get the equation 1 4+ 32 α− 13 λ

1 2+ 64 α− 16 λ

a1

1 2+ 64 α− 16 λ

1 2 4+ 128 α− 15 λ

a2

0 =

0

The characteristic equation is obtained by setting the determinant of the coefficient matrix equal to zero 1 λ−60  32λ−3α−384  3

= 0

The critical loads are given by 3 α, λ1 = 12 + 32

λ2 = 60

Therefore, the critical loads are P1 = λ1(α)EI∕ℓ 2 and P2 = λ2(α)EI∕ℓ 2. Notice that the second critical load does not depend upon the spring stiffness. The MATHEMATICAT commands needed to solve this problem are given below

322

Fundamentals of Structural Mechanics h = {x (1-x) , x^2 (1-x)} hp = D[h,x] hpp = D[hp,x] Kb = Integrate[ Outer[Times,hpp,hpp],{x,0,1}] G = Integrate[ Outer[Times,hp ,hp ],{x,0,1}] x = 1/2 Ks = a Outer[Times,h,h]/2 K = Kb + Ks F = K - m G d = Det[F] Solve[d==0,m]

268. Estimate the critical loads of the column in Problem 267 using the principle of virtual work in conjunction with the Ritz method. Use the eigenbasis of the problem without the spring, that is wn(x) = sin m n x, where m n = nπ∕ℓ.

Let the real and virtual displacement fields be N

w(x) =

 a w (x), n

n

N

w(x) =

n=1

 a w (x) n

n

i=1

where N is the number of functions used in the approximation. The virtual work functional is G(a, a) = a T  Ka−λGa  where the ijth components of the coefficient matrices are give by Kij = ℓ

3



ℓ

w i′′ wj′′ dx + 12 α w i(ℓ∕2) wj(ℓ∕2),

 w ′ w ′ dx ℓ

Gij = ℓ

0

i

j

0

where α ≡ 2kℓ 3∕EI and λ ≡ Pℓ 2∕EI. Note that EI∕ℓ3 has been factored out of the virtual work functional. Since wi (x) and wj (x) are eigenfunctions of the problem without the spring, they satisfy the orthogonality conditions

 w ′′ w ′′ dx = ℓ

ℓ3

i

j

 w ′ w ′ dx = ℓ

1 4 4 m ℓ δij , 2 i

ℓ

0

i

j

1 2 2 m ℓ δ ij 2 i

0

where the usual summation convention is suspended. For the spring contribution to the matrix K we note that wi (ℓ∕2) = sin(iπ∕2). Thus, 1

w i (ℓ∕2) wj (ℓ∕2) =

(−1) 2

(i+j)−1

0

Noting that m nℓ = nπ, the principle of virtual work yields

i, j both odd otherwise

Chapter 11 π2  π2−λ +α

0

−α



a1

0



a2

4π2  4π2−λ 

0

9π2  9π2−λ +α

0

−α ⋮

323


⋮

⋮

 ⋅

⋅

a3 ⋅

⋮

0 0 =

0 ⋮

Since the even rows and columns are not affected by the spring stiffness, the critical loads associated with the even terms are the same as those for the beam without spring support (by Gershgorin’s theorem). However, the critical loads associated with the odd terms increase due to the presence of the spring. For example, for a two-term approximation (the relevant terms are shown shaded above), the characteristic equation reduces to 4π2 4π2−λ   π2  π 2−λ +α  = 0

The critical loads are λ1 = π2 + α2 , π

λ 2 = 4π 2

The critical loads for the beam without the spring are λ1 = π2 and λ2 = 4π2. As α increases λ1 increases and can pass λ2 . In fact, for α > 3π 4 we have λ2 > λ1 and buckling is controlled by the “second” mode.

269. The prismatic beam shown below has flexuEI P ral modulus EI. Axial and shear deformations can k be neglected. The beam is supported on an elastic x foundation of modulus k. The force developed, ℓ per unit length, by the foundation is related to the deflection at that point by f (x) = kw(x). A beam on an elastic foundation with axial thrust is governed by the following (linearized) differential equation and boundary conditions EIwiv + Pw′′ + kw = 0 w(0) = 0,

w′′(0) = 0,

w(ℓ) = 0,

w′′(ℓ) = 0

The eigenfunctions of the beam without the elastic foundation are wn = sin nπx∕ℓ. Verify that the virtual-work functional, accounting for the elastic foundation, is G(w, w) =   EI w′′w′′ − Pw′w′ + k ww  dx ℓ

0

Does the presence of the elastic foundation affect the boundary conditions? Find the buckling loads of the system using the Ritz method, assuming that the real and virtual displacements have the shape of the nth eigenfunction w(x) = a sin nπx , ℓ

w(x) = a sin nπx ℓ

324


How does the buckling load vary with the elastic properties of the system, namely EI and k? Express your result in terms of P1 ≡ π2EI∕ℓ 2 and the ratio of foundation stiffness to beam stiffness, given by the dimensionless parameter β ≡ kℓ 4∕π4EI. (Hint: the critical buckling mode depends upon β.) Is your answer exact?

Consider the following weighted residual of the differential equation

  EIw ℓ

G(w, w) =

iv

+ Pw′′ + kw  w dx

0

By the fundamental theorem of the calculus of variations, G(w, w) = 0 for all w is clearly equivalent to the original classical differential equation. Integrating the first term by part twice and the second term once we obtain G(w, w) =   EIw′′w′′−Pw′w′+kww  dx ℓ

0



ℓ

+  EIw′′′+Pw′  w − EIw′′w′ 0



ℓ 0

Note that the classical boundary conditions for M = 0 and V = 0 can be expressed as EIw′′ = 0 and EIw′′′+ Pw′ = 0, respectively. If we insist that our virtual displacements satisfy homogeneous essential boundary conditions, then the virtual work functional is the one given. (b) Does the presence of the elastic foundation affect the boundary conditions? The elastic foundation does not affect the boundary conditions because it is a distributed effect. The elastic foundation affects only the domain terms of both the classical differential equation and the virtual work functional. (c) Find the buckling loads of the system using the Ritz method, assuming that the real and virtual displacements have the shape of the nth eigenfunction w(x) = a sin nπx , ℓ

w(x) = a sin nπx ℓ

Let m n = nπ∕ℓ. Then w′ = am n cos m n x and w′′ = −am 2n sin m n x. The derivatives of the virtual displacement field is similar. Thus, the virtual work functional can be expressed as G(a, a) =  aa EIm sin m x − Pm cos m x + k sin m x  dx ℓ

4 n

2

n

2 n

2

n

2

n

0

=

1 EIℓ5 a a  m 4nℓ4 2

− λm 2nℓ2 + π 4β 

where λ ≡ Pℓ 2∕EI and β ≡ kℓ 4∕π4EI. Setting G(a, a) = 0 for any a implies that either a = 0 (no displacement) or

Chapter 11


λn = m 2nℓ 2 +



π 4β β = π2 n2 + 2 m 2nℓ 2 n

325



Note that this approach is valid since the base functions used are eigenfunctions are are orthogonal to each other. If we started the approximation by taking the sum of eigenfunctions times unknown constants the equations would uncouple and the results would be the same. Note that Pn = λn EI∕ℓ 2. (d) How does the buckling load vary with the elastic properties of the system, namely EI and k? Express your result in terms of P1 ≡ π2EI∕ℓ 2 and the ratio of foundation stiffness to beam stiffness, given by the dimensionless parameter β ≡ kℓ 4∕π4EI. (Hint: the critical buckling mode depends upon β.) Is your answer exact? The buckling loads are a function of the ratio of flexural stiffness to foundation stiffness. In particular, they are a function of the defined parameter β. In the limit as β → 0 the buckling loads reduce to those of the pinned-pinned beam without elastic foundation. For non-zero values of β, the change from the foundationless case is most apparent for n = 1 and reduces for the higher modes. Physically, the higher modes are more tortuous but the net deflection from the initial position is less. Consequently, more energy is vested in flexure and less in deforming the foundation. Ordinarily, λ1 would be the fundamental buckling mode, but in this case the lowest buckling load depends upon the foundation stiffness. For each value of β we must find the smallest λn over all values of n. This process is shown graphically in the sketch below. The value λ1 controls for 0 ≤ β ≤ 4, after which the value λ2 is minimal. In general, the value λn is minimal for n−1 ≤ β ∕n ≤ n+1. 1+β

9+ 19 β

4+ 14 β

9 λ min(β) 4 0

4

9

β

The solution is exact because the base functions are eigenfunctions of the classical problem. One can observe that these functions are eigenfunctions by substituting them into the differential equation, noting that wiv = am 4n sin m n x, we have, substituting into EIwiv+Pw′′+ kw = 0, with λ and β defined previously, aEIℓ4 m 4n − λm 2n + β  sin m n x = 0 Thus, the classical differential equation is satisfied for all x if λ = λ n.

326


qo 270. The prismatic beam shown has a cross section that is symmetric with respect to the plane of the page. The P cross section has flexural modulus EI. Axial and shear deformations can be neglected. The beam is fixed ℓ against transverse deflection and rotation at both ends, but the supports provide no resistance to the axial force P. The beam is also subjected to a uniform transverse load of magnitude qo. The linearized buckling theory for a beam with transverse load and axial thrust gives rise to the following differential equations and boundary conditions for the present configuration EIwiv + Pw′′ = q(x) w(0) = 0,

w′(0) = 0,

w(ℓ) = 0,

w′(ℓ) = 0

Solve the governing differential equations by the classical method to find an expression for the transverse deflection w(x) and the bending moment M(x). In the design of beams subjected to transverse load and thrust, sometimes called beamcolumns, the concept of the magnification factor is often used. The idea behind the magnification factor is that the influence of the axial thrust is to magnify the values of displacement and moment that would be present if the axial thrust were not (i.e., the solution if P = 0). Show that the maximum deflection and moment can be expressed as w(ℓ∕2) = w o

1  1−P∕P , 1

M(0) = M o

P∕P  1−0.4  1−P∕P 1

1

where wo and M o are the maximum deflection and moment that would occur if P = 0, and P1 would be the fundamental critical load of the column if qo = 0.

(a) Solve the governing differential equations by the classical method to find an expression for the transverse deflection w(x) and the bending moment M(x). A particular solution for q(x) = −q o is wp = − 12

qo 2 x m 2EI

where m 2 = P∕EI. Therefore, the general solution to the given differential equation is w(x) = a 0ℓ + a1 x + a 2ℓ sin m x + a3ℓ cos m x − 12

qo 2 x m 2EI

Substituting the general solution into the boundary conditions yields a0 + a 3 = 0 a1 + mℓ a2 = 0 a0 + a 1 + a2 sin mℓ + a 3 cos mℓ − qoℓ∕2m 2EI = 0 a1 + a 2 mℓ cos mℓ − a3 mℓ sin mℓ − q oℓ∕m 2EI = 0 The first two equations give a0 = −a3 and a1 = −mℓ a2. Substituting these into the third and fourth equations gives

Chapter 11 sin mℓ−mℓ

cos mℓ−1

mℓ cos mℓ−mℓ

327

The Planar Buckling of Beams a2 =

a3

−mℓ sin mℓ

1

q oℓ 2m 2EI

2

These equations can be solved to give a2 = − 12

qo , m 3EI

a3 = − 12



qo cos mℓ+1 sin mℓ m 3EI



Observing that tan θ = sin 2θ∕(1+ cos 2θ), and letting α ≡ mℓ∕2, the displacement can be expressed as

w(x) =



24w o 1− cos 2αξ + 2α ξ−ξ2  − sin 2αξ tan α α3



where wo ≡ qoℓ 4∕384EI is the maximum deflection that the beam would experience if it had no axial load and ξ ≡ x∕ℓ. The bending moment field is



4M cos 2αξ 1 M(x) = α o − α + sin 2αξ tan α



where M o ≡ qoℓ 2∕12 is the maximum moment that the beam would experience if it had no axial load. (b) Show that the maximum deflection and moment for this problem can be expressed in the form w(ℓ∕2) = w o

1 −1P∕P ,

M(0) = M o

1

1 1−−0.4P∕P  P∕P 1

1

where wo and M o are the maximum deflection and moment that would occur if P = 0, and P1 would be the fundamental critical load of the column if qo=0. The maximum deflection is at ξ = 1∕2 and the maximum moment is at ξ = 0. Thus, the exact solution for the maxima are given by



wmax 2 cos α−2+α sin α wo = 12 α 3 sin α





M max = 3 tan2 α−α Mo α tan α



where α = mℓ∕2. To compare the approximate and exact expressions we can expand both in Taylor series and examine the difference. Noting that the load ratio can be expressed as P∕P1 = α 2∕π 2, we can expand the approximate maximum deflection as wapprox 1 α2 α 4 α6 α 8 max wo = 1 − α 2∕π 2 = 1 + π2 + π 4 + π6 + π 8 +⋅⋅⋅ The Taylor expansion for the exact midspan deflection is wmax 31 691 1 2 17 4 6 8 wo = 1 + 10 α + 1680 α + 30240 α + 6652800 α +⋅⋅⋅

328


The difference between the approximate and the exact expressions is wapprox w max max 2 4 6 wo − wo = 0.01304β + 0.01431β + 0.01445β +⋅⋅⋅ where β ≡ P∕P 1 . The moments can be treated similarly. We can expand the approximate maximum moment as



2 4 6 8 M approx 1 − 0.4α 2∕π 2 max = = 1 + 35 α 2 + α 4 + α 6 + α 8 +⋅⋅⋅ Mo π π π π 1 − α 2∕π 2



The Taylor expansion for the exact end moment is M max 1 2 2 1 2 = 1 + 15 α + 315 α 4 + 1575 α 6 + 31185 α 8 +⋅⋅⋅ Mo The difference between the approximate and the exact expressions is M approx M max − max = 0.0578β 2 + 0.0185β 4 + 0.0104β 6 +⋅⋅⋅ Mo Mo As the axial load approaches the buckling load α → π and P → P 1. The approximation contains the most error in this vicinity. However, even at P = 0.99P 1 the error in displacement is 1.5% and the error in moment is 1.4%.

271. The column shown is subjected to axial forces at the midpoint and top, both of magnitude P. The column has variable flexural modulus given by the expression EI(x) = EI o  2−x∕2ℓ . Estimate the buckling load of the system using the Ritz method with a one-term polynomial basis. Is your estimate higher or lower than the actual buckling load? Explain your answer. Propose a function for a one-term Ritz approximation that will give better results than you got in the first part. Why do you think it is better? Estimate the buckling load of the system using the Ritz method with a two-term polynomial basis. Repeat with a threeterm polynomial basis.

P P x

ℓ ℓ

(a) Estimate the buckling load of the system using the Ritz method with a one-term polynomial basis. The function w(x) = a x 2 satisfies the essential boundary conditions of zero displacement and rotation at x = 0. Then w′ = 2a x and w′′ = 2a. The virtual displacement field and its derivatives are similar. Letting ξ ≡ x∕ℓ, the virtual work functional can be expressed as

Chapter 11

 aa 2EI ℓ  4−ξ  − 8Pℓ ξ  dξ +  aa 2EI ℓ  4−ξ  − 4Pℓ ξ  dξ 1

G(a, a) =

329

The Planar Buckling of Beams 2

3

o

2

3 2

o

0

1

= aa 7EI oℓ − 83 Pℓ3 + 5EI oℓ − 28 Pℓ 3  3 = aa 12EI oℓ − 12Pℓ 3  Setting G(a, a) = 0 for any a implies that either a = 0 (no displacement) or Pcr =

EI o ℓ2

(b) Is your estimate higher or lower than the actual buckling load? Explain your answer. The estimate is higher than the actual buckling load because the Ritz method always provides an upper bound on buckling loads. (c) Propose a function for a one-term Ritz approximation that will give better results than you got in part (a). Why do you think it is better? The function w(x) = a  x3−6x2ℓ  satisfies the essential boundary conditions of zero displacement and rotation at x = 0. It also satisfies the condition of vanishing moment at x = 2ℓ. It should, therefore, give better results than the previous one-term polynomial approximation. The derivatives of the displacement function are w′ = 3a  x2−4xℓ  and w′′ = 6a  x−2ℓ . The virtual displacement field and its derivatives are similar. Thus, the virtual work functional can be expressed as

 a a  18EI ℓ  4−ξ   ξ−2  − 18Pℓ  ξ −4ξ   dξ +  a a  18EI ℓ  4−ξ   ξ−2  − 9Pℓ  ξ −4ξ   dξ 1

G(a, a) =

o

0

3

2

5

2

2

5

2

2

2

o

3

2

1

EIℓ3 − 318 Pℓ 5 + 33 EIℓ 3 − 609 Pℓ 5  = a a  303 2 2 5 5 = a a  168EIℓ 3 − 927 Pℓ5  5 Setting G(a, a) = 0 for any a implies that either a = 0 (no displacement) or Pcr =

280 EI 309 ℓ 2

The coefficient is 0.906, which is less than the previous value of 1.0. (d) Estimate the buckling load of the system using the Ritz method with a two-term polynomial basis. Repeat with a three-term polynomial basis. The base functions h1(ξ) = ξ2ℓ, h2(ξ) = ξ3ℓ, h3(ξ) = ξ4ℓ, etc. satisfy the essential boundary conditions for the cantilever beam. The virtual work functional can be expressed as G(w, w) = a T  Ka−λGa  where the coefficients are computed as

330

Fundamentals of Structural Mechanics Kij =



2

 2h ′h ′ dξ +  h ′h ′ dξ 1

1 2

 4−ξ  h i′′ h j′′ dξ

Gij =

0

2

i

j

i

0

j

1

where λ ≡ Pℓ 2∕EI o . Notice that we divided out EI o∕ℓ from the virtual work functional. For the two-term approximation we can perform the indicated integrals to get the equation 12−12λ 32− 51 λ 2

32− 51 λ 2

a1

297 λ 5

a2

120−

0 =

0

The characteristic equation is obtained by setting the determinant of the coefficient matrix equal to zero 1251 2 λ 20

− 2604 λ + 416 = 0 5

The critical loads are given by λ1 = 0.89497,

λ 2 = 7.43117

Therefore, the critical loads are P1 = λ1 EI o∕ℓ2 and P2 = λ2 EI o∕ℓ2 . Notice that the first critical load is smaller than the ones computed with the one-term expansions. The three-term expansion is similar to the two-term expansion. The MATHEMATICAT commands needed to solve this problem are given below h = {x^2 ,x^3, x^4 } hp = D[h,x] hpp = D[hp,x] EI = (4 - x)/2 K = Integrate[ EI Outer[Times,hpp,hpp],{x,0,2}] G1 = Integrate[ 2 Outer[Times,hp ,hp ],{x,0,1}] G2 = Integrate[ Outer[Times,hp ,hp ],{x,1,2}] G = G1 + G2 d = Det[K - m G] Solve[d==0,m]

The critical loads for the three-term expansion are given by λ1 = 0.88749,

λ 2 = 5.85419,

λ 3 = 27.1855

272. A flexible beam of length ℓ and modulus P Rigid EI EI is welded to a rigid beam of length ℓ and rests k on an elastic foundation of modulus k (per unit x length). It is pinned at the left end and is subjected to a compressive axial load P at the right ℓ ℓ end. The elastic foundation accrues a transverse force in proportion to the transverse displacement w. Shear and axial deformations in the beam are negligible. Write the expression for the energy of the system. What are the essential and natural boundary conditions for the flexible beam? Find an approximate solution for the buckling loads and mode shapes using a two-term polynomial Ritz basis.

Chapter 11

331


(a) Write the expression for the energy of the system. Let us designate the deflection of the beam as w(x), applicable over the entire length [0, 2ℓ]. Let ∆ ℓ ≡ w(ℓ) be the deflection at the end of the flexible portion of the beam and let θ ℓ ≡ w′(ℓ) be the rotation at the end of the flexible portion of the beam. The potential energy of the system using the linearized theory is E(w) =



ℓ

1 EI (w′′) 2+k w2−P(w′) 2  dx 2

0

+



2ℓ

1 k w 2 dx 2

−

1 Pℓ θ 2ℓ 2

ℓ

The deflection of the rigid bar can be expressed as w(x) = ∆ ℓ +  x−ℓ  θ ℓ,

ℓ ≤ x ≤ 2ℓ

Therefore, the second integral can be evaluated explicitly as



2ℓ

1 k w 2 dx 2

=

1 2

kℓ  ∆ 2ℓ + 13 ℓ 2 θ 2ℓ + ℓ∆ ℓ θ ℓ 

ℓ

The virtual work functional can be obtained by taking the directional derivative of the energy expression

  EI w′′ w′′+k ww−Pw′ w′  dx +  ℓ

G(w, w) =

0

2ℓ

k wwdx − Pℓ θ ℓ θ ℓ

ℓ

where θ ℓ = w′(ℓ) and w = ∆ ℓ +  x−ℓ  θ ℓ in the rigid portion of the beam. The term associated with the elastic foundation under the rigid bar can be obtained from the above expression as



2ℓ

k wwdx =

1 2

kℓ  2∆ ℓ ∆ ℓ + 23 ℓ2 θ ℓ θ ℓ + ℓ∆ ℓ θ ℓ + ℓ∆ ℓ θ ℓ 

ℓ

The virtual work functional can be written as

  EI w′′ w′′ + k ww − Pw′ w′  dx − Pℓw′(ℓ) w′(ℓ) ℓ

G(w, w) =

0

+ 12 kℓ  2w(ℓ)w(ℓ) + 23 ℓ2 w′(ℓ) w′(ℓ) + ℓw(ℓ) w′(ℓ) + ℓw′(ℓ) w(ℓ)  (b) What are the essential and natural boundary conditions for the flexible beam? Integrating the virtual work functional by parts until all of the derivatives are shed from the w terms we can find an equivalent expression for the virtual work functional

  EI w +k w+Pw′′  wdx − M(0) w′(0) + Q(0) w(0) ℓ

G(w, w) =

iv

0

+  EIw′′(ℓ)−Pℓw′(ℓ)+ 12 kℓ 2 w(ℓ)+ 13 kℓ3 w′(ℓ)  w′(ℓ)

+  −EIw′′′(ℓ)−Pw′(ℓ)+kℓw(ℓ)+ 12 kℓ 2 w′(ℓ)  w(ℓ)

332


By the fundamental theorem of the calculus of variation we can determine that the governing differential equation for the beam is EI w iv + Pw′′ + k w = 0 and that the appropriate boundary conditions for this problem are w(0) = 0 w′′(0) = 0

EIw′′(ℓ) − Pℓw′(ℓ) + 12 kℓ 2 w(ℓ) + 13 kℓ3 w′(ℓ) = 0 EIw′′′(ℓ) + Pw′(ℓ) − kℓw(ℓ) − 12 kℓ 2 w′(ℓ) = 0

The first boundary condition is essential and the last three are natural. (c) Find an approximate solution for the buckling loads and mode shapes using a twoterm polynomial Ritz basis. The base functions h1(ξ) = ξℓ, h2(ξ) = ξ2ℓ satisfy the essential boundary conditions for the cantilever beam. Let w = a 1 h1+a 2 h2 and w = a 1 h1+a 2 h2. The virtual work functional can be expressed as G(w, w) = a T  Ka−λGa  where the coefficients are computed as

  h ′′h ′′+α h h  dξ 1

Kij =

i

j

i

j

0

+ 12 α  2h i(1) hj(1) + 23 hi′(1) h j′(1) + hi(1) hj′(1) + h i′(1) hj(1) 

 h ′h ′ dξ + h ′( ) h ′( ) 1

Gij =

i

j

i

1

j

1

0

where λ ≡ Pℓ 2∕EI and α ≡ kℓ 3∕EI. Notice that we divided out EI∕ℓ from the virtual work functional. For the two-term approximation we can perform the indicated integrals to get the equation 8 α−2λ 3 41 α−3λ 12

41 α−3λ 12

a1

68 α− 16 λ 3 15

a2

4+

0 =

0


−  8+ 251 αλ +  90

32 299 α+ 720 α 2 3



= 0

The critical loads can be computed numerically for specific values of α. For example, for the value α = 1 we get λ1 = 1.2804,

λ 2 = 5.1929

Therefore, the critical loads are P1 = λ1 EI∕ℓ 2 and P2 = λ2 EI∕ℓ 2. The MATHEMATICAT commands needed to solve this problem are given below

Chapter 11

333


h = { x, x^2} hp = D[h,x] hpp = D[hp,x] K1 = Integrate[Outer[Times,hpp,hpp],{x,0,1}] K2 = Integrate[Outer[Times,h ,h ],{x,0,1}] G1 = Integrate[Outer[Times,hp ,hp ],{x,0,1}] x = 1 K3 = 2 Outer[Times,h,h] K4 = 2 Outer[Times,hp,hp]/3 K5 = Outer[Times,hp,h] + Outer[Times,h,hp] G2 = Outer[Times,hp,hp] K = K1 + a ( K2 + (K3 + K4 + K5)/2 ) G = G1 + G2 d = Det[K - m G] Solve [ d==0 , m ]

In this script, a stands for α and m stands for λ.

273. The column shown has modulus EI and weight per unit length p. It is fixed at one end and free at the other. Shear and axial deformations can be neglected. Find the (classical) governing differential equations and boundary conditions for the transverse deflection w(x). Express the governing equations in virtual-work form. Estimate by the Ritz method the maximum length the column can have before it buckles under its own weight.

ℓ

p, EI

(a) Find the (classical) governing differential equations and boundary conditions for the transverse deflection w(x). M p(x) = p o

H V

EI

po

x

ℓ

From equilibrium of forces we have H(x) = −p o  ℓ−x  and V(x) = 0. The (linearized) classical equations are, therefore, EIwiv + p o   ℓ−x  w′ ′ = 0 The boundary conditions are w(0) = 0,

w′(0) = 0,

M(ℓ) = 0,

V(ℓ) = 0

(b) Express the governing equations in virtual work form. The virtual work functional is given by

  EIw′′w′′ − p  ℓ−x w′w′  dx ℓ

G(w, w) =

o

0

334


(c) Using a polynomial approximation for w, estimate the maximum length the column can have before it buckles under the axial force. Use the Ritz method. The function w(x) = a x 2 satisfies the essential boundary conditions of zero displacement and rotation at x = 0. Then w′ = 2a x and w′′ = 2a. The derivatives of the virtual displacement field are similar. Letting ξ ≡ x∕ℓ, the virtual work functional can be expressed as

 a a  4EIℓ − 4p ℓ  1−ξ  ξ  dξ 1

G(a, a) =

o

4

2

0

1 p oℓ 4  = 4 a a  EIℓ − 12

Setting G(a, a) = 0 for any a implies that either a = 0 (no displacement) or ℓcr =

 12pEI 

1∕3

o

274. The stepped column shown has a variable modulus and is subjected to vertical forces at two points. It is fixed at one end and free at the other. Shear and axial deformations can be neglected. Find the (classical) governing differential equations and boundary conditions for the transverse deflection w(x). Express the governing equations in virtual-work form. Using a two-term polynomial approximation for w, estimate the critical load using the Ritz method.

P ℓ

EI

ℓ

2EI

3P

(a) Since the modulus and axial load are discontinuous we must consider each segment separately and relate the two segments with continuity conditions. Let w1(x) and w2(x) be the transverse displacement in the first and second segments so that w(x) =

w1(x 1)

0 ≤ x1 ≤ ℓ

w2(x 2)

0 ≤ x2 ≤ ℓ

The segments are governed by the differential equations wivi + m 2i w i′′ = 0

i = 1, 2

where m 21 = 2m 22 = P∕EI. These equations have the solution wi = ai + b i xi + c i sin m i x i + di cos m i xi The boundary conditions are w1(0) = 0

w1′(0) = 0

w2′′(ℓ) = 0

w2′′′(ℓ) + m 22 w 2′(ℓ) = 0

Chapter 11


335

The continuity conditions are w1(ℓ) = w2(0) w1′(ℓ) = w2′(0) 2w1′′(ℓ) = w2′′(0)

2  w 1′′′(ℓ) + m 21 w 1′(ℓ)  = w 2′′′(0) + m 22 w 2′(ℓ)

(b) Express the governing equations in virtual work form. Assume that the real and virtual displacement functions are defined on the entire range [0, 2ℓ]. Then the virtual work functional can be expressed as G(w, w) =   2EIw′′w′′−4Pw′w′  dx +  ℓ

0

2ℓ

 EIw′′w′′−Pw′w′  dx

ℓ

(c) Using a two-term polynomial approximation for w, estimate the critical load using the Ritz method. The base functions h1(ξ) = ξ2, h2(ξ) = ξ3, h3(ξ) = ξ4, etc. satisfy the essential boundary conditions for the cantilever beam. Note that the derivatives are with respect to x so that, for example, h1′ = 2ξ∕ℓ and h1′′ = 2∕ℓ2. The virtual work functional can be expressed as G(w, w) = a T  Ka−λGa  where the coefficients are computed as K =  2EI h ′′h ′′ ℓ dξ +  EI h ′′h ′′ ℓ dξ 1

2

ij

i

j

i

0

j

1

G =  4h ′h ′ ℓ dξ +  h ′h ′ ℓ dξ 1

ij

2

i

j

i

0

j

1

where λ ≡ Pℓ 2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional. For the two-term approximation we can perform the indicated integrals to get the equation 12− 44 λ 3

30− 57 λ 2

30− 57 λ 2

108−63λ

a1 a2

0 =

0


− 630 λ + 396 = 0

The critical loads are given by λ1 = 0.72071,

λ 2 = 4.91688

Therefore, the critical loads are P1 = λ1 EI o∕ℓ2 and P2 = λ2 EI o∕ℓ2 . The MATHEMATICAT commands needed to solve this problem are given below

336

Fundamentals of Structural Mechanics h = {x^2 ,x^3 } hp = D[h,x] hpp = D[hp,x] K1 = Integrate[2 K2 = Integrate[ G1 = Integrate[4 G2 = Integrate[ K = K1 + K2 G = G1 + G2 d = Det[K - m G] Solve[d==0,m]

Outer[Times,hpp,hpp],{x,0,1}] Outer[Times,hpp,hpp],{x,1,2}] Outer[Times,hp ,hp ],{x,0,1}] Outer[Times,hp ,hp ],{x,1,2}]

275. Consider the bar of length ℓ with bending modulus EI and shear modulus GA, subjected to axial load P as shown. Show that the linearized virtual-work functional for the buckling of a beam with shear deformation is given by

EI, GA

P

ℓ

  EIθ′θ′ + GA(w′−θ) (w′−θ) − P  w′θ+w′θ−θ θ   dx ℓ

G(w, θ, w, θ) =

0

Make the assumption that the (generalized) shear strain in the beam is constant. Estimate the critical loads of the beam using a polynomial approximation with the Ritz method. For example, a three-parameter approximation would have the expression 2 2 3 θ = a 0 x + a 1 x 2 , w = a1 x + 12 a0 x + 13 a1 x 2 ℓ ℓ ℓ ℓ Describe at least two ways of improving the approximation, and rank them according to which is likely to give the most improvement (no calculations necessary). If EI is very large in comparison to GAℓ2 , what will the buckled shape of the beam look like?

Derivation. The inextensibility constraint Á 2o + β 2o = 1 can be maintained without sacrificing the shearing deformations if we introduce the angle φ such that w′ = sin φ,

1+u′ = cos φ

Noting that β o = sin φ cos θ− cos φ sin θ, and using a trigonometric identity, the constitutive equation for shear can be written as Q = V cos θ − H sin θ = GA sin(φ−θ) The equation can be linearized to give Hθ + GA  φ−θ  − V = 0,

(*)

Equilibrium of moments, Eqn. (549), can now be put into the (linearized) form M′ + V − Hφ = 0,

(**)

Let us construct a weighted residual expression for the moment equilibrium equation (**) and the transverse force equation (*). To wit, multiply the moment equation by θ

Chapter 11

337


and the transverse force equation by φ−θ and integrate over the length of the beam to get

  − M′+V−H φ  θ +  Hθ + GA  φ−θ  − V  φ−θ   dx ℓ

G(φ, θ, φ, θ) =

0

Integrating by parts and rearranging terms we get

  Mθ′ − Vφ + H  φθ+φθ−θθ  + GA  φ−θ   φ−θ   dx ℓ

G(φ, θ, φ, θ) =

0

Finally, noting that V = 0, H = −P , and M = EI θ′ we get the expression

  EI θ′θ′ + GA  φ−θ   φ−θ  − P  φθ+φθ−θθ   dx ℓ

G(φ, θ, φ, θ) =

0

Noting that φ = w′ we can see the difference in the proposed expression and the correct expression. Note that this form of the virtual work functional reduces to the one for the Bernoulli-Euler theory when φ = θ = w′.

(a) Make the assumption that the (generalized) shear strain in the beam is constant. Estimate the critical loads of the beam using a polynomial approximation with the Ritz method. For example, a three-parameter approximation would have the expression 2 θ = a0 x + a1 x 2 , ℓ ℓ

2 3 w = a2 x + 12 a0 x + 13 a1 x 2 ℓ ℓ

Note that φ = w′ = a 2+a0 ξ+a1 ξ 2 where ξ ≡ x∕ℓ. The shear deformation can be computed as φ−θ = a 2, which is constant, as desired. Assuming that the virtual displacements and rotations are interpolated the same way as the real ones we can write the virtual work functional as

  a + a ξa + a ξ + βa a 1

G(a, a) =

0

2

1

0

2

1

2

2

0

− λ  a2+a 0 ξ+a1 ξ2  a0 ξ+a 1 ξ2  − λ  a2+a 0 ξ+a1 ξ2  a0 ξ+a 1 ξ2 



+ λ  a0 ξ+a 1 ξ 2  a 0 ξ+a1 ξ 2  dx where β ≡ GAℓ 2∕EI and λ = Pℓ 2∕EI. Note that the virtual work functional has been divided through by EI∕ℓ. Carrying out the integrations we arrive at the system of equations

338

Fundamentals of Structural Mechanics 1− 13 λ 1− 14 λ

− 12 λ

1− 14 λ

− 12 λ

a0

4 − 15 λ 3 − 13 λ

− 13 λ

a1

β

a2

0 =

0 0

The characteristic equation is obtained by setting the determinant of the coefficient matrix equal to zero 8λ3 +  9β−240  λ2 − 312β λ + 720β = 0

The critical loads can be computed numerically for specific values of β. Some values are given in the table below β

λ1

λ2

1 10 100 1000

1.232 2.117 2.437 2.481

30.07 30.55 31.67 32.12

λ3 --2.429 --13.918 --116.60 --1129.6

λEXACT 1 1.149 2.048 2.409 2.461

The exact value of the buckling load is given by the equation = λEXACT 1

π2

2+2 1+π 2∕β

Notice that as β gets large the fundamental buckling load approaches the buckling load for the Bernoulli-Euler beam, that is λ1 → π2∕4 = 2.467. It is interesting to observe that λ3 < 0, suggesting the existence of a critical load for tensile loads. An explanation can be constructed as follows. The classical differential equation can be obtained by adding (*) and (**) to eliminate V and by subsequently eliminating φ =  1+λ∕β  θ by (*) with V = 0. Dividing the result by EI we finally obtain the classical equation ℓ2 θ′′+λ  1+λ∕β  θ = 0 where λ and β are as previously defined. For the load λ = −β this equation reduces to θ′′ = 0 along with φ = 0. Thus, θ = a 0+a1 x. The boundary condition θ(0) = 0 gives a0 = 0 and the boundary condition M(ℓ) = 0 gives a1 = 0. Thus, θ(x) = 0 for this value of the load parameter. The negative root to the characteristic equation is the best effort the approximation can make to finding this “other” solution in the context of the given functional. Note that, to within the roughness of the approximation, the numerical solution gives λ3 ≈ −β. (b) Describe at least two ways of improving the approximation, and rank them according to which is likely to give the most improvement (no calculations necessary). The approximation can be improved by increasing the order of the basis and by eliminating the constant shear assumption. Actually, you cannot do the latter without doing

Chapter 11

339


the former. The computation can be organized conveniently by defining the following matrices hT(ξ) =  ξ, ξ 2 , 0, 0 

a T =  a 1, a 2, a 3, a 4 

gT(ξ) =  0, 0, ξ, ξ 2 

Now θ = a T h(ξ), φ = a T g(ξ), θ = a T h(ξ), φ = a T g(ξ). The virtual work functional can be computed as G = a T  K−λG  a where the components of the coefficient matrices are given by

  h ′h ′+β  g −h   g −h   dξ 1

Kij =

i

j

i

i

j

j

  h g +g h −h h  dξ 1

Gij =

0

i

j

i

j

i j

0

The MATHEMATICAT commands needed to solve this problem are given below h = g = hp gmh K1 K2 G1 G2 G3 K = G = d =

{ x, x^2, 0, 0} { 0, 0, x, x^2} = D[h,x] = g - h = Integrate[Outer[Times,hp ,hp ],{x,0,1}] = Integrate[Outer[Times,gmh,gmh],{x,0,1}] = Integrate[Outer[Times,g ,h ],{x,0,1}] = Integrate[Outer[Times,h ,g ],{x,0,1}] = Integrate[Outer[Times,h ,h ],{x,0,1}] K1 + b K2 G1 + G2 - G3 Det[K - m G]

The result for the four-term solution for β = 100 is λ1 = 2.427,

λ 2 = 25.62,

λ 3 = −102.4,

λ 4 = −125.6

Observe that now we get two negative eigenvalues that are roughly equal to β. (c) If EI is very large in comparison to GAℓ2 , what will the buckled shape of the beam look like? The deformation will be dominated by shear, not flexure. Therefore, the shape of the beam should be a straight line in the limit.

276. Consider the beam of length ℓ, fixed at both ends, with constant modulus EI shown in the sketch. The beam is subjected to a compressive axial load P. When the beam is not loaded, the initial shape can be described as



2 3 wo(x) = co 3 x 2 − 2 x 3 ℓ ℓ

P x

wo(x) ℓ



where co % 1 is known as given data. Assume that shear and axial deformations are negligible. Find an expression for the energy functional E and the virtual-work functional G for this problem. Estimate the deflection of the beam as a function of load P using the Ritz method and a one-term approximation as follows



2 3 4 w(x) = w o(x) + a x 2 − 2 x 3 + x 4 ℓ ℓ ℓ



340


assuming that the displacements are small enough to use the linearized buckling theory.

(a) Find an expression for the energy functional E and the virtual work functional G for this problem. The energy functional E for the imperfect system is E(w) =

1 2

  EI  w′′−w ′′  − P w′   dx ℓ

2

o

2

0

The virtual work functional can be found by taking the directional derivative of the energy to get

  EI  w′′−w ′′  w′′ − Pw′ w′  dx ℓ

G(w, w) =

o

0

(b) Estimate the deflection of the beam as a function of load P using the Ritz method and a one-term approximation as follows



2 3 4 w(x) = w o(x) + a x 2 − 2 x 3 + x 4 ℓ ℓ ℓ



assuming that the displacements are small enough to use the linearized buckling theory. Let the base function h(ξ) and wo(ξ) be given by h(ξ) = ξ 2−2ξ3+ξ 4,

w o(ξ) = co  3ξ2−2ξ 3 

so that w(ξ) = w o(ξ)+a h(ξ) and w(ξ) = a h(ξ). Then the virtual work functional can be expressed as

 a a  EIℓ  h′′  − Pℓ (h′)  dξ = Pℓ  w ′ h′ dξ 1

1

2

2

o

0

0

Observe that the integral on the right side of the equation vanishes, i.e.,

 w ′ h′ dξ =  c  ξ− ξ  ξ− ξ + ξ  dξ = 0 1

1

o

0

o

6

6

2

2

6

2

4

3

0

Thus, we have



a a 4EI3 − 2P 105ℓ 5ℓ

=0

hence the solution is either a = 0, the trivial solution, or a ≠ 0 when the load reaches a critical value

Chapter 11

341

The Planar Buckling of Beams Pcr = 42 EI2 ℓ

P 277. A flexible beam of length 2ℓ and modulus EI rests EI 2P 2P on an elastic foundation of modulus k. The beam is compressed by a known fixed force 2P and is subjected x k to a transverse load P at its midpoint. The properties ℓ ℓ have values such that kℓ4 = EI. Axial and shear deformations of the flexible beam can be neglected. Estimate the deflection at the middle and ends of the beam using virtual work and the Ritz method. (Note: due to symmetry odd base functions need not be included.)

The appropriate virtual work functional is G(w, w) =



ℓ

 EIw′′w′′−2 Pw′w′+k ww  dx − Pw(0) w(0) = 0

−ℓ

Using base functions h = [ 1 , ξ2 ], h′ = [ 0 , 2 ξ], h′′ = [ 0 , 2 ] we get the following stiffness matrix and force vector

  EIℓ h′′  h′′ − 2Pℓ h′  h′ + k ℓ h  h  dξ, 1

K=2

3

f = Ph(0)

0

Carrying out the integrals, K=

2 EI ℓ3

0

0

0

4

− 4P ℓ

0

0

0

4∕3

+ 2kℓ

1

1∕3

1∕3

1∕5

f=P

1 0

Noting that k ℓ 4 = EI, and letting λ ≡ 40 Pℓ 2∕EI, we get

K=

2 EI 15 ℓ 3

15

5

5

63−λ

f=

2 EI 3 λℓ 15 ℓ 3 16

1 0

The coefficients of the displacement approximation can be found by solving K a = f. To wit, 15 a 0 + 5 a1 = 3 λℓ∕16 5 a 0 +  63−λ  a 1 = 0 from which we find a0 =

3 λℓ  63−λ  , 80  184−λ 

a1 =

−15 λℓ 80  184−λ 

342


Note that the system will buckle when the denominator of these coefficients becomes infinite or when λ = 184. This condition occurs at a critical load of Pcr = 1.53 EI2 ℓ The displacement is w(ξ) =

3 λℓ   63−λ  − 5 ξ2  80  184−λ 

from which it is easy to find w(0) and w(1).

278. A square frame of dimension ℓ is composed of two 2P P columns connected together by a beam (the beam can be considered rigid). The left column, which is rigid and pinned at both ends, is subject to a force 2P. The right Rigid EI ℓ x column, which has flexural modulus EI, is subjected to a force P. What are the essential and natural boundary conditions on the flexible beam. Express all boundary conditions in terms of the transverse displacement w(x) ℓ of the flexible beam. Find the classical characteristic equation that determines the buckling load of the system. Find the exact value of the primary buckling load from the characteristic equation. Recall that the classical differential equation wiv + m 2w′′ = 0 has the general solution in the form w(x) = a 0+a1 x+a 2 sin mx+a3 cos mx. Estimate the buckling capacity of the structure using the Ritz method in conjunction with the principle of virtual work. Compare the classical and variational solutions. [Note: The left column is often referred to as a “leaner” because it leans on the right column to find resistance to sway. By itself the left column has no lateral stiffness, but it carries a destabilizing force.]

(a) What are the essential and natural boundary conditions on the flexible beam. Express all boundary conditions in terms of the transverse displacement w(x) of the flexible beam. Equilibrium of the rigid link on the left gives 2Pw(ℓ) = N ℓ

⇒

N=

2Pw(ℓ) ℓ

This force acts, through the horizontal rigid link, on the flexible beam. Therefore the boundary conditions are Essential:

w(0) = 0, w′(0) = 0

Chapter 11

343

The Planar Buckling of Beams P

2P N

N

N

N x

w(ℓ)

w(ℓ) Natural:

w′′(ℓ) = 0, − EIw′′′(ℓ) + Pw′(ℓ) + 2Pw(ℓ)∕ℓ = 0

(b) Find the classical characteristic equation that determines the buckling load of the system. Find the exact value of the primary buckling load from the characteristic equation. Recall that the solution to the classical differential equation wiv + m 2w′′ = 0 is given by w(x) = a 0 + a1 x + a 2 sin mx + a3 cos mx Substituting the boundary conditions gives equations to determine the coefficients of the solution w(0) = a 0 + a3 = 0 w′(0) = a 1 + ma2 = 0 w′′(ℓ) = −m 2  a2 sin mℓ + a3 cos mℓ  = 0 V = −EIm 3  a2 cos mℓ − a3 sin mℓ  + P  a 1ℓ + mℓ  a 2 cos mℓ − a 3 sin mℓ  ℓ + 2  a 0 + a1ℓ + a 2 sin mℓ + a3 cos mℓ   = 0 Let S ≡ sin mℓ and C ≡ cos mℓ and T ≡ S∕C. Substituting the first equations into the last two give, noting that m 2 = P∕EI, and dividing through by P∕ℓ a2 S + a3 C = 0

⇒

a3 = −a 2 T

−mℓ  a2C − a 3S  + mℓ  −a 2 + a2C − a 3S  + 2 −a3−mℓ a2 + a 2S + a3C  = 0 Substituting the first into the second gives −mℓ a 2 + 2  a2T−mℓ a 2  = 0 or 2 tan mℓ = 3mℓ This equation has a solution at mℓ = 0.968. Therefore, the critical load is Pcr = 0.937 EI2 ℓ (c) Estimate the buckling capacity of the structure using the Ritz method in conjunction with the principle of virtual work. Use a two term polynomial approximation. Compare the solutions from parts (b) and (c). The appropriate virtual work functional is

344


  EIw′′w′′ − Pw′w′  dx − 2Pw(ℓ)w(ℓ)∕ℓ = 0 ℓ

G(w, w) =

0

Using base functions h = [ ξ 2, ξ 3 ] we get the following stiffness and geometric stiffness matrices

 EIℓ h′′  h′′ dx, ℓ

K=

G=P ℓ

3

0

 h′  h′ dξ + 2Pℓ h(ℓ)  h(ℓ) ℓ

0

Carrying out the integrals gives the results K = EI3 ℓ

4

6

6

12

G=P ℓ

10 3

7 2

7 2

19 5

Letting λ = Pℓ 2∕EI we have the equation  K−λG  a = 0 governing the response of the system this equation has a nontrivial solution only if the determinant of the coefficient matrix is zero. Hence, we can determine the value of the critical load from the condition

det

4 − 10 λ 3

6 − 72 λ

6 − 72 λ

λ 12 − 19 5

=0

or

 4 − 103 λ  12 − 195 λ  −  6 − 72 λ  2 = 0 This equation can be simplified to 0.416667 λ 2 − 13.2 λ + 12 = 0 which has solution λ = 0.936. Therefore, the critical load is Pcr = 0.936 EI2 ℓ which is nearly identical to the classical solution found previously.

Chapter 11

345


P EI 279. A flexible beam of length 2ℓ and modulus EI is stuck on an elastic foundation of modulus k k (per unit length) over half of its length. It is x pinned at the left end and is subjected to a compressive axial load P at the right end. The elastic ℓ ℓ foundation accrues a transverse force in proportion to the transverse displacement w. Shear and axial deformations in the beam are negligible. Write the expression for the energy of the system. What are the essential and natural boundary conditions for the beam? Find an approximate solution for the buckling loads using a polynomial Ritz basis.

(a) Write the expression for the energy of the system. This problem is essentially a standard beam buckling problem but with and elastic foundation only over half of the length. Therefore, the energy is E(w) = 

2ℓ

1 EI(w′′) 2 − P(w′) 2  2

0

dx + 

ℓ

1 k w2 2

dx

0

(b) What are the essential and natural boundary conditions for the beam? The displacement and moment are zero at the left end and the moment and shear are zero at the right end. Therefore, the boundary conditions are Essential:

w(0) = 0

Natural:

w′′(0) = w′′(2ℓ) = 0 (Moment), w′′′(2ℓ) + m 2w′(2ℓ) = 0 (Shear)

(c) Find an approximate solution for the buckling loads using a two-term polynomial Ritz basis. Let ξ = x∕ℓ so that dx = ℓ dξ. A two-term polynomial that satisfies the essential boundary condition w(0) = 0 is w = a 1 ξ + a2 ξ2. The derivatives of the displacement (with respect to x) is w′′ = 12  2 a 2  ℓ

w′ = 1  a1 + a 2 2ξ  ℓ

with similar expressions for the virtual displacement. The virtual work functional is G(w, w) =



 kww dx = 0

2ℓ

ℓ

 EIw′′w′′−Pw′w′  dx +

0

0

Substituting the approximation into the virtual work functional

  EI a a 2∕ℓ  − ℓP  a +2a ξ   a +2a ξ   ℓ dξ +  k  a ξ+a ξ  a ξ+a ξ  ℓ dξ = 0 2

G(w, w) =

2

0

2 2

2

1

2

2

1

2

1

1

0

2

2

1

2

2

346


Carrying out the integrals suggests defining the matrices K = EI3 ℓ

0

0

0

8

+k ℓ

1 3

1 4

1 4

1 5

G = −P ℓ

2

4

4

32 3

Let β ≡ kℓ 4∕EI and λ ≡ Pℓ 2∕EI. The equation governing the response of the system is  K−λG  a = 0, which has nontrivial solution only if the determinant of the coefficient matrix is zero. Hence, we can find the critical loads from the condition

det

β −2λ 3 β −4λ 4

β −4λ 4 β 8 + − 32λ 3 5

=0

This equation can be solved for the critical load as a function of the stiffness.

P 280. A flexible beam of length ℓ and moduRigid Rigid EI lus EI is connected to rigid beams of length ℓ k k at both ends. The beams are supported by two x linear springs with modulus k = βEIℓ 3, where β is a given constant. The beam is supℓ ℓ ℓ ported as shown and is subjected to an end load P. Shear and axial deformations in the beam are negligible. Write the (quadratic) energy functional E and the virtual-work functional G for the system. What are the essential and natural boundary conditions for the flexible beam? Find an approximate solution for the lowest buckling load using a two-term polynomial Ritz basis. Express the result in terms of β, i.e. Pcr(β). What is the buckling load for very large spring stiffnesses (i.e., as β → 1)? Does the approximation appear to make sense in the limit? Explain.

(a) Write the (quadratic) energy functional E and the virtual-work functional G for the system. The figure below shows the deformed position of the system (with all end displacements shown positive). P

w(x) θ1

θ2

w(0) x

The geometry of the rigid bars gives the relations θ2 = w′(ℓ)

θ1 = w′(0) =

w(0) ℓ

Chapter 11

347


The energy of the system is the elastic energy (with the axial load term) for the flexible beam in the middle plus the energy of the two rigid portions. The axial load is constant and equal to P along the entire length. Hence, E(w) =



ℓ

1  EI w′′  2 2

0



2 − P w′  2 dx + 12 k  w(0)  +ℓ w′(ℓ) 2



− Pℓ  1− cos θ 1  − P ℓ  1− cos θ 2 

Or, recognizing that θ 1 = w′(0) = w(0)∕ℓ and θ 2 = w′(ℓ) we have E(w) =



ℓ

1  EI w′′  2 2

0



2 − P w′  2 dx + 12 k  w(0)  +ℓ w′(ℓ) 2



− 12 Pℓ   w′(0)  2 +  w′(ℓ)  2 

The virtual work functional can be found by taking the directional derivative of the energy in the direction of a virtual displacement w(x). Hence, G(w, w) =



ℓ

 EIw′′w′′ − Pw′w′ dx

0

+ k ℓ2 w′(0)w′(0)+w′(ℓ)w′(ℓ)  − Pℓ w′(0)w′(0)+w′(ℓ)w′(ℓ)  (b) What are the essential and natural boundary conditions for the flexible beam? To obtain the natural boundary conditions we must take freebody diagrams of the two rigid segments, representing the section forces with the positive sign convention and including the forces in the springs. V(0)

M(0) V(ℓ)

P

P

M(ℓ) P

P

kw(0)

R2

R1 ℓ V(0) + M(0) − ℓ kw(0) + P w(0) = 0

kℓ w′(ℓ)

−M(ℓ) + P ℓ w′(ℓ)−kℓ w′(ℓ) = 0

Hence the boundary conditions are Essential:

w(0)−ℓ w′(0) = 0,

w(ℓ) = 0

Natural:

−EI ℓ w′′′(0)+Pℓ w′(0) + EIw′′(0) − k ℓ w(0) + Pw(0) = 0 −EI w′′(ℓ) + Pℓ w′(ℓ) − k ℓ w′(ℓ) = 0

Note that the first essential boundary condition is the result of the kinematic constrain imposed by the rigid part on the left. It is the manifestation at the left end of the flexible beam of the pin at the left end of the rigid part.

348


(c) Find an approximate solution for the lowest buckling load using a two-term polynomial Ritz basis. Express the result in terms of β, i.e., Pcr(β). Let ξ = x∕ℓ so that dx = ℓ dξ. A two-term polynomial that satisfies the essential boundary condition can be found by starting with a quartic polynomial and substituting the two essential boundary conditions. To wit, w = a + bξ + cξ 2 + dξ3 w′ = 1  b + 2cξ + 3dξ2  ℓ The essential boundary conditions are EBC 1: w(ℓ) = 0

a+b+c+d = 0

⇒

EBC 2: w(0)−ℓ w′(0)

a−b = 0

⇒

Therefore, we get a = b and c = −2b−d. Substitution these into the original expression gives w = b + bξ + −2b−d ξ 2 + dξ3 = b 1 + ξ − 2ξ2  + d  ξ3 − ξ 2  ≡ b h 1(ξ) + d h2(ξ) The base functions and their derivatives are h1 = 1 + ξ − 2ξ2

h 2 = ξ3 − ξ 2

h1′ = 1  1−4ξ  ℓ

h2′ = 1  3ξ 2−2ξ  ℓ

h1′′ = − 42 ℓ

h2′′ = 12  6ξ−2  ℓ

Thus, the components of the stiffness and geometric stiffness matrices can be computed as Kij =



ℓ

EI h i′′hj′′ dx + kℓ 2  h i′(0)hj′(0) + h i′(ℓ)hj′(ℓ)

0

G =P 

ℓ

ij

hi′h j′ dx + Pℓ  hi′(0)hj′(0) + h i′(ℓ)hj′(ℓ) 

0

Substituting the base functions yields K = EI ℓ



ℓ

0

−4  6ξ−2 

16 --4( 6ξ--2 )

( 6ξ--2)

2

dξ + kℓ2

Let λ ≡ Pℓ 2∕EI and noting that β ≡ kℓ 3∕EI we have K = EI ℓ

16 −4 −4

4

+ β EI ℓ

10 −3 −3

1

10

--3

--3

1

Chapter 11

349


Similarly, for the geometric stiffness we get G = Pℓ



 1−4ξ  2

ℓ

 1−4ξ   3ξ 2−2ξ 

 1−4ξ  3ξ −2ξ 

 3ξ −2ξ 

2

0

2

2

dξ + Pℓ

10

−3

−3

1

or, carrying out the integrals G = Pℓ

7∕3

−1∕3

−1∕3

2∕15

EI  K− G  = EI ℓ ℓ

+ Pℓ

10

−3

−3

1

λ 16 + 10β − 37 3

− 4 − 3β + 10 λ 3

= Pℓ

37∕3

−10∕3

−10∕3

17∕15

− 4 − 3β + 10 λ 3 4 + β − 17 λ 15

Critical loads are computed from det A = 0 det A = 0

⇒

≡ EI A  λ, β  ℓ

Pcr(β).

(d) What is the buckling load for very large spring stiffnesses (i.e., as β → 1)? Does the approximation appear to make sense in the limit? Explain. From a physical point of view, if the springs are infinitely rigid then the buckling load is exactly the same as a beam fixed at both ends Pcr  β ⇒ ∞ =

P

⇒

2 Pcr = 4π 2EI ℓ

The numerical approximation does not make sense in the limit because quadratic approximation is needed?

Rigid 281. A flexible beam of length ℓ and modulus EI is EI P welded to a rigid beam of length ℓ which rests on a k spring of modulus k=2EI/ℓ3. The beam is supported x as shown and is subjected to an end load P. Shear and axial deformations in the beam are negligible. Write ℓ ℓ the energy functional E and the virtual-work functional G for the system. What are the essential and natural boundary conditions for the flexible beam? Find an approximate solution for the displacement w(x) using a two-term polynomial Ritz basis.

(a) Write the energy functional E and the virtual work functional G for the system. The energy functional of the system includes the standard part for the flexible beam, the energy of the spring, and the potential energy of the load for the rigid part E(w) =

1 2

  EI(w′′) −P(w′)  dx + ℓ

2

0

2

1 k ℓ2  w′(ℓ)  2 − 12 Pℓ  w′(ℓ)  2 2

350


The corresponding virtual work functional can be found by taking the directional derivative of the energy in the direction of a virtual displacement

  EIw′′w′′−Pw′w′  dx + kℓ w′(ℓ)w′(ℓ) − Pℓ w′(ℓ)w′(ℓ) ℓ

2

G(w, w) =

0

(b) What are the essential and natural boundary conditions for the flexible beam? The natural boundary conditions can be found by considering a freebody diagram of the rigid piece as shown below. P

M(ℓ)

M(ℓ) = Pℓw′(ℓ) − kℓ 2w′(ℓ)

kℓ w′(ℓ) R

Hence, the boundary conditions are Essential: w(0) = 0, w(ℓ) = 0 Natural: w′′(0) = 0, EIw′′(ℓ) − Pℓw′(ℓ) + kℓ2w′(ℓ) = 0 (c) Find an approximate solution for the displacement w(x) using a two-term polynomial Ritz basis. Let ξ = x∕ℓ so that dx = ℓ dξ. A two-term polynomial that satisfies the essential boundary condition is w = a 1 ξ 1−ξ +a 2 ξ2 1−ξ  . The base functions are, therefore, h1 = ξ−ξ2

h2 = ξ2−ξ 3

h1′ = 1  1−2ξ  ℓ h1′′ = − 22 ℓ

h2′ = 1  2ξ−3ξ 2  ℓ h2′′ = 12  2−6ξ  ℓ

Note that h1′(ℓ) = −1 and h2′(0) = −1. Substituting the base functions into the virtual work functional gives K=



1

0

4

EI ℓ ℓ2

−22−6ξ 

−22−6ξ 

dξ + kℓ2

 2−6ξ  2

1 −1 1

−1

Observing that k = 2EI∕ℓ 3 and carrying out the integrals we get K=

EI ℓ

4

2

2

4

+ k ℓ2

1 −1 −1

1

=

EI ℓ

6

0

0

6

Similarly, for the geometric stiffness we get

 Pℓ 1

G=

0

 1−2ξ  2

 1−2ξ   2ξ−3ξ 2 

 1−2ξ   2ξ−3ξ 2 

 2ξ−3ξ 2  2

dξ + Pℓ

1 −1 −1

1

Chapter 11

351


or G = Pℓ 30

10

5

5

4

+ Pℓ 30

30 −30 −30

30

= Pℓ 30

40 −25 −25

34

Let λ ≡ Pℓ 2∕30EI. The governing equations  K−G  a = 0 has a nontrivial solution only if the determinant of the coefficient matrix is zero, Thus, det

6−40 λ

−25 λ

−25 λ

6−34 λ

=  6−40 λ   6−34 λ −25 λ2 = 0

which can be simplified to give the quadratic equation 36−444 λ+1360 λ 2−25 λ = 1335 λ 2−444 λ+36 = 0 that has the solution(s) λ=

444  (444) 2 − (4)(1335)(36) = 0.192, 0.140 (2)(1335)

The displacement of the structure is only possible in a buckling mode. The first buckling mode is associated with the lowest buckling eigenvalue 0.140. Take a1 = 1 and compute a2 from the equilibrium equations to give 6−40 (0.140)−25 (0.140) a 2 = 0 ⇒ 0.4 − 3.5 a2 = 0 ⇒ a 2 = 0.114 Therefore the first mode shape is [ 1.000, 0.114].

282. A flexible beam of length ℓ and moduRigid EI lus EI is welded to a rigid beam of length ℓ P and rests on an elastic foundation of modulus k x k=20EI/ℓ4. The beam is simply supported and is subjected to an end load P. The elastic ℓ ℓ foundation accrues a transverse force in proportion to the transverse displacement w. Shear and axial deformations in the beam are negligible. Write the energy functional E and the virtual-work functional G for the system. What are the essential and natural boundary conditions for the flexible beam? Find an approximate solution for the displacement w(x) using a polynomial Ritz basis.

(a) Write the energy functional E and the virtual work functional G for the system. The energy can be established by considering the sketch below, which includes the forces due to the elastic foundation.

352

Fundamentals of Structural Mechanics P

P θ ≡ w′(ℓ) ξ

E(w) =



ℓ

H(x)

kθ ξ

V(x)

 12 EI  w′′  2 + 12 k w2 − 12 P w′  2  dx +

0

=

1 2



ℓ

1  k θξ  2 2

dξ − 12 Pℓ θ 2

0



ℓ

 EI  w′′  2 + k w2 − P w′  2  dx + 1 k ℓ 3 θ 2 − 1 Pℓ θ 2 6

2

0

The virtual work functional can be found by taking the directional derivative of the energy in the direction of a virtual displacement. To wit, G(w, w) =



ℓ

 EI w′′w′′ + k ww − Pw′ w′  dx + 1 k ℓ3 θ θ − Pℓ θ θ 3

0

(b) What are the essential and natural boundary conditions for the flexible beam? P

M

 ξ ⋅ k θ ξ dξ ℓ

M(ℓ) = Pℓθ −

0

kθ ξ V

Essential:

w(0) = 0,

Natural:

w′′(0) = 0,

= Pℓθ − 13 kθℓ 3

w(ℓ) = 0 EIw′′(ℓ) + 13 k ℓ3w′(ℓ) − Pℓw′(ℓ) = 0

Note that the shear boundary condition includes the effects of the elastic foundation under the rigid part. (c) Find an approximate solution for the displacement w(x) using a two-term polynomial Ritz basis. Let ξ = x∕ℓ so that dx = ℓ dξ. A two-term polynomial that satisfies the essential boundary condition is w = a 1 ξ 1−ξ  + a 2 ξ2 1−ξ . The base functions are, therefore, h1 = ξ−ξ2

h2 = ξ2−ξ 3

h1′ = 1  1−2ξ  ℓ h1′′ = − 22 ℓ

h2′ = 1  2ξ−3ξ 2  ℓ h2′′ = 12  2−6ξ  ℓ

The stiffness matrix can be computed from the base functions as

Chapter 11

353


  EI h′′  h′′ + k h  h  ℓdξ + 1

K=

1 k ℓ 3 h′(1)  h′(1) 3

0

K = EI3 ℓ = EI3 ℓ

4

2

2

10

+ 20EI ℓ3

11.333

9.000

9.000

16.857

1∕30

1∕60

1∕30 1∕105

+ 20EI ℓ3

1∕3

1∕3

1∕3

1∕3

Similarly, the geometric stiffness can be computed as

 Ph′  h′ ℓ dξ + Pℓ h′(1)  h′(1) 1

G=

0

G = Pℓ



1

0

G = Pℓ

−2ξ  2 ξ−3 ξ 2 

−2ξ  2 −2ξ  2 ξ−3 ξ 2  1∕3

1∕6

1∕6

2∕15

 2 ξ−3 ξ 2  2

+ Pℓ

1

1

1

1

ℓ dξ + Pℓ

= Pℓ

4∕3

1

1

1

1

7∕6

7∕6 17∕15

The governing equations for the system are  K−λG  a = 0 which has a nontrivial solution only if the determinant of the coefficient matrix is zero.

283. A flexible bar of length ℓ and bending modulus EI is welded to a rigid bar of length ℓ. The structure is fixed at the bottom and subjected to a compressive axial load P at the top as shown. What are the appropriate essential and natural boundary conditions for this problem? Find an appropriate energy functional E(w) for the system, where w(x) is the transverse deflection of the flexible bar. Compute an approximation of the critical load of the system using the Ritz method and a polynomial basis function.

P ℓ ℓ

rigid x

EI

(a) What are the appropriate essential and natural boundary conditions for this problem? Essential:

w(0) = 0,

w′(0) = 0

Natural:

V(ℓ) = −EIw′′′(ℓ) + Pw′(ℓ) = 0 M(ℓ) = EIw′′(ℓ) = P ℓ w′(ℓ)

(b) Find an appropriate energy functional E(w) for the system, where w(x) is the transverse deflection of the flexible bar.

354

Fundamentals of Structural Mechanics E(w) = 12

  EI  w′′  − P  w′  ℓ

2

2

 dx − 1 Pℓ  w′(ℓ)  2 2

0

(c) Compute an approximation of the critical load of the system using the Ritz method and a one-term polynomial basis function. G(w, w) =



ℓ

 EI w′′w′′ − P w′w′  dx − p ℓ w′(ℓ) w′(ℓ)

0

Let w(x) = a x 2. Carrying out the computation we get



ℓ

G(a, a) = a

 EI(4) − P(4x 2)  dx − P ℓ(4ℓ 2)

0



a =0

∀a

Carrying out the integrals and noting the fundamental theorem of the calculus of variations we have

 4 EI ℓ − Pℓ3 43 + 4   a = 0 This equation has a trivial solution for the value Pcr =

3 EI 4 ℓ2

which is (an approximation of) the critical buckling load of the system.

284. A beam of flexural modulus 4EI carries the load P to the frame as shown in the figure. The frame is made of two columns pinned together by a beam at midheight. The frame members all have length 2ℓ and flexural modulus EI as shown. The force P is applied directly above the left column. The members have axial modulus EA≫EI/ℓ2. Estimate the smallest buckling capacity Pcr of the structure using the Ritz method. Resolve the problem assuming that the load can be placed anywhere along the top beam.

P 4EI roller

ℓ EI ℓ

EI

EI

2ℓ Note that the right column has no axial load on it and hence only adds lateral stiffness to the left column. The stiffness of the right beam is essentially equivalent to an

Chapter 11

355


equivalent spring of modulus k = 3EI∕ℓ 3 (i.e., the stiffness of a cantilever beam).The appropriate virtual work functional is G(w, w) =



2ℓ

0

 EIw′′w′′ − Pw′w′  dx + 3EI w(ℓ) w(ℓ) ℓ3

Let ξ = x∕ℓ and consider a cubic approximation of the displacement that satisfies the essential boundary conditions w′ = 1  2a1 ξ+3a2 ξ 2  , ℓ

w = a 1 ξ2 + a 2 ξ3 , K = EI3 ℓ



2

4

12 ξ 36 ξ

12 ξ

0

w′′ = 12  2a 1+6a2 ξ  ℓ

dξ + 3EI ℓ3

2

1

1

1

1

Carrying out the integrals we get K = EI3 ℓ

8

24

24

96

+ EI3 ℓ

3

3

3

3

= EI3 ℓ

11 27 27 99

Similarly, for the geometric stiffness we get G=1 ℓ



2

0

4 ξ2

6 ξ3

6 ξ3

9 ξ4

dξ = 1 ℓ

32∕3 24 24 288∕5

Let λ ≡ Pℓ 2∕EI K − λG = EI3 ℓ

11− 32 λ 3

27−24 λ

27−24 λ

λ 99− 288 5

The condition for a nontrivial solution to the governing equation is







det  K−λG  = 11 − 32 λ 99 − 288 λ −  27 − 24λ  2 = 0 3 5 which simplifies to λ2−10.25 λ+9.375 = 0, which gives λ = 1.015, 9.235. Therefore, the lowest critical load is Pcr = 1.015 EI2 ℓ If the load can be placed at any point along the top, the both columns have axial load and we must formulate the problem a bit differently. Assume that the load is placed a distance βℓ from the right column line. Then the axial force in the left column is βP and the axial force in the right column is  1−β  P. We can again take a polynomial approximation for the displacement in the two columns, wL = a1 ξ 2+a2 ξ 3 and wR = a3 ξ 2+a4 ξ 3 . The constraint on the two displacement fields is that

356


wL(1) = wR(1). Thus, a1+a 2 = a 3+a4. We can, therefore, eliminate one variable with this constraint, a4 = a1+a 2−a3. This allows us to write wR = a1 ξ 3+a2 ξ 3+a3  ξ2−ξ 3 . Let a =  a 1, a 2, a 3  and define base functions as follows h L =  ξ 2, ξ 3, 0 

hR =  ξ3, ξ3, ξ2−ξ 3 

hL′ =  2 ξ, 3 ξ2, 0 ∕ℓ

hR′ =  3 ξ2, 3 ξ 2, 2 ξ−3 ξ 2 ∕ℓ

hL′′ =  2, 6 ξ, 0 ∕ℓ2

hR′′ =  6 ξ, 6 ξ, 2−6 ξ ∕ℓ 2

Now, wL = a ⋅ hL(ξ) and wR = a ⋅ hR(ξ) and the displacement functions automatically satisfy the kinematic constraint. The virtual work functional is G(wL, wR, wL, wR) =



2ℓ

 EIw L′′wL′′ − β Pw L′wL′  dx

0

+



2ℓ

 EIw R′′wR′′ −  1−β  PwR′w R′  dx

0

Substituting the Ritz approximations into the virtual work functional yields the following stiffness matrix

 EI  h ′′  h ′′+ h ′′  h ′′  ℓ dξ 2

K=

L

L

R

R

0

and geometric stiffness matrices GL =



2

GR =

0

K = EI3 ℓ

 P  h ′  h ′  ℓ dξ 2

P  h L′  hL′  ℓ dξ,

R

R

0



2

4+36 ξ 2 12 ξ  1+3 ξ  12 ξ  1−3 ξ  12 ξ  1+3 ξ  12 ξ  1−3 ξ  72 ξ 2 12 ξ  1−3 ξ  12 ξ  1−3 ξ 

0

dξ

 2−6 ξ  2

and G = β G L+ 1−β  G R with

G L = Pℓ



2

0

G R = Pℓ



2

0

4 ξ2

6 ξ3

0

3

4

0

6ξ 0

9ξ 0

9 ξ4

9 ξ4

9 ξ4

9 ξ4

3 ξ 3  2−3 ξ  3 ξ 3  2−3 ξ 

dξ

0 3 ξ 3  2−3 ξ  3 ξ 3  2−3 ξ  2

ξ 2−3 ξ  2

dξ

Chapter 11


357

Now, the critical load Pcr(β) can be found by solving the equation det  K−G(P, β)  = 0

285. Resolve Problem 284 by solving the classical differential equations and boundary conditions. No solution available.

358


Chapter 12 Numerical Computations for Nonlinear Problems

286. Modify the program NEWTON to account for initial geometric imperfections in the two-bar rigid linkage connected by rotational springs.

The governing equations for the rigid linkage with imperfections is given as g1(θ) ≡ 2  θ 1−θ o1  −  θ 2−θ o2  − λ sin θ 1 = 0 g2(θ) ≡ − θ 1−θ o1  +  θ 2−θ o2  − λ sin θ 2 = 0 where θo = {θ o1, θ o2 } represent the imperfections. Because g(θ, λ) = 0 is linear in the imperfections, the Hessian matrix is identical to the one for the system without imperfections, given in the text. The modification to the program Newton are sketched below: The input data must be altered. We read in xo(1) and xo(2) as the values of the imperfections. The initial value of the load xo(3) is assumed to be zero. c.... Read problem parameters read(5,1000) tol,alpha read(5,1001) maxsteps,maxit read(5,1000) xo(1),xo(2) read(5,1000) x(1), x(2) xo(3) = 0.0 write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3)

Before starting the march along the load path the imperfection values are stored in the array because the array is subsequently used as an estimate of the next equilibrium state. c.... Initialize values for zero load,set direction to next trial state [code that is omitted is unchanged] y(1) = xo(1) y(2) = xo(2) write(6,2001) n,xo(1),xo(2),xo(3)

360

Fundamentals of Structural Mechanics c.... Compute MAXSTEPS points along the Equilibrium Path

Finally, the expression for the residual of the equilibrium equations must be altered to account for the imperfections. c.... Compute equilibrium and constraint at current state b(1) = 2.0*(x(1)-y(1)) - (x(2)-y(2)) - x(3)*dsin(x(1)) b(2) = -(x(1)-y(1)) + (x(2)-y(2)) - x(3)*dsin(x(2)) b(3) = 0.0

The following example is indicative of the revised program. Convergence tolerance Arc length parameter Number of load steps Maximum number of iterations Initial theta 1, theta 2, load Next theta 1, theta 2, load n 0 1 2 3 4 5 6 7 8 9 10

Theta 1 .10000 .31675 .58101 .85255 1.13333 1.42284 1.70943 1.96366 2.16156 2.30716 2.41533

Theta 2 .10000 .44218 .85997 1.27355 1.67414 2.04864 2.37312 2.61807 2.77878 2.87932 2.94387

Load .00000 .29314 .36811 .44032 .54372 .70474 .95490 1.30898 1.73910 2.20675 2.69062

: : : : : :

.1000E-07 .5000 10 50 .1000 .2000

.1000 .2000

.0000 .2000

EV 1

EV 2

NU

|| b ||

.11324 .12271 .20617 .35113 .57000 .88476 1.30065 1.78270 2.29141 2.80765

2.34325 2.32940 2.37511 2.47461 2.65020 2.93375 3.33413 3.81187 4.32193 4.84222

6 5 3 4 4 5 5 5 4 4

.1412E-13 .2429E-08 .6349E-08 .2429E-11 .3419E-09 .9155E-15 .4957E-13 .1212E-13 .4459E-08 .8173E-10

287. Modify the program NEWTON to analyze the three-bar rigid linkage shown. The bars are hinged together and are restrained by elastic springs that resist vertical motion. The springs accrue force in proportion to their extension, with modulus k. The system is subjected to an axial force P.

P k ℓ

k ℓ

ℓ

ℓ ℓ

φ1

φ2

γ

ℓ

P

ℓ  cos φ1+ cos φ2+ cos γ 

Let us describe the motion of the linkage with the independent angles φ1 and φ2 as shown in the sketch. The angle γ(φ) depends upon φ1 and φ2 through the constraint relationship sin γ = sin φ 1 + sin φ2

Chapter 12

Numerical Computation for Nonlinear Problems

361

The energy of the system is E(φ) =

1 2 kℓ sin 2 φ 1+ sin 2 γ  + Pℓ  cos φ 1+ cos φ 2+ cos γ  2

The virtual work functional can be computed by taking the directional derivative of the energy functional. The result, after dividing by kℓ2 and defining λ ≡ P∕kℓ is G(φ, φ) = sin φ 1 cos φ 1 φ1 + sin γ cos γDγ ⋅ φ − λ  sin φ 1 φ1+ sin φ 2 φ2+ sin γDγ ⋅ φ  where Dγ ⋅ φ can be computed from the constraint equation as cos γ Dγ(φ) ⋅ φ = cos φ 1 φ1 + cos φ2 φ 2 Thus, the equilibrium equations are g1(φ) ≡  2 sin φ1+ sin φ2  cos φ1 − λ  sin φ1 + tan γ cos φ1  = 0 g2(φ) ≡  sin φ 1+ sin φ2  cos φ 2 − λ  sin φ 2 + tan γ cos φ 2  = 0 The Hessian A(φ, λ) = ∇φ g(φ, λ) can be computed by taking the derivatives of the function g(φ) the components are λ  cos(φ +γ)+ cos 2 φ sec2 γ  A11 = cos 2φ 1 − sin φ1 sin φ 2 − cos 1 1 γ λ  cos φ cos φ  = A 1 2 21 cos 3 γ λ  cos(φ +γ)+ cos 2 φ sec2 γ  = cos 2φ 2 − sin φ1 sin φ 2 − cos 2 2 γ

A12 = cos φ 1 cos φ 2 − A22

The components of ∇λ g(φ, λ) can be computed easily from the equilibrium equations as A13 = − sin φ1 − tan γ cos φ 1 A23 = − sin φ2 − tan γ cos φ 2 The equations associated with the arc-length constraint are identical to the example problem in the text and need not be modified. These equations can be directly implemented into the program NEWTON. program Newton implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3),xo(3) C C C

*-----------------------------------------------------------* | Problem 287. Fundamentals of Structural Mechanics | *-----------------------------------------------------------*

c.... Read problem parameters read(5,1000) tol,alpha read(5,1001) maxsteps,maxit read(5,1000) xo(1),xo(2),xo(3) read(5,1000) x(1), x(2), x(3) write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3) c.... Initialize values for load step zero,set next trial state n = 0

362

Fundamentals of Structural Mechanics sum = 0.0 do j = 1,3 b(j) = x(j) - xo(j) sum = sum + b(j)*b(j) end do sum = alpha/dsqrt(sum) do j = 1,3 x(j) = xo(j) + sum*b(j) end do write(6,2001) n,xo(1),xo(2),xo(3) c.... Compute MAXSTEPS points along the Equilibrium Path do 2 n = 1,maxsteps c.... Perform Newton iteration at each load step nu = 0 1 nu = nu + 1 c.... Compute equilibrium and constraint at current state s1 = dsin(x(1)) c1 = dcos(x(1)) s2 = dsin(x(2)) c2 = dcos(x(2)) sgamma = s2 + s1 if (dabs(sgamma).lt.1.d0) then gamma = dasin(sgamma) else stop ’Arcsine of value greater than 1’ end if tang = dtan(gamma) b(1) = 2.0*s1*c1 + s2*c1 - x(3)*(s1 + tang*c1) b(2) = s1*c2 + s2*c2 - x(3)*(s2 + tang*c2) b(3) = 0.0 do k = 1,3 b(3) = b(3) + (x(k) - xo(k))**2 end do b(3) = b(3) - alpha**2 c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2) c.... Compute eigenvalues of Tangent stiffness matrix c2p1 = dcos(2.d0*x(1)) c2p2 = dcos(2.d0*x(2)) secg = 1.d0 + tang**2 cosg = dcos(gamma) f1 = dcos(x(1) + gamma) f2 = dcos(x(2) + gamma) aa = 2.0*c2p1 - s1*s2 - x(3)*(f1 + c1*c1*secg)/cosg bb = c2p2 - s1*s2 - x(3)*(f2 + c2*c2*secg)/cosg ab = c1*c2 - x(3)*c1*c2/(cosg**3) det = aa*bb - ab*ab eig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det)) eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det)) c.... Compute Hessian matrix A(1,1) = aa A(1,2) = ab A(1,3) = -s1 - tang*c1 A(2,1) = ab A(2,2) = bb A(2,3) = -s2 - tang*c2 A(3,1) = 2.0*(x(1) - xo(1)) A(3,2) = 2.0*(x(2) - xo(2)) A(3,3) = 2.0*(x(3) - xo(3)) c... Update state vector call invert(A,3,3) do j=1,3 do k=1,3 x(j) = x(j) - A(j,k)*b(k) end do end do

Chapter 12


363

c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test c.... Set values for previous converged state and guess at next state do j=1,3 temp = xo(j) xo(j) = x(j) x(j) = 2.0*x(j) - temp end do 2 continue stop ’Maximum number of steps exhausted’ 1000 format(3f10.0) 1001 format(2i10) 2000 format(’ Problem 287. Fundamentals of Structural Mechanics’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Arc length parameter : ’,f12.4/ * 5x, ’ Number of load steps : ’,i6/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Initial Phi 1, Phi 2, load : ’,3f12.4/ * 5x, ’ Next Phi 1, Phi 2, load : ’,3f12.4// * ’ n’ 5x,’ Phi 1’,5x,’ Phi 2’,8x,’Load’, * 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’) 2001 format(i5,5f12.5,i5,e12.4) end

The first branch can be computed starting at the point ( 0, 0, 13) and moving in the direction of the first mode (1, --2). Problem 287. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .2000 Number of load steps : 15 Maximum number of iterations : 50 Initial Phi 1, Phi 2, load : .0000 Next Phi 1, Phi 2, load : .1000

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Phi 1 .00000 .08914 .17638 .25966 .33663 .40446 .45984 .49935 .52030 .52208 .50661 .47733 .43761 .38979 .33504 .27347

Phi 2 .00000 -.17899 -.35856 -.53927 -.72164 -.90600 -1.09215 -1.27893 -1.46382 -1.64335 -1.81458 -1.97650 -2.13024 -2.27830 -2.42396 -2.57097

Load .33333 .32933 .31730 .29714 .26856 .23096 .18321 .12361 .05029 -.03783 -.14001 -.25370 -.37531 -.50097 -.62661 -.74742

.0000 -.2000

.3333 .3000

EV 1

EV 2

NU

|| b ||

-.00957 -.03789 -.08382 -.14541 -.21979 -.30276 -.38780 -.46444 -.51836 -.53690 -.51619 -.46183 -.38421 -.29412 -.20121

1.64334 1.57813 1.48476 1.38399 1.30067 1.26028 1.28566 1.39372 1.59201 1.87669 2.23466 2.64809 3.09761 3.56292 4.02101

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

.9411E-10 .5885E-14 .1491E-13 .6271E-13 .3693E-12 .2486E-11 .1457E-10 .5250E-10 .8426E-10 .5235E-10 .1345E-10 .1611E-11 .9048E-13 .3292E-14 .6629E-14

The computation of the equilibrium path emanating from the second bifurcation point can fail under the previous parameterization because φ2 = 0 (by symmetry) and γ eventually exceeds π∕2 making the arcsine function ambiguous. We can parameterize the problem to be more favorable for this case, but then the algorithm fails when computing the equilibrium path emanating from the first bifurcation point.

364

Fundamentals of Structural Mechanics ℓ ℓ

φ1

γ

φ2

ℓ

P

ℓ  cos φ1+ cos φ2+ cos γ 

Let us describe the motion of the linkage with the independent angles φ1 and φ2 . The angle γ(φ) depends upon φ1 and φ2 through the constraint relationship sin γ = sin φ 2 − sin φ1 The energy of the system is E(φ) =

1 2 kℓ sin 2 φ 1+ sin 2 φ 2  + Pℓ  cos φ1+ cos φ2+ cos γ  2

The virtual work functional can be computed by taking the directional derivative of the energy functional. The result, after dividing by kℓ2 and defining λ ≡ P∕kℓ is G(φ, φ) = sin φ 1 cos φ 1 φ1 + sin φ2 cos φ 2 φ2 − λ  sin φ 1 φ1+ sin φ 2 φ2+ sin γDγ ⋅ φ  where Dγ ⋅ φ can be computed from the constraint equation as cos γ Dγ(φ) ⋅ φ ≡ cos φ 2 φ2 − cos φ1 φ 1 Thus, the equilibrium equations are g1(φ) ≡ sin φ 1 cos φ 1 − λ  sin φ1 − tan γ cos φ 1  = 0 g2(φ) ≡ sin φ 2 cos φ 2 − λ  sin φ2 + tan γ cos φ 2  = 0 The Hessian A(φ, λ) = ∇φ g(φ, λ) can be computed by taking the derivatives of the function g(φ) the components are λ  cos(φ −γ)+ cos 2 φ sec 2 γ  A11 = cos 2φ 1 − cos 1 1 γ λ  cos φ cos φ  = A 1 2 21 cos 3 γ λ  cos(φ +γ)+ cos 2 φ sec 2 γ  = cos 2φ 2 − cos 2 2 γ

A12 = A22

The components of ∇λ g(φ, λ) can be computed easily from the equilibrium equations as A13 = − sin φ1 + tan γ cos φ 1 A23 = − sin φ2 − tan γ cos φ 2 The equations associated with the arc-length constraint are identical to the example problem in the text and need not be modified. These equations can be directly imple-

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365

mented into the program NEWTON. The differences from the previous parameterization can be observed in the code segment given below. All of the remaining code is the same. c.... Compute equilibrium and constraint at current state s1 = dsin(x(1)) c1 = dcos(x(1)) s2 = dsin(x(2)) c2 = dcos(x(2)) sgamma = s2 - s1 if (dabs(sgamma).lt.1.d0) then gamma = dasin(sgamma) else stop ’Arcsine of value greater than 1’ end if tang = dtan(gamma) b(1) = s1*c1 - x(3)*(s1 - tang*c1) b(2) = s2*c2 - x(3)*(s2 + tang*c2) b(3) = 0.0 do k = 1,3 b(3) = b(3) + (x(k) - xo(k))**2 end do b(3) = b(3) - alpha**2 c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2) c.... Compute eigenvalues of Tangent stiffness matrix c2p1 = dcos(2.d0*x(1)) c2p2 = dcos(2.d0*x(2)) secg = 1.d0 + tang**2 cosg = dcos(gamma) aa = c2p1 - x(3)*(dcos(x(1)-gamma) + c1*c1*secg)/cosg bb = c2p2 - x(3)*(dcos(x(2)+gamma) + c2*c2*secg)/cosg ab = x(3)*c1*c2/(cosg**3) det = aa*bb - ab*ab eig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det)) eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det)) c.... Compute Hessian matrix A(1,1) = aa A(1,2) = ab A(1,3) = -s1 + tang*c1 A(2,1) = ab A(2,2) = bb A(2,3) = -s2 - tang*c2 A(3,1) = 2.0*(x(1) - xo(1)) A(3,2) = 2.0*(x(2) - xo(2)) A(3,3) = 2.0*(x(3) - xo(3))

The first branch can be computed starting at the point ( 0, 0, 13) and moving in the direction of the first mode (--1, 1). Note that the eigenvectors of the linearized problem are slightly different for this parameterization. As you can see, the solution fails when the center bar goes to vertical because the constraint causes a computational error trying to find the arcsine of a number greater than one. Problem 287. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .0500 Number of load steps : 40 Maximum number of iterations : 50 Initial Phi 1, Phi 2, load : .0000 Next Phi 1, Phi 2, load : -.1000 n 0 1 2 3

Phi 1 .00000 -.03535 -.07068 -.10597

Phi 2 .00000 .03535 .07068 .10597

Load .33333 .33271 .33082 .32766

.0000 .1000

.3333 .3000

EV 1

EV 2

NU

|| b ||

-.00376 -.01519 -.03474

.66583 .66336 .65931

4 3 3

.1263E-08 .2531E-09 .2754E-09

366


4 -.14118 .14118 5 -.17629 .17629 6 -.21127 .21127 7 -.24606 .24606 8 -.28060 .28060 9 -.31482 .31482 10 -.34857 .34857 11 -.38166 .38166 12 -.41376 .41376 13 -.44427 .44427 14 -.47217 .47217 15 -.49571 .49571 16 -.51267 .51267 STOP Arcsine of value greater than

.32318 .31732 .30999 .30110 .29046 .27786 .26298 .24538 .22440 .19914 .16843 .13113 .08726 1

-.06325 -.10206 -.15330 -.22026 -.30826 -.42611 -.58924 -.82657 -1.19748 -1.83910 -3.12288 -6.30658 -17.51073

.65378 .64693 .63898 .63022 .62102 .61190 .60353 .59687 .59328 .59472 .60402 .62467 .65927

3 3 3 3 3 3 3 4 4 4 4 5 6

.3174E-09 .3902E-09 .5179E-09 .7562E-09 .1247E-08 .2409E-08 .5721E-08 .1250E-13 .1398E-12 .2833E-11 .1378E-09 .6830E-14 .4016E-10

The second branch is easily computed with this parameterization as shown below. Note that the solution starts at the bifurcation point (0,0,1) and moves in the direction of the second mode (1,1). Problem 287. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .2000 Number of load steps : 15 Maximum number of iterations : 50 Initial Phi 1, Phi 2, load : .0000 Next Phi 1, Phi 2, load : .1000 n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Phi 1 .00000 .14125 .28114 .41859 .55289 .68376 .81124 .93559 1.05722 1.17659 1.29423 1.41063 1.52633 1.64182 1.75763 1.87427

Phi 2 .00000 .14125 .28114 .41859 .55289 .68376 .81124 .93559 1.05722 1.17659 1.29423 1.41063 1.52633 1.64182 1.75763 1.87427

Load 1.00000 .99004 .96074 .91366 .85101 .77520 .68860 .59335 .49130 .38407 .27306 .15948 .04446 -.07096 -.18575 -.29884

.0000 .1000

1.0000 .9000

EV 1

EV 2

NU

|| b ||

-1.96066 -1.85054 -1.69064 -1.50841 -1.33076 -1.17886 -1.06573 -.99580 -.96580 -.96616 -.98268 -.99820 -.99496 -.96550 -.91069

-.01982 -.07698 -.16522 -.27578 -.39906 -.52583 -.64794 -.75862 -.85249 -.92544 -.97457 -.99802 -.99425 -.95268 -.85732

6 4 4 4 4 4 4 4 3 3 3 3 3 3 3

.8184E-09 .5193E-11 .2933E-11 .1198E-11 .3751E-12 .9459E-13 .1975E-13 .3361E-14 .8605E-08 .2620E-08 .5739E-09 .6039E-10 .3481E-12 .2276E-11 .1135E-09

288. Implement the constraint c(θ, λ) = det A(θ, λ) = 0 into the program NEWTON to locate bifurcation points exactly. At the bifurcation point, compute the eigenvectors of the tangent stiffness matrix and switch to another equilibrium branch. Let us implement the constraint for the example problem in the text. The determinant of the Hessian matrix can be computed as c(θ, λ) ≡ det A(θ, λ) =  2−λ cos θ 1   1−λ cos θ 2  − 1 The derivatives with respect to θ 1, θ 2, and λ are, respectively, A31 = λ sin θ 1  1−λ cos θ 2  A32 = λ sin θ 2  2−λ cos θ 1  A33 = − cos θ 1  1−λ cos θ 2  − cos θ 2  2−λ cos θ 1 

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367

These equations can be directly implemented into the program NEWTON. The program listing is given below. Note that we do not march along a load path, but rather search directly for a bifurcation point. program Newton implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3) C C C

*-----------------------------------------------------------* | Problem 288. Fundamentals of Structural Mechanics | *-----------------------------------------------------------*

c.... Read problem parameters read(5,1000) tol read(5,1001) maxit read(5,1000) x(1),x(2),x(3) write(6,2000) tol,maxit,(x(j),j=1,3) c.... Perform Newton iteration at each load step nu = 0 1 nu = nu + 1 c.... Compute equilibrium and constraint at current state aa = 2.0 - x(3)*dcos(x(1)) bb = 1.0 - x(3)*dcos(x(2)) det = aa*bb - 1.0 b(1) = 2.0*x(1) - x(2) - x(3)*dsin(x(1)) b(2) = -x(1) + x(2) - x(3)*dsin(x(2)) b(3) = det c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2) c.... Compute eigenvalues of Tangent stiffness matrix eig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det)) eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det)) c.... Compute Hessian matrix A(1,1) = aa A(1,2) = -1.0 A(1,3) = -dsin(x(1)) A(2,1) = -1.0 A(2,2) = bb A(2,3) = -dsin(x(2)) A(3,1) = x(3)*dsin(x(1))*bb A(3,2) = x(3)*dsin(x(2))*aa A(3,3) = - dcos(x(1))*bb - dcos(x(2))*aa c.... Update state vector call invert(A,3,3) do j=1,3 do k=1,3 x(j) = x(j) - A(j,k)*b(k) end do end do c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test stop ’Critical point found’ 1000 format(3f10.0) 1001 format(2i10) 2000 format(’ Problem 288. Fundamentals of Structural Mechanics’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Initial theta 1, theta 2, load : ’,3f12.4// * ’ n’ 5x,’Theta 1’,5x,’Theta 2’,8x,’Load’, * 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’) 2001 format(i5,5f12.5,i5,e12.4) end

368


The program can, of course, be used to find the critical points on the trivial equilibrium path. However, there are other critical points that are less obvious. Using the starting values given below, we find one of those critical points. Problem 288. Fundamentals of Structural Mechanics Convergence tolerance Maximum number of iterations Initial theta 1, theta 2, load n 0

Theta 1 .84928

Theta 2 4.50308

: : :

Load -3.73532

.1000E-07 50 .2000 EV 1 .00000

-1.5000 EV 2 4.69111

5.0000 NU 10

|| b || .7174E-14

This solution corresponds to the position shown in the sketch. The top bar has rotated around nearly three quarters of a complete circle. λ = 3.73 θ2 = 4.5 θ1 = 0.85

P 289. Modify the program NEWTON to P analyze the three-bar rigid linkage shown below. This structure has three ℓ θ3 degrees of freedom. The bars are hinged k together and are restrained by elastic springs that resist relative rotation. The ℓ θ2 k springs accrue force in proportion to ℓ( cos θ1 + cos θ 2 + cos θ3) their extension, with modulus k. The θ1 ℓ system is subjected to an axial force P. k Lateral loads can be viewed as imperfections to a purely axial loading system. Modify the equations of equilibrium to allow the applications of the lateral loads Á 1 P, Á 2 P, and Á 3 P at locations ℓ, 2ℓ, and 3ℓ respectively, where Á i is a fixed value recording the ratio of the lateral load to the axial load. Implement the load imperfections in the program.

The energy of the system, including the transverse loads (acting to the left), is E(θ) =

1  2 k θ1 2

+  θ 2−θ 1  2 +  θ 3−θ 2  2  + Pℓ  cos θ 1+ cos θ 2+ cos θ 3 

− Pℓ   Á 1+Á 2+Á 3  sin θ 1 +  Á 2+Á 3  sin θ 2 + Á 3 sin θ 3  The virtual work functional can be computed by taking the directional derivative of the energy functional. The result, after dividing by k and defining λ ≡ Pℓ∕k is

Chapter 12


369

G(θ, θ) = θ 1 θ 1 +  θ 2−θ 1  θ 2−θ 1  +  θ 3−θ 2  θ 3−θ 2  − λ  sin θ 1 θ 1 + sin θ 2 θ 2 + sin θ 3 θ 3  − λ   Á 1+Á 2+Á 3  cos θ 1 θ 1 +  Á 2+Á 3  cos θ 2 θ 2 + Á 3 cos θ 3 θ 3  The equilibrium equations can be obtained as g1(θ) ≡ 2 θ 1 − θ 2 − λ sin θ 1 + Á 123 cos θ 1  = 0 g2(θ) ≡ −θ 1 + 2θ 2 − θ 3 − λ sin θ 2 + Á 23 cos θ 2  = 0 g3(θ) ≡ −θ 2 + θ 3 − λ  sin θ 3 + Á 3 cos θ 3  = 0 where Á 123 ≡ Á 1+Á 2+Á 3 and Á 23 ≡ Á 2+Á 3. The constraint is similar to the example in the text, except that there are three degrees of freedom c(θ, λ) ≡ ‖ θ−θn ‖ 2 +  λ−λn  2 + α 2 = 0 The components of the matrix A are given by A11 = 2−λ  cos θ 1−Á 123 sin θ 1  A22 = 2−λ  cos θ 2−Á 23 sin θ 2  A33 = 1−λ  cos θ 3−Á 3 sin θ 3  The off-diagonal terms are A12 = A23 = −1 and A13 = 0, with corresponding symmetric terms below the diagonal. The derivatives with respect to λ give A14 = − sin θ 1+Á 123 cos θ 1  A24 = − sin θ 2+Á 23 cos θ 2  A34 = − sin θ 3+Á 3 cos θ 3  The terms associated with the constraint are the same as before, extended to one more degree of freedom. These equations can be directly implemented into the program NEWTON. The program listing is given below. program Newton parameter (ndm=4) implicit double precision (a-h,o-z) dimension A(ndm,ndm),b(ndm),x(ndm),xo(ndm),sn(3),cn(3) dimension H(3,3),V(3,3),e(3) C C C

*-----------------------------------------------------------* | Problem 289. Fundamentals of Structural mechanics | *-----------------------------------------------------------*

c.... Read problem parameters read(5,1000) tol,alpha read(5,1001) maxsteps,maxit read(5,1000) eps1, eps2, eps3 read(5,1000) xo(1),xo(2),xo(3),xo(4) read(5,1000) x(1), x(2), x(3), x(4) write(6,2000) tol,alpha,maxsteps,maxit,eps1,eps2,eps3, * (xo(j),j=1,4),(x(j),j=1,4)

370

Fundamentals of Structural Mechanics c.... Initialize values for load step zero,set next trial state n = 0 sum = 0.0 do j = 1,ndm b(j) = x(j) - xo(j) sum = sum + b(j)*b(j) end do sum = alpha/dsqrt(sum) do j = 1,ndm x(j) = xo(j) + sum*b(j) end do eps123 = eps1 + eps2 + eps3 eps23 = eps2 + eps3 write(6,2001) n,(xo(i),i=1,4) c.... Compute MAXSTEPS points along the Equilibrium Path do 2 n = 1,maxsteps c.... Perform Newton iteration at each load step nu = 0 1 nu = nu + 1 c.... Compute equilibrium and constraint at do j=1,3 sn(j) = dsin(x(j)) cn(j) = dcos(x(j)) end do b(1) = 2.0*x(1) - x(2) b(2) = -x(1) + 2.0*x(2) - x(3) b(3) = - x(2) + x(3) b(4) = 0.0 do k = 1,ndm b(4) = b(4) + (x(k)-xo(k))**2 end do b(4) = b(4) - alpha**2

current state

x(4)*(sn(1)+ eps123*cn(1)) x(4)*(sn(2)+ eps23*cn(2)) x(4)*(sn(3)+ eps3*cn(3))

c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2 + b(4)**2) c.... Compute eigenvalues of Tangent stiffness matrix A(1,1) = 2.0 - x(4)*(cn(1)- eps123*sn(1)) A(1,2) = -1.0 A(1,3) = 0.0 A(2,1) = -1.0 A(2,2) = 2.0 - x(4)*(cn(2)- eps23*sn(2)) A(2,3) = -1.0 A(3,1) = 0.0 A(3,2) = -1.0 A(3,3) = 1.0 - x(4)*(cn(3)- eps3*sn(3)) do j=1,3 do k=1,3 H(j,k) = A(j,k) end do e(j) = 1.0 end do call eigens(H,V,e,3,6,nr) c.... Compute border of Hessian matrix A(1,4) = -(sn(1) + eps123*cn(1)) A(2,4) = -(sn(2) + eps23*cn(2)) A(3,4) = -(sn(3) + eps3*cn(3)) A(4,1) = 2.0*(x(1)-xo(1)) A(4,2) = 2.0*(x(2)-xo(2)) A(4,3) = 2.0*(x(3)-xo(3)) A(4,4) = 2.0*(x(4)-xo(4)) c... Update state vector call invert(A,4,4) do j=1,4 do k=1,4 x(j) = x(j) - A(j,k)*b(k) end do end do

Chapter 12


371

c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,(x(i),i=1,4),(e(i),i=1,3),nu,test c.... Set values for previous converged state and guess at next state do j=1,4 temp = xo(j) xo(j) = x(j) x(j) = 2.0*x(j) - temp end do 2 continue stop ’Maximum number of steps exhausted’ 1000 format(4f10.0) 1001 format(2i10) 2000 format(’ Problem 289. Fundamentals of Structural Mechanics’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Arc length parameter : ’,f12.4/ * 5x, ’ Number of load steps : ’,i6/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Eps 1, Eps 2, and Eps 3 : ’,3f12.4/ * 5x, ’ Initial theta 1, 2, 3, and load : ’,4f12.4/ * 5x, ’ Next theta 1, 2, 3, and load : ’,4f12.4// * 4x, ’n’,5x,’Theta 1’,5x,’Theta 2’,5x,’Theta 3’,8x,’Load’, * 8x, ’EV 1’,8x,’EV 2’,8x,’EV 3’,’ NU’,4x,’|| b ||’) 2001 format(i5,7f12.5,i5,e12.4) end

Moving from a two by two system to a three by three system requires that we get a little more general with our solution of the eigenvalue problem at each step. The subroutine EIGENS solves the eigenvalue problem H φ = m Eφ where H is a symmetric N × N matrix and E is a diagonal N × N matrix. The diagonal elements of E are stored as a vector. For our purposes E = I. The eigenvalue problem is solved by the Jacobi algorithm wherein off-diagonal elements are zeroed by performing a rotation. The eigenvectors are returned in the array U and the eigenvalues are returned in the array E. C C C C C C C C

subroutine eigens(h,u,e,n,ns,nr) *---------------------------------------------------------------* | Eigenvalues and eigenvectors by Jacobi’s method | | solve H.u = m.E.u where E is a diagonal matrix | | the eigenvectors are stored in u, eigenvalues in E | | n is the dimension of the matrices H and u | | ns is the number of digits of accuracy required (input) | | nr is the number of Jacobi rotations performed (output) | *---------------------------------------------------------------* implicit double precision (a-h,o-z) dimension h(n,n),u(n,n),e(n) test=1.0/10.**(2*ns) nn=n-1 nr=0 nrlm=5*n**2 toler=0.1

c.... Normalize to standard eigenvalue problem do 10 i=1,n 10 e(i)=1.0/dsqrt(e(i)) do 30 i=1,n do 20 j=1,n h(i,j)=e(i)*h(i,j)*e(j) 20 u(i,j)=0.0 30 u(i,i)=1.0

372

Fundamentals of Structural Mechanics c.... Reduce matrix 50 xmax=0.0 do 700 ii=1,nn jl=ii+1 do 600 jj=jl,n c.... Check if rotation is required hii=h(ii,ii) hij=h(ii,jj) hjj=h(jj,jj) d=dabs(hii*hjj) h2=hij*hij if(h2.gt.xmax*d) xmax=h2/d if(h2.lt.toler*d) go to 600 c.... Compute tan, sin and cos nr=nr+1 ht=.5*(hii-hjj)/hij tn=-ht-dsign(dsqrt(ht*ht+1.0),ht) cs =1.0/dsqrt (1.+tn **2) sn=cs*tn c2=cs**2 s2=sn**2 c.... Reduce ii,jj element to zero ht=2.*hij*cs*sn h(ii,jj)=0.0 h(ii,ii)= hii*c2 + ht + hjj*s2 h(jj,jj)= hii*s2 - ht + hjj*c2 do 530 i=1,n if(i-ii) 370,530,420 370 ht=h(i,ii) h(i,ii)= cs*ht + sn*h(i,jj) h(i,jj)=-sn*ht + cs*h(i,jj) go to 530 420 if(i-jj) 430,530,480 430 ht=h(ii,i) h(ii,i)= cs*ht + sn*h(i,jj) h(i,jj)=-sn*ht + cs*h(i,jj) go to 530 480 ht=h(ii,i) h(ii,i)= cs*ht + sn*h(jj,i) h(jj,i)=-sn*ht + cs*h(jj,i) 530 continue c.... Operate on eigenvectors 540 do 550 i=1,n ht=u(i,ii) u(i,ii)= cs*ht + sn*u(i,jj) 550 u(i,jj)=-sn*ht + cs*u(i,jj) 600 continue 700 continue c.... Test for end iteration set new tolerance if(nrlm.lt.nr) go to 1000 if(xmax.lt.test) go to 710 toler=0.10*xmax go to 50 c.... Normalize and order eigenvectors 710 do 800 i=1,n do 750 j=1,n 750 u(i,j)=u(i,j)*e(i) 800 e(i)=h(i,i) c.... Order eigenvalues and eigenvectors do 900 i=1,nn jl=i+1 ht=e(i) im=i do 850 j=jl,n if(ht.lt.e(j)) go to 850 ht=e(j) im=j

Chapter 12


373

850 continue e(im)=e(i) e(i)=ht do 900 j=1,n ht=u(j,i) u(j,i)=u(j,im) 900 u(j,im)=ht return 1000 write (not,2000) return 2000 format(’ **** warning: iteration terminated without convergence’/) end

The program is demonstrated by finding the equilibrium path emanating from the critical load λ1 = 0.198 moving in the direction of the buckling mode φ1 ={0.328, 0.591, 0.737}. The imperfections are zero for the bifurcation problem. Problem 289. Fundamentals of Structural Convergence tolerance : Arc length parameter : Number of load steps : Maximum number of iterations : Eps 1, Eps 2, and Eps 3 : Initial theta 1, 2, 3, and load : Next theta 1, 2, 3, and load : n 0 1 2 3 4 5 6 7 8 9 10

Theta 1 Theta 2 Theta 3 .0000 .0000 .0000 .0820 .1478 .1842 .1643 .2956 .3682 .2471 .4437 .5518 .3306 .5921 .7348 .4151 .7408 .9168 .5010 .8900 1.0977 .5887 1.0398 1.2771 .6786 1.1901 1.4545 .7712 1.3411 1.6295 .8672 1.4925 1.8014

Load .1980 .1989 .2016 .2062 .2129 .2218 .2333 .2479 .2662 .2890 .3173

Mechanics .1000E-07 .2500 10 20 .0000 .0000 .0000 .0000 .3300 .5900

.0000 .0000 .7400

.1980 .1980

EV 1

EV 2

EV 3

NU

|| b ||

.0018 .0071 .0161 .0288 .0454 .0663 .0916 .1220 .1581 .2007

1.3578 1.3605 1.3650 1.3715 1.3801 1.3910 1.4046 1.4213 1.4416 1.4665

3.0498 3.0525 3.0571 3.0639 3.0730 3.0850 3.1003 3.1197 3.1442 3.1752

3 3 3 3 3 3 3 3 3 4

.881E-11 .395E-10 .656E-10 .112E-09 .213E-09 .433E-09 .920E-09 .203E-08 .467E-08 .576E-15

The problem was also run with the imperfections values Á 1 = Á 2 = Á 3 = 0.01. Note that in this case one can specify the direction in which to move as {0, 0, 0, 1}. The results of both cases are shown in the figure below. λ

λ1 Á 1 = Á 2 = Á 3 = 0.01

θ3

374


290. Modify the program ELASTICA to incorporate a distributed transverse loading q(x) on the cantilever column in addition to the load P. Examine the case where the transverse load is proportional to the axial load, as well as the case where the transverse load is fixed and the axial load is increased.

The expression for the virtual work functional for the elastica with transverse load was derived in Problem 263 in Chapter 11. The result for q(x) = q o is

  θ′θ′ − λ  sin θ+γ  1−ξ  cos θ  θ  dξ ℓ

G(λ, θ, θ) =

0

where ξ ≡ x∕ℓ, λ ≡ P∕EI, and γ ≡ q o∕P. The directional derivatives of the virtual work functional can be computed as DG ⋅ ∆θ DG ⋅ ∆λ

  ∆θ′θ′ − λ  cos θ −γ  1−ξ  sin θ  ∆θ θ  dx ℓ

=

λo, θ o



o

o

o

0

 − ∆λ  sin θ +γ  1−ξ  cos θ  θ dx ℓ

λo, θ o

=

o

o

o

0

Interpolating the functions θ, θ, and ∆θ with base functions hi(x) we can put the linea^ rized virtual work functional in the form G(λ, a, a) = a T  K ν ∆a+k ν ∆ λ+g ν  where the components of the coefficient matrices are given by

  h ′h ′ − λ  cos θ −γ  1−ξ  sin θ  h h  dx ℓ

Kνij

=

i

ν

j

ν

ν

i j

0

 − sin θ +γ  1−ξ  cos θ  h dx ℓ

kνi =

ν

ν

i

0

   θ ′h ′ − λ  sin θ −γ  1−ξ  cos θ  h  dx ℓ

gνi =

ν

j

ν

ν

ν

i

0

The arc-length constraint is exactly the same as the example in the text. The modifications to the program ELASTICA are manifested in the following changes to the program. First, the ratio of the transverse load to the axial load must be input. The input segment of the code is modified as follows c.... Read problem parameters read(5,1000) tol,alpha,xlength,gamma

The output of the input values is also appropriately modified. The call to the subroutine FCN is changed to pass the value of γ as call fcn(A,b,x,z,xlength,gamma,factor)

Finally, the main changes are to the subroutine FCN to incorporate the new definitions of the matrices K ν , kν, and gν as given above. Recall, that this routine is called for

Chapter 12


375

each integration point z and is accumulated in the arrays. It should be clear how to alter the functional form of the transverse loading in this subroutine. C C C

subroutine fcn(A,b,x,z,xlength,gamma,factor) *------------------------------------------------------------------* | Compute Simpson contribution to Hessian and residual matrices | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3) data ev1/1.570796327d0/,ev2/4.712388981d0/

c.... Basis h1 = h2 = dh1 = dh2 =

functions and their derivatives dsin(ev1*z) dsin(ev2*z) ev1*dcos(ev1*z)/xlength ev2*dcos(ev2*z)/xlength

c.... Transverse loading function q = (1.0 - z)*gamma c.... Compute rotation and first derivative at current point theta = x(1)*h1 + x(2)*h2 dtheta = x(1)*dh1 + x(2)*dh2 c1 = dcos(theta)- dsin(theta)*q c2 = dsin(theta)+ dcos(theta)*q c.... Compute integral part of residual vector b(1) = b(1) + (dtheta*dh1 - x(3)*c2*h1)*factor b(2) = b(2) + (dtheta*dh2 - x(3)*c2*h2)*factor c.... Compute integral A(1,1) = A(1,1) A(1,2) = A(1,2) A(1,3) = A(1,3) A(2,1) = A(2,1) A(2,2) = A(2,2) A(2,3) = A(2,3)

part of Hessian matrix + (dh1*dh1 - x(3)*c1*h1*h1)*factor + (dh1*dh2 - x(3)*c1*h1*h2)*factor - (c2*h1)*factor + (dh2*dh1 - x(3)*c1*h2*h1)*factor + (dh2*dh2 - x(3)*c1*h2*h2)*factor - (c2*h2)*factor

return end

The program we run with γ = 0.1 with the following results. Note that the direction for the next state is simply {0, 0, 1} because the imperfection can drive the problem. Problem 290. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 10 Maximum number of iterations : 30 Number of integration intervals : 10 Column Length : 1.0000 Transverse load ratio to P : .1000 Initial a1, a2, load : .0000 Next a1, a2, load : .0000 n 0 1 2 3 4 5 6 7 8 9 10

a1 .00000 .01175 .03149 .07132 .18546 .52441 .93486 1.29972 1.60012 1.84204 2.03768

a2 .00000 .00118 .00242 .00372 .00505 .00659 .01045 .01843 .03019 .04453 .06039

P/EI .00000 .49986 .99947 1.49788 1.98468 2.35225 2.63776 2.97953 3.37906 3.81640 4.27627

.0000 .0000

.0000 1.0000

EV 1

EV 2

NU

|| b ||

.98385 .73450 .48733 .25795 .18865 .34207 .60593 .91886 1.24719 1.57544

10.85346 10.60411 10.35669 10.12403 10.02147 10.08710 10.22891 10.40964 10.60985 10.81942

4 3 4 4 7 5 4 4 4 4

.3886E-15 .4089E-08 .2939E-13 .3451E-08 .5684E-14 .2003E-13 .9325E-09 .3600E-08 .6387E-09 .5783E-10

The problem was also run with γ = 0 to compute the equilibrium path emanating from the critical load λ1 = 2.467 with the direction input as {1, 0, 2.467}. Both results are plotted in the following figure.

376

Fundamentals of Structural Mechanics λ

λ1 γ = 0.1 a1 The case of transverse loading that is not proportional to the axial load is very simple to implement. The components of the coefficient matrices are given by

  h ′h ′ −  λ cos θ −γ  1−ξ  sin θ  h h  dx ℓ

Kνij =

i

ν

j

ν

ν

i

j

0

 − sin θ h dx ℓ

kνi =

ν

i

0

=    θ ′h ′ −  λ sin θ −γ  1−ξ  cos θ  h  dx ℓ

gνi

ν

j

ν

ν

ν

i

0

where γ ≡ q o∕EI. The modifications to the program are trivial.

291. Modify the program ELASTICA to incorporate N base functions. Examine the performance of the system as the number of base functions is increased.

The exercise is primarily one in generalizing the program. The program from the previous problem (including the transverse load) is generalized to more base functions. The base functions used are the eigenfunctions hi(ξ) = sin (2i−1)πξ  The main program and the subroutine FCN are the only code segments that change. These are listed below. C C C C

program Elastica *-----------------------------------------------------------* | Problem 291. Fundamentals of Structural Mechanics | | NDM is the number of base functions | *-----------------------------------------------------------* parameter(ndm=4,nn=ndm+1) implicit double precision (a-h,o-z) dimension A(nn,nn),b(nn),x(nn),xo(nn),h(ndm),dh(ndm) dimension AA(ndm,ndm),U(ndm,ndm),e(ndm) data zero/0.d0/,one/1.d0/,two/2.d0/

c.... Read problem parameters read(5,1000) tol,alpha,xlength,gamma

Chapter 12


read(5,1001) maxsteps,maxit,inter read(5,1000) (xo(i),i=1,nn) read(5,1000) (x(i),i=1,nn) write(6,2000) tol,alpha,maxsteps,maxit,inter,xlength,gamma write(6,2004) (j,xo(j),x(j),j=1,nn) c.... Initialize constants of integration dz = one/(2*inter) npoints = 2*inter + 1 c.... Initialize values for load step zero,set next trial state n = 0 sum = zero do j = 1,nn b(j) = x(j) - xo(j) sum = sum + b(j)*b(j) end do sum = alpha/dsqrt(sum) do j = 1,nn x(j) = xo(j) + sum*b(j) end do write(6,2001) n,nn,sum write(6,2002) (i,xo(i),i=1,nn) c.... Compute MAXSTEPS points along the Equilibrium Path do 2 n = 1,maxsteps c.... Perform Newton iteration at each load step nu = 0 1 nu = nu + 1 c.... Execute Simpson integration of Hessian and residual components call zerovec(b,nn) call zerovec(A,nn*nn) z = zero do m=1,npoints call simpson(m,npoints,dz,xlength,factor) call fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm) z = z + dz end do c.... Finish computation of Hessian and residual do k=1,nn b(nn) = b(nn) + (x(k) - xo(k))**2 A(nn,k) = two*(x(k) - xo(k)) end do b(nn) = b(nn) - alpha**2 c.... Compute norm of residual for convergence test test = 0.d0 do k=1,nn test = test + b(k)**2 end do test = sqrt(test) c.... Compute eigenvalues of Tangent stiffness matrix do j=1,ndm do k=1,ndm AA(j,k) = A(j,k) end do e(j) = 1.d0 end do call eigens(AA,U,e,ndm,8,NR) c... Update state vector call invert(A,nn,nn) do j=1,nn do k=1,nn x(j) = x(j) - A(j,k)*b(k) end do end do c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1

377

378

Fundamentals of Structural Mechanics if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,nu,test write(6,2005) ’ Displacement Coefficients’ write(6,2002) (i,x(i),i=1,nn) write(6,2005) ’ Eigenvalues of tangent stiffness matrix’ write(6,2002) (i,e(i),i=1,ndm) c.... Set values for previous converged state and guess at next state do j=1,nn temp = xo(j) xo(j) = x(j) x(j) = two*x(j) - temp end do 2 continue stop ’Maximum number of steps exhausted’ 1000 format(8f10.0) 1001 format(3i10) 2000 format(’ Problem 291. Fundamentals of Structural Mechanics’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Arc length parameter : ’,f12.4/ * 5x, ’ Number of load steps : ’,i6/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Number of integration interval s : ’,i6/ * 5x, ’ Column Length : ’,f12.4/ * 5x, ’ Transverse load ratio to P : ’,f12.4/ * 5x, ’ Initial - Next values of a1,..., a(ndm), load : ’,/) 2004 format(15x,i3,5x,2f12.4) 2001 format(/’ Step number > ’,i4/, * ’ No. Iterations > ’,i4/, * ’ || b || > ’,e12.4) 2002 format(5x,i3,5x,f12.5) 2005 format(/a) end C----------------------------------------------------------------------FCN subroutine fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm) C C C

*------------------------------------------------------------------* | Compute Simpson contribution to Hessian and residual matrices | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension A(nn,nn),b(nn),x(nn),h(ndm),dh(ndm) data zero,one/0.d0,1.d0/

c.... Basis functions and their derivatives ev = 2.d0*atan(1.d0) do i=1,ndm cev = (2*i-1)*ev h(i) = dsin(cev*z) dh(i) = cev*dcos(cev*z)/xlength end do c.... Transverse loading function q = (one - z)*gamma c.... Compute rotation and first derivative at current point theta = zero dtheta = zero do i=1,ndm theta = theta + x(i)* h(i) dtheta = dtheta + x(i)*dh(i) end do c1 = dcos(theta)- dsin(theta)*q c2 = dsin(theta)+ dcos(theta)*q c.... Compute integral part of Hessian matrix and residual vector do i=1,ndm do j=1,ndm A(i,j) = A(i,j) + (dh(i)*dh(j) - x(nn)*c1*h(i)*h(j))*factor end do A(i,nn) = A(i,nn) - c2*h(i)*factor b(i) = b(i) + (dtheta*dh(i) - x(nn)*c2*h(i))*factor end do

Chapter 12


379

return end

The program was run for the load case of Problem 290 with γ = 0.1 using a one, two, three, and four term expansion of the displacement function. The results at the end of 10 steps are summarized below a1 a2 a3 a4 Load

Displacement Coefficients 1 Term 2 terms 3 terms 4 terms 2.01914 2.03768 2.03806 2.03807 .06039 .06073 .06074 .00480 .00482 .00087 4.28929 4.27627 4.2760 4.27603

Eigenvalues of tangent stiffness 1 Term 2 terms 3 terms 4 terms 1.60159 1.57544 1.57438 1.57435 10.81942 10.79868 10.79777 30.53156 30.51467 60.13066 4.28929 4.27627 4.2760 4.27603

It is clear that the response is dominated by the first mode for this structure. The contributions of the higher modes decreases with mode number.

292. Modify the program ELASTICA to use polynomial base functions. Examine the performance of the system as the number of base functions is increased. Compare the performance of the polynomial functions with the eigenfunctions.

The modification to the program in Problem 291 involves only the definition of the base functions in the subroutine FCN. We have implemented the base functions hn(ξ) = ξn where ξ = x∕ℓ. The modified segment of subroutine FCN is given below. c.... Base functions and their derivatives do i=1,ndm h(i) = z**i dh(i) = i*z**(i-1)/xlength end do

The program was run for the load case of Problem 290 with γ = 0.1 using a one, two, three, and four term expansion of the displacement function. The results at the end of 10 steps are summarized below. a1 a2 a3 a4 Load Rot

Displacement Coefficients 1 Term 2 terms 3 terms 4 terms 1.90046 2.77087 2.79818 2.63404 -1.31778 -.81545 -.25533 -.41310 -1.27835 .43191 4.30798 3.19684 3.37669 3.31565 1.90046 1.45309 1.56963 1.53227

Eigenvalues of tangent stiffness 1 Term 2 terms 3 terms 4 terms .84553 .10743 .01261 .00083 1.84271 .24839 .03437 3.66093 .37639 5.69782 4.30798 3.19684 3.37669 3.31565

One can clearly see the differences between the polynomial basis and the eigenbasis. At four terms the solution still changes quite a bit with the addition of another base function. The last row, called Rot, is the value θ(ℓ) computed from the approximation. It is evident that the load and rotation are converging as the number of base functions increases. For a point of reference we can interpolate the values from Problem 290 to give a load value of 3.289 at a displacement of 1.532, an error of less than 1%. The conditioning of the system of equations (as measured by the ratio of the largest to the

380


smallest eigenvalues of the tangent stiffness matrix) is considerably worse for the polynomial basis.

293. Modify the program ELASTICA to account for a nonlinear moment curvature relationship of the form M(κ o) =

EI o κ o

1+mo κ 2o

where EI o and m o are material constants. Note that for small values of m o, the constitutive model reduces to the linear model originally used. The material is hyperelastic because an energy function exists such that M = ∂W(κ o)∕∂κ o . What is the strain energy function W? Plot the bifurcation diagrams for various values of the material constants.

The strain energy function can be computed by finding the anti-derivative (indefinite integral) of the moment W(κ o) =

 M(κ ) dκ o

o

EI = mo o

 1+m κ o

2 o

−1



where the constant of integration has been set to give zero energy when the curvature is zero. The validity of this expression can be demonstrated by differentiating the result to give the original relationship between moment and curvature. The curvature is related to the rotation as κ o = θ′. The virtual work functional from Problem 290 can be modified to read

  m(θ′) θ′ − λ  sin θ+γ  1−ξ  cos θ  θ  dξ ℓ

G(λ, θ, θ) =

0

where the function m(θ′) = M(θ′)∕EI o and the parameter λ ≡ P∕EI o. We can compute the directional derivative of G in the direction of ∆ θ as

  m′(θ′)∆ θ′θ′ − λ  cos θ−γ  1−ξ  sin θ  ∆ θ θ  dξ ℓ

DG ⋅ ∆ θ =

0

where the derivative of the function m(θ′) is given by −3∕2 dm(κ o) =  1+m o κ 2o  dκ o

The components of the matrices K ν , kν, and gν are modified from those of Problem 290 to be

Chapter 12


381

  m′(κ ) h ′h ′ − λ  cos θ −γ  1−ξ  sin θ  h h  dx ℓ

Kνij =

ν o

i

ν

j

ν

ν

i

j

0

 − sin θ +γ  1−ξ  cos θ  h dx ℓ

kνi

=

ν

ν

i

0

  m(κ ) h ′ − λ  sin θ −γ  1−ξ  cos θ  h  dx ℓ

gνi

=

ν o

j

ν

ν

ν

i

0

The modifications are shown in the following code segment. The remaining parts of the code are unchanged from the previous problem. In the main program the value of m o must be input (it is called rmo) and passed to the subroutine FCN. c.... Read problem parameters read(5,1000) tol,alpha,xlength,gamma,rmo [ code between is unchanged ] call fcn(A,b,x,z,xlength,gamma,rmo,factor,h,dh,nn,ndm)

The main changes are in the subroutine FCN as shown below. C C C

subroutine fcn(A,b,x,z,xlength,gamma,rmo,factor,h,dh,nn,ndm) *------------------------------------------------------------------* | Compute Simpson contribution to Hessian and residual matrices | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension A(nn,nn),b(nn),x(nn),h(ndm),dh(ndm) data zero,one/0.d0,1.d0/

c.... Basis functions and their derivatives ev = 2.d0*atan(1.d0) do i=1,ndm cev = (2*i-1)*ev h(i) = dsin(cev*z) dh(i) = cev*dcos(cev*z)/xlength end do c.... Transverse loading function q = (one - z)*gamma c.... Compute rotation and first derivative at current point theta = zero dtheta = zero do i=1,ndm theta = theta + x(i)* h(i) dtheta = dtheta + x(i)*dh(i) end do c1 = dcos(theta)- dsin(theta)*q c2 = dsin(theta)+ dcos(theta)*q f = dsqrt(one + rmo*dtheta**2) fmu = dtheta/f fdmu = 1.0/f**3 c.... Compute integral part of Hessian matrix and residual vector do i=1,ndm do j=1,ndm A(i,j) = A(i,j)+(fdmu*dh(i)*dh(j)-x(nn)*c1*h(i)*h(j))*factor end do A(i,nn) = A(i,nn) - c2*h(i)*factor b(i) = b(i) + (fmu*dh(i) - x(nn)*c2*h(i))*factor end do return end

The problem was also run with γ = 0 to compute the equilibrium path emanating from the critical load λ1 = 2.467 with the direction input as {1, 0, 2.467}. Both results are plotted in the following figure.

382


The program was run for the problem with γ = 0.1 and m o = 5 with two base functions (eigenfunctions) starting at the state (0,0,0) and moving in the direction (0,0,1). The arc length was set at α = 0.05. Problem 293. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .0500 Number of load steps : 10 Maximum number of iterations : 50 Number of integration intervals : 10 Column Length : 1.0000 Transverse load ratio to P : .1000 Ratio (EIo/Mo)^2 : 5.0000 Initial - Next values : Initial a1 .0000 a2 .0000 Load .0000

Next .0000 .0000 1.0000

The problem was run for 100 steps (10 of which are given below). The results are plotted in the figure below. (Note: the program writes the output in a slightly different form). n 0 1 2 3 4 5 6 7 8 9 10

a1 .00000E+00 .95678E-03 .19540E-02 .29943E-02 .40807E-02 .52162E-02 .64044E-02 .76492E-02 .89548E-02 .10326E-01 .11768E-01

a2 .00000E+00 .11608E-03 .23269E-03 .34985E-03 .46758E-03 .58589E-03 .70481E-03 .82436E-03 .94458E-03 .10655E-02 .11872E-02

P/EIo .00000E+00 .49991E-01 .99981E-01 .14997E+00 .19996E+00 .24994E+00 .29993E+00 .34991E+00 .39990E+00 .44988E+00 .49986E+00

EV1

EV2

.12087E+01 .11836E+01 .11585E+01 .11334E+01 .11081E+01 .10829E+01 .10575E+01 .10320E+01 .10065E+01 .98086E+00

.11078E+02 .11053E+02 .11027E+02 .11001E+02 .10975E+02 .10948E+02 .10921E+02 .10894E+02 .10866E+02 .10837E+02

2

λ

1

0

γ = 0.1 mo = 5.0 α = 0.05

1

2

a1

One can observe that the nonlinear moment-curvature relationship causes the column to exhibit a limit point under the applied loading.

294. Modify the program ELASTICA to use piecewise linear finite element base functions. Examine the performance of the system as the number of base functions is increased. Compare the performance of the finite element base functions with the polynomial functions. The modification to the program involves only the definition of the base functions in the subroutine FCN. We have implemented the base functions given in the text, specialized for equidistant spacing of the nodes. To wit,

Chapter 12

hn(x) =

N x−xn−1  ℓ

xn−1 ≤ x ≤ xn

N x −x  ℓ n+1

xn ≤ x ≤ xn+1

hn′(x) =

otherwise

0

383


−

N ℓ

xn−1 ≤ x ≤ xn

N ℓ

xn ≤ x ≤ xn+1

otherwise

0

where xn is the coordinate of node n. The change requires only the modification of the subroutine FCN, which computes the basis functions. c.... Base functions (finite element) and their derivatives do i=1,ndm za = dfloat(i-1)/dfloat(ndm) zb = dfloat(i)/dfloat(ndm) zc = dfloat(i+1)/dfloat(ndm) if ((z.ge.za).and.(z.lt.zb)) then h(i) = (z-za)/(zb-za) dh(i) = dfloat(ndm)/xlength elseif ((z.ge.zb).and.(z.le.zc)) then h(i) = (zc-z)/(zc-zb) dh(i) = -dfloat(ndm)/xlength else h(i) = 0.0 dh(i) = 0.0 end if end do

The program was run for the problem with γ = 0.1 with from two to nine base functions starting at the state a = 0 and λ = 0 moving in the direction a = 0 and λ = 1. The arc length was set at α = 0.5 and the problem was run for 10 steps. The number of integration points was set to 200 to be certain that the integration was accurate for the finest discretization. (See Problem 300 for more discussion on this point.) The values of the input parameters are given below. Problem 294. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 10 Maximum number of iterations : 30 Number of integration intervals : 200 Column Length : 1.0000 Transverse load ratio to P : .1000

The results are summarized below. The values of the displacement coefficients and load at the end of step 10 are presented for each approximation. Since the solution is controlled by the arc-length, the place on the path that the solutions falls is different for each case. One can assess the accuracy by interpolating the results given in Problem 290. 2 Terms a1 1.39358 a2 1.88422 a3 a4 a5 a6 a7 a8 a9 Load 4.00767 Cond. No. 6.3

3 Terms .95416 1.58194 1.78957

3.79835 16.6

Displacement Coefficients 4 Terms 5 Terms 6 Terms 7 Terms .70394 .54302 .43433 .35197 1.25495 1.00551 .81769 .68215 1.59065 1.34622 1.13719 .96381 1.70350 1.55441 1.36803 1.19218 1.62413 1.50611 1.35576 1.55289 1.45565 1.48843

3.61925 33.0

3.47365 56.9

3.35279 89.2

3.25393 131.3

8 Terms .30091 .58129 .83174 1.04387 1.21005 1.33166 1.40501 1.42966 3.16795 186.1

9 Terms .25640 .50175 .71951 .91666 1.07958 1.21023 1.30163 1.35838 1.37718 3.09819 251.8

384


The condition number (ratio of maximum to minimum eigenvalues of the tangent stiffness matrix) at step 10 is presented for each approximation. One can observe that the finite element equations are very well conditioned. For the 4-term approximation we can compare with the polynomial basis (Problem 292) and the eigenbasis (Problem 291). The polynomial basis had a condition number of 6855, the eigenbasis had 38, and the finite element basis had 33.

295. Describe a method for using the program NONLINEARBEAM to locate the bifurcation points of a system without imperfections.

Technically, the constraint c(x) ≡ det K(x) = 0, where K(x) is given in the text and x ≡ ( λ, a ), would be a suitable constraint for homing in on a critical point. However, observing that there are three fields in the approximation, the dimension of the matrix K(x) gets large as the order of the approximation (i.e., the number of base functions) increases. Computation of the determinant is not very practical for large systems of equations. Linearization of the constraint is also difficult. We can observe that an eigenvalue of the tangent stiffness matrix goes to zero at a critical point let the eigenvalue problem be stated as K(x) φ = m φ Because the tangent stiffness matrix is a function of the state x so are the eigenvalue m and the eigenvector φ (implicitly). The constraint equation can be stated as c(x) = m(x) = 0 The eigenvalue problem at any given state can be solved by any of a variety of algorithms. Thus, let us assume that m(xo) and φ(xo) can be computed at the state xo. The constraint equation can be linearized to be c^(x) = ∇m(x o) ∆x + m(xo) = 0 where [∇m] i = ∂m∕∂xi. The gradient of the eigenvalue with respect to x can be computed from the original eigenvalue problem. Taking the derivative we find  ∇K−∇m I  φ + K−m I  ∇φ = 0

Multiplying the above equation by φT annihilates the second term, and hence obviates the need to compute the rate of change of the eigenvector. Rearranging terms we arrive at the result ∇m ∆x = φT  ∇K ∆x  φ if φ is normalized such that φT φ = 1. The above expression can be computed in components to be φi Kij, k φ j ∆ xk. This equation gives us what we need to form the bordered

Chapter 12


385

Hessian matrix. The main issue in this computation is how to compute ∂K∕∂ai and ∂K∕∂λ. It is possible, albeit tedious, to compute these terms analytically. The main difficulty is that K depends implicitly on the load parameter through the internal moment, shear, and axial force. One can also compute these derivatives numerically by augmenting x and taking the difference, essentially replacing ∇K with ∂K ≈ K(x+δx i ei ) − K(x ) ∂xi δxi where ei is a vector with a one in the ith slot and zeros everywhere else. The value of δxi must be chosen small enough to give a good approximation but large enough to avoid an ill-conditioned operation.

296. Explore the features of the program NONLINEARBEAM by using it to solve the cantilever beam problem under a variety of loading scenarios.

This problem is open-ended and, as such, has no identifiable “solution” to present. Some suggestions on loading scenarios include investigation of the effects of proportional versus non-proportional transverse loading on the buckling behavior of the beam and the effects of different transverse load distributions. One can make comparisons with some of the classical results that are available in the literature. 297. Modify the program NONLINEARBEAM to account for initial geometric imperfections in the column. Is it sufficient to specify imperfections only in the field w(x)?

It is not sufficient to consider only an imperfection in the transverse displacement w(x). One also needs to consider an imperfection in the rotation field. It is not really necessary to consider imperfections in the axial displacement u, but we can. Let the imperfection fields be called uo(x), wo(x), and θ o(x). The strains associated with the initial imperfections are κ oi = θ o′ β oi = wo′ cos θ o −  1+uo′  sin θ o Á oi = wo′ sin θ o +  1+uo′  cos θ o − 1 The initial strains are accounted for in the constitutive equations M = EI  κ o − κ oi  Q = GA  β o − β oi  N = EA  Á o − Á oi 

386


Since the initial strains are not a function of the subsequent displacements, which are measured from a nominal straight configuration, there are no further complications from initial imperfections. We must be specific about the variation of the initial imperfection fields. We can, for example consider a shear-free initial imperfection by taking uo(ξ) = 0



wo(ξ) = co 1− cos

πξ 2



πξ θ o(ξ) = π c o sin 2 2ℓ

where ξ ≡ x∕ℓ, so that wo′−θ o = 0. If the constant co is small there should be only an effect on the moment equation. The changes to the program are minor. First, the value of co must be input. In the main program we can read in this value as cimperf c.... Read problem parameters read(5,1000) tol,alpha,xlength [ code between is unchanged ] read(5,1000) (dpo(i),i=1,3) read(5,1000) cimperf

The call to subroutine FCN must pass the value of co . The functions given above have been implemented in the subroutine FCN as follows C C C

subroutine fcn(A,b,D,x,z,xlength,cimperf,factor,ndm,nbasis) *------------------------------------------------------------------* | Compute contribution to K and g at current integration point | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension A(ndm,ndm),b(ndm),x(ndm),D(3) dimension G(4,4),GB(4,3),BGB(3,3) data zero/0.d0/,one/1.d0/

c.... Compute displacements, derivatives, and current load factor clf du = zero dw = zero dtheta = zero theta = zero do i=1,nbasis call basis(i,z,xlength,h,dh,nbasis) du = du + x(3*i-2)*dh dw = dw + x(3*i-1)*dh dtheta = dtheta + x(3*i)*dh theta = theta + x(3*i)*h end do ct = dcos(theta) st = dsin(theta) c.... Compute epsilon beta curv

axial strain, shear strain, and curvature = dw*st + (one + du)*ct - one = dw*ct - (one + du)*st = dtheta

c.... Compute initial axial strain, shear strain, and curvature twopi = 2.0*atan(1.0) dw0 = (cimperf*twopi/xlength)*dsin(twopi*z) du0 = 0.0 th0 = dw0 ct0 = dcos(th0) st0 = dsin(th0) eps0 = dw0*st0 + (one + du0)*ct0 - one beta0 = dw0*ct0 - (one + du0)*st0 curv0 = (cimperf*(twopi/xlength)**2)*dcos(twopi*z) c.... Compute bend = shear = axial =

axial force, shear force, bending moment, and other forces D(1)*(curv-curv0) D(2)*(beta-beta0) D(3)*(epsilon-eps0)

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[ Remaining part of subroutine is unchanged from text ]

The program was run for a cantilever beam subjected to a compressive end load with an imperfection of co = 0.1. The input values are given below Problem 297. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 40 Maximum number of iterations : 40 Number of integration intervals : 10 Number of points in plotted shape : 20 Output shape every N points, N = : 10 Number of basis functions : 6 Beam properties, end load and dist. load amplitudes Column Length : 10.00 EI, GA, EA : 1000. .1000E+07 .1000E+07 Mo, Qo, Po : .0000E+00 .0000E+00 -2.460 mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00 mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00 mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00 Imperfection : .1000

The problem was run for 40 steps. The output is given below for selected steps (the output of the program has been edited) Step 1 2 3 4 5 10 15 20 25 30 35 40

Load .48855E+00 .98851E+00 .14885E+01 .19884E+01 .24883E+01 .49874E+01 .74787E+01 .95113E+01 .10322E+02 .11317E+02 .12762E+02 .14639E+02

u(L) -.76802E-04 -.16657E-03 -.27032E-03 -.39166E-03 -.53539E-03 -.19457E-02 -.90667E-02 -.16797E+00 -.10142E+01 -.24660E+01 -.42030E+01 -.58404E+01

w(L) .10512E+00 .11093E+00 .11743E+00 .12472E+00 .13299E+00 .19887E+00 .39230E+00 .16425E+01 .39219E+01 .58131E+01 .70895E+01 .77553E+01

theta(L) .16512E-01 .17425E-01 .18445E-01 .19592E-01 .20890E-01 .31240E-01 .61647E-01 .25994E+00 .64369E+00 .10184E+01 .13524E+01 .16192E+01

NU 8 6 4 4 4 5 6 9 5 6 6 6

|| b || .2498E-08 .8651E-08 .2784E-09 .4601E-09 .1660E-09 .1699E-09 .2205E-09 .6065E-09 .2608E-08 .1000E-08 .2343E-08 .3906E-08

The results are plotted below for two values of the imperfection. One can observe that the equilibrium path does not start at the origin because of the imperfection. 15

λ

15

co = 0.1

co = 0.1

co = 0.3

co = 0.3 0

w(ℓ)

10

0

θ(ℓ)

10

388


298. Modify the subroutine basis in the program NONLINEARBEAM to use piecewise linear finite-element base functions. Examine the performance of the system as the number of base functions is increased. Compare the performance of the finite element base functions with the sinusoids used in the original program.

The finite element base functions are simple to implement and are identical to those implemented in Problem 294. The program NONLINEARBEAM was already structured to make the change to a new basis simple. The new subroutine BASIS is given as C C C

subroutine basis(i,z,xlength,h,dh,nbasis) *------------------------------------------------------------------* | Evaluate ith basis function h and derivative dh at point z | *------------------------------------------------------------------* implicit double precision (a-h,o-z) za zb zc if

= dfloat(i-1)/dfloat(nbasis) = dfloat(i) /dfloat(nbasis) = dfloat(i+1)/dfloat(nbasis) ((z.ge.za).and.(z.le.zb)) then h = (z-za)/(zb-za) dh = dfloat(nbasis)/xlength elseif ((z.gt.zb).and.(z.le.zc)) then h = (zc-z)/(zc-zb) dh = -dfloat(nbasis)/xlength else h = 0.0 dh = 0.0 end if return end

One can observe that the solution with the linear finite element base functions shows a tendency to shear lock, that is, displacements are constrained to very small values because the contribution to the tangent stiffness from the shear term does not have the appropriate rank. This phenomenon is well-known and can be cured by “reduced integration” of the system matrices. One point integration, with the sampling point at the element center, is known to cure the shear-locking problem. We can trick the program into doing one-point integration by making a modification to the SIMPSON subroutine. We have adjusted the weights in the Simpson rule to give all of the weight to the midpoint of a segment and none to the ends. If the number of integration points is equal to the number of base functions then we achieve the appropriate reduced integration scheme. C C C

subroutine simpson(m,npoints,dz,xlength,factor) *------------------------------------------------------------------* | Evaluate the weight factor for the current integration point m | *------------------------------------------------------------------* implicit double precision (a-h,o-z) c = xlength*dz/3.d0 n = mod(m,2) if((m.eq.1).or.(m.eq.npoints)) then factor = 0.0 else if (n.eq.0) then factor = 6.0*c else factor = 0.0 endif return end

Chapter 12


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The program was run on the example presented in the text (cantilever beam subjected to a uniform transverse load and a compressive axial load). The properties of the system are given below. Problem 298. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 10 Maximum number of iterations : 10 Number of integration intervals : 8 Number of points in plotted shape : 20 Output shape every N points, N = : 10 Number of basis functions : 8 Beam properties, end load and dist. load amplitudes Column Length : 10.00 EI, GA, EA : 1000. .1000E+07 .1000E+07 Mo, Qo, Po : .0000E+00 .0000E+00 -2.460 mo1, qo1, po1 : .0000E+00 .2000 .0000E+00 mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00 mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The results for the first 10 steps are present below. One can compare the result at step 10 with the example in the text from the output. Note that the transverse tip displacement with 6 eigenfunctions was w(ℓ) = 0.478 compared to w(ℓ) = 0.463 for 8 finite element functions. Step 1 2 3 4 5 6 7 8 9 10

Load .27270E+00 .77227E+00 .12717E+01 .17711E+01 .22702E+01 .27691E+01 .32677E+01 .37658E+01 .42632E+01 .47595E+01

u(L) -.37735E-02 -.41948E-02 -.46873E-02 -.52683E-02 -.59610E-02 -.67964E-02 -.78173E-02 -.90833E-02 -.10680E-01 -.12731E-01

w(L) .25655E+00 .26975E+00 .28444E+00 .30088E+00 .31940E+00 .34041E+00 .36445E+00 .39222E+00 .42464E+00 .46296E+00

theta(L) .34614E-01 .36665E-01 .38950E-01 .41510E-01 .44398E-01 .47679E-01 .51438E-01 .55786E-01 .60868E-01 .66882E-01

NU 7 7 4 4 4 5 5 5 5 5

|| b || .5542E-09 .7040E-09 .4582E-09 .1080E-08 .4007E-08 .4267E-09 .3071E-09 .3231E-09 .4285E-09 .4386E-09

50

λ

0

w(ℓ)

10

390


299. Modify the basis subroutine in the program NONLINEARBEAM to use the polynomials hi (x) ∈  x, x2,. . ., x N . Examine the performance of the system as the number of base functions is increased using the pure bending problem. Are these functions able to capture the exact solution, which is a circular shape, as shown in the text example? Why is convergence so difficult with a large number of base functions? Implement the orthogonal base functions described in Chapter 6. Do these base functions work better than the original polynomials?

The modification of the subroutine BASIS is straightforward. For the polynomials hi given above the routine is C C C

subroutine basis(i,z,xlength,h,dh,nbasis) *------------------------------------------------------------------* | Evaluate ith basis function h and derivative dh at point z | *------------------------------------------------------------------* implicit double precision (a-h,o-z) h = z**i dh = i*z**(i-1)/xlength return end

The orthogonal polynomials can also be easily implemented, but are a bit more cumbersome. The expressions for the first four are on page 212 of the text. The fifth and sixth orthogonal polynomials are given by g5(ξ) = 210 ξ5 − 504 ξ4 + 420 ξ 3 − 140 ξ2 + 15 ξ g6(ξ) = 792 ξ6 − 2310 ξ5 + 2520 ξ 4 − 1260 ξ3 + 280 ξ 2 − 21 ξ The orthogonal polynomials up to order six are implemented in the routine as follows: C C C

subroutine basis(i,z,xlength,h,dh,nbasis) *------------------------------------------------------------------* | Evaluate ith basis function h and derivative dh at point z | *------------------------------------------------------------------* implicit double precision (a-h,o-z) if (nbasis.gt.6) Stop ’Not enough basis functions in BASIS’ go to (1,2,3,4,5,6),i

c.... Basis function h1(z) = z 1 h =z dh=1./xlength return c.... Basis function h2(z) = (4z-3)z 2 h =(4.*z-3.)*z dh=(8.*z-3.)/xlength return c.... Basis function h3(z) = ((15z-20)z+6)z 3 h =((15.*z-20.)*z+6.)*z dh=((45.*z-40.)*z+6.)/xlength return c.... Basis function h4(z) = (((56z-105)z+60)z-10)z 4 h =((( 56.*z-105.)*z+ 60.)*z-10.)*z dh=(((224.*z-315.)*z+120.)*z-10.)/xlength return c.... Basis function h5(z) = ((((210z-504)z+420)z-140)z+15)z 5 h =(((( 210.*z- 504.)*z+ 420.)*z-140.)*z+15.)*z dh=((((1050.*z-2016.)*z+1260.)*z-280.)*z+15.)/xlength return

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c.... Basis function h6(z) = (((((792z-2310)z+2520)z-1260)z+280)z-21)z 6 h =((((( 792.*z- 2310.)*z+ 2520.)*z-1260.)*z+280.)*z-21.)*z dh=(((((4752.*z-11550.)*z+10080.)*z-3780.)*z+560.)*z-21.) dh=dh/xlength return end

The program was run on the example problem in the text wherein a beam is subjected to an end moment of magnitude sufficient to curl it into a circle with circumference equal to its original length of 10. The input for the problem is Problem 299. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : 10.0000 Number of load steps : 10 Maximum number of iterations : 40 Number of integration intervals : 10 Number of points in plotted shape : 20 Output shape every N points, N = : 10 Number of basis functions : 6 Beam properties, end load and dist. load amplitudes Column Length : 10.00 EI, GA, EA : 1000. .1000E+07 .1000E+07 Mo, Qo, Po : 6.280 .0000E+00 .0000E+00 mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00 mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00 mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The output for the case of orthogonal polynomials with 6 base functions is given below as an example. Step 0 1 2 3 4 5 6 7 8 9 10

Load 0 .96915E+01 .19390E+02 .29101E+02 .38830E+02 .48583E+02 .58364E+02 .68174E+02 .78013E+02 .87884E+02 .97790E+02

u(L) 0 -.60604E+00 -.22944E+01 -.47075E+01 -.73477E+01 -.96970E+01 -.11360E+02 -.12192E+02 -.12265E+02 -.11817E+02 -.11171E+02

w(L) 0 .29503E+01 .53723E+01 .68613E+01 .72304E+01 .65572E+01 .51758E+01 .35257E+01 .19918E+01 .85711E+00 .22152E+00

theta(L) 0 .60863E+00 .12177E+01 .18275E+01 .24384E+01 .30492E+01 .36555E+01 .42540E+01 .48407E+01 .54015E+01 .59092E+01

NU 0 7 5 5 5 6 7 8 8 8 8

|| b || 0 .3324E-09 .1436E-08 .9216E-09 .5750E-08 .2784E-08 .1646E-08 .5047E-08 .2023E-08 .1992E-08 .3493E-08

The problem was run with 3, 4, 5, and 6 base functions for both the original polynomials h and the orthogonal polynomials g. Below we plot the final deflected shape of the beam for each of these approximations along with the exact shape and the shape obtained with the 6-eigenfunction basis (presented in the text). The approximation using the 6 eigenfunctions is labeled e6. The exact solution is shown as a dotted line. 3

4

3

4

5 5 e6

e6 6 polynomials

orthogonal polynomials

392


We also present a plot of the load factor versus the end moment for each approximation. The exact solution is the diagonal line. 100

100 3

4 5

3

e6

λ

5

4

6 e6

λ orthogonal polynomials

polynomials

0

θ(ℓ)

2π

0

θ(ℓ)

2π

One can clearly see the difference between the original and orthogonal polynomials. For the 3-term approximation the results are essentially identical. As the order increases the original polynomials give out while the orthogonal polynomials do very well. The solution with 6 original polynomial base functions failed to converge on the 6th load step out of 10. Ill-conditioning is the culprit for the failure to converge. The solution for the orthogonal polynomials is still robust at 6 terms. It is interesting to observe the reluctance to rotate beyond a certain point in the low-order polynomials (both original and orthogonal).

xo ν 300. The arc-length constraint forces the next equix g(x) = 0 librium configuration to be a fixed distance from the previous converged state. Therefore, all iterates do must lie on a sphere of radius α centered on the converged state xn. One of the problems with this stratexn x n+1 gy is that the equilibrium path pierces the sphere at α two points (at least). For a highly nonlinear equilibrium path the Newton iterations can converge to the other point on the sphere, which causes the loading x direction to change. We can observe this phenomenon in the program NONLINEARBEAM if the step size is not judiciously chosen. Once the loading direction has turned around, it is not likely to change back. Consider another possible constraint pictured on the following page. The difference between the estimate xν and xo is forced to lie in a hyperplane normal to the tangent direction do ≡ xo − x n. The normal plane is set a distance α from xn. Therefore, ‖ do ‖= α. How can one compute the tangent direction do ? Show that the normality condition is dν ⋅ d o = α 2, where the νth increment in state is dν ≡ xν − x n. Develop a method based upon a secant direction where do ≡ xn − x n−1 (the previous two converged states). How would you start a method based on this definition? Implement

Chapter 12


393

these constraints in the program NONLINEARBEAM and assess their performance. Are there advantages over the arc-length constraint? Are there disadvantages?

(a) How can one compute the tangent direction do ? Observe that x = ( λ, a ), where a is the vector of displacement coefficients and λ is the load parameter. The tangent direction can be obtained by examining the linearized equilibrium equations g^ (λ, a) = 0. We have g^ (λ, a) = K(a) ∆a + k(λ) ∆ λ + g(λ, a) = 0 At the state xn = ( λn, an ) equilibrium is satisfied. Therefore g(λn, an) = 0. To find the tangent direction we need only assume a unit increment in the load parameter and compute the corresponding change in displacement. To wit, let do ≡ ( 1, ∆a o ). The increment in displacement can then be computed by solving the system of equations K(an) ∆ao = −k(λ n) The vector do can subsequently be normalized to have length ‖ do ‖= α. Note that if the current state is on a downhill branch of the equilibrium path, as evidenced by negative eigenvalues of the tangent stiffness K(an), then the choice do ≡ ( −1, ∆a o ) is appropriate. Clearly, the computation will fail if K(an) is singular, as it is at a limit point or bifurcation point. (b) Show that the normality condition is dν ⋅ d o = α 2, where the νth increment in state is dν ≡ xν−x n. By definition of the method the increment xν−x o is orthogonal to the tangent line do . Thus, 0 = d o ⋅  xν−x o  0 = d o ⋅  xν−x n + xn−x o  0 = d o ⋅  xν−x n  − do ⋅  xo−x n  0 = d o ⋅  xν−x n  − do ⋅ d o Since do ⋅ d o = α 2 we get the result do ⋅  xν−x n  = α 2 . Letting dν ≡ xν−x n then gives the desired result. (c) Develop a method based upon a secant direction where do ≡ xn−x n−1 (the previous two converged states). How would you start a method based on this definition? The method is identical to the previous one except that it avoids the computation of the tangent direction from the condition given in part (b). Since there is no state x−1 the method cannot be started with a secant. Because most problems start out linear, one might start the secant method with the tangent direction do ≡ ( 1, ∆a o ) with ∆ao computed from the equation K(0) ∆a o = −k(0) The secant method can commence on the second step. As an alternative, a guess at the next direction can be an input to the algorithm (as it is for most of the programs in the

394


chapter). This approach is necessary if starting from a bifurcation point since the tangent computation fails there. (d) Implement these constraints in the program NONLINEARBEAM and assess their performance. Are there advantages over the arc-length constraint? Are there disadvantages? The tangent line algorithm appears to have little advantage over the regular arc-length constraint.

301. When using finite element base functions in the programs ELASTICA and NONLINEARBEAM, most of the integration points contribute nothing to the integrals because the base functions are zero over much of the region. Restructure the order of the programs to make them more efficient by putting the loop over integration points inside the loop over base functions.

The modification to the program in Problem 294 involves a change in the way the computations are organized. In the main program the dimension of the arrays for the base functions is changed to 2 because only two base functions will be non-zero at any one time. dimension A(nn,nn),b(nn),x(nn),xo(nn),h(2),dh(2)

Also in the main program, the integration of the Hessian and residual components is changed to have a loop over “elements” with an inner loop over integration points within that element. c.... Execute Simpson integration of Hessian and residual components call zerovec(b,nn) call zerovec(A,nn*nn) do nel=1,ndm z = dfloat(nel-1)/dfloat(ndm) do m=1,npoints call basis(z,xlength,h,dh,ndm,nel) call simpson(m,npoints,dz,xlength,factor) call fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm,nel) z = z + dz end do end do

The subroutine BASIS is modified to return the values of the base functions and derivatives for only the ith and (i--1)th base functions. All of the remain functions are known to be zero at this point. C C C C

subroutine basis(z,xlength,h,dh,ndm,i) *------------------------------------------------------------------* | Compute the ith and i-1th basis functions at point z | | h(1) is i-1th (ramping down), h(2) is ith (ramping up) | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension h(2),dh(2)

c.... Compute basis function (finite element) and its derivative za = dfloat(i-1)/dfloat(ndm) zb = dfloat(i)/dfloat(ndm) h(1) = (zb-z)/(zb-za)

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h(2) = (z-za)/(zb-za) dh(1) = -dfloat(ndm)/xlength dh(2) = dfloat(ndm)/xlength return end

The subroutine FCN is modified to recognize that only two of the base functions are non-zero. Hence the computation of θ and θ′ only involve two of the coefficients. Furthermore, the contributions to A and b come only from the two base functions. C C C

subroutine fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm,nel) *------------------------------------------------------------------* | Compute Simpson contribution to Hessian and residual matrices | *------------------------------------------------------------------* implicit double precision (a-h,o-z) dimension A(nn,nn),b(nn),x(nn),h(2),dh(2) data zero,one/0.d0,1.d0/

c.... Transverse loading function q = (one - z)*gamma c.... Compute rotation and first derivative at current point if(nel.eq.1) then x0 = 0.0 else x0 = x(nel-1) endif x1 = x(nel) theta = x0*h(1) + x1*h(2) dtheta = x0*dh(1) + x1*dh(2) c1 = dcos(theta)- dsin(theta)*q c2 = dsin(theta)+ dcos(theta)*q c.... Compute integral part of Hessian matrix and residual vector do i=1,2 ii = nel+i-2 do j=1,2 jj = nel+j-2 if((ii.gt.0).and.(jj.gt.0)) * A(ii,jj)= A(ii,jj)+(dh(i)*dh(j)-x(nn)*c1*h(i)*h(j))*factor end do if(ii.gt.0) then A(ii,nn) = A(ii,nn) - c2*h(i)*factor b(ii) = b(ii) + (dtheta*dh(i) - x(nn)*c2*h(i))*factor end if end do return end

The program was run for the problem with γ = 0.1 with from two to nine base functions starting at the state a = 0 and λ = 0 moving in the direction a = 0 and λ = 1, exactly as was done in Problem 294. The arc length was set at α = 0.5 and the problem was run for 10 steps. The number of integration points were set to 2 (per element). The values of the input parameters are given below. Problem 183. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 10 Maximum number of iterations : 10 Number of integration intervals : 2 Column Length : 1.0000 Transverse load ratio to P : .1000

The results are summarized below. The values of the displacement coefficients and load at the end of step 10 are presented for each approximation. Since the solution is controlled by the arc-length the place on the path that the solutions falls is different for

396


each case. Again, one can assess the accuracy by interpolating the results in Problem 290. a1 a2 a3 a4 a5 a6 a7 a8 a9 Load

2 Terms 1.39142 1.88316

3 Terms .95682 1.58142 1.78954

4.00905

3.79776

Displacement Coefficients 4 Terms 5 Terms 6 Terms 7 Terms .70144 .54050 .43216 .35538 1.25254 1.00296 .81944 .68286 1.59063 1.34658 1.13578 .96545 1.70316 1.55483 1.36665 1.19188 1.62427 1.50615 1.35597 1.55271 1.45501 1.48807

3.62091

3.47457

3.35367

3.25335

8 Terms .29878 .57910 .82956 1.04176 1.21024 1.33192 1.40530 1.42980 3.16951

9 Terms .25569 .49855 .72071 .91591 1.07966 1.20903 1.30227 1.35847 1.37723 3.09879

The condition number (ratio of maximum to minimum eigenvalues of the tangent stiffness matrix) at step 10 was the same as those reported in Problem 294, as expected. The small differences in the values reported here and those reported in Problem 294 are due to the accuracy of numerical integration. Ironically, the numerical integration is more accurate in the present implementation than in the former implementation. The finite element base functions are not smooth. When we do Simpson quadrature, which approximates the curve as a quadratic, over the entire segment (Problem 294) we suffer under the inaccuracy of the method trying to integrate over the peaks of the base functions and the jumps in their derivatives (and the ambiguity if a sampling point is right at the jump). The numerical integration can be made arbitrarily accurate by taking more segments, but that number is unacceptably large. In the present implementation we integrate within each element and sum over all elements. In doing so we explicitly account for the peaks and jumps in the functions. A two-segment (5 sampling points) Simpson quadrature is very accurate in this case. Also, with this organization, the number of integration points increases as the order of approximation increases (i.e., two segments per element times nine elements is 18 segments, far below the 200 used in Problem 294).

302. Add a subroutine to find the eigenvalues and eigenvectors of K(a) at each point on the equilibrium path in the program NONLINEARBEAM to examine the stability of equilibrium. Implement a procedure for branch switching so that the program will trace the bifurcation diagram when there are no imperfections.

Finding the eigenvalues and eigenvectors of the tangent stiffness matrix is simply a matter of a call to the subroutine EIGENS that we have used in the other programs. The implementation involves providing storage space, as indicated in these commands in the main program, parameter (nn=ndm-1) dimension AA(nn,nn),U(nn,nn),e(nn)

and preparing for the eigenvalue problem, and calling the subroutine, again in the main program c.... Compute eigenvalues of Tangent stiffness matrix do j=1,nn

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do k=1,nn AA(j,k) = A(j,k) end do e(j) = 1.d0 end do call eigens(AA,U,e,nn,8,NR) enext = e(1)

To get the solution to switch from the trivial branch to a buckled configuration we simply watch for a change in sign of the first eigenvalue. When detected we provide a guess at the next state in the direction of the first eigenvector, scaled to have length α, as the following code segment indicates (again in the main program). c.... Update values for previous converged state and guess at next state if ((elast.gt.zero).and.(enext.lt.zero)) then sum = zero do j = 1,nn sum = sum + U(j,1)*U(j,1) end do sum = alpha/dsqrt(sum) do j = 1,nn x(j) = x(j) + sum*U(j,1) end do else do j=1,ndm temp = xo(j) xo(j) = x(j) x(j) = two*x(j) - temp end do end if elast = enext

The problem was run for the same beam as used in the example in the text. The input properties are given below Problem 302. Fundamentals of Structural Mechanics Convergence tolerance : .1000E-07 Arc length parameter : .5000 Number of load steps : 50 Maximum number of iterations : 50 Number of integration intervals : 10 Number of points in plotted shape : 20 Output shape every N points, N = : 1 Number of basis functions : 2 Beam properties, end load and dist. load amplitudes Column Length : 10.00 EI, GA, EA : 1000. .1000E+07 .1000E+07 Mo, Qo, Po : .0000E+00 .0000E+00 -2.470 mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00 mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00 mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The computation was run for 50 steps, some of which are included below. The results are shown in the accompanying figure. Step 1 2 3 19 20 21 22 23 24 25 26 27

Load .50000E+00 .10000E+01 .15000E+01 [ .95000E+01 .10000E+02 .10500E+02 .10488E+02 .10985E+02 .11483E+02 .11982E+02 .12481E+02 .12981E+02

u(L) w(L) theta(L) -.12233E-04 .00000E+00 .00000E+00 -.24466E-04 .00000E+00 .00000E+00 -.36699E-04 .00000E+00 .00000E+00 steps omitted for brevity ] -.23243E-03 .00000E+00 .00000E+00 -.24466E-03 .00000E+00 .00000E+00 -.25690E-03 .00000E+00 .00000E+00 -.96550E-03 .10771E+00 .16919E-01 -.18518E-02 .16095E+00 .25284E-01 -.27403E-02 .20061E+00 .31514E-01 -.36297E-02 .23366E+00 .36707E-01 -.45194E-02 .26258E+00 .41252E-01 -.54094E-02 .28863E+00 .45346E-01

NU 2 1 1

|| b || .7486E-10 .1692E-10 .2108E-10

1 1 1 15 8 6 6 5 5

.3641E-10 .2708E-11 .1348E-10 .6174E-10 .6422E-08 .6705E-09 .6114E-10 .4401E-09 .5623E-10

398


28 29 30

.13480E+02 -.62994E-02 .31251E+00 .49099E-01 .13980E+02 -.71895E-02 .33470E+00 .52586E-01 .14479E+02 -.80797E-02 .35550E+00 .55856E-01 [ steps omitted for brevity ]

Step 1 2 3

Load EV1 EV2 EV3 .50000E+00 .28233E+01 .21087E+05 .44830E+05 .10000E+01 .26760E+01 .21087E+05 .44830E+05 .15000E+01 .25287E+01 .21087E+05 .44830E+05 [ steps omitted for brevity ] .95000E+01 .17229E+00 .21086E+05 .44830E+05 .10000E+02 .25003E-01 .21085E+05 .44830E+05 .10500E+02 -.12228E+00 .21085E+05 .44830E+05 .10488E+02 .23747E+00 .21086E+05 .44831E+05 .10985E+02 .53030E+00 .21086E+05 .44833E+05 .11483E+02 .82388E+00 .21086E+05 .44835E+05 .11982E+02 .11177E+01 .21086E+05 .44837E+05 [ steps omitted for brevity ]

19 20 21 22 23 24 25

5 5 4

.9440E-10 .6043E-10 .7212E-08

One can see that the first occurrence of a negative first eigenvalue is step 21. Thus, the bifurcation path is sought in step 22. The actual bifurcation point is λ = 10, which is approximately reached in step 20. However, in step 20 the first eigenvalue is still positive (albeit very near zero). This worst case scenario gives us about a two step lag in getting on the right path. We show the jump onto the path with a dotted line in the figure. 20

λ

10

0

0.2

w(ℓ)

0.4

Chapter 12


399

Appendix A. Fortran Programs The solutions to the problems in this chapter were done for the first edition of the text, in which the programs were written in Fortran. The original Fortran codes are listed here for reference because the solutions only indicate small code segments that need to be inserted into this code. Perhaps I will get the energy to redo these solutions in MatLab, but for now these will have to do.

The Program NEWTON program NEWTON implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3),xo(3) c.... Read problem parameters read(5,1000) tol,alpha read(5,1001) maxsteps,maxit read(5,1000) xo(1),xo(2),xo(3) read(5,1000) x(1), x(2), x(3) write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3) c.... Initialize values for load step zero,set next trial state n = 0 sum = 0.0 do j = 1,3 b(j) = x(j) - xo(j) sum = sum + b(j)*b(j) end do sum = alpha/dsqrt(sum) do j = 1,3 x(j) = xo(j) + sum*b(j) end do write(6,2001) n,xo(1),xo(2),xo(3) c.... Compute MAXSTEPS points along the Equilibrium Path do 2 n = 1,maxsteps c.... Perform Newton iteration at each load step 1

nu = 0 nu = nu + 1

c.... Compute equilibrium and constraint at current state b(1) = 2.0*x(1) - x(2) - x(3)*dsin(x(1)) b(2) = -x(1) + x(2) - x(3)*dsin(x(2)) b(3) = 0.0 do k = 1,3 b(3) = b(3) + (x(k) - xo(k))**2 end do b(3) = b(3) - alpha**2 c.... Compute Hessian matrix A A(1,1) A(1,2) A(1,3) A(2,1) A(2,2) A(2,3) A(3,1) A(3,2) A(3,3)

= = = = = = = = =

2.0 - x(3)*dcos(x(1)) -1.0 -dsin(x(1)) -1.0 1.0 - x(3)*dcos(x(2)) -dsin(x(2)) 2.0*(x(1) - xo(1)) 2.0*(x(2) - xo(2)) 2.0*(x(3) - xo(3))

c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2)

400

Fundamentals of Structural Mechanics c.... Compute eigenvalues of Tangent stiffness matrix tr det eig1 eig2

= = = =

A(1,1) + A(2,2) A(1,1)*A(2,2) - A(2,1)*A(1,2) 0.5*(tr - dsqrt(tr**2 - 4*det)) 0.5*(tr + dsqrt(tr**2 - 4*det))

c... Update state vector call invert(A,3,3) do j=1,3 do k=1,3 x(j) = x(j) - A(j,k)*b(k) end do end do c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test c.... Set values for previous converged state and guess at next state do j=1,3 temp = xo(j) xo(j) = x(j) x(j) = 2.0*x(j) - temp end do 2 continue stop ’Maximum number of steps exhausted’

The Program ELASTICA program Elastica implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3),xo(3) data zero/0.d0/,one/1.d0/,two/2.d0/ c.... Read problem parameters read(5,1000) tol,alpha,xlength read(5,1001) maxsteps,maxit,inter read(5,1000) xo(1),xo(2),xo(3) read(5,1000) x(1), x(2), x(3) write(6,2000) tol,alpha,maxsteps,maxit,inter,xlength, * (xo(j),j=1,3),(x(j),j=1,3) c.... Initialize constants of integration dz = one/(2*inter) npoints = 2*inter + 1 c.... Initialize values for load step zero,set next trial state n = 0 sum = zero do j = 1,3 b(j) = x(j) - xo(j) sum = sum + b(j)*b(j) end do sum = alpha/dsqrt(sum) do j = 1,3 x(j) = xo(j) + sum*b(j) end do write(6,2001) n,xo(1),xo(2),xo(3) c.... Compute MAXSTEPS points along the Equilibrium Path do 2 n = 1,maxsteps c.... Perform Newton iteration at each load step 1

nu = 0 nu = nu + 1

Chapter 12


c.... Execute Simpson integration of Hessian and residual components call zerovec(b,3) call zerovec(A,9) z = zero do m=1,npoints call simpson(m,npoints,dz,xlength,factor) call fcn(A,b,x,z,xlength,factor) z = z + dz end do c.... Finish computation of Hessian and residual do k=1,3 b(3) = b(3)+(x(k)-xo(k))**2 A(3,k) = two*(x(k)-xo(k)) end do b(3) = b(3) - alpha**2 c.... Compute norm of residual for convergence test test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2) c.... Compute eigenvalues of Tangent stiffness matrix tr det eig1 eig2

= = = =

A(1,1) + A(2,2) A(1,1)*A(2,2) - A(2,1)*A(1,2) 0.5d0*(tr - dsqrt(tr**2 - 4.d0*det)) 0.5d0*(tr + dsqrt(tr**2 - 4.d0*det))

c... Update state vector call invert(A,3,3) do j=1,3 do k=1,3 x(j) = x(j) - A(j,k)*b(k) end do end do c.... Test for convergence, if successful output values and update if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test c.... Set values for previous converged state and guess at next state do j=1,3 temp = xo(j) xo(j) = x(j) x(j) = two*x(j) - temp end do 2 continue stop ’Maximum number of steps exhausted’ c.... IO Formats 1000 format(3f10.0) 1001 format(3i10) 2000 format(’ Equilibrium Path for Cantilever Elastica’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Arc length parameter : ’,f12.4/ * 5x, ’ Number of load steps : ’,i6/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Number of integration intervals : ’,i6/ * 5x, ’ Column Length : ’,f12.4/ * 5x, ’ Initial a1, a2, load : ’,3f12.4/ * 5x, ’ Next a1, a2, load : ’,3f12.4// * ’ n’ 10x,’a1’,10x,’a2’,8x,’P/EI’, * 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’) 2001 format(i5,5f12.5,i5,e12.4) end C---Compute Simpson contribution to Hessian and residual matrices--------FCN subroutine fcn(A,b,x,z,xlength,factor) implicit double precision (a-h,o-z) dimension A(3,3),b(3),x(3) data ev1/1.570796327d0/,ev2/4.712388981d0/

401

402

Fundamentals of Structural Mechanics c.... Base functions and their derivatives h1 h2 dh1 dh2

= = = =

dsin(ev1*z) dsin(ev2*z) ev1*dcos(ev1*z)/xlength ev2*dcos(ev2*z)/xlength

c.... Compute rotation and first derivative at current point theta = x(1)*h1 + x(2)*h2 dtheta = x(1)*dh1 + x(2)*dh2 c1 = x(3)*dcos(theta) c2 = x(3)*dsin(theta) c.... Compute integral part of residual vector b(1) = b(1) - (c2*h1 - dtheta*dh1)*factor b(2) = b(2) - (c2*h2 - dtheta*dh2)*factor c.... Compute integral part of Hessian matrix A(1,1) A(1,2) A(1,3) A(2,1) A(2,2) A(2,3)

= = = = = =

A(1,1) A(1,2) A(1,3) A(2,1) A(2,2) A(2,3)

+ + + + -

(dh1*dh1 - c1*h1*h1)*factor (dh1*dh2 - c1*h1*h2)*factor (dsin(theta)*h1)*factor (dh2*dh1 - c1*h2*h1)*factor (dh2*dh2 - c1*h2*h2)*factor (dsin(theta)*h2)*factor

return end C----Evaluate Simpson integration weight factors---------------------SIMPSON subroutine simpson(m,npoints,dz,xlength,factor) implicit double precision (a-h,o-z) c = xlength*dz/3.d0 n = mod(m,2) if((m.eq.1).or.(m.eq.npoints)) then factor = c else if (n.eq.0) then factor = 4.d0*c else factor = 2.d0*c endif return end C---Initialize the array V to zero-----------------------------------ZEROVEC subroutine zerovec(v,n) implicit double precision (a-h,o-z) dimension v(n) do i=1,n v(i) = 0.d0 end do return end c.... Output formats 1000 format(3f10.0) 1001 format(2i10) 2000 format(’ Equilibrium Path for Chapter 10 Example Problem’// * 5x, ’ Convergence tolerance : ’,e12.4/ * 5x, ’ Arc length parameter : ’,f12.4/ * 5x, ’ Number of load steps : ’,i6/ * 5x, ’ Maximum number of iterations : ’,i6/ * 5x, ’ Initial theta 1, theta 2, load : ’,3f12.4/ * 5x, ’ Next theta 1, theta 2, load : ’,3f12.4// * 10x,’Theta 1’,5x,’Theta 2’,8x,’Pl/k’,

Chapter 12


* 8x,’EV 1’,8x,’EV 2’,’ 2001 format(i5,5f12.5,i5,e12.4) end

NU’,4x,’|| b ||’)

C---- Invert matrix A (nmax,nmax) when array dimension is NDM ---INVERT subroutine invert(a,nmax,ndm) implicit double precision (a-h,o-z) dimension a(ndm,ndm) do 200 n = 1,nmax d = a(n,n) do 100 j = 1,nmax a(n,j) = -a(n,j)/d 100 continue

140

do 150 i = 1,nmax if(n.eq.i) go to 150 do 140 j = 1,nmax if(n.ne.j) a(i,j) = a(i,j) + a(i,n)*a(n,j) continue

150

a(i,n) = a(i,n)/d continue

a(n,n) = 1.0/d 200 continue return end

The Program NONLINEARBEAM program NonlinearBeam implicit double precision (a-h,o-z) parameter (nbasis=6) parameter (ndm=3*nbasis+1) dimension A(ndm,ndm),b(ndm),x(ndm),xo(ndm),D(3) common /loads/ dmo(3),dqo(3),dpo(3) common /out/ npts,nnstep data zero/0.d0/,one/1.d0/,two/2.d0/ c.... Read problem parameters read(5,1000) tol,alpha,xlength read(5,1001) maxsteps,maxit,inter read(5,1001) npts,nnstep read(5,1000) D(1),D(2),D(3) read(5,1000) Cm,Cq,Cp read(5,1000) (dmo(i),i=1,3) read(5,1000) (dqo(i),i=1,3) read(5,1000) (dpo(i),i=1,3) write(6,2000) tol,alpha,maxsteps,maxit,inter,npts,nnstep,nbasis write(6,2001) xlength,D(1),D(2),D(3),Cm,Cq,Cp, * (dmo(i),dqo(i),dpo(i),i=1,3) c.... Open output files and write header for summary output open(3,file=’disp.beam’,status=’unknown’) open(4,file=’coef.beam’,status=’unknown’) write(6,2002) c.... Initialize integration increment, change intervals to Simpson points dz = one/(2*inter) npoints = 2*inter + 1 c.... Initialize values for load step zero,set first load level call zerovec(x,ndm) call zerovec(xo,ndm) x(ndm) = alpha

403

404

Fundamentals of Structural Mechanics c.... Compute MAXSTEPS points along the Equilibrium Path do 2 nstep = 1,maxsteps c.... Perform Newton iteration at each load step 1

nu = 0 nu = nu + 1

c.... Execute numerical integration of Hessian and residual components call zerovec(b,ndm) call zerovec(A,ndm*ndm) z = 0.d0 do mpoint = 1,npoints call simpson(mpoint,npoints,dz,xlength,factor) call fcn(A,b,D,x,z,xlength,factor,ndm,nbasis) z = z + dz end do C.... Add end load terms to the residual and coefficient matrix do i=1,nbasis call basis(i,one,xlength,h,dh,nbasis) mm = 3*(i-1) b(mm+1) = b(mm+1) - h*Cp*x(ndm) b(mm+2) = b(mm+2) - h*Cq*x(ndm) b(mm+3) = b(mm+3) - h*Cm*x(ndm) A(mm+1,ndm) = A(mm+1,ndm) - h*Cp A(mm+2,ndm) = A(mm+2,ndm) - h*Cq A(mm+3,ndm) = A(mm+3,ndm) - h*Cm end do c.... Add arc-length constraint terms to Hessian and residual do k=1,ndm b(ndm) = b(ndm) + (x(k) - xo(k))**2 A(ndm,k) = two*(x(k) - xo(k)) end do b(ndm) = b(ndm) - alpha**2 c.... Compute norm of residual for convergence test test = zero do k=1,ndm test = test + b(k)**2 end do test = dsqrt(test) c.... Update state vector call invert(A,ndm,ndm) do j=1,ndm do k=1,ndm x(j) = x(j) - A(j,k)*b(k) end do end do c.... Test for convergence, if successful output values if ((test.gt.tol).and.(nu.lt.maxit)) go to 1 if (nu.ge.maxit) stop ’Iteration limit exceeded’ call results(x,xlength,nstep,nu,test,ndm,nbasis) c.... Update values for previous converged state and guess at next state do j=1,ndm temp = xo(j) xo(j) = x(j) x(j) = two*x(j) - temp end do 2 continue close(3) close(4) stop ’Maximum number of steps exhausted’ c.... IO Formats 1000 format(3f10.0) 1001 format(3i10)

Chapter 12


2000 format(/’ * 5x, * 5x, * 5x, * 5x, * 5x, * 5x, * 5x, * 5x,

Fully Nonlinear Beam Analysis’// ’ Convergence tolerance ’ Arc length parameter ’ Number of load steps ’ Maximum number of iterations ’ Number of integration intervals ’ Number of points in plotted shape ’ Output shape every N points, N = ’ Number of basis functions

2001 format(/’ * 5x, * 5x, * 5x, * 5x, * 5x, * 5x,

Beam properties, ’ Column Length ’ EI, GA, EA ’ Mo, Qo, Po ’ mo1, qo1, po1 ’ mo2, qo2, po2 ’ mo3, qo3, po3

: : : : : : : :

’,e12.4/ ’,f12.4/ ’,i6/ ’,i6/ ’,i6/ ’,i6/ ’,i6/ ’,i6/)

end load and dist. load amplitudes’/ : ’,g12.4/ : ’,3g12.4/ : ’,3g12.4/ : ’,3g12.4/ : ’,3g12.4/ : ’,3g12.4//)

2002 format(’ Step’,8x,’Load’,8x,’u(L)’,8x,’w(L)’,4x,’theta(L)’, * ’ NU’,4x,’|| b ||’) end C----Compute contribution to

A

and

b

at current integration point-------FCN

subroutine fcn(A,b,D,x,z,xlength,factor,ndm,nbasis) implicit double precision (a-h,o-z) dimension A(ndm,ndm),b(ndm),x(ndm),D(3) dimension G(4,4),GB(4,3),BGB(3,3) data zero/0.d0/,one/1.d0/ c.... Compute displacements and derivatives du = zero dw = zero dtheta = zero theta = zero do i=1,nbasis call basis(i,z,xlength,h,dh,nbasis) du = du + x(3*i-2)*dh dw = dw + x(3*i-1)*dh dtheta = dtheta + x(3*i)*dh theta = theta + x(3*i)*h end do ct = dcos(theta) st = dsin(theta)

Áo, shear strain epsilon = dw*st + (one + du)*ct - one beta = dw*ct - (one + du)*st curv = dtheta

c.... Compute axial strain

βo ,

and curvature

κo

c.... Compute axial force, shear force, bending moment, and other forces bend shear axial Hor = Ver = Xi = Yi =

= D(1)*curv = D(2)*beta = D(3)*epsilon axial*ct - shear*st axial*st + shear*ct (one + du)*Hor + dw*Ver dw*Hor - (one + du)*Ver

c.... Compute components of G(1,1) G(1,2) G(1,3) G(1,4) G(2,2) G(2,3) G(2,4) G(3,3) G(3,4) G(4,4)

= = = = = = = = = =

[E TDE + G]

store in matrix G

D(2)*st*st + D(3)*ct*ct ct*st*(D(3)-D(2)) D(2)*st*(one+epsilon) + D(3)*ct*beta - Ver zero D(2)*ct*ct + D(3)*st*st D(3)*st*beta - D(2)*ct*(one+epsilon) + Hor zero D(2)*(one+epsilon)**2 + D(3)*beta**2 - Xi zero D(1)

405

406

Fundamentals of Structural Mechanics c.... Compute the rest of

[E TDE + G]

by symmetry

do i=1,3 do j=i+1,4 G(j,i) = G(i,j) end do end do c.... Form stiffness matrix

K

and store it in matrix

A

do i=1,nbasis call basis(i,z,xlength,hi,dhi,nbasis) do j=1,nbasis call basis(j,z,xlength,hj,dhj,nbasis) c.... Compute

BTi[E TDE + G]B j

noting the sparse structure of

B

do k=1,4 GB(k,1) = dhj*G(k,1) GB(k,2) = dhj*G(k,2) GB(k,3) = hj*G(k,3) + dhj*G(k,4) end do do k=1,3 BGB(1,k) = dhi*GB(1,k) BGB(2,k) = dhi*GB(2,k) BGB(3,k) = hi*GB(3,k) + dhi*GB(4,k) end do c.... Assemble the result into the matrix

A

do m=1,3 mm = 3*(i-1) do n=1,3 nn = 3*(j-1) A(mm+m,nn+n) = A(mm+m,nn+n)+ BGB(m,n)*factor end do end do end do c.... Form integral part of residual force and assemble into matrix

b

mm = 3*(i-1) call applied(z,dm,dq,dp,x(ndm),1) b(mm+1) = b(mm+1) + (dhi*Hor - hi*dp)*factor b(mm+2) = b(mm+2) + (dhi*Ver - hi*dq)*factor b(mm+3) = b(mm+3) + (dhi*bend + hi*(Yi - dm))*factor c.... Form the integral part of load factor part of matrix A call applied(z,dm,dq,dp,x(ndm),2) A(mm+1,ndm) = A(mm+1,ndm) - hi*dp*factor A(mm+2,ndm) = A(mm+2,ndm) - hi*dq*factor A(mm+3,ndm) = A(mm+3,ndm) - hi*dm*factor end do return end C---- Evaluate ith basis function h and derivative dh----------------BASIS subroutine basis(i,z,xlength,h,dh,nbasis) implicit double precision (a-h,o-z) n = mod(i,2) m = i/2 + n a = dble(2*m-1)*2.d0*datan(1.d0) if (n.eq.0) then h = dsin(a*z) dh = a*dcos(a*z)/xlength else h = 1.d0 - dcos(a*z) dh = a*dsin(a*z)/xlength end if return end

Chapter 12


C----Evaluate the distributed load functions at point z--------------APPLIED subroutine applied(z,dm,dq,dp,clf,n) implicit double precision (a-h,o-z) common /loads/ dmo(3),dqo(3),dpo(3) c.... Compute the nominal values of the applied forces at point z f = 1.d0 - z go to (1,2), n c.... Compute total transverse loads 1 dm = dmo(1) + (dmo(2) + dmo(3)*f)*clf dq = dqo(1) + (dqo(2) + dqo(3)*f)*clf dp = dpo(1) + (dpo(2) + dpo(3)*f)*clf return c.... Compute transverse loads associated with load factor only 2 dm = dmo(2) + dmo(3)*f dq = dqo(2) + dqo(3)*f dp = dpo(2) + dpo(3)*f return end C----Print results of current step to various files----------------RESULTS subroutine results(x,xlength,nstep,nu,test,ndm,nbasis) implicit double precision (a-h,o-z) logical plot dimension x(ndm),disp(3) common /out/ npts,nnstep c.... Determine if current step is an output step if(mod(nstep,nnstep).eq.0) then plot = .true. nplot = npts else plot = .false. nplot = 1 end if c.... Write coefficients for current step if (plot) then write(4,2001) nstep, x(ndm), nu, test do i=1,nbasis ii = 3*(i-1) write(4,2002) i,(x(ii+k),k=1,3) end do end if c.... Compute and print current geometry of beam if (plot) write(3,3001) nstep,x(ndm) z = 0.d0 dz = 1.0/nplot do ii=1,nplot+1 do j=1,3 disp(j) = 0.d0 do i=1,nbasis call basis(i,z,xlength,h,dh,nbasis) disp(j) = disp(j) + h*x(3*(i-1)+j) end do end do if (plot) write(3,3000) z*xlength,z*xlength+disp(1),disp(2) z = z + dz end do write(6,2003) nstep,x(ndm),(disp(j),j=1,3),nu,test return 2001 format(/’ Load Step :’,i4,5x,’ Load factor :’,e12.5, * /’ Iterations :’,i4,5x,’ Norm of Residual :’,e12.5/

407

408

Fundamentals of Structural Mechanics 2002 2003 3000 3001

* /’ i’,15x,’ a(i)’,15x,’ b(i)’,15x,’ c(i)’) format(i5,3e20.5) format(i5,4e12.5,i5,e12.4) format(4e15.5) format(’ Step : ’,i5,5x,’ Load : ’,e15.5) end

Solution

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