CONTENTS CHAPTER (1)
PROGRAM OF MATRICES 1-1 1-2 1-3 1-3-1 1-3-2 1-3-3 1-3-4 1-3-5 1-3-6 1-3-7 1-3-8
Introduction ………………………………... PROGRAMMING LANGUAGES………...
1 2
ELEMENTARY MATRIX OPERATION . Definitions and Rules………………………………. Transpose of Rectangular Matrix ……………….. Transpose of Square Matrix ……………………… Addition and Subtraction of Matrices……………. Program of Addition and Subtraction of Matrices Multiplication of Matrices Program of Multiplication of Matrices Program of Inverse of Matrices Solved Problems Sheet ( 1 )
3 3 7 8 8 9 10 11 12 13 16
CHAPTER (2)
ANALYSIS OF INDETERMINATE STRUCTURES 2-1 2-2 2-3 2-4 2-5 2-6
INTRODUCTION DEFINITION DEGREE OF INDETERMANCY KINEMATIC INDETERMANCY ANALYSIS OF STATICALLY INDETERMINATE STR. FORCE – DISPLACEMENT RELATIONS
17 17 20 23 24 25
2-7
DESCRIPTION OF METHODS 2-7-1 Solution by flexibility method 2-7-2 Solution by stiffness method 2-7-3 Comparison between the flexibility and stiffness method Sheet ( 2 )
25 26 27 28
CHAPTER (3)
SLOPE DEFLECTION METHOD 3-1 3-2 3-2-1 3-3 3-4 3-5 3-5-1 3-6 3-7
Introduction …………………………………. Assumption in slope deflection ……………... Degree of Freedom ………………………….. Sign Conventions …………………………… Fixed End Moments ………………………… Derivation of Slope Deflection Equation …… Examples ……………………………………. Frames with a single degree of freedom in translation (Examples) ………………………. Frames with Multiple degrees of freedom in translation (Examples) ……………………….. Sheet ( 3 )
30 31 31 34 35 36 38 51 58 74
CHAPTER (4)
MATRIX APPROCH 1 (SLOPE DEFLECTION IN MATRIX FORM) 4-1 4-2 4-3 4-4
Introduction ………………………………….. Members Without Relative Displacements……... Solved Examples…………………………… Effect of Members With Moments Releases...…. Sheet ( 4 ) ……………………………………..
76 76 79 85 86
CHAPTER (5)
STIFNESS MATRIX METHOD 5-1 5-2 5-3 5-4 5-5 5-6 5-7
Definition……………….…………………... Step 1: Modeling………………………………. Step 2: Load Vector………………………….. Step 3: Stiffness Matrix……………………… Step 4: Equilibrium Equation ……… Step 5: Solve The Equilibrium Equation … Step 6 : Internal Forces……… ……….
87 89 94 95 97 97 98
CHAPTER (6)
ANALYSI OF PLANE TRUSSES BY USING STIFNESS MATRIX METHOD 6-1 6-2 6-3 6-4 6-5
Modeling………………………………. Load Vector………………………….. Stiffness Matrix……………………… Equilibrium Equation And Internal Forces Solved Examples Sheet ( 5 )……… ……….
99 102 104 106 107 118
CHAPTER (7)
ANALYSI OF PLANE FRAMES BY USING STIFNESS MATRIX METHOD 7-1 7-2 7-3 7-4 7-5
Modeling………………………………. Load Vector………………………….. Stiffness Matrix……………………… Equilibrium Equation And Internal Forces Solved Examples Sheet ( 6 )……… ………. References …………………………………… Examinations
120 123 133 137 138 158 160 161
CHAPTER 1
PROGRAM OF MATRICES 1-1- INTRODUCTION A definition of engineering is the science by which the properties of matter and sources of energy in nature are made useful to man. Thus an engineer will have to study the properties and behavior of physical systems, applying his knowledge to attain purposes useful to society. Such activity can be considered as consisting of the study and solution of real physical. The complexities involved in dealing with all the parameters relative to the properties of a real problem will, on many occasions, force him to study an equivalent engineering problem which can be mathematically defined. He will try to solve such a problem according to his knowledge and experience, and using the computational tools at hand. When we think of today’s engineering problems, such as the construction of nuclear reactors, space satellites, oil rigs in the sea, etc. it is clear that they must be analyzed using very relined discrete models, so as to closely approximate their real behavior. Fortunately, with the help of the modern computer, it is possible today to treat discrete models with up to several thousands of degrees of freedom. Thus, the need for formulating larger and more complex discrete models, and the availability of computers which make feasible their analysis, are the causes of the considerable interest produced in relation to numerical methods and computer analysis techniques.
1-2- PROGRAMMING LANGUAGES A computer can perform elementary operations using stored information. The user can order the computer to perform an operation, giving the proper instruction, as a part of his program. The set of all the instructions which a computers accepts constitutes the machine language, or absolute language of that computer. The machine instructions are always expressed by numeric codes. Each instruction normally consists of one operation code, and one or more parameters or arguments. For instance, the sequence can represent two instructions written according to a given machine language. He first step in programming the solution of a problem, is to make an analysis of the problem. The problem is studied in all its detail, and an appropriate solution scheme is selected. Then the computer solution to be implemented is schematized graphically, by means of a diagram called a flow chart. These diagrams can be detailed to a greater or leaser extent, according to circumstances, but they are fundamental to the analysis and understanding of all aspects of the computer solution. The person in charge of this task is normally called an analyst. Once the flow-chart is completed, the computer solution is coded, writing the computer program using the language selected. This task is carried out by a programmer. With the program registered in some information support, and the data and proper control commands added, the program is tested by submitting it for processing. At the beginning the program will include some errors. These can be syntax errors and logic errors. Syntax errors appear because some of the rules of the language are violated. Normally These can be easily detected and corrected. Logic errors correspond to the case that a syntax errors-free program is submitted for execution, and it does not produce the desired results.
The following situations can be encountered. 1. Execution of the program stops before producing results. 2. Execution of the program does not stop and results are not produced. The program is finally discontinued after exceeding the allowed time. 3. Results are produced, but they are incorrect. 4. The program works successfully.
1-3- ELEMENTARY MATRIX OPERATION 1-3-1 Definitions and Rules Matrix algebra is the set of rules which defines operations performed with matrices. A matrix is a bi-dimensional array of numbers organized according to “ n “ rows and “ m “ columns, which we refer to as a matrix of order “ n x m “. In what follows, a matrix will be indicated by bold type. Thus, R will be a matrix, while T is a scalar number. The elements or coefficients of a matrix will be two sub-indices, indicating row and column position, respectively. For instance
[A] = ⎡⎢
a11 ⎣ a 21
a12 a 22
a13 ⎤ a 23 ⎥⎦
(1-1)
is a rectangular matrix , of order 2 x 3, that is having 2 rows and 3 columns. The sub-indices of the coefficients indicate their position within the matrix. Thus, the coefficient a12 is placed in the first row and the second column. A matrix having the same number of rows and columns is called a square matrix, such as
⎡ b11 [B] = ⎢⎢b21 ⎢⎣b31
b13 ⎤ b23 ⎥⎥ b33 ⎥⎦
b12 b22 b32
(1-2)
which is a square matrix of 3 rows and 3 columns, or of order 3 x 3. In case such as this, when speaking specifically of square matrices, we can simply say that B is of order 3. The coefficients b11, b22, and b33 or in general bij for i = j, are called the coefficients of the principal or main diagonal of the matrix. Several types of square matrices can be considered. When the coefficients of a square matrix B satisfy the condition bij = bji ,
for i = 1,2,….,n and j = 1,2,….,n
(1-3)
this called a symmetric matrix. In particular, if B is a symmetric matrix of order 3, it will have the following structure ⎡ b11 [B] = ⎢⎢b12 ⎢⎣ b13
b13 ⎤ b23 ⎥⎥ b33 ⎥⎦
b12 b22 b23
(1-4)
On the other hand, skew-symmetric matrix is such that bij = -bji , for i ≠ j & bij = 0, for i = j (1-5) for example, ⎡ 0 [B] = ⎢⎢ − b12 ⎢⎣ − b13
b12 0 − b23
b13 ⎤ b23 ⎥⎥ 0 ⎥⎦
(1-6)
when all the coefficients of a square matrix B are null, except for the main diagonal, such as, ⎡ b11 [B] = ⎢⎢ 0 ⎢⎣ 0
0 b22 0
0⎤ 0 ⎥⎥ b33 ⎥⎦
(1-7)
this is called a diagonal matrix. If, in particular, all the coefficients of the main diagonal are equal to 1, the matrix is called a unit matrix, and is symbolized I. For instance, the following is the unit matrix of order 3. ⎡1 0 0 ⎤ [I ] = ⎢⎢0 1 0⎥⎥ ⎢⎣0 0 1⎥⎦
(1-8)
A matrix having only one column is called a column vector. Normally, the coefficients of a column vector will be given only one subscript, corresponding to the row position, and will be enclosed by curled brackets instead of square brackets, placed vertically, as in ⎧ c1 ⎫ {C} = ⎪⎨c 2 ⎪⎬ ⎪c ⎪ ⎩ 3⎭
(1-9)
where C is a column vector of order 3. When a matrix has only one row it will be called a row vector and again, each of its coefficients will be given only one subscript. For instance, {D} = {d1
d2
d3
d4}
(1-10)
is a row vector of order 3. In what follows the use of the term “vector” alone will always be taken to mean “column vector”. The transpose of a matrix is obtained by interchanging rows and columns, of the matrix, and is indicated by an upper-index T. For instance, the transpose of A, given by eq. (1-1) is
[A]
T
⎡ a11 = ⎢⎢ a12 ⎢⎣ a13
a 21 ⎤ a 22 ⎥⎥ a 23 ⎥⎦
(1-11)
while the transpose of matrix B, given by eq. (1-2) is
[B]
T
⎡b11 = ⎢⎢b12 ⎢⎣b13
b21 b22 b23
b31 ⎤ b32 ⎥⎥ b33 ⎥⎦
(1-12)
It can be noticed that the transpose of a column vector is a row vector, and vice versa. Thus, the transpose of column vector, given by eq. (1-9) is {C}T = {c1
c2
c3 }
(1-13)
While the transpose of row vector, given by eq. (1-10) is
{D}T
⎧ d1 ⎫ ⎪d ⎪ ⎪ ⎪ = ⎨ 2⎬ ⎪d 3 ⎪ ⎪⎩d 4 ⎪⎭
(1-14)
The transpose of the transpose gives the original matrix, that is
([A] )
T T
= [A]
(1-15)
For symmetric matrix, the transpose of the matrix is the original matrix,
[B] = [B]T
(1-16)
Another special matrix which can be defined as the null matrix, represented by 0, whose coefficients are all equal to zero. For instance,
[0] = ⎡⎢
0 0 0⎤ ⎥ ⎣0 0 0 ⎦
(1-17)
is a null matrix of order 2 x 3
1-3-2 Transpose of Rectangular Matrix The following routine is a very simple example of a computer program operating with matrices which can be used to obtain the transpose of a rectangular matrix.
C This program computes the transpose of a matrix “A” of order n xm DIMENSION A(10,10),B(10,10) DO 10 I=1,n DO 10 J=1,m 10 B(J,I) = A(I,J) END where the array A contains the matrix to be transposed, the integer n contains the number of rows, and the integer m contains the number of columns.
1-3-3 Transpose of Square Matrix In the case of a square matrix it is possible to transpose the matrix in itself, without creating a new one. The following routine receives the original matrix in the array A, of order n, and after operating returns the transposed matrix in the same array A. Note that an auxiliary variable S is used to allow for the transposition in place.
C This program computes the transpose of a square matrix “A” of order n DIMENSION A(10,10) n1 = n - 1 DO 10 I=1,n1 I1 = I + 1 DO 10 J=I1,n S = A(1,J) A(I,J) = A(J,I) A(J,I) = S END
10
1-3-4 Addition and Subtraction of Matrices Two different matrices can be added or subtracted, provided they are of the same order. Let us consider the following two matrices
[A] = ⎡⎢
a11 ⎣ a 21
a12 a 22
a13 ⎤ a 23 ⎥⎦
,
[B] = ⎡⎢
b11 ⎣b21
b12 b22
b13 ⎤ b23 ⎥⎦
(1-18)
By definition the sum of A and B will give another matrix C, also of the same order, i.e.
[C ] = ⎡⎢
c11 ⎣ c 21
c12 c 22
c13 ⎤ c 23 ⎥⎦
(1-19)
where coefficients are c11 = a11 + b11 c12 = a12 + b12 b13 c21 = a21 + b21 c22 = a22 + b22 b23 In general, cij = aij + bij for i = 1,2,….,n and j = 1,2,….,m
c13 = a13 + c23 = a23 + (1-20)
1-3-5 Program of Addition and Subtraction of Matrices The following routine can be used for addition and subtraction of matrices. The arrays A and B contain the matrices to be added or subtracted, which of order n x m. The array C will contain the result matrix. When the integer L = 1 is specified the routine will perform the addition operation, but when L = 2 the routine will perform the subtraction operation. C C C C C
This program computes the matrix operation C=A+B when L = 1 and C=A-B when L = 2 n : number of rows m : number of columns DIMENSION A(10,10), B(10,10), C(10,10) DO 10 I=1,n DO 10 J=1,m GO TO (2,4), L 2 C(I,J) = A(I,J) + B(I,J) GO TO 10 4 C(I,J) = A(I,J) - B(I,J) 10 CONTINUE END
1-3-6 Multiplication of Matrices Multiplication of matrices can also be performed. Let us consider a matrix A of order n x m, and matrix B of order m x n, such as a [A] = ⎡⎢ 11 ⎣ a 21
a12 a 22
a13 ⎤ a 23 ⎥⎦
,
⎡ b11 [B ] = ⎢⎢b21 ⎣⎢b31
b12 ⎤ b22 ⎥⎥ b32 ⎦⎥
(1-21)
The matrix multiplication equation
[C ] = [A][B]
(1-22)
gives as a result a new matrix with coefficients: c11 = a11 b11 + a12 b21 + a13 b31 b22 + a13 b32 c21 = a21 b11 + a22 b21 + a23 b31 b22 + a23 b32
c12 = a11 b12 + a12 c22 = a21 b12 + a22
or, in general: m
cij = ∑ aik bkj
(1-23)
k =1
It is easily seen that two matrices can be multiplied only if the number of columns of the first matrix is equal to the number of rows of the second matrix. The result matrix will have the same number of rows of the first matrix and the same number of columns of the second matrix. Notice that according to its definition, matrix multiplication is not commutative, i.e.,
[A] [B] ≠ [B] [A]
(1-24)
and that, in particular, if the left–hand side is defined, the right-hand side may be undefined unless the number of rows of matrix A is equal to the number of columns of matrix B. Even in that cases, the results of the left-hand side, and the right-hand side will be different, excluding very particular cases.
1-3-7 Program of Multiplication of Matrices The multiplication of two matrices can be readily programmed. The following routine receives the first matrix of order n x m, in the array A, and the second matrix of order m x h, in the array B. The result matrix , of order n x h, will be stored in the array C. Note that since the implementation of eq. (1-23) requires accumulating the partial multiplications in the array C, we have been careful in previously setting cij = 0. C C C C C
This program computes the matrix operation C=A*B n : number of rows of A and C m : number of columns of A and rows of B h : number of columns of B and C DIMENSION A(10,10), B(10,10), C(10,10) DO 10 I=1,n DO 10 J=1,h C(I,J) = 0 DO 10 K=1,m 10 C(I,J) = C(I,J) + A(I,K) * B(K,J) END
1-3-8 Program of Inverse of Matrices The inverse of matrix C is the matrix which satisfy the following relation
[C] [C]-1 = [I]
The following routine can be used for computing the matrix inverse [C]-1. C
This program computes the matrix operation DIMENSION C(5,5), ST(5,10), C1(5,5) DO 2 I=1,5 DO 2 J=1,5 2 ST(I,J) = C(I,J) M1 = 6 K= DO 3 I=1,5 DO 3 J=6,10 IF (I.EQ.(J-5)) GO TO 4 ST(I,J) = 0.0 GO TO 3 4 ST(I,J) = 1 3 CONTINUE L=1 10 DO 5 J=1,K IF (L.EQ.J) GO TO 6 ST(L,J) = ST(L,J) / ST(L,L) 5 CONTINUE ST(L,L) = 1 DO 6 J=1,5 IF(J.EQ.L) GO TO 6 IF(ST(I,J).EQ.0.0) GO TO 6 DO 7 M=1,K IF(L.EQ.M) GO TO 7 ST(J,M) = ST(J,M) – ST(J,L) * ST(L,M) 7 CONTINUE ST(J,L) = 0.0 6 CONTINUE IF(L.EQ.6) GO TO 8 GO TO 10 8 DO 9 I=1,5 DO 9 J=1,5 9 C1(I,J) = ST(I,J+5) END
Solved Problems 1- For the following matrices, evaluate the following expressions if possible. If not possible, state the reason: a) 3[D] – 4[E] ] b) [A] x [C] c)
[A] x 0.5[B]
⎡ 7 − 5⎤ [A] = ⎢⎢ 1 5 ⎥⎥ ⎢⎣ − 1 4 ⎥⎦
−2 3
4 [B] = ⎡⎢ ⎣ 6
7 ⎤ − 0.5⎥⎦
⎡ 7 −5 3 ⎤ ⎢1 5 − 1⎥⎥ [D] = ⎢ ⎢−1 4 0⎥ ⎢ ⎥ 3 1⎦ ⎣0
⎡ 7 −6 3 [C ] = ⎢⎢ 2 8 − 6 ⎢⎣ − 1 3 0
⎡ 6 −5 ⎢ 8 −4 [E ] = ⎢ ⎢− 3 0 ⎢ 3 ⎣0
Solution a)
3 [D] – 4 [E] ⎡ 6 −5 ⎡ 7 −5 3 ⎤ ⎢ 8 −4 ⎢1 ⎥ 5 − 1⎥ 3 [ D] − 4 [ E ] = 3 ⎢ −4⎢ ⎢− 3 0 ⎢−1 4 0⎥ ⎢ ⎢ ⎥ 3 1⎦ 3 ⎣0 ⎣0 ⎡ 21 ⎢ 3 =⎢ ⎢ −3 ⎢ ⎣ 0
− 15 15 12 9
9 ⎤ − 3 ⎥⎥ − 0 ⎥ ⎥ 3 ⎦
⎡ −3 ⎢− 29 =⎢ ⎢ 9 ⎢ ⎣ 0
5 31
1 ⎤ 1 ⎥⎥ −8 ⎥ ⎥ −1 ⎦
12 −3
2⎤ − 1⎥⎥ 2⎥ ⎥ 1 ⎦
⎡ 24 − 20 ⎢ 32 − 16 ⎢ ⎢ − 12 0 ⎢ 12 ⎣ 0
8 ⎤ − 4 ⎥⎥ 8 ⎥ ⎥ 4 ⎦
2⎤ − 1⎥⎥ 2⎥ ⎥ 1 ⎦
6 4 5
− 1⎤ 2 ⎥⎥ 3 ⎥⎦
b)
[A]3x2 x [C]3x5
ﻏﻴﺮ ﻣﻤﻜﻨﺔ ﻟﻌﺪم ﺗﺴﺎوي ﻋﺪد أﻋﻤﺪة اﻟﻤﺼﻔﻮﻓﺔ اﻷوﻟﻰ ﻣﻊ ﻋﺪد ﺻﻔﻮف اﻟﻤﺼﻔﻮﻓﺔ اﻟﺜﺎﻧﻴﺔ
c)
5[A] x [B]
⎡ 7 − 5⎤ 7 ⎤ −2 ⎡ 4 5 [ A] x 0.5 [ B ] = 5 ⎢⎢ 1 5 ⎥⎥ x 0.5 ⎢ 6 3 − 0.5⎥⎦ ⎣ ⎢⎣ − 1 4 ⎥⎦ ⎡ 35 − 25⎤ 3.5 ⎤ −1 ⎡ 2 25 ⎥⎥ x ⎢ = ⎢⎢ 5 3 1.5 − 0.25⎥⎦ ⎣ ⎢⎣ − 5 20 ⎥⎦ − 150 128.75⎤ ⎡ −5 ⎢ = ⎢ 85 − 72.5 11.25 ⎥⎥ ⎢⎣ 50 35 − 22.5 ⎥⎦
2)
[A]6 x 6
Find the inverse of the following matrix: ⎡ 7 −5 ⎢1 5 ⎢ ⎢−1 4 =⎢ 3 ⎢0 ⎢ 0 −1 ⎢ ⎣⎢ 2 − 2
3 0 −1 7 0 5 1 2 7 3 3 −2
−1 2 1 0 0 1
2⎤ 2 ⎥⎥ 3⎥ ⎥ 4⎥ − 1⎥ ⎥ 0 ⎦⎥
Solution
[A]−1 =
[ ]
1 Aadj A
By using Saraus method, we can get A 7 −5 1 5 −1 4 0 3 0 −1 −
2 −
−2 −
3 −1 0 1 7
0 7 5 2 3
−1 2 1 0 0
2 2 3 4 −1
7 −5 1 5 −1 4 0 3 0 −1
3 −
−2 −
1 −
0
2 +
−2 +
3 −1 0 1 7
0 7 5 2 3
−1 2 1 0 0
3 +
−2 +
1 +
+
∴ A = 0 + 0 + 0 + 0 – 84 + 24 – (-40) – (-392) – 0 – 0 – 0 – 0 = 372 Q A ≠0
⎡ − 10.45 ⎢ − 116.58 ⎢ ⎢− 209.06 ∴ Matrix of min ors = ⎢ ⎢ − 94.41 ⎢ 4.46 ⎢ ⎢⎣ 13.39
∴ [A] is nonsingular matrix.
54.68 141.36 305.78 193.47 − 29.64 − 31.62
− 16.74 − 15.54 − 66.58 − 62.27 45.79 − 18.56
− 314.34 − 16.36 − 29.64 ⎤ 41.29 98.58 ⎥⎥ − 21.94 − 116.51 − 240.08 − 187.78 ⎥ ⎥ 165.91 − 88.53 − 23.06 ⎥ 40.17 ⎥ − 13.46 − 59.89 ⎥ 39.5 ⎥⎦ − 40.54 − 198.46
− 54.68 ⎡ − 10.45 ⎢ 116.58 141.36 ⎢ ⎢− 209.06 − 305.78 ∴ Co − factor matrix = ⎢ 193.47 ⎢ 94.41 ⎢ 4.46 29.64 ⎢ − 31.62 ⎣⎢ − 13.39
− 16.74 15.54 − 66.58 62.27 45.79 18.56
314.34 − 21.94 116.51 − 88.53 13.46 − 40.54
− 16.36 − 41.29 240.08 − 165.91 − 59.89 198.46
29.64 ⎤ − 98.58 ⎥⎥ − 187.78 ⎥ ⎥ − 23.06 ⎥ − 40.17 ⎥ ⎥ 39.5 ⎦⎥
[
⎡ − 10.45 ⎢ − 54.68 ⎢ ⎢ − 16.74 =⎢ ⎢ 314.34 ⎢ − 16.36 ⎢ ⎢⎣ 29.64
116.58 141.36 15.54 − 21.94 − 41.29 − 98.58
− 209.06 − 305.78 − 66.58 116.51 240.08 − 187.78
94.41 193.47 62.27 − 88.53 − 165.91 − 23.06
4.46 29.64 45.79 13.46 − 59.89 − 40.17
− 13.39 ⎤ − 31.62 ⎥⎥ 18.56 ⎥ ⎥ − 40.54 ⎥ 198.46 ⎥ ⎥ 39.5 ⎥⎦
⎡ − 10.45 ⎢ − 54.68 ⎢ 1 ⎢ − 16.74 = ⎢ 372 ⎢ 314.34 ⎢ − 16.36 ⎢ ⎣⎢ 29.64
116.58 141.36 15.54 − 21.94 − 41.29 − 98.58
− 209.06 − 305.78 − 66.58 116.51 240.08 − 187.78
94.41 193.47 62.27 − 88.53 − 165.91 − 23.06
4.46 29.64 45.79 13.46 − 59.89 − 40.17
− 13.39 ⎤ − 31.62 ⎥⎥ 18.56 ⎥ ⎥ − 40.54 ⎥ 198.46 ⎥ ⎥ 39.5 ⎦⎥
0.3134 0.38 0.0418 − 0.059 − 0.111 − 0.265
− 0.562 − 0.822 − 0.179 0.3132 0.6454 − 0.5048
0.2538 0.5201 0.1674 − 0.238 − 0.446 − 0.062
0.012 0.0797 0.1231 0.0362 − 0.161 − 0.108
− 0.036 ⎤ − 0.085 ⎥⎥ 0.0499 ⎥ ⎥ − 0.109 ⎥ 0.5335 ⎥ ⎥ 0.1062 ⎦⎥
]
∴ Aadj = [ A
∴[ A
]−1
∴[ A
]T
]−1
⎡ − 0.0281 ⎢ − 0.147 ⎢ ⎢ − 0.045 =⎢ ⎢ 0.845 ⎢ − 0.044 ⎢ ⎣⎢ 0.0797
SHEET (1) 1) Identify the properties of each of the following matrices: ⎡ 0 .5 [A] = ⎢⎢ 0.2 ⎢⎣ 0
⎡ 2 [D ] = ⎢⎢ 3 ⎢⎣ 7
0 ⎤ 0.4 − 0.5⎥⎥ − 0 .5 1 ⎥⎦
⎡8 [B ] = ⎢⎢− 1 ⎢⎣ 5
0 .2
0 ⎤ − 2⎥⎥ − 3⎥⎦
⎡ ⎢ [E ] = ⎢ ⎢ ⎢ ⎣
0.5 0 0 0
6 −3 0 0
0.4 2 1 0
0⎤ 4 ⎥⎥ 0 ⎥⎦
− 0.3⎤ 0 ⎥⎥ 0 ⎥ ⎥ 0.2 ⎦
⎡8 [C ] = ⎢⎢ 4 ⎢⎣− 1
0⎤ 3 ⎥⎥ 7 ⎥⎦
⎡1 [F ] = ⎢⎢ 3 ⎢⎣ 5
0⎤ 5 ⎥⎥ − 6⎥⎦
2) Evaluate the following expressions if possible. If not possible, state the reason. [C] + [B]
[A] + [C]
[F]T
2 x [D] + [F] x [E] x [F]T
[E] x [F]
[A]-1
[B]-1
[B] x [C] – [A]-1
[D]-1 – [C] x [B] 3) Solve the following simultaneous equations: 5X1 – 2X2 + X3 = 9 X1 + 6X2 – 2X3 = 4 X1 + X3 = 1
[F]T x [C]
Chapter (2) ANALYSIS OF INDETERMINATE STRUCTRES 1.1 Statically Indeterminate Structures A structure of any type is classified as statically indeterminate when the number of unknown reaction or internal forces exceeds the number of equilibrium equations available for its analysis. Most of structures are statically indeterminate. This indeterminacy may arise as a result of added support or members, or by the general form of the structure. For example, reinforced concrete buildings are almost always statically indeterminate since the columns and beams are poured as continuous members through the joints and over supports. Although the analysis of a statically indeterminate structure is more involved than that of one that is statically determinate, there are usually several very important reasons for choosing this type of structures for design. Most important reasons for a given loading the maximum stress and deflection of an indeterminate structure are generally smaller than those of its statically determinate counterpart. For example, the shown fixed beam in Fig.(l.l-a), the maximum moment of Mmax= PL/8, whereas the same beam when simply supported Fig.(l.l-b), will
Fig.(l.l-a)
Fig.(l.l-b),
be subjected to twice the moment, that is, Mmax = Pl/4 . As a result, the fixed support beam has one fourth the deflection and one half the stress. At its center of the one that is simplysupported. Another important reason for selecting a statically indeterminate structures is that it has a tendency to redistribute its load to its redundant supports in cases where faulty design or overloading occurs. In these cases, the structure maintains its stability and collapse is prevented. This is particularly important when sudden lateral loads, such as wind or earthquake, are imposed on the structure. To illustrate, consider again the fixed beam loaded at its center, Fig.(l-a), As load P is increased, the beam’s matenal at the fixations and at the center of the beam begins to yield and forms a localized “Plastic Hinge”, which causes the beam to deflect. Although the deflection becomes
large, the fixations will develop horizontal force and moment reactions that will be hold the beam and thus prevent it from totally collapsing. In the case of the simply supported beam, an excessive load P will cause the “Plastic Hinge” to form only at the center of the beam, and due to the large vertical deflection, the supports will not develop ;the horizontal force and moment reactions that may be necessary to prevent total collapse. Although statically indeterminate structures can support a loading with thinner members and with increased stability compared to their determinate counterparts, there are cases when this advantages may instead become disadvantages. The cost savings in matenal must be compared with the added cost necessary to fabricate the structure, since often time it becomes more costly to construct the supports and joints of an indeterminate structure compared to one that is determinate. More important, though, because statically lndeterrmante structures have redundant support reactions, one has to be very careful to prevent differential settlement of the supports, smce this effect will introduce internal stress in the structure. For example, if the fixed support at one end of fixed beam Fig.(l-a) were to settle, additional moment would be occur. On the other hand, if the beam was simply supported, then any settlement of its end would not cause the beam to deform, and therefore no stress or bending moment would be developed in the beam. In
general, then, any deformation, such as that caused by relative support displacement, or changes in member lengths caused by temperature or fabrication errors, will introduce additional stresses in the structure, which must be acknowledged when designing indeterminate structures. 1-2 Method of Analyses When analyzing any indeterminate structures, it is necessary to satisfy equilibrium, compatibility, and forcedisplacement requirements for the structure. Equilibrium is satisfied when the reactive forces hold the structure at rest, and compatibility is satisfied when the various segments of the structure fit together without intentional breaks or overlaps. The force -displacement requirement depend upon the way the material responds, which assumed linear-elastic response, hi general there are two different ways to satisfy these requirements when analysing a statically indeterminate structure, the force or displacement method. 1.2.1 Flexibility or Force Method The force method was originally developed by J.G. Maxwell and refined by Otto Mohr and Muller-Breslau. This method was one of the first available for the analysis of statically indeterminate structures. The force method consists of writing equations that satisfy the compatibility and forces displacement
requirements for the structure and involve redundant forces as the unknowns. The coefficients of these unknowns are called flexibility coefficients. Since compatibility forms the basis for this method, it has sometimes been referred to as the compatibility method or the method of consistent deformations. Once the redundant forces have been determined, the structure are determined by satisfying the equilibrium requirements for the structure. The fundamental principles involved in applying this method are easy to understand and develop. 1.2.2 Stiffness or Displacement Method The displacement method of analysis is based on first writing force displacement relations for the members and then satisfying the equilibrium requirements for the structures. In this case the unknowns in the equations are displacements and their coefficients
are
called
stiffness
coefficients.
Once
the
displacements are obtained the forces are determmed from the compatibility
and
force-displacement
equations.
A
matrix
formulation of this method will given. 1.2.3 Comparison between the Flexibility and Stiffness Methods 1.2.3.1 Flexibility = deformation per unit force.
δ
Cm/t
The deflection (d) of the shown
Spring Fig. (1-2) is; d = δ .P 1.2.3.2. Stiffness: Stiffness = force per unit deformation =K
(t/cm
or
t.m/rad.)
For the shown spring in Fig. (2), the force required to produce a displacement d is; P=K.d The two methods can be compared as shown in the ollowing table.
Flexibility Method
Stiffness Method
- Unknown redundants X
Unknown displacements d (θ,∆)
- Flexibility matrix
Stiffness matrix K
8
- Displacements due to X =
δ .X
- Displacements due to load =
δ
o
- For Compatibility δo + δ X = 0
Force due d = K . d Force due to loads =Ro Equilibrium Ro + K.d = 0 Solve for d
- Solve for X Indeterminate released
Constraints
and
the
deformation calculated.
are Additional restrains are added to
resulting fix all degrees of freedom and the
discontinuities values These
of
these
restraints
redundant calculated . The restrains are then
actions are then replaced to removed to allow deformations restore the continuity.
and
restore
equilibrium.
The
resulting
equilibrium
are solved to get
equations
displacements
and subsequently the forces are determined.
The force or flexibility method of analyses discussed in the previous course of second year civil;. For example the methods of consistent deformation, three moment equation, and column analogy methods are considered a flexibility or force method of analysis whereas the slope deflection, moment distribution, and stiffness methods are a displacement or stiffness method of analysis.
Chapter (3) SLOPE DEFLECTION METHOD 3.1 Introduction:The methods of three moment equation, and consistent deformation method are represent the FORCE METHOD of structural analysis, The slope deflection method use displacements as unknowns, hence this method is the displacement method. In this method, if the slopes at the ends and the relative displacement of the ends are known, the end moment can be found in terms of slopes, deflection, stiffness and length of the members. In- the slope-deflection method the rotations of the joints are treated as unknowns. For any one member bounded by two joints the end moments can be expressed in terms of rotations. In this method all joints are considered rigid; i.e the angle between members at the joints are considered not-to change in value as loads are applied, as shown in fig 1. joint conditions:-
to get θB & θC
MBA+MBC+MBD = 0
………….
(1)
MCB+MCE
………….
(2)
=0
Figure (1) 3.2 ASSUMPTIONS IN THE SLOPE DEFLECTION METHOD
This method is based on the following simplified assumptions. 1- All the joints of the frame are rigid, i.e, the angle between the members at the joints do not change, when the members of frame are loaded. 2- Distortion, due to axial and shear stresses, being very small, are neglected. 3.2.1 Degree of freedom:-
The number of joints rotation and independent joint translation in a structure is called the degrees of freedom. Two types for degrees of freedom.
In rotation:For beam or frame is equal to Dr.
Dr = j-f
Where: Dr
= degree of freedom.
j
= no. of joints including supports.
F
= no. of fixed support.
In translation:For frame is equal to the number of independent joint translation which can be give in a frame. Each joint has two joint translation, the total number or possible joint translation = 2j. Since on other hand each fixed or hinged support prevents two of these translations, and each roller or connecting member prevent one these translations, the total number of the available translational restraints is; 2f + 2h + r + m
where
f = no. of fixed supports. h = no. of hinged supports. r = no. of roller supports. m = no. of supports. The degree of freedom in translation, Dt, is given by:Dt = 2j-(2f+2h+r+m)
The combined degree of freedom for frame is:D = Dr + Dt = j-f + 2j – (2f + 2h + r + m)
D = 3j – 3j – 2h – r - m
The slop defection method is applicable for beams and frames. It is useful for the analysis of highly statically indeterminate structures which have a low degree of kinematical indeterminacy. For example the frame shown in fig. 2.a p
(b)
The frame (a) is nine times statically indeterminate. On other hand only tow unknown rotations, θb and θc i.e Kinematically
indeterminate to second degree- if the slope deflection is used. The frame (b) is once indeterminate.
3.3 Sign Conventions:Joint rotation & Fixed and moments are considered positive when occurring in a clockwise direction.
θ Al =
2 M A.L
θ BI =
hence
=
M A.L 3 EI
3 2 EI
1 M A.L − M A.L = = 3 2 EI 6 EI θ B1 =
1 θ 2 A1
θ A2 =
1 M B.L − M B.L = 3 2 EI 6 EI
θ B2 =
2 MB.L MB.L = 3 2 EI 3 EI
θB1 + θB2 = 0
Hence: MA = 2MB and θA = θA1 -θA2
=
M A .L 3 EL
−
M A .L 12 EL
θA =
3 MA. L 12 EI
4EI .θA L 2EI MB = .θA L
MA =
Relation between Δ & M R =
Δ L
by moment area method or by conjugate beammethod. Δ = ∑ M at B M.L 2L = ( ) 4EI 3 =
M.L2 6 EI
6 EI Δ L2 6EI = .R L
M =
R (+ ve) when the rotation of member AB with clockwise. 3.4Fixed and moments: As given in the chapter of Moment distribution method.
3.5 Derivation of slope deflection equation:-
Ma1
=
4 EI θA L
Mb1
=
2 EI θA L
Ma2
=
2 EI θB L
Mb2
=
4 EI θB L
Required Mab & Mba in term of
(1) θA, θB at joint B
(2) rotation of member (R) (3) loads acting on member First assume:Get Mfab & Mfba due to acting loads. These fixed and moment must be corrected to allow for the end rotations θA,θB and the member rotation R. The effect of these rotations will be found separately.
Ma1
=
4EI .θ A L
Mb1
=
2EI .θ A L
Mb2
=
4EI .θ B L
Ma2
=
2EI .θ B L
Mb3
= M a3 =
− 6 EI .Δ L2
=
− 6 EI .R L
by Superposition; Mab = Mfab + Ma1 + Ma2 + Ma3 Mf ab +
4EI 2EI − 6EI .θ A + θB + .R L L L
M ab = Mf ab +
2EI (2θ A + θ B − 3R ) L
In case of relative displacement between the ends of members, equal to zero (R = 0)
The term (
M ab = Mf ab +
2EI (2θ a + θb) L
M ba = Mf ba +
2 EI (2θ b + θ A ) L
2 EI ) represents the relative stiffness of member say L
(K) hence: M ab = Mf ab + K ab (2θ A + θ b )
M ba = Mf ba + K ba (2θ B + θ a )
Note: Δ = R is (+ ve) If the rotation of member with clockwise. L
And (– ve) If anti clockwise. M
=
− 6 EI .Δ ( with + ve R ) L2
M
=
− 6 EI .Δ ( with − ve R ) L2
3-5-1 Example 1 Draw B.M.D. S.F.D Solution:1- Relative stiffness:-
KAB : KBC =
1 2.66 : 1: 2 6 8
2- Fixed and Moment:MFBA = MFBA = +
3 × 62 = − 9 t.m. 12
3 × 62 =+ 9 , 12 MFCB = +
MFBC = +
3 × 82 12
3 × 82 = − 18 12
= + 18
3- Two unknown θB + θC then two static equations are required.
1) ∑ MB = 0 2)
MC = 0
Hence: MBA + MBC = 0 ………………
(1)
MBC
= 0……………….
(2)
But: MAB = - 9 + θB MBA =
9
+ 1 (2θB)
MBC = -16 + 2 (2θB + θC) MCB = +16 + 2 (2θC + θB) B
From eqns. (1&2) 9 + 2θB + (- 16 + 2 (2θB + θC) = 0 B
B
6θB + 2θC
=7
B
and
4θC + 2θB
= - 16
2θC + θB
=-8
B
…….(3) …….. (4)
from 3 & 4 5 θB B
1.e.
= 15 15 = 3.0 5
θB
=
θC
= - 5.5
MAB = - 9 + 3.4 = 5.6 t.m MBA = 9 + 2 × 3.4 = 15.8 t.m MBC = - 18 + 2 (2 × 3.4) + (- 5.5) = - 15.0 t.m MCB = 16 + 2 (2.3 – 5.7 + 3.4)
= 0.0 (0.k)
1- Unknowns θA , θB , & θC 2- Fixed end Moment MFAB = MFBC = MFCD =
2 × 62 = - 6 t.m … etc 12
3- Condition eqns. MAB = - 4 t.m, MBA + MBC = 0, & MCB + MCD = 0
4- Slope deflection equations MAB = - 6 + (2θA + θB) = - 4 B
2θA + θB
=2
MBA + MBC
=0
B
……….(1)
+ 6 + (2θB + θA) –6 + (2θB + θC) = 0 4θB + θA + θC
=0
MCB + MCD
=0
……… (2)
= 6 + 2θC + θB – 6 + 2θC = 0 B
4 θC + θB
=0
B
……….(3)
From eqn.3 θC = - θB 4
Substitute in eqn. (2) Hence:
3.75 θB + θA = 0 B
0.5θB + θA B
=1
3.25θB
=-1
θB
=-1
θA
= 1.15
θC
= 0.077
B
B
................(2) ………… (2)
Hence: MAB
= - 6 + 2xl.15 + (- .307) = - 4 t.m
0.K
MBA
= 6 + 2x (- .307) + 1.15 = 6.536 t.m
MCB
= 6 + 2x .77 + (- .307)
= 5.85 t.m
MDC
= 6 + .077
= 6.077 t.m
Solution:1- Unknown displacements are θB & θD B
2- Equations of equilibrium are:MDB = 0
……….(1)
MBA + MBD + MBC = 0
………(2)
3- Relative Stiffness:KAB: KBC: KBD = 35:31. 5:22 ; 51. 56:1. 4:1.0.
4- Fixed and Moments: MFAB =
− 9 × 6 × 3× 3 = − 6 t.m 9×9
MFBA =
9 × 6 × 3× 6 =12 t.m 9×9
MFBD =
− 3× 7 2 = − 12.25 t.m 12
MFDB =
− 3× 7 2 = − 12.25 t.m 12
From the equations 1 & 2 hence; MDB = MFDB + (2θD + θB) B
= 12.25 + 1 (2θD + θB)
=0
B
2θD + θB + 12.25
= 0 ------- (3)
B
and
MBA
= 12 + 1.56 (2θB)
MBD
= 12.25 + 1.0 (2θB + θD)
MBC
= 0 + 1.4
B
B
(2θB + 0) B
i.e. 12+1.56 (2θB) – 12.25 + 2θB + θD + 1.4 (2θB) = o B
B
B
7.92θB + θD – .25
= 0 ------------- (4)
0.5θB + θD + 6.125
= 0 ------------- (3)
B
B
i.e
7.42 θB
- 6.375
θB
= 0.86
θD
= - 6.55
B
B
=o
Hence: MBA = 12 + 1.56 (2× .86) = 14.68 t.m
MBD = - 12.25 + (2 × .86 – 6.55 × 1) = - 17.08 MBC = 1.4 (2 × .86)
= 2.41
MDB = 12.25 + (2 × -6.55)
= zero
1 MBC 2
= 1.205
MAB = -6 + 1.56 (.86)
= - 4.66
MCB =
Two equilibrium eqns. MAB + MAA
= 0
………………… (1)
MBB + MBA + 4= 0
………………….(2)
Slope deflection eqns. MAB = o + 1.6 (2θA + θB) B
− 10 ×16 + (2θ + θ ) A A 8
MAA =
MAA = - 20 + θA MBA = o + 1.6 (2θB + θA) B
MBB = - 42.67 + (2θB + θB) B
B
= - 42.67 + θB B
Hence: 3.2θA + 1.6θB + θA – 20 = o B
4.2θA + 1.6θB
= 20
B
……………… (1)
- 42.67 + 4.2 θB + 1.6θA + 4 = 0 B
1.6θA + 4.2θB
= 38.67 ……………….(2)
B
1.6θA + 0.61θB
= 7.62 ……………….(1)
B
3.59θB
= 31.05
θB
= 8.65
θA
= 1.46
MAB
= -18.52
B
B
MBA
= 30
MBB
= - 34
Example 5 Draw B.M.D for the shown frame Solution:- Two condition equations. MAA + MAB
=0
…….. …………...(1)
MBA + MBB + 8 = 0
…………………..(2)
- Relative stiffness
1 1 : = 1:1.6 16 10
- Slope deflection equations: MAA = (2θA – θA) = θA MAB = (2θA – θB) × 1.6 B
MBA = (2θB – θA) × θA B
MBB = 42.67 + (2θB - θB) B
B
Hence: θA + 3.2θA + 1.6VB = 0 B
4.2θA + 1.6θB
= 0 .… (1)
B
3.2θB + 1.6θA + θB – 42.67 + 8 = 0 B
B
4.2θB + 1.6θA B
= 34.67… (2)
By Solving 1 & 2 Hence
θA = - 3.68 , θB = 9.66 B
MAA = - 3.68 , MAB = 3.68 t.m MBA = 25
MBB = 33
Example 6: - Draw B.M.D for the given structure.
Solution:- once statically indeterminate. 1- Fixed end moments MFAB = -
8 × 20 = − 20 t.m 8
MFBA = -
8 × 20 = − 20 t.m 8
MFBC = -
4 × 10 = − 5 t.m 8
MFCB = -
10 × 8 = −10 t.m 8
MFDB 2- From Static:-
= 10 t.m ∑ MB = o B
MBA + MBC + MBD = 0 MBA = MFBA + (2θB) B
MBA = 20 + 2θB
……………….. (1)
B
MBC = - 5 + 2θB
……………….. (2)
B
MBD = - 10 + 2θB B
Hence:
………………. (3)
5 + 6θB = o B
θB B
= - o.833
Hence: MBA = 18.34 t.m , MBC = -6.67, MBD = -11.67 t.m MAB = - 20
= - 20.833 t.m
MCB = 5 + θB
= - 4.167 t.m
MDB = 10 + θB
=
Example 7:
9.167 t.m
Draw B.M.D for the shown frame Solution: “ 3 time statically ind.” θA , θB , & θC B
1- Fixed end moments: MFAB = - 10 MFBA = + 10 MFBC = - 25
MFCD = MFDC = zero 2- Relative Stiffness MAB
1:1:1
=0
…………………….(1)
MBA + MBC = 0
…………………….(2)
MCB + MCD = 0
…………………….(3)
Equs. MAB
= - 10 + (2θA + θB)
MBA
= 10 + (2θB + θA)
MBC
= - 25 + 2θB + θC
MCB
= 25 + 2θC + θB
MCD
= 2θC
MDC
= θC
B
B
B
B
From 1,2 & 3 2θA + θB
= 10
…………….(1)
4θB + θA + θC = 15
…………….(2)
B
B
4θC + θB B
= - 25
……………(3)
By solving the three eqns. hence; θA = 2.5
θB = 5 B
θC = - 7.5
Substitute in eqns of moments hence; MAB = - 10 + 5
= zero (o.k)
MBA = 10 + 10 + 2.5
= 22.5 t.m
MBC = - 25 +10 – 7.5
= - 22.5 t.m
MCB = 25 – 15 + 5
= 15 t.m
MCD
= - 15
t.m
MDC
= - 7.5 t.m
3-6 Frames with Translation Examples to frames with a single degree of freedom in translation.
Example 8: Draw B.M.D for the shown frame. 1- Unknowns: θB , θC , Δ
2- Relative stiffness KAB : KBA : KCD 1 2 1.5 : : 4 8 6
1: 1 : 1 3- Fixed end moments MFAB = o MFBC = MCB = zero MFCD = - 6 t.m MFDC = + 6 t.m
MBA = o
4- From Statics the equilibrium eqns MBA + MBC = 0
…………………(1)
MCB + MCD = 0
………………….(2)
5- Shear equation (In direction of X, ∑ × = o) 6 + XA + XD – 8 6+ hence
XA +
=o
M + M DC M BA + M AB + CD − 4 = 0 (3) 4 6
MBA + MAB MCD + MDC and xD = +4 4 6
6- Slope deflection eqns: MBA – 0 + 1 (2θB – 3 B
Δ Δ ), MAB = 0 + 1 (θB – 3. ) 4 4 B
MBC = 0 + 1 (2θB + θC) B
Hence:
4θB – 0.75Δ + θC = 0
(1)
B
MCB = 0 + 1 (2θC + θB) B
MCD = - 6 + 1 (2θC – 3 MDC = + 6 + 1 (θC-3
Δ ), 6
Δ ) 4
Hence: 4 θC + θB B
2+
1 Δ =6 2
(2)
(2θ B − .75 Δ ) + (1θ B − .75 Δ ) (− 6 + 2θ C − Δ ) + (6 + 1θ C − + 4 6
2 + 0.75 θB - .375 Δ + B
Δ
=0
1 θC – 0.1667Δ = o 2
θB + .67 θC - 072Δ = - 2.66 B
(3)
Subtract (3) from (2) 1 2
3.33 θ + 0.288 Δ = 8.33 θC – 0.067 Δ = 2.6
(4)
Subtract (1) from (2) × (4) 15 θC – 1.25 Δ
= 24
θC – 0.08 Δ
= 1.6
From (4) & (5)
0.147Δ = 1 Δ = 6.80 θC = 2.149 θB = 0.799 B
(5)
MBA = - 3.5 t.m
,MAB = - 4.301 t,m, MBC = 3.79
MCB = 5.1 t.m
,MCD = - 5.1 t.m ,
MDC = 4.744
Example 9:Write the shear & condition eqns for the following frame.
Solution:Three unknowns:
θB , θC, Δ
Condition equations: MBA + MBC = o
(1)
MCB + MCD = o
(2)
Shear eqn. XA +XB + P1 + P2 = o B
(−
M + M DC P1 M AB + M BA ) + ( CD ) P1 + P2) = o + 2 h1 h2
Example 10: Find the B.M.D for the shown structure.
(3)
Solution:θD = θE = o θC = - θC θB = - θB B
B
1- Unknown displacements are: 2- Relative Stiffness: AB : BE : BC : CD : ED 1 2 1 1 1 : : : : 5 3 5 3 3
3 : 10: 3 : 5: 5
3- Fixed end moment:-
θB , θC , Δ
MFBE = -
4 × 36 = − 12 t.m 12
MFEB = + 12 t.m MFCD =
1.5 × 36 = − 4.5 t.m 12
MFDC = + 4.5 4- Equilibrium equations:1- MCD + MCB = o 2- MBC + MBA + MBE = o 3- Shear condition:(33–16.5)+
M CD + M DC M DE + M ED + 6 6
MCD = - 4.5 + 5 (2θC + θD – 3R) MCB = 0 3 (2θC + θB) B
MBC = 0 + 3 (2θB + θC) B
MBA = 0 + 3 (2θB) B
MBE = - 12 + 10 (2θB – 3R) B
Hence - 4.5 + 10 θC – 15 R + 6 θC + 3 θB = 0 B
16θC + 3θB – 15R – 4.5 = 0 B
(1)
And 16θB + 3θC + 6θB – 12 + θB - 3θR = 0 B
B
B
3θC + 32θB – 30R – 12 = 0 B
and 16.5(
15θ C − 30 R 30 θ B − 60 R + )=0 6 6
(2)
2.5 θC + 5θC + 17R + 16.5
=0
(3)
by solving equation 1,2 & 3 get MAB = + 6.66 t.m MBA = + 13.32 t.m MBC = + 19.0
t.m
MCB = + 18
t.m
MBE = - 32.32 t.m MEB = - 30.53 t.m MCD = - 18
t.m
MDC = - 18.43 t.m
3-7 Frame with multiple degree of freedom in translation. Example 11: Write the shown equations and condition eqns for the given frame. Solution Unknowns:
θB , θC , θD , θE , ∆1 , ∆1
Condition eqns MBE + MBA + MBC
=0
MCB + MCD
=0
(2)
MDC + MDE
=0
(3)
MEB + MEF + MED = 0
(4)
(1)
Shear eqns : Equilibrium of the two stories. At sec (1) – (1) :(Level CD) P2 + Xc + XE P2 +
=0
M CB + M BC M + M ED =0 + DE h2 h2
At sec. (2) – (2):(Level BE) or ∑ x = 0 P 1 + P2 + x A + xF = 0 P 1 + P2 +
M BA + M AB M + M FE =0 + EF h1 h1
Example 12:Draw B.M.D for the given structure.
S olution:1- Relative Stiffness:-
2- Equilibrium equations:MAB + MAC
=0
(1)
MBA + MBD
=0
(2)
MCA + MCD + MCE = 0
(3)
MDB + MDF + MDC = 0
(4)
Σx = 0 at Level A-B 2 + (6-3) +
M AC + M CA M + M DB = 0 (5) + BD 6 6
Σx = 0 at Level CD 11 +
M CE + M EC M + M FD + DF 6 6
= 0 (6)
MAB = - 8 + 1 (2θA + θB) B
MAC = 3 + (2θA + θC – 3R1) MAC = - 3 + (2θC + θA – 3R1) MCA = 16 + (2θB + θA) B
MBD = 0 + (2θB + θD – 3R1) B
MDB = 0 + (2θD + θB - 3R1) B
MDF = 0 + (2θD + 0 - 3R2) MFD = 0 + (θD -
3R2)
MCD = - 48 + 2 (2θC + θC) MCD = + 48 + 2 (2θD + θC) MCE = - 8 + (2θC - 3R2) MEC = + (θC - 3R2) 3- Fixed end moment:MFAB = −
9× 4×8 × 4 = − 8 t.m 12 ×12
MFBA= −
9×82 × 4 = + 16 t.m 12 2
MFAC =
1× 6 2 = + 3 t.m 12
MFCA=
1× 6 2 = − 3 t.m 12
MFCD =
4 ×12 2 = − 48 12
t.m
= + 48
t.m
MFDC
4- Unknown displacement: θA , θB , θC , θD , ∆1 , ∆2 by Solving the six equations one can get; MAB
= - 3.84
t.m
MBA
= + 18.39
t.m
MAC
= 3.84
t.m
MCA
= + 7.29
t.m
MBD
= - 18.39
t.m
MDB
= - 22.97
t.m
MCD
= - 11.15
t.m
MDC
= - 53.44
t.m
MCE
=
t.m
MEC
= - 13.44
t.m
MDF
= - 30.47
t.m
MFD
= - 26.15
t.m
3.87
Example (13):Write the shear equations & equilibrium equations for the shown frame. Solution: Shear eqns: XCE + XBA + P1
= 0 ……………………....(1)
M EC + M CE M + M BA + AB + P1 = o h1 h1 + h2
xD + xG + xE + P2 = 0 --- (2) M DE + M ED h2
+
M EC + M CE M GF + M FG − + P2 = 0 h2 h1
Or: XA + XD + XG + P1 + P2
=0
M + M FG M AB M + M ED + DE + GF + P 1 + P2 = 0 h1 + h2 h2 h2
Example 14:a- Write the equations of equilibrium including the shear equations for the frame. b- Write the slope deflection equations in matrix for members CE & GH. c- By using the slope - deflection method; sketch elastic curve. d- Sketch your expected B.M.D
Solution:-
(Unknowns = θC , θD , θE , θF , θG , θA+θK , θL , ∆1 , ∆2 , ∆3 , ∆4 Relative stiffness: 1 : 1 a- equilibrium equations MKL + MKG
=0
(1)
MLK + MLH
=0
(2)
MGK + MGH + MGE = 0
(3)
MHG + MHL + MHF = 0
(4)
MEG + MEC + MFF = 0
(5)
MFE + MFD + MFH = 0
(6)
MCE + MCD + MCA = 0
(7)
MDC + MDB + MDF = 0
(8)
Shear equations:a- at Level GH 5 + 10 + (XG – 5) + XH = 0
(9)
Where: XG = XH =
M GK + M KG 5 M HL + M LH 5
b- at Level EF 5 + 10 + 20 + (XE – 5) + XF
=0
30 + XE + XF
=0
Where: XE = XF =
M EG + M GE 5 M FH + M HF 5
(10)
c- at Level CD 5 + 10 + 10 + 30 + (XC – 5) + XD
=0
50 + XC + XD
= 0 ………(11)
Where: XC =
M CE + M EC 5
XD =
M DF + M FD 5
d- at Sec AB:5 + 10 + 10 + 10 + 40 + (XA – 5) + XB B
=0
70 + XA + XB
= 0 …..(12)
B
XA = XB =
M AC + M CA 5 M CE + M EC
B
5
3-8 Slope deflection eqns in matrix form: 1- Member CE MCD = MFCE +
Δ 2 − Δ1 2EI (2θC + θE –3 ) 5 5
MEC = MFEC +
Δ 2 − Δ1 2EI (2θE + θC –3 ) 5 5
Where: 2× 52 MFCE = 12
= - 4.16
t.m
MFEC
= + 4.16
t.m
In Matrix form: MCE
- 4.16 +
= MEC
4.16
Where: R2 =
Δ 2 − Δ1 5
2
1–3
2EI 5
θC θE
1
2–3
R2
2- member GH MGH
- 16.67 +
=
1
θG
1
2
θH
26I 5
+ 16.67
MHG
2
d- B.M.D
Example 15:By using slope deflection method; 1- Draw B.M.D for the shown frame. 2- Sketch elastic curve. Solution: 1- Relative stiffness 1: 1 2- unknowns:
θB = - θB (From symmetry) B
B
3- Equilibrium eqns MBA + MBC + MBD + MBB = o
(1)
4-Fixed end moments MFAB =
4× 62 12
MFBA =
= - 12 t.m = + 12
MFBC = MFCB = MFBD = MFDE = o MFBB = -
2 ×12 2 8 ×12 − 12 8
4- Slope deflection eqns MAB = - 12 +(θB) MBA = 12 + 2θB
= - 36 t.m
MBC = 2θB MBD = 2θB MBB = - 36 + θB MCB = θB MDB = θB
From eqn (1) (12 + 2θB) + (2θB) + (2θB) + (- 36 + θB) B
B
B
7θB - 24
B
=0 =0
B
θB
= 3.4286
B
hence MAB = - 8.57
t.m
MBA = 18.86
t.m
MBC = 6.86
t.m
MBB = - 32.58
t.m
MCB = 3.428
t.m
MDB = 3.428 MBD =
6.86
t.m
The Free Body Diagram to find the S. F. & N. F.
SHEET (3) 1) Draw S.F.D. and B.M.D. for the statically indeterminate beams shown in figs. From 1 to 10.
2) Draw N.F.D., S.F.D. for the statically indeterminate frames shown in figs. 11 to 17. Using matrix approach 1.
THE STIFNESS MATRIX METHOD: APROACH 1 ( SLOP DEFLECTION IN MATRIX FORM ) In this method, the slop deflection equations method is formulated using matrix operations. The method can be formulated for weather the members have relative displacements or not Members without relative displacements In this approach, the slop deflection equations for each member are collected in a matrix form. For example, the slop deflection equations for member AB in a matrix form become
⎡ 4 EI ⎡M AB ⎤ ⎡M FAB ⎤ ⎢ L ⎢M ⎥ = ⎢M ⎥ + ⎢ 2 EI ⎣ BA ⎦ ⎣ FBA ⎦ ⎢ ⎣ L
2 EI ⎤ L ⎥ ⎡θ AB ⎤ 2 EI ⎥ ⎢⎣θ BA ⎥⎦ ⎥ L ⎦ AB
(4.108)
In a short form, one may write equation 4.108 as MAB = MFBA + SAB DAB
(4.109)
For a structure consists of a set members, Equation 4.109 can be repeated For each member. The structure shown in Figure 4.76, for example, the slope Deflection equation are collected in a matrix form as
C
B
E
A
D
Figure 4.76 ⎡ M AB ⎢M ⎢ BC ⎢ M AC ⎢ ⎢ M CD ⎢⎣ M CD
⎤ ⎡ M FAB ⎥ ⎢M ⎥ ⎢ FBC ⎥ = ⎢ M FAC ⎥ ⎢ ⎥ ⎢ M FCD ⎥⎦ ⎢⎣ M FCD
⎤ ⎡ S AB ⎥ ⎢ 0 ⎥ ⎢ ⎥+⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥⎦ ⎢⎣ 0
0 S BC
0 0
0 0
0 0 0
S AC 0 0
0 S CD 0
0 ⎤ ⎡ D AB 0 ⎥⎥ ⎢⎢ D BC 0 ⎥ ⎢ D AC ⎥⎢ 0 ⎥ ⎢ D CD S CE ⎥⎦ ⎢⎣ D CD
⎤ ⎥ ⎥ ⎥ (4.110) ⎥ ⎥ ⎥⎦
Equation 4.110 can be written for any structure whose members are not subjected to relative displacements. It can be expressed in a general form as M m = M Fm + [S m ]D m (4.111)
In which M m member end moments, MFM contains the members fixed end moments And [Sm] is the augmented member's stiffness matrices The compatibility condition for the frame of figure 4.76 can be stated as Follows:
θ
AB
= θ
AC
= θ
A
θ BA = θ BC = θ B
θCA =θCB =θCD =θCE =θC θ DC = θ D (4.112) Where θA , θB , θC ,and θD are the cinematic variables at the joints of this structure . Equation 4.112 can be expressed in a matrix form as
⎡θ AB ⎤ ⎡1 ⎢θ ⎥ ⎢ ⎢ BA ⎥ ⎢0 ⎢θ BC ⎥ ⎢0 ⎥ ⎢ ⎢ θ CB ⎥ ⎢0 ⎢ ⎢θ ⎥ ⎢1 ⎢ AC ⎥ = ⎢ ⎢θ CA ⎥ ⎢0 ⎢θ ⎥ ⎢0 ⎢ DC ⎥ ⎢ ⎢θ CD ⎥ ⎢0 ⎥ ⎢ ⎢ ⎢θ CE ⎥ ⎢0 ⎢⎣θ EC ⎥⎦ ⎢⎣0
0 0 0⎤ 1 0 0⎥⎥ 1 0 0⎥ ⎥ 0 1 0⎥ ⎡θ A ⎤ ⎢ ⎥ 0 0 0⎥ ⎢θ B ⎥ ⎥ 0 1 0⎥ ⎢θ C ⎥ (4.113) ⎢ ⎥ 0 0 1⎥ ⎣θ D ⎦ ⎥ 0 1 0⎥ 0 1 0⎥ ⎥ 0 0 0⎥⎦
Equation 4.113 can be expressed in a short form by ( 4.114 ) Dm = C D Where C is called the compatibility or connectivity matrix. In order to solve the problem one has to apply the equilibrium conditions at the free joints. For the structure of figer 4.76 one has the following equilibrium condition: MA=MAB + MAC MB=MBC + MBA MC=MCB + MCD+MCD+MCE MD=MDC Equation 4.115 can also be expressed in a matrix form as
⎡M A ⎤ ⎡1 ⎢ ⎥ ⎢ ⎢ M B ⎥ = ⎢0 ⎢ M C ⎥ ⎢0 ⎢ ⎥ ⎢ ⎣ M D ⎦ ⎣0
0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 1 0 0
⎡M AB ⎤ ⎢M ⎥ ⎢ BA ⎥ ⎢M BC ⎥ ⎢ ⎥ 0⎤ ⎢M CB ⎥ 0⎥⎥ ⎢M AC ⎥ ⎢ ⎥ 0⎥ ⎢M CA ⎥ ⎥ 0⎦ ⎢M DC ⎥ ⎢ ⎥ ⎢M CD ⎥ ⎢ ⎥ ⎢M CE ⎥ ⎢⎣M EC ⎥⎦
(4.116)
Comparing equation 4.116 with equation 4.113, One may write equation 4.116 (4.117) M=CT Mm Where M is a vector which contains the external moments at the free joints Substituting equation 4.111 and 4.114 into equation 4.117 one obtains M=CT MFm+CT[Sm] C D
(4.118)
Equation 4.118 can also be written in a shorter form as M=MF + [S] D
(4.119)
Where MF and [S] are, respectively, obtained from MF =CT MFm
(4.120)
[S] = CT [Sm] C
(4.121)
Equation 4.119 is now in a suitable form to be solved for the unknown displacements D as follows D = [S]-1(M-MF )
(4.122)
It is a common practice to call the terms (-MF) in equation 4.122 by the equivalent joint moment due to the direct loadings on the members. It is obvious that this matrix approach is suitable for computer applications. The analyst stories the structural data in the form of [Sm], the loading data in the form of M and MFm, and finally the compatibility matrix C which is obtained from boundary and connectivity conditions. The matrices MF and S are then calculated by matrix multiplication according to equations 4.120 and 4.121. The free displacements D are obtained according to equation 4.122. The member end moments are determined from equation 4.111 using equation 4.114
Example 4.26 Determined the bending moment diagram for the beam shown in figure 4.77 using the stiffness matrix method – abrosh I (EI=105 KN.m2 ,α=10-5/ οc )
50KN
A
C
B
10KN/m
Figure 4.77 Solution The stiffness matrices SAB and SBC, are determined as follows:
S AB = S BC
⎡ 4 EI ⎢ =⎢ 5 2 EI ⎢ ⎣ 5
2 EI ⎤ 5 ⎥ = EI ⎡0.80 0.40⎤ ⎢0.40 0.80⎥ 4 EI ⎥ ⎦ ⎣ ⎥ 5 ⎦
The members stiffness matrix [Sm] is composed as follows:
0⎤ ⎡0.8 0.4 0 ⎢0.4 0.8 0 0 ⎥⎥ [S m ] = EI ⎢ ⎢0 0 0.8 0.4⎥ ⎢ ⎥ 0 0.4 0.8⎦ ⎣0 The degree of freedom in this structure is two, which represents the angles of Rotation at B and C . The compatibility equations are put in a matrix form as
⎡θ AB ⎤ ⎡0 ⎢θ ⎥ ⎢ ⎢ BA ⎥ = ⎢1 ⎢θ BC ⎥ ⎢1 ⎥ ⎢ ⎢ ⎣θ CB ⎦ ⎣0
0⎤ 0⎥⎥ ⎡θ B ⎤ = CD 0⎥ ⎢⎣θ C ⎥⎦ ⎥ 1⎦
Figure 4.78
The fixed end moments are determined for the cases shown in Figure 4.78 using the Tables and according to the signs of the slope deflection equation as follows: M FAB =
4 EI 50 * 5 θA + = 191.25 KN .m 5 8
M FBA =
2 EI 50 * 5 θA − = 48.75 KN .m 5 8
M FBC = EIα
T1 − T2 = 50 KN .m h
M FBC = −50 KN.m The fixed end moments are collected to form matrix MFm as follows: T M Fm = [191.25 48.75 50 -50] The matrix MF and [S] are obtained according to equations 4.120 and 4.121 as Follows: ⎡0 1 1 0 ⎤ ⎡98.75⎤ T M F = C M Fm = ⎢ M Fm = ⎢ ⎥ ⎥ ⎣0 0 0 1 ⎦ ⎣− 50 ⎦
[S ] = C T [S m ]C = EI ⎡⎢
1.6 ⎣ 0 .4
0.4 ⎤ 0 . 8 ⎥⎦
Substituting into the equilibrium equation one obtains
M=MF + [S] D
⎡M B ⎤ ⎡98.75⎤ ⎡1.6 0.4⎤ ⎡θ B ⎤ ⎡0 ⎤ = + E ⎢M ⎥ ⎢− 50 ⎥ ⎢0.4 0.8⎥ ⎢θ ⎥ = ⎢− 20⎥ ⎦ ⎦ ⎣ ⎦⎣ C ⎦ ⎣ ⎣ C⎦ ⎣ The deformation D is solved to have
⎡θ B ⎤ 1 ⎡− 81.25⎤ D=⎢ ⎥ ⎢ ⎥ ⎣θ C ⎦ EI ⎣78.125 ⎦ The members end moments are calculated from equation 4.111as follow:
M m = M Fm + [S m ]D m ⎤ ⎡0 0 ⎤⎢ ⎡M AB ⎤ ⎡191.25⎤ ⎡0.8 0.4 0 ⎥ ⎡158.75 ⎤ 81 . 25 − ⎢M ⎥ ⎢ ⎢ ⎥ EI ⎥ ⎢− 16.25⎥ 0 ⎥⎥ ⎢ ⎢ BA ⎥ = ⎢48.75 ⎥ + EI ⎢0.4 0.8 0 ⎥ KN .m ⎥=⎢ ⎢ 81 . 25 − ⎢M BC ⎥ ⎢50 ⎢ ⎥ ⎢0 ⎥ 16.25 ⎥ 0 0.8 0.4 ⎢ EI ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ⎥ 0 0.4 0.8⎦ 78.125 ⎥ ⎣− 20 ⎦ ⎣0 ⎣M CB ⎦ ⎣− 50 ⎦ ⎥ ⎢ EI ⎦ ⎣ Which gives the same results obtained previously. The bending moment diagram is Given again Figure 4.79 Example 4.27
Determine the bending moment diagram for the frame shown in Figure 4.80 using the Stiffness matrix method approach I, where Support B has displaced down 1cm, and EI= 105 KN.m2 Solution
The stiffness matrices SAC and SCB are determined as follows:
1 5 8 .7 5 K N .m
2 5 K N .m
1 6 .2 5 K N .m
Figure 4.79
2 0 K N .m
20KN
B
C 50KN
A Figure 4.80
S AC
⎡ 4 EI ⎢ =⎢ 6 2 EI ⎢ ⎣ 6
2 EI ⎤ 6 ⎥, 4 EI ⎥ ⎥ 6 ⎦
S CB
⎡ 4 EI ⎢ =⎢ 8 2 EI ⎢ ⎣ 8
2 EI ⎤ 8 ⎥ 4 EI ⎥ ⎥ 8 ⎦
The member's stiffness matrix [Sm] is obtained form
[S m ] = ⎡⎢
S AC ⎣ 0
0 0 ⎤ ⎡0.667 0.333 ⎢ 0 ⎤ 0 0 ⎥⎥ ⎢0.333 0.667 EI = ⎢ 0 S CB ⎥⎦ 0 0.5 0.25⎥ ⎥ ⎢ 0 0.25 0.5 ⎦ ⎣ 0
The degree of freedom in this structure is two, which represents θC and θB . The compatibility equations are θAC =0, θCA = θC, and θBC = θB . These relations are Put in a matrix form as follows: B
⎡θ AB ⎤ ⎡0 ⎢θ ⎥ ⎢ ⎢ CA ⎥ = ⎢1 ⎢θ CB ⎥ ⎢1 ⎥ ⎢ ⎢ ⎣θ BC ⎦ ⎣0
0⎤ 0⎥⎥ 0⎥ ⎥ 1⎦
⎡θ C ⎤ ⎢θ ⎥ = C D ⎣ B⎦
The fixed end moments are determined for each member using the tables for the cases Shown in Figure 4.81, and in the singes of the slope deflection equation. 50 * 6 = 37.5 KN.m 8 MFAC = -37.5 KN.m M FAC =
50KN
C
C
B
50 KN
A Figure 4.81
M FCB = +
20 * 8 6EI ⎛ 1 ⎞ + 2 ⎜ ⎟ = 113.75 KN.m ; M FBC = −20 + 9375 = 73.75 KN.m 8 8 ⎝ 100 ⎠
The calculations of MF and S follow equations 4.120 and 4.121 as follows: M
T FM
= [37.5 − 37.5 113.75 73.75]
⎡0 1 1 0⎤ ⎡76.25⎤ T M F = C M FM = ⎢ M FM = ⎢ ⎥ ⎥ KN .m ⎣0 0 0 1 ⎦ ⎣73.75⎦ 0 ⎤ ⎡ 0.333 ⎢0.667 0 ⎥⎥ ⎡1.167 0.25⎤ = EI ⎢ [S] = C T [S m ] C = EI⎢ ⎥ ⎢ 0.5 0.25⎥ ⎣ 0.25 0.5 ⎦ ⎥ ⎢ ⎣ 0.25 0.5 ⎦ Substituting into Equation 4.119, one obtains
⎡M C ⎤ ⎡76.25⎤ ⎡1.167 0.25⎤ ⎡θ C ⎤ ⎡0⎤ ⎢M ⎥ = ⎢73.75⎥ + EI⎢ 0.25 0.5 ⎥ ⎢θ ⎥ = ⎢0⎥ ⎦ ⎣ ⎦⎣ B ⎦ ⎣ ⎦ ⎣ B⎦ ⎣ The solution for θC and θB gives
⎡θ C ⎤ 1 ⎡− 37.788 ⎤ ⎢θ ⎥ = ⎢ ⎥ rad ⎣ B ⎦ EI ⎣− 128.606⎦ The end moments are now calculated using Equation 4.111 as follows:
M m = M FM + [S m ] D m
⎤ ⎡0 0 0 ⎤⎢ ⎡M AC ⎤ ⎡37.5 ⎤ ⎡0.667 0.333 ⎥ ⎡24.916 ⎤ 37 . 788 − ⎢M ⎥ ⎢ ⎥ ⎢ ⎥ EI ⎥ ⎢− 62.704⎥ 0 0 ⎥⎢ ⎢ CA ⎥ = ⎢− 37.5 ⎥ + EI ⎢ 0.333 0.667 ⎥ KN.m ⎥=⎢ ⎢ ⎢M CB ⎥ ⎢113.75⎥ ⎢62.704 ⎥ ⎢ 0 0 0.5 0.25⎥ ⎢− 37.788 ⎥ EI ⎥ ⎢ ⎢ ⎥ ⎥⎢ ⎢ ⎥ ⎥ ⎢0 0 0 0 . 25 0 . 5 73 . 75 M ⎦ ⎦ ⎣ ⎦ ⎣ ⎣ BC ⎦ ⎢− 128.606 ⎥ ⎣ EI⎦ ⎣ The bending moment diagram is given in Figure 4.82, which is the same as the Results of Example 4.7.
BMD Figure 4.82 4.6.2 Effect of Members with Moments Releases if the moment in a member is released at one end, one may use the modified Slope deflection equation, given in Equation 4.15 For any member AB which has a hinge at B, one may write Equation 4.15 in a matrix form as follows:
SHEET (4) Draw the bending moment diagram for the following structures using the matrix approach 1 method (EI=104 t.m2 ,α=10-5/ οc )
CHAPTER 3 Stiffness Matrix Method 3-1- DEFINITION The stiffness method is a method of analysis, where the main unknowns are the displacements of joints. These unknowns are determined from equilibrium. The method can be used for determination of displacements and internal forces due to • external loads, • environmental changes (temperature and shrinkage), and • support movement. The stiffness method is applicable to skeletal structures (beams, plane and space frames and trusses, and grids), and continuum structures (plates, shells, and three dimensional solids). (Question: what is the difference between skeletal and non-skeletal structures (continuum structures)?) The steps of this method are: 1- Modeling • Model the structure with a number of members and joints, • Define types of members and connections, • Identify the relevant unknown components of joint displacements. 2- Load vector • For each member determine the member fixed end forces (which are the reactions due to loads if all the relevant joint displacements were prevented). 3- Stiffness matrix • For each member determine the member end forces due to joint displacements (i.e. member stiffness equations). 4- Equilibrium equations • Write the equations of equilibrium of all joints (under the forces in steps 2 and 3, and joint loads). 5- Solving equilibrium equation
• Solve equilibrium displacements.
eq.
to
determine
joints
6- Internal forces • Use the resulting joint displacements to determine the total member end forces. The above steps can be illustrated by the following example:
EXAMPLE Determine the displacements and member end forces in the shown frame due to the given loads.
3t 2t/m
3.00
2.00
6.00
STEP 1: Modeling.
D2 D1
D3 2
D4 3
1 1
2
STEP 2: Load vector Fixed-end forces {F}
2t/m
6
6
0
0
6
6
STEP 3: Stiffness matrix Stiffness relations [K]
STEP 4: Equilibrium STEP 5: Equations solution STEP 6: Internal forces
{F} = [K] {D} Get {D}
Q, N, and M
3-2- STEP 1: MODELING a- Numbering Organization of the solution requires numbering the members, joints, and forces and displacements components. Several numbering schemes are possible. Choice of the most appropriate scheme depends on several considerations including the method of assembling and solving equations of equilibrium (banded matrix methods, frontal method, etc.). When banded matrix methods are used for solving equilibrium equations, members may be numbered in any convenient order, but joints should be numbered in such an order that the maximum difference between the numbers of the two joints of each member is as small as possible. For example, the numbering in Fig. Is more appropriate than that in Fig. The unknown displacement components are usually numbered in a certain sequence (for example in plane frame; x, y, and rotation) starting at joint 1 and proceeding in ascending order through the joints. These displacement components are called degrees of freedom.
b- Degrees of Freedoms Several models of skeletal structures may be used depending on the nature of the structure and the loads. Common skeletal models include space frames, space trusses, plane frames, plane trusses, beams, and grids. * In space frames, each joint has six degrees of freedom: three translations (in X, Y, and Z directions) and three rotations (about X, Y, and Z axes). * In space trusses, all member ends are assumed to have hinged connections, i.e. three degrees of freedom at each joint (rotation about X, Y, and Z axes). * In plane frames, a joint has three degrees of freedom since it has two translations and one rotation. * In plane trusses, a joint has two degrees of freedom (the translation only) since rotation are not considered. * In beams, axial and transverse forces and displacements are uncoupled, and separate analyses may be carried out for axial effects and transverse effects. * In grids, in-plane displacements are not considered; therefore, also rotations about an axis normal to the plane of the grid are not considered. Thus, a grid joint has only three degrees of freedom (an out-of-plane translation, and two rotations about in-plane axes).
20
19
18
17
16
15
14
13
12
11
1
2
3
4
5
6
7
8
9
10
5 4
6 7 1
3 2
6
7
18 19
5
8
17 20
4
9
16 21
3
10 15 22
2
11 14 23
1
12 13 24
2
3
6
7
10
11
14
15
18
19
1
4
5
8
9
12
13
16
17
20
5 6
3 2 1
7 4
21
22 23 24
17
18 19 20
13
14 15 16
9
10 11 12
5
6
7
8
1
2
3
4
c- Local and Global Axes The axes which are convenient in dealing with members individually are called local (member) axes, but the axes which are convenient in dealing with the structure as a whole are called global (structure) axes. Displacement and force components may be expressed using one of the previous two systems. The relation between the components in the two systems of axes is expressed in matrix form which called transformation matrix [T].
Y
Y
End 2 ( X2,Y2,Z2)
End 2 End 1 Z ( X1,Y1,Z1)
( X2,Y2) Length =L
X
End 1
Y
( X1,Y1)
X
X
Plane or grid member
Space member
Z
a) Geometry g g f5 d5
l l f5 d5
g g f4 d4
g g f6 d6
g g f2 d2
Global
g g f1 d1 g g f3 d3
l l f4 d4
l l f6 d6 l l f2 d2 l l f1 d1
Local l l f3 d3
b) Plane frame member g g f4 d4
g g f2 d2 g g f1 d1
l l f4 d4
g g f3 d3
Global
l l f2 d2
l l f3 d3
Local
l l f1 d1
c) Plane truss member
Global and local force and displacement components
• In the case of plane frame, if the components of member end forces in global direction are: { f g } = { f 1g f 2g f 3g f 4g f 5g f 6g }
and the corresponding components in local directions are: { f l } = { f 1l f 2l f 3l f 4l f 5l f 6l }
as shown in Fig.
, then;
{ f g } = [T ] { f l }
where; c s 0 [T ] = 0 0 0
−s c 0 0 0 0
0 0 1 0 0 0
0 0 0 c s 0
0 0 0 −s c 0
0 0 0 0 0 1
in which; c = cos θ = (X2 – X1) / L , and s = sin θ = (Y2 – Y1) / L where (X1 , Y1) and (X2 , Y2) are the coordinates of the joints at the start and the end of member respectively with respect to global axes, and L is the length of the member. Similarly, the relation between the displacement components in global directions {dg} and the displacement components in local directions {dl} for the joints of plane frame members is: {d g } = [T ] { d l }
where [T] as defined above. • In the case of plane trusses, each of {fg}, {fl}, {dg}, and {dl} has only four elements and the transformation matrix reduces to:
−s c 0 0
c s [T ] = 0 0
0 0 c s
0 0 −s c
• In the case of grids, the transformation matrix is: 1 0 0 [T ] = 0 0 0
0 s −c 0 0 0
0 c s 0 0 0
0 0 0 1 0 0
0 0 0 0 s −c
0 0 0 0 c s
• In the case of space frames, each joint has six degrees of freedom then the transformation matrix is:
[T ] =
t*
φ
φ φ φ
*
t
φ φ
φ φ
t*
φ φ φ
φ
t*
where [Ø] = [0]3*3 , and lx
ly
lz
[t ] = m x
my
mz
nx
ny
nz
*
where (lx, mx, nx), (ly, my, ny), and (lz, mz, nz) are the direction cosines of the local axes x, y, and z with respect to the global axes X, Y, and Z. • In the case of space trusses, all member ends are hinged. Therefore the transformation matrix is:
[T ] =
lx
*
*
0
0
0
mx nx
* *
* *
0 0
0 0
0 0
0
0
0
lx
*
*
0 0
0 0
0 0
mx nx
* *
* *
where the asterisks refer to unneeded elements, and (lx, mx, nx) are the direction cosines which given by: lx = (X2 – X1) / L,
mx = (Y2 – Y1) / L, and
nx = (Z2 – Z1) / L
3-2- STEP 2: LOAD VECTOR (MEMBER FIXED-END FORCES) Member fixed-end forces means the reactions at the ends of members due to loads, environmental changes, or support movement, when all the unknown displacements at member end joints are prevented. These reactions can be determined by classical methods such as column analogy or consistent deformations. Components of fixed-end forces in local directions may be arranged in a vector { f ml } , which is called member load vector in local directions. The corresponding components in global directions may be arranged in a vector { f mg } which is called member load vector in global directions, and can be determined from the transformation relation { f mg }
Fig.
= [T] { f ml }
Shows some cases of member fixed-end forces.
2
WL /12
2
Wt/m
WL /12
PL/8
L WL/2
PL/8
P
L WL/2
P/2
P/2
Wt/m 2
2
WL /30
WL /20
Wt/m
2
5WL /96
L 0.167 WL
2
5WL /96
L
0.333 WL
WL/4 2
(
2
WL 20
2
WL ) 30
(
L
2PL/9
2
WL ) 30
WL 20
L
P
P
2
P
2
Pb/L
a
2
2
2
P.b.a /L
P
b
Pa/L
P
(
M
M/4
L 1.5 M / L
2
P.a.b /L
2PL/9
L
M/4
WL/4
1.5 M / L
P.a.b P.b.a ) L2 L2 L
2
(
2
P.a.b P.b.a ) L2 L2 L
(1)
(2) P1
W P1 P2
P2
(3)
(4 ) P1
W
P3
P2 W
W
(5 )
3-3- STEP 3: STIFFNESS MATRIX (MEMBER END FORCES DUE TO JOINT DISPLACEMENTS) Displacements of joints cause member end displacements and hence member deformations and internal forces between joints and member ends. These member end forces depend on the type of connection between joints and member end. a- Force-Displacement Relation in Local Directions • In the case of plane truss member shown in Fig. , let the joints undergo displacements whose components in local directions are d1l , d 2l , d 3l , and d 4l . The lateral displacements can occur without deformation (small displacements and hinged connections). On the other hand, the axial displacements d1l and d 3l result in an elongation ( d 3l - d1l ) and hence tension N = ( d 3l - d1l ) EA / L. The corresponding forces from joints to member ends are: f dl1 = -N = ( d 3l - d 1l ) EA / L f dl 2 = f dl 4 = 0 f dl3 = N = ( d 3l - d 1l ) EA / L
or in matrix form; { f dl } = [ k l ] { d l }
where 1 0 [ kl ] = −1 0
0 0 0 0
−1 0 1 0
0 0 0 0
The matrix [ k l ] is known as the member stiffness matrix in local directions. It can be noticed that the elements of the ith column of [ k l ] are the forces { f dl } when d il = 1 while all other components of { d l } are zeros. This observation is usually used as a convenient basis for deriving the matrix [ k l ] for members of different types. Consider a plane frame member with fixed ends as shown in Fig. . To derive elements of the first column of [ k l ] , let d1l = 1 while all other components of { d l } are zeros. The corresponding end forces are the elements of the first column of [ k l ] , and these elements are: (EA/L) { 1 0 0 -1 0 0 }T
Elements of the second column of [ k l ] are the forces correspond to d 2l = 1, while all other components of { d l } are zeros. Then the elements of the second column of [ k l ] are: 1 }T (6EI/L2) { 0 2/L 1 0 -2/L Note that these forces can be determined using the method of consistent deformations. Similarly the components of the third, fourth, fifth, and sixth columns can be determined and the complete stiffness matrix of the member in local axes directions is: EA / L 0 0 [k ]= − EA / L l
0 0
0
0
− EA / L
0
0
12 EI / L
6 EI / L
0
− 12 EI / L
6 EI / L2
6 EI / L2 0
4 EI / L 0
0 EA / L
− 6 EI / L2 0
2 EI / L 0
− 12 EI / L3 6 EI / L2
− 6 EI / L2 2 EI / L
0 0
12 EI / L3 − 6 EI / L2
− 6 EI / L2 4 EI / L
3
2
3
It may be observed that each column of [ k l ] represent a set of forces in equilibrium. It may also be observed that the matrix [ k l ] is symmetric, i.e. k ijl = k lji . These two observations may be used for deriving some elements of [ k l ] .
D 3 D 4 D 2 2
P1 W1
P2
2
3
W2
1
Fixed
D 1
Structure and loads 1
0 0
2 3
4
1
2 3
4 5
6
1 Hinged
Members, joints and degrees of freedom (D.O.F.)
global position numbers
1 1
2
3 4
0 0
0
1
2 3
4 5
6
local column numbers
2 1
0 2
3 2
0 3
4 3
2 4
0 4
3 5
0 5
4 6
0 6
K
1
g
K
local row numbers
g
2
Member stiffness matrices 1 1
2 1
0 2
3 2
0 3
4 3
2 4
0 4
3 5
0 5
4 6
0 6 g fm
1
g fm
2
Member load vectors 1 2 3 4
1
2
g1
g1
k11 g1
3
k14 g1
4
g1
g1
k15 g2
g1
k16 g2
g1
k41
k44+k11 k45+k12 k46+k13
g1
k54+k21 k55+k22 k56+k23
k51 g1
k61
g1
0 - fm1
D1 g2
g1
g2
g1
g2
g1
g2
g1
g2
g1
g2
g1
g2
k64+k31 k65+k32 k66+k33
D2 D3 D4
=
3 -5 0 -
g1 fm4 g1 fm5 g1 fm6
g2
- fm1
g2
- f m2
g2
- fm3
Equilibrium equations
Example for assembly of equilibrium equations To find the overall stiffness matrix
=1
2
EA/L
4EI/L
2
6EI/L
6EI/L
2EI/L =1
=1
EA/L
3
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎥ ⎢ ⎢ 0 ⎥ ⎥ ⎢ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎢⎣ 0 ⎥⎦
2
3
12EI/L
2
6EI/L
12EI/L
6EI/L
⎡ ⎤ 0 ⎢ 3 ⎥ ⎢ 12 EI / L ⎥ ⎢ 6 EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 12 EI / L3 ⎥ ⎢ ⎥ 2 ⎣⎢ 6 EI / L ⎦⎥
⎡ ⎤ 0 ⎢ 2 ⎥ ⎢ 6 EI / L ⎥ ⎢ 4 EI / L ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 6 EI / L2 ⎥ ⎢ ⎥ ⎣⎢ 2 EI / L ⎦⎥
2
2
= 1 6EI/L
6EI/L
2EI/L
4EI/L =1
EA/L
=1
EA/L 3
3
12EI/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎣⎢ 0 ⎦⎥
[k ]= l
12EI/L
2
2
6EI/L
⎡ ⎤ 0 ⎢ 3⎥ ⎢− 12 EI / L ⎥ ⎢ − 6 EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ 12 EI / L3 ⎥ ⎢ ⎥ 2 ⎢⎣ − 6 EI / L ⎥⎦
6EI/L
⎡ ⎤ 0 ⎢ 2 ⎥ ⎢ 6 EI / L ⎥ ⎢ 2 EI / L ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 6 EI / L2 ⎥ ⎢ ⎥ ⎢⎣ 4 EI / L ⎥⎦
EA / L
0
0
− EA / L
0
0
0
12 EI / L3
6 EI / L2
0
− 12 EI / L3
6 EI / L2
0
6 EI / L2
4 EI / L
0
− 6 EI / L2
2 EI / L
− EA / L 0
0 − 12 EI / L3
0 − 6 EI / L2
EA / L 0
0 12 EI / L3
0 − 6 EI / L2
0
6 EI / L2
2 EI / L
0
− 6 EI / L2
4 EI / L
=1
3EI/L
2
3EI/L
EA/L
=1
EA/L
=1 2
3
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
3EI/L
⎡ ⎤ 0 ⎢ 3 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 3EI / L3 ⎥ ⎢ ⎥ 0 ⎣⎢ ⎦⎥
=1 EA/L
2
3EI/L
3
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
[ kl ] =
=1
=1 3
3EI/L
3EI/L
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
⎡ ⎤ 0 ⎢ 3⎥ ⎢ − 3EI / L ⎥ ⎢− 3EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ 3EI / L3 ⎥ ⎢ ⎥ 0 ⎣⎢ ⎦⎥
0
3EI/L
⎡ ⎤ 0 ⎢ 2 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 3EI / L2 ⎥ ⎢ ⎥ 0 ⎣⎢ ⎦⎥
EA/L
EA / L
2
3EI/L
3
3EI/L
0
0 3
3EI / L
2
3EI / L
− EA / L
0
0
− 3EI / L
0
0 3
0 − EA / L
3EI / L 0
3EI / L 0
0 EA / L
− 3EI / L 0
0 0
0 0
− 3EI / L3 0
− 3EI / L2 0
0 0
3EI / L3 0
0 0
2
2
2
=1
3EI/L
3EI/L =1
=1
3
3
3EI/L
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎣⎢0⎦⎥
EA/L
⎡ ⎤ 0 ⎢ 2 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 3EI / L2 ⎥ ⎢ ⎥ 0 ⎢⎣ ⎥⎦
2
3EI/L
=1
=1
EA/L
3
3
3EI/L
3EI/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
⎡ ⎤ 0 ⎢ 3⎥ ⎢ − 3EI / L ⎥ ⎢− 3EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ 3EI / L3 ⎥ ⎢ ⎥ 0 ⎣⎢ ⎦⎥
0 0
2
3EI/L
3EI/L
⎡ ⎤ 0 ⎢ 3 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L2 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢− 3EI / L3 ⎥ ⎢ ⎥ 0 ⎢⎣ ⎥⎦
=1
[ kl ] =
2
3EI/L
0
0 3
3EI / L
2
3EI / L
− EA / L
0
0
− 3EI / L
0
0 3
0
3EI / L
3EI / L
0
− 3EI / L
0
0 0
0 − 3EI / L3
0 − 3EI / L2
EA / L 0
0 3EI / L3
0 0
0
0
0
0
0
0
2
2
=1 EA/L
EA/L
=1
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
=1
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
=1 EA/L
=1
EA/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
EA / L 0 0 [ kl ] = − EA / L 0 0
=1
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
0 0 0 0 0 0
0 0 0 0 0 0
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
− EA / L 0 0 EA / L 0 0
0 0 0 0 0 0
0 0 0 0 0 0
3-4- STEP 4 : The Overall Equilibrium Equation From the steps number 3 & 4 we can construct the overall equilibrium equation as follow:
{ F } nx1 = [ K ] nxn * { D }nx1 Where: (n) the number of degree of freedom (DOF) which defined from step no 1( modeling ).
3-5- STEP 5 : Solve The Equilibrium Equation By using back Guss elimination we can solve the equilibrium equation and find the overall global displacement { D }g We can find the local displacement from the relation { D }l = [ T ]T * { D }g Where [ T ] the transformation matrix for member For plane truss member is: c s [T ] = 0 0
−s c 0 0
0 0 c s
0 0 −s c
−s c 0 0 0 0
0 0 1 0 0 0
0 0 0 c s 0
For plane frame member is: c s 0 [T ] = 0 0 0
0 0 0 −s c 0
0 0 0 0 0 1
3-6- STEP 6 : Find The Internal Forces For plane truss member: N=
EA Δ L
Δ = (d 3 − d1 ) * cosθ + (d 4 − d 2 ) * sin θ d4 d3
d2
For plane frame member:
d1
Dg1 = D1 * cosθ + D2 * sin θ
Dg 2 = D2 * cosθ − D1 * sin θ
Dg 3 = D3 Dg 4 = D4 * cos θ + D5 * sin θ Dg 5 = D5 * cos θ − D4 * sin θ Dg 6 = D6
N1 = N 2 = ( Dg 4 − Dg1 ) S1 = S 2 = M1 =
EA L
12 EI 12 EI ( Dg − Dg ) + ( Dg 6 + Dg3 ) 2 5 L3 L2
6 EI 2 EI ( Dg − Dg ) − ( Dg 6 + 2 Dg3 ) 5 2 L2 L
M2 =
6 EI 2 EI ( Dg − Dg ) + ( Dg 3 + 2 Dg 6 ) 2 5 L2 L
CHAPTER (6)
Analysis of Plane Trusses By Using Stiffness Matrix Method 6-1- STEP 1: MODELING a- Numbering Organization of the solution requires numbering the members, joints, and forces and displacements components. Several numbering schemes are possible. Choice of the most appropriate scheme depends on several considerations including the method of assembling and solving equations of equilibrium (banded matrix methods, frontal method, etc.). When banded matrix methods are used for solving equilibrium equations, members may be numbered in any convenient order, but joints should be numbered in such an order that the maximum difference between the numbers of the two joints of each member is as small as possible. For example, the numbering in Figure (b) is more appropriate than that in Figure (a)
20
19
18
17
16
15
14
13
12
11
1
2
3
4
5
6
7
8
9
10
(a) 2
3
6
7
10
11
14
15
18
19
1
4
5
8
9
12
13
16
17
20
(b)
6
1
2
2
1
3
7
8
3 (a)
4
5
6
4 (b)
7
5
8
The unknown displacement components are usually numbered in a certain sequence starting at joint 1 and proceeding in ascending order through the joints. These displacement components are called degrees of freedom.
b- Degrees of Freedoms Several models of skeletal structures may be used depending on the nature of the structure and the loads. Common skeletal models include space frames, space trusses, plane frames, plane trusses, beams, and grids. * In plane trusses, a joint has two degrees of freedom (the translation only) since rotation are not considered. * In beams, axial and transverse forces and displacements are uncoupled, and separate analyses may be carried out for axial effects and transverse effects.
c- Local and Global Axes The axes which are convenient in dealing with members individually are called local (member) axes, but the axes which are convenient in dealing with the structure as a whole are called global (structure) axes. Displacement and force components may be expressed using one of the previous two systems. The relation between the components in the two systems of axes is expressed in matrix form which called transformation matrix [T]. In the case of plane trusses, each of {fg}, {fl}, {dg}, and {dl} has only four elements and the transformation matrix reduces to: dg4
[T ] =
c
−s
0
0
s 0 0
c 0 0
0 c s
0 −s c
dl4
dl3 dg3
dg2 dl2
dl1
dg1
in which; , and
c = cos θ = (X2 – X1) / L s = sin θ = (Y2 – Y1) / L
where (X1 , Y1) and (X2 , Y2) are the coordinates of the joints at the start and the end of member respectively with respect to global axes, and L is the length of the member.
{d }l
⎧ d1 ⎫ ⎪d ⎪ ⎪ ⎪ = ⎨ 2⎬ ⎪d 3 ⎪ ⎪⎩d 4 ⎪⎭
c s [ T] = 0
−s c 0
0 0 c
0 0 −s
0
0
s
c
Similarly, the relation between the displacement components in global directions {dg} and the displacement components in local directions {dl} for the joints of plane frame members is: {d g } = [T ] { d l }
where [T] as defined above.
6-2- STEP 2: LOAD VECTOR (MEMBER FIXED-END FORCES) A member fixed-end force means the reactions at the ends of members due to loads, environmental changes, or support movement, when all the unknown displacements at member end joints are prevented. These reactions can be determined by classical methods such as consistent deformations. Components of fixed-end forces in local directions may be arranged in a vector { f ml } , which is called member load vector in local directions. The corresponding components in global directions may be arranged in a vector { f mg } which is called member load vector in global directions, and can be determined from the transformation relation. { f mg } = [T] { f ml }
Example (1) Construct the modeling and find the over all load vector for the given plane trusses. 10 t
4.00
8t 5t 4.00
Solution
10 t
D3 D2
Q D.O.F. = 4
4
4
2
4.00
5
1
2
6
∴ {F }4 x1 = [K ]4 x 4 {D}4 x1
D4
∴ {F }4 x1
⎡ 8 ⎤ ⎢ 0 ⎥ ⎥ =⎢ ⎢− 10⎥ ⎢ ⎥ ⎣ −5⎦
8t
1
3
3 D1
5t 4.00
Example (2) Construct the modeling and find the over all load vector for the given 8t
8t
8t 6t
3.00
6t 4.00
4.00
plane trusses. Solution
8t
8t 2
8t
3
3
4
5 6t
10
1
1
7
6
5
3.00
9
8
4
2
6 6t
4.00
4.00
D. O.F. = 4
∴ {F }4 x1 = [K ]4 x 4 {D}4 x1
Overall Load Vector:
{ F }4x1 =
6-3- STEP 3: STIFFNESS MATRIX (MEMBER END FORCES DUE TO JOINT DISPLACEMENTS) Displacements of joints cause member end displacements and hence member deformations and internal forces between joints and member ends. These member end forces depend on the type of connection between joints and member end.
a- Force-Displacement Relation in Local Directions • In the case of plane truss member shown in Fig., let the joints undergo displacements whose components in local directions are d1l , d 2l , d 3l , and d 4l . The lateral displacements can occur without deformation (small displacements and hinged connections). On the other hand, the axial displacements d1l and d 3l result in an elongation ( d 3l - d1l ) and hence tension N = ( d 3l - d1l ) EA / L. The corresponding forces from joints to member ends are: f dl1 = -N = ( d 3l - d1l ) EA / L f dl 2 = f dl 4 = 0 f dl3 = N = ( d 3l - d1l ) EA / L
or in matrix form;
{ f dl }
Where:
l l [ k ] { d } =
⎡ EA/ L ⎤ ⎡0⎤ ⎡− EA/ L⎤ ⎡0⎤ ⎢ 0 ⎥ ⎢ ⎥⎢ 0 ⎥ ⎢ ⎥ ⎥ ⎢0⎥ ⎢ ⎥ ⎢0⎥ ⎢ ⎢− EA/ L⎥ ⎢0⎥ ⎢ EA/ L ⎥ ⎢0⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ 0 0 ⎦ ⎣0⎦ ⎣ ⎦ ⎣0⎦ ⎣
[ kl ] =
=1 EA/L
EA/L
⎡ EA/ L ⎤ ⎢ 0 ⎥ ⎥ ⎢ ⎢− EA/ L⎥ ⎥ ⎢ 0 ⎦ ⎣
=1
⎡0⎤ ⎢0⎥ ⎢ ⎥ ⎢0⎥ ⎢ ⎥ ⎣0⎦ =1
EA/L
=1
EA/L
⎡− EA/ L⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA/ L ⎥ ⎢ ⎥ 0 ⎣ ⎦
⎡0⎤ ⎢0⎥ ⎢ ⎥ ⎢0⎥ ⎢ ⎥ ⎣0⎦
Then; The local stiffness Matrix for any plane truss member equal:
⎡1 ⎢0 EA ⎢ kl = L ⎢− 1 ⎢ ⎣0
[ ]
0 − 1 0⎤ 0 0 0⎥⎥ 0 1 0⎥ ⎥ 0 0 0⎦
The matrix [ k l ] is known as the member stiffness matrix in local directions. It can be noticed that the elements of the ith column of [ k l ] are the forces { f dl } when d il = 1 while all other components of { d l } are zeros. This observation is usually used as a convenient basis for deriving the matrix [ k l ] for members of different types. It may be observed that each column of [ k l ] represent a set of forces in equilibrium. It may also be observed that the matrix [ k l ] is symmetric, i.e. k ijl = k lji . These two observations may be used for deriving some elements of [ k l ] .
6-4- STEP 4: The Overall Equilibrium Equation From the step no 2 & 3 we can construct the equation as follow:
{ F }nx1 = [ K ] n x n { D } nx1
……..Where n is the DOF by solution the overall equilibrium equation we get the values of {D}.
6-5- STEP 6: The internal forces for plane truss member By using the known global displacement we can determined the internal forces for plane trusses as follows: N=
EA Δ L
where:
Δ = (d 3 − d1 ) * cos θ + (d 4 − d 2 ) * sin θ
Solved Examples 10 t
Example3: The following truss has constant EA, determine the reactions and member forces.
4.00
8t 5t 4.00
Solution
10 t
D3 D2
Q D.O.F. = 4
4
4
2
4.00
5
1
2
6
∴ {F }4 x1 = [K ]4 x 4 {D}4 x1
D4
∴ {F }4 x1
⎡ 8 ⎤ ⎢ 0 ⎥ ⎥ =⎢ ⎢− 10⎥ ⎢ ⎥ ⎣ −5⎦
8t
[k1 ]g
5t 4.00
0 1 0 −1
For element 2:
0⎤ − 1⎥⎥ 0⎥ ⎥ 1⎦
0 0 0 0
[k 2 ]g
For element 3: L = 4.0 m, θ = 0.0˚
[k 3 ]g = [k 3 ]l
3
3 D1
For element 1: L = 4.0 m, θ = 90˚ ⎡0 ⎢ EA ⎢ 0 = 4 ⎢0 ⎢ ⎣0
1
⎡1 ⎢ EA ⎢ 0 = 4 ⎢− 1 ⎢ ⎣0
0 0 0 0
−1 0
1 0
0 0 0 0
L = 4.0 m, θ = 90˚
⎡0 ⎢ EA ⎢ 0 = 4 ⎢0 ⎢ ⎣0
0 1 0 −1
0⎤ − 1⎥⎥ 0⎥ ⎥ 1⎦
0 0 0 0
For element 4: L = 4.0 m, θ = 0.0˚ ⎤ ⎥ ⎥ ⎥ ⎥ ⎦
[k 4 ]g = [k 4 ]l
⎡1 ⎢ EA ⎢ 0 = 4 ⎢− 1 ⎢ ⎣0
0 0 0 0
−1 0
1 0
0 0 0 0
⎤ ⎥ ⎥ ⎥ ⎥ ⎦
For element 5: L = 4 / 2 m, θ = 45˚
[k 5 ]g
⎡0.5 ⎢ EA ⎢ − = 4/ 2 ⎢ − ⎢ ⎣−
− − − −
− − − −
− − − −
For element 6: L = 4 / 2 m, θ = 315˚
⎤ ⎥ ⎥ ⎥ ⎥ ⎦
Overall equilibrium equation: ⎧ 8 ⎫ ⎡ 0.338 ⎪ 0 ⎪ ⎢ 0 ⎪ ⎢ ⎪ ⎨ ⎬= − 10 ⎪ ⎪ ⎢ 0 ⎪⎩ − 5 ⎪⎭ ⎢⎣ 0
0 ⎤ ⎧ D1 ⎫ 0.088 ⎥⎥ ⎪⎪ D2 ⎪⎪ ⎨ ⎬ − 0.088 0.338 − 0.088⎥ ⎪ D3 ⎪ ⎥ 0.088 − 0.088 0.338 ⎦ ⎪⎩ D4 ⎪⎭ 0 0.338
0 − 0.088
By solving 4 equations; then ⎧ 23.66 ⎫ ⎪ EA ⎪ ⎪ ⎧ D1 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − 3.42 ⎪ ⎪ ⎪D ⎪ ⎪ EA ⎪ 2 ⎪ ⎪⎪ ⎪⎪ ⎪ ⎨ ⎬=⎨ ⎬ ⎪ D3 ⎪ ⎪ − 36.57 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ EA ⎪ ⎪ D4 ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭ ⎪ − 23.42 ⎪ ⎪ ⎪ ⎪⎩ EA ⎪⎭
[k 6 ]g
⎡ 0.5 − 0.5 − 0.5 0.5 ⎤ ⎢ 0.5 − 0.5⎥⎥ EA ⎢− 0.5 0.5 = 0.5 − 0.5⎥ 4 / 2 ⎢− 0.5 0.5 ⎢ ⎥ ⎣ 0.5 − 0.5 − 0.5 0.5 ⎦
For elements:
− 36.57 −0) EA − 36.57 Δ1 = EA − 23.66 23.42 , Δ3 = , Δ2 = EA EA
Δ 1 = ( − 3.42 − 23.66 ) × cos 90 + (
then; similarly; Δ6 =
− 6.88 EA
Δ4 =
3.42 , EA
then;
F=Δ
EA L
− 36.57 EA * = −9.14 t EA 4 23.42 EA F2 = * = 5.85 t EA 4 − 23 . 66 EA = − 5 . 91 t F3 = * EA 4 3.42 EA F4 = * = 0.855 t EA 4 F1 =
F5 =
− 16.73 EA * = −2.95 t EA 4 2
F6 =
− 6.88 EA * = −1.21 t EA 4 2
5 -2 .9
t .21 -1
-9.14 t
t
-5.91 t
5.85 t
0.855 t
The Final internal forces
Δ5 =
− 16.73 , EA
CHAPTER (7)
Analysis of Plane Frames By Using Stiffness Matrix Method 7-1- STEP 1: MODELING a- Numbering As we indicated in the previous chapter, organization of the solution requires numbering the members, joints, and forces and displacements components. Several numbering schemes are possible. Choice of the most appropriate scheme depends on several considerations including the method of assembling and solving equations of equilibrium (banded matrix methods, frontal method, etc.). When banded matrix methods are used for solving equilibrium equations, members may be numbered in any convenient order, but joints should be numbered in such an order that the maximum difference between the numbers of the two joints of each member is as small as possible. For example, the numbering in Figure (b) is more appropriate than that in Figure (a)
6
1 19
11 20
7 17
18
8
3
15
16
9
4
13
13
14
14
11
12
11
10
17
18 8
7 15
16 5
4 13
9 6
5
4
12 9
8
7
6
5
4
12
15 20
10
9
8
7
19
12
11
10 2
14
13
14
6
1
2
3
1
2
3
5
10
15
1
2
3
(a)
(b)
The unknown displacement components are usually numbered in a certain sequence (for example in plane frame; x, y, and rotation) starting at joint 1 and proceeding in ascending order through the joints. These displacement components are called degrees of freedom.
b- Degrees of Freedoms Several models of skeletal structures may be used depending on the nature of the structure and the loads. Common skeletal models include space frames, space trusses, plane frames, plane trusses, beams, and grids. * In plane frames, a free node has three degrees of freedom since it has two translations and one rotation ( Δx , Δy & θ ). * In beams, axial and transverse forces and displacements are uncoupled, and separate analyses may be carried out for axial effects and transverse effects.
c- Local and Global Axes The axes which are convenient in dealing with members individually are called local (member) axes, but the axes which are convenient in dealing with the structure as a whole are called global (structure) axes. Displacement and force components may be expressed using one of the previous two systems. dg4 dl4
dl3 dg3
dg2 dl2
dl1
dg1
The relation between the components in the two systems of axes is expressed in matrix form which called transformation matrix [T].
• In the case of plane frame, if the components of member end forces in global direction are: { f g } = { f1g f 2g f 3g f 4g f 5g f 6g }
and the corresponding components in local directions are: { f l } = { f1l f 2l f 3l f 4l f 5l f 6l }
as shown in Fig.
, then;
{ f g } = [T ] { f l }
where; c s 0 [T ] = 0 0 0
−s c 0 0 0 0
0 0 1 0 0 0
0 0 0 c s 0
0 0 0 −s c 0
0 0 0 0 0 1
in which; c = cos θ = (X2 – X1) / L , and
s = sin θ = (Y2 – Y1) / L
where (X1 , Y1) and (X2 , Y2) are the coordinates of the joints at the start and the end of member respectively with respect to global axes, and L is the length of the member. Similarly, the relation between the displacement components in global directions {dg} and the displacement components in local directions {dl} for the joints of plane frame members is:
{d g } = [T ] { d l } where [T] as defined above.
7-2- STEP 2: LOAD VECTOR (MEMBER FIXED-END FORCES) A member fixed-end force means the reactions at the ends of members due to loads, environmental changes, or support movement, when all the unknown displacements at member end joints are prevented. These reactions can be determined by classical methods such as column analogy or consistent deformations. Components of fixed-end forces in local directions may be arranged in a vector { f ml } , which is called member load vector in local directions. The corresponding components in global directions may be arranged in a vector { f mg } which is called member load vector in global directions, and can be determined from the transformation relation
{ f mg } = [T] { f ml } Some cases of member fixed-end forces. 2
2
WL /12
WL /12
Wt/m
PL/8
L
L
WL/2
WL/2
P/2
P/2
Wt/m 2
2
WL /30
WL /20
Wt/m
2
WL/4 (
WL 20
WL ) 30 L
2
(
2
WL ) 30
WL 20
L
5WL /96
L
0.333 WL
2
2
5WL /96
L 0.167 WL
2
PL/8
P
WL/4
No7
CASE OF LOADING
LOCAL LOAD {f} = [0 1/6wL-wL / 60 0 1/6wL wL /60 VECTFOR MEMBER l
L
2
m
2
]T
8 1
{f}lm = [0 wL/30 -wL2/60 0 wL/60 wL2/30 ]T {f}lm = [ 0 wL/2 -wL2/12 0 wL/2 wL2/12]T
L
L
9
{f}lm = [0 wL/60 -wL2/30 0 wL/30 wL2/60 ]T L
2
b
a
0 wb/2 wbL/24 (3-b2/L2 ) ]T {f}lm = [0 wL/15 -wL2/15 0 wL/20 wL2/20 ]T
10 3
11
a
{fm}l = [ 0 wb/2 - wbL/24 (3-b2/L2 )
L b
a
a
{f}lm = [ 0 wa -wa2/6 (3-2a /L ) T 2 wa -wL wa22/20 /6 ×(3-2a /L ) ]wL {f}lm = [0 0wL/20 0 wL/15 /15 ]T
L
4 12
{f}lm=[ 0 1/4wL -5/96wL2 0 1/4wL 5/96w L2 ] L/2
{f}lm = [0 wc*a/a+b -wc/L2 (ab2 –c2/6(b-a/2))
L/2
a-c/2
T
b-c/2
c
0
5
wc*b/a+b
wc/L2 (ab2 –c2/6 (a-b/2))]T
{f}lm = [ 0 (b+L/4 )× w – w/12 (L2-a2 (2-a/L) L/2
L/2
0 ( b+L/4 )×w w/12 (L2-a2 (2-a/L))]T
13
[
{f}lm = 0
6 b
a L
wa* (a/2+b)/a+b -wa2/4[2-a/L(8/3-
]T
3 [2-a/L(8/3a/L) (a/2)/(a+b) {f}lm =0[0wa* 1/3wL -wL2/15wa0 /4 1/3wL wL2/15 ]T
14 {f}lm = [0
wb/15 -wL2/15 0 wL/20 wL2/20 ]T
b
a
15
{f}lm = [0 1/3 wL -wL2/20
0 1/6 wL wL2/30]T
L
16
{f}lm = [0 1/6 wL -wL2/30 0 1/3 wL wL2/20 ]T L
17 {f}lm = [0 1/6 wL -wL2/30 0 1/3 wL wL2/20 ]T b
a
18 {f}lm = [0 q*a/2(b+2q/3L) –q*a2/6(1a/L+3/10*a2/L2) 0 q*a/2 (q-3a) q*a2/60 *a/L(5.3a/L) ]T
b
a
19 {f}lm = [0 q*a/2(b+a/3L) –q*a2(1/32 2 a/2L+a /5L 0 q*a/2(2q/3L) –q*a2(1/3-
b
a
20 P
{f}lm = [0 P/2
L/2
-PL/8 0 P/2
PL/8]T
L/2
21 P a
b
{f}lm = [0 P b/L -Pa b 2/L2 0 P a/L Pb a 2/L2 ]T
22 P
{f }lm =[0 P -P*a/L(L-a)
P L-2a
a
0 P -P*a/L(L-a) ]T
a
23 P
P
a
{f }lm =[0 3P/2 -5PL/16
P
a
a
0
a
24 P
P
P
a
a
{f }lm =[0
P
a
a
2P
0
a
3P/2
5PL/16) ]T
-5PL/16
2P
5PL/16) ]T
25 P
P a
a/2
P a
P a
P a
P a
a/2
{ f }lm =[0
6P/2
-P*L/24(2n+1/n)
0
6P/2
P*L/24(2n+1/n) ]T
7-3- STEP 3: STIFFNESS MATRIX (MEMBER END FORCES DUE TO JOINT DISPLACEMENTS) Displacements of joints cause member end displacements and hence member deformations and internal forces between joints and member ends. These member end forces depend on the type of connection between joints and member end.
a- Force-Displacement Relation in Local Directions Consider a plane frame member with fixed ends as shown in Fig. . To derive elements of the first column of [ k l ] , let d1l = 1 while all other components of { d l } are zeros. The corresponding end forces are the elements of the first column of [ k l ] , and these elements are: (EA/L) { 1
0
0
-1
0
0 }T
Elements of the second column of [ k l ] are the forces correspond to d 2l = 1, while all other components of { d l } are zeros. Then the elements of the second column of [ k l ] are: (6EI/L2) { 0
2/L
1
0
-2/L
1 }T
Note that these forces can be determined using the method of consistent deformations. Similarly the components of the third, fourth, fifth, and sixth columns can be determined and the complete stiffness matrix of the member in local axes directions is:
[ kl ] =
EA / L 0
0 12 EI / L3
0 6 EI / L2
− EA / L 0
0 − 12 EI / L3
0 6 EI / L2
0
6 EI / L2
4 EI / L
0
− 6 EI / L2
2 EI / L
− EA / L
0
EA / L
0
0
0
− 12 EI / L
− 6 EI / L
0
12 EI / L
− 6 EI / L2
0
6 EI / L2
2 EI / L
0
− 6 EI / L2
4 EI / L
0 3
2
3
It may be observed that each column of [ k l ] represent a set of forces in equilibrium. It may also be observed that the matrix [ k l ] is symmetric, i.e. k ijl = k lji . These two observations may be used for deriving some elements of [ k l ] .
=1
2
EA/L
4EI/L
2
6EI/L
6EI/L
2EI/L =1
=1
EA/L
3
12EI/L
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
2
3
2
6EI/L
12EI/L
6EI/L
⎤ ⎡ 0 ⎢ 3 ⎥ ⎢ 12 EI / L ⎥ ⎢ 6 EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 12 EI / L3 ⎥ ⎥ ⎢ 2 ⎣⎢ 6 EI / L ⎦⎥
⎤ ⎡ 0 ⎢ 2 ⎥ ⎢ 6 EI / L ⎥ ⎢ 4 EI / L ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 6 EI / L2 ⎥ ⎥ ⎢ ⎣⎢ 2 EI / L ⎦⎥
2
2
= 1 6EI/L
6EI/L
2EI/L
4EI/L =1
EA/L
=1
EA/L 3
3
12EI/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎥ ⎢ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎣⎢ 0 ⎦⎥
[ kl ] =
12EI/L
2
2
6EI/L
⎤ ⎡ 0 ⎢ 3⎥ ⎢− 12 EI / L ⎥ ⎢ − 6 EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢ 12 EI / L3 ⎥ ⎥ ⎢ 2 ⎢⎣ − 6 EI / L ⎥⎦
6EI/L
⎤ ⎡ 0 ⎢ 2 ⎥ ⎢ 6 EI / L ⎥ ⎢ 2 EI / L ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 6 EI / L2 ⎥ ⎥ ⎢ ⎢⎣ 4 EI / L ⎥⎦
EA / L 0
0 12 EI / L3
0 6 EI / L2
− EA / L 0
0 − 12 EI / L3
0 6 EI / L2
0 − EA / L 0
6 EI / L2 0 − 12 EI / L3
4 EI / L 0 − 6 EI / L2
0 EA / L 0
− 6 EI / L2 0 12 EI / L3
2 EI / L 0 − 6 EI / L2
0
6 EI / L2
2 EI / L
0
− 6 EI / L2
4 EI / L
=1
3EI/L
2
3EI/L
EA/L
EA/L
=1
=1 2
3
3EI/L
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎣⎢ 0 ⎦⎥
3EI/L
⎤ ⎡ 0 ⎢ 3 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 3EI / L3 ⎥ ⎥ ⎢ 0 ⎥⎦ ⎢⎣
=1 EA/L
2
3EI/L
=1
=1 3
3
3EI/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎣⎢ 0 ⎦⎥
[ kl ] =
3EI/L
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎣⎢0⎦⎥
⎤ ⎡ 0 ⎢ 3⎥ ⎢ − 3EI / L ⎥ ⎢− 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢ 3EI / L3 ⎥ ⎥ ⎢ 0 ⎦⎥ ⎣⎢
− EA / L 0
0 3EI / L3 3EI / L2
0 3EI / L2 3EI / L
0
0 − 3EI / L3
0 − 3EI / L2
0 EA / L 0
0
0
0
0
0 − EA / L
3EI/L
⎤ ⎡ 0 ⎢ 2 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥⎦ ⎢⎣
EA/L
EA / L 0
2
3EI/L
3
0 − 3EI / L3 − 3EI / L2
0 0
0 3EI / L3
0 0 0
0
0
2
=1
3EI/L
3EI/L =1
=1
3
3
3EI/L
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎣⎢0⎦⎥
3EI/L
⎤ ⎡ 0 ⎢ 3 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 3EI / L3 ⎥ ⎥ ⎢ 0 ⎥⎦ ⎢⎣
=1 EA/L
2
3EI/L
⎤ ⎡ 0 ⎢ 2 ⎥ ⎢ 3EI / L ⎥ ⎢ 3EI / L ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢− 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥⎦ ⎢⎣
2
3EI/L
=1
=1
EA/L
3
3
3EI/L
3EI/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎥ ⎢ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
[ kl ] =
2
3EI/L
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
⎤ ⎡ 0 ⎢ 3⎥ ⎢ − 3EI / L ⎥ ⎢− 3EI / L2 ⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎢ 3EI / L3 ⎥ ⎥ ⎢ 0 ⎦⎥ ⎣⎢
0 0 0
0 3EI / L3 3EI / L2
0 3EI / L2 3EI / L
0 0 0
0 − 3EI / L3
0 − 3EI / L2
0
0
− EA / L 0 0 EA / L 0 0
0 − 3EI / L3 − 3EI / L2
0 0 0
0 3EI / L3 0
0 0 0
=1 EA/L
EA/L
=1
⎡ EA / L ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢− EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
=1
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
=1 EA/L
=1
EA/L
⎡− EA / L ⎤ ⎢ 0 ⎥ ⎥ ⎢ ⎢ 0 ⎥ ⎢ ⎥ ⎢ EA / L ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦
EA / L 0 0 [ kl ] = − EA / L 0 0
=1
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
0 0 0 0 0 0
0 0 0 0 0 0
⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢⎣0⎥⎦
− EA / L 0 0 EA / L 0 0
0 0 0 0 0 0
0 0 0 0 0 0
SOLVED EXAMPLES Example (1) By using stiffness matrix method, draw internal forces diagrams for the shown frame.
D3
3t/m
D2 D1 2
2
3 3.00
6t
EA = 4 * 103 t EI = 5 * 104 t.m2
1 3.00 1 2.00
3t/m
Solution Q D.O.F. = 3
3.6
∴ {F }3 x1 = [K ]3 x 3 {D}3 x1 ∴ {F }3x1
6.00
5.4
3.0 3.0
⎡ −3 ⎤ = ⎢⎢− 5.7 ⎥⎥ ⎢⎣− 6.6⎥⎦
2.7
6.3
3.0
12 6.0
3.0
3.0 3.0
Element 1 2
L
Ends F-F 6.0 F-F 6.0
θ
c
s
EA/L 12EI/L3 6EI/L2 4EI/L
90 0
0 1
1 0
666.7 2777.78 8333.3 33333.3 16666.7 666.7 2777.78 8333.3 33333.3 16666.7
2EI/L
For element 1:
[k1 ]l
0 0 − 666.7 0 0 ⎤ ⎡ 666.7 ⎢ 0 2777.78 8333.3 0 − 2777.78 8333.3 ⎥⎥ ⎢ ⎢ 0 8333.3 33333.3 0 − 8333.3 16666.7 ⎥ =⎢ ⎥ 0 0 666.7 0 0 ⎥ ⎢− 666.7 ⎢ 0 − 2777.78 − 8333.3 0 2777.78 − 8333.3⎥ ⎢ ⎥ 8333.3 16666.7 0 − 8333.3 33333.3 ⎥⎦ ⎢⎣ 0
[k1 ]g
∴ [k1 ]
g
0 0 1 0 0 0
0 0 0 0 0 0 0 −1 1 0 0 0
0⎤ 0⎥⎥ 0⎥ l ⎥ [k1 ] 0⎥ 0⎥ ⎥ 1⎥⎦
⎡0 ⎢− 1 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0
1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 −1 0 0
0 0 0 1 0 0
0⎤ 0⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎥⎦
0 0 8333.3 ⎤ − 8333.3 − 2777.78 ⎡ 2777.78 ⎢ 0 666.7 0 0 0 ⎥⎥ − 666.7 ⎢ ⎢ − 8333.3 0 33333.3 8333.3 0 16666.7 ⎥ =⎢ ⎥ 0 8333.3 2777.78 0 8333.3 ⎥ ⎢− 2777.78 ⎢ 0 0 0 666.7 0 ⎥ − 666.7 ⎥ ⎢ 0 16666.7 8333.3 0 33333.3⎥⎦ ⎢⎣ 8333.3
∴ [k 2 ] = [k 2 ] g
⎡0 − 1 ⎢1 0 ⎢ ⎢0 0 =⎢ ⎢0 0 ⎢0 0 ⎢ ⎢⎣0 0
l
0 0 − 666.7 0 0 ⎤ ⎡ 666.7 ⎢ 0 2777.78 8333.3 0 − 2777.78 8333.3 ⎥⎥ ⎢ ⎢ 0 8333.3 33333.3 0 − 8333.3 16666.7 ⎥ ⎥ ⎢ 0 0 666.7 0 0 ⎥ ⎢− 666.7 ⎢ 0 − 2777.78 − 8333.3 0 2777.78 − 8333.3⎥ ⎢ ⎥ 8333.3 16666.7 0 − 8333.3 33333.3 ⎥⎦ ⎢⎣ 0
Overall equilibrium equation: 0 8333.3 ⎤ ⎧ D1 ⎫ ⎧ − 3 ⎫ ⎡3444.48 ⎪ ⎪ ⎪ ⎢ ⎪ 3444.48 8333.3 ⎥⎥ ⎨ D2 ⎬ ⎨− 5.7 ⎬ = ⎢ 0 ⎪− 6.6⎪ ⎢ 8333.3 8333.3 66666.6⎥ ⎪ D ⎪ ⎭ ⎣ ⎦ ⎩ 3⎭ ⎩
By solving 3 equations; then −3 ⎧ D1 ⎫ ⎧− 2.19 × 10 ⎫ ⎪ ⎪ ⎪ −3 ⎪ ⎨ D2 ⎬ = ⎨− 2.98 × 10 ⎬ ⎪ D ⎪ ⎪ 5.48 × 10 − 4 ⎪ ⎩ 3⎭ ⎩ ⎭
For member 1: 0 ⎧ ⎫ ⎧ 3 ⎫ ⎧ 4.516 ⎫ ⎪ ⎪ ⎪ 3 ⎪ ⎪ 4.986 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ − 0 3 6 . 11 ⎪ ⎪ ⎪ ⎪ ⎪ +⎨ ⎬=⎨ { F1 } g = [ k1 ] g ⎨ ⎬ −3 ⎬ ⎪− 2.19 × 10 ⎪ ⎪− 3⎪ ⎪− 4.516⎪ ⎪− 2.98 × 10 −3 ⎪ ⎪ 3 ⎪ ⎪ 1.014 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 5.48 × 10 − 4 ⎪⎭ ⎪⎩ 3 ⎪⎭ ⎪⎩ 3.016 ⎪⎭
For member 2: ⎧− 2.19 × 10 −3 ⎫ ⎧ 0 ⎫ ⎧ − 1.46 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −3 ⎪ ⎪− 2.98 × 10 ⎪ ⎪ 2.7 ⎪ ⎪ − 1.01 ⎪ −4 ⎪ ⎪ ⎪ ⎪⎪ 3.6 ⎪⎪ ⎪⎪− 2.96⎪⎪ g g ⎪ 5.48 × 10 { F2 } = [ k 2 ] ⎨ ⎬ ⎬=⎨ ⎬+⎨ 0 ⎪ ⎪ 0 ⎪ ⎪ 1.46 ⎪ ⎪ ⎪ ⎪ 6.3 ⎪ ⎪ 10.01 ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩− 5.4⎪⎭ ⎪⎩ 21.09 ⎪⎭ ⎪⎩ 0
Example 2:
D3
By using stiffness matrix method, draw internal forces diagrams for the shown frame.
D1
6.0 t.m
4
4t
2.00
3 4t
2.00 3 3.00
2.00
1.5
Solution Q D.O.F. = 3
2.00 4t
2.0
2.0
6.0 t.m 5.33
1.5
⎡ 4 ⎤ = ⎢⎢− 0.5⎥⎥ ⎢⎣ 1.83 ⎥⎦
2.0
1.5
1.5
∴ {F }3 x1 = [K ]3 x 3 {D}3 x1
L
Ends F-F 6.32 F-F 4.0 F-F 6.0
2.0
4.0
4t
4t
5.33
1 2 3
2
2
1
3.00
Element
4t
1
2.00
EA = 4 * 103 t EI = 5 * 104 t.m2
∴ {F }3 x1
D2
4.0
θ
c
s
EA/L
12EI/L3
6EI/L2
4EI/L
18.43 0 270
0.99 1 0
0.316 0 -1
632.91 1000 666.67
2376.8 9375 2777.78
7510.81 18750 8333.3
31645.56 50000 33333.3
For element 1:
[k1 ]l
⎡− ⎢− ⎢ ⎢− =⎢ ⎢− ⎢− ⎢ ⎢⎣−
− − − − − −
− − − − ⎤ ⎥ − − − − ⎥ ⎥ − − − − ⎥ − 632.91 0 0 ⎥ − 0 2376.8 − 7510.81⎥ ⎥ − 0 − 7510.81 31645.56 ⎥⎦
2EI/L
15822.78 25000 16666.7
[k1 ]g
⎡ 0.99 − 0.316 ⎢0.316 0.99 ⎢ ⎢ 0 0 =⎢ 0 ⎢ 0 ⎢ 0 0 ⎢ 0 ⎢⎣ 0
0 0 0 0 0 0 1 0 0 0 0.99 − 0.316 0 0.316 0.99 0 0 0
∴ [k1 ]
g
∴ [k 2 ] = [k 2 ] g
l
⎡− ⎢− ⎢ ⎢− =⎢ ⎢− ⎢− ⎢ ⎢⎣−
− − − − − −
0⎤ 0⎥⎥ 0⎥ l ⎥ [k1 ] 0⎥ 0⎥ ⎥ 1⎥⎦
0.316 ⎡ 0.99 ⎢− 0.316 0.99 ⎢ ⎢ 0 0 ⎢ 0 ⎢ 0 ⎢ 0 0 ⎢ 0 ⎢⎣ 0
0 0 0 0 0 0 1 0 0 0 0.99 0.316 0 − 0.316 0.99 0 0 0
− − − − ⎤ ⎥ − − − − ⎥ ⎥ − − − − ⎥ − 857.65 − 545.55 2373.41 ⎥ − − 545.55 2392.7 − 7435.7 ⎥ ⎥ − 2373.41 − 7435.7 31645.56⎥⎦
− − −⎤ 0 0 0 8333.3 ⎡1000 ⎡2777.78 ⎢ 0 ⎢ ⎥ 9375 18750 − − −⎥ 666.67 0 ⎢ ⎢ 0 ⎢ 0 ⎢ 8333.3 18750 50000 − − −⎥ 0 33333.33 g ⎢ ⎥ ∴ [k 3 ] = ⎢ − − − − −⎥ − − ⎢ − ⎢ − ⎢ − ⎢ − − − − − −⎥ − − ⎢ ⎢ ⎥ − − − − −⎥⎦ − − ⎢⎣ − ⎢⎣ −
Overall equilibrium equation: ⎧ 4 ⎫ ⎡ 4635.35 − 545.55 10706.74 ⎤ ⎧ D1 ⎫ ⎪ ⎢ ⎪ ⎥⎪ ⎪ ⎨− 0.5⎬ = ⎢ − 545.55 12434.37 11314.3 ⎥ ⎨ D2 ⎬ ⎪ 1.83 ⎪ ⎢10706.74 11314.3 114978.89⎥ ⎪ D ⎪ ⎭ ⎣ ⎦ ⎩ 3⎭ ⎩
By solving 3 equations; then −3 ⎧ D1 ⎫ ⎧ 1.09 × 10 ⎫ ⎪ ⎪ ⎪ −5 ⎪ ⎨ D2 ⎬ = ⎨ 9.44 × 10 ⎬ ⎪ D ⎪ ⎪− 9.52 × 10 −5 ⎪ ⎩ 3⎭ ⎩ ⎭
0⎤ 0⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎥⎦
− − − − − −
− − − − − −
−⎤ −⎥⎥ −⎥ ⎥ −⎥ −⎥ ⎥ −⎥⎦
For member 1: 0 ⎧ ⎫ ⎧ 0 ⎫ ⎧− 0.675⎫ ⎪ ⎪ ⎪ 1.5 ⎪ ⎪ 1.161 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ 1.5 ⎪⎪ ⎪⎪ 1.878 ⎪⎪ 0 g g ⎪ +⎨ { F1 } = [ k1 ] ⎨ ⎬=⎨ ⎬ −3 ⎬ ⎪ 1.09 × 10 ⎪ ⎪ 0 ⎪ ⎪ 0.675 ⎪ ⎪ 9.44 × 10 −5 ⎪ ⎪− 1.5⎪ ⎪ − 1.61 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 9.52 × 10 −5 ⎪⎭ ⎪⎩ 1.5 ⎪⎭ ⎪⎩ 0.373 ⎪⎭
For member 2: ⎧ 1.09 × 10 −3 ⎫ ⎧ 0 ⎫ ⎧ 1.09 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −5 ⎪ ⎪ 9.44 × 10 ⎪ ⎪ 2.0 ⎪ ⎪ 1.10 ⎪ −5 ⎪⎪− 9.52 × 10 ⎪⎪ ⎪⎪ 2.0 ⎪⎪ ⎪⎪− 0.99⎪⎪ { F2 } g = [ k 2 ] g ⎨ ⎬ ⎬=⎨ ⎬+⎨ 0 ⎪ ⎪ 0 ⎪ ⎪ − 1.09 ⎪ ⎪ ⎪ ⎪ 2.0 ⎪ ⎪ 2.90 ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩− 2.0⎪⎭ ⎪⎩− 6.15⎪⎭ ⎪⎩ 0
For member 3: ⎧ 1.09 × 10 −3 ⎫ ⎧ − 4 ⎫ ⎧ − 1.77 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −5 ⎪ ⎪ 9.44 × 10 ⎪ ⎪ 0 ⎪ ⎪ 0.062 ⎪ −5 ⎪ ⎪ ⎪⎪− 5.33⎪⎪ ⎪⎪ 0.57 ⎪⎪ g g ⎪− 9.52 × 10 ⎪ { F3 } = [ k 3 ] ⎨ ⎬ ⎬=⎨ ⎬+⎨ 0 ⎪ ⎪ − 4 ⎪ ⎪ − 6.23 ⎪ ⎪ ⎪ ⎪ 0 ⎪ ⎪− 0.062⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩ 5.33 ⎪⎭ ⎪⎩ 12.82 ⎪⎭ ⎪⎩ 0
Example 3: By using stiffness matrix method, draw internal forces diagrams for the shown frame.
D6
2t/m
D5 D4
4
3 8.0 t.m 3.00 D3 8t
3
EA = 4 * 10 t EI = 5 * 104 t.m2
D2
2
4
4t
D1 2
3.00
1
3 5
1 2.00
2.00
2.00
4.00
Solution Q D.O.F. = 3
2t/m
∴ {F }3 x1
2.0
2.0
∴ {F }3 x1 = [K ]3 x 3 {D}3 x1
8.0 t.m
4.8
3.0
2.0
2.0
4.8
4.0
4.0
2.0
⎡ 4 ⎤ = ⎢⎢− 0.5⎥⎥ ⎢⎣ 1.83 ⎥⎦
4.0
3.0
8.0
0.0 2.0
2.0 2.0
Element
L
θ
c
s
EA/L
12EI/L3
6EI/L2
4EI/L
2EI/L
3.0 5.0 6.0 6.0
90 36.87 270 0
0 0.8 0 1
1 0.6 -1 0
1333.34 800 666.67 666.67
22222.2 4800 2777.78 2777.78
33333.34 12000 8333.34 8333.34
66666.67 40000 33333.34 33333.34
33333.34 20000 16666.67 16666.67
Ends 1 2 3 4
F-F F-F F-F F-F
For element 1:
[k1 ]l
0 0 − 1333.34 0 0 ⎡ 1333.34 ⎤ ⎢ 0 22222.2 33333.34 0 − 22222.2 33333.34 ⎥⎥ ⎢ ⎢ 0 33333.34 66666.67 0 − 33333.34 33333.34 ⎥ =⎢ ⎥ 0 0 1333.34 0 0 ⎢− 1333.34 ⎥ ⎢ 0 − 22222.2 − 33333.34 0 22222.2 − 33333.34⎥ ⎢ ⎥ 0 33333.34 33333.34 0 − 33333.34 66666.67 ⎥⎦ ⎢⎣
[k1 ]g
⎡0 − 1 ⎢1 0 ⎢ ⎢0 0 =⎢ ⎢0 0 ⎢0 0 ⎢ ⎢⎣0 0
0 0 1 0 0 0
0 0 0 0 0 0 0 −1 1 0 0 0
0⎤ 0⎥⎥ 0⎥ l ⎥ [k1 ] 0⎥ 0⎥ ⎥ 1⎥⎦
⎡0 ⎢− 1 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0
1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 −1 0 0
0 0 0 1 0 0
0⎤ 0⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎥⎦
∴ [k1 ]
g
− 33333.34 − 22222.2 − 33333.34⎤ 0 0 ⎡ 22222.2 ⎢ ⎥ − 1333.34 0 1333.34 0 0 0 ⎢ ⎥ ⎢− 33333.34 0 66666.67 33333.34 0 33333.34 ⎥ =⎢ ⎥ 0 33333.3 22222.2 0 33333.34 ⎥ ⎢ − 22222.2 ⎢ ⎥ − 1333.34 0 0 0 1333.34 0 ⎢ ⎥ 0 33333.34 33333.34 0 66666.67 ⎥⎦ ⎢⎣− 33333.34
For element 2: 0 8333.34 − 2777.78 0 8333.34 ⎤ ⎡ 2777.78 ⎢ ⎥ − 666.67 0 666.67 0 0 0 ⎢ ⎥ ⎢ 8333.34 0 33333.34 − 8333.34 0 16666.67 ⎥ =⎢ ⎥ − 8333.34 2777.78 − 8333.34⎥ 0 0 ⎢− 2777.78 ⎢ ⎥ − 666.67 0 0 0 666.67 0 ⎢ ⎥ 0 16666.67 − 8333.34 0 33333.34 ⎥⎦ ⎢⎣ 8333.34
∴ [k 2 ]
g
For element 3:
∴ [k 3 ]
g
0 8333.34 − 2777.78 0 8333.34 ⎤ ⎡ 2777.78 ⎢ ⎥ − 666.67 0 666.67 0 0 0 ⎢ ⎥ ⎢ 8333.34 0 33333.34 − 8333.34 0 16666.67 ⎥ =⎢ ⎥ − 8333.34 2777.78 − 8333.34⎥ 0 0 ⎢− 2777.78 ⎢ ⎥ − 666.67 0 0 0 666.67 0 ⎢ ⎥ 0 16666.67 − 8333.34 0 33333.34 ⎥⎦ ⎢⎣ 8333.34
For element 4:
∴ [k 4 ] = [k 4 ] g
l
− 666.67 0 0 0 0 ⎡ 666.67 ⎤ ⎢ − 2777.78 8333.34 ⎥⎥ 0 2777.78 8333.34 0 ⎢ ⎢ 0 − 8333.34 16666.67 ⎥ 8333.34 33333.34 0 =⎢ ⎥ 0 0 666.67 0 0 ⎢− 666.67 ⎥ ⎢ 0 − 2777.78 − 8333.34 0 2777.78 − 8333.34⎥ ⎢ ⎥ − 8333.34 33333.34 ⎥⎦ 8333.34 16666.67 0 ⎢⎣ 0
Overall equilibrium equation: 1920 − 2240 − 7200 ⎤ ⎧ D1 ⎫ ⎧ 8 ⎫ ⎡ 24462.23 − 1920 26133.34 ⎪ − 3 ⎪ ⎢ − 1920 4693.34 9600 1920 9600 ⎥⎥ ⎪⎪ D2 ⎪⎪ − 3360 ⎪ ⎢ ⎪ 9600 106666.67 7200 20000 ⎥ ⎪⎪ D3 ⎪⎪ − 9600 ⎪⎪ − 2 ⎪⎪ ⎢26133.34 ⎥⎨ ⎬ ⎬=⎢ ⎨ 1920 7200 5684.44 15533.34 ⎥ ⎪ D4 ⎪ − 1920 ⎪ 2 ⎪ ⎢ − 2240 ⎪ − 3 ⎪ ⎢ 1920 6804.44 − 1266.66 ⎥ ⎪ D5 ⎪ − 3360 − 9600 − 1920 ⎥⎪ ⎪ ⎪ ⎢ ⎪ ⎪⎩− 4.8⎪⎭ ⎣⎢ 7200 20000 15533.34 − 1266.66 106666.67 ⎦⎥ ⎪⎩ D6 ⎪⎭ − 9600
By solving 6 equations; then ⎧ D1 ⎫ ⎧ 7.29 × 10 −4 ⎫ ⎪D ⎪ ⎪ −3 ⎪ ⎪ 2 ⎪ ⎪ − 1.02 × 10 ⎪ ⎪⎪ D3 ⎪⎪ ⎪⎪ − 2.8 × 10 − 4 ⎪⎪ ⎬ ⎨ ⎬=⎨ −4 ⎪ ⎪ D4 ⎪ ⎪ 9.9 × 10 ⎪ D5 ⎪ ⎪ − 1.275 × 10 −3 ⎪ ⎪ ⎪ ⎪ ⎪ −5 ⎩⎪ D6 ⎭⎪ ⎪⎩− 2.929 × 10 ⎪⎭
For member 1: 0 ⎧ ⎫ ⎧0⎫ ⎧− 6.84⎫ ⎪ ⎪ ⎪0⎪ ⎪ 1.36 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ 0 0 14 . 94 ⎪ ⎪ ⎪ ⎪ ⎪ + = { F1 } g = [ k1 ] g ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ −4 ⎪ 7.29 × 10 ⎪ ⎪0⎪ ⎪ 6.84 ⎪ ⎪ − 1.02 × 10 −3 ⎪ ⎪0⎪ ⎪ − 1.36 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 2.81 × 10 − 4 ⎪⎭ ⎪⎩0⎪⎭ ⎪⎩ 5.56 ⎪⎭
For member 2: ⎧ 7.29 × 10 −4 ⎫ ⎧ 0 ⎫ ⎧ 1.16 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −3 ⎪ ⎪ − 1.02 × 10 ⎪ ⎪ 3 ⎪ ⎪ 1.379 ⎪ ⎪⎪ − 2.81 × 10 − 4 ⎪⎪ ⎪⎪ 2 ⎪⎪ ⎪⎪ − 5.49 ⎪⎪ { F2 } g = [ k 2 ] g ⎨ ⎬ ⎬+⎨ ⎬=⎨ −4 ⎪ ⎪ 0 ⎪ ⎪ − 1.16 ⎪ ⎪ 9.9 × 10 ⎪ − 1.275 × 10 −3 ⎪ ⎪− 3⎪ ⎪ − 1.379 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 2.929 × 10 −5 ⎪⎭ ⎪⎩ 2 ⎪⎭ ⎪⎩− 0.4644⎪⎭
For member 3: ⎧ 9.9 × 10 −4 ⎫ ⎧− 2⎫ ⎧ 0.5 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −3 ⎪ ⎪ − 1.275 × 10 ⎪ ⎪ 2 ⎪ ⎪ 1.15 ⎪ −5 ⎪⎪− 2.929 × 10 ⎪⎪ ⎪⎪− 2⎪⎪ ⎪⎪ 5.27 ⎪⎪ { F3 } g = [ k 3 ] g ⎨ ⎬ ⎬+⎨ ⎬=⎨ 0 ⎪ ⎪ 2 ⎪ ⎪ − 0.5 ⎪ ⎪ ⎪ ⎪− 2⎪ ⎪− 1.15⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩− 2⎪⎭ ⎪⎩ 5.76 ⎪⎭ ⎪⎩ 0
For member 4: ⎧ 9.9 × 10 −4 ⎫ ⎧ 0 ⎫ ⎧ 0.66 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ −3 ⎪ ⎪ − 1.275 × 10 ⎪ ⎪ 4 ⎪ ⎪ 0.215 ⎪ −5 ⎪ ⎪ ⎪⎪ 4.8 ⎪⎪ ⎪⎪ − 6.8 ⎪⎪ g g ⎪− 2.929 × 10 ⎪ { F4 } = [ k 4 ] ⎨ ⎬ ⎬=⎨ ⎬+⎨ 0 ⎪ ⎪ 0 ⎪ ⎪− 0.66⎪ ⎪ ⎪ ⎪ 4 ⎪ ⎪ 7.78 ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭ ⎪⎩− 4.8⎪⎭ ⎪⎩ − 15.9 ⎪⎭ ⎪⎩ 0
Example 4: By using stiffness matrix method, draw internal forces diagrams for the shown frame; if: EA = 4 *103 t. EI = 5 * 104 t.m2. Δ2 = 2 cm. Δt = 20° for member 2. α = 1 * 10-5. h = 60 cm.
D5
D6
1 1
6.00
D4
3 2
4
3
2t/m 4.00
2 4.00
Solution
2.4
0.0
Q D.O.F. = 3
0.0
1.8
∴ {F }3 x1 = [K ]3 x 3 {D}3 x1
∴ { f }3 x1
⎧1.8 ⎫ ⎪ ⎪ = ⎨ 0 ⎬ (due to loads) ⎪2.4⎪ ⎩ ⎭
2t/m
4.2 3.6
∴ { f }3 x1
⎧ − 0.8 ⎫ ⎪ ⎪ = ⎨ 0 ⎬ (due to temp.) ⎪− 16.67 ⎪ ⎭ ⎩
16.67 0.8
0.0
t=0 t = 20
16.67 0.8
0.0
∴ { f }3 x1
⎧ 0 ⎫ ⎪ ⎪ = ⎨361.67 ⎬ (due to settle.) ⎪ 0 ⎪ ⎭ ⎩
13.33 375
375
187.5
∴ {F }3 x1
375
375
187.5
187.5
⎧ 1 ⎫ ⎪ ⎪ = ⎨ 361.67 ⎬ ⎪− 14.27 ⎪ ⎭ ⎩
13.33
187.5
Elemen t 1 2 3
L
θ
C
s
EA/L
12EI/L3
6EI/L2
4EI/L
2EI/L
4.0 4.0 6.0
0 180 90
1 -1 0
0 0 1
1000 1000 666.67
9375 9375 2777.78
18750 18750 8333.33
50000 50000 33333.33
25000 25000 16666.7
Ends F-F F-F F-F
For element 1:
[k1 ]g = [k1 ]l
0 0 − 1000 0 0 ⎤ ⎡ 1000 ⎢ 0 9375 18750 0 − 9375 18750 ⎥⎥ ⎢ ⎢ 0 18750 50000 0 − 18750 25000 ⎥ =⎢ ⎥ 0 0 1000 0 0 ⎥ ⎢− 1000 ⎢ 0 − 9375 − 18750 0 9375 − 18750⎥ ⎢ ⎥ 18750 25000 0 − 18750 50000 ⎦⎥ ⎣⎢ 0
For element 2:
[k 2 ]g = [k 2 ]l
0 0 − 1000 0 0 ⎤ ⎡ 1000 ⎢ 0 9375 − 18750 0 − 9375 − 18750⎥⎥ ⎢ ⎢ 0 − 18750 50000 0 18750 25000 ⎥ =⎢ ⎥ 0 0 1000 0 0 ⎥ ⎢− 1000 ⎢ 0 − 9375 18750 0 9375 18750 ⎥ ⎢ ⎥ − 18750 25000 0 18750 50000 ⎥⎦ ⎢⎣ 0
For element 3:
∴ [k 3 ]
g
− 8333.3 − 2777.78 − 8333.3⎤ 0 0 ⎡ 2777.78 ⎢ − 666.7 0 666.7 0 0 0 ⎥⎥ ⎢ ⎢ − 8333.3 0 33333.3 8333.3 0 16666.7 ⎥ =⎢ ⎥ 0 8333.3 2777.78 0 8333.3 ⎥ ⎢− 2777.78 ⎢ − 666.7 0 0 0 666.7 0 ⎥ ⎢ ⎥ 0 16666.7 8333.3 0 33333.3 ⎥⎦ ⎢⎣ − 8333.3
Overall equilibrium equation: 0 8333.3 ⎤ ⎧ D1 ⎫ ⎧ 1 ⎫ ⎡4777.78 ⎪ ⎢ ⎪ ⎥ ⎪D ⎪ 194166.67 0 ⎨ 361.67 ⎬ = ⎢ 0 ⎥ ⎨ 2⎬ ⎪− 14.27 ⎪ ⎢ 8333.3 0 133333.33⎥⎦ ⎪⎩ D3 ⎪⎭ ⎭ ⎣ ⎩
By solving 3 equations; then
−4 ⎧ D1 ⎫ ⎧ 4.44 × 10 ⎫ ⎪ ⎪ ⎪ −3 ⎪ ⎨ D2 ⎬ = ⎨ 1.86 × 10 ⎬ ⎪ D ⎪ ⎪− 1.348 × 10 − 4 ⎪ ⎩ 3⎭ ⎩ ⎭
For member 1: 0 ⎧ ⎫ ⎧ 0 ⎫ ⎧ − 0.44 ⎫ ⎪ ⎪ ⎪ 187.5 ⎪ ⎪ 167.51 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ 0 375 336 . 7 ⎪ ⎪ ⎪ ⎪ ⎪ { F1 } g = [ k1 ] g ⎨ +⎨ ⎬=⎨ ⎬ −4 ⎬ ⎪ 4.44 × 10 ⎪ ⎪ 0 ⎪ ⎪ 0.44 ⎪ ⎪ 1.86 × 10 −3 ⎪ ⎪− 187.5⎪ ⎪− 167.51⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 1.348 × 10 − 4 ⎪⎭ ⎪⎩ 375 ⎪⎭ ⎪⎩ 333.33 ⎪⎭
For member 2: 0 ⎧ ⎫ ⎧ − 0.8 ⎫ ⎧ 0 ⎫ ⎧ − 1.24 ⎫ ⎪ ⎪ ⎪ 0 ⎪ ⎪ 187.5 ⎪ ⎪ 172.56 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ 0 16 . 67 − 375 − − 360 . 11 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ { F2 } g = [ k 2 ] g ⎨ +⎨ ⎬+⎨ ⎬=⎨ ⎬ −4 ⎬ ⎪ 4.44 × 10 ⎪ ⎪ 0.8 ⎪ ⎪ 0 ⎪ ⎪ 1.24 ⎪ ⎪ 1.86 × 10 −3 ⎪ ⎪ 0 ⎪ ⎪− 187.5⎪ ⎪ − 172.56 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 1.348 × 10 − 4 ⎪⎭ ⎪⎩ 16.67 ⎪⎭ ⎪⎩ − 375 ⎪⎭ ⎪⎩− 330.14⎪⎭
For member 3: 0 ⎧ ⎫ ⎧ − 3.8⎫ ⎧ 0 ⎫ ⎧ − 3.9 ⎫ ⎪ ⎪ ⎪ 0 ⎪ ⎪− 13.33⎪ ⎪− 14.53⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ 0 3 . 6 0 5 . 06 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ { F3 } g = [ k 3 ] g ⎨ + + = ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ −4 ⎪ 4.44 × 10 ⎪ ⎪ − 1.8 ⎪ ⎪ 0 ⎪ ⎪ − 1.69 ⎪ ⎪ 1.86 × 10 −3 ⎪ ⎪ 0 ⎪ ⎪ 13.33 ⎪ ⎪ 14.57 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩− 1.348 × 10 − 4 ⎪⎭ ⎪⎩− 2.4⎪⎭ ⎪⎩ 0 ⎪⎭ ⎪⎩ − 3.19 ⎪⎭
Example 5: By using stiffness matrix method, draw internal forces diagrams for the shown frame.
6 t.m
1 1
3t
2t
3 2
2
3
5.00
4 4.00
4.00
Solution ∴ {F }3 x1 = [K ]3 x 3 {D}3 x1
Q D.O.F. = 3 Elemen t 1 2 3
∴ {F }3 x1
⎧2⎫ ⎪ ⎪ = ⎨− 3⎬ ⎪6⎪ ⎩ ⎭
L
Θ
c
s
EA/L
12EI/L3
6EI/L2
4EI/L
2EI/L
4.0 4.0 5.0
0 0 270
1 1 0
0 0 -1
1000 1000 800
9375 9375 4800
18750 18750 12000
50000 50000 40000
25000 25000 20000
[k1 ]g = [k1 ]l
⎡− ⎢− ⎢ ⎢− =⎢ ⎢− ⎢− ⎢ ⎢⎣−
Ends F-F F-F F-F
For element 1: − − − − − −
− − − ⎤ − − − ⎥⎥ − − − ⎥ ⎥ 0 0 ⎥ − − 1000 0 9375 − 18750⎥ − − ⎥ 0 − − − 18750 50000 ⎥⎦
For element 2:
[k 2 ]g = [k 2 ]l
0 0 − − −⎤ ⎡1000 ⎢ 0 9375 18750 − − −⎥⎥ ⎢ ⎢ 0 18750 50000 − − −⎥ =⎢ ⎥ − − − − −⎥ ⎢ − ⎢ − − − − − −⎥ ⎢ ⎥ − − − − −⎥⎦ ⎢⎣ −
For element 3:
∴ [k 3 ]
l
∴ [k 3 ]
g
⎡0 ⎢− 1 ⎢ ⎢0 =⎢ ⎢0 ⎢0 ⎢ ⎣⎢ 0
− − −⎤ 0 0 ⎡800 ⎢ 0 9375 18750 − − −⎥⎥ ⎢ ⎢ 0 18750 50000 − − −⎥ =⎢ ⎥ − − − − −⎥ ⎢ − ⎢ − − − − − −⎥ ⎢ ⎥ − − − − −⎥⎦ ⎢⎣ −
1 0 0 0 0 1 0 0 0 0 0 0
∴ [k 3 ]
g
0 0⎤ 0 0 0⎥⎥ 0 0 0⎥ l ⎥ [k 3 ] 0 1 0⎥ − 1 0 0⎥ ⎥ 0 0 1⎦⎥ 0
⎡0 − 1 ⎢1 0 ⎢ ⎢0 0 ⎢ ⎢0 0 ⎢0 0 ⎢ ⎣⎢0 0
0 0 0 1 0 0
0 0 0 1
0 0
0 12000 − − −⎤ ⎡ 4800 ⎢ 0 − − −⎥⎥ 800 0 ⎢ ⎢12000 0 40000 − − −⎥ =⎢ ⎥ − − − − −⎥ ⎢ − ⎢ − − − − − −⎥ ⎢ ⎥ − − − − −⎥⎦ ⎢⎣ −
Overall equilibrium equation: 0 12000 ⎤ ⎧ D1 ⎫ ⎧ 2 ⎫ ⎡ 6800 ⎪ ⎪ ⎪ ⎪ ⎢ 19550 0 ⎥⎥ ⎨ D2 ⎬ ⎨− 3⎬ = ⎢ 0 ⎪ 6 ⎪ ⎢12000 0 140000⎥⎦ ⎪⎩ D3 ⎪⎭ ⎩ ⎭ ⎣
By solving 3 equations; then
−4 ⎧ D1 ⎫ ⎧ 2.57 × 10 ⎫ ⎪ ⎪ ⎪ −4 ⎪ ⎨ D2 ⎬ = ⎨− 1.53 × 10 ⎬ ⎪ D ⎪ ⎪ 2.07 × 10 −5 ⎪ ⎩ 3⎭ ⎩ ⎭
0⎤ 0 0⎥⎥ 0 0⎥ ⎥ − 1 0⎥ 0 0⎥ ⎥ 0 1⎦⎥ 0
For member 1: ⎡− ⎢− ⎢ ⎢− g { F1 } = ⎢ ⎢− ⎢− ⎢ ⎢⎣−
− − − −
− − 1000 0 0 ⎤⎧ 0 ⎫ ⎧0⎫ ⎧− 0.257 ⎫ ⎥ ⎪ ⎪ ⎪0⎪ ⎪ 1.822 ⎪ − − 9375 18750 ⎥ ⎪ 0 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪0⎪⎪ ⎪⎪ 3.38 ⎪⎪ − − 18750 25000 ⎥ ⎪⎪ 0 0 ⎥⎨ ⎬+⎨ ⎬=⎨ ⎬ − 1000 0 0 ⎥ ⎪ 2.57 × 10 − 4 ⎪ ⎪0⎪ ⎪ 0.257 ⎪ − − − 18750⎥ ⎪− 1.53 × 10 − 4 ⎪ ⎪0⎪ ⎪ − 1.822 ⎪ 0 9375 ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − − − 18750 50000 ⎥⎦ ⎪⎩ 2.07 × 10 −5 ⎪⎭ ⎪⎩0⎪⎭ ⎪⎩ 3.9 ⎪⎭ 0
For member 2: 0 0 ⎡ 1000 ⎢ 0 9375 18750 ⎢ ⎢ 0 18750 50000 { F2 } g = ⎢ 0 0 ⎢− 1000 ⎢ 0 − 9375 − 18750 ⎢ 18750 25000 ⎣⎢ 0
− − − ⎤ ⎧ 2.57 × 10 −4 ⎫ ⎧0⎫ ⎧ 0.257 ⎫ ⎪ ⎪ − − − ⎥⎥ ⎪− 1.53 × 10 − 4 ⎪ ⎪⎪0⎪⎪ ⎪⎪ − 1.046 ⎪⎪ − − − ⎥ ⎪⎪ 2.07 × 10 −5 ⎪⎪ ⎪⎪0⎪⎪ ⎪⎪ − 1.833 ⎪⎪ ⎬ ⎬+⎨ ⎬=⎨ ⎥⎨ − − −⎥ ⎪ 0 ⎪ ⎪0⎪ ⎪− 0.257 ⎪ ⎪ ⎪0⎪ ⎪ 1.04 ⎪ − − −⎥ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ − − − ⎦⎥ ⎪⎩ ⎪⎭ ⎪⎩0⎪⎭ ⎪⎩ − 2.35 ⎪⎭ 0
For member 3: 0 12000 ⎡ 4800 ⎢ 0 800 0 ⎢ ⎢ 12000 0 40000 { F3 } g = ⎢ 0 − 12000 ⎢− 4800 ⎢ 0 0 − 800 ⎢ 0 20000 ⎢⎣ 12000
− − − ⎤ ⎧ 2.57 × 10 −4 ⎫ ⎧0⎫ ⎧ 1.482 ⎫ ⎪ ⎪ − − − ⎥⎥ ⎪− 1.53 × 10 − 4 ⎪ ⎪⎪0⎪⎪ ⎪⎪− 0.122⎪⎪ − − − ⎥ ⎪⎪ 2.07 × 10 −5 ⎪⎪ ⎪⎪0⎪⎪ ⎪⎪ 3.91 ⎪⎪ ⎬ ⎬+⎨ ⎬=⎨ ⎥⎨ − − −⎥ ⎪ 0 ⎪ ⎪0⎪ ⎪ − 1.48 ⎪ ⎪ ⎪0⎪ ⎪ 0.122 ⎪ − − −⎥ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎥⎪ − − − ⎥⎦ ⎪⎩ ⎪⎭ ⎪⎩0⎪⎭ ⎪⎩ 3.49 ⎪⎭ 0
Example 6: 1
By using stiffness matrix method, draw internal forces diagrams for the shown frame.
2t/m
5t 1
2t/m 2
3
4
2.00 4t
4 3
2.00
2 3.00
Solution Q D.O.F. = 5
∴ {F }5×1
Elemen t 1 2 3
5t
3.75
∴ {F }5×1 = [K ]5×5 {D}5×1
3.00
2.5
3.00
2t/m 3.75
2.5
⎧− 3.75⎫ ⎪ 2 ⎪ ⎪⎪ ⎪⎪ = ⎨ −2 ⎬ ⎪ − 5.5 ⎪ ⎪ ⎪ ⎪⎩ − 0.5 ⎪⎭
3.00
2t/m
2.25 2.0
2.0
2.25 3.0
3.0
4t 0.0 2.0 2.0
L
Θ
c
s
EA/L
12EI/L3
6EI/L2
6 6 4
0 0 90
1 1 0
0 0 1
666.67 666.67 1000
2777.78 8333.33 33333.33 16666.67 2777.78 8333.33 33333.33 16666.67 9375 18750 50000 25000
4EI/L
2EI/L
Ends F-F F-F F-F
For element 1:
[k1 ]g = [k1 ]l
0 0 0 0 − 666.67 ⎡ 666.67 ⎤ ⎢ 0 2777.78 8333.33 0 − 2777.78 8333.33 ⎥⎥ ⎢ ⎢ 0 8333.33 33333.33 0 − 8333.33 16666.67 ⎥ =⎢ ⎥ 0 0 666.67 0 0 ⎢− 666.67 ⎥ ⎢ 0 0 2777.78 − 8333.33⎥ − 2777.78 − 8333.33 ⎢ ⎥ 0 8333.33 16666.67 0 − 8333.33 33333.33 ⎥⎦ ⎢⎣
For element 2:
[k1 ]g = [k1 ]l
0 0 0 0 − 666.67 ⎡ 666.67 ⎤ ⎢ 0 2777.78 8333.33 0 − 2777.78 8333.33 ⎥⎥ ⎢ ⎢ 0 8333.33 33333.33 0 − 8333.33 16666.67 ⎥ =⎢ ⎥ 0 0 666.67 0 0 ⎢− 666.67 ⎥ ⎢ 0 0 2777.78 − 8333.33⎥ − 2777.78 − 8333.33 ⎢ ⎥ 8333.33 16666.67 0 − 8333.33 33333.33 ⎥⎦ ⎢⎣ 0
For element 3:
∴ [k 3 ]
g
− 18750 − 9375 − 18750⎤ 0 0 ⎡ 9375 ⎢ 0 − 1000 1000 0 0 0 ⎥⎥ ⎢ ⎢− 18750 0 50000 18750 0 25000 ⎥ =⎢ ⎥ 0 18750 9375 0 18750 ⎥ ⎢ − 9375 ⎢ 0 − 1000 0 0 1000 0 ⎥ ⎢ ⎥ − 0 25000 18750 50000 ⎥⎦ ⎢⎣ 18750
Overall equilibrium equation: − 8333.33 16666.67 ⎤ ⎧ D1 ⎫ 0 0 ⎧− 3.75⎫ ⎡ 33333.33 ⎪ 2 ⎪ ⎢ 0 50000 18750 0 25000 ⎥⎥ ⎪⎪ D2 ⎪⎪ ⎪⎪ ⎪⎪ ⎢ ⎪ ⎪ 0 18750 10708.34 0 18750 ⎥ ⎨ D3 ⎬ ⎨ −2 ⎬= ⎢ ⎥⎪ ⎪ ⎪ − 5.5 ⎪ ⎢− 8333.33 0 0 6555.56 0 ⎥ ⎪ D4 ⎪ ⎪ ⎪ ⎢ ⎪⎩ − 0.5 ⎪⎭ ⎢⎣ 16666.67 25000 18750 0 16666.67 ⎥⎦ ⎪⎩ D5 ⎪⎭
By solving 5 equations; then 0.0 ⎧ D1 ⎫ ⎧ ⎫ ⎪D ⎪ ⎪ ⎪ 0.0 ⎪⎪ 2 ⎪⎪ ⎪⎪ ⎪ −3 ⎪ = D − × 1 . 6559 10 ⎨ 3⎬ ⎨ ⎬ ⎪ D ⎪ ⎪ − 1.96 × 10 −3 ⎪ ⎪ ⎪ 4⎪ ⎪ ⎪⎩ D5 ⎪⎭ ⎪⎩ 2.76 × 10 −3 ⎪⎭
For member 1:
{ F1 } g = [k1 ]
g
0 ⎧ ⎫ ⎧ 1.104 ⎫ ⎪ ⎪ ⎪ 13.455 ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪⎪ 0 ⎪⎪ 0 l + {F1 } = ⎨ ⎨ ⎬ −3 ⎬ ⎪− 1.6559 × 10 ⎪ ⎪ − 1.104 ⎪ ⎪ − 1.96 × 10 −3 ⎪ ⎪− 8.455⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 2.76 × 10 −3 ⎪⎭ ⎪⎩ 65.74 ⎪⎭
Similarly: ⎧ − 1.104 ⎫ ⎪− 20.72⎪ ⎪ ⎪ ⎪ ⎪⎪ 78 . 42 ⎪ { F2 } g = ⎨ ⎬ ⎪ 1.104 ⎪ ⎪ 14.72 ⎪ ⎪ ⎪ ⎪⎩ 27.89 ⎪⎭
⎧ − 21.5 ⎫ ⎪1.9064⎪ ⎪ ⎪ ⎪ ⎪⎪ 0 ⎪ { F3 } g = ⎨ ⎬ ⎪ 25.5 ⎪ ⎪1.9064⎪ ⎪ ⎪ ⎪⎩ 94 ⎪⎭
Example 7: Construct the equilibrium equation of the following frame.
6t
1
2
3
D2
D3
2.00
D1 2 t / m
D6
2t
1
3
D5 D4
2t
4
EA = 3 x 10 t. EI = 4 x 104 t.m2. 6.00
4t/m 3.00
3.00
3.0
3.0
Q D.O.F. = 6 ∴ {F }6×1 = [K ]6×6 {D}6×1
∴ {F }6×1
1
3.00 12.0
3.0
Solution
⎧ 6.6 ⎫ ⎪− 10⎪ ⎪ ⎪ ⎪⎪ 0.3 ⎪⎪ =⎨ ⎬ ⎪ 0 ⎪ ⎪−9 ⎪ ⎪ ⎪ ⎪⎩ 3.5 ⎪⎭
4
3
3.00
2t/m
3.0 7.5
2.00
2t
7.5
6.0 3.0
3.0
4.8
7.0 3.6
4t/m
8.4 7.2
7.0
Element
L
Θ
c
S
EA/L
12EI/L3
6EI/L2
4EI/L
2EI/L
6 6 6
0 0 90
1 1 0
0 0 1
500 500 500
222.22 222.22 222.22
666.67 666.67 666.67
2666.67 2666.67 2666.67
1333.33 1333.33 1333.33
Ends 1 2 3
F-F F-F F-F
For element 1:
[k1 ]g = [k1 ]l
⎡− ⎢− ⎢ ⎢− =⎢ ⎢− ⎢− ⎢ ⎣⎢−
− −
−
−
−
⎤ − − − − − ⎥⎥ − − − − − ⎥ ⎥ 0 0 ⎥ − − 500 222.22 − 666.67⎥ − − 0 ⎥ − − 0 − 666.67 2666.67 ⎦⎥
For element 2:
[k 2 ]g = [k 2 ]l
0 0 0 0 − 500 ⎡ 500 ⎤ ⎢ 0 222.22 666.67 0 − 222.22 666.67 ⎥⎥ ⎢ ⎢ 0 666.67 2666.67 0 − 666.67 1333.33 ⎥ =⎢ ⎥ 0 0 500 0 0 ⎢− 500 ⎥ ⎢ 0 0 222.22 − 666.67⎥ − 222.22 − 666.67 ⎢ ⎥ 666.67 1333.33 0 − 666.67 2666.67 ⎦⎥ ⎣⎢ 0
∴ [k 4 ]
g
⎡ − ⎢ − ⎢ ⎢ − = ⎢− ⎢ − ⎢ − ⎢ ⎣⎢ −
− − − − −
− − − − ⎤ − − − − ⎥⎥ − − − − ⎥ ⎥ − 215.84 137.42 235.40 ⎥ − 137.42 215.84 − 235.40⎥ ⎥ − − 235.40 − 235.40 1884.57 ⎦⎥
Overall equilibrium equation: 0 666.67 0 0 ⎤ ⎧ D1 ⎫ − 500 ⎧ 6.6 ⎫ ⎡1222.22 ⎪− 10⎪ ⎢ 0 944.44 0 0 − 222.22 666.67 ⎥⎥ ⎪⎪ D2 ⎪⎪ ⎪ ⎢ ⎪ 0 8000.01 0 − 666.67 1333.33 ⎥ ⎪⎪ D3 ⎪⎪ ⎪⎪ 0.3 ⎪⎪ ⎢ 666.67 ⎥⎨ ⎬ ⎬=⎢ ⎨ 0 0 853.36 0 0 ⎥ ⎪ D4 ⎪ ⎪ 0 ⎪ ⎢ − 500 ⎪−9 ⎪ ⎢ 0 0 300.66 − 999.63⎥ ⎪ D5 ⎪ − 222.22 − 666.67 ⎥⎪ ⎪ ⎪ ⎢ ⎪ ⎪⎩ 3.5 ⎪⎭ ⎢⎣ 0 666.67 1333.33 0 − 999.63 4551.24 ⎥⎦ ⎪⎩ D6 ⎪⎭
