Strength of Materials

CHAPTER 6

PROBLEM 6.1 Three boards, each of 1.5 × 3.5-in. rectangular cross section, are nailed together to form a beam that is subjected to a vertical shear of 250 lb. Knowing that the spacing between each pair of nails is 2.5 in., determine the shearing force in each nail.

SOLUTION I =

1 3 1 (3.5)(4.5)3 = 26.578 in 4 bh = 12 12

A = (3.5)(1.5) = 5.25 in 2 y1 = 1.5 in. Q = Ay1 = 7.875 in 3 q=

VQ (250)(7.875) = = 74.074 lb/in I 26.578

qs = 2 Fnail

Fnail =

qs (74.074)(2.5) = 2 2

Fnail = 92.6 lb 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

PROBLEM 6.2 Three boards, each 2 in. thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 150 lb, determine the allowable shear if the spacing s between the nails is 3 in.

SOLUTION 1 3 bh + Ad 2 12 1 (6)(2)3 + (6)(2)(3) 2 = 112 in 4 = 12 1 3 1 I2 = bh = (2)(4)3 = 10.667 in 4 12 12 I1 =

I 3 = I1 = 112 in 4 I = I1 + I 2 + I 3 = 234.667 in 4 Q = A1 y1 = (6)(2)(3) = 36 in 3 qs = Fnail

(1)

VQ I

(2)

q=

Dividing Eq. (2) by Eq. (1),

1 VQ = s Fnail I V =

Fnail I (150)(234.667) = Qs (36)(3)

V = 326 lb 


PROBLEM 6.3 Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is s = 75 mm and that the allowable shearing force in each nail is 400 N, determine the allowable shear when w = 120 mm.

SOLUTION Part

A (mm 2 )

d (mm)

7200

Middle Plank Bottom Plank

Top plank

Ad 2 (106 mm 4 )

I (106 mm 4 )

60

25.92

2.16

12,000

0

0

3.60

7200

60

25.92

2.16

51.84

7.92

Σ

I = Ad 2 +  I = 59.76 × 106 mm 4 = 59.76 × 10−6 m 4 Q = (7200)(60) = 432 × 103 mm3 = 432 × 10−6 m3 VQ I Fnail q= s

q=

V=

Fnail = qs V =

Iq IFnail = Q Qs

(59.76 × 10−6 )(400) (432 × 10−6 )(75 × 10−3 )

V = 738 N 


PROBLEM 6.4 Solve Prob. 6.3, assuming that the width of the top and bottom boards is changed to w = 100 mm . PROBLEM 6.3 Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is s = 75 mm and that the allowable shearing force in each nail is 400 N, determine the allowable shear when w = 120 mm.

SOLUTION A (mm 2 )

d (mm)

6000

Middle Plank Bottom Plank

Part Top plank

Ad 2 (106 mm 4 )

I (106 mm 4 )

60

21.6

1.80

12,000

0

0

3.60

6000

60

21.6

1.80

43.2

7.20

Σ

I = Ad 2 +  I = 50.4 × 106 mm 4 = 50.4 × 10−6 m 4 Q = (6000)(60) = 360 × 103 mm3 = 360 × 10−6 m3 VQ I Fnail q= s

q=

V=

Fnail = qs V =

Iq IFnail = Q Qs

(50.4 × 10−6 )(400) (360 × 10−6 )(75 × 10−3 )

V = 747 N 


PROBLEM 6.5 The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 × 200-mm plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

SOLUTION Calculate moment of inertia: A (mm 2 ) 3200 6650 3200

Part Top plate S310 × 52 Bot. plate Σ *d

=

d (mm) *

160.5 0

*

160.5

Ad 2 (106 mm 4 ) 82.43 82.43 164.86

I (106 mm 4 ) 0.07 95.3 0.07 95.44

305 16 + = 160.5 mm 2 2

I = ΣAd 2 + ΣI = 260.3 × 106 mm 4 = 260.3 × 10−6 m 4 Q = Aplate d plate = (3200)(160.5) = 513.6 × 103 mm3 = 513.6 × 10−6 m3 Abolt =

π 4

2 d bolt =

π 4

(18 × 10−3 ) 2 = 254.47 × 10−6 m 2

Fbolt = τ all Abolt = (90 × 106 )(254.47 × 10−6 ) = 22.90 × 103 N qs = 2 Fbolt q=

VQ I

q = V =

2Fbolt (2)(22.90 × 103 ) = = 381.7 × 103 N/m −3 s 120 × 10

Iq (260.3 × 10−6 )(381.7 × 103 ) = = 193.5 × 103 N Q 513.6 × 10−6 V = 193.5 kN 


PROBLEM 6.6 Solve Prob. 6.5, assuming that the reinforcing plates are only 12 mm thick. PROBLEM 6.5 The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 × 200-mm plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force.

SOLUTION Calculate moment of inertia: A(mm 2 ) 2400 6650 2400

Part Top plate S310 × 52 Bot. plate Σ *

d =

d (mm) *158.5

0 *158.5

Ad 2 (106 mm 4 ) 60.29 0 60.29 120.58

I (106 mm 4 ) 0.03 95.3 0.03 95.36

305 12 + = 158.5 mm 2 2

I = ΣAd 2 + ΣI = 215.94 × 106 mm 4 = 215.94 × 10−6 m 4 Q = Aplate d plate = (200)(12)(158.5) = 380.4 × 103 mm3 = 380.4 × 10−6 m3 Abolt =

π 4

2 d bolt =

π 4

(18 × 10−3 ) 2 = 254.47 × 10−6 m 2

Fbolt = τ all Abolt = (90 × 106 )(254.47 × 10−6 ) = 22.902 × 103 N qs = 2 Fbolt q=

VQ I

q = V =

2Fbolt (2)(22.903 × 103 ) = = 381.7 × 103 N/m s 120 × 10−3

Iq (215.94 × 10−6 )(381.7 × 103 ) = = 217 × 103 N Q 380.4 × 10−6 V = 217 kN 


PROBLEM 6.7 A columm is fabricated by connecting the rolled-steel members shown by bolts of 34 -in. diameter spaced longitudinally every 5 in. Determine the average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis.

SOLUTION Geometry: d  f =   + (t w )C  2 s 10.0 = + 0.303 = 5.303 in. 2 x = 0.534 in. y1 = f − x = 5.303 − 0.534 = 4.769 in.

Determine moment of inertia. Part C8 × 13.7

S 10 × 25.4 C8 × 13.7 Σ

d (in.) 4.769 0 4.769

A(in 2 ) 4.04 7.4 4.04

Ad 2 (in 4 ) 91.88 0 91.88 183.76

I (in 4 ) 1.52 123 1.52 126.04

I = ΣAd 2 + ΣI = 183.76 + 126.04 = 309.8 in 4 Q = A y1 = (4.04)(4.769) = 19.267 in 3 VQ (30)(19.267) = = 1.8658 kip/in 309.8 I 1 1 = qs =   (1.8658)(5) = 4.664 kips 2 2

q= Fbolt

Abolt =

τ bolt =

π 4

2 = d bolt

π 3

2

2   = 0.4418 in 4 4

4.664 Fbolt = = 10.56 ksi Abolt 0.4418

τ bolt = 10.56 ksi 


PROBLEM 6.8 The composite beam shown is fabricated by connecting two W6 × 20 rolled-steel members, using bolts of 58 -in. diameter spaced longitudinally every 6 in. Knowing that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest allowable vertical shear in the beam.

SOLUTION W6 × 20: A = 5.87 in 2 , d = 6.20 in., I x = 41.4 in 4 y =

Composite:

1 d = 3.1 in. 2

I = 2[41.4 + (5.87)(3.1)2 ] = 195.621 in 4 Q = A y = (5.87)(3.1) = 18.197 in 3

Bolts:

d = Abolt =

5 in., τ all = 10.5 ksi, s = 6 in. 8

π 5

2

2   = 0.30680 in 4 8

Fbolt = τ all Abolt = (10.5)(0.30680) = 3.2214 kips

Shear:

q=

2Fbolt (2)(3.2214) = = 1.07380 kips/in s 6

q =

VQ I

V =

Iq (195.621)(1.0780) = Q 18.197

V = 11.54 kips 


PROBLEM 6.9 For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

SOLUTION By symmetry, RA = RB .

 Fy = 0: RA + RB − 15 − 20 − 15 = 0 RA = RB = 25 k ips V = 30 kips at n-n.

From shear diagram, Determine moment of inertia. Part

A(in 2 )

Top Flng

6

Web

3.30

Bot. Flng

6

d (in.)

Ad 2 (in 4 )

I (in 4 )

4.7

132.54

0.18

0

0

21.30

4.7

132.54

0.18

265.08

21.66

Σ

I =  Ad 2 +  I = 286.74 in 4

(a)

Part

A(in 2 )



6



1.65

Σ

y (in.)

Ay (in 3 )

4.7

28.2

2.2

3.63 31.83

Q =  Ay = 31.83 in 3 t = 0.375 in.

τ max =

VQmax (25)(31.83) = It (286.74)(0.375)

τ max = 7.40 ksi 


PROBLEM 6.9 (Continued)

(b)

A(in 2 )

Part 

6



0.15

Σ

y (in.)

Ay (in 3 )

4.7

28.2

4.2

0.63 28.83

Q =  Ay = 28.83 in 3 t = 0.375 in.

τ =

VQ (23)(28.83) = It (286.74)(0.375)

τ = 6.70 ksi 



SOLUTION At section n-n, V = 10 kN. I = I1 + 4I 2 =

1  1  b1h13 + 4  b2h23 + A2d 22  12  12 

=

 1   1 (100)(150)3 + 4   (50)(12)3 + (50)(12)(69) 2  12  12  

= 28.125 × 106 + 4 0.0072 × 106 + 2.8566 × 106  = 39.58 × 106 mm 4 = 39.58 × 10−6 m 4

(a)

Q = A1 y1 + 2 A2 y2 = (100)(75)(37.5) + (2)(50)(12)(69) = 364.05 × 103 mm3 = 364.05 × 10−6 m3 t = 100 mm = 0.100 m

τ max =

(b)

VQ (10 × 103 )(364.05 × 10−6 ) = = 920 × 103 Pa −6 It (39.58 × 10 )(0.100)

τ max = 920 kPa 

Q = A1 y1 + 2 A2 y2 = (100)(40)(55) + (2)(50)(12)(69) = 302.8 × 103 mm3 = 302.8 × 10−6 m3 t = 100 mm = 0.100 m

τa =

VQ (10 × 103 )(302.8 × 10−6 ) = = 765 × 103 Pa It (39.58 × 10−6 )(0.100)

τ a = 765 kPa 



SOLUTION

V = 80 kN

At section n-n,

Consider cross section as composed of rectangles of types , , and . 1 (12)(80)3 + (12)(80)(90)2 = 8.288 × 106 mm 4 12 1 (180)(16)3 + (180)(16)(42)2 = 5.14176 × 106 mm 4 I2 = 12 1 (16)(68)3 = 419.24 × 103 mm 4 I3 = 12 I1 =

I = 4 I1 + 2 I 2 + 2I 3 = 44.274 × 106 mm 4 = 44.274 × 10−6 m 4

(a)

Calculate Q at neutral axis. Q1 = (12)(80)(90) = 86.4 × 103 mm 4 Q2 = (180)(16)(42) = 120.96 × 103 mm 4 Q3 = (16)(34)(17) = 9.248 × 103 mm 4 Q = 2Q1 + Q2 + 2Q3 = 312.256 × 103 mm3 = 312.256 × 10−6 m3

τ = (b)

At point a,

VQ (80 × 103 )(312.256 × 10−6 ) = = 17.63 × 106 Pa It (44.274 × 10−6 )(2 × 16 × 10−3 )

τ = 17.63 MPa 

Q = Q1 = 86.4 × 103 mm 4 = 86.4 × 10−6 m 4

τ =

VQ (80 × 103 )(86.4 × 10−6 ) = = 13.01 × 106 Pa −6 −3 It (44.274 × 10 )(12 × 10 )

τ = 13.01 MPa 



SOLUTION By symmetry, RA = RB . ΣFy = 0: RA + RB − 10 − 10 = 0 RA = RB = 10 kips

From the shear diagram,

V = 10 kips at n-n. 1 1 b2h23 − b1h13 12 12 1 1 = (4)(4)3 − (3)(3)3 = 14.583 in 4 12 12

I =

(a)

1 1 Q = A1 y1 + A2 y2 = (3)   (1.75) + (2)   (2)(1) = 4.625 in 3 2 2

t =

τ max =

(b)

1 1 + = 1 in. 2 2 VQ (10)(4.625) = It (14.583)(1)

τ max = 3.17 ksi 

1 Q = Ay = (4)   (1.75) = 3.5 in 3 2 1 1 t = + = 1 in. 2 2

τ =

VQ (10)(3.5) = It (14.583)(1)

τ a = 2.40 ksi 


PROBLEM 6.13 For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION

Calculate moment of inertia. I =

1 1  (50 mm)(120 mm)3 − 2  (30 mm) 4 + (30 mm × 30 mm)(35 mm) 2  12 12 

I = 7.2 × 106 mm 4 − 2[1.170 × 106 mm 4 ] = 4.86 × 106 mm 4 = 4.86 × 10−6 m 4

Assume that τ m occurs at point a.

t = 2(10 mm) = 0.02 m Q = (10 mm × 50 mm)(55mm) + 2[(10 mm × 30 mm)(35 mm)] = 48.5 × 103 mm3 = 48.5 × 10−6 m3



For τ all = 60 MPa,

τ m = τ all = 60 × 106 Pa =

Check τ at neutral axis:

VQ It

V (48.5 × 10−6 m3 ) (4.86 × 10−6 m 4 )(0.02 m)

V = 120.3 kN 

t = 50 mm = 0.05 m Q = (50 × 60)(30) − (30 × 30)(35) = 58.5 × 103 mm3

τ =

VQ (120.3 kN)(58.5 × 10−6 m3 ) = = 29.0 MPa < 60 MPa It (4.86 × 10−6 m 4 )(0.05m)

OK


PROBLEM 6.14 For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION Calculate moment of inertia.

1  1 I = 2  (10 mm)(120 mm)3  + (30 mm)(40 mm)3 12  12 = 2[1.440 × 106 mm 4 ] + 0.160 × 106 mm 4 = 3.04 × 106 mm 4 I = 3.04 × 10−6 m 4

Assume that τ m occurs at point a. t = 10 mm = 0.01m Q = (10 mm × 40 mm)(40 mm) = 16 × 103 mm3 Q = 16 × 10−6 m3

For

τ all = 60 MPa, 60 × 106 Pa =

τ m = τ all = V (16 × 10−6 m3 ) (3.04 × 10−6 m 4 )(0.01m)

VQ It

V = 114.0 kN 

Check τ at neutral axis: t = 50 mm = 0.05 m Q = 2[(10 × 60)(30)] + (30 × 20)(10) = 42 × 103 m3 = 42 × 10−6 m3

τ NA =

VQ (114.0 kN)(42 × 10−6 m3 ) = = 31.5 MPa < 60 MPa It (3.04 × 10−6 m 4 )(0.05 m)

OK


PROBLEM 6.15 For the beam and loading shown, determine the minimum required depth h, knowing that for the grade of timber used, σ all = 1750 psi and τ all = 130 psi.

SOLUTION (750 lb/ft)(16 ft) = 12 × 103 lb

Total load: Reaction at A:

RA = 6 × 103 lb ↑

Vmax = 6 × 103 lb 1 (8 ft)(6 × 103 ) = 24 × 103 lb ⋅ ft 2

M max =

= 288 × 103 lb ⋅ in

Bending: S =

1 2 bh for rectangular section. 6 S =

M max

σ all

h=

Shear: I =

=

6S = b

288 × 103 = 164.57 in 3 1750 (6)(164.57) = 14.05 in. 5

1 3 bh for rectangular section. 12 1 bh 2 1 y = h 4

A=

 1  1  1 Q = Ay = (b)  h  h  = bh 2  2  4  8 VQ 3Vmax τ max = = 2bh Ib

h=

3Vmax (3)(6 × 103 ) = = 13.85 in. 2bτ max (2)(5)(130)

The larger value of h is the minimum required depth.

h = 14.05 in. 


PROBLEM 6.16 For the beam and loading shown, determine the minimum required width b, knowing that for the grade of timber used, σ all = 12 MPa and τ all = 825 kPa.

SOLUTION

M D = 0: − 3RA + (2)(2.4) + (1)(4.8) = 0 RA = 3.2 kN

Draw shear and bending moment diagrams. V

max

= 4.0 kN

Bending: S =

M

max

σ all

M

=

max

= 4.0 kN ⋅ m

4.0 × 103 12 × 106

= 333.33 × 10−6 m3 = 333.33 × 103 mm3

For a rectangular cross section, 1 3 bh I 12 1 = bh 2 S = = 1 c 6 h 2 6S (6)(333.33 × 103 ) = = 88.9 mm h2 1502 1 1 A= bh, y = h 2 4 1 1 Q = Ay = bh 2 , I = bh3 8 12 VQ 3V = τ= It 2bh b=

bh =

3 V 3 4.0 × 103 = 2 τ 2 825 × 103

= 7.2727 × 10−3 m 2 = 7.2727 × 103 mm 2 b=

The required value for b is the larger one.

bh 7.2727 × 103 = = 48.5 mm h 150 b = 88.9 mm 


PROBLEM 6.17 A timber beam AB of length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio τ m /σ m of the maximum values of the shearing and normal stresses in the beam is equal to 2h/L, where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L = 5 m, w = 8 kN/m, τ m = 1.08 MPa, and σ m = 12 MPa.

SOLUTION RA = RB =

From shear diagram, For rectangular section,

wL 4

(1)

A = bh

(2)

|V |m =

τm = From bending moment diagram,

wL 2

3 Vm 3wL = 2 A 8bh

(3)

|M |m =

wL2 32

(4)

S =

1 2 bh 6

(5)

|M |m 3wL2 = S 16 bh2

(6)

For a rectangular cross section,

σm = (a)

Dividing Eq. (3) by Eq. (6),

(b)

Solving for h: h=

Lτ m (5)(1.08 × 106 ) = = 225 × 10−3 m 6 2σ m (2)(12 × 10 )

τm 2h  = L σm

h = 225 mm 

Solving Eq. (3) for b: b=

3wL (3)(8 × 103 )(5) = 8hτ m (8)(225 × 10−3 )(1.08 × 106 )

= 61.7 × 10−3 m

b = 61.7 mm 


PROBLEM 6.18 A timber beam AB of length L and rectangular cross section carries a single concentrated load P at its midpoint C. (a) Show that the ratio τ m /σ m of the maximum values of the shearing and normal stresses in the beam is equal to 2h/L, where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L = 2 m, P = 40 kN, τ m = 960 kPa, and σ m = 12 MPa.

SOLUTION RA = RB = P/2 ↑

Reactions: P 2

(1)

Vmax = RA =

(2)

A = bh for rectangular section.

(3)

τm =

(4)

M max =

(5)

S =

(6)

σm =

3 Vmax 3P = for rectangular section. 2 A 4bh PL 4

1 2 bh for rectangular section. 6 M max 3PL = S 2bh 2

τm h  = σ m 2L

(a)

(b)

Solving for h,

h=

(2)(2)(960 × 103 ) = 320 × 10−3 m 6 12 × 10

h = 320 mm 

3P (3)(40 × 103 ) = = 97.7 × 10−3 m 4hτ m (4)(320 × 10−3 )(960 × 103 )

b = 97.7 mm 

2Lτ m

σm

=

Solving Eq. (3) for b, b=


PROBLEM 6.19 For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 24 ksi and the largest shearing stress using the approximation τ m = V/Aweb is 14.5 ksi.

SOLUTION

 M C = 0: − 15RA + qP = 0 RA = 0.6 P

Draw shear and bending moment diagrams. V

= 0.6 P

max

M

max

= 0.6 PLAB

LAB = 6 ft = 72 in. S = 258 in 3

For W 24 × 104 ,

Bending.

M

S = P=

max

σ all

σ all S 0.6LAB

= =

0.6 PLAB

σ all (24)(258) = 143.3 kips (0.6)(72)

Aweb = dtw

Shear.

= (24.1)(0.500) = 12.05 in 2

τ = P=

V

max

Aweb

τ Aweb 0.6

=

0.6 P Aweb

=

(14.5)(12.05) = 291 kips 0.6

The smaller value of P is the allowable value.

P = 143.3 kips 


PROBLEM 6.20 For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 160 MPa and the largest shearing stress using the approximation τ m = V/Aweb is 100 MPa.

SOLUTION

 M E = 0: − 3.6RA + 3.0P + 2.4P + 1.8P = 0 RA = 2 P ↑

Draw shear and bending moment diagrams. M B = 2PLAB , M C = M D = 3PLAB V

max

= 2P

M

max

= 3PLAB

Bending. For W 360 × 122, S = 2020 × 103 mm3 = 2020 × 10−6 m3

M

max

σ all P=

Shear.

3PLAB

=

σ all

σ all S

=

3LAB

=S

(160 × 106 )(2020 × 103 ) = 179.6 × 103 N (3)(0.6)

Aweb = dt w = (363)(13.0) = 4.719 × 103 mm 2 = 4.719 × 10−3 m 2

τ = P=

The smaller value of P is the allowable one.

V

max

Aweb

τ Aweb 2

=

2P Aweb

=

(100 × 106 )(4.719 × 10−3 ) = 236 × 103 N 2 P = 179.6 kN 


PROBLEM 6.21 For the beam and loading shown, consider section n-n and determine the shearing stress at (a) point a, (b) point b.

SOLUTION |V |max = 90 kN

Draw the shear diagram. Part

A(mm 2 )

y (mm)

A y (103 mm3 )

Ad 2 (106 mm 4 )

I (106 mm 4 )



3200

90

288

25

2.000

0.1067



1600

40

64

−25

1.000

0.8533



1600

40

64

−25

1.000

0.8533

Σ

6400

4.000

1.8133

d(mm)

416 ΣAy 416 × 103 = = 65 mm ΣA 6400

Y =

I = ΣAd 2 + ΣI = (4.000 + 1.8133) × 106 mm 4 = 5.8133 × 106 mm 4 = 5.8133 × 10−6 m 4

(a)

A = (80)(20) = 1600 mm 2 y = 25 mm Qa = Ay = 40 × 103 mm3 = 40 × 10−6 m3

τa =

VQa (90 × 103 )(40 × 10−6 ) = = 31.0 × 106 Pa −6 −3 It (5.8133 × 10 )(20 × 10 )

τ a = 31.0 MPa  (b)

A = (30)(20) = 600 mm 2

y = 65 − 15 = 50 mm

Qb = Ay = 30 × 10 mm = 30 × 10−6 m3 3

τb =

3

VQb (90 × 103 )(30 × 10−6 ) = = 23.2 × 106 Pa It (5.8133 × 10−6 )(20 × 10−3 )

τ b = 23.2 MPa 



SOLUTION RA = RB = 12 kips

Draw shear diagram. V = 12 kips

Determine section properties. Part

A(in 2 )

y (in.)

Ay (in 3 )

d(in.)

Ad 2 (in 4 )

I (in 4 )



4

4

16

2

16

5.333



8

1

8

−1

8

2.667

Σ

12

24

8.000

24 Y =

ΣAy 24 = = 2 in. ΣA 12

I = ΣAd 2 + ΣI = 32 in 4

(a)

A = 1 in 2

y = 3.5 in. Qa = Ay = 3.5 in 3

t = 1 in.

τa = (b)

VQa (12)(3.5) = It (32)(1)

A = 2 in 2

τ a = 1.3125 ksi 

y = 3 in. Qb = Ay = 6 in 3

t = 1 in.

τb =

VQb (12)(6) = It (32)(1)

τ b = 2.25 ksi 


PROBLEM 6.23 For the beam and loading shown, determine the largest shearing stress in section n-n.

SOLUTION |V |max = 90 kN

Draw the shear diagram. Part

A (mm 2 )

x′

Ay (103 mm3 ) d(mm) Ad 2 (106 mm 4 ) I (106 mm 4 )



3200

90

288

25

2.000

0.1067



1600

40

64

−25

1.000

0.8533



1600

40

64

−25

1.000

0.8533

Σ

6400

4.000

1.8133

416

ΣAy 416 × 103 = = 65 mm ΣA 6400

Y =

I = ΣAd 2 + ΣI = (4.000 + 1.8133) × 106 mm 4 = 5.8133 × 106 mm 4 = 5.8133 × 10−6 m 4

Part

A(mm 2 )



3200



300

7.5

2.25



300

7.5

2.25

y (mm)

25

Σ

Ay (103 mm3 )

80

84.5 Q = ΣAy = 84.5 × 103 mm3 = 84.5 × 10−6 m3 t = (2)(20) = 40 mm = 40 × 10−3 m

τ max =

VQ (90 × 103 )(84.5 × 10−6 ) = It (5.8133 × 10−6 )(40 × 10−3 )

= 32.7 × 106 Pa

τ m = 32.7 MPa 



SOLUTION RA = RB = 12 kips

Draw shear diagram. V = 12 kips

Determine section properties. Part

A(in 2 )

y (in.)

Ay (in 3 )

d(in.)

Ad 2 (in 4 )

I (in 4 )



4

4

16

2

16

5.333



8

1

8

−1

8

2.667

Σ

12

24

8.000

24 Y =

ΣAy 24 = = 2 in. ΣA 12

I = ΣAd 2 + ΣI = 32 in 4 Q = A1 y1 = (4)(2) = 8 in 3 t = 1 in.

τ max =

VQ (12)(8) = It (32)(1)

τ max = 3.00 ksi 


PROBLEM 6.25 A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

τ max = k

V A

where A is the cross-sectional area of the beam.

SOLUTION I =

For semicircle,

As =

π 4

π 2

c4

c2

Q = As y =

(a)

τ max occurs at center where

(b)

τ max =

V ⋅ 2 c3 VQ 4V 4V = π 43 = = 2 3A It c ⋅ 2c 3π c 4

and

A = π c2

y =

π 2

c2 ⋅

4c 3π 4c 2 = c3 3π 3 t = 2c.  k =

4 = 1.333  3



τ max = k

V A


SOLUTION A=

1 bh 2

I =

1 3 bh 36

For a cut at location y, A( y) =

1  by  by 2 = y   2 h  2h

y ( y) =

2 2 h− y 3 3

Q( y) = Ay = t ( y) =

by 2 (h − y ) 3

by h

2

by VQ V 3 (h − y) 12 Vy(h − y) 12V τ ( y) = = = = (hy − y 2 ) 3 3 by 3 1 It bh bh ( 36 bh ) h

(a)

To find location of maximum of τ , set

dτ = 0. dy

dτ 12V ( h − 2 ym ) = 0 = dy bh3

(b)

τm

2 12V 12V  1 2  1   3V 3V 2 (hym − ym ) = = =  h −  h  = 3 3 2A bh bh  2  2   bh

ym =

1 h, i.e., at mid-height  2 k =

3 = 1.500  2



τ max = k

V A


SOLUTION A = 2π rmtm

For a thin-walled circular section,

J = Arm2 = 2π rm3tm ,

For a semicircular arc,

y =

I =

1 J = π rm3tm 2

2rm

π

As = π rmtm Q = As y = π rmtm

(a)

t = 2tm

(b)

τ max =

2rm

π

= 2rm2tm

at neutral axis where maximum occurs.  VQ V (2rm2tm ) V 2V = = = 3 It A (π rmtm )(2tm ) π rmtm

k = 2.00 



τ max = k

V A


SOLUTION 1  A = 2  bh  = bh 2 

 1  1 I = 2  bh3  = bh3  12  6

For a cut at location y, where y ≤ h, A( y) =

1  by  by 2  y = 2 h  2h

y ( y) = h −

2 y 3

Q( y) = Ay = t ( y) =

τ ( y) = (a)

by 2 by 3 − 2 3h

by h

2 VQ 6 h by 2 by 3 V   y  y  =V 3 ⋅ ⋅ − = 3   − 2    3h It bh   h  bh by 2  h  

To find location of maximum of τ , set

dτ = 0. dy

dτ V y = 2 [3 − 4 m ] = 0 dy h bh

(b)

τ ( ym ) =

2 V  3 V 3  9 V = 1.125 3   − 2    = bh   4  A  4   8 bh

ym 3 1 = , i.e., ± h from neutral axis.  4 h 4 k = 1.125 


PROBLEM 6.29 The built-up beam shown is made by gluing together five planks. Knowing that in the glued joints the average allowable shearing stress is 350 kPa, determine the largest permissible vertical shear in the beam.

SOLUTION

1 (240 mm)(160 mm)3 12 1 − (200 mm)(80 mm)3 12

I=

= 73.4 × 106 mm 4 I = 73.4 × 10−6 m 4 t = 40 mm = 0.04 m Q = (100 mm × 40 mm)(60 mm) = 240 × 103 mm3 Q = 240 × 10−6 m3

For

τ m = 350 kPa, τm =

VQ : Tt

350 × 10 3 Pa =

V (240 × 10−6 m3 ) (73.4 × 10−6 m 4 )0.04 m

V = 4.28 kN 


PROBLEM 6.30 For the beam of Prob. 6.29, determine the largest permissible horizontal shear. PROBLEM 6.29 The built-up beam shown is made by gluing together five planks. Knowing that in the glued joints the average allowable shearing stress is 350 kPa, determine the largest permissible vertical shear in the beam.

SOLUTION

I =2

1 1 (40)(240)3 + (80)(40)3 12 12

I = 92.6 × 10+6 mm 4 = 92.6 × 10−6 m 4

For

Q = (40 × 100)70 = 280 × 103 mm3 Q = (280 × 10−6 ) m3

τ = 350 kPa, τ =

VQ ; It

350 × 103 Pa =

V (280 × 10−6 m3 ) (92.6 × 10−6 m 4 )(0.04 m)

V = 4630 N

V = 4.63 kN 


PROBLEM 6.31 Several wooden planks are glued together to form the box beam shown. Knowing that the beam is subjected to a vertical shear of 3 kN, determine the average shearing stress in the glued joint (a) at A, (b) at B.

SOLUTION IA =

1 3 1 (60)(20)3 + (60)(20)(50) 2 bh + Ad 2 = 12 12

= 3.04 × 106 mm 4

1 3 1 (60)(20)3 = 0.04 × 106 mm4 bh = 12 12 1 3 1 (20)(120)3 = 2.88 × 106 mm 4 IC = bh = 12 12 IB =

I = 2I A + I B + 2 I C = 11.88 × 106 mm 4 = 11.88 × 10−6 m 4 QA = Ay = (60)(20)(50) = 60 × 103 mm3 = 60 × 10−6 m3 t = 20 mm + 20 mm = 40 mm = 40 × 10−3 m

(a)

τA =

VQA (3 × 103 )(60 × 10−6 ) = = 379 × 103 Pa It (11.88 × 10−6 )(40 × 10−3 )

QB = 0

(b)

τB =

VQB =0 It

τ A = 379 kPa  τB = 0 


PROBLEM 6.32 The built-up timber beam is subjected to a 1500-lb vertical shear. Knowing that the longitudinal spacing of the nails is s = 2.5 in. and that each nail is 3.5 in. long, determine the shearing force in each nail.

SOLUTION I1 =

1 (2)(4)3 + (2)(4)(3) 2 12

= 82.6667 in 4

1 (2)(6)3 = 36 in 4 12 I = 2 I1 + 2I 2

I2 =

= 237.333 in 4 Q = A1 y1 = (2)(4)(3) = 24 in 3 q=

VQ (1500)(24) = = 151.685 lb/in 237.333 I

2 Fnail = qs

Fnail =

1 1 qs =   (151.685)(2.5) 2 2

Fnail = 189.6 lb 


PROBLEM 6.33 The built-up wooden beam shown is subjected to a vertical shear of 8 kN. Knowing that the nails are spaced longitudinally every 60 mm at A and every 25 mm at B, determine the shearing force in the nails (a) at A, (b) at B. (Given: I x = 1.504 × 109 mm 4.)

SOLUTION I x = 1.504 × 109 mm 4 = 1504 × 10−6 m 4 s A = 60 mm = 0.060 m sB = 25 mm = 0.025 m

(a)

QA = Q1 = A1 y1 = (50)(100)(150) = 750 × 103 mm3 = 750 × 10−6 m3 FA = q A s A =

(8 × 103 )(750 × 10−6 )(0.060) VQ1s A = I 1504 × 10−6 FA = 239 N 

(b)

Q2 = A2 y2 = (300)(50)(175) = 2625 × 103 mm3 QB = 2Q1 + Q2 = 4125 × 103 mm3 = 4125 × 10−6 m3 FB = qB sB =

VQB sB (8 × 103 )(4125 × 10−6 )(0.025) = I 1504 × 10−6

FB = 549 N 


PROBLEM 6.34 Knowing that a vertical shear V of 50 kips is exerted on W14 × 82 rolled-steel beam, determine the shearing stress (a) at point a, (b) at the centroid C.

SOLUTION For W14 × 82, d = 14.3 in., b f = 10.1 in., t f = 0.855 in., tw = 0.510 in., I = 881 in 4 (a)

Aa = (4.15)(0.855) = 3.5482 in 2 d t f 14.3 0.855 − = − = 6.7225 in. 2 2 2 2

ya =

Qa = Aa ya = 23.853 in 3 ta = t f = 0.855 in.

τa = (b)

VQa (50)(23.853) = Ita (881)(0.855)

τ a = 1.583 ksi 

A1 = b f t f (10.1)(0.855) = 8.6355 in 2 d tf − = 6.7225 in. 2 2 d  A2 = tw  − t f  = (0.510)(6.295) = 3.2105 in 2 2   y1 =

y2 =

1d  1  − t f  = ( 6.295 ) = 3.1475 in. 2 2  2

QC = A1 y1 + A2 y2 (8.6355)(6.7225) + (3.2105)(3.1475) = 68.157 in 3 tC = t w = 0.510 in.

τC =

VQC (50)(68.157) = ItC (881)(0.510)

τ C = 7.59 ksi 


PROBLEM 6.35 An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

SOLUTION I=

1 1 (80)(80)3 − (56)(68)3 = 1.9460 × 106 mm 4 12 12

= 1.946 × 10−6 m 4

(a)

Qa = A1 y1 + 2 A2 y2 = (56)(6)(37) + (2)(12)(40)(20) = 31.632 × 103 mm3 = 31.632 × 10−6 m3 ta = (2)(12) = 24 mm = 0.024 m

τa =

VQa (150 × 103 )(31.632 × 10−6 ) = = 101.6 × 106 Pa −6 Ita (1.946 × 10 )(0.024)

τ a = 101.6 MPa  (b)

Qb = A1 y1 = (56)(6)(37) = 12.432 × 103 mm3 = 12.432 × 10−6 m3 tb = (2)(6) = 12 mm = 0.012 m

τb =

VQb (150 × 103 )(12.432 × 10−6 ) = = 79.9 × 106 Pa −6 Itb (1.946 × 10 )(0.012)

τ b = 79.9 MPa  


PROBLEM 6.36 Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in the hat-shaped extrusion shown, determine the corresponding shearing stress (a) at point a, (b) at point b.

SOLUTION Neutral axis lies 30 mm above bottom.

τc =

VQc It

τ a Qatc = τ c Qcta

τa =

VQa Ita

τb =

VQb Itb

τ b Qbtc = τ c Qctb

Qc = (6)(30)(15) + (14)(4)(28) = 4260 mm3 tc = 6 mm Qa = (14)(4)(28) = 1568 mm3 ta = 4 mm Qb = (14)(4)(28) = 1568 mm3 tb = 4 mm

τ c = 75 MPa (a)

τa =

Qa tc 1568 6 ⋅ τc = ⋅ ⋅ 75 Qc ta 4260 4

(b)

τb =

Qb tc 1568 6 ⋅ τc = ⋅ ⋅ 75 Qc tb 4260 4

τ a = 41.4 MPa  τ b = 41.4 MPa 


PROBLEM 6.37 Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having the cross section shown, determine the shearing stress at the three points indicated.

SOLUTION

τ =

VQ It

τ is proportional to Q/t.

Qc = (30)(10)(75)

Point c:

= 22.5 × 103 mm3 tc = 10 mm Qc /tc = 2250 mm 2 Qb = Qc + (20)(50)(55)

Point b:

= 77.5 × 103 mm3 tb = 20 mm Qb /tb = 3875 mm 2 Qa = 2Qb + (120)(30)(15)

Point a:

= 209 × 103 mm3 ta = 120 mm Qa /ta = 1741.67 mm 2

τ m = τ b = 75 MPa

(Q/t ) m occurs at b.

τa Qa /ta

τa 1741.67 mm

2

= =

τb Qb /tb

=

τc Qc /tc

75 MPa τa = 2 3875 mm 2250 mm 2

τ a = 33.7 MPa  τ b = 75.0 MPa  τ c = 43.5 MPa 


PROBLEM 6.38 An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stress τ = 9 ksi, determine the shearing stress at the four points indicated.

SOLUTION Qa = (0.2)(0.5)(0.5 − 0.25) = 0.125 in 3 Qb = (0.2)(0.5)(0.3 + 0.25) = 0.055 in 3 Qc = Qa + Qb + (1.4)(0.2)(0.9) = 0.432 in 3 Qd = 2Qa + 2Qb + (3.0)(0.2)(0.9) = 0.900 in.3 Qm = Qd + (0.2)(0.8)(0.4) = 0.964 in 3

τ =

VQ It

Since V, I, and t are constant, τ is proportional to Q.

τa 0.125

=

τb 0.055

=

τc 0.432

=

τd 0.900

=

τm 0.964

=

9 0.964

τ a = 1.167 ksi; τ b = 0.513 ksi; τ c = 4.03 ksi; τ d = 8.40 ksi 


PROBLEM 6.39 Solve Prob. 6.38, assuming that the beam is subjected to a horizontal shear V. PROBLEM 6.38 An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stress τ = 9 ksi, determine the shearing stress at the four points indicated.

SOLUTION Qa = (0.5)(0.2)(1.4) = 0.140 in 3 Qb = (0.5)(0.2)(1.4) = 0.140 in 3 Qc = Qa + Qb + (0.2)(1.4)(0.8) = 0.504 in 3 Qd = 0 Qm = Qc = 0.504 in 3

τ =

VQ Since V, I, and t are constant, τ is proportional to Q. It

τa 0.140

=

τb 0.140

=

τc 0.504

=

τd 0

=

τm Qm

=

9 0.504

τ a = 2.50 ksi; τ b = 2.50 ksi, τ c = 9.00 ksi, τ d = 0 


PROBLEM 6.40 Knowing that a given vertical shear V causes a maximum shearing stress of 50 MPa in a thin-walled member having the cross section shown, determine the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.

SOLUTION Qa = (12)(30)(25 + 10 + 15) = 18 × 103 mm3 Qb = (40)(10)(25 + 5) = 12 × 103 mm3 Qc = Qa + 2Qb + (12)(10)(25 + 5) = 45.6 × 103 mm3  25  Qm = Qc + (12)(25)   = 49.35 × 103 mm3  2  ta = tc = tm = 12 mm tb = 10 mm

τ m = 50 MPa (a)

τ a Qa tm 18 12 = ⋅ = ⋅ = 0.3647 τ m Qm ta 49.35 12

τ a = 18.23 MPa 

(b)

τ b Qb tm 12 12 = ⋅ = ⋅ = 0.2918 τ m Qm tb 49.35 10

τ b = 14.59 MPa 

(c)

τc Q t 45.6 12 = c ⋅ m = ⋅ = 0.9240 τ m Qm tc 49.35 12

τ c = 46.2 MPa 


PROBLEM 6.41 The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION I=

1 1 (2.50)(2.50)3 − (2.125)(2.25)3 = 1.2382 in 4  12 12

t = 0.125 in. at all sections. V = 2 kips  Qa = 0

τa =

VQa It

τ a = 0 

 1.25  3 Qb = (0.125)(1.25)   = 0.09766 in  2  

τb =

VQb (2)(0.09766) = It (1.2382)(0.125)

 τ b = 1.26 ksi 

Qc = Qb + (1.0625)(0.125)(1.1875) = 0.25537 in.2 

τc =

VQc (2)(0.25537) =  It (1.2382)(0.125)

τ c = 3.30 ksi 

Qd = 2Qc + (0.125) 2 (1.1875) = 0.52929 

τd =

VQd (2)(0.52929) = It (1.2382)(0.125)

 τ d = 6.84 ksi 

 1.125  Qe = Qd + (0.125)(1.125)   = 0.60839   2 

τe =

VQ (2)(0.60839) = It (1.2382)(0.125)

τ e = 7.86 ksi 


PROBLEM 6.42 The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION I=

1 1 (2.50)(2.50)3 − (2.125)(2.25)3 = 1.2382 in 4  12 2

Add symmetric points c’, b’, and a’. Qe = 0   1.125  3 Qd = (0.125)(1.125)   = 0.07910 in 2  

td = 0.125 in. 

Qc = Qe = (0.125)2 (1.1875) = 0.09765 in 4

tc = 0.25 in. 

Qb = Qc + (2)(1.0625)(0.125)(1.1875) = 0.41308 in 3

tb = 0.25 in. 

 1.25  Qa = Qb + (2)(0.125)(1.25)  = 0.60839 in 3   2 

ta = 0.25 in. 

τa =

VQa (2)(0.60839) = Ita (1.2382)(0.25)

τ a = 3.93 ksi 

τb =

VQb (2)(0.41308) = Itb (1.2382)(0.25)

τ b = 2.67 ksi 

τc =

VQc (2)(0.09765) = Itc (1.2382)(0.25)

τ c = 0.63 ksi 

τd =

VQd (2)(0.07910) = Itd (1.2382)(0.25)

τ d = 1.02 ksi 

τe =

VQe Ite

τ e = 0 


PROBLEM 6.43 Three 1 × 18-in. steel plates are bolted to four L6 × 6 × 1 angles to form a beam with the cross section shown. The bolts have a 78 -in. diameter and are spaced longitudinally every 5 in. Knowing that the allowable average shearing stress in the bolts is 12 ksi, determine the largest permissible vertical shear in the beam. (Given: I x = 6123 in 4 )

SOLUTION Flange:

Web: Angle:

If =

1 (18)(1)3 + (18)(1)(9.5) 2 = 1626 in 4 12

Iw =

1 (1)(18)3 = 486 in 4 12

I = 35.5 in 4 , y = 1.86 in.

A = 11.0 in 2 d = 9 − 1.86 = 7.14 in.

2

I a = I + Ad = 596.18 in 4 I = 2I f + I w + 4I a = 6123 in 4 , which agrees with the given value.

Flange:

Q f = (18)(1)(9.5) = 171 in 3

Angle:

Qa = Ad = (11.0)(7.14) = 78.54 in 3 Q = Q f + 2Qa = 328.08 in 3

Abolt =

π 7

2

2   = 0.60132 in 48

Fbolt = 2τ bolt Abolt = (2)(12)(0.60132) = 14.4317 kips qall = q =

Fbolt 14.4317 = = 2.8863 kip/s s 5

VQ I

Vall =

qall I (2.8863)(6123) = Q 328.08

Vall = 53.9 kips 


PROBLEM 6.44 Three planks are connected as shown by bolts of 14-mm diameter spaced every 150 mm along the longitudinal axis of the beam. For a vertical shear of 10 kN, determine the average shearing stress in the bolts.

SOLUTION Locate neutral axis and compute moment of inertia. Part

A(mm 2 )

y (mm)



12500

200



25000



12500

Σ

50000

Ay (mm3 )

d (mm)

Ad 2 (mm 4 )

I (mm 4 )

2.5 × 106

37.5

17.5781 × 106

10.4167 × 106

125

3.125 × 106

37.5

35.156 × 106

130.208 × 106

200

2.5 × 106

37.5

17.5781 × 106

10.4167 × 106

70.312 × 106

151.04 × 106

8.125 × 106 Y =

ΣAy 8.125 × 106 = = 162.5 mm ΣA 50 × 103

I = ΣAd 2 + ΣI = 221.35 × 106 mm 4 = 221.35 × 10−6 m 4 Q = A1 y1 = (12500)(37.5) = 468.75 × 103 mm3 = 468.75 × 10−6 m3 q =

VQ (10 × 103 )(468.75 × 10−6 ) = I 221.35 × 10−6

= 21.177 × 103 N/m Fbolt = qs = (21.177 × 103 )(150 × 10−3 ) = 3.1765 × 103 N Abolt =

τ bolt =

π 4

(14) 2 = 153.938 mm 2 = 153.938 × 10−6 m 2

Fbolt 3.1765 × 103 = = 20.6 × 106 Pa Abolt 153.938 × 10−6

τ bolt = 20.6 MPa 


PROBLEM 6.45 A beam consists of three planks connected as shown by steel bolts with a longitudinal spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN and that the allowable average shearing stress in each bolt is 60 MPa, determine the smallest permissible bolt diameter that can be used.

SOLUTION Part

A(mm 2 )

y (mm)

Ay 2 (106mm 4 )

I (106 mm 4 )



7500

50

18.75

14.06



7500

50

18.75

14.06



15,000

−50

37.50

28.12

75.00

56.25

Σ

I =  Ay 2 +  I = 131.25 × 106 mm 4 = 131.25 × 10−6 m 4 Q = A1 y1 = (7500)(50) = 375 × 103 mm3 = 375 × 10−6 m3 Fbolt = τ bolt Abolt = qs = Abolt =

VQs I

VQs (6 × 103 )(375 × 10−6 )(0.225) = = 64.286 × 10−6 m 2 τ bolt I (6 × 106 )(131.25 × 106 ) = 64.286 mm 2

d bolt =

4 Abolt

π

=

(4)(64.286)

π

d bolt = 9.05 mm 


PROBLEM 6.46 A beam consists of five planks of 1.5 × 6-in. cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi, determine the smallest permissible bolt diameter that can be used.

SOLUTION Part

A(in 2 )

y0 (in.)

Ay0 (in 3 )

y (in.)

Ay 2 (in 4 )

I (in 4 )



9

5

45

0.8

5.76

27



9

4

36

−0.2

0.36

27



9

3

27

−1.2

12.96

27



9

4

36

−0.2

0.36

27



9

5

45

0.8

5.76

27

Σ

45

25.20

135

189 Y0 =

ΣAy 189 = = 4.2 in. ΣA 45

I = ΣAd 2 + ΣI = 160.2 in 4

Between  and :

Q12 = Q1 = Ay1 = (9)(0.8) = 7.2 in 3

Between  and :

Q23 = Q1 + Ay2 = 7.2 + (9)(−0.2) = 5.4 in 3 q =

VQ I

Maximum q is based on Q12 = 7.2 in 3. (2000)(7.2) = 89.888 lb/in 160.2 = qs = (89.888)(9) = 809 lb

q= Fbolt

τ bolt = Abolt =

Fbolt Abolt

π 4

2 d bolt

Abolt = d bolt =

Fbolt

τ bolt

=

4 Abolt

π

809 = 0.1079 in 2 7500 =

(4)(0.1079)

π

d bolt = 0.371 in. 


PROBLEM 6.47 A plate of 14 -in. thickness is corrugated as shown and then used as a beam. For a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section, (b) the shearing stress at point B. Also, sketch the shear flow in the cross section.

SOLUTION LBD =

(1.2)2 + (1.6)2 = 2.0 in. ABD = (0.25)(2.0) = 0.5 in 2

Locate neutral axis and compute moment of inertia. Part

A(in 2 )

d (in.)

Ad 2 (in 4 )

I (in 4 )

AB

0.5

0

0

0.4

0.080

neglect

BD

0.5

0.8

0.4

0.4

0.080

*0.1067

DE

0.5

0.8

0.4

0.4

0.080

*0.1067

EF

0.5

0

0

0.4

0.080

neglect

Σ

2.0

0.320

0.2133

*

y (in.)

Ay (in 3 )

0.8

1 1 ABD h 2 = (0.5)(1.6)2 = 0.1067 in 4 12 12

Y =

ΣAy 0.8 = = 0.4 in. ΣA 2.0

I = ΣAd 2 + ΣI = 0.5333 in 4

(a)

Qm = QAB + QBC QAB = (2)(0.25)(0.4) = 0.2 in 3 QBC = (0.5)(0.25)(0.2) = 0.025 in 3 Qm = 0.225 in 3

τm = (b)

VQm (1.2)(0.225) = It (0.5333)(0.25)

τ m = 2.03 ksi 

QB = QAB = 0.2 in 3

τB =

VQB (1.2)(0.2) = It (0.5333)(0.25)

τ B = 1.800 ksi 

τD = 0 


PROBLEM 6.48 A plate of 4-mm thickness is bent as shown and then used as a beam. For a vertical shear of 12 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam. Also, sketch the shear flow in the cross section.

SOLUTION tan α =

20 48

α = 22.62°

As = (4 sec α )(48) = 208 mm 2

Slanted side:

Is =

1 (4 sec α )(48)3 = 39.936 × 103 mm 4 12

Top:

IT =

1 (50)(4)3 + (50)(4)(24)2 = 115.46 × 103 mm 4 12

Bottom:

I B = IT = 115.46 × 103 mm 4 I = 2I s + IT + I B = 310.8 × 103 mm 4 = 310.8 × 10−9 m 4

(a)

QA = (25)(4)(24) = 2.4 × 103 mm3 = 2.4 × 10−6 m3 t = 4 mm = 4 × 10−3 m

τA = (b)

VQA (12 × 103 )(2.4 × 10−6 ) = = 23.2 × 106 Pa It (310.8 × 10−9 )(4 × 10−3 )

τ A = 23.2 MPa 

Maximum shearing occurs at point M, 24 mm above the bottom QM = QA + (4 sec α )(24)(12) = 2.4 × 103 + 1.248 × 103 = 3.648 × 103 mm3 = 3.648 × 10−6 m3

τM =

VQM (12 × 103 )(3.648 × 10−6 ) = = 35.2 × 106 Pa It (310.8 × 10−9 )(4 × 10−3 )

QB = QA

τ M = 35.2 MPa 

τ B = τ A = 23.2 MPa




PROBLEM 6.49 A plate of 2-mm thickness is bent as shown and then used as a beam. For a vertical shear of 5 kN, determine the shearing stress at the five points indicated and sketch the shear flow in the cross section.

SOLUTION 1 1 1  I = 2  (2)(48)3 + (2)(52)3 + (20)(2)3 + (20)(2)(25) 2  12 12 12   = 133.75 × 103 mm 4 = 133.75 × 10−9 mm 4 Qa = (2)(24)(12) = 576 mm3 = 576 × 10−9 mm3 Qa = 0 Qc = Qb − (12)(2)(25) = −600 mm3 = −600 × 10−9 m3 Qd = Qc − (2)(24)(12) = −1.176 × 103 mm3 = −1.176 × 10−6 m3 Qe = Qd + (2)(26)(13) = −600 mm3 = −500 × 10−9 m3

τa =

VQa (5 × 103 )(576 × 10−9 ) = = 10.77 × 106 Pa It (133.75 × 10−9 )(2 × 10−3 )

τb =

VQb It

τc =

VQc (5 × 103 )(600 × 10−9 ) = = 11.21 × 106 Pa It (133.75 × 10−9 )(2 × 10−3 )

τ c = 11.21 MPa 

τd =

VQd (5 × 103 )(1.176 × 10−6 ) = = 22.0 × 106 Pa It (133.75 × 10−9 )(2 × 10−3 )

τ d = 22.0 MPa 

τe =

VQe (5 × 103 )(500 × 10−9 ) = = 9.35 × 106 Pa It (133.75 × 10−9 )(2 × 10−3 )

τ e = 9.35 MPa 

τ a = 10.76 MPa  τb = 0 




PROBLEM 6.50 A plate of thickness t is bent as shown and then used as a beam. For a vertical shear of 600 lb, determine (a) the thickness t for which the maximum shearing stress is 300 psi, (b) the corresponding shearing stress at point E. Also, sketch the shear flow in the cross section.

SOLUTION LBD = LEF =

4.82 + 22 = 5.2 in.

Neutral axis lies at 2.4 in. above AB. Calculate I. I AB = (3t )(2.4)2 = 17.28t I BD =

1 (5.2t )(4.8)2 = 9.984t 12

I DE = (6t )(2.4) 2 = 34.56t I EF = I DB = 9.984t I FG = I AB = 17.28t I = ΣI = 89.09t

(a)

At point C,

QC = QAB + QBC = (3t )(2.4) + (2.6t )(1.2) = 10.32t

τ =

(b)

VQC It

∴ t =

VQ (600)(10.32t ) = = 0.23168 in. τI (300)(89.09t )

t = 0.232 in. 

I = (89.09)(0.23168) = 20.64 in 4 QE = QEF + QFG = 0 + (3)(0.23168)(2.4) = 1.668 in 3

τE =

VQE (600)(1.668) = It (20.64)(0.23168)

τ E = 209 psi 


PROBLEM 6.51 The design of a beam calls for connecting two vertical rectangular 83 × 4 -in. plates by welding them to horizontal 12 × 2 -in. plates as shown. For a vertical shear V, determine the dimension a for which the shear flow through the welded surface is maximum.

SOLUTION 3

 1  3   1  1 1 I = (2)    (4)3 + (2)   (2)   + (2)(2)   a 2  12  8   12   2  2

= 4.041667 + 2a 2 in 4 1 Q = (2)   a = a in 3 2 VQ Va = q= I 4.041667 + 2a 2

Set

dq = 0. da

dq  (4.041667 + 2a 2 ) − (a)(4a)  = V = 0 da  (4.041667 + 2a 2 ) 2  2a 2 = 4.041667 a = 1.422 in. 




PROBLEM 6.52 An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as τ max = k (V/A) and determine the constant k for each of the two orientations shown.

SOLUTION

(a)

3 a 2 A1 = A2 = at h=

(b)

3 3 at 4 1 1 3 1 I 2 = A2 h2 = at a 2 = a3t 3 3 4 4 5 3 I = 2 I1 + 4 I 2 = a t 2 I1 = A1h2 = ath 2 =

A2 =

2

1 a h ath 2 + at  +  12 2 2 1 3 9 7 a t + a 3t = a 3t = 48 16 12 =

3 2 a t 2 3 2 h Q2 = A2 = a t 2 4 Qm = Q1 + 2Q2 = 3a 2 t

= k=

3

1 a 1 3 I2 = t   = at 3 2 24 5 I = 4 I1 + 4 I 2 = a3t 2 a h 3 2 Q1 = at  +  = a t 2 2 4

3V VQ V 3a 2t = = 3 5 5 at I (2t ) a t 2t 2

(

)

6 3 V 6 3V V = =k 5 6at 5 A A 6 3 5

1 at 2

I1 = I1 + A1d 2

Q1 = A1h =

τm =

a 2 A1 = at h=

k = 2.08 

 1  a  1 Q2 =  at   = a 2 t  2  4  8 7 Q = 2Q1 + 2Q2 = a 3t 4 V ⋅ 74 a3t VQ τm = = 5 3 I (2t ) a t (2t ) 2

(

=

)

7 V 42 V 21 V = = 20 at 20 6at 10 A

=k

V A

k=

21 = 2.10  10


PROBLEM 6.53 An extruded beam has a uniform wall thickness t. Denoting by V the vertical shear and by A the cross-sectional area of the beam, express the maximum shearing stress as τ max = k (V/A) and determine the constant k for each of the two orientations shown.

SOLUTION

2

(a)

1 a I1 = ( at )   = a 3t 2 4  

(b)

3

1 a 1 3 I2 = t   = at 3 2 24 2 I = 2 I1 + 4 I 2 = a3t 3 a 1 Q1 = ( at )   = a 2 t 2 2

(

1  1  2  A1h 2 =  at   a 3  3   2  1 = a 3t 6 2 I = 4 I1 = a3t 3

h 1 Q1 = at   = 2 a 2t 2 4   1 2a 2t Q = 2Q1 = 2

)

V 34 a 2 t VQ = 2 3 I (2t ) a t (2t ) 3

(

)

1 2a 2 2

I1 =

 1  a  1 Q2 =  at   = a 2 t  2  4  8 3 Q = Q1 + 2Q2 = a 2 t 4

τ max =

h=

9 V 9 V 9V = = = 16 at 4 4at 4 A V =k A

τ max

(

3 2 8 3 2 = 2 =

k=

9 = 2.25  4

(

)

V 12 2 a 2 t VQ = = 2 3 I (2t ) a t (2t ) 3

)

V 3 2 V = at 2 4at V V =k A A

k=

3 2 = 2.12  2


PROBLEM 6.54 (a) Determine the shearing stress at point P of a thin-walled pipe of the cross section shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for θ = 90° and is equal to 2V/A, where A is the cross-sectional area of the pipe.

SOLUTION A = 2π rm t

J = Arm2 = 2π rm3 t

sin θ

for a circular arc.

r=

θ

I=

1 J = π rm3t 2

AP = 2rθ t QP = AP r = 2rt sin θ

(a)

τP =

(b)

τm =

VQP (V )(2rt sin θ ) = I (2t ) π rm3t (2t )

(

2V sin π2 2π rm t

)

τP =

V sin θ  π rm t

τm =

2V  A


PROBLEM 6.55 For a beam made of two or more materials with different moduli of elasticity, show that Eq. (6.6)

τ ave =

VQ It

remains valid provided that both Q and I are computed by using the transformed section of the beam (see Sec. 4.6), and provided further that t is the actual width of the beam where τ ave is computed.

SOLUTION Let Eref be a reference modulus of elasticity. n1 =

E1 E , n2 = 2 , etc. Eref Eref

Widths b of actual section are multiplied by n’s to obtain the transformed section. The bending stress distribution in the cross section is given by

σx = −

nMy I

where I is the moment of inertia of the transformed cross section and y is measured from the centroid of the transformed section. The horizontal shearing force over length Δ x is



ΔH = − (Δσ x ) dA =



n(ΔM ) y (ΔM ) Q ( ΔM ) dA = ny dA = I I I





Q = ny dA = first moment of transformed section.

Shear flow:

q=

Δ H ΔM Q VQ = = Δx Δx I I

q is distributed over actual width t, thus

q τ= . t VQ τ= It


PROBLEM 6.56 A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29 × 106 psi for the steel and 10.6 × 106 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION n = 1 in aluminum. n=

29 × 106 psi = 2.7358 in steel. 10.6 × 106 psi

Part

nA (in 2 )

y (in.)

nA y (in 3 )

d (in.)

nAd 2 (in 2 )

nI (in 4 )

Steel

8.2074

2.0

16.4148

0.2318

0.4410

2.7358

Alum.

1.5

0.5

0.75

1.2682

2.4125

0.1250

Σ

9.7074

2.8535

2.8608

17.1648 ΣnA y 17.1648 = = 1.7682 in. ΣA 9.7074 I = ΣnAd 2 + ΣnI = 5.7143 in 4

Y =

(a)

At the bonded surface,

Q = (1.5)(1.2682) = 1.9023 in 3

τ= (b)

At the neutral axis,

VQ (4)(1.9023) = It (5.7143)(1.5)

τ = 0.888 ksi 

 1.2318  Q = (2.7358)(1.5)(1.2318)  = 3.1133 in 3   2 

τ max =

VQ (4)(3.1133) = It (5.7143)(1.5)

τ max = 1.453 ksi 


PROBLEM 6.57 A steel bar and an aluminum bar are bonded together as shown to form a composite beam. Knowing that the vertical shear in the beam is 4 kips and that the modulus of elasticity is 29 × 106 psi for the steel and 10.6 × 106 psi for the aluminum, determine (a) the average stress at the bonded surface, (b) the maximum shearing stress in the beam. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION n = 1 in aluminum. n=

29 × 106 psi = 2.7358 in steel. 10.6 × 106 psi

Part

nA (in 2 )

Alum.

3.0

2.0

Steel

4.1038

0.5

Σ

7.1038

nA y (in 3 )

y (in.)

d (in.)

nAd 2 (in 2 )

6.0

0.8665

2.2525

1.0

2.0519

0.6335

1.6469

0.3420

3.8994

1.3420

8.0519

Y =

nI (in 4 )

ΣnA y 8.0519 = = 1.1335 in. ΣnA 7.1038

I = ΣnAd 2 + ΣnI = 5.2414 in 4

(a)

At the bonded surface,

Q = (1.5)(2)(0.8665) = 2.5995 in 3

τ = (b)

At the neutral axis,

VQ (4)(2.5995) = It (5.2414)(1.5)

τ = 1.323 ksi 

 1.8665  3 Q = (1.5)(1.8665)   = 2.6129 in 2  

τ max =

VQ (4)(2.6129) = It (5.2814)(1.5)

τ max = 1.329 ksi 


PROBLEM 6.58 A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION Eref = Es = 200 GPa

Let

ns = 1

nw =

Ew 10 GPa 1 = = Es 200 GPa 20

Widths of transformed section:  1  bw =   (150) = 7.5 mm  20  1  1 I = 2  (150)(12)3 + (150)(12)(125 + 6) 2  + (7.5)(250)3 12  12

bs = 150 mm

= 2[0.0216 × 106 + 30.890 × 106 ] + 9.766 × 106 = 71.589 × 106 mm 4 = 71.589 × 10−6 m 4

(a)

Q1 = (150)(12)(125 + 6) = 235.8 × 103 mm3 = 235.8 × 10−6 m3 VQ1 (4 × 103 )(235.8 × 10−6 ) = = 13.175 × 103 N/m −6 I 71.589 × 10 = qs = (23.187 × 103 )(200 × 10−3 ) = 2.635 × 103 N

q= Fbolt

Abolt =

τ bolt = (b)

π 4

π  2 d bolt =   (12)2 = 113.1 mm 2 = 113.1 × 10−6 m 2 4

Fbolt 2.635 × 103 = = 23.3 × 106 Pa −6 Abolt 113.1 × 10

τ bolt = 23.3 MPa 

Q2 = Q1 + (7.5)(125)(62.5) = 235.8 × 103 + 58.594 × 103 = 294.4 × 103 mm3 = 294.4 × 10−6 m3 t = 150 mm = 150 × 10−3 m

τc =

VQ2 (4 × 103 )(294.4 × 10−6 ) = = 109.7 × 103 Pa It (71.589 × 10−6 )(150 × 10−3 )

τ c = 109.7 kPa 


PROBLEM 6.59 A composite beam is made by attaching the timber and steel portions shown with bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of 4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)

SOLUTION Let steel be the reference material. ns = 1.0

nw =

Ew 10 GPa = = 0.05 200 GPa Es

Depth of section:

d = 90 + 84 + 90 = 264 mm

For steel portion:

Is = 2

For the wooden portion:

Iw =

For the transformed section:

1  1  bd 3 = (2)   (6)(264)3 = 18.400 × 106 mm 4 12  12 

(

)

1 1 (140)(2643 − 843 ) = 207.75 × 106 mm 4 b d13 − d 23 = 12 12

I = ns I s + nw I w

I = (1.0)(18.400 × 106 ) + (0.05)(207.75 × 106 ) = 28.787 × 106 mm 4 = 28.787 × 10−6 m 4

(a)

Shearing stress in the bolts.

For the upper wooden portion

Qw = (90)(140)(42 + 45) = 1.0962 × 106 mm3

For the transformed wooden portion Q = nwQw = (0.05)(1.0962 × 106 ) = 54.81 × 103 mm3 = 54.81 × 10−6 m3

Shear flow on upper wooden portion. q =

VQ (4000)(54.81 × 10−6 ) = = 7616 N/m I 28.787 × 10−6

Fbolt = qs = (7616)(0.200) = 1523.2 N Abolt =

Double shear:

π 4

2 d bolt =

π 4

τ bolt =

(12)2 = 113.1 mm 2 = 113.1 × 10−6 m 2

Fbolt 1523.2 = 2 Abolt (2)(113.1 × 10−6 )

= 6.73 × 106 Pa

τ bolt = 6.73 MPa 

  PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.


(b)

Shearing stress at the center of the cross section.

For two steel plates

Qs = (2)(6)(90 + 42)(90 − 42) = 76.032 × 103 mm3 = 76.032 × 10−6 m3

For the neutral axis

Q = 54.81 × 10−6 + 76.032 × 10−6 = 130.842 × 10−6 m3

Shear flow across the neutral axis q =

VQ (4000)(130.842 × 10−6 ) = = 18.181 × 103 N/m −6 I 28.787 × 10

Double thickness

2t = 12 mm = 0.012 m

Shearing stress

τ =

q 18.181 × 103 = = 1.515 × 106 Pa 2t 0.012

τ = 1.515 MPa 


PROBLEM 6.60 Consider the cantilever beam AB discussed in Sec. 6.8 and the portion ACKJ of the beam that is located to the left of the transverse section CC′ and above the horizontal plane JK, where K is a point at a distance y < yY above the neutral axis. (See Figure). (a) Recalling that σ x = σ Y between C and E and σ x = (σ Y /yY ) y between E and K, show that the magnitude of the horizontal shearing force H exerted on the lower face of the portion of beam ACKJ is H=

 1 y2  bσ Y  2c − yY −  yY  2 

(b) Observing that the shearing stress at K is ΔH 1 ΔH 1 ∂ H = lim = Δ A→ 0 Δ A Δ x →0 b Δ x b ∂x

τ xy = lim

and recalling that yY is a function of x defined by Eq. (6.14), derive Eq. (6.15).

SOLUTION Point K is located a distance y above the neutral axis. The stress distribution is given by

σ = σY

y yY

for 0 ≤ y < yY

and σ = σ Y

for

yY ≤ y ≤ c.



For equilibrium of horizontal forces acting on ACKJ,



H = σ dA =



yY y

σ Y yb dy + yY



c yY

σ Y bdy

  + σ y b(c − yY )   1 y2  H = bσ Y  2c − yY −  yY  2 

=

σ Y b  yY2 − y 2  yY  2

(a)

Note that yY is a function of x.  ∂ yY y 2 dyY  + −   ∂x yY 2 dx   1  y 2  dyY = − σ Y 1 −  yY 2  dx 2 

τ xy =

But

Differentiating,

1∂H 1 = σY b ∂x 2

M = Px =

 1 yY2  3 M y 1 −  3 c 2  2  

dM 3  2 yY dyY  = P = MY  −  2 dx 2  3 c dx  dyY Pc 2 Pc 2 3 P =− =− =− 2 2 2 σ Y byY dx yY M Y yY 3 σ Y bc

Then

1 2

  

τ xy = σ Y 1 −

y2 yY 2

3 P 3P =   2 σ Y bσ Y 4byY 

 y2  1 −   yY 2  

(b)


PROBLEM 6.61 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown.

SOLUTION 2

1 3 7  3a  ta + (ta)   = ta3 12 3  2  1 = I EF = (2at ) a 2 + (2a) t 3 ≈ 2a3t 12 1 2 28 I = ΣI = ta3 = t (2a )3 = ta 3 12 3 3

I AB = I FG = I DB I DE

Part AB:

A = t (2a − y ); Q = Ay =

y=

2a + y 3

1 t (2a − y ) (2a + y ) 2

1 t (4a 2 − y 2 ) 2 VQ V (4a 2 − y 2 ) τ= = 2I It

=

V (4a 2 − y 2 ) tdy a 2I Vt  2 y 3  2 a Vta3  (2)3  1  = − (4)(1) +     4a y −  = (4)(2) − 2I  3 a 2I  3  3 



F1 = τ dA =

=



2a

5 Vta 3 5 = V 6 I 56



Part DB:

3a + txa 2  3a  = ta  + x  2  

Q = (ta )

τ=

VQ Va  3a  = + x It I  2 



F2 = τ dA = =

0

Vta  3ax x 2  +   I  2 2 

=5 Σ MH =



Vta  3a   + x  tdx = I  2 I 

2 a Va

2a

= 0



2 a  3a

0

 + x  dx   2 

Vta3  (3) (2) (2) 2  +   I  2 2 

3

Vta 15 = V I 28 Σ M H:

Ve = F2 (2a ) − 2 F1 (2a) =

30 20 5 Va − Va = Va 28 56 7

e=

5 a = 0.714a  7



SOLUTION 2

1 9  3a  I AB = I HJ = at   + at 3 ≈ ta3 2 12 4   2

1 1 a I DE = I FG = at   + at 3 ≈ ta 2 4  2  12 1 9 I AH = t (3a)3 = ta 3 12 4 29 3 I = ΣI = ta 4

Part AB:

3a 3 Q = atx 2 2 3 V atx ⋅ VQ 6Vx 2 = = τ= 3 29 It ta t 29a 2 t 4

A = tx

y=



F1 = τ dA =

Part DE:



a

0

6Vx 6V tdx = 2 29a t 29a 2



a

0

x dx =

3 V 29

xdx =

1 V 29

a 1 Q = atx 2 2 1 VQ V ⋅ 2 atx 2Vx τ= = = 3 29 It ta t 29 a 2t 4 A = tx



y=

F2 = τ dA = ΣM K =



a 0

2Vx 2V t dx = 2 29a t 29a 2



a 0

ΣM K : Ve = F1 (3a ) + F2 (a) =

9 1 10 Va + Va = Va 29 29 29 e=

10 a 29

e = 0.345a 


PROBLEM 6.63 An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

SOLUTION 2  1   192   1 3 I = 2   (72) (12)3 + (72) (12)    + (6) (192)  2   12  12 

= 19.4849 × 106 mm 4 = 19.4849 × 10−6 m 4

Part AB:

 192  Q = A y = (12 x)   = 1152 x  2  VQ 1152Vx = q= I I

A = 12 x

x = 0 at point A. x = l AB = 72 mm at point B. 1152V (72)2 x 72 1152Vx F1 =  xAB qdx =  0 dx = I I 2 2 (576) (72) V = 0.153246V = 19.4849 × 106 Σ MC =

(a)

e = 29.423 mm

(b)

Point A: x = 0

M C: Ve = (0.153246V ) (192) e = 29.4 mm 

Q = 0,

τA = 0 

q=0 x = 72 mm

Point B in part AB:

Q = (1152) (72) = 82.944 × 103 mm3 − 82.944 × 10−6 m3 t = 12 mm = 0.012 m

τB =

VQ (110 × 103 )(82.944 × 10−6 ) = It (19.4849 × 106 )(0.012)

= 39.0 × 106 Pa

τ B = 39.0 MPa in AB 

PROBLEM 6.63 (Continued) PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

Part BD: y = 96 mm Q = 82.944 × 103 mm3 = 82.944 × 10−6 m3 t = 6 mm = 0.006 m

Point B:

τB =

VQ (110 × 103 ) (82.944 × 10−6 ) = It (19.4849 × 10−6 ) (0.006)

= 78.0 × 106 Pa

Point C:

τ B = 78.0 MPa in BD 

y = 0, t = 6 mm = 0.006 m  96  Q = 82.944 × 103 + (6) (96)   = 110.592 × 103 mm3 = 110.592 × 10−6 m3  2 

τ=

VQ (110.103 ) (110.592 × 10−6 ) = = 104.1 × 106 Pa −6 It (19.4849 × 10 ) (0.006)

τ C = 104.1 MPa 



SOLUTION 2  1   192   1 3 3 I = 2   (72) (6) + (72) (6)    + (12) (192)  2   12  12 

= 15.0431 × 106 mm 4 = 15.0431 × 10−6 m 4

Part AB:

 192  Q = Ay = (6 x)   = 576 x  2  VQ 576Vx = q= I I

A = 6x

x = 0 at point A. x = l AB = 72 mm at point B. F1 = = MC =

(a)

e = 19.0555 mm

(b)

Point A: x = 0



xB xA

qdx =



72 0

576Vx 576V (72)2 dx = I I 2

2

(288) (72) V = 0.099247 V 15.0431 × 106 M C : Ve = (0.099247 ) V (192) e = 19.06 mm 

Q = 0,

τA = 0 

q=0 x = 72 mm

Point B in part AB:

Q = (576) (72) = 41.472 × 103 mm3 = 41.472 × 10−6 m3 t = 6 mm = 0.006 m

τB =

VQ (110 × 103 )(41.472 × 10−6 ) = It (15.0431 × 10−6 )(0.006)

= 50.5 × 106 Pa

τ B = 50.5 MPa 



Part BD: Point B:

y = 96 mm Q = 41.472 × 103 mm3 = 41.472 × 10−6 m3 t = 12 mm = 0.012 m

τB =

VQ (110 × 103 ) (41.472 × 10−6 ) = It (15.0431 × 10−6 ) (0.012)

= 25.271 × 106 Pa

Point C:

τ B = 25.3 MPa 

y = 0, t = 0.012 m  96  Q = 41.472 × 103 + (12) (96)   = 96.768 × 103 mm3 = 96.768 × 10−6 m3  2 

τ=

VQ (110 × 103 ) (96.768 × 10−6 ) = = 58.967 × 106 Pa −6 It (15.0431 × 10 ) (0.012)

τ C = 59.0 MPa 



SOLUTION Part

A (in2)

d (in.)

Ad 2 (in 4 )

I (in 4 )

BD

0.50

3

4.50

≈0

ABEG

1.25

0

0

EF

0.50

3

4.50

≈0

Σ

2.25

9.00

10.417

10.417

I = Σ Ad 2 + Σ I = 19.417 in 4

(a)

Q( x) = 3 tx VQ ( x) V q ( x) = = (3tx) I I

Part BD:

FBD =



4 0

xdx =

3Vt 24Vt (8) = I I

( M BD ) H = 3 FBD =

Its moment about H: Part EF:

3Vt I

FEF =

By same method,

24Vt I

72Vt I ( M EF ) H =

72Vt I

Moments of FAB , FBE , and FEG about H are zero. Ve = Σ M H = e=

72Vt 72Vt 144Vt + = I I I

144 t (144) (0.125) = I 19.417

e = 0.927 in. 



(b)

At A, D, F , and G : Just above B:

τ A = τD = τF = τG = 0 

Q=0 Q1 = QAB = (2t )(4) = 8 t

τ1 = Just to the right of B:



τ 2 = 1.700 ksi 

VQ3 (2.75)(20t ) = It (19.417) t

τ 3 = 2.83 ksi 

QH = Q3 + QBH = 20t + t (3)(1.5) = 24.5 t

τH = By symmetry:

VQ2 (2.75)(12t ) = It (19.417) t

Q3 = Q1 + Q2 = 20t

τ3 = At H (neutral axis):

τ1 = 1.133 ksi 

Q2 = QBD = (3) t (4) = 12t

τ2 = Just below B:

VQ1 (2.75) (8t ) = It (19.417) t

VQH (2.75) (24.5t ) = It (19.417) t

τ H = 3.47 ksi  τ 4 = τ 3 = 2.83 ksi  τ 5 = τ 2 = 1.700 ksi  τ 6 = τ1 = 1.133 ksi 



SOLUTION 1 I AB = (0.125)(3)3 = 1.125 in 4 3 1 I BD = (4) (0.125)3 + (4)(0.125)(3) 2 = 4.50065 in 4 12 1 I DE = (0.125) (6)3 = 2.25 in 4 12 I EF = I BD = 4.50065 in 4 I FG = I AB = 1.125 in 4 I = Σ I = 13.50 in 4

(a)

Part AB:

y = 0.5ty 2 2 VQ( y ) 0.5Vt 2 = q( y ) = y I I 3 0.5Vt FAB = q ( y ) dy = 0 I

Q ( y ) = ty



Its moment about H is

4 FAB = 18



3 0

y 2 dy = 4.5

Vt ↑ I

Vt I

QB = (0.5) (t )(3)2 = 4.5 t

Part BD:

Q( x) = QB + xt (3) = (4.5 + 3x) t Vq( x) Vt = (4.5 + 3x) I I 4 ∇t 4 Vt = q( x) dx = (4.5 + 3x) dx = 42 ← 0 I 0 I

q ( x) = FBD

Its moment about H : 3FBD = 126





Vt I

QD = [4.5 + (3)(4)] t = 16.5 t



Part EF:

By symmetry with part BD,

Its moment about H is 3 FEF = 126

FEF = 42

Vt I

By symmetry with part AB, FFG = 4.5

Part FG:

Its moment about H is 4. FFG = 18

Vt → I

VT ↑ I

Vt I

Moment about H of force in part DE is zero. Vt 144Vt (18 + 126 + 0 + 126 + 18) = I I 144 t (2.88) (0.125) e= = I 13.50

Ve = Σ M H =

(b)

e = 2.67 in. 

τ A = τG = 0 

QA = QG = 0 QB = QF = 4.5t

τB =τF =

VQB (2.75) (4.5 t ) = It 13.50 t



QD = QE = 16.5t 



τD =τE =

τ B = τ F = 0.917 ksi 

VQ0 (2.75) (16.5t ) =  It 13.50 t

At H (neutral axis):

τ D = τ E = 3.36 ksi 

QH = QD + t (3)(1.5) = 21t

τH =

VQH (2.75) (21t ) = It 13.50t

τ H = 4.28 ksi 



SOLUTION 1 (30)(6)3 + (30)(6)(45)2 = 0.365 × 106 mm 4 12 1 = (30)(4)3 + (30)(4)(15)2 = 0.02716 × 106 mm 4 12

I AB = I HJ = I DE = I FG

1 (6)(90)3 = 0.3645 × 106 mm 4 12 I = ΣI = 1.14882 × 106 mm 4

I AH =

(a)

For a typical flange,

A( s ) = ts Q( s ) = yts VQ ( s ) Vyts q( s ) = = I I b Vytb 2 F = q( s )ds = 0 2I



Flange AB:

FAB =

V (45)(6)(302 ) = 0.10576V ← (2)(1.14882 × 10−6 )

Flange DE:

FDE =

V (15)(4)(30) 2 = 0.023502V ← (2)(1.14882 × 106 )

Flange FG:

FFG = 0.023502V →

Flange HJ:

FHJ = 0.10576V → ΣM K =

ΣM K : Ve = 45 FAB + 15FDE + 15FFG + 45 FHJ = 10.223V

Dividing by V,

e = 10.22 mm 



(b)

Calculation of shearing stresses. V = 35 × 103 N

I = 1.14882 × 10−6 m 4

τ =0 

At B, E, G, and J, At A and H, Q = (30)(6)(45) = 8.1 × 103 mm3 = 8.1 × 10−6 m3 t = 6 × 10−3 m

τ=

(35 × 103 )(8.1 × 10−6 ) VQ = = 41.1 × 106 Pa It (1.14882 × 10−6 )(6 × 10−3 )

τ = 41.1 MPa 

Just above D and just below F: Q = 8.1 × 103 + (6)(30)(30) = 13.5 × 103 mm3 = 13.5 × 10−6 m3 t = 6 × 10−3 m

τ=

VQ (35 × 103 )(13.5 × 10−6 ) = = 68.5 × 106 Pa It (1.14882 × 10−6 )(6 × 10−3 )

τ = 68.5 MPa 

Just to right of D and just to the right of F: Q = (30)(4)(15) = 1.8 × 103 mm3 = 1.8 × 10−6 m3

τ=

t = 4 × 10−3 m

VQ (35 × 103 )(1.8 × 10−6 ) = = 13.71 × 106 Pa It (1.14882 × 10−6 )(4 × 10−3 )

τ = 13.71 MPa 

Just below D and just above F: Q = 13.5 × 103 + 1.8 × 103 = 15.3 × 103 mm3 = 15.3 × 10−6 m3 t = 6 × 10−3 m

τ= At K,

VQ (35 × 103 )(15.3 × 10−6 ) = = 77.7 × 106 Pa It (1.14882 × 10−6 )(6 × 10−3 )

τ = 77.7 MPa 

Q = 15.3 × 103 + (6)(15)(7.5) = 15.975 × 103 mm3 = 15.975 × 10−6 m3

τ=

VQ (35 × 103 )(15.975 × 10−6 ) = = 81.1 × 106 Pa It (1.14882 × 10−6 )(6 × 10−3 )

τ = 81.1 MPa 



SOLUTION 1 (30)(4)3 + (30)(4)(45)2 = 0.24316 × 106 mm 4 12 1 = (30)(6)3 + (30)(6)(15)2 = 0.04104 × 106 mm 4 12

I AB = I HJ = I DE = I FG

1 (6)(90)3 = 0.3645 × 106 mm 4 12 I = ΣI = 0.9329 × 106 mm 4

I AH =

(a)

For a typical flange,

A( s ) = ts Q( s ) = yts VQ ( s ) Vyts q( s ) = = I I b Vytb 2 F = q( s )ds = 0 2I



Flange AB:

FAB =

V (45)(4)(30) 2 = 0.086826V ← (2)(0.9329 × 106 )

Flange DE:

FDE =

V (15)(6)(30) 2 = 0.043413 V ← (2)(0.9329 × 106 )

Flange FG:

FFG = 0.043413 V →

Flange HJ:

FHJ = 0.086826 V → ΣM K =

ΣM K : Ve = 45 FAB + 15FDE + 15FFG + 45 FHJ = 9.1167 V

Dividing by V,

e = 9.12 mm 

 PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.


(b)

Calculation of shearing stresses. V = 35 × 103 N I = 0.9329 × 10−6 m 4

τ =0 

At B, E, G, and J, At A and H,

Q = (30)(4)(45) = 5.4 × 103 mm3 = 5.4 × 10−6 m3 q=

Just to the right of A and H:

VQ (35 × 103 )(5.4 × 10−6 ) = = 202.59 × 103 N/m −6 I 0.9329 × 10

t = 4 × 10−3 m

τ=

q 202.59 × 103 = = 50.6 × 106 Pa t 4 × 10−3

τ = 50.6 MPa 

Just below A and just above H: t = 6 × 10−3 m

τ=

q 202.59 × 103 = = 33.8 × 106 Pa t 6 × 10−3

τ = 33.8 MPa 

Just above D and just below F: t = 6 × 10−3 m Q = 5.4 × 103 + (6)(30)(30) = 10.8 × 103 mm3 = 10.8 × 10−6 m3

τ= Just to the right of D and E:

(35 × 103 )(10.8 × 10−6 ) VQ = = 67.5 × 106 Pa −6 −6 It (0.9329 × 10 )(6 × 10 )

τ = 67.5 MPa 

t = 6 × 10−3 m Q = (30)(6)(15) = 2.7 × 103 mm 2 = 2.7 × 10−6 m3

τ=

VQ (35 × 103 )(2.7 × 10−6 ) = = 16.88 × 106 Pa It (0.9329 × 10−6 )(6 × 10−3 )

τ = 16.88 MPa 

Just below D and just above F: t = 6 × 10−3 Q = 10.8 × 103 + 2.7 × 103 = 13.5 × 103 mm3 = 13.5 × 10−6 m3

τ= At K,

VQ (35 × 103 )(13.5 × 10−6 ) = = 84.4 × 106 Pa It (0.9329 × 10−6 )(6 × 10−6 )

τ = 84.4 MPa 

Q = 13.5 × 103 + (6)(15)(7.5) = 14.175 × 103 mm3 = 14.175 × 10−6 m3

τ=

VQ (35 × 103 )(14.175 × 10−6 ) = = 88.6 × 106 Pa It (0.9329 × 10−6 )(6 × 10−3 )

τ = 88.6 MPa 



SOLUTION

LAB = 42 + 32 = 5 in.

AAB = 5t

1 1 AAB h 2 + AAB d 2 = (5t )(3)2 + (5t )(4)2 = 83.75 t in 4 12 12 1 I BD = (t )(5)3 = 10.417t in 4 12 I = 2 I AB + I BD = 177.917t in 4 I AB =

Q = QAB + QBY

In part BD,

1 Q = (5t )(4) + (2.5 − y )t   (2.5 + y ) 2 1 1   = 20t + 3.125t − ty 2 =  23.125 − y 2  t 2 2  

τ=

VQ It



FBD = τ dA =

2.5

V (23.125 − 12 y 2 )t

−2.5

It



⋅ tdy 2.5

Vt  1 2 1 3   23.125 − 2 y  dy = I  23.125 y − 6 y  − 2.5     −2.5

=

Vt I

=

Vt  (2.5)3  Vt (110.417) ⋅ 2 (23.125)(2.5) − = 0.62061V = I 6  177.917 t 



2.5

ΣM K = e=

10  10  ΣM K : −V  − e  = − (0.62061V ) 3  3 

10 [1 − 0.62061] 3

e = 1.265 in. 

Note that the lines of action of FAB and FDE pass through point K. Thus, these forces have zero moment about point K.



SOLUTION 1 I DB = (6) (35)3 = 85.75 × 103 mm 4 3 LAB = 70 mm AAB = (70) (6) = 420 mm 2 I AB =

1 1 AAB h 2 =   (420)(35)2 = 171.5 × 103 mm 4 3 3

I = (2)(85.75 × 103 ) + (2)(171.5 × 103 ) = 514.5 × 103 mm 4

Part AB:

A = ts = 6s 1 1 s sin 30° = s 2 4 3 2 Q = Ay = s 2 VQ 3Vs 2 τ= = It It y=



F1 = τ dA = = ΣM D =



70 0

3Vs 2 3V tds = I 2 It



70 0

s 2 ds

(3)(70)3 1 V= V (2)(3) I 3 Σ M D:

Ve = 2[( F1 cos 60°) (70 sin 60°)] = 20.2V

Dividing by V,

e = 20.2 mm 



SOLUTION I AB = (40t ) (60) 2 = 144 × 103 t LDB = 802 + 602 = 100 mm

ADB = 100 t

1 1 ADB h 2 = (100t )(60) 2 = 120 × 103 t 3 3 I = 2 I AB + 2 I DB = 528 × 103 t

I DB =

Part AB:

A = tx y = 60 mm Q = A y = 60 tx mm3 VQ V (60 tx) 60 Vx τ= = = It It I 40 60Vx 60Vt F1 = τ dA = tdx = 0 I I



= ΣM D =



60Vt x 2 I 2 Σ M D:

30

= 0



40

0

xdx

(60) (30) 2 Vt = 0.051136V (2) (528 × 103 ) t

Ve = (0.051136V ) (120)

e = 6.14 mm 



SOLUTION I AB =

1 (1.5) (0.1)3 + (1.5) (0.1)(2) 2 = 0.600125 in 4 12

LBD = 1.52 + 22 = 2.5 in. ABD = (2.5) (0.1) = 0.25 in 2 1 1 ABD h 2 = (0.25)(2)2 = 0.33333 in 4 3 3 I = 2 I AB + 2 I BD = 1.86692 in 4

I BD =

Part AB:

A ( x) = tx = 0.1 x,

y = 2 in.

Q ( x) = A( x) y = 0.2 x in 3 VQ( x) 0.2Vx q ( x) = = I I 1.5 0.2V 1.5 F1 = q ( x) dx = xdx 0 0 I (0.2) (1.5) 2 V V = = 0.225 I I 2



Likewise, by symmetry in part EF:



F1 = 0.225 ΣM D =

Dividing by V,

V I

Σ M D:

Ve = 4 F1 = 0.9

V = 0.482V I e = 0.482 in. 



SOLUTION For a thin-walled hollow circular cross section, A = 2π at J = a 2 A = 2π a3t I=

For the half-pipe section,

π 2

I=

1 J = π a 3t 2

a 3t

Use polar coordinate θ for partial cross section. A = st = a θ t s = arc length sin α θ r =a where α = 2 α sin α cos α y = r cos α = a

α

Q = Ay = aθ t a

sin α cos α

α

= a 2 t (2 sin α cos α )

= a 2 t sin 2α = a 2 t sin θ

τ=

VQ Va 2 = sin θ It I



M H = a τ dA = =2

But M H = Ve, hence



π 0

a

Va 2 Va 4t sin θ tadθ = − cos θ I I

π 0

4

Va t 4 = Va I π e=

4

π

a = 1.273a 



SOLUTION For whole cross section,

A = 2π at J = Aa 2 = 2π a3t

I=

1 J = π a 3t 2

Use polar coordinate θ for partial cross section. A = st = aθ t s = arc length sin α 1 r =a where α = θ 2 α 2 sin α y = r sin α = a

α

sin 2 α

Q = Ay = aθ t a = a 2 t 2sin 2

τ=



But

M C = Ve,

hence

2

= a 2 t 2 sin 2 α

= a 2 t (1 − cos θ )

VQ Va 2 = (1 − cos θ ) It I

M C = a τ dA = =

θ

α



2π 0

Va3 Va 4 t (1 − cos θ ) tadθ = (θ − sin θ ) I I

2π 0

4

2π Va t = 2aV π a 3t e = 2a 


PROBLEM 6.75 A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION I=

Right flange:

1 3 1 t1h1 + t2 h23 12 12

1  A =  h2 − y  t2 2  11  y =  h2 + y  t2 22  Q = Ay

11  1  h2 − y  h2 + y  t2  22  2  11  =  h22 − y 2  t2 24  =

τ=

VQ V = It2 2 It2



F2 = τ dA = = MH = e=

Vt2 2I

1 2 2  4 h2 − y  t2  

Vt2  1 2 Vt  h2 − y 2  t2 dy = 2  − h2 / 2 2 I t  4 2I  2



h2 / 2

1 2 y3   h2 y −  3  4

h2 /2 h2 /2

 1 2 h 1  h  1 2 h 1  h   Vt h3 V t h3 2 −  2  + h2 2 −  2   = 2 2 = 3 2 2 3  h2 2 3 2  4 2 3  2   12 I t1h1 + t2 h2  4

M H: t2 h23b t1h13 + t2 h23

3

3

− Ve = − F2 b = −V =

t2 h23b t1h13 + t2 h23

(0.75)(6)3 (8) (0.75) (8)3 + (0.75) (6)3

e = 2.37 in. 


PROBLEM 6.76 A thin-walled beam has the cross section shown. Determine the location of the shear center O of the cross section.

SOLUTION Let

h1 = AB, h2 = DE , and h3 = FG.

I=

1 t (h13 + h23 + h33 ) 12

11 1   A =  h1 − y  t y =  h1 + y  22 2   1 1  1  1 1  Q = Ay = t  h1 − y  h1 + y  = t  h12 − y 2  2 2  2  2 4 

Part AB:

τ=

VQ V  1 2  = h1 − y 2   It 2I  4 



F1 = τ dA = =



1h 2 1

1h 2 1

V 2I

Vt  1 2 y3  1 2 2  4 h1 − y  tdy = 2 I  4 h1 y − 3     

Vt  1 2 1 1 h   h1 h1 −  1  I  4 2 3 2 

Likewise, for part DE,

F2 =

and for part FG,

F3 = Σ MH =

Dividing by V,

e=

1h 2 1 − 1 h1 2

3

Vt h13 h3V = = 3 13  12 I h1 + h2 + h33  h23 V

h13 + h23 + h33 h33 V h13 + h23 + h33 ΣM H : Ve = a2 F2 + a3 F3 = a2 h23 + a3 h33 h13 + h23 + h32

=

(3)(5)3 + (5)(4)3 53 + 53 + 43

a2 h23 + a3 h33 h13 + h23 + h3

V

e = 2.21 in. 


PROBLEM 6.77 A thin-walled beam of uniform thickness has the cross section shown. Determine the dimension b for which the shear center O of the cross section is located at the point indicated.

SOLUTION Part AB:

1 s 2 1 Q( s ) = A( s ) y ( s ) = ty A s − ts 2 2 VQ( s ) Vt  1  q( s ) = =  yAs − s2  I I  2  A( s ) = ts

FAB = =



y ( s) = y A −

l AB 0

q ( s ) ds

2 l3 Vt  y Al AB − AB  I  2 6

By symmetry,

1 2 QB = ty Al AB − t l AB 2 FFG = FAB

Part BD:

A( x) = tx

At B:

  ↓ 

Q( x) = QB + yB A( x) = QB + t yB x V Q( x) V = (QB + tyB x) I I b V 1  = q ( x) dx =  QB b + t yB b 2  → 0 I 2 

q ( x) = FBD

By symmetry,



FEF = FBD

FDE is not required, since its moment about O is zero.



Σ M O = 0 : b( FAB + FFG ) − yB FBD + yF FEF = 0 2b FAB − 2 yB FBD = 0 2b ⋅

2 l3  Vt  y Al AB V − AB  − 2 yB   I  2 6  I

1  2  QB b + tyB b  2  

2Vt  1 1 3  2Vt  1 2  1 2 2 2  y A l AB − l AB  b −  y Al AB − l AB  yB b − yB b = 0 I 2 6  I  2 2 

Dividing by

2Vt and substituting numerical data, I 1 1 1 1  2 3 2 2 2  (90) (60) − (60)  b − (90)(60) − (60)  (30)b + (30) b = 0 6 2 2 2    126 × 103 b − 108 × 103 b + 450 b 2 = 0 18 × 103 b − 450 b 2 = 0

b = 0 and b = 40.0 mm 



SOLUTION Part AB:

A = tx

y = 60 mm

Q = Ay = 60tx mm3

τ =

60 Vx VQ = It I



F1 = τ dA = 60 Vt x 2 = 2 I

Part DE:

A = tx



30

0

60 Vx 60Vt t dx = I I

30

= 0



30

0

x dx

Vt (60)(30)2 Vt = 27 × 103 2 I I

y = 45 mm

Q = Ay = 45tx

τ =

45 Vx VQ = It I b

F2 =  τ dA = O  MO =

 MO :

45Vx 45Vt b 45b 2Vt t dx = xdx =  I I O 2I 0 = (2)(45) F2 − (2)(60) F1

(45) 2 b2 − (2)(60)(27 × 103 )  Vt = 0   I

b2 =

(2)(60)(27 × 103 ) = 1600 mm 2 2 45

b = 40 mm 

Note that the pair of F1 forces form a couple. Likewise, the pair of F2 forces form a couple. The lines of action of the forces in BDOGK pass through point O.


PROBLEM 6.79 For the angle shape and loading of Sample Prob. 6.6, check that angle and qdy = P along its vertical leg.



 qdz = 0

along the horizontal leg of the

SOLUTION Refer to Sample Prob. 6.6.

τf =

Along horizontal leg:



qdz =

3P(a − z )(a − 3z ) 3P 2 ( a − 4az + 3z 2 ) = 4ta3 4ta3



a 0

τ f t dz =

3P 4a 3



a 0

(a 2 − 4az + 3z 2 )dz

3P  z2 z2 = 3  a 2 z − 4a + 3 2 3 4a  =

τe =

Along vertical leg:



qdy = = =

  

a

0

3P 3 ( a − 2a 3 + a 3 ) = 0 4a 3



3P(a − y )(a + 5 y ) 3P 2 (a + 4ay − 5 y 2 ) = 4ta3 4ta3



a 0

τ e t dy =

3P 4a 3



a 0

(a 2 + 4ay − 5 y 2 )dy

3P  2 y2 y3  a y a 4 5 + −   2 3  4a 3 

a

0

3P  3 5  3P 4 a + 2a 3 − a 3  = 3 ⋅ a 3 = P 3  3  4a 3 4a 




PROBLEM 6.80 For the angle shape and loading of Sample Prob. 6.6, (a) determine the points where the shearing stress is maximum and the corresponding values of the stress, (b) verify that the points obtained are located on the neutral axis corresponding to the given loading.

SOLUTION Refer to Sample Prob. 6.6. (a)

Along vertical leg:

τe =

3P(a − y )(a + 5 y ) 3P 2 (a + 4ay − 5 y 2 ) = 4ta3 4ta3

dτ e 3P (4a − 10 y ) = 0 = dy 4ta3

τm = Along horizontal leg:

τf = dτ f dz

=

τm = 

At the corner:

(b)

1 I y′ = ta 3 3

3P 4ta3

2  2 2   2   3P  9 2  a   a + (4a)  a  − (5)  a   = 3  5   5   4ta  5  

tan ϕ =

τm =

2 a  5

27 P  20 ta

3P(a − z )(a − 3z ) 3P 2 = ( a − 4az + 3z 2 ) 4ta3 4ta3 3P (−4a + 6 z ) = 0 4ta3 3P 4ta3

2  2 2   2   3P  5 2  − a   a − (4a)  a  + (3)  a   = 3  3   3   4ta  3  

y = 0, z = 0,

I z′ =

y=

z=

2 a  3

τm = −

1 P  4 ta

τ=

3P  4 ta

1 3 ta θ = 45° 12 I z′ 1 tan θ = ϕ = 14.036° I y′ 4

θ − ϕ = 45 − 14.036 = 30.964° ΣAy at (a/2) 1 = = a ΣA 2at 4 ΣAz at ( a/2) 1 = = a z= ΣA 2at 4

y=



Neutral axis intersects vertical leg at y = y + z tan 30.964° 1 1  =  + tan 30.964°  a = 0.400a 4 4 

y=

2 a  5

z=

2 a  3

Neutral axis intersects horizontal leg at z = z + y tan (45° + ϕ ) 1 1  =  + tan 59.036°  a = 0.667a 4 4 


PROBLEM 6.81* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

SOLUTION Use results of Example 6.06 with e=

b = 30 mm,

h = 100 mm, and t = 8 mm.

b 30 = = 9.6429 mm = 9.6429 × 10−3 m h 100 2 + 3b 2 + (3)(30)

1 2 1 t h (6b + h) = (8)(100) 2 [(16)(30) + 100] = 1.86667 × 106 mm 4 = 1.86667 × 10−6 m 4 12 12 V = 15 × 103 N I=

(a)

T = Ve = (15 × 103 )(9.6429 × 10−3 )

T = 144.64 N ⋅ m 

Stress at neutral axis due to V: Q = bt

h  h  h  1 + t    = th(h + 4b) 2  2  4  8

1 = (8)(100) [100 + (4)(30) ] = 22 × 103 mm3 = 22 × 10−6 m3 8 t = 8 × 10−3 m

τV =

VQ (15 × 103 )(22 × 10−6 ) = = 22.10 × 106 Pa = 22.10 MPa −6 −3 It (1.86667 × 10 )(8 × 10 )

Stress due to T:

a = 2b + h = 160 mm = 0.160 m 1 t  1 8  c1 = 1 − 0.630  = 1 − (0.630) = 0.3228 3 a 3 160  T 144.64 = = 43.76 × 106 Pa = 43.76 MPa τV = 2 (0.3228)(0.160)(8 × 10−3 )2 c1at

(b)

By superposition,

τ max = τ V + τ T

τ max = 65.9 MPa 


PROBLEM 6.82* Solve Prob. 6.81, assuming that a 6-mm-thick plate is bent to form the channel shown. PROBLEM 6.81* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

SOLUTION Use results of Example 6.06 with b = 30 mm, h = 100 mm, and t = 6 mm.

e=

b 30 = = 9.6429 mm = 9.6429 × 10−3 m h 100 2 + 3b 2 + (3)(30)

1 2 1 t h (6b + h) = (6)(100) 2 [(6)(30) + 100] = 1.400 × 106 mm 4 = 1.400 × 10−6 m 4 12 12 3 V = 15 × 10 N I=

(a)

T = Ve = (15 × 103 )(9.6429 × 10−3 )

T = 144.64 N ⋅ m 

Stress at neutral axis due to V: Q = bt

h  h  h  1 + t    = th(h + 4b) 2  2  4  8

1 = (6)(100) [100 + (4)(30) ] = 16.5 × 103 mm3 = 16.5 × 10−6 m3 8 t = 6 × 10−3 m

τV =

VQ (15 × 103 )(16.5 × 10−6 ) = = 29.46 × 106 Pa = 29.46 MPa −6 −6 It (1.400 × 10 )(6 × 10 )

Stress due to T:

a = 2b + h = 160 mm = 0.160 m 1 t  1  6  c1 = 1 − 0.630  = 1 − (0.630)    = 0.32546 3 a 3  160   T 144.64 = = 77.16 × 106 Pa = 77.16 MPa τV = c1a t 2 (0.32546)(0.160)(6 × 10−3 ) 2

(b)

By superposition,

τ max = τ V + τ T

τ max = 106.6 MPa 


PROBLEM 6.83∗ The cantilever beam AB, consisting of half of a thinwalled pipe of 1.25-in. mean radius and 83 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.73 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION

From the solution to Prob. 6.73,



I=

 

e=



π 2 4

π

a 3t

Q = a 2 t sin θ Qmax = a 2t

a

For a half-pipe section, the distance from the center of the semi-circle to the centroid is

 

x=



2

π

a

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is d =e−x =

(a)

4

π

a−

2

π

a=

2

π

a

Equivalent force-couple system at O. V =P

Data: P = 500 lb

M 0 = Vd =

2

π

Pa

a = 1.25 in. V = 500 lb  2 M 0 =   (500)(1.25) π 

 

M 0 = 398 lb ⋅ in 




PROBLEM 6.83* (Continued)

(b)

Shearing stresses. (1)

Due to V : τ V =

τV =





(2)

VQmax  It ( P)( a 2 t ) 2P (2)(500) = = = 679 psi  π at π (1.25)(0.375) π 3  a t ( t ) 2   

Due to the torque. M 0 , For a long rectangular section of length l and width t, the shearing stress due to torque M 0 is

τM = Data:

c1lt

2

t 1 where c1 = 1 − 0.630  l 3

l = π a = π (1.25) = 3.927 in. t = 0.375 in. c1 = 0.31328

τM = By superposition,

M0

397.9 = 2300 psi (0.31328)(3.927)(0.375)2

τ = τ V + τ M = 679 psi + 2300 psi

τ = 2980 psi 


PROBLEM 6.84∗ Solve Prob. 6.83, assuming that the thickness of the beam is reduced to 14 in. PROBLEM 6.83∗ The cantilever beam AB, consisting of half of a thin-walled pipe of 1.25-in. mean radius and 83 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.73 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION 

From the solution to Prob. 6.73,



I = π a 3t 4 e= a



π



Q = a 2 t sin θ Qmax = a 2 t

For a half-pipe section, the distance from the center of the semi-circle to the centroid is



x=

2

π

a

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is d =e−x =

(a)

Data:



π

a−

2

π

a=

2

π

a

Equivalent force-couple system at O. V =P



4

M 0 = Vd =

P = 500 lb

2

π

Pa

a = 1.25 in. V = 500 lb  M 0 = 398 lb ⋅ in 





(b)

Shearing stresses. (1)

Due to V , τ V =

τV =

(2)

VQmax It ( P)(a 2 t )

=

π 3   2 a t (t )   

2P (2)(500) = = 1019 psi π at π (1.25)(0.250)

Due to the torque. M 0 , For a long rectangular section of length l and width t, the shearing stress due to torque M 0 is

τM = Data:

c1lt

2

t 1 where c1 = 1 − 0.630  l 3

l = π a = π (1.25) = 3.927 in. t = 0.250 in. c1 = 0.31996

τM = By superposition

M0

397.9 = 5067 psi (0.31996)(3.927)(0.250) 2

τ = τ V + τ M = 1019 psi + 5067 psi

τ = 6090 psi 


PROBLEM 6.85∗ The cantilever beam shown consists of a Z-shape of 14 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A′B′ in the upper horizontal leg of the Z-shape. The x′ and y ′ axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x′ = 166.3 in 4 and I y′ = 13.61 in 4 .

SOLUTION V = 3 kips β = 22.5° Vx′ = V sin β Vy ′ = V cos β

In upper horizontal leg, use coordinate x : (−6 in ⭐ x ⭐ 0) 1 (6 + x) in. 4 1 x = (−6 + x) in. 2 y = 6 in. x′ = x cos β + y sin β y ′ = y cos β − x sin β A=

τ1 =

Due to Vx′ :

Vx′ Ax′ I yt

1 1  (V sin β )   (6 + x)  ( −6 + x) cos β + 6sin β  4 2    τ1 = 1 (13.61)   4 = 0.084353(6 + x)(−0.47554 + 0.46194 x) 1 1   (V cos β )   (6 + x) 6 cos β − (−6 + x)sin β  2 4   τ2 = = 1 I xt   (166.3)   4 = 0.0166665(6 + x)[6.69132 − 0.19134 x] Vy′ A y ′

Due to Vy′ :

τ1 + τ 2 = (6 + x)[−0.07141 + 0.035396 x]

Total: x (in)

−6

−5

−4

−3

−2

−1

0

τ (ksi)

0

−0.105

−0.140

−0.104

0.003

0.180

0.428

 


PROBLEM 6.86∗ For the cantilever beam and loading of Prob. 6.85, determine the distribution of the shearing stress along line B′D′ in the vertical web of the Z-shape. PROBLEM 6.85∗ The cantilever beam shown consists of a Z-shape of 14 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A′B′ in the upper horizontal leg of the Z-shape. The x′ and y ′ axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x′ = 166.3 in 4 and I y′ = 13.61 in 4 .

SOLUTION V = 3 kips β = 22.5° Vx′ = V sin β Vy ′ = V cos β

For part AB’

For part B′Y,

Due to Vx′ :

1 A =   (6) = 1.5 in 2 4 x = −3 in., y = 6 in.

1 (6 − y ) 4 1 x = 0 y = (6 + y ) 2 x′ = x cos β + y sin β y ′ = y cos β − x sin β A=

τ1 = τ1 = =

′ ) Vx′ ( AAB x AB + ABY xBY I y ′t (V sin β )[(1.5)(−3cos β + 6sin β ) + 14 (6 − y ) 12 (6 + y ) sin β ] (13.61) ( 14 )

(V sin β )[−0.7133 + 1.7221 − 0.047835 y 2 ] = 0.3404 − 0.01614 y 2 3.4025



τ2 =

Due to Vy′ :

τ2 = =

I x′ t (V cos β )[(1.5)(6 cos β + 3 sin β ) + 14 (6 − y ) 12 (6 + y ) cos β ] (166.3) ( 14 )

(V cos β )[10.037 + 4.1575 − 0.11548 y 2 ] = 0.9463 − 0.00770 y 2 1 (166.3) ( 4 )

τ1 + τ 2 = 1.2867 − 0.02384 y 2

Total:



Vy′ ( AAB y ′AB + ABY y ′)

y (in.)

0

±2

±4

±6

τ (ksi)

1.287

1.191

0.905

0.428






PROBLEM 6.87∗ Determine the distribution of the shearing stresses along line D ′B′ in the horizontal leg of the angle shape for the loading shown. The x′ and y ′ axes are the principal centroidal axes of the cross section.

SOLUTION

β = 15.8°

Vx′ = P cos β

A( y ) = (2a − y )t

Coordinate transformation.

In particular,

y=

Vy′ = − P sin β

1 (2a + y ), 2

x =0

2  1    y ′ =  y − a  cos β −  x − a  sin β 3  6    1  2    x′ =  x − a  cos β +  y − a  sin β 6  3    2  1    y ′ =  y − a  cos β −  x − a  sin β 3  6    1  1  1  =  y + a  cos β −  − a  sin β 3  2  6  = 0.48111y + 0.36612a 1  2    x′ =  x − a  cos β +  y − a  sin β 6  3    1   1  1 =  − a  cos β +  y + a  sin β 3   6  2 = 0.13614 y − 0.06961a


PROBLEM 6.87∗ (Continued)

+→ τ =

Vx′ Ax′ Vy′ A y ′ + I y′t I x′ t

( P cos β )(2a − y )(t )(0.13614 y − 0.06961a ) (0.1557 ta3 )(t ) (− P sin β )(2a − y )(0.48111 y + 0.36612a) + (1.428 a3t )(t ) P(2a − y )(0.750 y − 0.500a ) = ta3 =

y (a ) P −τ    at  

0

1 3

2 3

1

4 3

5 3

2

−1.000

−0.417

0

0.250

0.333

0.250

0




PROBLEM 6.88∗ For the angle shape and loading of Prob. 6.87, determine the distribution of the shearing stresses along line D ′A′ in the vertical leg. PROBLEM 6.87* Determine the distribution of the shearing stresses along line D ′B′ in the horizontal leg of the angle shape for the loading shown. The x′ and y ′ axes are the principal centroidal axes of the cross section.

SOLUTION

β = 15.8°

Vx′ = P cos β

Vx′ = − P sin β x=

1 (a + x), 2

A( x) − (a − x)t y =0

Coordinate transformation. 2  1    y ′ =  y − a  cos β −  x − a  sin β 3  6    1  2    x′ =  x − a  cos β +  y − a  sin β 6  3   

In particular, 2  1    y ′ =  y − a  cos β −  x − a  sin β 3 6     1   2  1 =  − a  cos β −  x + a  sin β 3 2 3     = −0.13614 x − 0.73224a 1  2    x′ =  x − a  cos β +  y − a  sin β 6  3    1  1  2  =  x + a  cos β +  − a  sin β 3  2  3  = 0.48111x + 0.13922a



τ= =

Vx′ A( x) x′ Vy′ A( x) y ′ − I ′y t I x′ t ( P cos β )(a − x)(t )(0.48111x + 0.13922a ) (0.1557 ta 3 )(t )

(− P sin β )(a − x)(t )(−0.13614 x − 0.73224 a) (1.428 ta3 )(t ) P(a − x)(3.00 x + 1.000a) = ta3 +

x(a ) P

τ ↓   at  

0

1 6

1 3

1 2

2 3

5 6

1

1.000

1.250

1.333

1.250

1.000

0.583

0




PROBLEM 6.89 A square box beam is made of two 20 × 80-mm planks and two 20 × 120-mm planks nailed together as shown. Knowing that the spacing between the nails is s = 30 mm and that the vertical shear in the beam is V = 1200 N, determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

SOLUTION 1 1 b2 h23 − b1h13 12 12 1 1 = (120)(120)3 − (80)(80)3 = 13.8667 × 106 mm 4 12 12 −6 4 = 13.8667 × 10 m

I=

(a)

A1 = (120)(20) = 2400 mm 2 y1 = 50 mm Q1 = A1 y1 = 120 × 103 mm3 = 120 × 10−6 m3 VQ (1200)(120 × 10−6 ) = = 10.385 × 103 N/m I 13.8667 × 10−6 qs = 2 Fnail q=

Fnail =

(b)

qs (10.385 × 103 )(30 × 10−3 ) = 2 2

Fnail = 155.8 N 

Q = Q1 + (2)(20)(40)(20) = 120 × 103 + 32 × 103 = 152 × 103 mm3 = 152 × 10−6 m3

τ max =

VQ (1200)(152 × 10−6 ) = It (13.8667 × 10−6 )(2 × 20 × 10−3 )

= 329 × 103 Pa

τ max = 329 kPa 


PROBLEM 6.90 The beam shown is fabricated by connecting two channel shapes and two plates, using bolts of 34 -in. diameter spaced longitudinally every 7.5 in. Determine the average shearing stress in the bolts caused by a shearing force of 25 kips parallel to the y-axis.

SOLUTION C12 × 20.7: d = 12.00 in., I x = 129 in 4

For top plate,

y=

12.00  1  1  +    = 6.25 in. 2  2  2  3

1 1 1 I t = (16)   + (16)   (6.25) 2 = 312.667 in 4 12 2 2

For bottom plate,

I b = 312.667 in 4

Moment of inertia of fabricated beam: I = (2)(129) + 312.667 + 312.667 = 883.33 in 4 1 Q = Aplate yplate = (16)   (6.25) = 50 in 3 2 VQ (25)(50) = = 1.41510 kips/in q= 883.33 I 1 1 Fbolt = qs =   (1.41510)(7.5) = 5.3066 kips 2 2

τ bolt

π

(d bolt ) 2 =

π 3

2

= 0.44179 in 2 4 4  4  F 5.3066 = bolt = = 12.01 ksi Abolt 0.44179

Abolt =

τ bolt = 12.01 ksi 



SOLUTION ΣM B = 0: −2.3 A + (1.5)(72) = 0 A = 46.957 kN ↑

V = A = 46.957 kN

At section n-n,

Calculate moment of inertia: 1  1  1 I = 2  (15)(40)3  + 2  (15)(80)3  + (30)(1203 ) 12  12  12 = 5.76 × 106 mm 4 = 5.76 × 10−6 m 4 ta = 30 mm = 0.030 m

At a,

Qa = (30 × 20)(50) = 30 × 103 mm3 = 30 × 10−6 m3

τa =

VQa (46.957 × 103 )(30 × 10−6 ) = Ita (5.76 × 10−6 )(0.030)

= 8.15 × 106 Pa = 8.15 MPa tb = 60 mm = 0.060 m

At b,

Qb = Qa + (60 × 20)(30) = 30 × 103 + 36 × 103 = 66 × 103 mm3 = 66 × 10−6 m 4

τb =

VQb (46.957 × 103 )(66 × 10−6 ) = = 8.97 × 106 Pa = 8.97 MPa Itb (5.76 × 10−6 )(0.060)

t NA = 90 mm = 0.090 m

At NA,

QNA = Qb + (90 × 20)(10) = 66 × 103 + 18 × 103 = 84 × 103 mm3 = 84 × 10−6 m3

τ NA = (a) (b)

τ max occurs at b.

VQNA (46.957 × 103 )(84 × 10−6 ) = = 7.61 × 106 Pa = 7.61 MPa It NA (5.76 × 10−6 )(0.090)

τ max = 8.97 MPa  τ a = 8.15 MPa 


PROBLEM 6.92 For the beam and loading shown, determine the minimum required width b, knowing that for the grade of timber used, σ all = 12 MPa and τ all = 825 kPa.

SOLUTION ΣM D = 0: −3 A + (2)(2.4) + (1)(4.8) − (0.5)(7.2) = 0 A = 2 kN ↑

Draw the shear and bending moment diagrams. |V |max = 7.2 kN = 7.2 × 103 N |M |max = 3.6 kN ⋅ m = 3.6 × 103 N ⋅ m

σ=

Bending:

Smin =

M S |M |max

σ

3.6 × 103 12 × 106 = 300 × 10−6 m3 = 300 × 103 mm3 =

1 S = bh 2 6 6 S (6)(300 × 103 ) = 80 mm b= 2 = h (150)2

For a rectangular section,

Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section. 1 1 1 bh, y = h, Q = Ay = bh 2 2 4 8 1 3 I = bh t = b 12 V ( 18 bh 2 ) VQ 3V τ= = = 3 1 It ( 12 bh )(b) 2 bh

A=

b=

3V (3)(7.2 × 103 ) = = 87.3 × 10−3 m 2hτ (2)(150 × 10−3 )(825 × 103 )

The required value of b is the larger one.

b = 87.3 mm 



SOLUTION RA = RB = 25 kips V = 25 kips

At section n-n,

Locate centroid and calculate moment of inertia. Part

A(in 2 )

y (in.)

A y (in 3 )

d(in.)

Ad 2 (in 4 )

I (in 4 )



4.875

6.875

33.52

2.244

24.55

0.23



10.875

3.625

39.42

1.006

11.01

47.68

Σ

15.75

35.56

47.86

72.94

ΣAy 72.94 = = 4.631 in. ΣA 15.75 I = ΣAd 2 + ΣI = 35.56 + 47.86 = 83.42 in 4

Y =

(a)

3 Qa = Ay =   (1.5)(4.631 − 0.75) = 4.366 in 3 4 3 t = = 0.75 in. 4

τa = (b)

VQ (25)(4.366) = It (83.42)(0.75)

τ a = 1.745 ksi 

3 Qb = Ay =   (3)(4.631 − 1.5) = 7.045 in 3 4 t = 0.75 in.

τb =

VQ (25)(7.045) = It (83.42)(0.75)

τ b = 2.82 ksi 



SOLUTION RA = RB = 25 kips V = 25 kips.

At section n-n,

Locate centroid and calculate moment of inertia.

Part

A(in 2 )

y (in.)

Ay (in 3 )

d(in.)

Ad 2 (in 4 )



4.875

6.875

33.52

2.244

24.55

0.23



10.875

3.625

39.42

1.006

11.01

47.68

Σ

15.75

35.56

47.86

72.94

I (in 4 )

ΣAy 72.94 = = 4.631 in. ΣA 15.75 I = ΣAd 2 + ΣI = 35.56 + 47.86 = 83.42 in 4

Y =

Largest shearing stress occurs on section through centroid of entire cross section.

3  4.631  3 Q = Ay =   (4.631)   = 8.042 in 4  2  3 t = = 0.75 in. 4 VQ (25)(8.042) = τ= It (83.42)(0.75)

τ m = 3.21 ksi 


PROBLEM 6.95 The composite beam shown is made by welding C 200 × 17.1 rolled-steel channels to the flanges of a W250 × 80 wide-flange rolled-steel shape. Knowing that the beam is subjected to a vertical shear of 200 kN, determine (a) the horizontal shearing force per meter at each weld, (b) the shearing stress at point a of the flange of the wide-flange shape.

SOLUTION For W250 × 80, d = 257 mm, t f = 15.6 mm, I x = 126 × 106 mm 4 For C200 × 17.1,

A = 2170 mm 2 , b f = 57.4 mm, t f = 9.91 mm

I y = 0.545 × 106 mm 4 , x = 14.5 mm

For the channel in the composite beam, yc =

257 + 57.4 − 14.5 = 171.4 mm 2

For the composite beam, I = 126 × 106 +2 0.545 × 106 + (2170)(171.4)2  = 254.59 × 106 mm 4 = 254.59 × 10−6 m 4

(a) For the two welds, Qw = Ayc = (2170) (171.4) = 371.94 × 103 mm3 = 371.94 × 10−6 m3  q =

VQ (200 × 103 ) (371.94 × 10−6 ) = = 292.2 × 103 N/m I 254.59 × 10−6

For one weld,

q = 146.1 × 103 N m 2

Shearing force per meter of weld: (b)

146.1 kN m 

For cuts at a and a ' together, Aa = 2(112) (15.6) = 3494.4 mm 2

ya =

257 15.6 − = 120.7 mm 2 2

Qa = 371.94 × 103 + (3494.4)(120.7) = 793.71 × 103 mm3 = 793.71 × 10−6 m3

Since there are cuts at a and a ', t = 2t f = (2)(15.6) = 31.2 mm = 0.0312 m.

τa =

VQa (200 × 103 )(793.71 × 10−6 ) = = 19.99 × 106 Pa It (254.59 × 106 )(0.0312)

τ a = 19.99 MPa 


PROBLEM 6.96 An extruded beam has the cross section shown and a uniform wall thickness of 3 mm. For a vertical shear of 10 kN, determine (a) the shearing stress at point A, (b) the maximum shearing stress in the beam. Also, sketch the shear flow in the cross section.

SOLUTION tan α =

16 30

α = 28.07°

A = (3 sec α )(30) = 102 mm 2 1 I = (3 sec α )(30)3 = 7.6498 × 103 mm 4 12

Side:

Part

A (mm 2 )

y0 (mm)

Top

180

30

5.4

11.932

25.627

neglect

Side

102

15

1.53

3.077

0.966

7.6498

Side

102

15

1.53

3.077

0.966

7.6498

Bot

84

0

18.077

27.449

neglect

55.008

15.2996

Σ

468

A y (103 mm3 )

0

d (mm)

8.46

Ad 2 (103 mm 4 )

I (103 mm 4 )

ΣAy 8.46 × 103 = = 18.077 mm ΣA 468 I = ΣAd 2 + ΣI = 70.31 × 103 mm 4 = 70.31 × 10−9 m 4

Y0 =

(a)

QA = (180)(11.932) = 2.14776 × 103 mm3 = 2.14776 × 10−6 m3 t = (2)(3 × 10−3 ) = 6 × 10−3 m

τA = (b)

VQ (10 × 103 )(2.14776 × 10−6 ) = = 50.9 × 106 Pa It (70.31 × 10−9 )(6 × 10−6 )

τ A = 50.9 MPa 

1  Qm = QA + (2)(3 sec α )(11.932)  × 11.932  2  = 2.14776 × 103 + 484.06 = 2.6318 × 103 mm3 = 2.6318 × 10−6 m3 t = 6 × 10−3 m

τm =

VQm (10 × 103 )(2.6318 × 10−6 ) = = 62.4 × 106 Pa −9 −3 It (70.31 × 10 )(6 × 10 )

τ m = 62.4 MPa 



QB = (28)(3)(18.077) = 1.51847 × 103 mm3 QB 1.51847 × 103 τA = (50.9) QA 2.14776 × 103 = 36.0 MPa

τB =

Multiply shearing stresses by t (3 mm = 0.003 m) to get shear flow.


PROBLEM 6.97 The design of a beam requires welding four horizontal plates to a vertical 0.5 × 5-in. plate as shown. For a vertical shear V, determine the dimension h for which the shear flow through the welded surface is maximum.

SOLUTION Horizontal plate: 1 (4.5)(0.5)3 + (4.5)(0.5)h 2 12 = 0.046875 + 2.25h 2

Ih =

Vertical plate:

Iv =

1 (0.5)(5)3 = 5.2083 in 4 12

Whole section:

I = 4 I h + I v = 9h 2 + 5.39583 in 4

For one horizontal plate,

Q = (4.5)(0.5)h = 2.25 h in 3 q=

To maximize q, set

VQ 2.25Vh = 2 I 9h + 5.39583

dq = 0. dh 2.25V

(9h 2 + 5.39583) − 18h 2 =0 (9h2 + 5.39583)2

h = 0.774 in. 



SOLUTION 2

1 1 h I AB = I EF = ( a + b)t   + (a + b)t 3 ≈ t (a + b)h 2 4  2  12 1 1 I DG = th3 I = ΣI = t (6a + 6b + h)h 2 12 12

Part AD:

h 1 = thx 2 2 VQ Vhx = τ= It 2I a Vhx Vht F1 = τ dA = tdx = 0 2I 2I Q = tx





Vht x 2 = 2I 2

Part BD:

a

= 0



a 0

xdx

Vhta 2 4I

h 1 = thx 2 2 VQ Vhx = τ= 2I It

Q = tx



F2 = τ dA = Vht x 2 = 2I 2 ΣM H =



b Vhx 0

b

= 0

2I

tdx =

Vht 2I



b 0

xdx

Vhtb 2 4I

ΣM H :

Vh 2 t (b 2 − a 2 ) 4I Vh 2t (b 2 − a 2 ) 3V (b 2 − a 2 ) = = 4 ⋅ 121 t (6a + 6b + h)h 2 6a + 6b + h

Ve = F2 h − F1h =

e=

3(b 2 − a 2 )  6(a + b) + h



SOLUTION 1 1 I AB =   (1.5)3 = 0.28125 in 4 3 4  1 LBD = 3 in. ABD = (3)   = 0.75 in 2 4 1 1 I BD = ABD h 2 = (0.75)(1.5)2 = 0.5625 in 4 3 3 I = (2)(0.28125) + (2) (0.5625) = 1.6875 in 4

Part AB:

1 1 1 y y = y Q = Ay = y 2 4 2 8 2 VQ Vy Vy 2 = = τ= It (8) (1.6875) (0.25) 3.375 A=



F1 = τ dA = =



1.5

0

(0.25)V y 3 3.375 3

Vy 2 (0.25dy ) 3.375 1.5

= 0

(0.25) (1.5)3 (3.375) (3)

= 0.08333V MD =

M D : Ve = 2 F1 (3 sin 60°) Ve = (2) (0.08333) V (3 sin 60°) e = (2)(0.08333) (3 sin 60°)

e = 0.433 in. 



SOLUTION Part AB:

A = tx, q= F1 = =

Part DE:

y = 100 mm, Q = 100 tx

VQ 100Vtx = I I



xB xA

qdx = 100

Vt I



60 0

x dx

(100)(60)2 Vt Vt = 180 × 103 2 I I

A = tx,

y = 80 mm, Q = 80 tx

VQ 80 Vtx = I I x Vt F2 = qdx = 80 xD I Vt = (40 b 2 ) I q=





b 0

xdx

ΣM 0 = 0: (200)(180 × 103 ) b2 =

Vt Vt − (160)(40 b 2 ) = 0 I I

(200)(180 × 103 ) = 5625 mm 2 (160)(40)

b = 75.0 mm 


Strength of Materials

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