PROBLEM 4.1 SOLUTION. Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.

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PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

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PROBLEM 4.1

Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.

SOLUTION

For rectangle: 1 3 12

I= bh

For cross sectional area:

3 3 3 4 1 2 3 1 1 1 (2)(1.5) (2)(5.5) (2)(1.5) 28.854 in 12 12 12 I = I +I + I = + + = (a) yA = 2.75 in. (25)(2.75) 28.854 A A My I σ = − = − σA = −2.38 ksi (b) yB =0.75 in. (25)(0.75) 28.854 B B My I σ = − = − σB = −0.650 ksi

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PROBLEM 4.7

Two W4 13× rolled sections are welded together as shown. Knowing that for the steel alloy used σY =36 ksi and σU =58 ksi and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis.

SOLUTION

Properties of W4 × 13 rolled section. (See Appendix C.) 2 4 Area 3.83 in Depth 4.16 in. 11.3 in x I = = =

For one rolled section, moment of inertia about axis a-a is

2 11.3 (3.83)(2.08)2 27.87 in4

a x

I =I +Ad = + =

For both sections, 2 55.74 in4

depth 4.16 in. z a I I c = = = = all all all 58 19.333 ksi . . 3.0 (19.333) (55.74) 4.16 σ σ σ σ = = = = = = U Mc F S I I M c Mall=259 kip in⋅

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PROBLEM 4.9

Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam.

SOLUTION 2 2 2 1 1 2 2 2 4 (4)(25) (25) 981.7 mm 10.610 mm 2 2 3 3 25 (50)(25) 1250 mm 12.5 mm 2 2 r A r y h A bh y π π π π = = = = = = = = = = − = − = − 1 1 2 2 1 2 (981.7)(10.610) (1250)( 12.5) 2.334 mm 981.7 1250 A y A y y A A + + − = = = − + + 1 2 4 2 4 2 6 4 1 1 1 1 1 1 1 2 3 2 3 4 1 1 1 1 3 3 3 4 2 2 2 (25) (981.7)(10.610) 42.886 10 mm 8 8 10.610 ( 2.334) 12.944 mm 42.866 10 (981.7)(12.944) 207.35 10 mm 1 1 (50)(25) 65.104 10 mm 12 12 12.5 x I I A y r A y d y y I I A d I bh d y y π π = − = − = − = × = − = − − = = + = × + = × = = = × = − = − 2 3 2 3 4 2 2 2 2 3 4 9 4 1 2 top bot ( 2.334) 10.166 mm 65.104 10 (1250)(10.166) 194.288 10 mm 401.16 10 mm 401.16 10 m 25 2.334 27.334 mm 0.027334 m 25 2.334 22.666 mm 0.022666 m I I A d I I I y y − − − = = + = × + = × = + = × = × = + = = = − + = − = − 3 3 0 : (4 10 )(300 10 ) 1200 N m MPa = M = Pa = × × − = ⋅ top 6 top 9 (1200)(0.027334) 81.76 10 Pa 401.16 10 My I σ = − = − = − × × σtop = −81.8 MPa 6 bot bot 9 (1200)( 0.022666) 67.80 10 Pa 401.16 10 My I σ = − = − − = × × σbot =67.8 MPa

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PROBLEM 4.14

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 50 kip ⋅ in., determine the total force acting (a) on the top flange, (b) on the shaded portion of the web.

SOLUTION

The stress distribution over the entire cross-section is given by the bending stress formula: x My

I σ = −

where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area element dA, the force is

x My dF dA dA I σ = = −

The total force on the shaded area is then

* *

My M M

F dF dA ydA y A

I I I

= = − = − = −

where y* is the centroidal coordinate of the shaded portion and A* is its area.

Calculate the moment of inertia.

3 3 4

1 1

(6 in.)(7 in.) (4 in.)(4 in.) 150.17 in

12 12 50 kip in I M = − = = ⋅

(a) Top flange: A*=(6 in.)(1.5 in.) =9 in2 y*= 2 in. 0.75 in.+ = 2.75 in.

2 4 50 kip in (9 in )(2.75 in.) 8.24 kips 150.17 in F = ⋅ = F=8.24 kips

(b) Half web: A*=(2 in.)(2 in.) = 4 in2 y*=1 in.

2 4 50 kip in (4 in )(1 in.) 1.332 kips 150.17 in F = ⋅ = F=1.332 kips

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PROBLEM 4.24

A 60 N ⋅ m couple is applied to the steel bar shown. (a) Assuming that the couple is applied about the z axis as shown, determine the maximum stress and the radius of curvature of the bar. (b) Solve part a, assuming that the couple is applied about the y axis. Use E=200 GPa.

SOLUTION

(a) Bending about z-axis.

3 3 3 4 9 4 1 1 (12)(20) 8 10 mm 8 10 m 12 12 20 10 mm 0.010 m 2 − = = = × = × = = = I bh c 6 9 (60)(0.010) 75.0 10 Pa 8 10 Mc I σ = = = × × σ =75.0 MPa 3 1 9 9 1 60 37.5 10 m (200 10 )(8 10 ) M EI ρ = = × × − = × − − ρ=26.7 m

(b) Bending about y-axis.

3 3 3 4 9 4 6 9 1 1 (20)(12) 2.88 10 mm 2.88 10 m 12 12 12 6 mm 0.006 m 2 (60)(0.006) 125.0 10 Pa 2.88 10 I bh c Mc I σ − − = = = × = × = = = = = = × × σ =125.0 MPa 3 1 9 9 1 60 104.17 10 m (200 10 )(2.88 10 ) M EI ρ = = × × − = × − − ρ=9.60 m

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PROBLEM 4.34

A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Using the data given below, determine the largest permissible bending moment when the composite bar is bent about a horizontal axis.

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa Allowable stress 100 MPa 160 MPa

SOLUTION

Use aluminum as the reference material. For aluminum, n=1.0

For brass, n E E= b/ a =105/70 1.5= Values of n are shown on the sketch. For the transformed section,

(

)

3 3 3 4 1 1 1 1 3 3 3 3 3 4 2 2 2 2 2 3 4 3 1 3 4 9 4 1 2 3 1.5 (8)(32) 32.768 10 mm 12 12 1.0 (32)(32 16 ) 76.459 10 mm 12 12 32.768 10 mm 141.995 10 mm 141.995 10 m n I b h n I b H h I I I I I I − = = = × = − = − = × = = × = + + = × = × | |σ = nMy M = σI I ny Aluminum: 6 6 9 1.0, | | 16 mm 0.016 m, 100 10 Pa (100 10 )(141.995 10 ) 887.47 N m (1.0)(0.016) σ − = = = = × × × = = ⋅ n y M Brass: 6 6 9 1.5, | | 16 mm 0.016 m, 160 10 Pa (160 10 )(141.995 10 ) 946.63 N m (1.5)(0.016) σ − = = = = × × × = = ⋅ n y M

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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 4.41

The 6 12-in.× timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 10 psi× 6 and for

steel, 29 10 psi.× 6 Knowing that the beam is bent about a horizontal axis by a

couple of moment M =450 kip in.,⋅ determine the maximum stress in (a) the wood, (b) the steel.

SOLUTION

Use wood as the reference material. For wood, 1 For steel, s/ w 29 /1.8 16.1111 n n E E = = = =

Transformed section: = wood = steel

421.931 112.278 3.758 in. o Y = =

The neutral axis lies 3.758 in. above the wood-steel interface.

3 2 3 2 4 1 1 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 2 2 4 1 2 1 (6)(12) (72)(6 3.758) 1225.91 in 12 12 16.1111(5) (0.5) (40.278)(3.578 0.25) 647.87 in 12 12 1873.77 in 450 kip in n I b h n A d n I b h n A d I I I nMy M I σ = + = + − = = + = + + = = + = = ⋅ = − (a) Wood: n=1, y= −12 3.758 8.242 in= (1) (450) (8.242) 1.979 ksi 1873.77 w σ = − = − σw= −1.979 ksi (b) Steel: n=16.1111, y= −3.758 0.5− = −4.258 in (16.1111) (450) ( 4.258) 16.48 ksi 1873.77 s σ = − − = σs =16.48 ksi 2 , in A nA, in2 o y , in3 o nAy 72 72 6 432 2.5 40.278 −0.25 −10.069 112.278 421.931

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PROBLEM 4.42

The 6 12-in.× timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 10 psi× 6

and for steel, 29 10 psi.× 6 Knowing that the beam is bent about a

horizontal axis by a couple of moment M =450 kip in.,⋅ determine the maximum stress in (a) the wood, (b) the steel.

SOLUTION

Use wood as the reference material.

6 6 For wood, 1 29 10 For steel, 16.1111 1.8 10 s w n E n E = × = = = × For C8 11.5× channel section,

2 4

3.38 in , w 0.220 in., 0.571 in., y 1.32 in

A= t = x= I =

For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section. The centroid of the wood (part 2) lies 0.220+6.00 =6.22 in. above the bottom.

Transformed section: 3 0 2 478.93 in 3.787 in. 126.456 in Y = = d = y0 −Y0

The neutral axis lies 3.787 in. above the bottom of the section.

2 2 4 1 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 2 2 4 1 2 (16.1111)(1.32) (54.456)(3.216) 584.49 in 1 (6)(12) (72)(2.433) 1290.20 in 12 12 1874.69 in 450 kip in I n I n A d n I b h n A d I I I n My M I σ = + = + = = + = + = = + = = ⋅ = −

(a) Wood: n=1, y=12 0.220 3.787 8.433 in.+ − = (1)(450)(8.433) 2.02 ksi 1874.69 w σ = − = − σw = −2.02 ksi (b) Steel: n=16.1111, y= −3.787 in. (16.1111) (450) ( 3.787) 14.65 ksi 1874.67 s σ = − − = σs =14.65 ksi

Part A, in2 nA, in2 y, in. nAy, in3 d, in.

1 3.38 54.456 0.571 31.091 3.216

2 72 72 6.22 447.84 2.433

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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 4.51

A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 3 10 psi× 6 for the concrete and 29 10 psi× 6 for the steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the steel, determine the largest allowable positive bending moment in the beam.

SOLUTION 6 6 2 2 2 2 29 10 9.67 3 10 7 3 (3) 1.8040 in 17.438 in 4 4 8 s c s s E n E A π d π nA × = = = × = = = =

Locate the neutral axis:

2 8 (17.438)(14 ) 0 2 4 17.438 244.14 0 x x x x x − − = + − = Solve for x. 2 17.438 17.438 (4)(4)(244.14) 5.6326 in. (2)(4) x=− + + = 14− =x 8.3674 in. 3 2 3 2 4 1 1 8 (14 ) (8)(5.6326) (17.438)(8.3674) 1697.45 in 3 s 3 I = x +nAx = + = nMy I M I ny σ σ = ∴ =

Concrete: n=1.0, y =5.6326 in., σ =1350 psi

3

(1350)(1697.45)

406.835 10 lb in 407 kip in (1.0)(5.6326)

M = = × ⋅ = ⋅

Steel: n=9.67, y =8.3674 in., σ =20 10 psi× 3 3

(20 10 )(1697.45)

419.72 lb in 420 kip in (9.67)(8.3674)

M = × = ⋅ = ⋅

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PROBLEM 4.52

Knowing that the bending moment in the reinforced concrete beam is +100 kip ⋅ ft and that the modulus of elasticity is 3.625 10 psi× 6 for the

concrete and 29 10 psi× 6 for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 6 6 2 2 2 29 10 8.0 3.625 10 (4) (1) 3.1416 in 25.133 in 4 s c s s E n E A π nA × = = = × = = =

Locate the neutral axis.

(24)(4)( 2) (12 ) (25.133)(17.5 4 ) 0 2 x x+ + x − − −x = 2 2 96x+192+6x −339.3+25.133x = 0 or 6x +121.133x−147.3= 0 Solve for x. 2 121.133 (121.133) (4)(6)(147.3) 1.150 in. (2)(6) x = − + + = 3 17.5 4 12.350 in. d = − − =x 3 2 3 2 4 1 1 1 1 1 3 3 4 2 2 2 2 4 3 3 3 4 1 2 3 1 1 (24)(4) (24)(4)(3.150) 1080.6 in 12 12 1 1 (12)(1.150) 6.1 in 3 3 (25.133)(12.350) 3833.3 in 4920 in I b h A d I b x I nA d I I I I = + = + = = = = = = = = + + = nMy I

σ = − where M =100 kip ft⋅ =1200 kip in.⋅ (a) Steel: n =8.0 y = −12.350 in.

(8.0)(1200)( 12.350) 4920 s σ = − − σs = 24.1 ksi (b) Concrete: n =1.0, y = +4 1.150 =5.150 in. (1.0)(1200)(5.150) 4920 c σ = − σc = −1.256 ksi

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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 4.55

Five metal strips, each 40 mm wide, are bonded together to form the composite beam shown. The modulus of elasticity is 210 GPa for the steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅ m, determine (a) the maximum stress in each of the three metals, (b) the radius of curvature of the composite beam.

SOLUTION

Use aluminum as the reference material. 1 in aluminum. / 210 / 70 3 in steel. / 105 / 70 1.5 in brass. s a b a n n E E n E E = = = = = = =

Due to symmetry of both the material arrangement and the geometry, the neutral axis passes through the center of the steel portion.

For the transformed section,

3 2 3 2 3 4 1 1 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 2 2 3 3 3 4 3 3 3 3 3 4 3 4 4 2 5 1 1 (40)(10) (40)(10)(25) 253.33 10 mm 12 12 1.5 (40)(10) (1.5)(40)(10)(15) 140 10 mm 12 12 3.0 (40)(20) 80 10 mm 12 12 140 10 mm 253.33 10 mm n I b h n A d n I b h n A d n I b h I I I I = + = + = × = + = + = × = = = × = = × = = × 3 4 9 4 866.66 10 mm 866.66 10 m I = I = × = × − (a) nMy I σ = − where M =1800 N m⋅ Aluminum: n =1.0 y = −30 mm =0.030 m 6 9 (1.0)(1800)(0.030) 62.3 10 Pa 866.66 10 a σ = = × × σa =62.3 MPa Brass: n=1.5 y= −20 mm= −0.020 m 6 9 (1.5)(1800)(0.020) 62.3 10 Pa 866.66 10 b σ = = × × σb=62.3 MPa Steel: n=3.0 y= −10 mm= −0.010 m 6 9 (3.0)(1800)(0.010) 62.3 10 Pa 866.66 10 s σ = = × × σs =62.3 MPa (b) Radius of curvature. 1 9 1800 9 0.02967 m 1 (70 10 )(866.66 10 ) a M E I ρ = = × × − = − ρ=33.7 m

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PROBLEM 4.57

The composite beam shown is formed by bonding together a brass rod and an aluminum rod of semicircular cross sections. The modulus of elasticity is

6

15 10 psi× for the brass and 10 10 psi× 6 for the aluminum. Knowing that the composite beam is bent about a horizontal axis by couples of moment 8 kip in.,⋅ determine the maximum stress (a) in the brass, (b) in the aluminum.

SOLUTION

For each semicircle, 0.8 in. 2 1.00531 in2

2 r = A= πr = , 4 4 base 4 (4)(0.8) 0.33953 in. 0.160850 in 3 3 8 o r y I πr π π = = = = = 2 2 4 base o 0.160850 (1.00531)(0.33953) 0.044953 in I = IAy = − =

Use aluminum as the reference material.

6 6 1.0 in aluminum 15 10 1.5 in brass 10 10 b a n E n E = × = = = × Locate the neutral axis.

0.17067 0.06791 in. 2.51327

o

Y = =

The neutral axis lies 0.06791 in. above the material interface.

1 2 2 2 4 1 1 1 1 2 2 4 2 2 2 2 1 2 0.33953 0.06791 0.27162 in., 0.33953 0.06791 0.40744 in. (1.5)(0.044957) (1.5)(1.00531)(0.27162) 0.17869 in (1.0)(0.044957) (1.0)(1.00531)(0.40744) 0.21185 in d d I n I n Ad I n I n Ad I I I = − = = + = = + = + = = + = + = = + =0.39054 in4

(a) Brass: n =1.5, y =0.8−0.06791= 0.73209 in. (1.5)(8)(0.73209) 0.39054 nMy I σ = − = − σ= −22.5 ksi (b) Aluminium: n =1.0, y = −0.80.06791= −0.86791 in. (1.0)(8)( 0.86791) 0.39054 nMy I σ= − = − − σ =17.78 ksi A, in2 nA, in2 yo, in. , in3 o nAy 1.00531 1.50796 0.33953 0.51200 1.00531 1.00531 −0.33953 −0.34133 Σ 2.51327 0.17067

Figure

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