Mechanics of Materials 7th Edition Beer Johnson Chapter 4

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C

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M 5 15 kN · m Dimensions in mm B 20 40 20 20 20 80 A

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 Outside rectangle: 1 1 (80)(120)3 12 I 6 4 6 4 1 11.52 10 mm 11.52 10 m I Cutout: 3 2 1 (40)(80) 12 I 6 4 6 4 2 1.70667 10 mm 1.70667 10 m I Section: I I1 I2 9.81333 10 m6 4 (a) yA 40 mm 0.040 m 3 6 6 (15 10 )(0.040) 61.6 10 Pa 9.81333 10 A A My I 61.6 MPa A (b) yB 60 mm 0.060 m 3 6 6 (15 10 )( 0.060) 91.7 10 Pa 9.81333 10 B B My I 91.7 MPa B

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2 in. 2 in. 1.5 in. 2 in. 2 in. 2 in. A B

M ! 25 kip · in.

PROBLEM 4.2

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 A 2.38 ksi (b) yB 0.75 in. (25)(0.75) 28.854 B B My I B 0.650 ksi

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200 mm 220 mm 12 mm 12 mm 8 mm C x y M

PROBLEM 4.3

Using an allowable stress of 155 MPa, determine the largest bending moment M that can be applied to the wide-flange beam shown. Neglect the effect of fillets.

SOLUTION

Moment of inertia about x-axis:

3 2 1 6 4 3 6 4 2 6 4 3 1 1 (200)(12) (200)(12)(104) 12 25.9872 10 mm 1 (8)(196) 5.0197 10 mm 12 25.9872 10 mm I I I I 6 4 6 4 1 2 3 6 6 6 3 56.944 10 mm 56.944 10 m 1 with (220) 110 mm 0.110 m 2 with 155 10 Pa (56.944 10 )(155 10 ) 80.2 10 N m 0.110 x I I I I Mc c I I M c M 80.2 kN m x M

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200 mm 220 mm 12 mm 12 mm 8 mm C x y M

PROBLEM 4.4

Solve Prob. 4.3, assuming that the wide-flange beam is bent about the y axis by a couple of moment My.

PROBLEM 4.3. Using an allowable stress of 155 MPa, determine the

largest bending moment M that can be applied to the wide-flange beam shown. Neglect the effect of fillets.

SOLUTION

Moment of inertia about y axis:

3 6 4 1 3 3 4 2 6 4 3 1 1 (12)(200) 8 10 mm 12 1 (196)(8) 8.3627 10 mm 12 8 10 mm I I I I 6 4 6 4 1 2 3 6 6 6 3 16.0084 10 mm 16.0084 10 m 1 with (200) 100 mm 0.100 m 2 with 155 10 Pa (16.0084 10 )(155 10 ) 24.8 10 N m 0.100 y y I I I I Mc c I I M c M 24.8 kN m y M

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M2 M1 0.1 in. 0.2 in. 0.5 in. 0.5 in. (a) (b)

PROBLEM 4.5

Using an allowable stress of 16 ksi, determine the largest couple that can be applied to each pipe.

SOLUTION (a) 4 4 4 4 3 4 3 (0.6 0.5 ) 52.7 10 in 4 4 0.6 in. (16)(52.7 10 ) : 0.6 o i I r r c Mc I M I c 1.405 kip in. M (b) 4 4 3 4 3 (0.7 0.5 ) 139.49 10 in 4 0.7 in. (16)(139.49 10 ) : 0.7 I c Mc I M I c 3.19 kip in. M

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120 mm 30 mm 30 mmM = 2.8 kN · m r 5 20 mm A B

PROBLEM 4.6

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

SOLUTION 3 4 6 4 1 1 (0.120 m)(0.06 m) 2 (0.02 m) 12 12 4 2.1391 10 mm I (a) 3 6 4 (2.8 10 N m)(0.03 m) 2.1391 10 mm A A M y I 39.3 MPa A (b) 3 6 4 (2.8 10 N m)(0.02 m) 2.1391 10 m B B M y I 26.2 MPa B

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y

z C

PROBLEM 4.7

Two W4 13 rolled sections are welded together as shown. Knowing that for the steel alloy used Y 36 ksi and U 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 Width 4.060 in. 3.86 in y I

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

2 3.86 (3.83)(2.030)2 19.643 in4

b y

I I Ad

For both sections, 2 39.286 in4

width 4.060 in. z b I I c all all all 58 19.333 ksi . . 3.0 (19.333)(39.286) 4.060 U Mc F S I I M

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y

z

C

PROBLEM 4.8

Two W4 13 rolled sections are welded together as shown. Knowing that for the steel alloy used U 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 U Mc F S I I M

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D C B A 6 in. 2 in. 3 in. 3 in. 15 kips 15 kips 3 in.

40 in. 60 in. 40 in.

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 A y0 A y0 18 5 90 18 1 18 36 108 0 108 3 in. 36 Y

Neutral axis lies 3 in. above the base.

3 2 3 2 4 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 4 1 2 1 1 (3)(6) (18)(2) 126 in 12 12 1 1 (9)(2) (18)(2) 78 in 12 12 126 78 204 in I b h A d I b h A d I I I

top 5 in. bot 3 in.

y y 0 (15)(40) 600 kip in. M Pa M Pa top top (600)(5) 204 M y

I top 14.71 ksi (compression)

bot

bot (600)( 3)

204 M y

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D C B

A

25 kips 25 kips

20 in. 60 in. 20 in. 4 in. 1 in. 1 in. 1 in. 6 in. 8 in.

PROBLEM 4.10

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 A y0 A y0 8 7.5 60 6 4 24 4 0.5 2 18 86 86 4.778 in. 18 o Y Neutral axis lies 4.778 in. above the base.

3 2 3 2 4 1 1 1 1 1 3 2 3 2 4 2 2 2 2 2 3 2 3 2 4 3 3 3 3 3 4 1 2 3 top b 1 1 (8)(1) (8)(2.772) 59.94 in 12 12 1 1 (1)(6) (6)(0.778) 21.63 in 12 12 1 1 (4)(1) (4)(4.278) 73.54 in 12 12 59.94 21.63 73.57 155.16 in 3.222 in. I b h A d I b h A d I b h A d I I I I y yot 4.778 in. 0 (25)(20) 500 kip in. M Pa M Pa top top (500)(3.222) 155.16 My

I top 10.38 ksi (compression)

(500)( 4.778) My

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10 mm 10 mm 50 mm 10 mm 150 mm 150 mm A D B C 10 kN 10 kN 250 mm 50 mm

PROBLEM 4.11

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 , mm A y0, mm 3 0, mm Ay 600 30 18 103 600 30 18 103 300 5 1.5 103 1500 37.5 103 3 0 37.5 10 25mm 1500 Y

Neutral axis lies 25 mm above the base.

3 2 3 4 4 1 2 1 3 2 3 4 3 3 4 9 4 1 2 3 1 (10)(60) (600)(5) 195 10 mm 195 mm 12 1 (30)(10) (300)(20) 122.5 10 mm 12 512.5 10 mm 512.5 10 m I I I I I I I I top bot 3 3 3 35 mm 0.035 m 25 mm 0.025 m 150 mm 0.150 m 10 10 N (10 10 )(0.150) 1.5 10 N m y y a P M Pa 3 top 6 top 9 (1.5 10 )(0.035) 102.4 10 Pa 512.5 10 My

I top 102.4 MPa (compression)

3 6 bot bot 9 (1.5 10 )( 0.025) 73.2 10 Pa 512.5 10 M y

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72 mm 216 mm 36 mm 54 mm 108 mm y z C

PROBLEM 4.12

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN m, determine the total force acting on the shaded portion of the web.

SOLUTION

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

M y I

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

M y

dF dA dA

I

The total force on the shaded area is then

* *

M y M M

F dF dA y dA y A

I I I

where y is the centroidal coordinate of the shaded portion and A* * is its area. 1 2 54 18 36 mm 54 36 54 36 mm d d

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PROBLEM 4.12 (Continued)

Moment of inertia of entire cross section:

3 2 3 2 6 4 1 1 1 1 1 3 2 3 2 6 4 2 2 2 2 2 6 4 6 4 1 2 1 1 (216)(36) (216)(36)(36) 10.9175 10 mm 12 12 1 1 (72)(108) (72)(108)(36) 17.6360 10 mm 12 12 28.5535 10 mm 28.5535 10 m I b h A d I b h A d I I I

For the shaded area,

* 2 * * * 3 3 6 * * 3 6 6 3 (72)(90) 6480 mm 45 mm 291.6 10 mm 291.6 10 m (6 10 )(291.6 10 ) 28.5535 10 61.3 10 N A y A y MA y F I 61.3 kN F

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24 mm 12 mm 12 mm 20 mm 20 mm 20 mm 20 mm 24 mm z y C

PROBLEM 4.13

Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 4 kN m, determine the total force acting on the shaded portion of the beam.

SOLUTION Dimensions in mm: 3 3 6 6 6 4 6 4 1 1 (12 12)(88) (40)(40) 12 12 1.3629 10 0.213 10 1.5763 10 mm 1.5763 10 m z I

For use in Prob. 4.14,

3 3 6 6 6 4 6 4 1 1 (88)(64) (24 24)(40) 12 12 1.9224 10 0.256 10 1.6664 10 mm 1.6664 10 m y I

Bending about horizontal axis. Mz 4 kN m

6 4 6 4 (4 kN m)(0.044 m) 111.654 MPa 1.5763 10 m (4 kN m)(0.020 m) 50.752 MPa 1.5763 10 m z A z z B z M c I M c I

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PROBLEM 4.13 (Continued) Portion (1): A (44)(12) 528 mm2 528 10 m6 2 avg 1 1 (111.654) 55.83 MPa 2 A 2 6 2 1 avg Force A (55.83 MPa)(528 10 m ) 29.477 kN Portion (2): A (20)(20) 400 mm2 400 10 m6 2 avg 1 1 (50.752) 25.376 MPa 2 B 2 6 2 2 avg Force A (25.376 MPa)(400 10 m ) 10.150 kN

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24 mm 12 mm 12 mm 20 mm 20 mm 20 mm 20 mm 24 mm z y C

PROBLEM 4.14

Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a couple of moment 4 kN m.

PROBLEM 4.13. Knowing that a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 4 kN m, determine the total force acting on the shaded portion of the beam.

SOLUTION

Bending about vertical axis. My 4 kN m

See Prob. 4.13 for sketch and 1.6664 10 m6 4

y I 6 4 6 4 (4 kN m)(0.032 m) 76.81 MPa 1.6664 10 m (4 kN m)(0.020 m) 48.01 MPa 1.6664 10 m D y E y Mc I Mc I Portion (1): A (44)(12) 528 mm2 528 10 m6 2 avg 1 1 ( ) (76.81 48.01) 62.41 MPa 2 D E 2 6 2 1 avg Force A (62.41 MPa)(528 10 m ) 32.952 kN Portion (2): A (20)(20) 400 mm2 400 10 m6 2 avg 1 1 (48.01) 24.005 MPa 2 E 2 6 2 2 avg Force A (24.005 MPa)(400 10 m ) 9.602 kN

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M 1.5 in. 0.5 in. 1.5 in. 1.5 in. 0.5 in. 0.5 in. 0.5 in.

PROBLEM 4.15

Knowing that for the extruded beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied. SOLUTION A y0 Ay0 2.25 1.25 2.8125 2.25 0.25 0.5625 4.50 3.375 3.375 0.75 in. 4.50 Y The neutral axis lies 0.75 in. above bottom.

top bot 3 2 3 2 4 1 1 1 1 1 2 2 3 2 4 2 2 2 2 2 4 1 2 2.0 0.75 1.25 in., 0.75 in. 1 1 (1.5)(1.5) (2.25)(0.5) 0.984375 in 12 12 1 1 (4.5)(0.5) (2.25)(0.5) 0.609375 in 12 12 1.59375 in y y I b h A d I b h A d I I I My I M I y

Top: (compression) (16)(1.59375) 20.4 kip in.

1.25 M

Bottom: (tension) (12)(1.59375) 25.5 kip in.

0.75 M

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M

15 mm

d ! 30 mm

20 mm

40 mm

PROBLEM 4.16

The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam.

SOLUTION 2 , mm A y0, mm Ay0, mm3 600 22.5 13.5 10 3 300 7.5 2.25 10 3 Σ 900 15.75 10 3 3 0 15.5 10

17.5 mm The neutral axis lies 17.5 mm above the bottom. 900 Y top bot 3 2 3 2 3 4 1 1 1 1 1 3 2 3 2 3 4 2 2 2 2 2 3 4 9 4 1 2 30 17.5 12.5 mm 0.0125 m 17.5 mm 0.0175 m 1 1 (40)(15) (600)(5) 26.25 10 mm 12 12 1 1 (20)(15) (300)(10) 35.625 10 mm 12 12 61.875 10 mm 61.875 10 m y y I b h A d I b h A d I I I | | My M I I y

Top: (tension side)

6 9 (24 10 )(61.875 10 ) 118.8 N m 0.0125 M Bottom: (compression) 6 9 (30 10 )(61.875 10 ) 106.1 N m 0.0175 M

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M 15 mm d ! 30 mm 20 mm 40 mm

PROBLEM 4.17

Solve Prob. 4.16, assuming that d 40 mm.

PROBLEM 4.16 The beam shown is made of a nylon for which the

allowable stress is 24 MPa in tension and 30 MPa in compression. Determine the largest couple M that can be applied to the beam.

SOLUTION 2 , mm A y0, mm Ay0, mm3 600 32.5 19.5 10 3 500 12.5 6.25 10 3 Σ 1100 25.75 10 3 3

0 25.75 10 23.41 mm The neutral axis lies 23.41 mm above the bottom.

1100 Y top bot 3 2 3 2 3 4 1 1 1 1 1 2 2 3 2 3 4 2 2 2 2 2 3 4 1 2 40 23.41 16.59 mm 0.01659 m 23.41 mm 0.02341 m 1 1 (40)(15) (600)(9.09) 60.827 10 mm 12 12 1 1 (20)(25) (500)(10.91) 85.556 10 mm 12 12 146.383 10 mm 146.3 y y I b h A d I b h A d I I I 83 10 m9 4 | | My M I I y

Top: (tension side) (24 10 )(146.383 10 )6 9 212 N m

0.01659 M

Bottom: (compression) (30 10 )(146.383 10 )6 9 187.6 N m

0.02341 M

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1.2 in. 0.75 in.

2.4 in.

M

PROBLEM 4.18

Knowing that for the beam shown the allowable stress is 12 ksi in tension and 16 ksi in compression, determine the largest couple M that can be applied.

SOLUTION

rectangle semi-circular cutout

2 1 2 2 2 2 1 2 (2.4)(1.2) 2.88 in (0.75) 0.8836 in 2 2.88 0.8836 1.9964 in 0.6 in. 4 (4)(0.75) 0.3183 in. 3 3 (2.88)(0.6) (0.8836)(0.3183) 0.7247 in. 1.9964 A A A y r y Ay Y A

Neutral axis lies 0.7247 in. above the bottom. Moment of inertia about the base:

3 4 3 4 4 1 1 (2.4)(1.2) (0.75) 1.25815 in 3 8 3 8 b I bh r

Centroidal moment of inertia:

2 2 4 top bot 1.25815 (1.9964)(0.7247) 0.2097 in 1.2 0.7247 0.4753 in., 0.7247 in. b I I AY y y | | M y M I I y

Top: (tension side) (12)(0.2097) 5.29 kip in.

0.4753 M

Bottom: (compression) (16)(0.2097) 4.63 kip in.

0.7247 M

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54 mm

40 mm 80 mm

M

PROBLEM 4.19

Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

SOLUTION 2 , mm A y0, mm Ay0, mm3 d, mm 2160 27 58,320 3 1080 36 38,880 3 Σ 3240 97,200 97, 200

30 mm The neutral axis lies 30 mm above the bottom. 3240 Y top bot 3 2 3 2 3 4 1 1 1 1 1 2 2 3 2 3 4 2 2 2 2 2 3 4 9 4 1 2 54 30 24 mm 0.024 m 30 mm 0.030 m 1 1 (40)(54) (40)(54)(3) 544.32 10 mm 12 12 1 1 1 (40)(54) (40)(54)(6) 213.84 10 mm 36 36 2 758.16 10 mm 758.16 10 m y y I b h A d I b h A d I I I | | M y |M| I I y

Top: (tension side)

6 9 3 (120 10 )(758.16 10 ) 3.7908 10 N m 0.024 M Bottom: (compression) 6 9 3 (150 10 )(758.16 10 ) 3.7908 10 N m 0.030 M

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M 48 mm 48 mm 48 mm 36 mm 36 mm

PROBLEM 4.20

Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

SOLUTION 2 , mm A y0, mm Ay0, mm3 Solid rectangle 4608 48 221,184 Square cutout –1296 30 –38,880 Σ 3312 182,304 182,304 55.04 mm 3312

Y Neutral axis lies 55.04 mm above bottom.

top bot 3 2 3 2 6 4 1 1 1 1 1 3 2 3 2 6 4 2 2 2 2 2 6 4 1 2 96 55.04 40.96 mm 0.04096 m 55.04 mm 0.05504 m 1 1 (48)(96) (48)(96)(7.04) 3.7673 10 mm 12 12 1 1 (36)(36) (36)(36)(25.04) 0.9526 10 mm 12 12 2.8147 10 mm y y I b h A d I b h A d I I I 2.8147 10 m6 4 | | M y M I I y

Top: (tension side) (120 10 )(2.8147 10 )6 6 8.25 10 N m3

0.04096 M Bottom: (compression) 6 6 3 (150 10 )(2.8147 10 ) 7.67 10 N m 0.05504 M

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PROBLEM 4.21

Straight rods of 6-mm diameter and 30-m length are stored by coiling the rods inside a drum of 1.25-m inside diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a coiled rod, (b) the corresponding bending moment in the rod. Use E 200 GPa.

SOLUTION

Let inside diameter of the drum,

1 diameter of rod, ,

2

radius of curvature of center line of rods when bent. D d c d 3 4 4 12 4 1 1 1(1.25) 1(6 10 ) 0.622 m 2 2 2 2 (0.003) 63.617 10 m 4 4 D d I c (a) 9 6 max (200 10 )(0.003) 965 10 Pa 0.622 Ec 965 MPa (b) (200 10 )(63.617 109 12) 20.5 N m 0.622 EI M M 20.5 N m

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900 mm 8 mm t r M M'

PROBLEM 4.22

A 900-mm strip of steel is bent into a full circle by two couples applied as shown. Determine (a) the maximum thickness t of the strip if the allowable stress of the steel is 420 MPa, (b) the corresponding moment M of the couples. Use E 200 GPa.

SOLUTION

When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to 2 times , the radius of curvature, we get

900 mm

143.24 mm 0.14324 m

2

Stress: E Ec or c

E

For 420 MPa and E 200 GPa,

6 3 9 (0.14324)(420 10 ) 0.3008 10 m 200 10 c

(a) Maximum thickness: t 2c 0.6016 10 m3

0.602 mm

t

Moment of inertia for a rectangular section.

3 3 3 3 15 4 (8 10 )(0.6016 10 ) 145.16 10 m 12 12 bt I (b) Bending moment: M EI 9 15 (200 10 )(145.16 10 ) 0.203 N m 0.14324 M M 0.203 N m

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5 ft

PROBLEM 4.23

Straight rods of 0.30-in. diameter and 200-ft length are sometimes used to clear underground conduits of obstructions or to thread wires through a new conduit. The rods are made of high-strength steel and, for storage and transportation, are wrapped on spools of 5-ft diameter. Assuming that the yield strength is not exceeded, determine (a) the maximum stress in a rod, when the rod, which is initially straight, is wrapped on a spool, (b) the corresponding bending moment in the rod. Use E 29 10 psi6 .

SOLUTION

Radius of cross section: 1 1(0.30) 0.15 in.

2 2 r d Moment of inertia: 4 (0.15)4 397.61 10 in6 4 4 4 I r 1 5 ft 60 in. 30 in. 2 0.15 in. D D c r (a) 6 3 max (29 10 )(0.15) 145.0 10 psi 30 Ec max 145.0 ksi (b) 6 6 (29 10 )(397.61 10 ) 30 EI M M 384 lb in.

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20 mm 12 mm 60 N · m z y

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 75.0 MPa 3 1 9 9 1 60 37.5 10 m (200 10 )(8 10 ) M EI 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 125.0 MPa 3 1 9 9 1 60 104.17 10 m (200 10 )(2.88 10 ) M EI 9.60 m

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C 80 mm 5 mm 5 mm 10 mm 10 mm 80 mm M

PROBLEM 4.25

(a) Using an allowable stress of 120 MPa, determine the largest couple M that can be applied to a beam of the cross section shown. (b) Solve part a, assuming that the cross section of the beam is an 80-mm square.

SOLUTION

(a) I I1 4I2, where I1 is the moment of inertia of an 80-mm square and I2 is the moment of inertia of

one of the four protruding ears.

3 3 6 4 1 1 1 (80)(80) 3.4133 10 mm 12 12 I bh 3 2 3 2 3 4 2 6 4 6 4 1 2 6 6 1 1 (5)(10) (5)(10)(45) 101.667 10 mm 12 12 4 3.82 10 mm 3.82 10 mm , 50 mm 0.050 m (120 10 )(3.82 10 ) 0.050 I bh Ad I I I c Mc I M I c 3 9.168 10 N m 9.17 kN m (b) Without the ears:

6 2 1 3.4133 10 m , 40 mm 0.040 m I I c 6 6 3 (120 10 )(3.4133 10 ) 10.24 10 N m 0.040 I M c 10.24 kN m

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0.75 in. 0.2 in. 1.5 in. 0.1 in. M

PROBLEM 4.26

A thick-walled pipe is bent about a horizontal axis by a couple M. The pipe may be designed with or without four fins. (a) Using an allowable stress of 20 ksi, determine the largest couple that may be applied if the pipe is designed with four fins as shown. (b) Solve part a, assuming that the pipe is designed with no fins.

SOLUTION

x

I of hollow pipe: (1.5 in.)4 (0.75 in.)4 3.7276 in4

4 x I x I of fins: 2 1 (0.1)(0.2)3 (0.1 0.2)(1.6)2 2 1 (0.2)(0.1)3 12 12 x I 4 0.1026 in (a) Pipe as designed, with fins:

4 3.8302 in , 1.7 in. x I c 4 all all 3.8302 in 20 ksi, (20 ksi) 1.7 in. x I M c 45.1 kip in. M

(b) Pipe with no fins:

4

all 20 ksi, Ix 3.7276 in , c 1.5 in.

4 all 3.7276 in (20 ksi) 1.5 in. x I M c 49.7 kip in. M

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b

d

M' M

PROBLEM 4.27

A couple M will be applied to a beam of rectangular cross section that is to be sawed from a log of circular cross section. Determine the ratio d/b for which (a) the maximum stress m will be as small as possible, (b) the radius of curvature of the beam will be maximum.

SOLUTION

Let D be the diameter of the log.

2 2 2 2 2 2 3 2 1 1 1 12 2 6 D b d d D b I I bd c d bd c (a) m is the minimum when I

c is maximum. 2 2 2 3 2 2 1 1 1 ( ) 6 6 6 1 3 1 0 6 6 3 I b D b D b b c d I D b b D db c 2 1 2 2 3 3 d D D D d 2 b (b) EI M

is maximum when I is maximum, 1 3

12bd is maximum, or 2 6 b d is maximum. 2 2 6 (D d d is maximum. ) 2 5 7 6D d 8d 0 3 2 d D 2 3 2 1 4 2 b D D D d 3 b

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h h C h0 h0 M

PROBLEM 4.28

A portion of a square bar is removed by milling, so that its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h 0.9h0, express the maximum stress in the

bar in the form m k 0, where 0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k.

SOLUTION 1 2 3 3 0 4 3 3 3 4 0 0 2 3 4 0 0 4 2 1 1 (4) (2) (2 2 )( ) 12 3 1 4 4 4 3 3 3 3 3 4 (4 3 ) 3 I I I h h h h h h h h h h h h h c h Mc Mh M I h h h h h h

For the original square, 0 0

0 2 3 0 0 0 0 , . 3 3 (4 3 ) h h c h M M h h h h 3 3 0 0 2 2 0 0 0 0 0 0 0.950 (4 3 ) (4 (3)(0.9) )(0.9 ) 0.950 h h h h h h h h 0.950 k

(33)

h h C h0 h0 M

PROBLEM 4.29

In Prob. 4.28, determine (a) the value of h for which the maximum stress m is as small as possible, (b) the corresponding value of k.

PROBLEM 4.28 A portion of a square bar is removed by milling, so that

its cross section is as shown. The bar is then bent about its horizontal axis by a couple M. Considering the case where h 0.9h0, express the

maximum stress in the bar in the form m k 0,where 0 is the maximum stress that would have occurred if the original square bar had been bent by the same couple M, and determine the value of k.

SOLUTION 1 2 3 3 0 4 3 3 3 4 0 0 2 3 0 4 2 1 1 (4) (2) (2 2 ) 12 3 1 4 4 4 3 3 3 3 4 3 I I I hh h h h h h h h h h h I c h h h h c I c is maximum at 2 3 0 4 0. 3 d h h h dh 2 0 8 3 0 3h h h 0 8 9 h h 2 3 3 0 0 0 0 3 0 4 8 8 256 729 3 9 9 729 256 I Mc M h h h h c I h

For the original square, 0 3

0 0 0 0 1 3 I h h c h h c 0 0 2 0 0 3 Mc M I h 0 729 1 729 0.949 256 3 768 k 0.949

(34)

PROBLEM 4.30

For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature , (b) the radius of curvature of a transverse cross section, (c) the angle between the sides of the bar that were originally vertical. Use E 29 10 psi6 and v 0.29.

SOLUTION

From Example 4.01, M 30 kip in. I 1.042 in4

(a) 3 6 1 6 1 (30 10 ) 993 10 in. (29 10 )(1.042) M EI 1007 in. (b) 1 v vc v c 6 1 6 1 1 1 (0.29)(993 10 )in. 288 10 in. v 3470 in.

(c) length of arc 0.8 230 10 rad6

radius 3470

b

(35)

z x y C A M

PROBLEM 4.31

A W200 31.3 rolled-steel beam is subjected to a couple M of moment 45 kN m. Knowing that E 200 GPa and v 0.29, determine (a) the radius of curvature , (b) the radius of curvature of a transverse cross section.

SOLUTION

For W 200 31.3 rolled steel section,

6 4 6 4 31.3 10 mm 31.3 10 m I (a) 3 3 1 9 6 1 45 10 7.1885 10 m (200 10 )(31.3 10 ) M EI 139.1 m (b) 1 v1 (0.29)(7.1885 10 ) 2.0847 10 m3 3 1 480 m

(36)

2 ! 2! 2 ! 2! " y y # $c y # %c y "x "x "y

PROBLEM 4.32

It was assumed in Sec. 4.1B that the normal stresses y in a member in pure bending are negligible. For an initially straight elastic member of rectangular cross section, (a) derive an approximate expression for y as a function of y, (b) show that ( y)max ( /2 )(c x)max and, thus, that y can be neglected in all practical situations. (Hint: Consider the free-body diagram of the portion of beam located below the surface of ordinate y and assume that the distribution of the stress x is still linear.)

SOLUTION

Denote the width of the beam by b and the length by L. L

Using the free body diagram above, with cos 1 2 0 : 2 sin 0 2 2 1 sin 2 y y y c x y y y y c x c x c x F bL b dy dy dy dy L L But, x ( x)max y c (a) 2 max max ( ) ( ) 2 y y x x y c c y y dy c c 2 2 max ( ) ( ) 2 x y y c c The maximum value y occurs at y 0.

(b) 2 max max max ( ) ( ) ( ) 2 2 x x y c c c

(37)

30 mm 6 mm 6 mm 30 mm Aluminum Brass

PROBLEM 4.33

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.

1.0 in aluminum

/ 105/70 1.5 in brass

b a n

n E E

For the transformed section,

1 3 2 1 1 1 1 1 1 3 3 3 4 12 1.5 (30)(6) (1.5)(30)(6)(18) 88.29 10 mm 12 n I b h n A d 3 3 3 4 2 2 2 2 3 4 3 1 3 4 1 2 3 1.0 (30)(30) 67.5 10 mm 12 12 88.29 10 mm 244.08 10 mm n I b h I I I I I I 9 4 244.08 10 m nMy I M I ny Aluminum: n 1.0, y 15 mm 0.015 m, 100 10 Pa6 6 9 3 (100 10 )(244.08 10 ) 1.627 10 N m (1.0)(0.015) M Brass: n 1.5, y 21 mm 0.021 m, 160 10 Pa6 6 9 3 (160 10 )(244.08 10 ) 1.240 10 N m (1.5)(0.021) M

(38)

32 mm 32 mm 8 mm 8 mm 8 mm 8 mm Aluminum Brass

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 Eb/ 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 | | nMy M I 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) 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) n y M

(39)

30 mm 6 mm 6 mm 30 mm Aluminum Brass

PROBLEM 4.35

For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis.

PROBLEM 4.35. Bar of Prob. 4.33.

SOLUTION

Use aluminum as reference material.

1.0 in aluminum

/ 105/70 1.5 in brass

b a n

n E E

For transformed section,

3 1 1 1 1 3 3 4 12 1.5 (6)(30) 20.25 10 mm 12 n I b h 3 2 2 2 2 3 3 4 12 1.0 (30)(30) 67.5 10 mm 12 n I b h 3 4 3 1 20.25 10 mm I I 3 4 9 4 1 2 3 108 10 mm 108 10 m I I I I nMy M I I ny Aluminum: n 1.0, y 15 mm 0.015 m, 100 10 Pa6 6 9 (100 10 )(108 10 ) 720 N m (1.0)(0.015) M Brass: n 1.5, y 15 mm 0.015 m, 160 10 Pa6 6 9 (160 10 )(108 10 ) 768 N m (1.5)(0.015) M

Choose the smaller value. M 720 N m

Aluminum Brass

Modulus of elasticity 70 GPa 105 GPa

(40)

32 mm 32 mm 8 mm 8 mm 8 mm 8 mm Aluminum Brass

PROBLEM 4.36

For the composite bar indicated, determine the largest permissible bending moment when the bar is bent about a vertical axis.

PROBLEM 4.36 Bar of Prob. 4.34.

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 Eb/ a 105/70 1.5 Values of n are shown on the sketch. For the transformed section,

3 3 3 3 3 4 1 1 1 1 1 3 3 3 4 2 2 2 2 3 4 3 2 3 4 9 4 1 2 3 1.5 (32)(48 32 ) 311.296 10 mm 12 12 1.0 (8)(32) 21.8453 10 mm 12 12 21.8453 10 mm 354.99 10 mm 354.99 10 m n I h B b n I h b I I I I I I | | nMy M I I ny Aluminum: 6 6 9 3 1.0, | | 16 mm 0.016 m, 100 10 Pa (100 10 )(354.99 10 ) 2.2187 10 N m (1.0)(0.016) n y M Brass: 6 6 9 3 1.5 | | 24 mm 0.024 m 160 10 Pa (160 10 )(354.99 10 ) 1.57773 10 N m (1.5)(0.024) n y M Choose the smaller value.

3

1.57773 10 N m

(41)

10 in. 6 in. in. 1 2 5 3 in. 1 2 5 3

PROBLEM 4.37

Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis.

Wood Steel

Modulus of elasticity: 2 10 psi 6 29 10 psi 6

Allowable stress: 2000 psi 22 ksi

SOLUTION

Use wood as the reference material. 1.0 in wood

/ 29/2 14.5 in steel

s w

n n E E

For the transformed section,

3 2 1 1 1 1 1 1 1 3 2 4 2 3 4 2 2 2 2 4 3 1 4 1 2 3 12 14.5 1 1 (5) (14.5)(5) (5.25) 999.36 in 12 2 2 1.0 (6)(10) 500 in 12 12 999.36 in 2498.7 in n I b h n A d n I b h I I I I I I nMy I M I ny Wood: 3 1.0, 5 in., 2000 psi (2000)(2499) 999.5 10 lb in. (1.0)(5) n y M Steel: 3 3 3

14.5, 5.5 in., 22 ksi 22 10 psi

(22 10 )(2499)

689.3 10 lb in. (14.5)(5.5)

n y

M

(42)

10 in. 3 in. in. 3 in. 1 2

PROBLEM 4.38

Wooden beams and steel plates are securely bolted together to form the composite member shown. Using the data given below, determine the largest permissible bending moment when the member is bent about a horizontal axis.

Wood Steel

Modulus of elasticity: 2 10 psi 6 29 10 psi 6

Allowable stress: 2000 psi 22 ksi

SOLUTION

Use wood as the reference material. 1.0 in wood

/ 29/2 14.5 in steel

s w

n n E E

For the transformed section,

3 3 4 1 1 1 1 3 3 4 2 2 2 2 4 3 1 4 1 2 3 1.0 (3)(10) 250 in 12 12 14.5 1 (10) 604.17 in 12 12 2 250 in 1104.2 in n I b h n I b h I I I I I I nMy I M I ny Wood: 3 1.0, 5 in., 2000 psi (2000)(1104.2) 441.7 10 lb in. (1.0)(5) n y M Steel: 3 3 3

14.5, 5 in., 22 ksi 22 10 psi

(22 10 )(1104.2)

335.1 10 lb in. (14.5)(5)

n y

M

(43)

PROBLEM 4.39

A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa) are bonded together to form the composite beam shown. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 35 N m, determine the maximum stress in (a) the aluminum strip, (b) the copper strip.

SOLUTION

Use aluminum as the reference material.

1.0 in aluminum / 105/75 1.4 in copper c a n n E E Transformed section: 2 , mm A nA, mm2 0, mm y 3 0, mm nAy 144 144 9 1296 144 201.6 3 604.8 Σ 345.6 1900.8 0 1900.8 5.50 mm 345.6 Y

The neutral axis lies 5.50 mm above the bottom.

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 9 4 1 2 1.0 (24)(6) (1.0)(24)(6)(3.5) 2196 mm 12 12 1.4 (24)(6) (1.4)(24)(6)(2.5) 1864.8 mm 12 12 4060.8 mm 4.0608 10 m n I b h n A d n I b h n A d I I I (a) Aluminum: 6 9 1.0, 12 5.5 6.5 mm 0.0065 m (1.0)(35)(0.0065) 56.0 10 Pa 56.0 MPa 4.0608 10 n y nMy I 56.0 MPa (b) Copper: 6 9 1.4, 5.5 mm 0.0055 m (1.4)(35)( 0.0055) 66.4 10 Pa 66.4 MPa 4.0608 10 n y nMy I 66.4 MPa

(44)

PROBLEM 4.40

A copper strip (Ec = 105 GPa) and an aluminum strip (Ea = 75 GPa) are bonded together to form the composite beam shown. Knowing that the beam is bent about a horizontal axis by a couple of moment M = 35 N m, determine the maximum stress in (a) the aluminum strip, (b) the copper strip.

SOLUTION

Use aluminum as the reference material.

1.0 in aluminum / 105/75 1.4 in copper c a n n E E Transformed section: 2 , mm A nA, mm2 0, mm Ay 3 0, mm nAy 216 216 7.5 1620 72 100.8 1.5 151.8 Σ 316.8 1771.2 0 1771.2 5.5909 mm 316.8 Y

The neutral axis lies 5.5909 mm above the bottom.

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 9 4 1 2 1.0 (24)(9) (1.0)(24)(9)(1.9091) 2245.2 mm 12 12 1.4 (24)(3) (1.4)(24)(3)(4.0909) 1762.5 mm 12 12 4839 mm 4.008 10 m n I b h n A d n I b h n A d I I I (a) Aluminum: 6 9 1.0, 12 5.5909 6.4091 mm 0.0064091 (1.0)(35)(0.0064091) 56.0 10 Pa 4.008 10 n y nMy I 56.0 MPa 56.0 MPa (b) Copper: 6 9 1.4, 5.5909 mm 0.0055909 m (1.4)(35)( 0.0055909) 68.4 10 Pa 4.008 10 n y nMy I 68.4 MPa

(45)

in. 5 312 6 in. 12 in. M

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 and for 6

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

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

(a) Wood: n 1, y 12 3.758 8.242 in.

(1)(450)(8.242) 1.979 ksi 1873.77 w w 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 s 16.48 ksi 2 , in A nA, in2 0 y nAy0, in3 72 72 6 432 2.5 40.278 0.25 10.069 112.278 421.931

(46)

6 in. 12 in. C8 & 11.5 M

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 and for 6

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

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 Y 0

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

(a) Wood: n 1, y 12 0.220 3.787 8.433 in.

(1)(450)(8.433)

2.02 ksi 1874.69

w w 2.02 ksi

(b) Steel: n 16.1111, y 3.787 in.

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

(47)

24 mm 6 mm 6 mm Aluminum Copper

PROBLEM 4.43

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 35 N m.

Beam of Prob. 4.39.

SOLUTION

See solution to Prob. 4.39 for the calculation of I.

1 9 9 1 35 0.1149 m (75 10 )(4.0608 10 ) a M E I 8.70 m

(48)

24 mm 9 mm 3 mm Aluminum Copper

PROBLEM 4.44

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 35 N m.

Beam of Prob. 4.40.

SOLUTION

See solution to Prob. 4.40 for the calculation of I.

1 9 9 1 35 0.1164 m (75 10 )(4.008 10 ) a M E I 8.59 m

(49)

in. 5 312 6 in. 12 in. M

PROBLEM 4.45

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip in.

Beam of Prob. 4.41.

SOLUTION

See solution to Prob. 4.41 for calculation of I.

4 6 3 3 6 1 6 1873.77 in 1.8 10 psi 450 kip in 450 10 lb in. 1 450 10 133.421 10 in. (1.8 10 )(1873.77) w I E M M EI 7495 in. 625 ft

(50)

6 in.

12 in.

C8 & 11.5

M

PROBLEM 4.46

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip in.

Beam of Prob. 4.42.

SOLUTION

See solution to Prob. 4.42 for calculation of I.

4 6 3 3 6 1 6 1874.69 in 1.8 10 psi

450 kip in. 450 10 lb in.

1 450 10 133.355 10 in. (1.8 10 )(1874.69) w I E M M EI 7499 in. 625 ft

(51)

5.5 in. 6 in. 5.5 in. 4 in. 5.5 in. 5.5 in. -in. diameter 5 8

PROBLEM 4.47

A concrete slab is reinforced by 5

8-in.-diameter steel rods

placed on 5.5-in. centers as shown. The modulus of elasticity is 3 106 psi for the concrete and 29 106 psi for the steel. Using an allowable stress of 1400 psi for the concrete and 20 ksi for the steel, determine the largest bending moment in a portion of slab 1 ft wide.

SOLUTION 6 6 29 10 9.6667 3 10 s c E n E

Consider a section 5.5 in. wide.

2 2 5 0.3068 in2 4 4 8 s s A d 2 2.9657 in s nA

Locate the natural axis.

5.5 (4 )(2.9657) 0 2 x x x 2 2.75x 2.9657x 11.8628 0 Solve for x. 1.6066 in. 4 2.3934 in. x x 3 2 3 2 4 1 (5.5) (2.9657)(4 ) 3 1 (5.5)(1.6066) (2.9657)(2.3934) 24.591 in 3 I x x nMy I M I ny

Concrete: n 1, y 1.6066 in., 1400 psi

3 (24.591)(1400)

21.429 10 lb in. (1.0)(1.6066)

(52)

PROBLEM 4.47 (Continued)

Steel: n 9.6667, y 2.3934 in., 20 ksi=20 10 psi3

3 3 (24.591)(20 10 ) 21.258 10 lb in. (9.6667)(2.3934) M

Choose the smaller value as the allowable moment for a 5.5 in. width. 3

21.258 10 lb in. M

For a 1 ft = 12 in. width,

3 3 12 (21.258 10 ) 46.38 10 lb in. 5.5 M 46.38 kip in. M 3.87 kip ft

(53)

5.5 in. 6 in. 5.5 in. 4 in. 5.5 in. 5.5 in. -in. diameter 5 8

PROBLEM 4.48

Solve Prob. 4.47, assuming that the spacing of the 5

8-in.-diameter

steel rods is increased to 7.5 in.

PROBLEM 4.47 A concrete slab is reinforced by 5

8-in.-diameter

steel rods placed on 5.5-in. centers as shown. The modulus of elasticity is 3 × 106 psi for the concrete and 29 106 psi for the steel. Using an allowable stress of 1400 psi for the concrete and 20 ksi for the steel, determine the largest bending moment in a portion of slab 1 ft wide.

SOLUTION 6 6 29 10 psi 9.667 3 10 psi s c E n E Number of rails per foot:

12 in. 1.6 7.5 in. Area of 5

-in.-8 diameter bars per foot:

2

2

5

1.6 0.4909 in

4 8 As

Transformed section, all concrete. First moment of area:

12 4.745(4 ) 0 2 x x x 1.4266 in. x 2 9.667(0.4909) 4.745 in s nA 3 2 4 1 (12)(1.4266) 4.745(4 1.4266) 43.037 in 3 NA I

For concrete: all 1400 psi c x 1.4266 in.

4 43.037 in (1400 psi) 1.4266 in. I M c M 42.24 kip.in.

For steel: all 20 ksi c 4 x 4 1.4266 2.5734 in.

4 steel 20 ksi 43.042 in 9.667 2.5734 in. I M n c M 34.60 kip in.

We choose the smaller M. M 34.60 kip in.

(54)

300 mm 540 mm 60 mm 25-mm diameter

PROBLEM 4.49

The reinforced concrete beam shown is subjected to a positive bending moment of 175 kN m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 2 2 3 2 3 2 200 GPa 8.0 25 GPa 4 (4) (25) 1.9635 10 mm 4 4 15.708 10 mm s c s s E n E A d nA Locate the neutral axis.

3 2 3 6 300 (15.708 10 )(480 ) 0 2 150 15.708 10 7.5398 10 0 x x x x x Solve for x. 3 3 2 6 15.708 10 (15.708 10 ) (4)(150)(7.5398 10 ) (2)(150) 177.87 mm, 480 302.13 mm x x x 3 3 2 3 3 2 9 4 3 4 1 (300) (15.708 10 )(480 ) 3 1 (300)(177.87) (15.708 10 )(302.13) 3 1.9966 10 mm 1.9966 10 m I x x nMy I (a) Steel: y 302.45 mm 0.30245 m 3 6 3 (8.0)(175 10 )( 0.30245) 212 10 Pa 1.9966 10 212 MPa (b) Concrete: y 177.87 mm 0.17787 m 3 6 3 (1.0)(175 10 )(0.17787) 15.59 10 Pa 1.9966 10 15.59 MPa

(55)

300 mm 540 mm 60 mm 25-mm diameter

PROBLEM 4.50

Solve Prob. 4.49, assuming that the 300-mm width is increased to 350 mm.

PROBLEM 4.49 The reinforced concrete beam shown is subjected to a

positive bending moment of 175 kN m. Knowing that the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

SOLUTION 2 2 3 2 3 2 200 GPa 8.0 25 GPa 4 (4) (25) 4 4 1.9635 10 mm 15.708 10 mm s c s s E n E A d nA Locate the neutral axis.

3 2 3 6 350 (15.708 10 )(480 ) 0 2 175 15.708 10 7.5398 10 0 x x x x x Solve for x. 3 3 2 6 15.708 10 (15.708 10 ) (4)(175)(7.5398 10 ) (2)(175) 167.48 mm, 480 312.52 mm x x x 3 3 2 3 3 2 9 4 3 4 1 (350) (15.708 10 )(480 ) 3 1 (350)(167.48) (15.708 10 )(312.52) 3 2.0823 10 mm 2.0823 10 m I x x nMy I (a) Steel: y 312.52 mm 0.31252 m 3 6 3 (8.0)(175 10 )( 0.31252) 210 10 Pa 2.0823 10 210 MPa (b) Concrete: y 167.48 mm 0.16748 m 3 6 3 (1.0)(175 10 )(0.16748) 14.08 10 Pa 2.0823 10 14.08 MPa

(56)

12 in. 2.5 in. 20 in. 4 in. 24 in. 1-in. diameter

PROBLEM 4.51

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 for the 6 concrete and 29 10 psi for the steel, determine (a) the stress in the 6 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 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 where M 100 kip ft 1200 kip in.

(a) Steel: n 8.0 y 12.350 in.

(8.0)(1200)( 12.350) 4920

s s 24.1 ksi

(57)

8 in.

2 in.

16 in. 7-in. diameter

8

PROBLEM 4.52

A concrete beam is reinforced by three steel rods placed as shown. The modulus of elasticity is 3 10 psi for the concrete and 6 29 10 psi for the 6 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 d 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 nA x nMy I M I ny

Concrete: n 1.0, y 5.6326 in., 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., 20 10 psi3

3

(20 10 )(1697.45)

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

M

(58)

b

d

PROBLEM 4.53

The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal, respectively, to the allowable stresses s and c. Show that to achieve a balanced design the distance x from the top of the beam to the neutral axis must be

1 s c c s d x E E

where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d is the distance from the top of the beam to the reinforcing steel.

SOLUTION ( ) ( ) 1 1 1 1 s c s c s c s c s c c s s c nM d x Mx I I n d x d n n x x E d x n E d x E E

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b

d

PROBLEM 4.54

For the concrete beam shown, the modulus of elasticity is 25 GPa for the concrete and 200 GPa for the steel. Knowing that b = 200 mm and d = 450 mm, and using an allowable stress of 12.5 MPa for the concrete and 140 MPa for the steel, determine (a) the required area As of the steel reinforcement if the beam is to be balanced, (b) the largest allowable bending moment. (See Prob. 4.53 for definition of a balanced beam.)

SOLUTION 9 9 6 6 200 10 8.0 25 10 ( ) ( ) 1 1 140 10 1 1 2.40 8.0 12.5 10 0.41667 (0.41667)(450) 187.5 mm s c s c s c s c E n E nM d x Mx I I n d x d n n x x d x n x d

Locate neutral axis. ( )

2 s x bx nA d x (a) 2 (200)(187.5)2 1674 mm2 2 ( ) (2)(8.0)(262.5) s bx A n d x As 1674 mm2 3 2 3 2 9 4 3 4 1 1 ( ) (200)(187.5) (8.0)(1674)(262.5) 3 3 1.3623 10 mm 1.3623 10 m s I bx nA d x nMy I M I ny (b) Concrete: n 1.0 y 187.5 mm 0.1875 m 12.5 10 Pa6 3 6 3 (1.3623 10 )(12.5 10 ) 90.8 10 N m (1.0)(0.1875) M Steel: n 8.0 y 262.5 mm 0.2625 m 140 10 Pa6 3 6 3 (1.3623 10 )(140 10 ) 90.8 10 N m (8.0)(0.2625) M

Note that both values are the same for balanced design.

90.8 kN m

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Aluminum Brass Steel Brass Aluminum 1.5 in. 0.5 in. 0.5 in. 0.5 in. 0.5 in. 0.5 in.

PROBLEM 4.55

Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to form the composite beam shown. The modulus of elasticity is 30 × 106 psi for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 12 kip in., 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.

6 6 6 6 30 10 3.0 in steel 10 10 15 10 1.5 in brass 10 10 1.0 in aluminum s a b a E n E E n E n For the transformed section,

3 2 3 2 1 1 1 1 1 1 1 4 1 (1.5)(0.5) (0.75)(1.0) 12 12 0.7656 in n I b h n A d 3 2 3 2 4 2 2 2 2 2 2 2 3 3 4 3 3 3 3 4 4 4 2 5 1 5 4 1 1.5 (1.5)(0.5) (1.5)(0.75)(0.5) 0.3047 in 12 12 3.0 (1.5)(0.5) 0.0469 in 12 12 0.3047 in 0.7656 in 2.1875 in i n I b h n A d n I b h I I I I I I

(a) Aluminum: (1.0)(12)(1.25) 6.86 ksi

2.1875 nMy I Brass: (1.5)(12)(0.75) 6.17 ksi 2.1875 nMy I Steel: (3.0)(12)(0.25) 4.11 ksi 2.1875 nMy I (b) 3 6 1 6 1 12 10 548.57 10 in. (10 10 )(2.1875) a M E I

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Steel Aluminum Brass Aluminum Steel 1.5 in. 0.5 in. 0.5 in. 0.5 in. 0.5 in. 0.5 in.

PROBLEM 4.56

Five metal strips, each a 0.5 × 1.5-in. cross section, are bonded together to form the composite beam shown. The modulus of elasticity is 30 × 106 psi for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum. Knowing that the beam is bent about a horizontal axis by a couple of moment 12 kip · in., 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.

6 6 6 6 30 10 3.0 in steel 10 10 15 10 1.5 in brass 10 10 1.0 in aluminum s a b a E n E E n E n For the transformed section,

3 2 1 1 1 1 1 1 1 3 2 4 12 3.0 (1.5)(0.5) (3.0)(0.75)(1.0) 12 2.2969 in n I b h n A d 3 2 3 2 4 2 2 2 2 2 2 2 3 3 4 3 3 3 3 4 4 4 2 5 1 5 4 1 1.0 (1.5)(0.5) (1.0)(0.75)(0.5) 0.2031 in 12 12 1.5 (1.5)(0.5) 0.0234 in 12 12 0.2031 in 2.2969 in 5.0234 in i n I b h n A d n I b h I I I I I I

(a) Steel: (3.0)(12)(1.25) 8.96 ksi

5.0234 nMy I Aluminum: (1.0)(12)(0.75) 1.792 ksi 5.0234 nMy I Brass: (1.5)(12)(0.25) 0.896 ksi 5.0234 nMy I (b) 3 6 1 6 1 12 10 238.89 10 in. (10 10 )(5.0234) a M E I 4186 in. 349 ft

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Brass

Aluminum 0.8 in.

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 for the aluminum. Knowing that the 6 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 r , 4 4 0 base 4 (4)(0.8) 0.33953 in. 0.160850 in 3 3 8 r y I r 2 2 4 base 0 0.160850 (1.00531)(0.33953) 0.044953 in I I Ay

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

0.17067

0.06791 in. 2.51327

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 22.5 ksi (b) Aluminium: n 1.0, y 0.8 0.06791 0.86791 in. A, in2 nA, in2 0, in. y 3 0, in nAy 1.00531 1.50796 0.33953 0.51200 1.00531 1.00531 0.33953 0.34133 2.51327 0.17067

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Steel 38 mm 10 mm z y 3 mm 6 mm Aluminum

PROBLEM 4.58

A steel pipe and an aluminum pipe are securely bonded together to form the composite beam shown. The modulus of elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that the composite beam is bent by a couple of moment 500 N m, determine the maximum stress (a) in the aluminum, (b) in the steel.

SOLUTION

Use aluminum as the reference material. 1.0 in aluminum

/ 200 / 70 2.857 in steel

s a

n

n E E For the transformed section,

Steel: 4 4 (2.857) (164 10 ) 124.62 10 mm4 3 4 4 4 s s o i I n r r Aluminium: 4 4 (1.0) (194 16 ) 50.88 10 mm4 3 4 4 4 a a o i I n r r 3 4 9 4 175.50 10 mm 175.5 10 m s a I I I (a) Aluminum: c 19 mm 0.019 m 6 9 (1.0)(500)(0.019) 54.1 10 Pa 175.5 10 a a n Mc I 54.1 MPa a (b) Steel: c 16 mm 0.016 m 6 9 (2.857)(500)(0.016) 130.2 10 Pa 175.5 10 s s n Mc I 130.2 MPa s

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50 mm 100 mm " % ' % Et #12Ec Ec M

PROBLEM 4.59

The rectangular beam shown is made of a plastic for which the value of the modulus of elasticity in tension is one-half of its value in compression. For a bending moment M 600 N m, determine the maximum (a) tensile stress, (b) compressive stress.

SOLUTION

1 2

n on the tension side of neutral axis 1

n on the compression side Locate neutral axis.

1 2 2 2 ( ) 0 2 2 1 1 ( ) 0 2 4 x h x n bx n b h x bx b h x 2 2 3 3 6 4 1 1 3 3 6 4 2 2 6 4 6 1 2 1 1 ( ) ( ) 2 2 1 0.41421 41.421 mm 2 1 58.579 mm 1 1 (1) (50)(41.421) 1.1844 10 mm 3 3 1 1 1 ( ) (50)(58.579) 1.6751 10 mm 3 2 3 2.8595 10 mm 2.8595 10 m x h x x h x x h h h x I n bx I n b h x I I I 4

(a) Tensile stress: 1, 58.579 mm 0.058579 m

2 n y 6 6 (0.5)(600)( 0.058579) 6.15 10 Pa 2.8595 10 nMy I t 6.15 MPa (b) Compressive stress: n 1, y 41.421 mm 0.041421 m 6 6 (1.0)(600)(0.041421) 8.69 10 Pa 2.8595 10 nMy I c 8.69 MPa

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PROBLEM 4.60*

A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in compression. Show that the curvature of the beam in pure bending is

1 r M E I where 2 4 ( ) t c r t c E E E E E SOLUTION

Use Et as the reference modulus. Then Ec nEt.

Locate neutral axis.

2 2 ( ) 0 2 2 ( ) 0 ( ) 1 1 x h x nbx b h x nx h x nx h x h nh x h x n n 3 3 3 3 3 trans 3/ 2 3 3 3 3 3 2 1 1 ( ) 3 3 3 1 1 1 1 1 1 3 1 3 1 3 1 n h n I bx b h x bh n n n n n n n bh bh bh n n n 3 trans trans 3 trans 3 2 2 2 1 1 where 12 12 3 1 4 / 4 / 1 t r r t t r t t c t t c c t c t M M I bh E I E I E I E I E I n E E bh I bh n E E E E E E E E E

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r 80 mm 40 mm 8 mm M

PROBLEM 4.61

Knowing that M 250 N m, determine the maximum stress in the beam shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.

SOLUTION 3 3 3 4 9 4 1 1 (8)(40) 42.667 10 mm 42.667 10 m 12 12 20 mm 0.020 m 80 mm 2.00 40 mm I bh c D d (a) 4 mm 0.10 40 mm r d From Fig. 4.27, K 1.87 6 max 9 (1.87)(250)(0.020) 219 10 Pa 42.667 10 Mc K I max 219 MPa (b) 8 mm 0.20 40 mm r d From Fig. 4.27, K 1.50 6 max 9 (1.50)(250)(0.020) 176.0 10 Pa 42.667 10 Mc K I max 176.0 MPa

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r 80 mm 40 mm 8 mm M

PROBLEM 4.62

Knowing that the allowable stress for the beam shown is 90 MPa, determine the allowable bending moment M when the radius r of the fillets is (a) 8 mm, (b) 12 mm. SOLUTION 3 3 3 4 9 4 1 1 (8)(40) 42.667 10 mm 42.667 10 m 12 12 20 mm 0.020 m 80 mm 2.00 40 mm I bh c D d (a) 8 mm 0.2 40 mm r d From Fig. 4.27, K 1.50 max Mc K I 6 9 max (90 10 )(42.667 10 ) (1.50)(0.020) I M Kc 128.0 N m M (b) 12 mm 0.3 40 mm r d From Fig. 4.27, K 1.35 6 9 (90 10 )(42.667 10 ) (1.35)(0.020) M 142.0 N m M

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r M 4.5 in. in. 3 4

PROBLEM 4.63

Semicircular grooves of radius r must be milled as shown in the sides of a steel member. Using an allowable stress of 8 ksi, determine the largest bending moment that can be applied to the member when (a) r = 3

8 in., (b) r =3 4 in. SOLUTION (a) 2 4.5 (2) 3 3.75 in. 8 4.5 0.375 1.20 0.10 3.75 3.75 d D r D r d d From Fig. 4.28, K 2.07 3 3 4 1 1 3 1 (3.75) 3.296 in 1.875 in. 12 12 4 2 (8)(3.296) 6.79 kip in. (2.07)(1.875) I bh c Mc I K M I Kc 6.79 kip in. (b) 2 4.5 (2) 3 3.0 4.5 1.5 0.75 0.25 4 3.0 3.0 D r d D r d d From Fig. 4.28, 1.61 1 3 1 3 (3.0)3 1.6875 in4 12 12 4 K I bh 1 (8)(1.6875)

1.5 in. 5.59 kip in.

2 (1.61)(1.5)

I

c d M

Figure

Updating...

References

Updating...

Outline : Bar of Prob. 4.34