Mechanics of Materials 7th Edition Beer Johnson Chapter 4

CHAPTER 4

20

40

PROBLEM 4.1

20 20

A

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

M 5 15 kN · m

80

20

B

Dimensions in mm

SOLUTION For rectangle:

I

1 3 bh 12

Outside rectangle:

I1

1 (80)(120)3 12

I1

11.52 106 mm 4

I2

1 (40)(80)3 12

I2

1.70667 106 mm 4

Cutout:

Section: (a)

yA

40 mm

I

I1

I2

11.52 10

6

m4

1.70667 10

9.81333 10

0.040 m

6

A

6

m4

m4 My A I

(15 103 )(0.040) 9.81333 10 6

61.6 106 Pa

61.6 MPa

A

(b)

yB

60 mm

0.060 m

B

MyB I

(15 103 )( 0.060) 9.81333 10 6

91.7 106 Pa B

91.7 MPa

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 447

2 in. 2 in. 2 in.

PROBLEM 4.2

M ! 25 kip · in.

A


2 in.

B

1.5 in. 2 in.

SOLUTION

I

For rectangle:

1 3 bh 12

For cross sectional area: I

I1

I2

I3

1 (2)(1.5)3 12

1 (2)(5.5)3 12

1 (2)(1.5)3 12

(a)

yA

2.75 in.

A

My A I

(25)(2.75) 28.854

(b)

yB

0.75 in.

B

MyB I

(25)(0.75) 28.854

28.854 in 4

A

B

2.38 ksi 0.650 ksi


200 mm

PROBLEM 4.3

12 mm

y

C M

x

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.

220 mm

8 mm 12 mm

SOLUTION Moment of inertia about x-axis:

I1

1 (200)(12)3 12

(200)(12)(104)2

25.9872 106 mm 4 I2

1 (8)(196)3 12

I3

I1

25.9872 106 mm 4

I

I1

I2

M Mx

I3

Mc with c I I with c

5.0197 106 mm 4

56.944 106 mm 4 1 (220) 2

110 mm

56.944 10

6

m4

0.110 m

155 106 Pa

(56.944 10 6 )(155 106 ) 0.110

80.2 103 N m

Mx

80.2 kN m


200 mm

PROBLEM 4.4

12 mm

y

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

x

220 mm

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.

8 mm 12 mm

SOLUTION Moment of inertia about y axis: I1 I2

1 (12)(200)3 8 106 mm 4 12 1 (196)(8)3 8.3627 103 mm 4 12

I3

I1

8 106 mm 4

I

I1

I2

My My

I3

Mc with c I I with c

16.0084 106 mm 4 1 (200) 2

100 mm

16.0084 10

6

m4

0.100 m

155 106 Pa

(16.0084 10 6 )(155 106 ) 0.100

24.8 103 N m

My

24.8 kN m


PROBLEM 4.5

0.1 in. 0.5 in. M1

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

(a) 0.2 in. 0.5 in. M2

(b)

SOLUTION (a)

I c

ro4 ri4 4 0.6 in. Mc : I

M

4

(0.64

I c

0.54 )

52.7 10

3

in 4

(16)(52.7 10 3 ) 0.6

M (b)

I c

(0.74 4 0.7 in. Mc : I

0.54 )

M

139.49 10

I c

3

1.405 kip in.

in 4

(16)(139.49 10 3 ) 0.7 M

3.19 kip in.


PROBLEM 4.6

r 5 20 mm

A

30 mm B

M = 2.8 kN · m


30 mm

120 mm

SOLUTION I

1 (0.120 m)(0.06 m)3 12 2.1391 10

(a)

A

6

2

1 12

4

(0.02 m) 4

mm 4

(2.8 103 N m)(0.03 m) 2.1391 10 6 mm 4

M yA I

39.3 MPa

A

(b)

B

M yB I

(2.8 103 N m)(0.02 m) 2.1391 10 6 m 4 B

26.2 MPa


PROBLEM 4.7

y

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. z

C

SOLUTION Properties of W4

13 rolled section.

(See Appendix C.) Area

3.83 in 2

Width

4.060 in.

Iy

3.86 in 4

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

For both sections,

Ad 2

Ib

Iy

Iz

2 Ib

c all

M all

width U

F .S . all I c

3.86 (3.83)(2.030)2

19.643 in 4

39.286 in 4 4.060 in. 58 19.333 ksi 3.0 (19.333)(39.286) 4.060

Mc I

M all 187.1 kip in.


PROBLEM 4.8

y

C

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.

z

SOLUTION Properties of W4

13 rolled section.

(See Appendix C.) Area

3.83 in 2

Depth

4.16 in.

Ix

11.3 in 4

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

For both sections,

Ad 2

Ia

Ix

Iz

2Ia

c all

M all

depth U

F .S . all I c

11.3 (3.83)(2.08)2

27.87 in 4

55.74 in 4 4.16 in.

58 19.333 ksi 3.0 (19.333)(55.74) 4.16

Mc I

M all

259 kip in.


PROBLEM 4.9

3 in. 3 in. 3 in.

6 in.

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.

2 in.

A

15 kips

15 kips

B

C

40 in.

60 in.

D

40 in.

SOLUTION A

y0

A y0

18

5

90

18

1

18

36 Y0

108 36

108 3 in.

Neutral axis lies 3 in. above the base. I1 I2 I

ytop M M

1 1 b1h13 A1d12 (3)(6)3 (18)(2) 2 126 in 4 12 12 1 1 3 2 b2 h2 A2 d 2 (9)(2)3 (18)(2) 2 78 in 4 12 12 I1 I 2 126 78 204 in 4

5 in. ybot

Pa 0 Pa (15)(40) 600 kip in. M ytop

top

bot

3 in.

I

(600)(5) 204

M ybot I

(600)( 3) 204

top

14.71 ksi (compression)

bot

8.82 ksi (tension)


PROBLEM 4.10

8 in. 1 in.


6 in.

1 in.

1 in. 4 in.

A

25 kips

25 kips

B

C

20 in.

60 in.

D

20 in.

SOLUTION

A

y0

A y0

8

7.5

60

6

4

24

4

0.5

18

86

86 18

4.778 in.

(8)(2.772) 2

59.94 in 4

(6)(0.778) 2

21.63 in 4

(4)(4.278) 2

73.54 in 4

Yo

2

Neutral axis lies 4.778 in. above the base. I1 I2 I3 I ytop

1 1 (8)(1)3 b1h13 A1d12 12 12 1 1 (1)(6)3 b2 h23 A2 d 22 12 12 1 1 (4)(1)3 b3 h33 A3 d32 12 12 I1 I 2 I 3 59.94 21.63 4.778 in. 3.222 in. ybot

M M

Pa 0 Pa (25)(20) 500 kip in. Mytop

top

bot

73.57 155.16 in 4

I Mybot I

(500)(3.222) 155.16 (500)( 4.778) 155.16

top

10.38 ksi (compression) bot

15.40 ksi (tension)


10 mm

PROBLEM 4.11

10 mm 10 kN

10 kN

B

50 mm

C

A

D


10 mm 50 mm

250 mm

150 mm

150 mm

SOLUTION

A, mm 2

y0 , mm

A y0 , mm3

600

30

18 103

600

30

18 103

300

5

1.5 103 37.5 103

1500 Y0

37.5 103 1500

25mm

Neutral axis lies 25 mm above the base. I1 I3 I

ytop

35 mm 0.035 m

a 150 mm M

bot

ybot

25 mm

0.025 m

0.150 m P 10 103 N

(10 103 )(0.150) 1.5 103 N m

Pa

Mytop top

1 (10)(60)3 (600)(5)2 195 103 mm 4 I 2 I1 195 mm 4 12 1 (30)(10)3 (300)(20) 2 122.5 103 mm 4 12 I1 I 2 I 3 512.5 103 mm 4 512.5 10 9 m 4

I M ybot I

(1.5 103 )(0.035) 512.5 10 9

102.4 106 Pa

(1.5 103 )( 0.025) 512.5 10 9

73.2 106 Pa

top

102.4 MPa (compression)

bot

73.2 MPa (tension)


PROBLEM 4.12

216 mm y

z

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.

36 mm

54 mm C

108 mm

72 mm

SOLUTION The stress distribution over the entire cross section is given by the bending stress formula: My I

x

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 dF

My dA I

x dA

The total force on the shaded area is then F

My dA I

dF

M I

y dA

M * * y A I

where y * is the centroidal coordinate of the shaded portion and A* is its area. d1

54 18 36 mm

d2

54 36 54 36 mm


PROBLEM 4.12 (Continued) Moment of inertia of entire cross section:

I1 I2 I

1 1 (216)(36)3 (216)(36)(36)2 10.9175 106 mm4 b1h13 A1d12 12 12 1 1 b2 h23 A2 d 22 (72)(108)3 (72)(108)(36)2 17.6360 106 mm 4 12 12 I1 I 2 28.5535 106 mm 4 28.5535 10 6 m 4

For the shaded area, A*

(72)(90)

y*

45 mm

A* y * F

6480 mm 2

291.6 103 mm3 MA* y * I

291.6 10 6 m

(6 103 )(291.6 10 6 ) 28.5535 10 6

61.3 103 N

F

61.3 kN


20 mm 12 mm

PROBLEM 4.13

20 mm y

12 mm

24 mm

z

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.

20 mm C

20 mm 24 mm

SOLUTION Dimensions in mm: Iz

1 1 (12 12)(88)3 (40)(40)3 12 12 1.3629 106 0.213 106 1.5763 106 mm 4

1.5763 10 6 m 4

For use in Prob. 4.14, Iy

1 1 (88)(64)3 (24 24)(40)3 12 12 1.9224 106 0.256 106 1.6664 106 mm 4

Bending about horizontal axis.

Mz

1.6664 10 6 m 4

4 kN m

A

M zc Iz

(4 kN m)(0.044 m) 1.5763 10 6 m 4

111.654 MPa

B

M zc Iz

(4 kN m)(0.020 m) 1.5763 10 6 m 4

50.752 MPa


PROBLEM 4.13 (Continued)

A (44)(12) 528 mm 2

Portion (1):

avg

1 2

Force1

A avg A

avg

Force2 Total force on shaded area

1 2

B avg A

6

m2

1 (111.654) 55.83 MPa 2 (55.83 MPa)(528 10

A (20)(20)

Portion (2):

528 10

400 mm 2

1 (50.752) 2

6

400 10

m2 ) 6

29.477 kN

m2

25.376 MPa

(25.376 MPa)(400 10

6

m 2 ) 10.150 kN

29.477 10.150 39.6 kN


20 mm 12 mm

PROBLEM 4.14

20 mm y

12 mm

Solve Prob. 4.13, assuming that the beam is bent about a vertical axis by a couple of moment 4 kN m. 24 mm

z

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.

20 mm C

20 mm 24 mm

SOLUTION My

Bending about vertical axis.

Iy

See Prob. 4.13 for sketch and

4 kN m

1.6664 10

Mc Iy

(4 kN m)(0.032 m) 1.6664 10 6 m 4

76.81 MPa

E

Mc Iy

(4 kN m)(0.020 m) 1.6664 10 6 m 4

48.01 MPa

avg

Force1

1 ( 2

E)

D

avg A

avg

Force2 Total force on shaded area

1 2

E avg A

528 10

6

1 (76.81 48.01) 2

(62.41 MPa)(528 10

A (20)(20)

Portion (2):

m4

D

A (44)(12) 528 mm 2

Portion (1):

6

400 mm 2

1 (48.01) 2

6

400 10

62.41 MPa m 2 ) 32.952 kN 6

m2

6

m 2 ) 9.602 kN

24.005 MPa

(24.005 MPa)(400 10

32.952 9.602

m2

42.6 kN


0.5 in.

0.5 in.

0.5 in.

1.5 in.

1.5 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.

1.5 in. 0.5 in.

M

SOLUTION

A

y0

Ay0

2.25

1.25

2.8125

2.25

0.25

0.5625

4.50 Y

3.375 3.375 4.50

0.75 in.

The neutral axis lies 0.75 in. above bottom.

ytop I1 I2 I

2.0 0.75 1.25 in.,

ybot

0.75 in.

1 1 b1h13 A1d12 (1.5)(1.5)3 (2.25)(0.5)2 0.984375 in 4 12 12 1 1 2 2 b2 h2 A2 d 2 (4.5)(0.5)3 (2.25)(0.5)2 0.609375 in 4 12 12 I1 I 2 1.59375 in 4 My I

M

I y

Top: (compression)

M

(16)(1.59375) 1.25

20.4 kip in.

Bottom: (tension)

M

(12)(1.59375) 0.75

25.5 kip in.

M all

Choose the smaller as Mall.

20.4 kip in.


PROBLEM 4.16

40 mm

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.

15 mm d ! 30 mm

20 mm M

SOLUTION

Σ Y0

A, mm2

y0 , mm

A y0 , mm3

600

22.5

13.5 103

300

7.5

2.25 103 15.75 103

900 15.5 103 900

17.5 mm

ytop ybot I1 I2 I | |

The neutral axis lies 17.5 mm above the bottom.

30 17.5 12.5 mm 17.5 mm

0.0175 m

1 b1h13 A1d12 12 1 b2 h23 A2 d 22 12 I1 I 2 61.875 My I

M

0.0125 m

1 (40)(15)3 (600)(5)2 26.25 103 mm 4 12 1 (20)(15)3 (300)(10)2 35.625 103 mm 4 12 103 mm 4 61.875 10 9 m 4 I y

Top: (tension side)

M

(24 106 )(61.875 10 9 ) 0.0125

Bottom: (compression)

M

(30 106 )(61.875 10 9 ) 106.1 N m 0.0175

Choose smaller value.

118.8 N m

M

106.1 N m


PROBLEM 4.17

40 mm

Solve Prob. 4.16, assuming that d

15 mm d ! 30 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.

20 mm M

SOLUTION A, mm2

y0 , mm

A y0 , mm3

600

32.5

19.5 103

500

12.5

6.25 103

Σ Y0

25.75 103

1100 25.75 103 1100 ytop ybot I1 I2 I | |

23.41 mm


40 23.41 16.59 mm 23.41 mm

0.01659 m

0.02341 m

1 1 b1h13 A1d12 (40)(15)3 (600)(9.09) 2 60.827 103 mm 4 12 12 1 1 b2 h22 A2 d 22 (20)(25)3 (500)(10.91) 2 85.556 103 mm 4 12 12 I1 I 2 146.383 103 mm 4 146.383 10 9 m 4 My I

M

I y

Top: (tension side)

M

(24 106 )(146.383 10 9 ) 0.01659

212 N m


M

(30 106 )(146.383 10 9 ) 0.02341

187.6 N m

Choose smaller value.

M

187.6 N m


PROBLEM 4.18

2.4 in.

0.75 in.

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.

1.2 in.

M

SOLUTION rectangle

semi-circular cutout A1

2.88 in 2

(2.4)(1.2)

A2

2

(0.75)2

A

2.88

y1

0.6 in.

0.8836 in 2 1.9964 in 2

0.8836

4r (4)(0.75) 0.3183 in. 3 3 Ay (2.88)(0.6) (0.8836)(0.3183) A 1.9964

y2 Y

0.7247 in.

Neutral axis lies 0.7247 in. above the bottom. Moment of inertia about the base: 1 3 bh 3

Ib

8

r4

1 (2.4)(1.2)3 3

(0.75) 4

1.25815 in 4

1.25815 (1.9964)(0.7247) 2

0.2097 in 4

8

Centroidal moment of inertia: I

Ib

AY 2

ytop

1.2 0.7247

ybot

0.7247 in.

| |

My I

M

0.4753 in.,

I y

Top: (tension side)

M

(12)(0.2097) 0.4753

5.29 kip in.


M

(16)(0.2097) 0.7247

4.63 kip in. M

Choose the smaller value.

4.63 kip in.


PROBLEM 4.19

80 mm

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.

54 mm

40 mm M

SOLUTION

Σ

Y

A, mm2

y0 , mm

2160

27

58,320

3

1080

36

38,880

3

A y0 , mm3

3240

d, mm

97,200

97, 200 3240

30 mm

The neutral axis lies 30 mm above the bottom.

ytop

54 30

24 mm

I1 I2 I | |

1 b1h13 A1d12 12 1 b2 h22 A2 d 22 36 I1 I 2 758.16 My I

0.024 m

ybot

30 mm

0.030 m

1 (40)(54)3 (40)(54)(3) 2 544.32 103 mm 4 12 1 1 (40)(54)3 (40)(54)(6)2 213.84 103 mm 4 36 2 103 mm 4 758.16 10 9 m 4 I y

|M |

Top: (tension side)

M

(120 106 )(758.16 10 9 ) 0.024

3.7908 103 N m


M

(150 106 )(758.16 10 9 ) 0.030

3.7908 103 N m

Choose the smaller as Mall.

M all

3.7908 103 N m

M all

3.79 kN m


PROBLEM 4.20

48 mm

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.

48 mm 36 mm

48 mm

36 mm M

SOLUTION A, mm2

y0 , mm

Solid rectangle

4608

48

221,184

Square cutout

–1296

30

–38,880

Σ

Y

3312

182,304 3312

ytop ybot I1 I2 I

| |

55.04 mm

182,304

Neutral axis lies 55.04 mm above bottom.

96 55.04 40.96 mm 55.04 mm

0.04096 m

0.05504 m

1 b1h13 A1d12 12 1 b2 h32 A2 d 22 12 I1 I 2 2.8147

My I

A y0 , mm3

M

1 (48)(96)3 (48)(96)(7.04)2 3.7673 106 mm 4 12 1 (36)(36)3 (36)(36)(25.04) 2 0.9526 106 mm 4 12 106 mm 4 2.8147 10 6 m 4

I y

Top: (tension side)

M

(120 106 )(2.8147 10 6 ) 0.04096


M

(150 106 )(2.8147 106 ) 0.05504

Mall is the smaller value.

M

7.67 103 N m

8.25 103 N m 7.67 103 N m 7.67 kN m


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

D

inside diameter of the drum,

d

diameter of rod,

1 d, 2 radius of curvature of center line of rods when bent. 1 D 2 I

(a)

(b)

Ec max

M

EI

4

c4

c

1 d 2 4

(200 109 )(0.003) 0.622 (200 109 )(63.617 10 0.622

1 (1.25) 2

(0.003) 4

1 (6 10 3 ) 2 63.617 10

12

0.622 m m4

965 106 Pa 12

)

965 MPa

20.5 N m

M

20.5 N m


M'

PROBLEM 4.22

M 8 mm

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.

t

r 900 mm

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 2 Stress:

E

Ec

420 MPa and E

For c (a)

143.24 mm

0.14324 m

or

c

E

200 GPa,

(0.14324)(420 106 ) 200 109

t

Maximum thickness:

0.3008 10

2c

3

m

0.6016 10

3

m t

0.602 mm

Moment of inertia for a rectangular section. I (b)

bt 3 12

Bending moment:

M

(8 10 3 )(0.6016 10 3 )3 12 M

145.16 10

15

m4

EI

(200 109 )(145.16 10 0.14324

15

)

0.203 N m

M

0.203 N m


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 29 106 psi .

5 ft

SOLUTION Radius of cross section:

r

Moment of inertia:

I

(a)

(b)

D

5 ft

c

r

60 in.

4

r4

1 (0.30) 2 4

(0.15) 4

1 D 2

30 in.

106 )(0.15) 30

145.0

0.15 in. 397.61 10

6

in 4

0.15 in.

Ec

(29

max

M

1 d 2

EI

103 psi

(29 106 )(397.61 10 6 ) 30

max

M

145.0 ksi 384 lb in.


PROBLEM 4.24 12 mm y 60 N · m

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.

20 mm z

SOLUTION (a)

Bending about z-axis.

I c

1

(b)

1 3 1 bh (12)(20)3 8 103 mm 4 12 12 20 10 mm 0.010 m 2

Mc I

(60)(0.010) 8 10 9

M EI

60 (200 109 )(8 10 9 )

8 10 9 m 4

75.0 106 Pa 37.5 10 3 m

75.0 MPa 1

26.7 m

Bending about y-axis.

I c

1

1 3 1 bh (20)(12)3 2.88 103 mm 4 12 12 12 6 mm 0.006 m 2 Mc (60)(0.006) 125.0 106 Pa 9 I 2.88 10 M EI

60 (200 10 )(2.88 10 9 ) 9

2.88 10 9 m 4

104.17 10 3 m

125.0 MPa 1

9.60 m


10 mm M

(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.

80 mm

C

PROBLEM 4.25

10 mm 5 mm

80 mm

5 mm

SOLUTION (a)

I I1 4 I 2 , 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. I1 I2 I

1 3 bh 12

1 (80)(80)3 12

3.4133 106 mm 4

1 3 1 (5)(10)3 (5)(10)(45)2 101.667 103 mm 4 bh Ad 2 12 12 I1 4 I 2 3.82 106 mm 4 3.82 10 6 mm 4 , c 50 mm 0.050 m Mc I

I c

M

(120 106 )(3.82 10 6 ) 0.050

9.168 103 N m 9.17 kN m

(b)

Without the ears:

I M

I1 I c

3.4133 10

6

m2 ,

c

40 mm

0.040 m

(120 106 )(3.4133 10 6 ) 10.24 103 N m 0.040

10.24 kN m


0.1 in.

PROBLEM 4.26 0.2 in.

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.

1.5 in. M 0.75 in.

SOLUTION

I x of hollow pipe:

Ix

I x of fins:

Ix

(1.5 in.) 4

4

2

(0.75 in.) 4

1 (0.1)(0.2)3 12

3.7276 in 4

(0.1 0.2)(1.6) 2

2

1 (0.2)(0.1)3 12

0.1026 in 4 (a)

(b)

Pipe as designed, with fins:

Ix

3.8302 in 4 ,

all

20 ksi,

c 1.7 in. M

all

Ix c

(20 ksi)

3.8302 in 4 1.7 in. M

45.1 kip in.

M

49.7 kip in.

Pipe with no fins: all

M

20 ksi, all

Ix c

Ix

3.7276 in 4 ,

(20 ksi)

c 1.5 in.

3.7276 in 4 1.5 in.


PROBLEM 4.27 M

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.

M'

d b

SOLUTION Let D be the diameter of the log. D2

m

d2

1 bd 3 12

I

(a)

b2

c

is the minimum when I c d db

D2

1 d 2

I c

b2

D2

1 2 bd 6

I is maximum. c

1 1 2 b( D 2 b 2 ) Db 6 6 I 1 2 3 2 0 D b 6 6 c

d

(b)

d2

1 2 D 3

1 3 b 6 1 D 3

b

2 D 3

d b

2

d b

3

EI M

is maximum when I is maximum,

(D2

d 2 )d 6 is maximum.

6 D 2d 5

b

1 bd 3 is maximum, or b2d 6 is maximum. 12

D2

8d 7

0

3 2 D 4

d

1 D 2

3 D 2


PROBLEM 4.28 h0 M

h C

h0

h

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 I

4 I1

2I2

1 h h3 12 1 4 4 h h 0 h3 3 3 h Mc Mh 4 h h3 I 3 0

(4)

c

For the original square,

h

h0 , c

0

3M (4h0 3h0 )h02

1 (2h0 2h)(h3 ) 3 4 4 h h3 h 0 h3 h 4 3 3

(2)

h

4

3M 3h) h 2

h0 . 3M h03

h03 0

(4h 0

(4h0 0.950

h03

3h)h 2

(4h0

(3)(0.9)h0 )(0.9 h02 )

0.950 k

0

0.950


PROBLEM 4.29

h0 M

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.

h

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.

C h

h0

SOLUTION I

4 I1

2I 2

1 1 hh3 (2) (2h0 2h) h3 12 3 1 4 4 4 3 4 h h 0 h3 h h 0 h3 h 4 3 3 3 3 I 4 2 3 h h0 h h c 3

(4)

c

d 4 I is maximum at h0 h2 c dh 3

h3

0. 8 h 0 h 3h 2 3

I c For the original square,

4 8 h0 h0 3 9 h

h0

2

8 h0 9

3

256 3 h0 729

c

h0

I0 c0

0

Mc0 I0

3M h02

0

0

729 1 256 3

h

Mc I

8 h0 9

729M 256h03

1 3 h0 3

729 768

0.949

k

0.949


PROBLEM 4.30 For the bar and loading of Concept Application 4.1, determine (a) the radius of curvature , (b) the radius of of a transverse cross section, (c) the angle between the sides of the bar that were originally curvature vertical. Use E 29 106 psi and v 0.29.

SOLUTION M

From Example 4.01, (a) (b)

1

1

1

(c)

M EI

vc

v v

(30 103 ) (29 106 )(1.042)

1

v

993 10

6

in.

1.042 in 4

I

1

1007 in.

c

(0.29)(993 10 6 )in.

length of arc radius

30 kip in.

b

0.8 3470

1

288 10

6

in.

230 10 6 rad

1

3470 in.

0.01320


PROBLEM 4.31

y

A z

M

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 of a transverse cross radius of curvature , (b) the radius of curvature section.

C x

SOLUTION For W 200 31.3 rolled steel section, I

31.3 106 mm 4 31.3 10 6 m 4

(a)

(b)

1 1

M EI v

1

45 103 (200 109 )(31.3 10 6 ) (0.29)(7.1885 10 3 )

7.1885 10 3 m 2.0847 10 3 m

1

1

139.1 m 480 m


PROBLEM 4.32

y ! 2

y # %c

"y

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 (c /2 )( x ) max and, thus, that y can be neglected in ( y )max 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.)

! 2

"y

"x

"x

! 2

! 2

y # $c

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

cos

Using the free body diagram above, with Fy

0:

2 sin L 2

y

But, (a)

( y

x ) max

y

c

c

y dy

The maximum value

y bL

x

(

x ) max

(

x ) max

y2 2

c y

occurs at y

y

2

c

y c

2

x

1 x b dy

sin

0

2 y

dy

L

c

x

1

dy

y c

x

dy

y c y

( y

c

x ) max

2 c

( y2

c2 )

0.

(b)

(

y ) max

(

x ) max c

2 c

2

(

x ) max c

2


PROBLEM 4.33

Brass 6 mm

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 30 mm

6 mm

Aluminum

Brass

70 GPa

105 GPa

100 MPa

160 MPa

Modulus of elasticity

30 mm

Allowable stress

SOLUTION Use aluminum as the reference material.

n

1.0 in aluminum

n

Eb /Ea

105/70

1.5 in brass

For the transformed section, I1

n1 b1h13 n1 A1d 12 12 1.5 (30)(6)3 (1.5)(30)(6)(18)3 12

I2

n2 b2h23 12

I3

I1

88.29 103 mm 4

I

I1

I2

1.0 (30)(30)3 12 I3

Aluminum:

n 1.0, M

Brass:

Choose the smaller value

67.5 103 mm 4

244.08 103 mm 4

244.08 10 nMy I

M y 15 mm

9

y

21 mm

m4

I ny 0.015 m,

(100 106 )(244.08 10 9 ) (1.0)(0.015)

n 1.5,

88.29 103 mm 4

0.021 m,

M

(160 106 )(244.08 10 9 ) (1.5)(0.021)

M

1.240 103 N m

100 106 Pa 1.627 103 N m

160 106 Pa 1.240 103 N m

1.240 kN m


8 mm

PROBLEM 4.34

8 mm

32 mm

8 mm 32 mm

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.

8 mm Brass

Aluminum

Brass

70 GPa

105 GPa

100 MPa

160 MPa


Aluminum

Allowable stress

SOLUTION Use aluminum as the reference material. For aluminum, n 1.0 For brass, n

Eb /Ea

105/70 1.5

Values of n are shown on the sketch. For the transformed section, I1 I2 I3 I | | Aluminum:

I1

I2

nMy I

I 3 141.995 103 mm 4 M

100 106 Pa

0.016 m,

(100 106 )(141.995 10 9 ) (1.0)(0.016)

n 1.5, | y | 16 mm M

141.995 10 9 m 4

I ny

n 1.0, | y | 16 mm M

Brass:

n1 1.5 b1h13 (8)(32)3 32.768 103 mm 4 12 12 n2 1.0 b2 H 23 h23 (32)(323 163 ) 76.459 103 mm 4 12 12 I1 32.768 103 mm 4

887.47 N m 160 106 Pa

0.016 m,

(160 106 )(141.995 10 9 ) (1.5)(0.016)

946.63 N m


M

887 N m


PROBLEM 4.35 Brass 6 mm

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

30 mm

PROBLEM 4.35. Bar of Prob. 4.33.

Aluminum


6 mm

Allowable stress

30 mm

Aluminum

Brass

70 GPa

105 GPa

100 MPa

160 MPa

SOLUTION Use aluminum as reference material.

n

1.0 in aluminum

n

Eb /Ea

105/70

1.5 in brass

For transformed section, I1

I2

n1 b1h13 12 1.5 (6)(30)3 12 n2 b2h23 12 1.0 (30)(30)3 12

I1

20.25 103 mm 4

I

I1

I2

n 1.0, M

Brass:

67.5 103 mm 4

I3

nMy I Aluminum:

20.25 103 mm 4

108

103 mm 4

108

10

9

m4

I ny

M y 15 mm

0.015 m,

(100 106 )(108 10 9 ) (1.0)(0.015)

n 1.5, M

I3

y 15 mm

720 N m

0.015 m,

(160 106 )(108 10 9 ) (1.5)(0.015)

100 106 Pa

160 106 Pa

768 N m

M


720 N m


8 mm

PROBLEM 4.36

8 mm

32 mm

8 mm

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.

32 mm 8 mm Brass

Modulus of elasticity Allowable stress

Aluminum

Aluminum 70 GPa 100 MPa

Brass 105 GPa 160 MPa

SOLUTION Use aluminum as the reference material. For aluminum, n 1.0 For brass, n

Eb /Ea

105/70 1.5

Values of n are shown on the sketch. For the transformed section, I1 I2 I3 I | | Aluminum:

I1

I2

nMy I

I3 M

354.99 103 mm 4

354.99 10 9 m 4

I ny

n 1.0, | y | 16 mm M

Brass:

n1 1.5 h1 B13 b13 (32)(483 323 ) 311.296 103 mm 4 12 12 n2 1.0 h2b23 (8)(32)3 21.8453 103 mm 4 12 12 I 2 21.8453 103 mm 4

0.016 m,

(100 106 )(354.99 10 9 ) (1.0)(0.016)

n 1.5 | y | 24 mm

0.024 m

M

(160 106 )(354.99 10 9 ) (1.5)(0.024)

M

1.57773 103 N m

100 106 Pa

2.2187 103 N m 160 106 Pa 1.57773 103 N m

Choose the smaller value. M

1.578 kN m


PROBLEM 4.37

1 5 3 2 in.

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. 10 in.

Modulus of elasticity:

1 5 3 2 in.

6 in.

Allowable stress:

Wood

Steel

2 106 psi

29 106 psi

2000 psi

22 ksi

SOLUTION Use wood as the reference material.

n 1.0 in wood n

E s /E w

29/2 14.5 in steel For the transformed section, I1

n1 b1h13 12

n1 A1d12 3

I2 I3 I

nMy I Wood:

n M

Steel:

n M

M

I ny

1.0,

y

14.5 1 1 (5) (14.5)(5) (5.25)2 12 2 2 n2 1.0 b2 h22 (6)(10)3 500 in 4 12 12 I1 999.36 in 4 I1

y

2498.7 in 4

2000 psi 999.5 103 lb in.

5.5 in.,

(22 103 )(2499) (14.5)(5.5)


I3

5 in.,

(2000)(2499) (1.0)(5) 14.5,

I2

999.36 in 4

22 ksi

22 103 psi

689.3 103 lb in.

M

689 103 lb in.

M

689 kip in.


PROBLEM 4.38

10 in.

3 in. 1 2

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.

3 in.

Modulus of elasticity:

in.

Allowable stress:

Wood

Steel

2 106 psi

29 106 psi

2000 psi

22 ksi


n 1.0 in wood n

E s /E w

29/2 14.5 in steel For the transformed section,

I1 I2 I3 I

nMy I Wood:

n M

Steel:

n M

M

I ny

1.0,

y

n1 1.0 (3)(10)3 250 in 4 b1h13 12 12 n2 14.5 1 b2 h23 (10)3 604.17 in 4 12 12 2 I1

y

I3

1104.2 in 4

2000 psi 441.7 103 lb in.

5 in.,

(22 103 )(1104.2) (14.5)(5)


I2

5 in.,

(2000)(1104.2) (1.0)(5) 14.5,

250 in 4

I1

22 ksi

22 103 psi

335.1 103 lb in.

M

335 103 lb in.

M

335 kip in.


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.


n 1.0 in aluminum n

Ec /Ea

105/75 1.4 in copper

Transformed section:

A, mm2

nA, mm 2

y0 , mm

nAy0 , mm3

144

144

9

1296

144

201.6

3

Σ

Y0

345.6 1900.8 345.6

604.8 1900.8

5.50 mm


I

n1 1.0 b1h13 n1 A1d12 (24)(6)3 (1.0)(24)(6)(3.5) 2 2196 mm 4 12 12 n2 1.4 (24)(6)3 (1.4)(24)(6)(2.5)2 1864.8 mm 4 b2 h23 n2 A2 d 22 12 12 I1 I 2 4060.8 mm 4 4.0608 10 9 m 4

n

1.0,

I1 I2

(a)

Aluminum:

y

nMy I

12

5.5

6.5 mm

0.0065 m

(1.0)(35)(0.0065) 4.0608 10 9

56.0 106 Pa

56.0 MPa

56.0 MPa (b)

Copper:

n

1.4,

y

nMy I

5.5 mm

0.0055 m

(1.4)(35)( 0.0055) 4.0608 10 9

66.4 106 Pa

66.4 MPa

66.4 MPa


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.


n 1.0 in aluminum n

Ec /Ea

105/75 1.4 in copper


A, mm2

nA, mm 2

Ay0 , mm

nAy0 , mm3

216

216

7.5

1620

72

100.8

1.5

Σ Y0

151.8

316.8 1771.2 316.8

1771.2

5.5909 mm


I

n1 1.0 b1h13 n1 A1d12 (24)(9)3 (1.0)(24)(9)(1.9091)2 2245.2 mm 4 12 12 n2 1.4 b2 h23 n2 A2 d 22 (24)(3)3 (1.4)(24)(3)(4.0909) 2 1762.5 mm 4 12 12 I1 I 2 4839 mm 4 4.008 10 9 m 4

n

1.0,

I1 I2

(a)

Aluminum:

y

nMy I

12

5.5909

6.4091 mm

(1.0)(35)(0.0064091) 4.008 10 9

0.0064091 56.0 10

6

Pa

56.0 MPa 56.0 MPa (b)

Copper:

n

1.4,

y

nMy I

5.5909 mm

0.0055909 m

(1.4)(35)( 0.0055909) 4.008 10 9

68.4 106 Pa

68.4 MPa

68.4 MPa


PROBLEM 4.41

6 in.

M

12 in.

53

1 2

The 6 12-in. timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for steel, 29 106 psi. 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.

in.


For wood,

n 1

For steel,

n

Es /Ew wood


Yo

29 / 1.8 16.1111

421.931 112.278 3.758 in.

steel A, in 2

nA, in 2

72

72

6

40.278

0.25

2.5

y0

112.278

nA y0 , in 3 432 10.069 421.931

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

I1 I2 I M (a)

n1 1 b1h13 n1 A1d12 (6)(12)3 (72)(6 3.758) 2 1225.91 in 4 12 12 n2 16.1111 b2 h23 n2 A2 d 22 (5)(0.5)3 (40.278)(3.578 0.25)2 12 12 I1 I 2 1873.77 in 4 nMy I

450 kip in.

Wood:

n 1,

y 12 3.758 8.242 in. w

(b)

Steel:

(1)(450)(8.242) 1873.77

n 16.1111, s

647.87 in 4

y

1.979 ksi

3.758 0.5

1.979 ksi

w

4.258 in.

(16.1111)(450)( 4.258) 16.48 ksi 1873.77

s

16.48 ksi


PROBLEM 4.42

6 in.

M

12 in.

The 6 12-in. timber beam has been strengthened by bolting to it the steel reinforcement shown. The modulus of elasticity for wood is 1.8 106 psi and for steel, 29 106 psi. 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.

C8 & 11.5


For wood,

n 1

For steel,

n

Es Ew

29 106 1.8 106

16.1111

For C8 11.5 channel section, A 3.38 in 2 , tw

0.220 in., x

0.571 in., I y

1.32 in 4

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:

A, in2 3.38

Part 1

72

2

nA, in2 54.456

y , in. 0.571

nAy , in 3 31.091

d, in. 3.216

72

6.22

447.84

2.433

478.93

126.456

Y0

478.93 in 3 126.456 in 2

d

3.787 in.

y0

Y0

The neutral axis lies 3.787 in. above the bottom of the section. I1 I2 I M (a)

n1 I1

n1 A1d12

(16.1111)(1.32) (54.456)(3.216) 2

n2 1 b2 h23 n2 A2 d 22 (6)(12)3 12 12 I1 I 2 1874.69 in 4 450 kip in

(72)(2.433) 2

1290.20 in 4

n My I y 12 0.220 3.787 8.433 in.

Wood:

n 1,

Steel:

(1)(450)(8.433) 2.02 ksi 1874.69 n 16.1111, y 3.787 in. w

(b)

s

584.49 in 4

(16.1111)(450)( 3.787) 14.65 ksi 1874.67

w

s

2.02 ksi

14.65 ksi


PROBLEM 4.43 6 mm

Aluminum

Copper

6 mm

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 35 N m. Beam of Prob. 4.39.

24 mm

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

1

M Ea I

35 (75 10 )(4.0608 10 9 ) 9

0.1149 m

1

8.70 m


PROBLEM 4.44 Aluminum

9 mm

Copper

3 mm

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 35 N m. Beam of Prob. 4.40.

24 mm

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

1

M Ea I

35 (75 10 )(4.008 10 9 ) 9

0.1164 m

1

8.59 m


6 in.

PROBLEM 4.45 For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip in.

M

12 in.

Beam of Prob. 4.41. 53

1 2

in.

SOLUTION See solution to Prob. 4.41 for calculation of I. I

1873.77 in 4

M

450 kip in

1

M EI

Ew

1.8 106 psi

450 103 lb in.

450 103 (1.8 106 )(1873.77)

133.421 10 6 in.

1

7495 in. 625 ft


6 in.

PROBLEM 4.46

M

12 in.

For the composite beam indicated, determine the radius of curvature caused by the couple of moment 450 kip in. Beam of Prob. 4.42.

C8 & 11.5

SOLUTION See solution to Prob. 4.42 for calculation of I. I M 1

1874.69 in 4

Ew

1.8 106 psi

450 kip in. 450 103 lb in. M EI

450 103 (1.8 106 )(1874.69)

133.355 10 6 in.

1

7499 in. 625 ft


PROBLEM 4.47 5 8

-in. diameter

4 in.

A concrete slab is reinforced by 85 -in.-diameter steel rods placed on 5.5-in. centers as shown. The modulus of elasticity is 106 psi for the steel. 3 106 psi for the concrete and 29 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.

5.5 in. 5.5 in.

5.5 in. 6 in.

5.5 in.

SOLUTION n

29 106 3 106

Es Ec

9.6667

Consider a section 5.5 in. wide. As

4

d s2

5 4 8

2

0.3068 in 2

2.9657 in 2

nAs

Locate the natural axis. 5.5 x

x 2

(4

2.75 x 2

x )(2.9657)

0

2.9657 x 11.8628 0

Solve for x. x 1.6066 in. I

M

2.3934 in.

1 (5.5) x3 (2.9657)(4 x)2 3 1 (5.5)(1.6066)3 (2.9657)(2.3934) 2 3

nMy I Concrete: n

4 x

1,

y

M

I ny

1.6066 in.,

(24.591)(1400) (1.0)(1.6066)

24.591 in 4

1400 psi 21.429

103 lb in.



Steel:

n M

9.6667,

y

2.3934 in.,

(24.591)(20 103 ) (9.6667)(2.3934)

21.258

20 ksi=20 103 psi 103 lb in.

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

M

21.258

103 lb in.

For a 1 ft = 12 in. width, M

12 (21.258 5.5

M

46.38 kip in.

103 )

46.38

103 lb in. 3.87 kip ft


PROBLEM 4.48 5 8

-in. diameter

Solve Prob. 4.47, assuming that the spacing of the 85 -in.-diameter steel rods is increased to 7.5 in. 4 in.

PROBLEM 4.47 A concrete slab is reinforced by 85 -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.

5.5 in. 5.5 in.

5.5 in. 6 in.

5.5 in.

SOLUTION 29 106 psi 3 106 psi

Es Ec

n

9.667

Number of rails per foot: 12 in. 1.6 7.5 in. 5 5 Area of -in.-diameter bars per foot: 1.6 8 4 8

2

0.4909 in 2

As

Transformed section, all concrete. First moment of area:

12 x

x 2

4.745(4 x) 0 x 1.4266 in.

nAs I NA

For concrete:

all

M

For steel:

all

M

We choose the smaller M.

M

1 (12)(1.4266)3 3

1400 psi c I c

steel

n

I c

4.745(4 1.4266)2

4.745 in 2 43.037 in 4

x 1.4266 in.

(1400 psi)

20 ksi c

9.667(0.4909)

4 x

43.037 in 4 1.4266 in.

M

42.24 kip.in.

M

34.60 kip in.

4 1.4266 2.5734 in.

20 ksi 43.042 in 4 9.667 2.5734 in.

34.60 kip in. M

Steel controls.

2.88 kip ft


PROBLEM 4.49 540 mm

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.

25-mm diameter 60 mm 300 mm

SOLUTION n As nAs

Es Ec

4

200 GPa 25 GPa 4

d2

(4)

8.0 (25) 2

4

1.9635 103 mm 2

15.708 103 mm 2

Locate the neutral axis.

x (15.708 103 )(480 x) 0 2 15.708 103 x 7.5398 106 0

300 x 150 x 2 Solve for x.

x

15.708 103

x 177.87 mm,

I

(15.708 103 ) 2 (4)(150)(7.5398 106 ) (2)(150) 480 x 302.13 mm

1 (300) x3 (15.708 103 )(480 x)2 3 1 (300)(177.87)3 (15.708 103 )(302.13)2 3 1.9966 109 mm 4 1.9966 10 3 m 4 nMy I

(a)

Steel:

y

302.45 mm

0.30245 m

(8.0)(175 103 )( 0.30245) 1.9966 10 3 (b)

Concrete:

y 177.87 mm

212 106 Pa

212 MPa

0.17787 m

(1.0)(175 103 )(0.17787) 1.9966 10 3

15.59 106 Pa

15.59 MPa


PROBLEM 4.50 540 mm

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

60 mm 300 mm

SOLUTION n As

Es Ec 4

4

200 GPa 25 GPa d2

(4)

4

8.0 (25) 2

1.9635 103 mm 2 nAs

15.708 103 mm 2

Locate the neutral axis. x (15.708 103 )(480 x) 0 2 15.708 103 x 7.5398 106 0

350 x 175 x 2 Solve for x.

(15.708 103 ) 2 (4)(175)(7.5398 106 ) (2)(175) x 167.48 mm, 480 x 312.52 mm x

I

(a)

Steel:

y

15.708 103

1 (350) x3 (15.708 103 )(480 x)2 3 1 (350)(167.48)3 (15.708 103 )(312.52) 2 3 2.0823 109 mm 4 2.0823 10 3 m 4 nMy I 312.52 mm

0.31252 m

(8.0)(175 103 )( 0.31252) 2.0823 10 3 (b)

Concrete:

y 167.48 mm

210 106 Pa

210 MPa

0.16748 m

(1.0)(175 103 )(0.16748) 2.0823 10 3

14.08 106 Pa

14.08 MPa


4 in.

24 in.

PROBLEM 4.51 Knowing that the bending moment in the reinforced concrete beam is 100 kip ft and that the modulus of elasticity is 3.625 106 psi for the concrete and 29 106 psi for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete.

20 in.

1-in. diameter

2.5 in.

12 in.

SOLUTION n

Es Ec

As

(4)

29 106 3.625 106 (1)2

4

8.0

3.1416 in 2

nAs

25.133 in 2


96 x

6x2

192

Solve for x.

(24)(4)( x

2)

339.3

25.133x

d3 I1 I2 I3 I

Steel:

17.5

nA3d32

(25.133)(12.350) 2

I1

I3

n

Concrete:

where M

8.0

n

1.0, c

6x2

4

121.133x

x) 147.3

0 0

1.150 in.

12.350 in. (24)(4)(3.150) 2

1080.6 in 4

6.1 in 4 3833.3 in 4

4920 in 4 100 kip ft y

y

1200 kip in.

12.350 in.

(8.0)(1200)( 12.350) 4920

s

(b)

x

1 (12)(1.150)3 3

I2

or

1 (24)(4)3 12

A1d12

(25.133)(17.5

(4)(6)(147.3)

4

1 b1h13 12 1 3 b2 x 3

nMy I (a)

0

(121.133)2 (2)(6)

121.133

x

x 2

(12 x)

4

1.150

s

24.1 ksi

5.150 in.

(1.0)(1200)(5.150) 4920

c

1.256 ksi


PROBLEM 4.52 7 8

16 in.

-in. diameter

2 in.

A concrete beam is reinforced by three steel rods placed 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 1350 psi for the concrete and 20 ksi for the steel, determine the largest allowable positive bending moment in the beam.

8 in.

SOLUTION

As

8x

Locate the neutral axis:

29 10 6 3 106

Es Ec

n

3

x 2

4

d2

(3)

9.67

4

2

7 8

1.8040 in 2

nAs

17.438 in 2

(17.438)(14 x) 0

4 x 2 17.438 x 244.14 0 Solve for x.

17.438

x

17.4382 (4)(4)(244.14) (2)(4)

5.6326 in.

14 x 8.3674 in. I

1 3 8x 3

nMy I

nAs (14

M

x) 2

1 (8)(5.6326)3 3

(17.438)(8.3674)2

1697.45 in 4

I ny n 1.0,

Concrete:

M Steel:

n

M Choose the smaller value.

5.6326 in.,

y

(1350)(1697.45) (1.0)(5.6326) 9.67,

y

(20 103 )(1697.45) (9.67)(8.3674) M

1350 psi

406.835 103 lb in.

8.3674 in.,

419.72 lb in. 407 kip in.

407 kip in.

20 103 psi

420 kip in. M

33.9 kip ft


PROBLEM 4.53

d

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 d

x 1 b

s Ec c Es

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 s s c

d x x

nM (d x) Mx c I I n(d x) d n n x x 1

1 n

d Ec 1 Es

s

1

c

Ec Es

s c

s c


PROBLEM 4.54 d

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.)

b

SOLUTION n s s c

d x

1

x

0.41667d

bx

Locate neutral axis. As

(a)

x 2

M

1

c

nAs (d

1 140 106 2.40 8.0 12.5 106 (0.41667)(450) 187.5 mm

x) (200)(187.5) 2 1674 mm 2 (2)(8.0)(262.5)

M

y 187.5 mm

0.1875 m

(1.3623 10 3 )(12.5 106 ) (1.0)(0.1875)

n 8.0

Steel:

s

bx 2 2n ( d x )

n 1.0

Concrete:

1 n

1 3 1 (200)(187.5)3 bx nAs (d x)2 3 3 9 4 1.3623 10 mm 1.3623 10 3 m 4 nMy I M I ny

I

(b)

200 109 8.0 25 109 nM (d x) Mx c I I n( d x ) d n n x x Es Ec

y

262.5 mm

As

1674 mm 2

(8.0)(1674)(262.5) 2

12.5 106 Pa

90.8 103 N m

0.2625 m

(1.3623 10 3 )(140 106 ) (8.0)(0.2625)

140 106 Pa

90.8 103 N m

Note that both values are the same for balanced design. M

90.8 kN m


Aluminum

0.5 in.

Brass

0.5 in.

Steel

0.5 in.

Brass

0.5 in.

Aluminum

0.5 in. 1.5 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. n

30 106 10 106

Es Ea

3.0 in steel

15 106 1.5 in brass 10 106 n 1.0 in aluminum n

Eb Ea


I2 I3 I4

n1 b1h13 n1 A1d12 12 0.7656 in 4

1 (1.5)(0.5)3 12

(0.75)(1.0) 2

n2 1.5 (1.5)(0.5)3 (1.5)(0.75)(0.5)2 b2 h23 n2 A2 d 22 12 12 n3 3.0 (1.5)(0.5)3 0.0469 in 4 b3 h33 12 12 I 2 0.3047 in 4 I 5 I1 0.7656 in 4 5

I

Ii

0.3047 in 4

2.1875 in 4

1

(a)

Aluminum:

nMy I

(1.0)(12)(1.25) 2.1875

6.86 ksi

Brass:

nMy I

(1.5)(12)(0.75) 2.1875

6.17 ksi

Steel:

nMy I

(3.0)(12)(0.25) 2.1875

4.11 ksi

(b)

1

M Ea I

12 103 (10 106 )(2.1875)

548.57 10

6

in.

1

1823 in. = 151.9 ft


Steel

0.5 in.

Aluminum

0.5 in.

Brass

0.5 in.

Aluminum

0.5 in.

Steel

0.5 in. 1.5 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. n

30 106 10 106

Es Ea

3.0 in steel

15 106 1.5 in brass 10 106 n 1.0 in aluminum n

Eb Ea


I2 I3 I4

n1 b1h13 n1 A1d12 12 3.0 (1.5)(0.5)3 (3.0)(0.75)(1.0)2 12 2.2969 in 4 n2 1.0 (1.5)(0.5)3 (1.0)(0.75)(0.5)2 b2 h23 n2 A2 d 22 12 12 n3 1.5 (1.5)(0.5)3 0.0234 in 4 b3 h33 12 12 I 2 0.2031 in 4 I 5 I1 2.2969 in 4 5

I

Ii

0.2031 in 4

5.0234 in 4

1

(a)

Steel:

nMy I

(3.0)(12)(1.25) 5.0234

Aluminum:

nMy I

(1.0)(12)(0.75) 1.792 ksi 5.0234

Brass:

nMy I

(1.5)(12)(0.25) 5.0234

(b)

1

M Ea I

8.96 ksi

12 103 (10 106 )(5.0234)

0.896 ksi 238.89 10

6

in.

1

4186 in. 349 ft PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 505

Brass

PROBLEM 4.57

Aluminum

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 15 106 psi for the brass and 10 106 psi 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.

0.8 in.

SOLUTION r

For each semicircle, y0

4r 3

I

I base

(4)(0.8) 3 Ay02

0.8 in.

A

r2

2

0.33953 in.

1.00531 in 2 ,

I base

8

r4

0.160850 in 4

(1.00531)(0.33953)2

0.160850

0.044953 in 4

Use aluminum as the reference material. n 1.0 in aluminum n

15 106 10 106

Eb Ea

1.5 in brass


A, in2

nA, in2

1.00531

1.50796

0.33953

0.51200

1.00531

1.00531

0.33953

0.34133

y0 , in.

nAy0 , in 3

2.51327 d1

0.06791

I1

n1I

n1 Ad12

I2

n2 I

n2 Ad 22

I (a)

0.33953

I1

I2

0.27162 in., d 2

(1.5)(0.044957)

0.33953

0.06791

(1.5)(1.00531)(0.27162)

(1.0)(0.044957)

0.17067 2.51327

0.06791 in.

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

0.17067

2

(1.0)(1.00531)(0.40744)2

0.40744 in. 0.17869 in 4 0.21185 in 4

0.39054 in 4 n

Brass:

1.5,

y

0.8

nMy I (b)

Y0

Aluminium:

n

1.0,

y

0.8 nMy I

0.06791

0.73209 in.

(1.5)(8)(0.73209) 0.39054 0.06791

22.5 ksi

0.86791 in.

(1.0)(8)( 0.86791) 0.39054

17.78 ksi


y Aluminum

PROBLEM 4.58 3 mm

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.

Steel 6 mm z

10 mm

38 mm


n 1.0 in aluminum n

Es / Ea

Steel:

Is

ns

Aluminium:

Ia

na

I

Is

200 / 70 2.857 in steel

For the transformed section,

(a)

Aluminum:

4

ro4

ri4

(2.857)

ro4

ri4

(1.0)

4

Ia

c 19 mm a

na Mc I

4

4

(164 104 ) 124.62 103 mm 4

(194 164 ) 50.88 103 mm 4

175.50 103 mm 4

175.5 10

9

m4

0.019 m (1.0)(500)(0.019) 175.5 10 9

54.1 106 Pa a

(b)

Steel:

c 16 mm s

ns Mc I

54.1 MPa

0.016 m (2.857)(500)(0.016) 175.5 10 9

130.2 106 Pa s

130.2 MPa


%" M

PROBLEM 4.59 Et #

100 mm

1 2

Ec

%'

50 mm

Ec

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 n

1 on the tension side of neutral axis 2

n 1 on the compression side Locate neutral axis. n1bx

x2 x h x

(a)

Tensile stress:

x h x n2 b(h x) 2 2 1 2 1 bx b( h x ) 2 2 4

Compressive stress:

0

1 1 (h x) 2 x (h x) 2 2 1 h 0.41421 h 41.421 mm 2 1 58.579 mm

I1

1 n1 bx3 3

I2

1 n2 b(h x)3 3

I

I1

I2

n

1 , 2

y

nMy I

(b)

0

n 1,

y nMy I

(1)

1 (50)(41.421)3 1.1844 106 mm 4 3 1 2

1 (50)(58.579)3 1.6751 106 mm 4 3

2.8595 106 mm 4 58.579 mm

2.8595 10 6 m 4

0.058579 m

(0.5)(600)( 0.058579) 2.8595 10 6

41.421 mm

6.15 106 Pa

t

6.15 MPa

0.041421 m

(1.0)(600)(0.041421) 2.8595 10 6

8.69 106 Pa

c

8.69 MPa


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

M Er I

where Er

4 Et Ec Ec ) 2

( Et

SOLUTION Use Et as the reference modulus. Then Ec

nEt .

Locate neutral axis. nbx nx 2 x

I trans

x 2

b( h x ) (h x) 2 h

nh

h x

n 1

n 3 bx 3

h x 0 2 0 nx (h x) n 1

1 b ( h x )3 3

M Et I trans

Er I

Et I trans

Er

Et I trans I

where I

12 bh3

Et

1

n 1 1 3

bh3

n n 1

2

bh3

1 3 bh 12

n 3

3

n

n 1

M Er I

4 Et Ec /Et Ec /Et

3

1

n 1n 1 bh3 3 n 13

1 n n3/ 2 bh3 3 n 13

1

h 3

n 1

2

bh3

4 Et Ec 2

Ec

Et

2


PROBLEM 4.61

8 mm

r M

80 mm

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.

40 mm

SOLUTION

c

1 3 bh 12 20 mm

D d

80 mm 40 mm

2.00

r d

4 mm 40 mm

0.10

I

(a)

max

(b)

r d

K

Mc I

8 mm 40 mm max

K

1 (8)(40)3 12 0.020 m

42.667 103 mm 4

0.20

Mc I

K

From Fig. 4.27,

(1.87)(250)(0.020) 42.667 10 9

K

176.0 106 Pa

9

m4

1.87

219 106 Pa

From Fig. 4.27,

(1.50)(250)(0.020) 42.667 10 9

42.667 10

max

219 MPa

1.50

max

176.0 MPa


PROBLEM 4.62

8 mm

r M

80 mm

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.

40 mm

SOLUTION

c

1 3 bh 12 20 mm

D d

80 mm 40 mm

2.00

r d

8 mm 40 mm

0.2

I

(a)

max

(b)

r d

K

12 mm 40 mm

1 (8)(40)3 12 0.020 m

Mc I

0.3

42.667 103 mm 4

K

From Fig. 4.27, M

max I

Kc

9

m4

1.50

(90 106 )(42.667 10 9 ) (1.50)(0.020)

K

From Fig. 4.27, M

42.667 10

M

128.0 N m

M

142.0 N m

1.35

(90 106 )(42.667 10 9 ) (1.35)(0.020)


3 4

r

PROBLEM 4.63

in.

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 = 83 in., (b) r = 34 in.

M

4.5 in.

SOLUTION (a)

d

D

2r

D d

4.5 3.75

4.5 1.20

From Fig. 4.28, I

1 3 bh 12 Mc K I

(2)

3 8

r d

3.75 in. 0.375 3.75

K

0.10

2.07

1 3 (3.75)3 12 4 I M Kc

3.296 in 4

1 2

c

(8)(3.296) (2.07)(1.875)

1.875 in.

6.79 kip in.

6.79 kip in. (b)

d

D

2r

4.5

From Fig. 4.28,

(2)

3 4

3.0

D d

4.5 3.0

K

1.61

I

c

1 d 2

1.5 in.

r d

1.5

1 3 bh 12 M

0.75 3.0

1 3 (3.0)3 12 4 I Kc

0.25 1.6875 in 4

(8)(1.6875) (1.61)(1.5)

5.59 kip in.


3 4

r

PROBLEM 4.64

in.

Semicircular grooves of radius r must be milled as shown in the sides of a steel member. Knowing that M = 4 kip · in., determine the maximum stress in the member when the radius r of the semicircular grooves is (a) r = 83 in., (b) r = 34 in.

M

4.5 in.

SOLUTION (a)

From Fig. 4.28,

d

D

D d

4.5 3.75

K

2.07

2r

4.5

r d

1.20

1 3 bh 12 Mc K I

I

3 8

(2)

3.75 in. 0.375 3.75

0.10

1 3 (3.75)3 3.296 in 4 12 4 (2.07)(4)(1.875) 4.71 ksi 3.296

c

1 2

1.875 in.

4.71 ksi (b)

d

D

2r

4.5

(2)

3 4

3.00 in.

D d

4.5 3.00

1.50

r d

0.75 3.0

c

1 d 2

1.5 in.

0.25

From Fig. 4.28, K

1.61

I

1 3 bh 12

1 3 (3.00)3 12 4

K

Mc I

1.6875 in 4

(1.61)(4)(1.5) 1.6875

5.72 ksi 5.72 ksi


M

PROBLEM 4.65

M

100 mm

100 mm

150 mm

150 mm

18 mm

(a)

18 mm

A couple of moment M = 2 kN · m is to be applied to the end of a steel bar. Determine the maximum stress in the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 10 mm, as shown in Fig. a, (b) if the bar is redesigned by removing the material to the left and right of the dashed lines as shown in Fig. b.

(b)

SOLUTION For both configurations, D

150 mm

d

100 mm

r

10 mm

D d r d

150 100 10 100

Fig. 4.28 gives

Ka

2.21.

For configuration (b),

Fig. 4.27 gives K b

1.50 0.10

For configuration (a),

I c

(a)

KMc I

1.79.

1 3 1 (18)(100)3 1.5 106 mm 4 bh 12 12 1 d 50 mm 0.05 m 2

(2.21)(2 103 )(0.05) 1.5 10 6

147.0 106 Pa

1.5 10

6

m4

147.0 MPa 147.0 MPa

(b)

KMc I

(1.79)(2 103 )(0.05) 1.5 10 6

119.0 106 Pa

119.0 MPa 119.0 MPa


M

100 mm 150 mm

PROBLEM 4.66

M

100 mm 150 mm

18 mm

(a)

18 mm

The allowable stress used in the design of a steel bar is 80 MPa. Determine the largest couple M that can be applied to the bar (a) if the bar is designed with grooves having semicircular portions of radius r 15 mm, as shown in Fig. a, (b) if the bar is redesigned by removing the material to the left and right of the dashed lines as shown in Fig. b.

(b)

SOLUTION For both configurations, D

150 mm

r

15 mm

D d r d

150 100 15 100

d

100 mm

1.50 0.15

For configuration (a), Fig. 4.28 gives K a

1.92.

For configuration (b), Fig. 4.27 gives Kb

1.57.

I c KMc I

(a)

M

I Kc

1 3 1 (18)(100)3 1.5 106 mm 4 bh 12 12 1 d 50 mm 0.050 m 2 (80 106 )(1.5 10 6 ) (1.92) (0.05)

1.5 10

6

m4

1.250 103 N m

1.250 kN m (a)

M

I Kc

(80 106 )(1.5 10 6 ) (1.57)(0.050)

1.530 103 N m

1.530 kN m


M

PROBLEM 4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic with Y 300 MPa and is subjected to a couple M parallel to the x axis. Determine the moment M of the couple for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.

x

z

12 mm

8 mm

SOLUTION (a)

I

1 3 bh 12

1 (12)(8)3 12 c

yY

1 (4) 2

2 mm

512 10 12 m 4 1 h 4 mm 0.004 m 2 (300 106 )(512 10 c 0.004 38.4 N m YI

MY

(b)

512 mm 4

yY c

2 4

M

3 1 yY MY 1 2 3 c

12

)

MY

38.4 N m

M

52.8 N m

0.5 2

3 1 (38.4) 1 (0.5) 2 2 3 52.8 N m


M

PROBLEM 4.68 Solve Prob. 4.67, assuming that the couple M is parallel to the z axis. PROBLEM 4.67 The prismatic bar shown is made of a steel that is assumed to be elastoplastic with Y 300 MPa and is subjected to a couple M parallel to the x axis. Determine the moment M of the couple for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.

x

z

12 mm

8 mm

SOLUTION (a)

I

1 3 bh 12

1 (8)(12)3 12 c

yY

1 (4) 2

2 mm

1.152 10 9 m 4 1 h 6 mm 0.006 m 2 (300 106 )(1.152 10 9 ) c 0.006 57.6 N m YI

MY

(b)

1.152 103 mm 4

yY c

2 6

M

3 1 yY MY 1 2 3 c

MY

57.6 N m

M

83.2 N m

1 3

3 1 1 (57.6) 1 2 3 3

2

2

83.2 N m


PROBLEM 4.69 A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.

SOLUTION I MY

1 (0.6)(0.6)3 12 I

Y

c c

yY c M

Y M

3

in 4

4.16667 10

1.65517 10 4.16667 10

3 MY 1 2

1 YY 3 c

2

3

1 h 2

c

(10.8 10 3 )(48 103 ) 0.3 0.3 72

max

10.8 10

0.3 in.

1728 lb in. Y

Y

E

48 103 29 106

6 ft

72 in.

1.65517 10

3

3 3

0.39724 3 (1728) 1 2

1 (0.39724)2 3

M

2460 lb in.


PROBLEM 4.70 For the solid square rod of Prob. 4.69, determine the moment M for which the radius of curvature is 3 ft. PROBLEM 4.69. A solid square rod of side 0.6 in. is made of a steel that is assumed to be elastoplastic with E = 29 × 106 psi and Y 48 ksi. Knowing that a couple M is applied and maintained about an axis parallel to a side of the cross section, determine the moment M of the couple for which the radius of curvature is 6 ft.

SOLUTION I MY

1 (0.6)(0.6)3 12 I

Y

c

yY c M

3

in 4

Y max

8.3333 10

1.65517 10 3 8.3333 10 3

3 MY 1 2

1 YY 3 c

2

3

1 h 2

c

(10.8 10 3 )(48 103 ) 0.3 0.3 36

c max

10.8 10

0.3 in.

1728 lb in. Y

48 103 29 106

Y

E

3 ft

36 in.

1.65517 10

3

0.19862 3 (1728) 1 2

1 (0.19862)2 3 M

2560 lb in.


y

PROBLEM 4.71 18 mm

The prismatic rod shown is made of a steel that is assumed to be elastoplastic with E 200 GPa and Y 280 MPa . Knowing that couples M and M of moment 525 N · m are applied and maintained about axes parallel to the y axis, determine (a) the thickness of the elastic core, (b) the radius of curvature of the bar.

M

24 mm

M!

x

SOLUTION I c MY M

1 3 1 bh (24)(18)3 11.664 103 mm 4 12 12 1 h 9 mm 0.009 m 2 YI

c

(280 106 )(11.664 10 9 ) 0.009

3 MY 1 2

yY c

3

1 yY 2 3 c2

(2)(525) 362.88

or

0.32632

yY c

yY

11.664 10

3

m4

362.88 N m 3

2

M MY

0.32632c

2.9368 mm

(a) (b)

tcore Y

yY

Y

E

EyY Y

(200 109 )(2.9368 10 3 ) 280 106

2 yY

5.87 mm

2.09 m


y

PROBLEM 4.72 18 mm

Solve Prob. 4.71, assuming that the couples M and M are applied and maintained about axes parallel to the x axis.

M

24 mm

PROBLEM 4.71 The prismatic rod shown is made of a steel that is assumed to be elastoplastic with E 200 GPa and Y 280 MPa . Knowing that couples M and M of moment 525 N · m are applied and maintained about axes parallel to the y axis, determine (a) the thickness of the elastic core, (b) the radius of curvature of the bar.

M!

x

SOLUTION I c MY M

yY c

1 3 1 bh (18)(24)3 20.736 103 mm 4 12 12 1 h 12 mm 0.012 m 2 YI

c

(280 106 )(20.736 10 9 ) 0.012

3 MY 1 2

3

1 yY 2 3 c2

(2)(525) 483.84

or

yY c

0.91097

yY

20.736 10

9

m4

483.84 N m 3

2

M MY

0.91097

c

10.932 mm

(a) (b)

tcore Y

yY

Y

E

EyY Y

(200 109 )(10.932 10 3 ) 280 106

2 yY

21.9 mm

7.81 m


y

z

PROBLEM 4.73

C

90 mm

A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E 200 GPa and Y 240 MPa. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 30 mm thick.

60 mm

SOLUTION (a)

I c MY

1 3 1 (60)(90)3 3.645 106 mm 4 3.645 10 6 m 4 bh 12 12 1 h 45 mm 0.045 m 2 (240 106 )(3.645 10 6 ) YI 19.44 10 N m 0.045 c

MY

R1

Y

19.44 kN m

(240 106 )(0.060)(0.030)

A1

432 103 N y1 15 mm 15 mm R2

1 2

Y

A2

0.030 m

1 (240 106 )(0.060)(0.015) 2

108 103 N y2 (b)

M

2( R1 y1

R2 y2 )

2 (15 mm) 10 mm 3

0.010 m

2[(432 103 )(0.030) (108 103 )(0.010)]

28.08 103 N m

M

28.1 kN m


PROBLEM 4.74

y

A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E 200 GPa and Y 240 MPa. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 30 mm thick.

30 mm 30 mm

C

z

30 mm 15 mm

30 mm

15 mm

SOLUTION (a)

1 3 1 bh (60)(90)3 3.645 106 mm 4 12 12 1 3 1 I cutout bh (30)(30)3 67.5 103 mm 4 12 12 I 3.645 106 67.5 103 3.5775 106 mm 4 I rect

3.5775 10

c MY

45 mm

y1

15 mm

M

0.045 m

(240 106 )(3.5775 10 6 ) 0.045 c 3 19.08 10 N m

Y A1

y2

mm 4

YI

R1

R2

(b)

1 h 2

6

(240 106 )(0.060)(0.030) 15 mm

30 mm

19.08 kN m

M

27.0 kN m

432 103 N

0.030 m

1 1 (240 106 )(0.030)(0.015) Y A2 2 2 2 (15 mm) 10 mm 0.010 m 3 2( R1 y1

MY

54 103 N

R2 y2 )

2[(432 103 )(0.030)

(54 103 )(0.010)]

27.00 103 N m


PROBLEM 4.75

y

3 in.

3 in.

C

z

A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E 29 106 psi and Y 42 ksi. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 3 in. thick.

3 in.

1.5 in.

1.5 in.

3 in.

SOLUTION (a)

I1 I2 I3

1 1 b1h13 A1d12 (3)(3)3 (3)(3)(3) 2 12 12 1 1 b2 h23 (6)(3)3 13.5 in 4 12 12 I1 87.75 in 4 I 3 188.5 in 4

I

I1

c

4.5 in. (42)(188.5) YI c 4.5

MY

I2

87.75 in 4

MY R1

Y

A1

1759 kip in.

(42)(3)(3) 378 kip

y1 1.5 1.5 3.0 in. R2

y2

(b)

M

2( R1 y1

R2 y2 )

1 1 (42)(6)(1.5) Y A2 2 2 189 kip 2 (1.5) 1.0 in. 3

2[(378)(3.0) (189)(1.0)] 2646 kip in.

M

2650 kip in.


PROBLEM 4.76

y

3 in.

3 in.

C

z

A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E 29 106 psi and Y 42 ksi. For bending about the z axis, determine the bending moment at which (a) yield first occurs, (b) the plastic zones at the top and bottom of the bar are 3 in. thick.

3 in.

1.5 in.

3 in.

1.5 in.

SOLUTION (a)

I1 I2 I3

1 1 (6)(3)3 (6)(3)(3)2 b1h13 A1d12 12 12 1 1 3 (3)(3)3 6.75 in 4 b2 h2 12 12 I1 175.5 in 4

I

I1

c

4.5 in.

MY

I2 YI

c

I3

175.5 in 4

357.75 in 4

(42)(357.75) 4.5

3339 kip in.

MY R1

Y

(42)(6)(3)

A1

3340 kip in.

756 kip

y1 1.5 1.5 3 in.

(b)

M

2( R1 y1

R2 y2 )

R2

1 2

y2

2 (1.5) 1.0 in. 3

2[(756)(3) (94.5)(1.0)] 4725 kip in.

Y

A2

1 (42)(3)(1.5) 2 94.5 kip

M 4730 kip in.


PROBLEM 4.77

y

For the beam indicated (of Prob. 4.73), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section. z

90 mm

C

60 mm

SOLUTION From Problem 4.73,

E

200 GPa and

A1

(60)(45)

R

2700 10 6 m 2 Y A1

Y

240 MPa

2700 mm 2

(240 106 )(2700 10 6 ) d

(a)

Mp

(b)

I c MY

Rd

(648 103 )(0.045)

1 3 1 bh (60)(90)3 12 12 45 mm 0.045 m YI

c

648 103 N 45 mm 0.045 m

29.16 103 N m

3.645 106 mm 4

(240 106 )(3.645 10 6 ) 0.045 k

Mp

29.2 kN m

3.645 10 6 m 4

19.44 103 N m Mp MY

29.16 19.44

k 1.500


PROBLEM 4.78

y

For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment M p , (b) the shape factor of the cross section.

30 mm 30 mm

C

z

30 mm 15 mm

15 mm

30 mm

SOLUTION From Problem 4.74, (a)

R1

y1

Y

200 GPa and

Y

240 MPa

A1

(240 106 )(0.060)(0.030) 432 103 N 15 mm 15 mm 30 mm 0.030 m

R2

y2

E

Y

A2

(240 106 )(0.030)(0.015) 108 103 N 1 (15) 7.5 mm 0.0075 m 2

2( R1 y1

Mp

2[( 432 103 )(0.030) (108 103 )(0.0075)]

R2 y2 )

27.54 103 N m (b)

I rect I cutout I

Mp

1 3 1 (60)(90)3 3.645 106 mm 4 bh 12 12 1 3 1 bh (30)(30)3 67.5 103 mm 4 12 12 I rect I cutout 3.645 106 67.5 103 3.5775 103 mm 4 3.5775 10

c MY k

27.5 kN m

1 h 2 YI c Mp MY

45 mm

9

m4

0.045 m

(240 106 )(3.5775 10 9 ) 0.045 27.54 19.08

19.08 103 N m k 1.443


PROBLEM 4.79

y

3 in.

3 in.

C

z


3 in.

1.5 in.

1.5 in.

3 in.


R1

Y

A1

E

29 106 psi and

Y

42 ksi.

(42)(3)(3) 378 kip

y1 1.5 1.5 3.0 in. R2

(b)

Y

A2

(42)(6)(1.5) 378 kip

y2

1 (1.5) 2

0.75 in.

Mp

2( R1 y1

R2 y2 )

I1 I2 I3

1 1 b1h13 A1d12 (3)(3)3 (3)(3)(3) 2 12 12 1 1 b2 h23 (6)(3)3 13.5 in 4 12 12 I1 87.75 in 4

I

I1

c

4.5 in.

MY k

2[( 378)( 3.0) ( 378)(0. 75)]

YI c Mp

MY

I2

Mp

2840 kip in.

87.75 in 4

I 3 188.5 in 4 (42)(188.5) 4.5 2835 1759.3

1759.3 kip in k 1.611


PROBLEM 4.80

y


3 in.

3 in.

C

z

3 in.

1.5 in.

3 in.

1.5 in.


R1

Y

A1

E

(42)(6)(3)

29 106 and

Y

42 ksi

756 kip

y1 1.5 1.5 3.0 in. R2

(b)

Y

A2

(42)(3)(1.5) 189 kip

y2

1 (1.5) 2

0.75 in.

Mp

2( R1 y1

R2 y2 )

I1 I2 I3

1 1 b1h13 A1d12 (6)(3)3 (6)(3)(3)2 12 12 1 1 3 b2 h2 (3)(3)3 6.75 in 4 12 12 I1 175.5 in 4

I

I1

c

4.5 in.

MY k

2[( 756)( 3.0) (189)(0. 75)]

I2 YI

c Mp MY

I3

Mp

4820 kip in.

175.5 in 4

357.75 in 4

(42)(357.75) 4.5 4819.5 3339

3339 kip in.

k 1.443


PROBLEM 4.81 r ! 18 mm

Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa.

SOLUTION

For a semicircle,

A

2

Resultant force on semicircular section:

r 2; R

r

4r 3

YA

Resultant moment on entire cross section: Mp Data:

Y

Mp

2 Rr

4 3

240 MPa

Yr

3

240 106 Pa,

4 (240 106 )(0.018)3 3

r

18 mm

0.018 m

1866 N m Mp

1.866 kN m


PROBLEM 4.82 50 mm

30 mm


10 mm 10 mm

30 mm

10 mm

SOLUTION A (50)(90) (30)(30) 3600 mm 2

Total area:

1 A 1800 mm 2 2 1 A 1800 x 2 50 b

A1

(50)(36) 1800 mm 2 ,

A2

(50)(14)

A3

(20)(30)

A4

36 mm

y1 18 mm,

A1 y1

32.4 103 mm3

700 mm 2 ,

y2

7 mm,

A2 y2

4.9 103 mm3

600 mm 2 ,

y3

29 mm,

A3 y3

17.4 103 mm3

(50)(10) 500 mm 2 ,

y4

49 mm,

A4 y4

24.5 103 mm3

A1 y1 Mp

A2 y2 Y

Ai yi

A3 y3

A4 y4

79.2 103 mm3

79.2 10 6 m3

(240 106 )(79.2 10 6 ) 19.008 103 N m

Mp

19.01 kN m


PROBLEM 4.83 36 mm


30 mm

SOLUTION Total area:

A

1 (30)(36) 540 mm 2 2

1 A 270 mm 2 2

Half area:

By similar triangles,

b y

30 36

b

Since

A1

1 by 2

5 2 y , 12

A1

5 y 6 y2

12 A1 5

12 (270) 25.4558 mm 5 b 21.2132 mm 1 (21.2132)(25.4558) 270 mm 2 270 10 6 m 2 A1 2 A2 (21.2132)(36 25.4558) 223.676 mm 2 223.676 10 6 m 2 y

A3 Ri R1 y1 y2 y3 Mp

A Y

A1 Ai

A2

46.324 mm 2

46.324 10 6 m 2

240 106 Ai

64.8 103 N, R2 53.6822 103 N, R3 11.1178 103 N 1 y 8.4853 mm 8.4853 10 3 m 3 1 (36 25.4558) 5.2721 mm 5.2721 10 3 m 2 2 (36 25.4558) 7.0295 mm 7.0295 10 3 m 3 R1 y1 R2 y2 R3 y3 911 N m Mp

911 N m


0.4 in.

PROBLEM 4.84

1.0 in.

Determine the plastic moment Mp of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 42 ksi.

1.0 in. 0.4 in.

0.4 in.

SOLUTION A (1.0 in.)(0.4 in.) 2(1.4 in.)(0.4 in.) 1.52 in 2 x R1

1 2

(1.52)

0.95 in.

2(0.4)

0.4 in 2

(1.0)(0.4)

y1 1.2 0.95 0.25 in. R2 y2 R3

2(0.4)(1.4 0.95)

0.36 in 2

1 (1.4 0.95) 0.225 in. 2 2(0.4)(0.95) 0.760 in 2

y3

1 (0.95) 2

Mp

( R1 y1

0.475 in.

R2 y 2

R3 y 3 )(

y)

[(0.4)(0.25) (0.36)(0.225) (0.760)(0.475)](42) Mp

22.8 kip in .


5 mm

PROBLEM 4.85

5 mm 80 mm t = 5 mm 120 mm

Determine the plastic moment Mp of the cross section shown when the beam is bent about a horizontal axis. Assume the material to be elastoplastic with a yield strength of 175 MPa.

SOLUTION For M p , the neutral axis divides the area into two equal parts. Total area

(100 100 120)t

320t

1 (320)t 2 a 80 mm 80 b (80 mm) 64 mm 100

Shaded area

6

2at

m2

A1

2at

A2

2(100 a )t

2(0.02 m)(0.005 m)

200 10

A3

(120 mm)t

(0.120 m)(0.005 m)

600 10

2(0.08 m)(0.005 m) 800 10

R1

Y

A1

(175 MPa)(800 10

6

m 2 ) 140 kN

R2

Y

A2

(175 MPa)(200 10

6

m 2 ) 35 kN

R3

Y

A3

(175 MPa)(600 10

6

m 2 ) 105 kN

Mz : M p

6 6

m2 m2

R1 (32 mm) R2 (8 mm) R3 (16 mm) (140 kN)(0.032 m) (35 kN)(0.008 m) (105 kN)(0.016 m) M p = 6.44 k N m


PROBLEM 4.86

4 in. 1 2

in.

1 2

in.

1 2

in.

3 in.

Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 36 ksi.

2 in.

SOLUTION Total area:

A (4)

1 2

1 1 (3) (2) 2 2

4.5 in 2

1 A 2.25 in 2 2

A1

2.00 in 2 ,

y1

0.75,

A1 y1 1.50 in 3

A2

0.25 in 2 ,

y2

0.25,

A2 y2

0.0625 in 3

A3 1.25 in 2 ,

y3

1.25,

A3 y3

1.5625 in 3

1.00 in 2 ,

y4

2.75,

A4 y4

2.75 in 3

A4 Mp

Y

( A1 y1

A2 y2

A3 y3

A4 y4 )

(36)(1.50 0.0625 1.5625 2.75)

Mp

212 kip in .


PROBLEM 4.87

y

z

C

90 mm

For the beam indicated (of Prob. 4.73), a couple of moment equal to the full plastic moment M p is applied and then removed. Using a yield strength of 240 MPa, determine the residual stress at y 45 mm .

60 mm

SOLUTION 29.16 103 N m

Mp

See solutions to Problems 4.73 and 4.77. I

3.645 10 6 m 4

c

0.045 m M max y I

LOADING

Mp c

UNLOADING (29.16 103 )(0.045) 3.645 10 6 res

Y

360 106

120 106 Pa

I

at y

c

45 mm

RESIDUAL STRESSES

360 106 Pa 240 106 res

120.0 MPa


PROBLEM 4.88

y

30 mm z

C

30 mm

For the beam indicated (of Prob. 4.74), a couple of moment equal to the full plastic moment M p is applied and then removed. Using a yield strength of 240 MPa, determine the residual stress at y 45mm .

30 mm 15 mm

30 mm

15 mm

SOLUTION Mp

I

27.54 103 N m (See solutions to Problems 4.74 and 4.78.)

3.5775 10 6 m 4 , M max y I

Mp c I

at

c

0.045 m

y

c

(27.54 103 )(0.045) 3.5775 10 6

LOADING

UNLOADING

res

Y

346.4 106

240 106

346.4 106 Pa

RESIDUAL STRESSES 106.4 106 Pa

res

106.4 MPa


PROBLEM 4.89

y

A bending couple is applied to the bar indicated, causing plastic zones 3 in. thick to develop at the top and bottom of the bar. After the couple has been removed, determine (a) the residual stress at y 4.5 in., (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the bar.

3 in.

3 in.

C

z

Beam of Prob. 4.75. 3 in.

1.5 in.

1.5 in.

3 in.

SOLUTION See solution to Problem 4.75 for bending couple and stress distribution during loading. M

2646 kip in. 42 ksi

Y

Mc I MyY I

(a)

At

y

c,

At

y

yY ,

res

(c)

y0

I Y M

At

y

188.5 in 4

I

63.167

Y

res

Y

c

E

29 106 psi

29 103 ksi

4.5 in.

42

21.056

21.167 ksi

42

20.944 ksi

UNLOADING My0 Y I (188.5)(42) 2646

0

res

1.5 in.

(2646)(4.5) 63.167 ksi 188.5 (2646)(1.5) 21.056 ksi 188.5

LOADING (b)

yY

yY ,

Ey

2.99 in.

20.9 ksi

res

RESIDUAL STRESSES

Answer:

y0

2.99 in.,

0,

2.99 in.

20.944 ksi

res

Ey

(29 103 )(1.5) 20.944

2077 in.

173.1 ft


PROBLEM 4.90

y

A bending couple is applied to the bar indicated, causing plastic zones 3 in. thick to develop at the top and bottom of the bar. After the couple has been removed, determine (a) the residual stress at y 4.5 in., (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the bar.

3 in.

3 in.

C

z

Beam of Prob. 4.76. 3 in.

1.5 in.

3 in.

1.5 in.

SOLUTION See solution to Problem 4.76 for bending couple and stress distribution. M Y

4725 kip in.

42 ksi

I

yY

1.5 in.

357.75 in 4

At At

y

yY ,

res

Y

I Y M

y0 (c)

At

My0 I

0

y

19.8113

59.4 ksi

17.4340 ksi 42

22.189 ksi

UNLOADING

LOADING res

29 103 ksi

4.5 in.

Mc (4725)(4.5) 59.434 ksi I 357.75 MyY (4725)(1.5) 19.8113 ksi I 357.75 y c, 59.434 42 res Y

(a)

(b)

c

29 106 psi

E

0

Y

(357.75)(42) 4725

yY , Ey

res

RESIDUAL STRESSES

3.18 in.

Answer:

y0

3.18 in., 0, 3.18 in.

22.189 ksi Ey

(29 103 )(1.5) 22.189

1960.43 in.

163.4 ft.


PROBLEM 4.91

y

z

A bending couple is applied to the beam of Prob. 4.73, causing plastic zones 30 mm thick to develop at the top and bottom of the beam. After the couple has been removed, determine (a) the residual stress at y 45 mm, (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the beam.

90 mm

C

60 mm

SOLUTION See solution to Problem 4.73 for bending couple and stress distribution during loading.

28.08 103 N m

M

Mc I

c,

At y

yY ,

res

y0

E

c

(28.08 103 )(0.045) 3.645 10 6

346.67 106 Pa

346.67 MPa

(28.08 103 )(0.015) 3.645 10 6 res

346.67

Y

res

115.556 106 Pa

Y

240

115.556

LOADING (b)

0.015 m

3.645 10 6 m 4

M yY I At y

15 mm

I

240 MPa

Y

(a)

yY

0.045 m

115.556 MPa

106.670 MPa

240

My0 I

I Y M

(3.645 10 6 )(240 106 ) 28.08 103

Y

res

124.444 MPa

UNLOADING

0

200 GPa

res

At

y

yY , Ey

res

124.4 MPa

RESIDUAL STRESSES

0 31.15 10 3 m

31.15 mm

Answer: y0 (c)

106.7 MPa

31.2 mm, 0, 31.2 mm

124.444 106 Pa Ey

(200 109 )(0.015) 124.444 106

24.1 m


PROBLEM 4.92

y

A beam of the cross section shown is made of a steel that is assumed to be elastoplastic with E 29 × 106 psi and Y = 42 ksi. A bending couple is applied to the beam about the z axis, causing plastic zones 2 in. thick to develop at the top and bottom of the beam. After the couple has been removed, determine (a) the residual stress at y 2 in., (b) the points where the residual stress is zero, (c) the radius of curvature corresponding to the permanent deformation of the beam.

1 in.

z

2 in.

C

1 in. 1 in.

1 in.

1 in.

SOLUTION Flange:

I1

Web:

I2

1 b1h13 12

1 (3)(1)3 12

A1d12

1 1 b2 h23 (1)(2)3 12 12 I1 7.0 in 4

I3 I

I1

I2

c

2 in. YI

MY

Y

7.0 in 4

0.6667 in 4

14.6667 in 4

(42)(14.6667) 2

c

R1

I3

(3)(1)(1.5)2

A1

MY

308 kip in .

M

406 kip in .

(42)(3)(1) 126 kips

y1 1.0 0.5 1.5 in. 1 Y A2 2 21 kips

R2

M Y

(a)

y2

2 (1.0) 0.6667 in. 3

M

2( R1 y1

406 kip in. yY 42 ksi

1 (42)(1)(1) 2

I

R2 y 2 )

1.0 in. E

14.6667 in 4 c

Mc (406)(2) I 14.6667

2[(126)(1.5) (21)(0.6667)] 29 106

29 103 ksi

2 in.

55.364

MyY (406)(1.0) I 14.662

27.682


PROBLEM 4.92 (Continued) At y

c,

res

55.364 42 13.3640 ksi

Y

res

At y

yY ,

res

Y

27.682 42

14.3180 ksi res

(b)

res

y0

0 I Y M

At y

yY ,

14.32 k si

My0 0 Y I (14.667)(42) 1.517 in 406

Answer : y0 (c)

13.36 k si

res

Ey

1.517 in., 0, 1.517 in.

14.3180 ksi Ey

(29 103 )(1.0) 14.3180

2025.42 in. 168.8 ft


PROBLEM 4.93* A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of moment M. After the couples are removed, it is observed that the radius of curvature of the bar is R . Denoting by Y the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy the following relation: 1

1

3 2

1

R

2

1 3

1 Y

Y

SOLUTION 1 Y

Let m denote

MY , EI

M

3 MY 1 2

1 3

2

2

2 Y

2 Y

2

,

2 Y

M . MY

m

M MY

1

1

M EI

R

1

3 1 2 1

1

mM Y EI 1

m Y

1

3

1

2m

m Y

3 2

1 Y

1 3

2 2 Y


PROBLEM 4.94 A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by M Y and Y , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the radius of curvature when a couple of moment M 1.25 M Y is applied to the bar, (b) the radius of curvature after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.

SOLUTION (a)

1 Y

m

MY , EI M MY

3 MY 1 2

M 3 1 2

1 3

1 3

2

Let m

2 Y

M MY

1.25

2 2 Y

3

2m

0.70711

Y

0.707 (b)

1

1

M EI

R

1

mM Y EI

1

m Y

1 0.70711

Y

1.25 Y

Y

0.16421 R

6.09

Y

Y


PROBLEM 4.95

M

The prismatic bar AB is made of a steel that is assumed to be elastoplastic and for which E 200 GPa . Knowing that the radius of curvature of the bar is 2.4 m when a couple of moment M 350 N m is applied as shown, determine (a) the yield strength of the steel, (b) the thickness of the elastic core of the bar.

B

A 20 mm

16 mm

SOLUTION M

1 3

3 YI 1 2 c

1 2 Y2 3 E 2c 2

3 2

Y b(2c)

bc 2

2 Y

1

2 Y 2 2

3E c

M

Data:

2 Y

3

1 2 Y2 3 E 2c 2

1

12c

1 2 Y2 3 E 2c 2

2 Y bc 1

(a)

2

3 MY 1 2

Cubic equation for

Y

200 109 Pa

E

420 N m

M

2.4 m b

20 mm

c

1 h 2

8 mm

Y

1

750 10

21

2 Y

350

Y

1

750 10

21

2 Y

273.44 106

(1.28 10 6 )

Solving by trial, (b)

yY

Y

E

(292 106 )(2.4) 200 109

Y

0.020 m 0.008 m

292 106 Pa

3.504 10 3 m

Y

292 MPa

2 yY

7.01 mm

3.504 mm thickness of elastic core


PROBLEM 4.96

" (MPa) 300

The prismatic bar AB is made of an aluminum alloy for which the tensile stress-strain diagram is as shown. Assuming that the - diagram is the same in compression as in tension, determine (a) the radius of curvature of the bar when the maximum stress is 250 MPa, (b) the corresponding value of the bending moment. versus y and use an (Hint: For part b, plot approximate method of integration.)

B

40 mm

M'

200

M 60 mm A

100

0

0.005

0.010

#

SOLUTION (a)

250 MPa

m

0.0064 from curve

c b

1 h 30 mm 0.030 m 2 40 mm 0.040 m

1

m

c

(b)

250 106 Pa

m

0.0064 0.030

0.21333 m

1

Strain distribution:

4.69 m y c

m

Bending couple:

c

M

where the integral J is given by

c 1 u| 0

where

mu

y b dy

2b

c 0

u

y 1

2bc 2 0 u | | du

y | | dy

2bc 2 J

| du

Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is J

u wu | | 3

where w is a weighting factor. Using u 0 0.25 0.5 0.75 1.00

u

0.25, we get the values given in the table below:

| | 0 0.0016 0.0032 0.0048 0.0064 J M

| |, (MPa) 0 110 180 225 250 (0.25)(1215) 3

u | |, (MPa) 0 27.5 90 168.75 250 101.25 MPa

w 1 4 2 4 1

wu | |, (MPa) 0 110 180 675 250 1215

wu | |

101.25 106 Pa

(2)(0.040)(0.030)2 (101.25 106 )

7.29 103 N m

M

7.29 kN m


PROBLEM 4.97 0.8 in. B

M

" (ksi)

1.2 in. A

50

The prismatic bar AB is made of a bronze alloy for which the tensile stress-strain diagram is as shown. Assuming that the - diagram is the same in compression as in tension, determine (a) the maximum stress in the bar when the radius of curvature of the bar is 100 in., (b) the corresponding value of the bending moment. (See hint given in Prob. 4.96.)

40 30 20 10 0

0.004

0.008

#

SOLUTION (a)

100 in., b c

0.6 100

m

(b)

0.8 in., c

0.6 in.

0.006

From the curve,

Strain distribution: Bending couple:

m c

M

where the integral J is given by

y c c

1 u 0

mu

y b dy

where u 2b

c 0

y | | dy

m

43.0 ksi

y 1

2bc 2 0 u | | du

2bc 2 J

| | du

Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is J

u wu| | 3

where w is a weighting factor. Using

u

0.25, we get the values given the table below:

u

| |

0 0.25 0.5 0.75 1.00

0 0.0015 0.003 0.0045 0.006

| |, ksi 0 25 36 40 43

u | |, ksi 0 6.25 18 30 43

J M

w 1 4 2 4 1

(0.25)(224) 18.67 ksi 3 (2)(0.8)(0.6) 2 (18.67)

wu | |, ksi 0 25 36 120 43 224

wu | |

M

10.75 kip in.


PROBLEM 4.98 "

A prismatic bar of rectangular cross section is made of an alloy for which the stress-strain diagram can be represented by the relation k n for 0 and k n for 0 . If a couple M is applied to the bar, show that the maximum stress is

# M

1 m

2n Mc I 3n

SOLUTION Strain distribution:

y c

m

where

mu

y c

u

Bending couple: c

M

c

1 0

2 bc 2 For

K

n

c 0

2b

y b dy

2bc 2

y | | dy

,

K

m

m

u m

m

M

2bc 2 0 u

1

2bc

Recall that Then

| |

2

m

mu

u 2

1 n

2 1n

m

2n 1 M 2 bc 2

I c

1 b(2c)3 12 c

m

y dy | | c c

u | | du

n

Then

c 0

1 n

du 1 0

2 2 bc 3

2bc 2

mu

1 n

1 1 1n du m 0u

2n bc 2 2n 1

1 bc 2

m

2 c 3 I

2n 1 Mc 3n I


PROBLEM 4.99

30 mm

Knowing that the magnitude of the horizontal force P is 8 kN, determine the stress at (a) point A, (b) point B. B

24 mm

A D

P

45 mm

15 mm

SOLUTION A

(30)(24)

720 mm 2

e

45

33 mm

I c M

(a)

A

P A

Mc I

12

720 10 6 m 2 0.033 m

1 3 1 bh (30)(24)3 34.56 103 mm 4 34.56 10 9 m 4 12 12 1 (24 mm) 12 mm 0.012 m P 8 103 N 2 Pe

(8 103 )(0.033)

8 103 720 10

6

264 N m

(264)(0.012) 34.56 10 9

102.8 106 Pa

102.8 MPa

A

(b)

B

P A

Mc I

8 103 720 10

6

(264)(0.012) 34.56 10 9

80.6 106 Pa B

80.6 MPa


PROBLEM 4.100

y b 3 in.

6 kips

A short wooden post supports a 6-kip axial load as shown. Determine the stress at point A when (a) b 0, (b) b 1.5 in., (c) b 3 in.

C

A z

x

SOLUTION A I S P

(a)

b

0

M P A

(b)

b

(3) 2

28.27 in 2

r4 (3) 4 63.62 in 4 4 4 I 63.62 21.206 in 3 c 3 6 kips M Pb

0 6 28.27

1.5 in. M

P A

r2

M S

0.212 ksi 212 psi

(6)(1.5)

9 kip in.

6 28.27

9 21.206

0.637 ksi 637 psi

(c)

b

3 in.

P A

M

M S

(6)(3)

18 kip in.

6 28.27

18 21.206

1.061 ksi 1061 psi


P

r

r

P

PROBLEM 4.101 Two forces P can be applied separately or at the same time to a plate that is welded to a solid circular bar of radius r. Determine the largest compressive stress in the circular bar, (a) when both forces are applied, (b) when only one of the forces is applied.

SOLUTION For a solid section,

A

Both forces applied.

4

r 4, c

r

F Mc A I F 4M 2 r r3

Compressive stress

(a)

r 2, I

F

2 P, M

0 2P r2

(b)

One force applied.

F

P, M F r2

Pr 4Pr r2

5P r2


PROBLEM 4.102

y

30 mm

60 mm

A short 120 × 180-mm column supports the three axial loads shown. Knowing that section ABD is sufficiently far from the loads to remain plane, determine the stress at (a) corner A, (b) corner B.

30 kN 20 kN 100 kN C z

x

A

D 90 mm 90 mm

B

120 mm

SOLUTION

A

S M

(a)

A

(0.120 m)(0.180 m) 21.6 10 3 m 2 1 (0.120 m)(0.180 m)2 6.48 10 4 m 2 6 (30 kN)(0.03 m) (100 kN)(0.06 m) 5.10 kN m P A

M S

150 103 N 21.6 10 3 m 2

5.10 103 N m 6.48 10 4 m3 A

(b)

B

P A

M S

3

150 10 N 21.6 10 3 m 2

0.926 MPa

3

5.10 10 N m 6.48 10 4 m3 B

14.81 MPa


PROBLEM 4.103

80 mm 80 mm

P

C 1

As many as three axial loads, each of magnitude P 50 kN, can be applied to the end of a W200 × 31.1 rolled-steel shape. Determine the stress at point A (a) for the loading shown, (b) if loads are applied at points 1 and 2 only.

P

P 2

3

A

SOLUTION For W200

(a)

31.3 rolled-steel shape.

Centric load:

A

3970 mm 2

c

1 d 2

I

31.3 106 mm 4

3P

50

1 (210) 2 50

3P A

(b)

Ececentric loading:

e

50

50

M

Pe

(50 103 )(0.080)

2P A

Mc I

80 mm

50

3

m2

105 mm

0.105 m

31.3 10 150 kN

150 103 3.970 10 3

6

m4

150 103 N 37.783 106 Pa

37.8 MPa

0.080 m

100 103 N

2P

A

100 kN

4.000 10

4.0 103 N m

100 103 3.970 10 3

(4.0 103 )(0.105) 31.3 10 6

38.607 106 Pa

38.6 MPa


PROBLEM 4.104

y 10 mm 10 mm

Two 10-kN forces are applied to a 20 × 60-mm rectangular bar as shown. Determine the stress at point A when (a) b 0, (b) b 15 mm, (c) b 25 mm.

A 30 mm z

30 mm

10 kN C b 10 kN

x 25 mm

SOLUTION

A

1.2 10

3

m2

b

1 (0.020 m)(0.060 m)2 12 10 6 m3 6 0, M (10 kN)(0.025 m) 250 N m

A

(20 kN)/(1.2 10

S

(a)

(0.060 m)(0.020 m)

16.667 MPa

3

m2 )

(250 N m)/(12 10

6

m3 )

20.833 MPa

4.17 MPa

A

(b)

b

15 mm, M

A

(20 kN)/(1.2 10 16.667 MPa

(10 kN)(0.025 m 3

m2 )

0.015 m)

100 N m

(100 N m)/(12 10

6

m3 )

8.333 MPa A

(c)

b

25 mm, M

0:

A

(20 kN)/(1.2 10

3

8.33 MPa

m2 ) A

16.67 MPa


P

P'

P

PROBLEM 4.105

P'

Portions of a 12 12 -in. square bar have been bent to form the two machine components shown. Knowing that the allowable stress is 15 ksi, determine the maximum load that can be applied to each component.

1 in.

(a)

(b)

SOLUTION The maximum stress occurs at point B. B B

where

ec I

15 103 psi

15 ksi P A

K

1 A

A

(0.5)(0.5)

I

1 (0.5)(0.5)3 12

Mc I

P A

Pec I

KP

1.0 in.

e

0.25 in 2 5.2083 10 3 in 4 for all centroidal axes.

(a) (a)

(b)

c

(b)

0.25 in.

K

1 0.25

(1.0)(0.25) 5.2083 10 3

P

B

( 15 103 ) 52

c

K 0.5 2

52 in

2

P

288 lb

P

209 lb

0.35355 in.

K

1 0.25

(1.0)(0.35355) 5.2083 10 3

P

B

( 15 103 ) 71.882

K

71.882 in

2


PROBLEM 4.106

P

Knowing that the allowable stress in section ABD is 80 MPa, determine the largest force P that can be applied to the bracket shown. A

D B

18 mm 40 mm

12 mm 12 mm

SOLUTION A

(24)(18)

432 mm 2

432 10

1 (24)(18)3 11.664 103 mm 4 12 1 (18) 9 mm 0.009 m c 2 1 (18) 40 49 mm 0.049 m e 2 On line BD, both axial and bending stresses are compressive. I

Therefore

Solving for P gives

max

P

P

P A

Mc I

P A

Pec I

6

m2 11.664 10

9

m4

max

1 A

ec I

80 106 Pa 1 432 10 6 m 2

(0.049 m)(0.009 m) 11.664 10 9 m 4 P

1.994 kN


PROBLEM 4.107

P'

a

d a

P

A milling operation was used to remove a portion of a solid bar of square cross section. Knowing that a 30 mm, d 20 mm, and all 60 MPa, determine the magnitude P of the largest forces that can be safely applied at the centers of the ends of the bar.

SOLUTION A e

Data:

ad ,

I

1 ad 3 , 12

a d 2 2 P Mc P 6 Ped A I ad ad 3 3P (a d ) P KP ad ad 2

a

30 mm

K

1 (0.030)(0.020)

P

c

K

0.030 m

d

1 d 2

K

20 mm

0.020 m

(3)(0.010) (0.030)(0.020)2

60 106 4.1667 103

1 ad

where

14.40 103 N

3(a d ) ad 2

4.1667 103 m

2

P

14.40 kN


PROBLEM 4.108

P'

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P 18 kN are applied at the centers of the ends of the bar. Knowing that a 30 mm and 135 MPa, determine the smallest allowable depth d of the all milled portion of the bar.

a

d

P

a

SOLUTION 1 ad 3 , 12

A

ad ,

e

a 2

d 2

P A

Mc I

3P d2

I

2P ad

Solving for d,

d

1 2

Data:

a

0.030 m,

d

1 (2)(135 106 )

16.04 10

3

P ad

2P a

c

1 d 2

Pec I

P ad

d2

or 2

18 103 N,

(2)(18 103 ) 0.030

3P

P ad

3P(a d ) ad 2

0

2P a

12 P P

2P d a

P 1 (a d ) 1 d 2 2 1 3 ad 12

135 106 Pa

2

12(18 103 )(135 106 )

m

(2)(18 103 ) 0.030

d

16.04 mm


12 kips

5 in.

PROBLEM 4.109

P

The two forces shown are applied to a rigid plate supported by a steel pipe of 8-in. outer diameter and 7-in. inner diameter. Determine the value of P for which the maximum compressive stress in the pipe is 15 ksi.

SOLUTION

15 ksi

all

I NA A

4

(4 in.)4

4

(4 in.) 2

(3.5 in.)4

(3.5 in.)2

83.2 in 4 11.78 in 2

Max. compressive stress is at point B. B

Q A

Mc I

15 ksi

1.019

13.981

0.325P

12 P 11.78 in 2

0.085P

(5P)(4.0 in.) 83.2 in 4

0.240 P P

43.0 kips


PROBLEM 4.110

d P'

P

h P'

P

An offset h must be introduced into a solid circular rod of diameter d. Knowing that the maximum stress after the offset is introduced must not exceed 5 times the stress in the rod when it is straight, determine the largest offset that can be used.

d

SOLUTION For centric loading,

c

P A

For eccentric loading,

e

P A

Given

e

P A

5

Phc I c

Phc I

5

P A

Phc I

P 4 A

h

4I cA

(4) d 2

64 4

d4 d

2

1 d 2

h

0.500 d


PROBLEM 4.111

d P'

P

An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used.

h P'

P d

SOLUTION c c1

1 d 0.375 in. 2 c t 0.375 0.08 c2

A

c12

0.295 in.

(0.3752

0.2952 )

0.168389 in 2 I

4

c4

c14

4

(0.3754

0.2954 )

9.5835 10 3 in 4

For centric loading,

cen

P A

For eccentric loading,

ecc

P A

ecc

4

hc I

3 A

Phc I or

cen

h

P A 3I Ac

Phc I

4

P A

(3)(9.5835 10 3 ) (0.168389)(0.375)

h

0.455 in.


PROBLEM 4.112

16 kips 2

4

1

A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in. timber. Using an allowable stress of 600 psi, determine the largest compressive load P that can be applied at the center of the top section of the timber column as shown if (a) the column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are removed, (d) planks 1, 2, and 3 are removed, (e) all planks are removed.

3

SOLUTION (a)

Centric loading:

M

P A

0

A

(4

4)

32 in 2

4(1)(4)

P A

(0.600 ksi)(32 in 2 )

P

19.20 kips (b)

Eccentric loading: M

P A

Pe

A

(4)(4)

y

Ay A

I

(I

(c)

Centric loading:

28 in 2

(3)(1)(3) (1)(4)(2.5) 28

e

y

0.35714 in.

Ad 2 )

1 (6)(4)3 12

P

Pec I

1 A

1 (4)(1)3 12

(6)(4)(0.35714)2

ec I

P

(4)(1)(2.14286)2

(0.600 ksi) (0.35714)(2.35714) 53.762

1 28

53.762 in 4

11.68 kips

P A

M

0

A

(6)(4)

P

(0.600 ksi)(24)

24 in 2 14.40 kips



(d)

Eccentric loading:

M

Pe

A

(4)(4)

x

2.5

P

Centric loading:

(1)(4)(1) 2

1 (4)(5)3 12

I

(e)

P A

M

Pec I 20 in 2

x

0.5 in. 41.667 in 4

(0.600 ksi) 1 (0.5)(2.5) 20 41.667

0

e

7.50 kips

P A 16 in 2

A

(4)(4)

P

(0.600 ksi)(16.0 in 2 )

9.60 kips


1 in. a 1.5 in.

PROBLEM 4.113

3 in.

a

0.75 in.

1.5 in.

A

A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are all 5 ksi and 12 ksi , determine the largest downward all force and the largest upward force that can be exerted by the rod.

3 in. 0.75 in.

B Section a–a

SOLUTION X

Ay A

X

12.75 in 3 7.5 in 2

A

7.5 in 2

all

5 ksi

(1 3)(0.5) (1 3)

all

1 (3)(1)3 12

0.25)(2.5) 0.75)

1.700 in.

1 3 bh 12

Ic

2(3 2(3

12 ksi

Ad 2 (3 1)(1.70

0.5)2

1 (1.5)(3)3 12

(1.5

3)(2.5

1.70)2

10.825 in 4

Ic Downward force.

M At D:

D

5 ksi 5 At E:

E

12 ksi 12

P(1.5 in.+1.70 in.) P A

Mc I

P (3.20) P(1.70) 7.5 10.825 P( 0.6359) P A

(3.20 in.) P

P

7.86 kips

P

21.95 kips

P

7.86 kips

Mc I

P (3.20) P(2.30) 7.5 10.825 P( 0.5466)

We choose the smaller value.



Upward force.

M At D:

D

12 ksi 12 At E:

E

5 ksi 5

P(1.5 in. P A

1.70 in.)

Mc I

P (3.20) P(1.70) 7.5 10.825 P( 0.6359) P A

(3.20 in.) P

P

18.87 kips

Mc I

P (3.20) P(2.30) 7.5 10.825 P( 0.5466)


P

9.15 kips

P

9.15 kips


PROBLEM 4.114 1 in. a 1.5 in.

3 in.

a 1.5 in.

A

Solve Prob. 4.113, assuming that the vertical rod is attached at point B instead of point A.

0.75 in. 3 in.

PROBLEM 4.113 A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stresses in the hanger are 5 ksi and all 12 ksi, determine the all largest downward force and the largest upward force that can be exerted by the rod.

0.75 in.

B Section a–a

SOLUTION X

Ay A

X

12.75 in 3 7.5 in 2

A

7.5 in 2

all

5 ksi

(1 3)(0.5) (1 3)

all

1 (3)(1)3 12

Ic

0.25)(2.5) 0.75)

1.700 in.

1 3 bh 12

Ic

2(3 2(3

12 ksi

Ad 2 (3 1)(1.70

0.5)2

1 (1.5)(3)3 12

(1.5

3)(2.5

1.70)2

10.825 in 4

Downward force.

all

M At D:

D

12 ksi 12 At E:

E

5 ksi 5

5 ksi (2.30 in.

P A

12 ksi

all

1.5 in.)

(3.80 in.) P

Mc I

P (3.80) P(1.70) 7.5 10.825 P( 0.4634) P A

P

25.9 kips

P

5.32 kips

P

5.32 kips

Mc I

P (3.80) P(2.30) 7.5 10.825 P( 0.9407)




Upward force.

all

M At D:

D

5 ksi 5 At E:

E

12 ksi 12

5 ksi (2.30 in.

P A

12 ksi

all

1.5 in.)P

Mc I

P (3.80) P(1.70) 7.5 10.825 P( 0.4634) P A

(3.80 in.)P

P

10.79 kips

P

12.76 kips

P

10.79 kips

Mc I

P (3.80) P(2.30) 7.5 10.825 P( 0.9407)



PROBLEM 4.115

2 mm radius A P

Knowing that the clamp shown has been tightened until P 400 N, determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis of section a a.

P'

32 mm

20 mm

a

B

a

4 mm

Section a–a

SOLUTION Cross section:

Rectangle

Circle (20 mm)(4 mm)

y1

1 (20 mm) 2

10 mm

A2

(2 mm)2

4 mm 2

y2

y

cA

20

d1

11.086

d2

18

y

2

18 mm

(80)(10) 80

(4 )(18) 4

1.086 mm

11.086

6.914 mm

I1

A1d12

1 (4)(20)3 12

I2

I2

A2d 22

(2) 4

I

I1

I2

A

A1

A2

(80)(1.086) 2

(4 )(6.914) 2

3.374 103 mm 4 92.566 mm 2

11.086 mm

8.914 mm 10

I1

4

20

Ay A

cB

80 mm 2

A1

2.761 103 mm 4 0.613 103 mm 4

3.374 10

92.566 10

6

9

m4

m2



(a)

e

32

8.914

M

Pe

(400 N)(0.040914 m)

16.3656 N m

Mc I

(16.3656)(8.914 10 3 ) 3.374 10 9

P A

4.321 106

400 92.566 10

6

43.23 106

47.55 106 Pa

A

47.6 MPa

Point B: B

P A

Mc I

4.321 106 (c)

0.040914 m

Point A: A

(b)

40.914 mm

Neutral axis:

400 92.566 10

6

53.72 106

(16.3656)(11.086) 3.374 10 9 49.45 106 Pa

B

49.4 MPa

By proportions, a 47.55 a

20 47.55 49.45 9.80 mm

9.80 mm below top of section


PROBLEM 4.116

P a

a

The shape shown was formed by bending a thin steel plate. Assuming that the thickness t is small compared to the length a of a side of the shape, determine the stress (a) at A, (b) at B, (c) at C.

90$ t B C

A

P'

SOLUTION Moment of inertia about centroid:

I

a

1 2 2t 12

3

2

1 3 ta 12 Area:

A

2 2t

a 2

2at , c

Pec I

P 2at

P

A

P A

(b)

Pec I

P 2at

P

B

P A

(c)

C

(a)

a 2 2 1 ta 12 a 2

a 2 2 a 2 2 3

A

P 2at

B

2P at

C

P 2at

a 2

1 ta3 12

A


PROBLEM 4.117

P

Three steel plates, each of 25 150-mm cross section, are welded together to form a short H-shaped column. Later, for architectural reasons, a 25-mm strip is removed from each side of one of the flanges. Knowing that the load remains centric with respect to the original cross section, and that the allowable stress is 100 MPa, determine the largest force P (a) that could be applied to the original column, (b) that can be applied to the modified column.

50 mm 50 mm

SOLUTION (a)

P A

Centric loading:

A

11.25 103 mm 2

(3)(150)(25)

P

11.25 10 3 m 2

( 100 106 )(11.25 10 3 )

A

1.125 106 N (b)

P

1125 kN

Eccentric loading (reduced cross section):

A, 103 mm 2

y , mm

A y (103 mm3 )

3.75

87.5

328.125

76.5625

3.75

0

0

10.9375

2.50

87.5

218.75

98.4375

10.00

d , mm

109.375

Y

Ay A

109.375 103 10.00 103

10.9375 mm


PROBLEM 4.117 (Continued) The centroid lies 10.9375 mm from the midpoint of the web.

I1 I2 I3

1 b1h13 12 1 b2h23 12 1 b3h33 12

A2d 22 A3d32

I2

I3

I

I1

c

10.9375

M

Pe P A

K P

75

K

54.012 106 mm 4 25

Mc I

P A

10.4375 mm

Pec I

1 10.00 10 ( 100 106 ) 122.465

3

22.177 106 mm 4 7.480 106 mm 4 24.355 106 mm 4

54.012 10 6 m 4

110.9375 mm

e

where

ec I

1 A

1 (150)(25)3 (3.75 103 )(76.5625)2 12 1 (25)(150)3 (3.75 103 )(10.9375)2 12 1 (100)(25)3 (2.50 103 )(98.4375) 2 12

A1d12

KP

0.1109375 m 10.4375 10 3 m

A

10.00 10 3 m 2

(101.9375 10 3 )(0.1109375) 54.012 10 6 817 103 N

122.465 m

2

P

817 kN


PROBLEM 4.118

y P

y B

3 in. y

x

3 in.

B

A vertical force P of magnitude 20 kips is applied at point C located on the axis of symmetry of the cross section of a short column. Knowing that y 5 in., determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis.

2 in.

C A

4 in. A 2 in.

x 2 in.

1 in. (a)

(b)

SOLUTION Locate centroid. A, in 2 12 8 20 Eccentricity of load: e 5

(a)

I1

1 (6)(2)3 12

I

I1

I2

Stress at A : c A

Ay , in 3 60 16 76

y , in. 5 2

Part

3.8

Stress at B : cB

(12)(1.2) 2

21.28 in 4

3.8 in.

1 (2)(4)3 12

I2

(8)(1.8)2

36.587 in 4

57.867 in 4

3.8 in.

P A

6

3.8

Pec A I

20 20

20(1.2)(3.8) 57.867

PecB I

20 20

20(1.2)(2.2) 57.867

A

0.576 ksi

2.2 in.

P A

B

(c)

76 20

1.2 in.

A

(b)

Ay A

y

Location of neutral axis:

0 a

I Ae

P Pea 0 A I 57.867 2.411 in. (20)(1.2)

Neutral axis lies 2.411 in. below centroid or 3.8

2.411

ea I

B

1.912 ksi

1 A

1.389 in. above point A.

Answer: 1.389 in. from point A


PROBLEM 4.119

y P

y B

3 in. y

x

3 in.

B

A vertical force P is applied at point C located on the axis of symmetry of the cross section of a short column. Determine the range of values of y for which tensile stresses do not occur in the column.

2 in.

C A

4 in. A

x

2 in.

2 in. 1 in.

(a)

(b)

SOLUTION Locate centroid. A, in 2

y , in.

Ay , in 3

5 2

60 16 76

12 8 20 Eccentricity of load:

e

y

I1

1 (6)(2)3 12

I

I1

If stress at A equals zero,

If stress at B equals zero,

y

3.8 in.

3.8 in.

A

P A

e

I Ac A

e y

3.8 in.

3.8 in.

(12)(1.2) 2

cA

B

76 20

21.28 in 4

1 (2)(4)3 12

I2

(8)(1.8) 2

36.587 in 4

57.867 in 4

I2

cB

e

Ai yi Ai

y

6

Pec A I

3.8

ec A I

0

57.867 (20)(3.8)

0.761 in.

y

0.761

3.8

4.561 in.

2.2 in.

P PecB 0 A I 57.867 I (20)(2.2) AcB 1.315

1 A

3.8

ecB I

1 A

1.315 in.

2.485 in.

Answer: 2.49 in.

y

4.56 in.


PROBLEM 4.120

P P

The four bars shown have the same cross-sectional area. For the given loadings, show that (a) the maximum compressive stresses are in the ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3. (Note: the cross section of the triangular bar is an equilateral triangle.)

P P

SOLUTION Stresses: At A,

A

P A

Pec A I

P 1 A

At B,

B

P A

PecB I

P AecB A I

A1

1 4 a , c A cB 12 1 1 a (a 2 ) a P 2 2 1 1 2 A a 12

a 2 , I1

A

B

A2

A

B

P A c2

(a 2 )

a2

1 1 a a 2 2 1 2 a 12 c

a

Aec A I 1

1 a, e 2

1 a 2 A

1

, I2

B

4

c4 , e

P ( c 2 )(c)(c) 1 A2 c4 4

P A1

2

P A1

5

P A2

3

P A2

c

A

P ( c 2 )(c)(c) 1 A2 4 c 4

4

B



A3

A

B

a2

2 a I3 2

c

(a 2 )

P 1 A3

P A3

1 4 a 12

e

c

2 2 a a 2 2 1 4 a 12

A

2 2 a a 2 2 1 4 a 12

1

A4

1 (s) 2

3 s 2

3 2 s 4

I4

1 s 36

3 s 2

cA

2 3 s 3 2

(a 2 )

A

B

3

s

P A4

cB

5

P A3

3 2 s 4

s 3

9

P A4

3

P A4

s s 3 A

3 4 s 96 3 2 s 4

P A3

3 4 s 96 e

3

P 1 A4

B

7

s 3 3 4 s 96

s 2 3

1

B


PROBLEM 4.121

25 mm

An eccentric force P is applied as shown to a steel bar of 25 90-mm cross section. The strains at A and B have been measured and found to be

30 mm A 90 mm

45 mm

B

350

A

P d

70

B

Knowing that E 200 GPa, determine (a) the distance d, (b) the magnitude of the force P.

15 mm

SOLUTION h 15 45 30 90 mm A bh (25)(90) I yA

1 h 45 mm 2 2.25 10 3 m 2

25 mm

b

2.25 103 mm 2

Subtracting,

A

A

E

A

(200 109 )(350 10 6 )

B

E

B

(200 109 )( 70 10 6 )

M

A( y A B yB y A yB Pd

d

MyA I

(1)

B

P A

MyB I

(2)

M ( y A yB ) I

M

I( A B) y A yB

M P

14 106 Pa

P A

B

A)

70 106 Pa

A

(1.51875 10 6 )(84 106 ) 0.045

Multiplying (2) by y A and (1) by yB and subtracting,

(a)

0.045 m

1 3 1 (25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4 bh 12 12 60 45 15 mm 0.015 m 30 mm 0.030 m yB 15 45

Stresses from strain gages at A and B:

P

c

yA

B

yB

A

( yA

2835 N m

yB )

P A

(2.25 10 3 )[(0.015)( 14 106 ) ( 0.030)(70 106 )] 0.045 2835 94.5 103

0.030 m

(b)

94.5 103 N

d P

30.0 mm 94.5 kN


PROBLEM 4.122 25 mm

Solve Prob. 4.121, assuming that the measured strains are

A 90 mm

B

600

A

30 mm

45 mm

P

PROBLEM 4.121 An eccentric force P is applied as shown to a steel bar of 25 90-mm cross section. The strains at A and B have been measured and found to be

d

350

A

15 mm

420

B

B

70

Knowing that E 200 GPa, determine (a) the distance d, (b) the magnitude of the force P.

SOLUTION h 15 45 30 90 mm A bh

b

1 h 45 mm 2 2.25 10 3 m 2

2.25 103 mm 2

(25)(90)

yA

Stresses from strain gages at A and B:

A

E

A

(200 109 )(600 10 6 ) 120 106 Pa

B

E

B

(200 109 )(420 10 6 ) 84 106 Pa

A

P A

My A I

(1)

B

P A

MyB I

(2)

Subtracting,

A

B

M

M ( y A yB ) I I( A B) y A yB

Multiplying (2) by y A and (1) by yB and subtracting,

(a)

M

A( y A B yB y A yB Pd

0.045 m

1 3 1 (25)(90)3 1.51875 106 mm 4 1.51875 10 6 m 4 bh 12 12 60 45 15 mm 0.015 m yB 15 45 30 mm 0.030 m

I

P

c

25 mm

d

A)

(1.51875 10 6 )(36 106 ) 0.045

yA

B

yB

A

( yA

yB )

(2.25 10 3 )[(0.015)(84 106 ) ( 0.030)(120 106 )] 0.045 M 1215 5 10 3 m 3 P 243 10

(b)

1215 N m

P A 243 103 N

d

5.00 mm

P

243 kN


PROBLEM 4.123

P'

The C-shaped steel bar is used as a dynamometer to determine the magnitude P of the forces shown. Knowing that the cross section of the bar is a square of side 40 mm and that strain on the inner edge was measured and found to be 450 , determine the magnitude P of the forces. Use E 200 GPa. 40 mm 80 mm

P

SOLUTION At the strain gage location, E A

(200 109 )(450 10 6 ) 1600 mm 2

(40)(40)

90 106 Pa

1600 10

e

1 (40)(40)3 213.33 103 mm 4 12 80 20 100 mm 0.100 m

c

20 mm

I

K P

6

m2 213.33 10

9

m4

0.020 m

P A

Mc I

P A

Pec I

1 A

ec I

K

90 106 10.00 103

KP

1 1600 10

6

(0.100)(0.020) 213.33 10 9

10.00 103 m

9.00 103 N

2

P

9.00 kN


P

y 6 in.

PROBLEM 4.124 6 in.

10 in.

Q B

A

A short length of a rolled-steel column supports a rigid plate on which two loads P and Q are applied as shown. The strains at two points A and B on the centerline of the outer faces of the flanges have been measured and found to be

x

x

z A

z A = 10.0 in2 Iz = 273 in4

6

in./in.

29

6

400 10

A

Knowing that E each load.

B

300 10

6

in./in.

10 psi, determine the magnitude of

SOLUTION Stresses at A and B from strain gages: A

E

A

(29 106 )( 400 10 6 )

11.6 103 psi

B

E

B

(29 106 )( 300 10 6 )

8.7 103 psi

Centric force:

F

P Q

Bending couple:

M

6 P 6Q

c F A

A

5 in. Mc I

11.6 103 F A

B

Mc I

8.7 103

P Q 10.0

(6 P 6Q)(5) 273

0.00989 P 0.20989Q P Q 10.0

(1)

(6 P 6Q )(5) 273

0.20989 P 0.00989Q

(2)

Solving (1) and (2) simultaneously, P

44.2 103 lb

Q

57.3 103 lb 57.3 kips

44.2 kips


PROBLEM 4.125

y P

A single vertical force P is applied to a short steel post as shown. Gages located at A, B, and C indicate the following strains: z

x

500

A

C

A

1000

200

C

Knowing that E 29 106 psi, determine (a) the magnitude of P, (b) the line of action of P, (c) the corresponding strain at the 1.5 in. hidden edge of the post, where x 2.5 in. and z

B 3 in.

5 in.

B

SOLUTION Ix Mx

1 (5)(3)3 11.25 in 4 12 Pz Mz Px 2.5 in., xB

xA zA

1.5 in., z B

2.5 in., xC 1.5 in., zC

1.5 in., z D

6

6

E

A

(29 10 )( 500 10 )

B

E

B

(29 106 )( 1000 10 6 )

A

31.25 in 4

2.5 in., xD

A

C

A

(5)(3)

15 in 2

2.5 in. 1.5 in.

14,500 psi 29, 000 psi

14.5 ksi 29 ksi

E C (29 106 )( 200 10 6 ) 5800 psi 5.8 ksi P M x z A M z xA 0.06667 P 0.13333M x 0.08M z A Ix Iz

(1)

B

P A

M x zB Ix

M z xB Iz

0.06667 P

0.13333M x

0.08M z

(2)

C

P A

M x zC Ix

M z xC Iz

0.06667 P

0.13333M x

0.08M z

(3)

Substituting the values for gives Mx

87 kip in.

x

Mz P Mx P

z D

A

,

Mz

B

, and

C

into (1), (2), and (3) and solving the simultaneous equations (a) P

90.625 kip in.

90.625 152.25 87 152.25

P A

M x zD Ix

M z xD Iz

(0.06667)(152.25)

(c)

1 (3)(5)3 12

Iz

Strain at hidden edge:

0.06667 P (0.13333)(87) D

E

0.13333M x

(b) x

0.595 in.

z

0.571 in.

0.08M z

(0.08)( 90.625)

8.70 103 29 106

152.3 kips

8.70 ksi 300


PROBLEM 4.126 b ! 40 mm A

a ! 25 mm d

B

D

P

The eccentric axial force P acts at point D, which must be located 25 mm below the top surface of the steel bar shown. For P 60 kN, determine (a) the depth d of the bar for which the tensile stress at point A is maximum, (b) the corresponding stress at point A.

C

20 mm

SOLUTION 1 bd 3 12 1 1 d e d a 2 2 P Pec A I 1 1 P 1 12 2 d a 2 d b d d3

A bd c A

A

(a)

Depth d for maximum

d A dd (b)

A

I

A:

P b

60 103 4 3 40 10 75 10

6a d2

Differentiate with respect to d.

4 d2 3

P 4 b d

12a d3

d

0

(6)(25 10 3 ) (75 10 3 )2

40 106 Pa

3a

d

A

75 mm

40 MPa


y

M ! 300 N · m A z

PROBLEM 4.127

" ! 60# B

16 mm C

16 mm

The couple M is applied to a beam of the cross section shown in a plane forming an angle with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

D 40 mm

40 mm

SOLUTION 1 (80)(32)3 218.45 103 mm 4 218.45 10 9 m 4 12 1 (32)(80)3 1.36533 106 mm 4 1.36533 10 6 m 4 12 yB yD 16 mm

Iz Iy yA zA My (a)

A

300cos 30 M z yA Iz

zB

zD

40 mm

259.81 N m

M y zA Iy

Mz

(150)(16 10 3 ) 218.45 10 9

300sin 30

150 N m

(259.81)(40 10 3 ) 1.36533 10 6

3.37 106 Pa (b)

B

M z yB Iz

M y zB Iy

(150)(16 10 3 ) 218.45 10 9

(259.81)( 40 10 3 ) 1.36533 10 6

18.60 106 Pa (c)

D

M z yD Iz

M y zD Iy

(150)( 16 10 3 ) 218.45 10 9

3.37 MPa

A

18.60 MPa

B

(259.81)( 40 10 3 ) 1.36533 10 6

3.37 106 Pa

D

3.37 MPa


PROBLEM 4.128

y

! " 30# A


B M " 400 lb · m

0.6 in. z

C 0.6 in. D 0.4 in.

SOLUTION Iz Iy yA zA My

(a)

A

400 cos 60

200 lb in.,

M y zA

M z yA Iz

Iy

1 (0.4)(1.2)3 57.6 10 3 in 4 12 1 (1.2)(0.4)3 6.40 10 3 in 4 12 yB yD 0.6 in. zB Mz

1 (0.4) 0.2 in. 2

zD

400 sin 60

( 346.41)(0.6) 57.6 10 3

346.41 lb in.

(200)(0.2) 6.40 10 3

9.86 103 psi 9.86 ksi

(b)

B

M y zB

M z yB Iz

Iy

( 346.41)(0.6) 57.6 10 6 2.64 103 psi

(c)

D

M z yD Iz

M y zD Iy

( 346.41)( 0.6) 57.6 10 3 9.86 103 psi

(200)( 0.2) 6.4 10 3 2.64 ksi

(200)( 0.2) 6.40 10 3 9.86 ksi


y A

M ! 25 kN · m

PROBLEM 4.129

" ! 15#

The couple M is applied to a beam of the cross section shown in a with the vertical. Determine the stress at plane forming an angle (a) point A, (b) point B, (c) point D.

B 80 mm C

z 20 mm

80 mm D

30 mm

SOLUTION My

25sin 15

6.4705 kN m

Mz

25cos15

24.148 kN m

1 1 (80)(90)3 (80)(30)3 12 12 5.04 10 6 m 4

Iy Iy

Iz

1 (90)(60)3 3

M yz

Stress:

(a)

A

1 (60)(20)3 3

Iy

5.04 106 mm 4

1 (30)(100)3 16.64 106 mm 4 3

16.64 10 6 m 4

Mzy Iz

(6.4705 kN m)(0.045 m) 5.04 10 6 m 4

(24.148 kN m)(0.060 m) 16.64 10 6 m 4

57.772 MPa 87.072 MPa (b)

B

(6.4705 kN m)( 0.045 m) 5.04 10 6 m 4

A

(24.148 kN m)(0.060 m) 16.64 10 6 m 4

57.772 MPa 87.072 MPa (c)

D

(6.4705 kN m)( 0.015 m) 5.04 10 6 m 4

29.3 MPa

B

144.8 MPa

D

125.9 MPa

(24.148 kN m)( 0.100 m) 16.64 10 6 m 4

19.257 MPa 145.12 MPa


y

PROBLEM 4.130

" ! 20#

M ! 10 kip · in. A z

2 in.

C B D 2 in.

3 in.


3 in.

4 in.


z , in.

Az , in 3

16

1

16

8

2

16

Σ

24

0

The centroid lies at point C.

yA

1 1 (2)(8)3 (4)(2)3 88 in 4 12 12 1 1 (8)(2)3 (2)(4)3 64 in 4 3 3 4 in. yB 1 in., yD

zA

zB

Iz Iy

(a)

M z yA Iz

M y zA

A

(b)

M z yB Iz

M y zB

B

(c)

M z yD Iz

M y zD

D

Iy

Iy

Iy

zD

4 in.,

0

Mz

10 cos 20

9.3969 kip in.

My

10 sin 20

3.4202 kip in.

(9.3969)(1) 88

(3.4202)( 4) 64

(9.3969)( 1) 88

(3.4202)( 4) 64

(9.3969)( 4) 88

(3.4202)(0) 64

A

0.321 ksi

0.107 ksi

B

D

0.427 ksi


PROBLEM 4.131

y M 5 60 kip · in.

b 5 508 A


B

3 in. z

C

3 in.

D 1 in.

2.5 in. 2.5 in. 5 in. 5 in.

1 in.

SOLUTION Mz My yA zA

(a)

A

M z yA Iz

60 sin 40 60 cos 40 yB

yD

zB

zD

38.567 kip in. 45.963 kip in. 3 in. 5 in.

Iz

1 (10)(6)3 12

2

Iy

1 (6)(10)3 12

2

M y zA Iy

4 4

( 38.567)(3) 178.429

(1) 2

178.429 in 4

(1) 4

(1) 2 (2.5) 2

459.16 in 4

(45.963)(5) 459.16

1.149 ksi (b)

B

M z yB Iz

M y zB Iy

( 38.567)(3) 178.429

A

(45.963)( 5) 459.16

0.1479 (c)

D

M z yD Iz

M y zD Iy

( 38.567)( 3) 178.429

1.149 ksi

B

0.1479 ksi

D

1.149 ksi

(45.963)( 5) 459.16

1.149 ksi


PROBLEM 4.132

y M 5 75 kip · in.


b 5 758 A 2.4 in.

B

1.6 in. z

C D 4 in. 4.8 in.

SOLUTION Iz Iy yA zA

(a)

M z yA Iz

M y zA

A

(b)

M z yB Iz

M y zB

B

(c)

M z yD Iz

M y zD

D

Iy

Iy

Iy

1 1 (4.8)(2.4)3 (4)(1.6)3 4.1643 in 4 12 12 1 1 (2.4)(4.8)3 (1.6)(4)3 13.5851 in 4 12 12 yB yD 1.2 in. zB

2.4 in.

zD

Mz

75sin15

19.4114 kip in.

My

75cos15

72.444 kip in.

(19.4114)(1.2) 4.1643

(72.444)(2.4) 13.5851

(19.4114)(1.2) 4.1643

(72.444)( 2.4) 13.5851

(19.4114)( 1.2) 4.1643

(72.444)( 2.4) 13.5851

A

B

D

7.20 ksi

18.39 ksi

7.20 ksi


PROBLEM 4.133

y

! " 30#

M " 100 N · m


B z

C D

A r " 20 mm

SOLUTION Iz

8

r4

2

4r 3

r2

(0.109757)(20) 4 Iy yA yB

A

8

8 9

17.5611 10

r4 3

mm 4

17.5611 10

(20) 4 62.832 10 3 mm 4 62.832 10 8 8 4r (4)(20) yD 8.4883 mm 3 3 20 8.4883 11.5117 mm r4

zA

(a)

2

20 mm

zD

zB

Mz

100 cos30

86.603 N m

My

100sin 30

50 N m

M z yA Iz

M y zA Iy

9

9

m4

m4

0

(86.603)( 8.4883 10 3 ) 17.5611 10 9

(50)(20 10 3 ) 62.832 10 9

57.8 106 Pa (b)

B

M z yB Iz

M y zB Iy

A

3

(86.603)(11.5117 10 ) 17.5611 10 9

(50)(0) 62.832 10

9

56.8 106 Pa (c)

D

M z yD Iz

M y zD Iy

56.8 MPa

B

3

(86.603)( 8.4883 10 ) 17.5611 10 9 25.9 106 Pa

57.8 MPa

3

(50)( 20 10 ) 62.832 10 9 D

25.9 MPa


W310 $ 38.7 15#

PROBLEM 4.134

B

A C

310 mm

The couple M is applied to a beam of the cross section shown in a plane with the vertical. Determine the stress at forming an angle (a) point A, (b) point B, (c) point D.

M " 16 kN · m D

E

165 mm

SOLUTION For W310 38.7 rolled steel shape,

(a)

tan

Iz tan I

84.9 10 7.20 10

Iz

84.9 106 mm 4

84.9 10 6 m 4

Iy

7.20 106 mm 4

7.20 10 6 m 4

yA

yB

1 yD 2

zA

zE

zB

yE zD

1 (310) 155 mm 2 1 (165) 82.5 mm 2

Mz

(16 103 ) cos 15

15.455 103 N m

My

(16 103 ) sin 15

4.1411 103 N m

6 6

tan 15

3.1596 72.4

72.4 15 57.4 (b)

Maximum tensile stress occurs at point E. E

M z yE Iz

M y zE Iy

(15.455 103 )( 155 10 3 ) 84.9 10 6 75.7 106 Pa

(4.1411 103 )(82.5 10 3 ) 7.20 10 6

75.7 MPa


PROBLEM 4.135

208

S6 3 12.5 A M 5 15 kip · in.

The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine (a) the angle that the neutral axis forms with the horizontal, (b) the maximum tensile stress in the beam.

B

C E

3.33 in. 6 in.

D

SOLUTION For S6 12.5 rolled steel shape, Iz

22.0 in 4

Iy

1.80 in 4

zE yA

(a)

tan

Iz tan Iy

zA

zB

yB

yD

zD yE

1 (3.33) 1.665 in. 2 1 (6) 3 in. 2

Mz

15 sin 20

5.1303 kip in.

My

15 cos 20

14.095 kip in.

22.0 tan (90 1.80

20 ) 33.58 88.29

88.29 (b)

70

18.29

Maximum tensile stress occurs at point D. D

M z yD Iz

M y zD Iy

(5.1303)( 3) 22.0

(14.095)(1.665) 13.74 ksi 1.80


PROBLEM 4.136

5#

C150 ! 12.2

B

A M " 6 kN · m C

152 mm


13 mm E

D

48.8 mm

SOLUTION My

6 sin 5

0.52293 kN m

Mz

6 cos 5

5.9772 kN m

C150 12.2

(a)

Iy

0.286 10

Iz

5.45 10

6

6

m4

m4

Neutral axis: tan

Iz tan Iy

tan

1.66718

5.45 10 6 m 4 tan 5 0.286 10 6 m 4 59.044

5 (b)

Maximum tensile stress at E: y M y zE E

Iy

Mz yE Iz

E

54.0

76 mm, z E

13 mm

(0.52293 kN m)(0.013 m) 0.286 10 6 m 4

(5.9772 kN m)( 0.076 m) 5.45 10 6 m 4 E

107.1 MPa


y'

30"

PROBLEM 4.137

B 50 mm

A

5 mm

5 mm C

M ! 400 N · m z'


D E

18.57 mm

5 mm 50 mm

Iy' ! 281 # 103 mm4 Iz' ! 176.9 # 103 mm4

SOLUTION Iz

176.9 103 mm 4

Iy

281 103 mm 4

yE

(a)

tan

Iz tan Iy

18.57 mm, z E

176.9 10 281 10

m4

m4

25 mm

Mz

400 cos 30

346.41 N m

My

400sin 30

200 N m

176.9 10 9 tan 30 281 10 9

9

9

0.36346

19.97 30 (b)

19.97

10.03

Maximum tensile stress occurs at point E. E

M z yE Iz

M y zE Iy

(346.41)( 18.57 10 3 ) 176.9 10 9 36.36 106

17.79 106

(200)(25 10 3 ) 281 10 9 54.2 106 Pa E

54.2 MPa


458

PROBLEM 4.138

y'

B

z'

0.859 in.

M 5 15 kip · in. C

A

1 2

in.

4 in.

D

4 in.


4 in.

Iy' 5 6.74 in4 Iz' 5 21.4 in4

SOLUTION Iz

21.4 in 4

zA

zB

yA Mz (a)

0.859 in.

4 in.

My

Iy yB

zD 4 in.

15 sin 45 15 cos 45

6.74 in 4 4 0.859 in. yD

3.141 in.

0.25 in.

10.6066 kip in. 10.6066 kip in.

Angle of neutral axis: tan

Iz tan Iy

21.4 tan ( 45 ) 3.1751 6.74

72.5 72.5

(b)

27.5

45

The maximum tensile stress occurs at point D. D

M z yD Iz

M y zD Iy

0.12391 4.9429

(10.6066)( 0.25) 21.4

( 10.6066)( 3.141) 6.74 D

5.07 ksi


y'

20"

PROBLEM 4.139

B 10 mm

A


6 mm

M ! 120 N · m C

z'

10 mm D

6 mm

Iy' ! 14.77 # 103 mm4 Iz' ! 53.6 # 103 mm4

E

10 mm 10 mm

SOLUTION

(a)

Iz

53.6 103 mm 4

Iy

14.77 103 mm 4

14.77 10 9 m 4

Mz

120 sin 70

112.763 N m

My

120 cos 70

41.042 N m

Angle of neutral axis:

20 tan

(b)

53.6 10 9 m 4

Iz tan Iy

53.6 10 9 tan 20 14.77 10 9

52.871 52.871

20

1.32084

32.9

The maximum tensile stress occurs at point E. yE zE E

16 mm 10 mm M z yE Iz

0.016 m 0.010 m M y zE Iy

(112.763)( 0.016) 53.6 10 9 61.448 106 Pa

(41.042)(0.010) 14.77 10 9 E

61.4 MPa


PROBLEM 4.140

208 A 90 mm

M 5 750 N · m C


B

30 mm

25 mm 25 mm

SOLUTION

(a)

My

750 sin 20

My

256.5 N m

Mz

750 cos 20

Mz

704.8 N m 1 (90)(25)3 12

0.2344 106 mm 4

Iy

2

Iy

0.2344 10 6 m 4

Iz

1 (50)(90)3 1.0125 106 mm 4 36

1.0125 10 6 m 4

Neutral axis: tan

Iz tan Iy

tan

1.5724 20

1.0125 10 6 m 4 tan 20 0.2344 106 m 4 57.5 57.5

30

37.5 37.5

(b)

Maximum tensile stress, at D: y D My zD D

Iy

Mz y D Iz

30 mm

z

(256.5 N m)( 0.030 m) 0.2344 10 6 m 4

32.83 MPa 17.40 MPa

50.23 MPa

25 mm (704.8 N m)(0.025 m) 1.0125 10 6 m 4 D

50.2 MPa


A

PROBLEM 4.141 y

z M ! 1.2 kN · m 10 mm

40 mm

The couple M acts in a vertical plane and is applied to a beam oriented as shown. Determine the stress at point A.

10 mm

C

40 mm

70 mm

10 mm

Iy ! 1.894 # 106 mm4 Iz ! 0.614 # 106 mm4 Iyz ! $0.800 # 106 mm4

SOLUTION Using Mohr’s circle, determine the principal axes and principal moments of inertia. Y : (1.894, 0.800) 106 mm 4 Z : (0.614, 0.800) 106 mm 4 E : (1.254, 0) 106 mm 4

R

tan 2

EF

2

FZ

2

0.6402

0.8002 10

6

1.0245 106 mm 4

Iv

(1.254 1.0245) 106 mm 4

0.2295 106 mm 4

0.2295 10 6 m 4

Iu

(1.254 1.0245) 106 mm 4

2.2785 106 mm 4

2.2785 10 6 m 4

m

Mv Mu

FZ 0.800 106 1.25 25.67 m FE 0.640 106 M cos m (1.2 103 ) cos 25.67 1.0816 103 N m M sin

(1.2 103 ) sin 25.67

m

0.5198 103 N m

uA

y A cos

m

z A sin

m

45 cos 25.67

45 sin 25.67

21.07 mm

vA

z A cos

m

y A sin

m

45 cos 25.67

45 sin 25.67

60.05 mm

A

M vuA Iv

M u vA Iu

(1.0816 103 )(21.07 10 3 ) 0.2295 10 6

113.0 106 Pa

( 0.5198 103 )(60.05 10 3 ) 2.2785 10 6 A

113.0 MPa


PROBLEM 4.142

y A 2.4 in.


2.4 in. z M ! 125 kip · in.

C

2.4 in. 2.4 in.

2.4 in. 2.4 in.

SOLUTION Iy

2

1 (7.2)(2.4)3 3

66.355 in 4

Iz

2

1 (2.4)(7.2)3 12

(2.4)(7.2)(1.2) 2

I yz

2 (2.4)(7.2)(1.2)(1.2)

199.066 in 4

49.766 in 4

Using Mohr’s circle, determine the principal axes and principal moments of inertia. Y : (66.355 in 4 , 49.766 in 4 ) Z : (199.066 in 4 ,

49.766 in 4 )

E : (132.710 in 4 , 0)



tan 2

m

2

m

R

DY 49.766 DE 66.355 36.87 18.435 m

DE

2

DY

2

82.944 in 4

Iu

132.710 82.944

49.766 in 4

Iv

132.710 82.944

215.654 in 4

Mu

125sin18.435

39.529 kip in.

Mv

125cos18.435

118.585 kip in.

uA

4.8 cos 18.435

2.4 sin 18.435

4.8 sin 18.435

A

M uA I

A

Mu Iu

2.4 cos 18.435

0.7589 in.

A

(118.585)(5.3126) 215.654

2.32 ksi

5.3126 in.

(39.529)(0.7589) 49.766 A

2.32 ksi


y

PROBLEM 4.143 1.08 in.

0.75 in.


2.08 in. z

C M 5 60 kip · in.

6 in. 0.75 in. A

4 in. Iy 5 8.7 in4 Iz 5 24.5 in4 Iyz 5 18.3 in4

SOLUTION Using Mohr’s circle, determine the principal axes and principal moments of inertia. Y : (8.7, 8.3) in 4 Z : (24.5,

8.3) in 4

E : (16.6, 0) in 4

EF

7.9 in 4

FZ

8.3 in 4

R

7.92

8.32

11.46 in 4

tan 2

m

FZ 8.3 1.0506 EF 7.9 I v 16.6 11.46 28.06 in 4

23.2

Iu

16.6 11.46 5.14 in 4

Mu

M sin

m

(60) sin 23.2

23.64 kip in.

Mv

M cos

m

(60) cos 23.2

55.15 kip in.

uA

y A cos

m

z A sin

m

3.92 cos 23.2

1.08 sin 23.2

4.03 in.

vA

z A cos

m

y A sin

m

1.08cos 23.2

3.92 sin 23.2

0.552 in.

m

A

M vuA Iv

M u vA Iu

(55.15)( 4.03) 28.06

(23.64)(0.552) 5.14

A

10.46 ksi


PROBLEM 4.144

D B

H 14 kN

G

E

A F

28 kN

The tube shown has a uniform wall thickness of 12 mm. For the loading given, determine (a) the stress at points A and B, (b) the point where the neutral axis intersects line ABD. 125 mm

75 mm

28 kN

SOLUTION Add y- and z-axes as shown. Cross section is a 75 mm cutout.

125-mm rectangle with a 51 mm

101-mm rectangular

1 1 (75)(125)3 (51)(101)3 7.8283 106 mm 4 7.8283 10 6 m 4 12 12 1 1 (125)(75)3 (101)(51)3 3.2781 103 mm 4 3.2781 10 6 m 4 Iy 12 12 A (75)(125) (51)(101) 4.224 103 mm 2 4.224 10 3 m 2 Iz

Resultant force and bending couples: 70 103 N

P 14 28 28 70 kN

(a)

A

Mz

(62.5 mm)(14 kN) (62.5 mm)(28kN) (62.5 mm)(28 kN)

2625 N m

My

(37.5 mm)(14 kN) (37.5 mm)(28 kN) (37.5 mm)(28 kN)

525 N m

P A

M z yA Iz

M y zA

70 103 4.224 10

Iy

3

(2625)( 0.0625) 7.8283 10 6

( 525)(0.0375) 3.2781 10 6

31.524 106 Pa B

P A

M z yB Iz

A

M y zB

70 103 4.224 10

Iy

3

(2625)(0.0625) 7.8283 10 6

( 525)(0.0375) 3.2781 10 6

10.39 106 Pa (b)

31.5 MPa

B

10.39 MPa

Let point H be the point where the neutral axis intersects AB. zH 0 yH

0.0375 m, P A

yH

M z yH Iz

Iz P Mz A

?,

H

0

M y zH

Mz H Iy

Iy 7.8283 10 2625

6

70 103 4.224 10

3

( 525)(0.0375) 3.2781 10 6

0.03151 m 31.51 mm

31.51 62.5 94.0 mm

Answer: 94.0 mm above point A.


PROBLEM 4.145 A horizontal load P of magnitude 100 kN is applied to the beam shown. Determine the largest distance a for which the maximum tensile stress in the beam does not exceed 75 MPa.

y

20 mm

a

20 mm O z

x

P

20 mm 60 mm 20 mm

SOLUTION Locate the centroid.

A, mm2

y , mm

2000

10

20 103

1200 3200

–10

12 103

Ay , mm3

Y

Ay A 8 103 3200 2.5 mm

8 103

Move coordinate origin to the centroid. Coordinates of load point:

XP

a,

Bending couples:

Mx

yP P

Ix

1 (100)(20)3 12

(2000)(7.5)2

yP

2.5 mm My

1 (60)(20)3 12

aP (1200)(12.5)2

0.40667 106 mm 4 0.40667 10

Iy

1 1 (20)(100)3 (20)(60)3 12 12 P M x y M yx A Ix Iy

2.0267 106 mm 4 A

6

m4

2.0267 10 6 m 4

75 106 Pa, P

100 103 N



My

Iy P x A

My

2.0267 10 50 10 3

Mxy Ix 6

For point A, 100 103 3200 10 6

2.0267 10 6 {31.25 50 10 3 a

My P

50 mm, y

( 2.5)(100 103 )( 2.5 10 3 ) 0.40667 10 6

1.537

(1.7111 103 ) 100 103

x

75} 106

2.5 mm 75 106

1.7111 103 N m

17.11 103 m

a

17.11 mm


1 in.

PROBLEM 4.146

1 in. 4 in. 1 in.

5 in.

P

Knowing that P 90 kips, determine the largest distance a for which the maximum compressive stress dose not exceed 18 ksi.

a 2.5 in.

SOLUTION

A Ix Iz

(5 in.)(6 in.)

2(2 in.)(4 in.)

14 in 2

1 1 (5 in.)(6 in.)3 2 (2 in.)(4 in.)3 12 12 1 1 2 (1 in.)(5 in.)3 (4 in.)(1 in.)3 12 12

21.17 in 4

Force-couple system at C:

P

P

For P

P

90 kips M x

90 kips:

Mx

68.67 in 4

Maximum compressive stress at B: B

18 ksi 18 1.741

P A

M x (3 in.) Ix

90 kips 14 in 2 6.429

B

Mz

Pa

(90 kips)(2.5 in.)

225 kip in.

Mz

(90 kips) a

18 ksi

M z (2.5 in.) Iz

(225 kip in.)(3 in.) 68.67 in 4 9.830

P(2.5 in.)

(90 kips) a (2.5 in.) 21.17 in 4

10.628a

10.628 a

a

0.1638 in.


1 in.

PROBLEM 4.147

1 in. 4 in. 1 in.

P

5 in.

Knowing that a 1.25 in., determine the largest value of P that can be applied without exceeding either of the following allowable stresses:

a 2.5 in.

10 ksi

ten

18 ksi

comp

SOLUTION A Ix Iz

(5 in.)(6 in.)

(2)(2 in.)(4 in.)

14 in 2

1 1 (5 in.)(6 in.)3 2 (2 in.)(4 in.)3 68.67 in 4 12 12 1 1 2 (1 in.)(5 in.)3 (4 in.)(1 in.)3 21.17 in 4 12 12 For a

Force-couple system at C: P My

P(2.5 in.)

P Mx Pa

1.25 in.,

(1.25 in.)

Maximum compressive stress at B: B

18 ksi 18 18

P A

M x (3 in.) Ix

P 14 in 2 0.0714P 0.3282P

P(2.5 in.)(3 in.) 68.67 in 4 0.1092 P P

10 ksi 10

P A

M x (3 in.) Ix

0.0714P 0.1854 P

P(1.25 in.)(2.5 in.) 21.17 in 4

0.1476 P

54.8 kips D

10 ksi

M z (2.5 in.) Iz

0.1092 P P

18 ksi

M z (2.5 in.) Iz

Maximum tensile stress at D: D

B

0.1476 P

53.9 kips

The smaller value of P is the largest allowable value. P

53.9 kips


y R ! 125 mm C z

P ! 4 kN

PROBLEM 4.148

E %

A rigid circular plate of 125-mm radius is attached to a solid 150 200-mm rectangular post, with the center of the plate directly above 30 , the center of the post. If a 4-kN force P is applied at E with determine (a) the stress at point A, (b) the stress at point B, (c) the point where the neutral axis intersects line ABD.

x A

D B 200 mm 150 mm

SOLUTION 4 103 N (compression)

P Mx Mz

Iz xA A

(a)

A

P A

56.25 106 mm 4 100 106 mm 4 zA

zB

4 103 30 10 3

M z xA Iz

56.25 10 100 10

6

6

B

P A

M x zB Ix

30 10

3

m2

( 433)( 100 10 3 ) 100 10 6

( 250)(75 10 3 ) 56.25 10 6

633 103 Pa

633 kPa

( 433)(100 10 3 ) 100 10 6 B

(c)

m4

75 mm

( 250)(75 10 3 ) 56.25 10 6

4 103 30 10 3

M z xB Iz

433 N m

m4

A

(b)

250 N m

3

(4 10 )(125 10 ) cos 30

30 103 mm 2

(200)(150)

M xzA Ix

3

PR cos 30

1 (200)(150)3 12 1 (150)(200)3 12 xB 100 mm

Ix

(4 103 )(125 10 3 )sin 30

PR sin 30

233 103 Pa

233 kPa

Let G be the point on AB where the neutral axis intersects. G

0 P A

G

xG

zG

75 mm

M x zG Ix

Iz P Mz A

46.2 10

M z xG Iz

M x zG Ix 3

m

xG

?

0

100 10 433

6

4 103 30 10 3

46.2 mm

( 250)(75 10 3 ) 56.25 10 6

Point G lies 146.2 mm from point A.


y R ! 125 mm C z

P ! 4 kN

PROBLEM 4.149

E %

In Prob. 4.148, determine (a) the value of θ for which the stress at D reaches it largest value, (b) the corresponding values of the stress at A, B, C, and D.

x A

D

PROBLEM 4.148 A rigid circular plate of 125-mm radius is attached to a solid 150 200-mm rectangular post, with the center of the plate directly above the center of the post. If a 4-kN force P is applied at E with 30 , determine (a) the stress at point A, (b) the stress at point B, (c) the point where the neutral axis intersects line ABD.

B 200 mm 150 mm

SOLUTION (a)

4 103 N

P

PR sin

Mx

Iz xD

For

M zx Iz

d d

to be a maximum, d d

D

P 0

RxD sin IZ

tan

I z zD I x xD

sin A

P A

0

Rz D cos Ix

sin cos

(b)

1 A

P

M xzA Ix

M z xA Iz

PR cos

Mx

30 103 mm 2

(200)(150)

A M xz Ix

500sin

500 N m 500 cos

1 (200)(150)3 56.25 106 mm 4 56.25 10 6 m 4 2 1 (150)(200)3 100 106 mm 4 100 10 6 m 4 2 100 mm 75 mm zD

Ix

P A

(4 103 )(125 10 3 )

PR

0.8,

Rz sin Ix

Rx cos Iz

with z

zD , x

30 10 3 m 2

xD

0

(100 10 6 )( 75 10 3 ) (56.25 10 6 )(100 10 3 ) cos

4 103 30 10 3

4 3 53.1

0.6

(500)(0.8)(75 10 3 ) 56.25 10 6

(500)(0.6)( 100 10 3 ) 100 10 6

( 0.13333

0.53333

0.300) 106 Pa

0.700 106 Pa

A

700 kPa

B

( 0.13333

0.53333

0.300) 106 Pa

0.100 106 Pa

B

100 kPa

C

( 0.13333

0

D

( 0.13333

0.53333

0) 106 Pa

133.3 kPa

C

0.300) 106 Pa

0.967 106 Pa

D

967 kPa


y

PROBLEM 4.150

0.5 in. 1.43 in.

z

M0

C

5 in.

0.5 in. 1.43 in.

A beam having the cross section shown is subjected to a couple M 0 that acts in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress in the beam is not to exceed 12 ksi. Given: I y I z 11.3 in 4 , A 4.75 in 2 , kmin 0.983 in. (Hint: By reason of symmetry, the principal axes form 2 an angle of 45 with the coordinate axes. Use the relations I min Akmin and I min I max I y I z .)

5 in.

SOLUTION Mu

M 0 sin 45

0.70711M 0

Mv

M 0 cos 45

0.70711M 0

I min

2 Akmin

I max

Iy

(4.75)(0.983)2

Iz

11.3

I min

11.3

uB

yB cos 45

z B sin 45

vB

z B cos 45

yB sin 45

B

M vu B Iv

M u vB Iu

0.70711M 0

M0

B

0.4124

4.59 in 4 3.57 cos 45 0.93 cos 45

0.70711M 0

( 1.866) 4.59

18.01 in 4

4.59

3.182 18.01

12 0.4124

uB I min

0.93 sin 45

1.866 in.

( 3.57) sin 45

3.182 in.

vB I max

0.4124M 0

M0

29.1 kip in.


PROBLEM 4.151

y

0.5 in.

Solve Prob. 4.150, assuming that the couple M 0 acts in a horizontal plane.

1.43 in. z

M0

C

5 in.

0.5 in. 1.43 in. 5 in.

PROBLEM 4.150 A beam having the cross section shown is subjected to a couple M 0 that acts in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress in the beam is not to exceed 12 ksi. Given: I y I z 11.3 in 4, A 4.75 in 2, kmin 0.983 in. (Hint: By reason of symmetry, the principal axes form an angle of 45 with the coordinate axes. Use the 2 relations I min Akmin and I min I max I y I z .)

SOLUTION Mu Mv

M 0 cos 45

0.70711M 0

M 0 sin 45

I min

2 Akmin

I max

Iy

0.70711M 0

(4.75)(0.983)2

Iz

11.3

I min

4.59 in 4 11.3

4.59

18.01 in 4

uD

yD cos 45

z D sin 45

0.93 cos 45

( 3.57 sin 45 )

vD

z D cos 45

yD sin 45

( 3.57) cos 45

(0.93) sin 45

D

M vu D Iv

M u vD Iu

0.70711M 0

M0

D

0.4124

0.70711M 0

( 1.866) 4.59

3.182 18.01

12 0.4124

uD I min

1.866 in. 3.182 in.

vD I max

0.4124M 0

M0

29.1 kip in.


PROBLEM 4.152

y z

M0

40 mm 10 mm

C

10 mm

40 mm

70 mm

10 mm

The Z section shown is subjected to a couple M 0 acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 80 MPa. Given: I max 2.28 10 6 mm 4 , I min 0.23 10 6 mm 4 , principal axes 25.7 and 64.3 .

SOLUTION Iv

I max

2.28 106 mm 4

2.28 10 6 m 4

Iu

I min

0.23 106 mm 4

0.23 10 6 m 4

Mv

M 0 cos 64.3

Mu

M 0 sin 64.3 64.3

tan

Iv tan Iu 2.28 10 0.23 10 20.597

6 6

tan 64.3

87.22 Points A and B are farthest from the neutral axis. uB

yB cos 64.3

z B sin 64.3

( 45) cos 64.3

( 35) sin 64.3

yB sin 64.3

( 35) cos 64.3

( 45) sin 64.3

51.05 mm vB

z B cos 64.3 25.37 mm

B

80 106

M vu B Iv

M u vB Iu

(M 0 cos 64.3 )( 51.05 10 3 ) 2.28 10 6

(M 0 sin 64.3 )(25.37 10 3 ) 0.23 10 6

109.1 103 M 0 M0

80 106 109.1 103

M0

733 N m


PROBLEM 4.153 y z

M0

40 mm 10 mm

C

10 mm

40 mm

70 mm

10 mm

Solve Prob. 4.152 assuming that the couple M 0 acts in a horizontal plane.

PROBLEM 4.152 The Z section shown is subjected to a couple M 0 acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 80 MPa. Given: I max 2.28 10 6 mm 4 , I min 0.23 10 6 mm 4 , principal axes 25.7 and 64.3 .

SOLUTION Iv

I min

0.23 106 mm 4

0.23 106 m 4

Iu

I max

2.28 106 mm 4

2.28 106 m 4

Mv

M 0 cos 64.3

Mu

M 0 sin 64.3 64.3

tan

Iv tan Iu 0.23 10 2.28 10 0.20961

6 6

tan 64.3

11.84 Points D and E are farthest from the neutral axis. uD

yD cos 25.7

z D sin 25.7

( 5) cos 25.7

yD sin 25.7

45cos 25.7

45 sin 25.7

24.02 mm vD

z D cos 25.7

( 5) sin 25.7

38.38 mm D

M vu D Iv

M u vD Iu

(M D cos 64.3 )( 24.02 10 3 ) 0.23 10 6

(M 0 sin 64.3 )(38.38 10 3 ) 2.28 10 6 80 106 M0

60.48 103 M 0 1.323 103 N m

M0

1.323 kN m


PROBLEM 4.154

y

0.3 in.

M0

1.5 in.

C

z 0.3 in.

An extruded aluminum member having the cross section shown is subjected to a couple acting in a vertical plane. Determine the largest permissible value of the moment M 0 of the couple if the maximum stress is not to exceed 12 ksi. Given: I max 0.957 in 4 , I min 0.427 in 4 , principal axes 29.4 and 60.6

0.6 in. 1.5 in. 0.6 in.

SOLUTION Iu

I max

0.957 in 4

Iv

I min

0.427 in 4

Mu

M 0 sin 29.4 , M v

M 0 cos 29.4

29.4 tan

Iv tan Iu

0.427 tan 29.4 0.957

0.2514

14.11

Point A is farthest from the neutral axis. yA

0.75 in., z A

uA

y A cos 29.4

z A sin 29.4

1.0216 in .

vA

z A cos 29.4

y A sin 29.4

0.2852 in .

A

M vu A Iv

M uVA Iu

(M 0 cos 29.4 )( 1.0216) 0.427

0.75 in.

(M 0 sin 29.4 )( 0.2852) 0.957

1.9381 M 0

M0

A

1.9381

12 1.9381

M0

6.19 kip in.


PROBLEM 4.155

y 20 mm z

A beam having the cross section shown is subjected to a couple M0 acting in a vertical plane. Determine the largest permissible value of the moment M0 of the couple if the maximum stress is not to exceed 100 MPa. Given: I y I z b 4 /36 and I yz b 4 /72.

M0 C

b = 60 mm

20 mm b = 60 mm

SOLUTION Iy I yz

b 4 604 0.360 106 mm 4 36 36 b 4 604 0.180 106 mm 4 72 72

Iz

Principal axes are symmetry axes. Using Mohr’s circle, determine the principal moments of inertia. R Iv

Iu

Mu

0.180 106 mm 4

I yz Iy

Iz

R 2 0.540 106 mm 4 I y Iz R 2 0.180 106 mm 4

M 0 sin 45 45

0.540 10 6 m 4

0.180 10 6 m 4

0.70711M 0 , Iv tan Iu

tan

Mv

0.540 10 0.180 10

M 0 cos 45

0.70711M 0

6 6

tan 45

3

71.56 Point A:

uA A

M0

0

vA M vuA Iv

20 2 mm M u vA Iu

A

111.11 103

0

(0.70711M 0 )( 20 2 103 ) 0.180 10

100 106 111.11 103

6

11.11 10 3 M 0

900 N m



Point B:

60

uB

2

mm,

M v uB Iv

B

vB M u vB Iu

20 2

mm

(0.70711M 0 )

60 2 6

0.540 10

10

3

(0.70711M 0 )

20 2

0.180 10

10

3

6

111.11 103 M 0 M0

B

111.11 10

3

100 106 111.11 10

3

900 N m M0


900 N m


b

PROBLEM 4.156

B

A

Show that, if a solid rectangular beam is bent by a couple applied in a plane containing one diagonal of a rectangular cross section, the neutral axis will lie along the other diagonal.

h

M C

D E

SOLUTION tan Mz Iz

tan

b h M cos , M z 1 3 bh 12

Iz tan Iy

M sin Iy

1 3 hb 12

1 3 bh b 12 1 3 h hb 12

h b

The neutral axis passes through corner A of the diagonal AD.


PROBLEM 4.157 y

(a) Show that the stress at corner A of the prismatic member shown in part a of the figure will be zero if the vertical force P is applied at a point located on the line

D

A A D

P B

z x

z

C

h 6

x

x b /6

h B

b

(a)

C

1

(b) Further show that, if no tensile stress is to occur in the member, the force P must be applied at a point located within the area bounded by the line found in part a and three similar lines corresponding to the condition of zero stress at B, C, and D, respectively. This area, shown in part b of the figure, is known as the kern of the cross section.

b 6

(b)

z h /6

SOLUTION 1 3 hb Ix 12 h xA 2

Iz zA

1 3 bh 12 b 2

A

bh

Let P be the load point. Mz A

x b/6

z h/6

0,

1

At point E:

z

0

xE

b /6

At point F:

x

0

zF

h/6

(a)

For

(b)

A

0

PxP

Mx

Pz P

P A

M z xA Iz

P bh

( PxP ) 1 12

M xzA Ix hb

b 2 3

( Pz P 1 bh 12

P 1 bh

xP b /6

zP h/6

x b/6

z h/6

1

3

h) 2

If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A. By considering

B

0,

C

0, and

D

0, the other portions producing tensile stresses are identified.


PROBLEM 4.158 y

A beam of unsymmetric cross section is subjected to a couple M 0 acting in the horizontal plane xz. Show that the stress at point A, of coordinates y and z, is

z

A y

C

z

A

zI z I yIz

yI yz 2 I yz

My

x

where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section with respect to the coordinate axes, and My the moment of the couple.

SOLUTION The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are centroidal axes: A

C1 y

C2 z

where C1 and C2 are constants. Mz

y I zC1

C1 My

I yz Iz z

I yzC2

AdA

I yz

C2 C1

A

C2 yz dA

0

C2

I yzC1

IzM y

C1 y 2dA

AdA

C1 yz dA

C2 z 2dA

I y C2 I yz Iz

C2

I y C2

2 I yz C2

I yIz IzM y I yIz

2 I yz

I yz M y I yIz Izz I yIz

2 I yz

I yz y 2 I yz

My


PROBLEM 4.159 y

A beam of unsymmetric cross section is subjected to a couple M 0 acting in the vertical plane xy. Show that the stress at point A, of coordinates y and z, is

z

A y

C

z

A

yI y

zI yz

I yIz

2 I yz

Mz

x

where I y , I z , and I yz denote the moments and product of inertia of the cross section with respect to the coordinate axes, and M z the moment of the couple.

SOLUTION The stress A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are centroidal axes: A

C1 y

C2 z

where C1 and C2 are constants. My

z

AdA

I yzC1 C2 Mz

I yz Iy

I y C2

C1 C2

A

C2 z 2dA

0

C1

y

C1 y 2dA

Adz

I z C1 I yM z

C1 yz dA

I yz

I yz Iy

C2 yz dA

C1

2 I yz C1

I yIz I yM z I yIz

2 I yz

I yz M z I yIz

2 I yz

Iyy

I yz 2

I yIz

2 I yz

Mz


PROBLEM 4.160

y

(a) Show that, if a vertical force P is applied at point A of the section shown, the equation of the neutral axis BD is

D

B

P

C A

xA

z

xA x rz2

x

zA

zA z rx2

1

where rz and rx denote the radius of gyration of the cross section with respect to the z axis and the x axis, respectively. (b) Further show that, if a vertical force Q is applied at any point located on line BD, the stress at point A will be zero.

SOLUTION Definitions: Ix , rz2 A

rx2 (a)

Mx

Pz A

E

P A

Mz

Px A

M z xE Iz

M x zE Ix

P 1 A

xA xE rz2

P A zA zE rx2

Px A xE Arz2

Iz A

Pz A z E Arx2

0

if E lies on neutral axis. 1

(b)

Mx

Pz E

A

P A

Mz M z xA Iz

xA x rz2

zA z rx2

0,

xA x rz2

P A

PxE x A Arz2

Pz E z A Arx2

zA z rx2

1

PxE M x zA Iy

0 by equation from part (a).


24 mm B

PROBLEM 4.161

B

For the curved bar shown, determine the stress at point A when (a) h 50 mm, (b) h 60 mm.

h A 600 N · m

A

50 mm

600 N · m

C

SOLUTION (a)

h

50 mm,

A

(24)(50) h r ln 2 r1

R

r e yA A

r1

50 mm,

r2

100 mm

1.200 103 mm 50 ln 100 50

72.13475 mm

1 (r1 r2 ) 75 mm 2 r R 2.8652 mm 72.13475

50

22.13475 mm

rA

50 mm 3

My A AerA

(600)(22.13475 10 ) (1.200 10 3 )(2.8652 10 3 )(50 10 3 )

77.3 106 Pa

77.3 MPa (b)

h R

r yA A

60 mm, h r ln r2 1

r1 60 ln 110 50

50 mm,

r2

110 mm,

A

(24)(60)

1.440 103 mm 2

76.09796 mm

1 (r1 r2 ) 80 mm e r R 3.90204 mm 2 76.09796 50 26.09796 mm rA 50 mm M yA AerA

(600)(26.09796 10 3 ) (1.440 10 3 )(3.90204 10 3 )(50 10 3 )

55.7 106 Pa 55.7 MPa


24 mm B

B

A

A

PROBLEM 4.162 For the curved bar shown, determine the stress at points A and B when h 55 mm.

h 600 N · m

C

50 mm

600 N · m

SOLUTION h A R

r e yA A

55 mm, (24)(55)

r1

50 mm, 3

1.320 10 mm

r2

105 mm

2

50 h 74.13025 mm 105 r2 ln ln 50 r1 1 (r1 r2 ) 77.5 mm 2 r R 3.36975 mm 74.13025 M yA AerA

50

24.13025 mm

rA

50 mm

(600)(24.13025 10 3 ) (1.320 10 3 )(3.36975 10 3 )(50 10 3 )

65.1 106 Pa

65.1 MPa

yB B

74.13025 MyB AerB

105

30.86975 mm

rB

105 mm 3

(600)( 30.8697 10 ) (1.320 10 3 )(3.36975 10 3 )(105 10 3 )

39.7 106 Pa 39.7 MPa


4 kip · in.

PROBLEM 4.163

C

4 kip · in.

3 in.

For the machine component and loading shown, determine the stress at point A when (a) h 2 in., (b) h 2.6 in.

h A

0.75 in.

B

SOLUTION M

4 kip in.

A

Rectangular cross section:

(a)

r

1 (r1 2

h

2 in.

r1

3

R

2 ln 13

2

(b)

M (r R ) Aer

h

2.6 in.

r1

3

R

2.6 3 ln 0.4

2.6

r1

e

2

r

r1

1)

1.8205

2 in.

0.1795 in.

r e

12.19 ksi A

12.19 ksi

A

11.15 ksi

1.95 in 2

0.4 in.

M (r R ) Aer

1 (3 2

1 in.

(0.75)(2.6)

1.2904 in.

h

R

( 4)(1 1.8205) (1.5)(0.1795)(1)

A

At point A: A

r

1.8205 in.

r2

1.5 in 2

(0.75)(2)

r

3 in. r1 r

1 in.

At point A: A

h , e r ln 2 r1

r2 ), R

A

bh r2

1.7

1 (3 2

0.4)

1.2904

1.7 in.

0.4906 in.

0.4 in.

( 4)(0.4 1.2904) (1.95)(0.4096)(0.4)

11.15 ksi


4 kip · in. 4 kip · in.

PROBLEM 4.164

C

3 in.

For the machine component and loading shown, determine the stress at points A and B when h 2.5 in.

h A

0.75 in.

B

SOLUTION M

4 kip in.

Rectangular cross section:

h

2.5 in. b

r2

3 in. r1

r R

At point A:

At point B:

1 (r1 2 h ln

r2

r2 ) 2.5 ln 3.0 0.5

r2 r1

1.75

e

r

R

r

r1

0.5 in.

A

M (r R) Aer

r

r2

B

M (r R ) Aer

1.875 in.2

0.75 in. A

h

0.5 in.

1 (0.5 2

3.0)

1.75 in.

1.3953 in. 1.3953

0.3547 in.

( 4 kip in.)(0.5 in. 1.3953 in.) (0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)

A

10.77 ksi

( 4 kip in.)(3 in. 1.3953 in.) (0.75 in. 2.5 in.)(0.3547 in.)(3 in.)

B

3.22 ksi

3 in.


PROBLEM 4.165 40 mm

r1

The curved bar shown has a cross section of 40 60 mm and an inner radius r1 15 mm. For the loading shown, determine the largest tensile and compressive stresses.

60 mm

120 N · m

SOLUTION h

40 mm, r1

A

(60)(40)

R

At r

h r ln 2 r1

r

1 (r1 2

e

r

2400 mm 2 40 55 ln 40

r2 )

R

15 mm,

15 mm, r2

55 mm

2400 10 6 m 2

30.786 mm

35 mm

My Aer

4.214 mm y

30.786

15

15.756 mm

(120)(15.786 10 3 ) (2400 10 6 )(4.214 10 3 )(15 10 3 )

12.49 10

6

Pa

12.49 MPa At r

55 mm,

y

30.786

55

24.214 mm

(120)( 24.214 10 3 ) (2400 10 6 )(4.214 10 3 )(55 10 3 )

5.22 106 Pa

5.22 MPa


PROBLEM 4.166 40 mm

r1

For the curved bar and loading shown, determine the percent error introduced in the computation of the maximum stress by assuming that the bar is straight. Consider the case when (a) r1 20 mm, (b) r1 200 mm, (c) r1 2 m.

60 mm

120 N · m

SOLUTION h

40 mm, A

I

1 3 bh 12

(60)(40)

1 (60)(40)3 12

2400 mm 2

2400 10

0.32 106 mm 4

6

m2 , M

120 N m

0.32 10 6 mm 4 ,

c

1 h 2

20 mm

Assuming that the bar is straight, s

(a)

r1 R

(120)(20 10 8 ) (0.32 10 6 )

Mc I

20 mm r2 h r ln 2 r1

1 (r1 2

a

M (r1 R) Aer

r2 )

% error

7.5 MPa

60 mm

40 60 ln 20

r

7.5 106 Pa

36.4096 mm

e

40 mm

r1

R

r

R

16.4096 mm

3.5904 mm

(120)( 16.4096 10 3 ) (2400 10 6 )(3.5904 10 3 )(20 10 3 )

11.426 ( 7.5) 11.426

100%

11.426 106 Pa

11.426 MPa

34.4%

For parts (b) and (c), we get the values in the table below:

(a) (b) (c)

r1, mm

r2 , mm

20 200 2000

60 240 2040

R, mm

36.4096 219.3926 2019.9340

r , mm 40 220 2020

e, mm 3.5904 0.6074 0.0660

, MPa

11.426 7.982 7.546

% error 34.4 % 6.0 % 0.6 %


0.3 in. B 0.4 in. P9

PROBLEM 4.167

B

P

0.4 in.

Steel links having the cross section shown are available with different central angles . Knowing that the allowable stress is 12 ksi, determine the largest force P that can be applied to a link for which 90 .

0.8 in.

A

A 0.8 in.

b 1.2 in. C

C

SOLUTION Reduce section force to a force-couple system at G, the centroid of the cross section AB. a The bending couple is M

r 1

cos

2

Pa.

For the rectangular section, the neutral axis for bending couple only lies at

h

R Also, e

r

.

r2 r1

ln

R

At point A, the tensile stress is where

A

K

1

r

1.2 in., r1

A

(0.3)(0.8)

e

1.2

a

1.2(1

K

1

P

(0.24)(12) 4.3940

My A Aer1

ay A er1

P A

Pay A Aer1

yA

and

R

P 1 A

ay A er1

K

P A

r1

A A K

P Data:

P A

0.8 in., r2

1.6 in., h

0.24 in 2

1.154156 cos 45 )

R

0.045844 in.,

yA

0.8 in., b 0.3 in. 0.8 1.154156 in. ln 1.6 0.8 1.154156

0.8

0.35416 in.

0.35147 in.

(0.35147)(0.35416) (0.045844)(0.8)

4.3940

P

0.65544 kips

655 lb


PROBLEM 4.168

0.3 in. B

B

0.4 in. P9

P

0.4 in. A

Solve Prob. 4.167, assuming that

PROBLEM 4.167 Steel links having the cross section shown are available with different central angles . Knowing that the allowable stress is 12 ksi, determine the largest force P that can be applied to a link for which 90 .

A 0.8 in.

b 1.2 in. C

60 .

0.8 in.

C

SOLUTION Reduce section force to a force-couple system at G, the centroid of the cross section AB. a The bending couple is M

r 1

cos

2

Pa.

For the rectangular section, the neutral axis for bending couple only lies at h

R Also, e

r

r2 r1

.

R

At point A, the tensile stress is

r

1.2 in., r1

A

(0.3)(0.8)

e

1.2

a

(1.2)(1

K

1

P

(0.24)(12) 2.5525

P A

A

where

Data:

ln

My A Aer1 ay A er1

K

1

P

A A K

0.8 in., r2

P A and

1.6 in., h

0.24 in 2

1.154156 cos 30 )

0.045844 in.

Pay A Aer1

P 1 A

yA

R

ay A er1

K

P A

r1

0.8 in., b 0.3 in. 0.8 1.154156 in. R ln 1.6 0.8 yA

1.154156

0.8

0.35416 in.

0.160770 in.

(0.160770)(0.35416) (0.045844)(0.8)

2.5525

P

1.128 kips

1128 lb


5 kN

PROBLEM 4.169

a

The curved bar shown has a cross section of 30 30 mm. Knowing that the allowable compressive stress is 175 MPa, determine the largest allowable distance a.

30 mm B

A

20 mm 20 mm

C

30 mm

5 kN

SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M

P(a

r)

Also, e

r

h . r2 ln r1

R

For the rectangular section, the neutral axis for bending couple only lies at

R

The maximum compressive stress occurs at point A . It is given by P A

A

K Thus, K Data:

h

30 mm, r1

e

35

a

r,

50 mm, r

2.2593 mm, b 6

175 10 Pa, P 6

a

r

r

(30.5)(2.2593)(20) 12.7407

a

108.17 mm

P A

35 mm, R

5 kN

P (a

P with y A R A (a r )( R r1) er1

30 mm, R

r1

30 ln 50 20

r ) yA Aer1

r1 (1)

32.7407 mm

12.7407 mm, a

?

3

5 10 N

6

(900 10 )( 175 10 ) 5 103

A A P

K

a

32.7407

175 MPa

A

Solving (1) for

20 mm, r2

1

My A Aer1

31.5

( K 1)er1 R r1

108.17 mm

35 mm

a

73.2 mm


PROBLEM 4.170

2500 N

For the split ring shown, determine the stress at (a) point A, (b) point B.

90 mm 40 mm B A 14 mm

SOLUTION r1

1 40 2

20 mm,

r2

A (14)(25) 350 mm 2

1 (90) 45 mm 2 25 h R r2 45 ln r ln 20

h

r2

r1

25 mm

30.8288 mm

1

r

1 (r1 2

r2 ) 32.5 mm

e

r

R 1.6712 mm

Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross section. The bending couple is

M

(a)

Pa

rA

Point A: A

P A

(2500)(32.5 10 3 ) 81.25 N m

Pr

20 mm

My A AeR

yA

2500 350 10

30.8288 20 10.8288 mm (81.25)(10.8288 10 3 ) (350 10 6 )(1.6712 10 3 )(20 10 3 )

6

82.4 106 Pa

(b)

Point B: B

P A

rB

45 mm

MyB AerB

82.4 MPa

A

yB

2500 350 10

30.8288 45 6

14.1712 mm

(81.25)( 14.1712 10 3 ) (350 10 6 )(1.6712 10 3 )(45 10 3 )

36.6 106 Pa

B

36.6 MPa


2 in.

PROBLEM 4.171

B

0.5 in. 0.5 in.

2 in. 0.5 in.

A

3 in.

M

M'

3 in.

Three plates are welded together to form the curved beam shown. For M 8 kip in., determine the stress at (a) point A, (b) point B, (c) the centroid of the cross section.

C

SOLUTION A dA

R

1 r

Part

b

3.5 5.5 6

(a)

yA

R

yB

R

yC C

R

3.25

4.875

1.0

0.225993

4.5

4.5

1.0

0.174023

5.75

5.75

3.5

0.862468

0.5

1.5

0.5

2

2

0.5

b ln

3.5 4.05812 in., 0.862468 r R 0.26331 in.

3

r M

15.125 15.125 4.32143 in. 3.5 8 kip in.

1.05812 in.

( 8)(1.05812) (3.5)(0.26331)(3)

4.05812

MyB Aer2

B

(c)

r2

0.462452

3

4.05812

My A Aer1

A

(b)

r1

Ar

A

e

ri 1 ri

r

h

R

A r bi ln i 1 ri

Ari A

r

r 3

bi hi r bi ln i 1 ri

r MyC Aer

6

A

3.06 ksi

1.94188 in.

( 8)( 1.94188) (3.5)(0.26331)(6)

B

2.81 ksi

C

0.529 ksi

e Me Aer

M Ar

8 (3.5)(4.32143)


2 in.

PROBLEM 4.172

B

0.5 in. 0.5 in.

2 in. 0.5 in.

A

3 in.

M

M'

3 in.

Three plates are welded together to form the curved beam shown. For the given loading, determine the distance e between the neutral axis and the centroid of the cross section.

C

SOLUTION A dA

R

b

Ar

0.462452

3.25

4.875

1.0

0.225993

4.5

4.5

1.0

0.174023

5.75

5.75

3.5

0.862468

A

3

0.5

1.5

5.5

0.5

2

6

2

0.5

3.5

R e

ri 1 ri

r

h

r 3

A r bi ln i 1 ri

Ari A

r

Part


1 r

b ln

3.5 4.05812 in., r 0.862468 r R 0.26331 in.

15.125 3.5

15.125 4.32143 in.

e

0.263 in.


PROBLEM 4.173

C M

A

Mʹ

150 mm

A

Knowing that the maximum allowable stress is 45 MPa, determine the magnitude of the largest moment M that can be applied to the components shown.

45 mm 135 mm

B

B

36 mm

SOLUTION A dA

R

1 r

bi hi r bi 2 ln i 1 ri

Ai r bi ln i 1 ri

Ai ri Ai

r

r, mm 150

Part bi , mm h, mm

195 330

R e yA

r1

ri 1 , mm ri

r , mm

Ar , mm3

108

45

4860

28.3353

172.5

838.35 103

36

135

4860

18.9394

262.5

1275.75 103

9720

47.2747

r

2114.1 103 9720

9720 205.606 mm 47.2747 r R 11.894 mm R

bi ln

A, mm 2

205.606

150

55.606 mm

A Aer1

M

yA (45 106 )(9720 10 6 )(11.894 10 3 )(150 10 3 ) (55.606 10 3 )

B

M

217.5 mm

My A Aer1

A

yB

2114.1 103

R

r2

205.606

330

14.03 kN m

124.394 mm

MyB Aer2 B Aer2

yB (45 106 )(9720 10 6 )(11.894 10 3 )(330 10 3 ) (124.394 10 3 ) 13.80 kN m

M

13.80 kN m


C M

A

PROBLEM 4.174 Mʹ

150 mm

A

B

B

Knowing that the maximum allowable stress is 45 MPa, determine the magnitude of the largest moment M that can be applied to the components shown.

135 mm 45 mm

36 mm

SOLUTION A dA

R

1 r


Ai r bi ln i 1 ri

Ai ri Ai

r

bi , mm h, mm

r, mm 150 285

bi ln

A, mm 2

ri 1 , mm ri2

135

4860

23.1067

217.5

1.05705 106

108

45

4860

15.8332

307.5

1.49445 106

9720

38.9399

R

9720 38.9399

e

r

R

12.885 mm,

yA

R

r1

249.615

249.615 mm,

M

150

2.5515 106 9720

r

262.5 mm

20 103 N m

99.615 mm

A Aer1

M

yA (45 106 )(9720 10 6 )(12.885 10 3 )(150 10 3 ) (99.615 10 3 )

M

2.5515 106

My A Aer1

A

B

Ar , mm3

36

330

yB

r , mm

R

r2

249.615

330

8.49 kN m

80.385 mm

MyB Aer2 B Aer2

yB (45 106 )(9720 10 6 )(12.885 10 3 )(330 10 3 ) (80.385 10 3 ) 23.1 kN m

M

8.49 kN m


120 lb

PROBLEM 4.175

120 lb

The split ring shown has an inner radius r1 0.8 in. and a circular cross section of diameter d 0.6 in. Knowing that each of the 120-lb forces is applied at the centroid of the cross section, determine the stress (a) at point A, (b) at point B.

r1 d

A

B

SOLUTION

rA

r1

0.8 in.

rB

rA

d

r

1 (rA 2

A

c2

e

1 r 2 r R

Q

120 lb

R

M0 (a)

0.8 rB )

0.6

1.4 in.

1.1 in.

(0.3)2

c

1 d 2

0.28274 in 2

0.3 in. for solid circular section

1 1.1 (1.1)2 (0.3)2 2 1.1 1.0791503 0.020850 in. r2

c2

Fx

0:

2rQ

r

rA

0.8 in.

A

P A

M (rA R) AerA

M

0: P

0 120 0.28274

Q

M

1.079150 in.

0

P

120 lb

2rQ

(2)(1.1)(120)

( 264)(0.8 1.079150) (0.28274)(0.020850)(0.8)

264 lb in.

16.05 103 psi A

(b)

r

rB

1.4 in.

B

P A

M (rB R) AerB

120 0.28274

16.05 ksi

( 264)(1.4 1.079150) (0.28274)(0.020850)(1.4)

9.84 103 psi B

9.84 ksi


120 lb

PROBLEM 4.176

120 lb

Solve Prob. 4.175, assuming that the ring has an inner radius r1 = 0.6 in. and a cross-sectional diameter d 0.8 in.

r1

PROBLEM 4.175 The split ring shown has an inner radius r1 0.8 in. and a circular cross section of diameter d 0.6 in. Knowing that each of the 120-lb forces is applied at the centroid of the cross section, determine the stress (a) at point A, (b) at point B.

d

A

B

SOLUTION

rA

r1

0.6 in.

rB

rA

d

r

1 (rA 2

A

c2

e

1 r 2 r R

Q

120 lb

R

0:

M0 (a)

0.6 rB )

0.8 1.0 in.

(0.4)2 r2 1.0

1.4 in. c

0.50265 in 2

0.4 in.

for solid circular section.

1 1.0 (1.0) 2 (0.4)2 2 0.958258 0.041742 in.

c2

Fx

2rQ

1 d 2

M

r

rA

0.6 in.

A

P A

M (rA R) AerA

0

0

P

M

120 0.50265

Q

2rQ

0

0.958258 in.

P

120 lb

(2)(1.0)(120)

240 lb in.

( 240)(0.6 0.958258) (0.50265)(0.041742)(0.6)

7.069 103 psi

(b)

r

rB

1.4 in.

B

P A

M (rB R) AerB

A

120 0.50265

7.07 ksi

( 240)(1.4 0.958258) (0.50265)(0.041742)(1.4)

3.37 103 psi

B

3.37 ksi


220 N

B

PROBLEM 4.177 The bar shown has a circular cross section of 14-mm diameter. Knowing that a 32 mm, determine the stress at (a) point A, (b) point B.

A C

220 N a 16 mm

12 mm

SOLUTION c R e

(a)

1 d 8 mm r 12 8 20 mm 2 1 1 r r 2 c2 20 202 82 2 2 r R 20 19.1652 0.83485 mm

A

r2

(8) 2

P

220 N

M

P (a

201.06 mm 2

r)

220(0.032

yA

R

r1

19.1652

12

yb

R

r2

19.1652

28

A

P A

My A Aer1

220 201.06 10

19.1652 mm

6

0.020)

11.44 N m

7.1652 mm 8.8348 mm

( 11.44)(7.1652 10 3 ) (201.06 10 6 )(0.83485 10 3 )(12 10 3 ) A

(b)

B

P A

41.8 MPa

MyB Aer2

220 201.06 10

6

( 11.44)(8.8348 10 3 ) (201.06 10 6 )(0.83485 10 3 )(28 10 3 ) B

20.4 MPa


220 N

B

PROBLEM 4.178 The bar shown has a circular cross section of 14-mm diameter. Knowing that the allowable stress is 38 MPa, determine the largest permissible distance a from the line of action of the 220-N forces to the plane containing the center of curvature of the bar.

A C

220 N a 16 mm

12 mm

SOLUTION c R e

1 d 8 mm r 12 8 20 mm 2 1 1 r r 2 c2 20 202 82 2 2 r R 20 19.1652 0.83485 mm

A

r2

(8) 2

P

220 N

M

P (a R

r1

A

P A

My A Aer1

K a

r

201.06 mm 2

r)

yA

KP A

19.1652 mm

19.1652 P A

where

12 P(a

K

7.1652 mm r ) yA Aer1

P 1 A

(a

r ) yA er1

1

(38 106 )(201.06 10 6 ) P 220 ( K 1)er1 yA AA

(34.729

(a

r ) yA er1

34.729

1)(0.83485 10 3 )(12 10 3 ) (7.1652 10 3 )

0.047158 m a

0.047158

0.020

0.027158

a

27.2 mm


C

M

PROBLEM 4.179

16 mm

The curved bar shown has a circular cross section of 32-mm diameter. Determine the largest couple M that can be applied to the bar about a horizontal axis if the maximum stress is not to exceed 60 MPa.

12 mm

SOLUTION c R

e max

16 mm

r

1 r 2 1 28 2

r2

r

28

16

28 mm

c2

282

R

12

162

25.4891 mm

25.4891

2.5109 mm

occurs at A, which lies at the inner radius.

It is given by Also,

A

Data:

yA M

My A from which Aer1

max

c2

R

(16) 2

r1

M

Aer1 yA

max

.

804.25 mm 2

25.4891 12

13.4891 mm

(804.25 10 6 )(2.5109 10 3 )(12 10 3 )(60 106 ) 13.4891 10 3

M

107.8 N m


P 90 mm

PROBLEM 4.180 Knowing that P

B

10 kN, determine the stress at (a) point A, (b) point B.

80 mm A 100 mm

SOLUTION Locate the centroid D of the cross section.

r

90 mm 3

100 mm

130 mm

Force-couple system at D.

P M

10 kN Pr

1300 N m

(10 kN)(130 mm)

Triangular cross section. A

1 bh 2

1 (90 mm)(80 mm) 2

3600 mm 2

R

(a) A

Point A: P A

M (rA R) AerA

2.778 MPa (b) B

Point B: P A

rA

1

R

126.752 mm

e

r

R

1 (90) 2 190 190 ln 90 100

130 mm

1

45 mm 0.355025

126.752 mm

3.248 mm

0.100 m

10 kN 3600 10 6 m 2

(1300 N m)(0.100 m 0.126752 m) (3600 10 6 m 2 )(3248 10 3 m)(0.100 m)

29.743 MPa

rB

M (rB R) AerB

2.778 MPa

100 mm

1 h 2 r2 r2 ln h r1

3600 10 6 m 2

190 mm

A

32.5 MPa

B

34.2 MPa

0.190 m

10 kN 3600 10 6 m 2

(1300 N m)(0.190 m 0.126752 m) (3600 10 6 m 2 )(3.248 10 3 m)(0.190 m)

37.01 MPa


M

2.5 in.

A

PROBLEM 4.181

B M

Knowing that M (b) point B.

C 3 in.

2 in.

2 in.

5 kip in., determine the stress at (a) point A,

3 in.

SOLUTION

r

1 bh 2 2 1

b1

2.5 in., r1

A

Use formula for trapezoid with

b2

1 (2.5)(3) 3.75 in 2 2 3.00000 in.

1 h 2 (b b ) 1 2 2 r2 b2r1) ln h(b1 r1

(b1r2

(a)

yA

R

yB B

R

5 in.

b2 )

(0.5)(3)2 (2.5 [(2.5)(5) (0)(2)] ln 52

0) (3)(2.5

r

M

R

0.15452 in.

0)

2.84548 in.

5 kip in.

0.84548 in.

My A Aer1

A

(b)

r1

0, r2

0.

R

e

2 in., b2

r2 MyB Aer2

(5)(0.84548) (3.75)(0.15452)(2)

3.65 ksi

A

2.15452 in. (5)( 2.15452) (3.75)(0.15452)(5)

B

3.72 ksi


M

PROBLEM 4.182

B

2.5 in. A 3 in.

Knowing that M (b) point B.

M

5 kip in., determine the stress at (a) point A,

C 2 in. 2 in. 3 in.

SOLUTION

r

1 (2.5)(3) 3.75 in 2 2 2 2 4.00000 in.

b1

0, r1

b1

0.

A

Use formula for trapezoid with

2 in.,

b2

1 h 2 (b b ) 1 2 2 r2 b2r1) ln h (b1 r1

R (b1r2

(0.5)(3) 2 (0 [(0)(5) e (a)

yA

R

yB B

R

R

(2.5)(2)] ln 5 2 0.14534 in.

r2

5 in.

b2 )

2.5) (3)(0 M

2.5)

3.85466 in.

5 kip in.

1.85466 in.

My A Aer1

A

(b)

r1

r

2.5 in.,

r2 MyB Aer2

(5)(1.85466) (3.75)(0.14534)(2)

A

8.51 ksi

1.14534 in. (5)( 1.14534) (3.75)(0.14534)(5)

B

2.10 ksi


PROBLEM 4.183

80 kip · in.

b

B

A

B

A

Knowing that the machine component shown has a trapezoidal cross section with a 3.5 in. and b 2.5 in., determine the stress at (a) point A, (b) point B.

C

a

6 in. 4 in.


r , in.

Ar , in 3

10.5

6

63

7.5

8

60

18 r R

123

123 18

6.8333 in. 1 2

(b1r2

h 2 (b1

b2 r1 ) ln

b2 )

r2 r1

h(b1 b2 )

(0.5)(6)2 (3.5 2.5) [(3.5)(10) (2.5)(4)]ln 104 (6)(3.5 2.5) e

(a)

yA A

(b)

yB B

r

R

0.4452 in.

M

6.3878 in.

80 kip in.

R r1 6.3878 4 2.3878 in. MyA (80)(2.3878) (18)(0.4452)(4) Aer1

5.96 ksi

A

R r2 6.3878 10 3.6122 in. MyB (80)( 3.6122) (18)(0.4452)(10) Aer2

B

3.61 ksi


PROBLEM 4.184

80 kip · in.

b

B

A

B

A

Knowing that the machine component shown has a trapezoidal cross section with a 2.5 in. and b 3.5 in., determine the stress at (a) point A, (b) point B.

C

a

6 in. 4 in.


r , in.

Ar , in 3

7.5

6

45

10.5

8

84

18 r R

e

(a)

yA A

(b)

yB B

R r1

129 18

7.1667 in. 1 2

(b1r2

h 2 (b1 b2 )

b2 r1 ) ln

r2 r1

h(b1 b2 )

(0.5)(6) 2 (2.5 3.5) [(2.5)(10) (3.5)(4)]ln 104 (6)(2.5 3.5)

6.7168 in.

r

80 kip in.

R

M

0.4499 in.

2.7168 in.

My A Aer1

R r2

129

(80)(2.7168) (18)(0.4499)(4)

A

6.71 ksi

3.2832 in.

MyB Aer2

(80)( 3.2832) (18)(0.4499)(10)

B

3.24 ksi


a

B

PROBLEM 4.185

20 mm B

A

A 250 N · m

250 N · m

For the curved beam and loading shown, determine the stress at (a) point A, (b) point B.

30 mm

a

35 mm

40 mm Section a–a

SOLUTION Locate centroid. A, mm 2

r , mm

Ar , mm3

600

45

27 103

300

55

16.5 103 43.5 103

900 r R

43.5 103 900 1 2

(b1r2

48.333 mm

h 2 (b1

b2r1) ln

r2 r1

b2 ) h(b2

b1)

(0.5)(30) 2 (40 20) 65 [(40)(65) (20)(35)]ln 35 (30)(40 e (a)

yA

R

yB B

R

R

1.4725 mm

M

46.8608 mm

250 N m

11.8608 mm

My A Aer1

A

(b)

r1

r

20)

r2 MyB Aer2

( 250)(11.8608 10 3 ) (900 10 6 )(1.4725 10 3 )(35 10 3 )

63.9 106 Pa

A

63.9 MPa

18.1392 mm ( 250)( 18.1392 10 3 ) (900 10 6 )(1.4725 10 3 )(65 10 3 )

52.6 106 Pa

B

52.6 MPa


PROBLEM 4.186 35 mm

25 mm

For the crane hook shown, determine the largest tensile stress in section a-a.

60 mm

40 mm a

a 60 mm Section a–a

15 kN

SOLUTION Locate centroid. A, mm 2

r , mm

Ar , mm3

1050

60

63 103

750

80

60 103 103 103

1800 r

103 103 1800

63.333 mm

P 15 103 N

Force-couple system at centroid:

M

(15 103 )(68.333 10 3 )

Pr

1 2

R

(b1r2

1.025 103 N m

h 2 (b1 b2 )

b2 r1 ) ln

r2 r1

h(b1 b2 )

(0.5)(60)2 (35 25) [(35)(100) (25)(40)]ln 100 (60)(35 25) 40 e

r

R

63.878 mm

4.452 mm

Maximum tensile stress occurs at point A. yA A

R r1 P A

23.878 mm

My A Aer1

15 103 1800 10

6

(1.025 103 )(23.878 10 3 ) (1800 10 6 )(4.452 10 3 )(40 10 3 )

84.7 106 Pa

A

84.7 MPa


PROBLEM 4.187 Using Eq. (4.66), derive the expression for R given in Fig. 4.61 for a circular cross section.

SOLUTION Use polar coordinate

as shown. Let w be the width as a function of w 2c sin r r c cos dr

c sin

d

dA

w dr

2c 2 sin 2

d

2

dA r

2c sin r c cos

0

dA r

c 2 (1 cos 2 ) d r c cos

0

r2

2

c 2 cos 2 (r 2 r c cos

0

2

d

0

(r

2r

2(r

2(r 2

c cos ) d 2c sin

0

2

c )

r2

c2

tan

1

0) 2c(0 0) 4 r 2

2r ( 2 r

r2

2

d

c2 )

0

r

dr c cos

0

2

2

c2 )

r2

c2

c 2 tan 1 2 r c 2

0

0

c2

c2

A

c2

A dA r

R

1 2

2 r

c2 r r

2

(r

r2

2

r2

c2

2

2

c )

c2 1 2

c2

1 2r

c2 r

r2 r2

c

2

c2

c2

r

r2

c2

r

r2

c2

1 r 2

r2

c2


PROBLEM 4.188 Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.

SOLUTION The section width w varies linearly with r. w w b1 b2 b2

b1

c1 r2b1

r1b2 c0

dA r

c0 c1r b1 at r r1 and w b2 at r c0 c1r1 c0 c1r2 c1 (r1 r2 ) c1h b1 b2 h (r2 r1 )c0 hc0 r2 b1 r1b2 h r2 w r2 c c1r 0 dr dr r1 r r1 r r2 r2 c0 ln r c1 r r1

r2

r1

r c0 ln 2 c1 (r2 r1 ) r1 r2 b1 r1b2 r2 b1 b2 h ln h r1 h r2 b1 r1b2 r2 ln (b1 b2 ) h r1 1 (b1 b2 ) h A 2 1 2 h (b1 b2 ) A 2 R r dA ( r2b1 r1b2 ) ln r2 h(b1 b2 ) r 1


PROBLEM 4.189 Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.

SOLUTION The section width w varies linearly with r. w w b 0 b c1 dA r

c0 c1r b at r r1 and w 0 at r r2 c0 c1r1 c0 c1r2 c1 (r1 r2 ) c1h br2 b and c0 c1r2 h h r2 w r2 c c1r 0 dr dr r1 r r1 r r2 r2 c0 ln r c1 r r1

r1

r c0 ln 2 c1 (r2 r1 ) r1 br2 r2 b h ln h r1 h

A R

br2 r2 ln h r1 1 bh 2 A dA r

b

b b

1 2 r2 h

r2 r2 ln h r1

bh

ln

r2 r1

1

r2 h

1

1 2

h

ln

r2 r1

1


PROBLEM 4.190 b2 b3

Show that if the cross section of a curved beam consists of two or more rectangles, the radius R of the neutral surface can be expressed as

b1

A

R

r1

r2 r1

ln

r2

b1

r3 r2

b2

r4 r3

b3

r3 r4

where A is the total area of the cross section.

SOLUTION A 1 dA r

R

A ri 1 ri

bi ln

A ln

Note that for each rectangle,

1 dA r

ri

1

ri

bi

ri r2

A

ri 1 ri bi 1

bi

r2 r1

ln

b1

r3 r2

b2

r4 r3

b3

dr r dr r

bi ln

ri

1

ri


PROBLEM 4.191

C

" 2

" 2

r1

!x

!x

For a curved bar of rectangular cross section subjected to a bending couple M, show that the radial stress at the neutral surface is r M R 1 1 ln r Ae R r1

R

b

and compute the value of r for the curved bar of Concept Applications 4.10 and 4.11. (Hint: consider the free-body diagram of the portion of the beam located above the neutral surface.)

!r !r

SOLUTION M (r R ) M Aer Ae For portion above the neutral axis, the resultant force is At radial distance r,

r

H

R

r dA

r b dr

r1

Mb R MRb R dr dr Ae r1 Ae r1 r Mb MRb R ( R r1 ) ln Ae Ae r1

Resultant of

n:

Fr /2 r

/2 r

For equilibrium: Fr

r

cos

/2

2 H sin

2 r bR sin

2

/2

2

r MbR 1 1 Ae R

ln

R r1

dA

cos (bR d )

bR sin

MR Aer

/2

r bR r

/2

bR sin

cos d

2

0

2 2

r MbR 1 1 Ae R

ln

R sin r1 2

r

r M 1 1 Ae R

ln

R r1

0

Using results of Examples 4.10 and 4.11 as data,

M r

8 kip in.,

A

3.75 in 2 , R

8 1 (3.75)(0.0314)

5.25 5.9686

5.9686 in., e ln

5.9686 5.25

0.0314 in., r1

5.25 in. r

0.536 ksi


PROBLEM 4.192 25 mm 25 mm

4 kN A


4 kN B

C

300 mm

300 mm

SOLUTION A1

2

A2

A2 y2 A2

r2

bh

2

(25)2

981.7 mm 2 1250 mm 2

(50)(25)

y

A1 y1 A1

(981.7)(10.610) (1250)( 12.5) 981.7 1250

I1

I x1

d1

y1

r 4 A1 y12 (25)4 (981.7)(10.610) 2 8 8 y 10.610 ( 2.334) 12.944 mm

I1

I1

A1d12

(981.7)(12.944)2

d2

1 3 1 bh (50)(25)3 65.104 103 mm 4 12 12 y2 y 12.5 ( 2.334) 10.166 mm

I2

I2

A2 d 22

I

I1

I2

I2

ytop ybot

65.104 103

(1250)(10.166) 2

401.16 103 mm 4

401.16 10

25 2.334 25 2.334

27.334 mm

bot

Mytop I

Mybot I

(4)(25) 10.610 mm 3 25 12.5 mm 2

42.886 106 mm 4

207.35 103 mm 4

194.288 103 mm 4 m4

0.027334 m

22.666 mm

0.022666 m M

top

9

y2

4r 3 h 2

2.334 mm

A1 y12

42.866 103

y1

Pa

0:

M

(1200)(0.027334) 401.16 10 9

81.76 106 Pa

(1200)( 0.022666) 401.16 10 9

67.80 106 Pa

Pa

(4 103 )(300 10 3 )

1200 N m

top

bot

81.8 MPa

67.8 MPa


PROBLEM 4.193 A steel band saw blade that was originally straight passes over 8-in.-diameter pulleys when mounted on a band saw. Determine the maximum stress in the blade, knowing that it is 0.018 in. thick and 0.625 in. wide. Use E 29 106 psi. 0.018 in.

SOLUTION Band blade thickness:

t

0.018 in.

Radius of pulley:

r

1 d 2

4.000 in.

Radius of curvature of centerline of blade: r c

1 t 2

1 t 2

4.009 in.

0.009 in.

c

0.009 4.009

Maximum strain:

m

Maximum stress:

m

E

m

65.1 103 psi

m

0.002245

(29 106 )(0.002245) m

65.1 ksi


PROBLEM 4.194

M

A couple of magnitude M is applied to a square bar of side a. For each of the orientations shown, determine the maximum stress and the curvature of the bar.

M

a (a)

(b)

SOLUTION

1 3 bh 12 a 2

I c

1

a4 12

Ma 2 a4 12

Mc I

max

1 3 aa 12

M EI

max

1

M 4 Ea 12

6M a3 12M Ea 4

For one triangle, the moment of inertia about its base is 1 3 bh 12

I1

1 12

2a

a

3

2

a4 24

4

c

I2

I1

a 24

I

I1

I2

a4 12

a

2

max

1

Mc I M EI

Ma/ 2 a 4 /12

6 2M a3

8.49M a3

max

1

M 4 Ea 12

12M Ea 4


40 mm

60 mm

PROBLEM 4.195 Determine the plastic moment M p of a steel beam of the cross section shown, assuming the steel to be elastoplastic with a yield strength of 240 MPa.

SOLUTION Let c1 be the outer radius and c2 the inner radius.

A1 y1

Aa ya 2

4c1 3

c12

2 3 c1 3

c2

A1 y1

2 3 c1 3

A2 y2

Y ( A1 y1

Mp Data:

Ab yb 2

3

c 22

A2 y2 )

4 3

Y

c13

c32

240 106 Pa

Y

240 MPa

c1

60 mm

0.060 m

c2

40 mm

0.040 m

Mp

4c2 3

c22

4 (240 106 )(0.0603 3 48.64 103 N m

0.0403 ) Mp

48.6 kN m


PROBLEM 4.196 M # 300 N · m

In order to increase corrosion resistance, a 2-mm-thick cladding of aluminum has been added to a steel bar as shown. The modulus of elasticity is 200 GPa for steel and 70 GPa for aluminum. For a bending moment of 300 N m, determine (a) the maximum stress in the steel, (b) the maximum stress in the aluminum, (c) the radius of curvature of the bar.

26 mm 30 mm

46 mm 50 mm


n

1 in aluminum

Es Ea

n

200 70

2.857 in steel

Cross section geometry: Steel: As

(46 mm)(26 mm)

Aluminum: Aa

1196 mm 2

1196 mm 2

(50 mm)(30 mm)

1 (50 mm)(30 mm3 ) 12

Ia

1 (46 mm)(26 mm)3 12

Is

67,375 mm 4

304 mm 2

67,375 mm 4

45,125 mm 4

Transformed section. I

na I a

ns I s

(1)(45,125)

M

300 N m

Bending moment. (a)

ns Mys I

s

(b)

(c)

(2.857)(300)(0.013) 237.615 10 9

Maximum stress in aluminum: a

2.857

ns

Maximum stress in steel:

na Mya I

(1)(300)(0.015) 237.615 10 9

Radius of curvative:

EI M

237, 615 mm 4

(2.857)(67,375)

na

ys

13 mm

237.615 10 9 m 4

0.013 m

46.9 106 Pa 1,

ya

s

15 mm

46.9 MPa

0.015 m

18.94 106 Pa (70 109 )(237.615 10 9 ) 300

a

18.94 MPa 55.4 m


PROBLEM 4.197 The vertical portion of the press shown consists of a rectangular tube of wall thickness t 10 mm. Knowing that the press has been tightened on wooden planks being glued together until P 20 kN, determine the stress at (a) point A, (b) point B.

t P a

P'

a

t A

200 mm

80 mm

60 mm 80 mm

B

Section a–a

SOLUTION Rectangular cutout is 60 mm

40 mm.

A

(80)(60)

I

1 (60)(80)3 12

2.4 103 mm 2

(60)(40)

1 (40)(60)3 12

2.4 10 3 m 2

1.84 106 mm 4

1.84 10 6 m 4 c P M (a) (b)

40 mm

0.040 m e

200

40

240 mm

0.240 m

3

20 10 N Pe

(20 103 )(0.240)

4.8 103 N m

A

P A

Mc I

20 103 2.4 10 3

(4.8 103 )(0.040) 1.84 10 6

112.7 106 Pa

A

112.7 MPa

B

P A

Mc I

20 103 2.4 10 3

(4.8 103 )(0.040) 1.84 10 6

96.0 106 Pa

B

96.0 MPa


PROBLEM 4.198

P

P y

P

P B

C D

a

A x

z

The four forces shown are applied to a rigid plate supported by a solid steel post of radius a. Knowing that P 24 kips and a 1.6 in., determine the maximum stress in the post when (a) the force at D is removed, (b) the forces at C and D are removed.

SOLUTION For a solid circular section of radius a,

Centric force: (a)

(b)

A

a2

I

F

4P,

Mx

F

3P,

Mx

F A

M xz I

2P,

Mx

M

M x2

4

a4 F A

4P a2

( Pa)( a) a2 4

7P a2

0

Mz

Force at D is removed.

Pa,

Mz

3P a2

0

Forces at C and D are removed.

F Resultant bending couple:

F A

2P a2

P

Numerical data: Answers:

Mc I

Pa,

Mz

M z2

2 Paa a2 4

2 Pa

2

24.0 kips,

Pa

4 2 P a2

a

2.437 P/a 2

1.6 in.

(a)

(7)(24.0) (1.6)2

20.9 ksi

(b)

(2.437)(24.0) (1.6)2

22.8 ksi


a

r # 20 mm

PROBLEM 4.199

P # 3 kN

The curved portion of the bar shown has an inner radius of 20 mm. Knowing that the allowable stress in the bar is 150 MPa, determine the largest permissible distance a from the line of action of the 3-kN force to the vertical plane containing the center of curvature of the bar.

25 mm

25 mm

SOLUTION Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the section. The bending couple is M

P(a

r)

For the rectangular section, the neutral axis for bending couple only lies at R Also, e

r

h r2 r1

ln

R

The maximum compressive stress occurs at point A. It is given by P A

A

My A Aer1

P A

with

yA

R

Thus,

K

1

Data:

25 mm, r1

R

25 45 ln 20

b

25 mm,

K a

r a

r ) yA Aer1

r )( R er1

P A

r1)

20 mm, r2

30.8288 mm, e A

3

K

r1 (a

h

P

P (a

3 10 N m,

bh

45 mm, r 32.5

(25)(25) A

30.8288

625 mm 2

32.5 mm 1.6712 mm 625 10 6 m 2

R

r1

10.8288 mm

6

150 10 Pa

( 150 106 )(625 10 6 ) 31.25 P 3 103 ( K 1)er1 (30.25)(1.6712)(20) 93.37 mm R r1 10.8288 AA

93.37

a

32.5

60.9 mm


Dimensions in inches

400 lb 400 lb

PROBLEM 4.200

400 lb

2.5

Determine the maximum stress in each of the two machine elements shown.

400 lb 2.5

3

0.5 3

r 5 0.3

r 5 0.3 1.5

0.5

0.5 1.5

0.5

SOLUTION For each case, M

(400)(2.5)

1000 lb in.

At the minimum section, 1 (0.5)(1.5)3 12 0.75 in.

I c

(a)

D/d r/d

3/1.5

0.140625 in 4

2

0.3/1.5

0.2 K

From Fig 4.32, max

(b)

D/d

KMc I 3/1.5

(1.75)(1000)(0.75) 0.140625 2

From Fig. 4.31, max

1.75

KMc I

r /d

0.3/1.5

K

9.33 103 psi

max

9.33 ksi

max

8.00 ksi

0.2

1.50

(1.50)(1000)(0.75) 0.140625

8.00 103 psi


PROBLEM 4.201

120 mm 10 mm

M

120 mm 10 mm

Three 120 10-mm steel plates have been welded together to form the beam shown. Assuming that the steel is elastoplastic with E 200 GPa and Y 300 MPa, determine (a) the bending moment for which the plastic zones at the top and bottom of the beam are 40 mm thick, (b) the corresponding radius of curvature of the beam.

10 mm

SOLUTION

A1 R1 A2 R2 A3 R3 y1

(a)

M

2( R1 y1

(120)(10) 1200 mm 2 Y

(300 106 )(1200 10 6 ) 360 103 N

A1

(30)(10) 300 mm 2 Y

(300 106 )(300 10 6 ) 90 103 N

A2

(30)(10) 1 Y A2 2 65 mm

R2 y2

300 mm 2 1 (300 106 )(300 10 6 ) 45 103 N 2 65 10 3 m y2 45 mm 45 10 3 m

R3 y3 )

yY

Y

E

EyY Y

20 mm

20 10 3 m

2{(360)(65) (90)(45) (45)(20)}

56.7 103 N m (b)

y3

M (200 109 )(30 10 3 ) 300 106

56.7 kN m 20.0 mm


P

4.5 in.

PROBLEM 4.202

Q

4.5 in.

A short length of a W8 31 rolled-steel shape supports a rigid plate on which two loads P and Q are applied as shown. The strains at two points A and B on the centerline of the outer faces of the flanges have been measured and found to be A

B

550 10

A

6

in./in.

B

680 10

6

in./in.

29 106 psi, determine the magnitude of each load.

Knowing that E

SOLUTION Strains: A

C

550 10 1 ( 2

6

B

1 ( 550 2

B)

A

680)10

6

615 10

6

E

A

(29 106 psi)( 550 10

6

in./in.)

15.95 ksi

C

E

C

(29 106 psi)( 615 10

6

in./in.)

17.835 ksi

W8 M

31: (4.5 in.)( P

P C

Q A

;

A

9.12 in 2

S

27.5 in 3

680 10

P A

15.95 ksi Solve simultaneously.

Q A

in./in.

Q) 17.835 ksi

P Q 9.12 in 2 P

At point A:

6

in./in.

A

Stresses:

At point C:

in./in.

Q

162.655 kips (1)

M S 17.835 ksi

P

(4.5 in.)( P Q) ; 27.5 in 3

75.6 kips Q

P

Q

11.5194 kips (2)

87.1 kips

P

75.6 kips

Q

87.1 kips


PROBLEM 4.203 !1 M1

A

M'1

M1

!1

M'1

B C

!1 !1

D

Two thin strips of the same material and same cross section are bent by couples of the same magnitude and glued together. After the two surfaces of contact have been securely bonded, the couples are removed. Denoting by 1 the maximum stress and by 1 the radius of curvature of each strip while the couples were applied, determine (a) the final stresses at points A, B, C, and D, (b) the final radius of curvature.

SOLUTION Let b

width and t

thickness of one strip.

Loading one strip, M

I1 1

1 1

M1

1 3 bt , c 12 M1c I M1 EI1

1 t 2

M1 bt 2 12 M 1 Et 3

After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are A

1,

1,

B

1,

C

D

1

The total bending couple is 2M1. After the strips are glued together, this couple is removed.

M

1 b(2t )3 12

2 M 1, I

2 3 bt 3

c

t

The stresses removed are My I A

2M 1 y 2 bt 3 3

3M1 bt 2

1 2

1,

3M1 y bt 2 B

C

0,

D

3M1 bt 2

1 2

1



(a)

Final stresses:

A

1

(

1 2

1)

1 2

A

B

1

C

D

1

(b)

M EI

2M 1 E 23 bt 3

Final radius:

3M 1 Et 3

1

1

1 2

D

1

1

1

1 2

1

4 3

1

1 1 4

1 1

1

1 1

1 1 4 1

3 1 4 1


PROBLEM 4.C1 Two aluminum strips and a steel strip are to be bonded together to form a composite member of width b 60 mm and depth h 40 mm. The modulus of elasticity is 200 GPa for the steel and 75 GPa for the aluminum. Knowing that M 1500 N m, write a computer program to calculate the maximum stress in the aluminum and in the steel for values of a from 0 to 20 mm using 2-mm increments. Using appropriate smaller increments, determine (a) the largest stress that can occur in the steel, (b) the corresponding value of a.

Aluminum a

Steel

h " 40 mm a b " 60 mm

SOLUTION Transformed section:

(all steel)

I

At Point 1:

At Point 2:

Esteel Ealum

n

1 3 bh 12 h 2

M alum

steel

1 1 2 (nb b) (h 2a )3 12 2

I

n

M

h 2

a

I


PROBLEM 4.C1 (Continued)

For a

0 to 20 mm using 2-mm intervals: compute: n, I , b

Moduli of elasticity:

Steel

60 mm h

alum ,

40 mm M

200 GPa Aluminum

steel .

1500 N m 75 GPa

Program Output

a mm

I m 4 /106

Sigma Aluminum MPa

Sigma Steel MPa

0.000

0.8533

35.156

93.750

2.000

0.7088

42.325

101.580

4.000

0.5931

50.585

107.914

6.000

0.5029

59.650

111.347

8.000

0.4352

68.934

110.294

10.000

0.3867

77.586

103.448

12.000

0.3541

84.714

90.361

14.000

0.3344

89.713

71.770

16.000

0.3243

92.516

49.342

18.000

0.3205

93.594

24.958

20.000

0.3200

93.750

0.000

Find ‘a’ for max. steel stress and the corresponding aluminum stress. 6.600

0.4804

62.447

111.572083

6.610

0.4800

62.494

111.572159

6.620

0.4797

62.540

111.572113

Max. steel stress 111.6 MPa occurs when a Corresponding aluminum stress

6.61 mm.

62.5 MPa


tf

PROBLEM 4.C2 y

x tw

d

bf

A beam of the cross section shown, made of a steel that is assumed to be elastoplastic with a yield strength Y and a modulus of elasticity E, is bent about the x axis. (a) Denoting by yY the half thickness of the elastic core, write a computer program to calculate the bending moment M and the radius of curvature for values of yY from 12 d to 16 d using decrements equal 1 to 2 t f . Neglect the effect of fillets. (b) Use this program to solve Prob. 4.201.

SOLUTION Compute moment of inertia I x . 1 bf d 3 12

Ix

Maximum elastic moment:

MY

Y

1 (b f 12

tw )(d

2t f )3

Ix (d/2)

For yielding in the flanges, (Consider upper half of cross section.)

d 2

c

Stress at junction of web and flange: (d/2) t f A

Y

yY

Resultant forces:

Detail of stress diagram:

a1

1 (c 2

a2

yY

a3

yY

a4

2 (c t f ) 3

yY ) 1 yY 3 2 [ yY 3

(c t f ) (c t f ) ]



Bending moment. 4

2

M

Rn an n 1

Radius of curvature. yY

Y

Y

E

yY E

;

Y

(Consider upper half of cross section.)

For yielding in the web,

a5 a6 a7

1 tf 2

c

1 [ yY 2 2 yY 3

(c t f )]

7

Bending moment.

M

2

Rn an n 5

Radius of curvature.

yY

Y

yY E

Y

E

Y

Program: Key in expressions for an and Rn for n 1 to 7. For yY

c to (c tf ) at

tf /2 decrements, compute M

2 Rn an for n 1 to 4 and

yY E

, then print.

Y

For yY

(c tw ) to c /3 at

tf /2 decrements, compute M

2 Rn an for n 5 to 7 and

yY E

, then print.

Y

Input numerical values and run program.



Program Output For a beam of Problem 4.201, Depth d

140.00 mm

Thickness of flange t f I

10.00 mm

Width of flange b f

120.00 mm

Thickness of web tw

10.00 mm

0.000011600 m to the 4th

Yield strength of steel sigmaY Yield moment M Y

300 MPa

49.71 kip in.

yY (mm)

M (kN m)

(m)

For yielding still in the flange, 70.000

49.71

46.67

65.000

52.59

43.33

60.000

54.00

40.00

For yielding in the web, 60.000

54.00

40.00

55.000

54.58

36.67

50.000

55.10

33.33

45.000

55.58

30.00

40.000

56.00

26.67

35.000

56.38

23.33

30.000

56.70

20.00

25.000

56.97

16.67


PROBLEM 4.C3

y

# 0.4

0.4

A

B

z

C

1.2 0.4

# M

1.2 D

E 0.8 0.4

1.6

An 8 kip in. couple M is applied to a beam of the cross section shown in a plane forming an angle with the vertical. Noting that the centroid of the cross section is located at C and that the y and z axes are principal axes, write a computer program to calculate the stress at A, B, C, and D for values of from 0 to 180° using 10° increments. (Given: I y 6.23 in 4 and I z 1.481 in 4 . )

0.4 0.8

Dimensions in inches

SOLUTION Input coordinates of A, B, C, D. zA

z (1)

2

yA

y (1) 1.4

zB

z (2)

2

yB

y (2) 1.4

zC

z (3)

1

yC

y (3)

1.4

zD

z (4) 1

yD

y (4)

1.4

My

M sin

Components of M. Mz Equation 4.55, Page 305: Program: For

( n)

M cos M z y ( n) Iz

M y z ( n) Iy

0 to 180 using 10° increments.

For n 1 to 4 using unit increments. Evaluate Equation 4.55 and print stresses. Return Return



Program Output Moment of couple:

M

8.00 kip in.

Moments of inertia:

Iy

6.23 in 4

Iz

1.481 in 4

Coordinates of Points A, B, D, and E: Point A: z (1) Point B: z (2)

2: y (1) 1.4 2: y (2) 1.4

Point D: z (3) 1: Po int E: z (4) 1:

Beta

y (3) y (4)

---Stress at Points--A B ksi ksi

1.4 1.4

D ksi

E ksi

0

–7.565

–7.565

7.565

7.565

10

–7.896

–7.004

7.673

7.227

20

–7.987

–6.230

7.548

6.669

30

–7.836

–5.267

7.193

5.909

40

–7.446

–4.144

6.621

4.970

50

–6.830

–2.895

5.846

3.879

60

–6.007

–1.558

4.895

2.670

70

–5.001

–0.174

3.794

1.381

80

–3.843

1.216

2.578

0.049

90

–2.569

2.569

1.284

–1.284

100

–1.216

3.843

–0.049

–2.578

110

0.174

5.001

–1.381

–3.794

120

1.558

6.007

–2.670

–4.895

130

2.895

6.830

–3.879

–5.846

140

4.144

7.446

–4.970

–6.621

150

5.267

7.836

–5.909

–7.193

160

6.230

7.987

–6.669

–7.548

170

7.004

7.896

–7.227

–7.673

180

7.565

7.565

–7.565

–7.565


PROBLEM 4.C4 b B

B

A

A

h M'

M

r1

C

Couples of moment M 2 kN m are applied as shown to a curved bar having a rectangular cross section with h 100 mm and b 25 mm. Write a computer program and use it to calculate the stresses at points A and B for values of the ratio r1/h from 10 to 1 using decrements of 1, and from 1 to 0.1 using decrements of 0.1. Using appropriate smaller increments, determine the ratio r1/h for which the maximum stress in the curved bar is 50% larger than the maximum stress in a straight bar of the same cross section.

SOLUTION Input:

h 100 mm, b 25 mm, M 2 kN m

For straight bar,

straight

M S 6M h 2b 48 MPa

Following notation of Section 4.15, key in the following: r2 Stresses:

A

h r1 ; R 1

M (r1

h /ln (r2

r1 ); r

R)( Aer1 )

r1 2

B

r2 : e M (r2

r

R; A bh

R)/(Aer2 )

2500

(I) (II)

Since h 100 mm, for r1/h 10, r1 1000 mm. Also, r1/h 10, r1 100 Program:

For r1 1000 to 100 at

100 decrements,

using equations of Lines I and II, evaluate r2 , R, r , e, Also evaluate ratio

1/

Return and repeat for r1

1,

and

2

straight

100 to 10 at

10 decrements.


PROBLEM 4.C4 (Continued) Program Output M

Bending moment

Stress in straight beam

2 kN m h 100.000 in. A 2500.00 mm 2

48.00 MPa

r1 mm

rbar mm

R mm

e mm

MPa

MPa

r1 /h –

ratio –

1000

1050

1049

0.794

–49.57

46.51

10.000

–1.033

900

950

949

0.878

–49.74

46.36

9.000

–1.036

800

850

849

0.981

–49.95

46.18

8.000

–1.041

700

750

749

1.112

–50.22

45.95

7.000

–1.046

600

650

649

1.284

–50.59

45.64

6.000

–1.054

500

550

548

1.518

–51.08

45.24

5.000

–1.064

400

450

448

1.858

–51.82

44.66

4.000

–1.080

300

350

348

2.394

–53.03

43.77

3.000

–1.105

200

250

247

3.370

–55.35

42.24

2.000

–1.153

100

150

144

5.730

–61.80

38.90

1.000

–1.288

1

2

===================================================== 100

150

144

5.730

–61.80

38.90

1.000

–1.288

90

140

134

6.170

–63.15

38.33

0.900

–1.316

80

130

123

6.685

–64.80

37.69

0.800

–1.350

70

120

113

7.299

–66.86

36.94

0.700

–1.393

60

110

102

8.045

–69.53

36.07

0.600

–1.449

50

100

91

8.976

–73.13

35.04

0.500

–1.523

40

90

80

10.176

–78.27

33.79

0.400

–1.631

30

80

68

11.803

–86.30

32.22

0.300

–1.798

20

70

56

14.189

–100.95

30.16

0.200

–2.103

10

60

42

18.297

–138.62

27.15

0.100

–2.888

=================================================================== Find r1/h for (

max ) /( straight )

1.5

52.70 52.80

103 103

94 94

8.703 8.693

–72.036 –71.998

35.34 35.35

0.527 0.528

–1.501 –1.500

52.90

103

94

8.683

–71.959

35.36

0.529

–1.499

Ratio of stresses is 1.5 for r1

52.8 mm or r1/h

0.529.

[Note: The desired ratio r1/h is valid for any beam having a rectangular cross section.]


bn hn h2

PROBLEM 4.C5 M

The couple M is applied to a beam of the cross section shown. (a) Write a computer program that, for loads expressed in either SI or U.S. customary units, can be used to calculate the maximum tensile and compressive stresses in the beam. (b) Use this program to solve Probs. 4.9, 4.10, and 4.11.

b2

h1 b1

SOLUTION Input:

Bending moment M.

For n 1 to n,

Enter bn and hn bn hn

Area an

an

(Print) (hn 1 )/2 hn /2

1

[Moment of rectangle about base] m

( Area)an

m

m

[For whole cross section] m; Area

Area

Area

Location of centroid above base. y

m/Area

(Print)

Moment of inertia about horizontal centroidal axis. For n 1 to n,

an

an

(hn 1 )/2 hn /2

1

I

bn hn3 /12 (bn hn )( y

I

I

I

an )2 (Print)

Computation of stresses. For n 1 to n,

Total height:

H

H

hn

Stress at top: M top

M

H

y I

(Print)

Stress at bottom: M bottom

M

y I

(Print)


PROBLEM 4.C5 (Continued) Problem 4.9 Summary of cross section dimensions: Width (in.)

Height (in.)

9.00

2.00

3.00

6.00

Bending moment 600.000 kip in. Centroid is 3.000 mm above lower edge. Centroidal moment of inertia is 204.000 in4. Stress at top of beam

14.706 ksi

Stress at bottom of beam 8.824 ksi Problem 4.10 Summary of cross section dimensions: Width (in.)

Height (in.)

4.00 1.00

1.00 6.00

8.00

1.00

Bending moment 500.000 kip in. Centroid is 4.778 in. above lower edge. Centroidal moment of inertia is 155.111 in4. Stress at top of beam

10.387 ksi

Stress at bottom of beam 15.401 ksi Problem 4.11 Summary of cross section dimensions: Width (mm)

Height (mm)

50

10

20

50

Bending moment 1500.0000 N m Centroid is 25.000 mm above lower edge. Centroidal moment of inertia is 512,500 mm4. Stress at top of beam

102.439 MPa

Stress at bottom of beam 72.171 MPa


PROBLEM 4.C6 y

Dy y c

M z

A solid rod of radius c 1.2 in. is made of a steel that is assumed to be elastoplastic with E 29,000 ksi and Y 42 ksi. The rod is subjected to a couple of moment M that increases from zero to the maximum elastic moment M Y and then to the plastic moment M p . Denoting by yY the half thickness of the elastic core, write a computer program and use it to calculate the bending moment M and the radius of curvature for values of yY from 1.2 in. to 0 using 0.2-in. decrements. (Hint: Divide the cross section into 80 horizontal elements of 0.03-in. height.)

SOLUTION

MY

Y

Mp

Y

c3 (42 ksi) (1.2 in.)3 57 kip in. 4 4 4 3 4 c (42 ksi) (1.2 in.)3 96.8 kip in. 3 3

Consider top half of rod. Let

i

Number of elements in top half. c Height of each element: h L

h For n

0 to i 1, Step 1: y

n( h) c 2 {(n 0.5) h}2

z If y

z at midheight of element

yY go to 100 (n 0.5) h

E

Y

Stress in elastic core

y

go to 200 100 E 200

Area Force Moment M P

Y

Stress in plastic zone

2 z ( h) E ( Area)

Repeat for yY 1.2 in. to yY 0 At 0.2-in. decrements

Force (n 0.5) h Moment M yY E / Y

Print yY , M , and . Next


PROBLEM 4.C6 (Continued) Program Output Radius of rod 1.2 in. Yield point of steel

Yield moment Plastic moment Number of elements in half of the rod

42 ksi

57.0 kip in. 96.8 kip in. 40

For yY

1.20 in.,

M

57.1 kip in.

Radius of curvature

828.57 in.

For yY

1.00 in.,

M

67.2 kip in.

Radius of curvature

690.48 in.

For yY

0.80 in.,

M

76.9 kip in.

Radius of curvature

552.38 in.

For yY

0.60 in.,

M

85.2 kip in.

Radius of curvature

414.29 in.

For yY

0.40 in.,

M

91.6 kip in.

Radius of curvature

276.19 in.

For yY

0.20 in.,

M

95.5 kip in.

Radius of curvature

138.10 in.

For yY

0.00 in.,

M

infinite

Radius of curvature

zero


PROBLEM 4.C7 2 in.

C

3 in.

The machine element of Prob 4.178 is to be redesigned by removing part of the triangular cross section. It is believed that the removal of a small triangular area of width a will lower the maximum stress in the element. In order to verify this design concept, write a computer program to calculate the maximum stress in the element for values of a from 0 to 1 in. using 0.1-in. increments. Using appropriate smaller increments, determine the distance a for which the maximum stress is as small as possible and the corresponding value of the maximum stress.

B

A

2.5 in. a

SOLUTION See Figure 4.79, Page 289. M

For a

5 kip in. r2

5 in. b2

2.5 in.

0 to 1.0 at 0.1 intervals, h 3 a r1 2 a b1

b2 (a /(h a ))

Area

(b1 b2 )(h/2)

x

a+

r

r2

R e

1 b1h(h/3) 2

Area

(h x ) 1 2

(b1r2 r

1 b2 h 2h/3 2

h 2 (b1 b2 )

b2 r1 ) ln

r2 r1

h(b1 b2 )

R

D

M (r1

R)/[Area (e)(r1 )]

B

M (r2

R) /[Area (e)(r2 )]

Print and Return



Program Output a in.

R in.

D

B

ksi

ksi

b1

r

e

0.00

3.855

8.5071

2.1014

0.00

4.00

0.145

0.10

3.858

7.7736

2.1197

0.08

4.00

0.144

0.20

3.869

7.2700

2.1689

0.17

4.01

0.140

0.30

3.884

6.9260

2.2438

0.25

4.02

0.134

0.40

3.904

6.7004

2.3423

0.33

4.03

0.127

0.50

3.928

6.5683

2.4641

0.42

4.05

0.119

0.60

3.956

6.5143

2.6102

0.50

4.07

0.111

0.70

3.985

6.5296

2.7828

0.58

4.09

0.103

0.80

4.018

6.6098

2.9852

0.67

4.11

0.094

0.90

4.052

6.7541

3.2220

0.75

4.14

0.086

1.00

4.089

6.9647

3.4992

0.83

4.17

0.078

Determination of the maximum compressive stress that is as small as possible. a in.

R in.

D

B

ksi

ksi

b1

r

e

0.620

3.961

6.51198

2.6425

0.52

4.07

0.109

0.625

3.963

6.51185

2.6507

0.52

4.07

0.109

0.630

3.964

6.51188

2.6591

0.52

4.07

0.109

Answer: When a

625 in., the compressive stress is 6.51 ksi.


Mechanics of Materials 7th Edition Beer Johnson Chapter 4

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