Mechanics of Materials 7th Edition Beer Johnson Chapter 8

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C

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A D C B

a 10 ft a

P P

PROBLEM 8.1

A W10 39 rolled-steel beam supports a load P as shown. Knowing that

P  45 kips, a10 in., and _all18 ksi, determine (a) the maximum value of the normal stress _m in the beam, (b) the maximum value of the principal stress

max

 at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned.

SOLUTION max max | | 90 kips | | (45)(10) 450 kip in.     V M

For W10 39 rolled steel section,

4 3

9.92 in., 7.99 in., 0.530 in.,

0.315 in., 209 in 42.1 in 1 4.96 in. 4.43 in. 2 f f w x x b f d b t t I S c d y c t            (a) | |max 450 42.1 m x M S    10.69 _m  ksi  2 3 max 2 2 4.43 (10.69) 9.55 ksi 4.96 4.2347 in 1₍ ₎ _{4.695 in.} 2 19.8819 in | | (45)(19.8819) 13.5898 ksi (209)(0.315) 14.4043 ksi 2 b b m f f f f b b f f b xy x w b xy y c A b t y c y Q A y V Q I t R     _    _ _                 _ _     (b) _max 19.18 ksi 2 b _R      _max 19.18 ksi 

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PROBLEM 8.2

Solve Prob. 8.1, assuming that P22.5 kips and a20 in.

PROBLEM 8.1 A W10 39 rolled-steel beam supports a load P as shown. Knowing that P  45 kips, all

10 in., and 18 ksi,

a   determine (a) the maximum value of the normal stress _m in the beam, (b) the maximum value of the principal stress _max at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned.

SOLUTION max max | | 22.5 kips | | (22.5)(20) 450 kip in.     V M

4 3

9.92 in., 7.99 in., 0.530 in.,

0.315 in., 209 in , 42.1 in 1 4.96 in. 4.43 in. 2 f f w x x b f d b t t I S c d y c t            (a) | |max 450 42.1 m x M S    10.69 _m ksi  2 3 max 2 2 4.43 (10.69) 9.55 ksi 4.96 4.2347 in 1 ( ) 4.695 in. 2 19.8819 in | | (22.5)(19.8819) 6.7949 ksi (209)(0.315) 8.3049 ksi 2 b b m f f f f b b f f b xy x w b xy y c A b t y c y Q A y V Q I t R         _ _                 _ _     (b) _max 13.08 ksi 2 b _R      _max 13.08 ksi 

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P B C A a a

PROBLEM 8.3

An overhanging W920 449 rolled-steel beam supports a load P as shown. Knowing that P700 kN, a2.5 m, and _all 100 MPa, determine (a) the maximum value of the normal stress _m in the beam, (b) the maximum value of the principal stress _max at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned.

SOLUTION 3 max 3 6 max | | 700 kN 700 10 N | | (700 10 )(2.5) 1.75 10 N m V M        

For W920 449 rolled steel beam,

6 4 3 3 947 mm, 424 mm, 42.7 mm, 24.0 mm, 8780 10 mm , 18,500 10 mm 1 473.5 mm, 430.8 mm 2 f f w x x b f d b t t I S c d y c t              (a) 6 max 6 | | 1.75 10 18,500 10 m x M S       94.595 MPa 94.6 MPa   m m    3 2 3 3 6 3 3 6 max 6 3 2 430.8 (94.595) 86.064 MPa 473.5 18.1048 10 mm 1 ( ) 452.15 mm 2 8186.1 10 mm 8186.1 10 m | | (700 10 )(8186.1 10 ) 27.194 MPa (8780 10 )(24.0 10 ) 2 b b m f f t f b b f f b xy x w b y c A b t y c y Q A y V Q I t R     _                             _ _    2 2 86.064 _27.1942 _{50.904 MPa} 2 xy        (b) _max 93.9 MPa 2 b _R      _max93.9 MPa 

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PROBLEM 8.4

Solve Prob. 8.3, assuming that P850 kN and a2.0 m.

PROBLEM 8.3 An overhanging W920 449 rolled-steel beam supports a load P as shown. Knowing that all

700 kN, 2.5 m, and 100 MPa,

P a   determine (a) the maximum value of the normal stress _m in the

beam, (b) the maximum value of the principal stress _max at the junction of the flange and web, (c) whether the specified shape is acceptable as far as these two stresses are concerned.

SOLUTION 3 max 3 6 max | | 850 kN 850 10 N | | (850 10 )(2.0) 1.70 10 N m V M        

6 4 3 3 947 mm, 424 mm, 42.7 mm, 24.0 mm, 8780 10 mm , 18,500 10 mm 1 473.5 mm 430.8 mm 2 f f w x x b f d b t t I S c d y c t              (a) max 6 6 | | 1.70 10 18,500 10 m x M S       91.892 MPa 91.9 MPa   m m    3 2 3 3 6 3 3 6 max 6 3 2 430.8 (91.892) 83.605 MPa 473.5 18.1048 10 mm 1 ( ) 452.15 mm 2 8186.1 10 mm 8186.1 10 m | | (850 10 )(8186.1 10 ) 33.021 MPa (8780 10 )(24.0 10 ) 2 b b m f f f f b b f f b xy x w b y c A b t y c y Q A y V Q I t R                                  _ _ 2 2 83.605 _33.0212 _{53.271 MPa} 2 xy      _ _     (b) _ _b _{ }_R 95.1 MPa _ __{95.1 MPa}_

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A B C 4.5 m 2.7 m 2.2 kN/m 40 kN

PROBLEM 8.5

(a) Knowing that _all 160 MPa and _all100 MPa, select the most economical metric wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for  _m, _m, and the principal stress _max at the junction of a flange and the web of the selected beam. SOLUTION 0:

_

M_C  7.2R_A (2.2)(7.2)(3.6) (40)(2.7)  0 22.92 kN 22.92 kN 22.92 (2.2)(4.5) 13.02 kN 13.02 40 26.98 kN 26.98 (2.2)(2.7) 32.92 kN 0 1 0 (22.92 13.02)(4.5) 80.865 kN m 2 0 A A A B B C A B C R V R V V V M M M                         3 6 3 max min 6 all 3 3 80.865 10 490 10 m 165 10 490 10 mm M S           (a) Use W310  38.7.   310 mm 9.65 mm 5.84 mm f w d t t     Shape S (103 _mm3₎ W360 39 578 W310 38.7 547  W250 44.8 531 W200 52 511

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PROBLEM 8.5 (Continued) (b) 3 6 6 80.865 10 147.834 10 Pa 547 10 B m M S    _    147.8 MPa m    3 6 max 3 3 32.92 10 18.1838 10 Pa (310 10 )(5.84 10 ) m w V dt    _  _     18.18 MPa m    1 155 mm 155 9.65 145.35 mm 2 145.35 (147.834) 138.630 MPa 155 b f b b m c d y c t y c              _ _    At point B, (26.98 10 )₃ 3 ₃ 14.9028 MPa (310 10 )(5.84 10 ) w w V dt    _  _    2 2 2 2 max (69.315) (14.9028) 70.899 MPa 2 69.315 70.899 2 b w b R R  _      _ _           _max 140.2 MPa 

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D B C A 1.5 m 3.6 m 1.5 m 275 kN 275 kN

PROBLEM 8.6

(a) Knowing that _all160 MPa and _all 100 MPa, select the most economical metric wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for

, ,

m m

  and the principal stress _max at the junction of a flange and the web of the selected beam.

SOLUTION max max 2 6 3 max min 6 all 3 3 504.17 kN 504.17 275 kN 412.5 kN m 412.5 10 2578 10 m 160 10 2578 10 mm B C R R V M M S                  (a) Use W690  125.   d678 mm t_f 16.3 mm t_w11.7 mm  (b) 3 6 max 6 412.5 10 118.195 10 Pa 3490 10 m M S     _     m 118.2 MPa   max max 3 6 3 3 275 10 34.667 10 Pa (678 10 )(11.7 10 ) m w w V V A dt     _  _      m 34.7 MPa   1 67.8 339 mm, 16.3 mm, 339 16.3 322.7 mm 2 2 f b f c  d   t  y  c t      2 2 2 2 322.7 (118.195) 112.512 MPa 339 (56.256) (34.667) 66.080 MPa 2 b b m b m y c R    _    _ _       _ _         _max 56.256 66.080 2 b _R        _max 122.3 MPa  Shape Sx(103 mm3) W760 147 4410  _{W690 125}_ ₃₄₉₀ W530 150 3720 W460 158 3340 W360 216 3800

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A B C D

10 ft 30 ft 10 ft

20 kips 20 kips

2 kips/ft

PROBLEM 8.7

(a) Knowing that _all 24 ksi and _all 14.5 ksi, select the most economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for  _m, _m, and the principal stress _max at the junction of a flange and the web of the selected beam. SOLUTION 50 kips 50 kips     A D R R

max 30 kips max  200 kip ft  2400 kip in.

V M 3 max min all 2400 100 in 24  M   S  Shape _S_{(in )}3 W24 68 154  _{W21 62}_ ₁₂₇ W18 76 146 W16 77 134 W12 96 131 W10 112 126 (a) Use W21 62. 

21.0 in. 0.615 in. 0.400 in.

 _f  _w d t t (b) max 2400 _{18.8976 ksi} 127    m M S  18.9 _m  ksi  max 30 _{3.5714 ksi} (21.0)(0.400)    m w V dt  3.57 ksi_m   1 21.0 10.50 in. 10.50 0.615 9.8850 in. 2 2    _b   _f    c d y c t

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A

B

C

12 ft 6 ft

1.5 kips/ft

PROBLEM 8.8

(a) Knowing that _all 24 ksi and _all 14.5 ksi, select the most economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for  _m, _m, and the principal stress _max at the junction of a flange and the web of the selected beam. SOLUTION 0: 12 (1.5)(18)(3) 0 6.75 kips B A A M R R       _ 0: 12 (1.5)(18)(9) 0 20.25 kips A B B M R R       max max 3 max min all 11.25 kips

27 kip ft 324 kip in. 324 13.5 in 24         V M M S  Shape S(in )3 W12 16 17.1  W10 15 13.8 W8 18 15.2 W6 20 13.4 (a) Use W10 15.   10.0 in. 0.270 in. 0.230 in. f w d t t     (b) max 324 _{23.478 ksi} 13.8 m M S     23.5 _m  ksi  max 11.25 _{4.8913 ksi} (10.0)(0.230)    m w V dt  4.89 _m  ksi   1 10.0 5.00 in. 2 2 c  d   

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PROBLEM 8.8 (Continued) 2 2 2 2 5.00 0.270 4.73 in. 4.73 (23.478) 22.210 ksi 5.00 22.210 (4.8913) 12.1345 ksi 2 2 b f b b m b m y c t y c R    _         _ _         _ _   _ _       max 22.210 12.1345 2 2 b _R       _max  23.2 ksi 

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PROBLEM 8.9

The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement _m_all. For the selected design (use the loading of Prob. 5.73 and selected W530  92 shape), determine (a) the actual value of _m in the beam, (b) the maximum value of the principal stress max at the junction of a flange and the web.

SOLUTION Reactions: R_A 97.5 kN  R_D  97.5 kN max max | | 97.5 kN | | 286 kN m V M   

For W530  92 rolled-steel section,

6 4 3 3 533 mm, 209 mm, 15.6 mm, 1 10.2 mm, 266.5 mm 2 554 10 mm 2080 10 mm f f w d b t t c d I S           (a) 3 6 max 6 | | 286 10 137.5 10 Pa 2080 10 m M S     _    137.5 MPa m    2 3 3 250.9 mm 3260.4 mm 1 ( ) 258.7 mm 2 843.47 10 mm b f f f f b f y c t A b t y c y Q A y            At midspan: V 0 _b  0 250.9 (137.5) 129.5 MPa 266.5 b b m y c      _max 129.5 MPa 

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PROBLEM 8.9 (Continued) At sections B and C: 3 6 3 6 3 6 3 270 10 129.808 MPa 2080 10 250.9 (129.808) 122.209 MPa 266.5 (82.5 10 )(3260.4 10 )(258.7 10 ) (554 10 )(10.2 10 ) m b b m f b w M S y c VA y VQ It It                          12.3143  MPa 2 2 max 62.333 MPa 2 2 b b b R R  _      _ _       _max 123.4 MPa 

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PROBLEM 8.10

The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement _m_all. For the selected design (use the loading of Prob. 5.74 and selected W250  28.4 shape), determine (a) the actual value of _m in the beam, (b) the maximum value of the principal stress max at the junction of a flange and the web.

SOLUTION

From Problem 5.74, _all 160 MPa

max  48 kN m

M at section E, which lies 1.6 m to the right of B. 0

V 

For W25028.4rolled-steel section, 259 mm  d b_f 102 mm t_f 10.0 mm 6.35 mmt_w  _I _ _{40.1 10 mm}_ 6 4 3 3 308 10 mm   S 1 129.5 mm 2   c d (a) 3 max 6 48 10 308 10     m M S  _m 155.8 MPa ◄ 119.5 mmy_b  c t_f  2 1 1020 mm 2   f f f A b t 1 ( ) 124.5 mm 2   _b  y c y 3 3 6 3 126.99 10 mm 126.99 10 m  _f     Q A y At section E, V 0 M  M_max 143.8 MPa  b  b m y c   0   b w VQ It 

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PROBLEM 8.10 (Continued) At section C, 40 kN 32 kN m    V M 3 3 6 (32 10 )(119.5 10 ) 95.6 MPa 40.0 10 b b My I          3 6 6 3 (40 10 )(126.99 10 ) 19.95 MPa (40.1 10 )(6.35 10 )           b w VQ It  2 2 _47.82 _19.952 _{51.08 MPa} 2    _ _       b b R   max  ₂b R 99.6 MPa 

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PROBLEM 8.11

The following problem refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement _m_all. For the selected design (use the loading of Prob. 5.75 and selected S12  31.8 shape), determine (a) the actual value of _m in the beam, (b) the maximum value of the principal stress max at the junction of a flange and the web.

SOLUTION

From Prob. 5.75, _all 24 ksi max

| |M 54 kip ft 648 kip in.   at C. At C, | | 18 V  kips

For S12 31.8 rolled-steel shape,

4 3

12.0 in., 5.00 in., 0.544 in.,

0.350 in. 217 in , 36.2 in 1 6.00 in. 2 f f w z z d b t t I S c d         | | 648 17.9006 ksi 36.2 m z M S     (a) 17.90 _m ksi  2 3 5.456 in. 16.2776 ksi 8.1388 ksi 2 2.72 in 1 ( ) 5.728 in. 2 15.5802 in b f b b b m f f f b f y c t y c A b t y c y Q A y                 2 2 2 2 (18)(15.5802) 3.6925 ksi (217)(0.350) 8.1388 3.6925 8.9373 ksi 2 b z w b b VQ I t R   _       _ _       max ₂b R 8.1388 8.9373       (b) _max17.08 ksi 

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PROBLEM 8.12

SOLUTION

From Prob. 5.76, _all 24 ksi max

| |M 96 kip ft 1152 kip in.   at D. At D, | | 38.4 V  kips

For S15 42.9 shape,

4 3

15.0 in., 5.50 in. 0.622 in.,

0.411 in., 446 in , 59.4 in 1 7.5 in. 2 f f w z z d b t t I S c d         | | 1152 19.3939 ksi 59.4 m M S     (a) 19.39 _m  ksi  2 3 6.878 in. 17.7855 ksi 8.8928 ksi 2 3.421 in 1 ( ) 7.189 in. 2 24.594 in b f b b b m f f f b f y c t y c A b t y c y Q A y                  2 2 2 2 (57.6)(24.594) 7.7281 ksi (446)(0.411) 8.8928 7.7281 11.7816 ksi 2 b z w b b VQ I t R   _       _ _         _ _b _{ }_R 8.8928 11.7816_ _ _(b) _ __{20.7 ksi}_

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PROBLEM 8.13

SOLUTION

From Problem 5.77, _all 160 MPa

max 256 kN m

M at point B

360 kN 

V at B

For S510 98.2 rolled-steel section,

508 mm, 159 mm, 20.2 mm  _f  _f  d b t 6 4 3 3 12.8 mm, 495 10 mm , 1950 10 mm      w x x t I S 1 254 mm 2   c d (a) max 3 6 256 10 131.3 MPa 1950 10      m x M S   (b) 2 3 3 3 6 6 3 2 2 2 2 max 233.8 120.9 MPa 60. 45 MPa 2 3212 mm 1 ( ) 243.9 mm 2 783.4 10 mm (360 10 )(783.4 10 ) 44.5 MPa (495 10 )(12.8 10 ) 60.45 44.5 75.06 MPa 2 2                            _ _         b f b b b m f f f b f b w b b b y c t y c A b t y c y Q A y VQ It R R      _   60.4575.06135.5 MPa max 135.5 MPa   ◄

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PROBLEM 8.14

SOLUTION Reactions: R_A  65 kN R_D 35 kN max max | | 65 kN | | 175 kN m V M   

For S460 81.4 rolled-steel section,

6 4 3 3 457 mm, 152 mm, 17.6 mm, 1 11.7 mm, 228.5 mm 2 333 10 mm 1460 10 mm f f w d b t t c d I S           (a) max 3 6 6 175 10 119.863 10 Pa 1460 10 m M S     _    _m 119.9 MPa  (b) 2 3 3 210.9 mm 2675.2 mm 1₍ ₎ _{219.7 mm} 2 587.74 10 mm b f f f f b f y c t A b t y c y Q A y            At section C, 210.9(119.863) 110.631 MPa 228.5 b b m y c      3 6 3 6 3 2 2 (35 10 )(2675.2 10 )(219.7 10 ) 5.2799 MPa (333 10 )(11.7 10 )   _                   f b w VA y VQ It It

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PROBLEM 8.14 (Continued) At section B, 162.5 103₆ 111.301 MPa 1460 10 m M S     _   3 6 3 6 3 2 2 210.9 (111.301) 102.728 MPa 228.5 (65 10 )(2675.2 10 )(219.7 10 ) 9.8055 MPa (333 10 )(11.7 10 ) 52.292 MPa 2 b b m f b w b b y c VA y VQ It It R                    _ _         _ max ₂b R     _max 103.7 MPa 

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150 mm T ⫽ 600 N · m P B C A D 150 mm r

PROBLEM 8.15

Determine the smallest allowable diameter of the solid shaft ABCD, knowing that _all  60 MPaand that the radius of disk B is r 80 mm.

SOLUTION 3 axis 3 3 3 3 600 0: 0 7.5 10 N 80 10 1 2 3.75 10 N (3.75 10 )(150 10 ) 562.5 N m A C B T M T Pr P r R R P M                    

Bending moment: (See sketch). Torque: (See sketch).

Critical section lies at point B.



2 2



3 max all 2 2 2 2 3 6 all 562.5 N m, 600 N m 2 (562.5) (600) 2 2 60 10 M T M T J c c M T c                  6 3 8.726 10 m   3 3 20.58 10 m 2 41.2 10 m c    d  c    41.2 d  mm 

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150 mm T ⫽ 600 N · m P B C A D 150 mm r

PROBLEM 8.16

Determine the smallest allowable diameter of the solid shaft ABCD, knowing that _all  60 MPaand that the radius of disk B is r 120 mm.

SOLUTION 3 3 600 0, 0 5 10 N 120 10          AD T M T Pr P r 3 3 3 1 2 2.5 10 N (2.5 10 )(0.150 10 ) 375 N m           A C B R R P M

Bending moment: (See sketch). Torque: (See sketch).

Critical section lies at point B.

375 N m, 600 N m     M T



2 2



3 max all 2    M T J c c   2 2 2 2 3 6 3 6 2 2 375 600 7.5073 10 m 60 10         all M T c    3 19.5807 10 m   c _d _₂_c __{39.2 10 m}_ 3 __{39.2 mm} _

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PROBLEM 8.17

Using the notation of Sec. 8.2 and neglecting the effect of shearing stresses caused by transverse loads, show that the maximum normal stress in a circular shaft can be expressed as follows:



₂ ₂

 

1/2 ₂ ₂ ₂



1/2 max max y z y z c M M M M T J   _     _   SOLUTION

Maximum bending stress:

2 2 | | y z m M M c M c I I    

Maximum torsional stress:

2 2 2 2 2 2 m y z m y z Tc J M M c _c M M I J       

Using Mohr’s circle,





2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 max 2 2 m m y z y z m y z y z c T c R M M J J c M M T J c c R M M M M T J J  _      _ _                



₂ ₂

 

1/2 ₂ ₂ ₂



1/2 y z y z c M M M M T J    _     _   

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D 100 mm 60 mm 90 mm 4 kN Q B C A y z x 80 mm 140 mm

PROBLEM 8.18

The 4-kN force is parallel to the x axis, and the force Q is parallel to the

z axis. The shaft AD is hollow. Knowing that the inner diameter is half the

outer diameter and that _all 60 MPa, determine the smallest permissible outer diameter of the shaft.

SOLUTION 3 3 3 3 0: 60 10 (90 10 )(4 10 ) 0 6 10 N 6 kN y M Q Q            

Bending moment and torque diagrams.

In xy plane: (M_z)_max 315 N m at C. In yz plane: (M_x)_max 412.5 N m at B.

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PROBLEM 8.18 (Continued) At B, 2 2 2 2 2 2 100 (315) 175 N m 180 175 412.5 360 574.79 N m z x z M M M T   _ _            At C, 2 2 2 2 2 2 140 (412.5) 262.5 N m 220 315 262.5 360 545.65 N m x x z M M M T   _ _            Largest value is 574.79 N m.





2 2 2 max 2 2 2 6 max 6 3 3 3 4 4 ₄ ₄ 3 3 4 3 max max 574.79 60 10 9.5798 10 m 9.5798 10 mm 1 2 ₁ ₁ 2 2 2 1.47262 x z x z o i _i o o o o o M M T c J M M T J c c c _c J c c c c c c                       _{ }    _  _  _{ }           _1.47262 3 _{9.5798 10}3 _{18.67 mm} o o c   c   d_o2c_o d_o37.3 mm 

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A 3 in. 10 in. 10 in. 8 in. B C D 6 in. P₁ P2

PROBLEM 8.19

The vertical force P1 and the horizontal force P2 are applied as shown

to disks welded to the solid shaft AD. Knowing that the diameter of the shaft is 1.75 in. and that _all 8 ksi, determine the largest permissible magnitude of the force P2.

SOLUTION Let P2 be in kips. shaft 1 2 1 2 4 0: 6 8 0 3    



M P P P P

Torque over portion ABC: T 8P₂

Bending in horizontal plane: 10 1 ₂ 5 ₂

2 Cy

M   P  P

Bending in vertical plane: ₁

2 1 3 4 3 3 4 Bz M P P P     Critical point is just to the left of point C.

2 2 2 8 _y 5 _z 2 T  P M  P M  P 4 4 2 2 2 all 2 2 2 2 2 2 2 1 1.75 in. 0.875 in. 2 (0.875) 0.92077 in 2 0.875 8 (8 ) (5 ) (2 ) 9.164 0.92077 y z d c d J c T M M J P P P P               2 0.873 kips P  P₂ 873 lb 

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B 7 in. 7 in. 7 in. 7 in. 4 in. 4 in. y A E x z B C 500 lb P 6 in. D 500 lb

PROBLEM 8.20

The two 500-lb forces are vertical and the force P is parallel to the z axis. Knowing that _all 8 ksi, determine the smallest permissible diameter of the solid shaft AE.

SOLUTION 0: (4)(500) 6 (4)(500) 0 666.67 lb x M P P       Torques: : 0 : (4)(500) 2000 lb in. : 4(500) 2000 lb in. : 0           AB T BC T CD T DE T

Critical sections are on either side of disk C. 2000 lb in. 3500 lb in. 4667 lb in.       z y T M M 2 2 2 all 3 2 2 2 2 y z y z c M M T J J c c M M T          

Forces in vertical plane:

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H 90⬚ O V M T 90⬚ (a) (b) O M T K ␤

PROBLEM 8.21

It was stated in Sec. 8.2 that the shearing stresses produced in a shaft by the transverse loads are usually much smaller than those produced by the torques. In the preceding problems, their effect was ignored and it was assumed that the maximum shearing stress in a given section occurred at point H (Fig. P8.21a) and was equal to the expression obtained in Eq. (8.5), namely,

2 2 H c M T J   

Show that the maximum shearing stress at point K (Fig. P8.21b), where the effect of the shear V is greatest, can be expressed as

2 2 2 ( cos ) 3 K c M cV T J    _  _  

where  is the angle between the vectors V and M. It is clear that the effect of the shear V cannot be ignored when _K _H. (Hint: Only the component of M along V contributes to the shearing stress at K.)

SOLUTION

Shearing stress at point K.

Due to V: For a semicircle, 2 3

3

Q c

For a circle cut across its diameter, t d 2c

For a circular section, 1

2 I  J

 

3 2 ₂ 3 1 2 ( ) ₂ 3 (2 ) xy V c VQ Vc It J c J     Due to T: _xy Tc J  

Since these shearing stresses have the same orientation, 2 3 xy c Vc T J   _  _  

(30)

PROBLEM 8.21 (Continued)

Bending stress at point K: _x Mu 2Mu

I J

  

Where u is the distance between point K and the neutral axis,

sin sin cos

2 2 cos x u c c c Mc J           _  _    Using Mohr’s circle,

2 2 2 2 2 2 ( cos ) 3 x K R xy c M Vc T J         _ _       _  _    Cross section

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A 3 in. 10 in. 10 in. 8 in. B C D 6 in. P₁ P2

PROBLEM 8.22

Assuming that the magnitudes of the forces applied to disks A and C of Prob. 8.19 are, respectively, P₁ 1080 lb and P₂ 810 lb,and using the expressions given in Prob. 8.21, determine the values _H and _K in a section (a) just to the left of B, (b) just to the left of C.

SOLUTION

From Prob. 8.19, shaft diameter  1.75 in.

4 1 0.875 in. 2 0.92077 2 c d J c    

Torque over portion ABC:

(6)(1080)T   (8)(810)6480 lb in.

(a) Just to the left of point B: V 1080 lb M 3240 lb in.

 90 T 6480 lb in. 2 2 0.875 ₍₃₂₄₀₎2 ₍₆₄₈₀₎2 0.92077 H c M T J      _H 6880 psi 





2 2 2 3 2 ( cos ) 3       _  _   K C C c c M V T V T J J   0.875 2 (1080)(0.875) 6480 0.92077 3     __{ }  _     6760 K  psi 

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(b) Just to the left of point C: 2 2

1 2 2 1 (162) (405) 436.2 lb 162 tan 21.80 405 (1620) (4050) 4362 lb in. 1620 tan 21.80 4050 90 21.8 21.8 46.4 V M                             0.875 ₍₆₄₈₀₎2 ₍₄₃₆₂₎2 0.92077 H     _H 7420 psi   2 2 (436.2)(0.875) 6480 6734 lb in. 3 3

cos 4362cos 46.4 3008 lb in. C V T M    _{ }             0.875 (3008)2 (6734)2 0.92077 K     _K  7010 psi 

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80 mm 120 mm 120 mm 160 mm 60 mm M A C D F E B

PROBLEM 8.23

The solid shaft AB rotates at 600 rpm and transmits 80 kW from the motor M to a machine tool connected to gear F. Knowing that _all 60 MPa, determine the smallest permissible diameter of shaft AB. SOLUTION 600 rpm 10 Hz 60 sec /min   f 3 80 10 1273.24 N m 2 (2 )(10) P T f        Gear C: _C  C T F r 3 3 1273.24 15.9155 10 N 80 10     C F Gear D: _D  D T F r 3 3 1273.24 21.221 10 N 60 10     D F

3 7 (120 10 ) 1336.90 N m 10      _ _     Cz c M F 120 572.96 N m 280    Dz Cz M M

Forces in horizontal plane:

3 7 (120 10 ) 1782.56 N.m 10      _ _    D y D M F 120 763.95 N m 280    Cy Dy M M

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PROBLEM 8.23 (Continued) 2 2 2 At ,C M_y M_z T 1998.01 N m 2 2 2 At ,D M_y M_z T  2264.3 N m



2 2 2



all max  c M_y M_z T J 



2 2 2



3 max 6 3 6 all 2264.3 37.738 10 m 2 60 10          y z M M T J c c   3 3 28.855 10 m 2 57.7 10 m c    d  c    57.7 mmd  ◄

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80 mm 120 mm 120 mm 160 mm 60 mm M A C D F E B

PROBLEM 8.24

Solve Prob. 8.23, assuming that shaft AB rotates at 720 rpm.

PROBLEM 8.23 The solid shaft AB rotates at 600 rpm and transmits 80 kW from the motor M to a machine tool connected to gear F. Knowing that

all  60 MPa,

 determine the smallest permissible diameter of shaft AB.

SOLUTION 720 rpm 12 Hz 60 sec / min   f 3 80 10 1061.03 N m 2 (2 )(12) P T f        Gear C: _C  C I F r 3 3 1061.03 13.2629 10 N 80 10     C F Gear D: _D  D T F r 3 3 1061.03 17.6838 10 N 60 10     D F

3 7 (120 10 ) 1114.08 N m 10 C z C M    _ F _     120 477.46 N.m 280   Dz Cz M M

Forces in horizontal plane:

3 7 (120 10 ) 1485.44 N m 10      _ _     D y D M F 120 _{636.62 N m} 280    C y D y M M

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PROBLEM 8.24 (Continued) 2 2 2 At ,C M_y M_z T 1665.01 N m 2 2 2 At ,D M_y M_z T 1886.87 N m









2 2 2 all max 2 2 2 3 max 6 3 6 all 3 3 1886.87 31.448 10 m 2 60 10 27.153 10 m 2 54.3 10                    y z y z c M M T J M M T J c c c d c    54.3 mmd  ◄

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M A B 3.5 in. D 6 in. 8 in. 4 in. E F C

PROBLEM 8.25

The solid shafts ABC and DEF and the gears shown are used to transmit 20 hp from the motor M to a machine tool connected to shaft DEF. Knowing that the motor rotates at 240 rpm and that _all7.5 ksi, determine the smallest permissible diameter of (a) shaft ABC, (b) shaft

DEF. SOLUTION 3 20 hp (20)(6600) 132 10 in. lb/s 240 240 rpm 4 Hz 60      

(a) Shaft ABC:

3 132 10 5252 lb in. 2 (2 )(4) P T f        Gear C: 5252 875.4 lb 6 CD C T F r   

Bending moment at B: M_B (8)(875.4) 7003 lb in. 

2 2 all 2 2 2 2 3 3 all (5252) (7003) 1.1671 in 2 7500 c M T J J M T c c            0.9057 c in. d2c d1.811 in. 

(b) Shaft DEF: (3.5)(875.4) 3064 Tr F_{D CD}  lb in.

Bending moment at E: M_E (4)(875.4) 3502 lb in. 

2 2 all 2 2 2 2 3 3 all (3502) (3064) 0.6204 in 2 7500 c M T J J M T c c            0.7337 c in. d2c 1.467 d in. 

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M A B 3.5 in. D 6 in. 8 in. 4 in. E F C

PROBLEM 8.26

Solve Prob. 8.25, assuming that the motor rotates at 360 rpm.

PROBLEM 8.25 The solid shafts ABC and DEF and the gears shown are used to transmit 20 hp from the motor M to a machine tool connected to shaft DEF. Knowing that the motor rotates at 240 rpm and that

all 7.5

  ksi, determine the smallest permissible diameter of (a) shaft

ABC, (b) shaft DEF.

SOLUTION 3 20 hp (20)(6600) 132 10 in. lb/s 360 360 rpm 6 Hz 60      

(a) Shaft ABC:

3 132 10 3501 lb in. 2 (2 )(6) P T f        Gear C: 3501 583.6 lb 6 CD C T F r   

Bending moment at B: (8)(583.6) 4669 lb in.M_B   

2 2 all 2 2 2 2 3 3 all 4669 3501 0.77806 in 2 7500 0.791 in. 2 c M T J J M T c c c d c              d1.582 in. 

(b) Shaft DEF: (3.5)(583.6) 2043 lb in.Tr F_{D CD}   Bending moment at E: (4)(583.6) 2334 M_E   lb in.

2 2 all 2 2 2 3 3 2334 2043 _{0.41362 in}3 2 7500 c M T J J M T c c           

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90 mm 100 mm M C B D E C A

PROBLEM 8.27

The solid shaft ABC and the gears shown are used to transmit 10 kW from the motor M to a machine tool connected to gear D. Knowing that the motor rotates at 240 rpm and that _all 60 MPa, determine the smallest permissible diameter of shaft ABC.

SOLUTION 3 240 rpm 4 Hz 60 sec/ min 10 10 397.89 N m 2 (2 )(4) f P T f          Gear A: 3 0 397.89 4421 N 90 10 A A Fr T T F r        Bending moment at B: 3 2 2 all 2 2 3 all 2 2 2 2 3 6 3 6 all (100 10 )(4421) 442.1 N m 2 2 (2) 442.1 397.89 6.3108 10 m (60 10 ) B AB M L F c M T J J M T c c M T c                          3 3 18.479 10 m 2 37.0 10 m c   d c   37.0 mmd  

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90 mm 100 mm M C B D E C A

PROBLEM 8.28

Assuming that shaft ABC of Prob. 8.27 is hollow and has an outer diameter of 50 mm, determine the largest permissible inner diameter of the shaft.

PROBLEM 8.27 The solid shaft ABC and the gears shown are used to transmit 10 kW from the motor M to a machine tool connected to gear D. Knowing that the motor rotates at 240 rpm and that

all 60 MPa,

  determine the smallest permissible diameter of shaft ABC. SOLUTION 3 240 rpm 4 Hz 60 sec/min 10 10 397.89 N m 2 (2 )(4) f P T f          Gear A: 3 0 397.89 4421 N 90 10 A A Fr T T F r        Bending moment at B: 3 2 2 3 all 4 4 2 2 all 2 2 4 4 all 3 2 2 3 4 6 9 9 9 (100 10 )(4421) 442.1 N m 1 25 10 m 2 ( ) 2 2 (2)(25 10 ) 442.1 397.89 (25 10 ) (60 10 ) 390.625 10 157.772 10 232.85 10 B AB o o o o i o o o i o M L F c M T c d J c c J M T c c c M T c c                                         

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M A 3 in. C F B 4 in. 6 in. 6 in. 8 in. C D H G 4 in. 4 in. E

PROBLEM 8.29

The solid shaft AE rotates at 600 rpm and transmits 60 hp from the motor M to machine tools connected to gears G and H. Knowing that _all 8 ksi and that 40 hp is taken off at gear G and 20 hp is taken off at gear H, determine the smallest permissible diameter of shaft AE.

SOLUTION 3 60 hp (60)(6600) 396 10 in. lb/ sec 600 rpm 10 Hz 60 sec/min       f Torque on gear B: 3 396 10 6302.5 lb in. 2 2 (10) B P T f       

Torques on gears C and D:

40 4201.7 lb in. 60 20 2100.8 lb in. 60       C B D B T T T T Shaft torques: : 0 : 6302.5 lb in. : 2100.8 lb in. : 0       AB BC CD DE AB T BC T CD T DE T Gear forces: 6302.5 2100.8 lb 3 4201.7 1050.4 lb 4 2100.8 525.2 lb 4 B B B C C C D D D T F r T F r T F r         

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PROBLEM 8.29 (Continued) At , 2 2 2 2450.92 6477.62 6302.52 9364 lb in.  _ _ _ _ _   z y B M M T At









2 2 2 2 2 2 2 2 2 all max 2 2 2 3 all 3 3 , 6127.3 3589.2 6302.5 9495 lb in. (maximum) 2 9495 1.1868 in 8 10  _ _ _ _ _             z y z y z y C M M T c M M T J M M T J c c    0.911 in.c d2c 1.822 in.d 

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M A 3 in. C F B 4 in. 6 in. 6 in. 8 in. C D H G 4 in. 4 in. E

PROBLEM 8.30

Solve Prob. 8.29, assuming that 30 hp is taken off at gear G and 30 hp is taken off at gear H.

PROBLEM 8.29 The solid shaft AE rotates at 600 rpm and transmits 60 hp from the motor M to machine tools connected to gears

G and H. Knowing that _all 8 ksi and that 40 hp is taken off at gear G and 20 hp is taken off at gear H, determine the smallest permissible diameter of shaft AE.

SOLUTION 3 60 hp (60)(6600) 396 10 in. lb/ sec 600 rpm 10 Hz 60 sec/min       f Torque on gear B: 3 396 10 6302.5 lb in. 2 2 (10) B P T f       

Torques on gears C and D: 30 3151.3 lb in. 60 30 3151.3 lb in. 60       C B D B T T T T Shaft torques: : 0 : 6302.5 lb in. : 3151.3 lb in. : 0       AB BC CD DE AB T BC T CD T DE T

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PROBLEM 8.30 (Continued) Gear forces: 6302.5 2100.8 lb 3 3151.3 787.8 lb 4 3151.3 787.8 lb 4 B B B C C C D D D T F r T F r T F r          At _, 2 2 2 _1838.22 _6214.92 _6302.52 9040.2 lb in. (maximum)  _ _ _ _ _   z y B M M T At









2 2 2 2 2 2 2 2 2 all max 2 2 2 3 all 3 3 , 4595.5 2932.4 6302.5 8333.0 lb in. 2 9040.3 1.1300 in 8 10  _ _ _ _ _             z y z y z y C M M T c M M T J M M T J c c    0.8960 in.c d2c 1.792 in.d  

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B d e f a c b A 12 in. 6 in. 1.2 kips 1.2 kips 3.5 in. 0.5 in. 1.0 in. 0.5 in. 1.8 in. 1.0 in.

PROBLEM 8.31

Two 1.2-kip forces are applied to an L-shaped machine element AB as shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

SOLUTION

Let  be the slope angle of line AB. 6

tan 26.565

12

  

 

Draw a free body sketch of the portion of the machine element lying above section abc.

P  (1.2)sin  0.53666 kips

V 1.2cos 1.07331 kips

M  (1.8)(1.2cos ) 1.93196 kip in.  

Section properties: _A __(1.0)2 __{1.0 in}2 3 4 1 (1.0)(1.0) 0.083333 in 12   I 0.5 in.  c (a) Point a: 0.53666 (1.93196)( 0.5) 1.0 0.083333 P Mx A I        11.06 ksi   ◄ 0   ◄ (b) Point b: 0.53666 1.0 P A     0.537 ksi   ◄ 3 (0.5)(1.0)(0.25) 0.125 in   Q (1.07331)(0.125) (0.083333)(1.0) VQ It    1.610 ksi  ◄ (c) Point c: 0.53660 (1.93196)(0.5) 1.0 0.083333 P Mx A I       12.13 ksi   ◄ 0   ◄

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B d e f a c b A 12 in. 6 in. 1.2 kips 1.2 kips 3.5 in. 0.5 in. 1.0 in. 0.5 in. 1.8 in. 1.0 in.

PROBLEM 8.32

Two 1.2-kip forces are applied to an L-shaped machine element AB as shown. Determine the normal and shearing stresses at (a) point d, (b) point e, (c) point f.

SOLUTION

Let  be the slope angle of line AB. 6

tan 26.535

12

  

 

Draw a free body sketch of the portion of the machine element lying to the right of section def.

P (1.2)cos

 1.07331 kips

V 1.2sin  0.53666 kips

M (3.5)(1.2sin ) 1.87831 kip in.

Section properties: _A__(1.0)2 __{1.0 in}2 3 4 1 (1.0)(1.0) 0.083333 in 12   I 0.5 in.  c (a) Point d: 1.07331 (1.87831)(0.5) 1.0 0.083333 P My A I       12.34 ksi    ◄ 0  ◄ (b) Point e: 1.07331 1.0 P A     1.073 ksi   ◄

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0.75 m 200 mm 14 kN 150 mm 0.3 m 0.6 m 0.9 m 100 mm 100 mm A D E B b b c a c a

PROBLEM 8.33

The cantilever beam AB has a rectangular cross section of 150  200 mm. Knowing that the tension in the cable BD is 10.4 kN and neglecting the weight of the beam, determine the normal and shearing stresses at the three points indicated.

SOLUTION 2 2 0.75 1.8   DB 1.95 m  Vertical component of T_DB: 0.75 (10.4) 4 kN 1.95   _     Horizontal components of T_DB: 1.8



10.4



9.6 kN 1.95   _    

At section containing points a, b, and c,

P 9.6 kN 14V   4 10 kN M (1.5)(4)(0.6)(14)  2.4 kN m Section properties: 2 (0.150)(0.200) 0.030 m   A 3 6 2 1 (0.150)(0.200) 100 10 m 12     I 0.100 m  c At point a, 9.6 103 (2.4 10 )(0.100)3 ₆ 2.08 MPa 0.030 100 10           x P Mc A I  _x  2.08 MPa ◄ 0 xy   ◄ At point b, 9.6 103 (2.4 10 )(0.100)3 ₆ 2.72 MPa 0.030 100 10            x P Mc A I  _x  2.72 MPa ◄ 0 xy   ◄

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PROBLEM 8.33 (Continued) At point c, 9.6 103 0.320 MPa 0.030        x P A  ◄ 3 3 6 3 (150)(100)(50) 750 10 mm 750 10 m      Q 3 6 6 (10 10 )(750 10 ) 0.500 MPa (100 10 )(0.150)            xy VQ It  ◄ 

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30⬚ 60 mm 60 mm K H G B A 12 mm 12 mm 40 mm 9 kN

PROBLEM 8.34

Member AB has a uniform rectangular cross section of 10 24 mm. For the loading shown, determine the normal and shearing stress at (a) point H, (b) point K.

SOLUTION 0: 0 x x F B    0: (120 sin 30 ) 9(60 sin 30 ) 0 A y M B       4.5 kNB_y 

At the section containing points H and K,

3 3 2 6 2 3 3 4 9 4 4.5 cos 30 3.897 kN 4.5 sin 30 2.25 kN (4.5 10 )(40 10 sin 30 ) 90 N m 10 24 240 mm 240 10 m 1 (10)(24) 11.52 10 mm 12 11.52 10 m P V M A I                          (a) At point H, 3.897 10₆3 240 10 x P A       _  16.24 MPa   3 6 3 3 2.25 10 2 2 240 10 xy V A     _  14.06 MPa   (b) At point K, 3.897 10₆3 (90)(12 10 )₉3 240 10 11.52 10 x P Mc A I        _   _   110.0 MPa    0   

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PROBLEM 8.35

Member AB has a uniform rectangular cross section of 10 24 mm. For the loading shown, determine the normal and shearing stress at (a) point H, (b) point K. SOLUTION 0: (120cos30 ) (60sin 30 )(9) 0 2.598 kN B A A M R R        0: 9 0 y y F B     9 kNB_y   0: 2.598 0 2.598 kN x x x F B B      

3 3 3 3 2 6 2 3 3 4 9 4 9 cos 30 2.598 sin 30 9.093 kN 9 sin 30 2.598 cos 30 2.25 kN (9 10 )(40 10 sin 30 ) (2.598 10 )(40 10 cos 30 ) 90 N m 10 240 240 mm 240 10 m 1 (10)(24) 11.52 10 mm 11.52 10 m 12 P V M A I                                   (a) At point H, 3 6 9.093 10 240 10 x P A       _  37.9 MPa   3 6 3 3 2.25 10 2 2 240 10 xy V A     _  14.06 MPa   (b) At point K,

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PROBLEM 8.36

Member AB has a uniform rectangular cross section of 10 24 mm. For the loading shown, determine the normal and shearing stress at (a) point H, (b) point K. SOLUTION 0: (9)(60sin 30 ) 120 0 2.25 kN B A A M R R       0: 2.25cos30 0 1.9486 kN x x x F B B        0: 2.25sin 30 9 0 7.875 kN y y y F B B        

3 3 3 3 2 6 2 3 3 4 9 4 7.875cos30 1.9486sin 30 7.794 kN 7.875sin 30 1.9486cos30 2.25 kN (7.875 10 )(40 10 sin 30 ) (1.9486 10 )(40 10 cos30 ) 90 N m 10 24 240 mm 240 10 m 1 (10)(24) 11.52 10 mm 11.52 10 m 12 P V M A I                                   (a) At point H, 3 6 7.794 10 240 10 x P A       _  32.5 MPax   3 6 3 3 2.25 10 2 2 240 10 xy V A     _  14.06 MPaxy   (b) At point K, 3 3 6 9 7.794 10 (90)(12 10 ) 240 10 11.52 10 x P Mc A I        _   _   126.2 MPa x     0 xy   

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C 9 in. 9 kip · in. 1.5 kips H K

PROBLEM 8.37

A 1.5-kip force and a 9-kip · in. couple are applied at the top of the 2.5-in.-diameter cast-iron post shown. Determine the normal and shearing stresses at (a) point H, (b) point K.

SOLUTION Diameter 2.5 in.

0 1.5 kips

 

P V

9 kip in. (1.5)(9) 13.5 kip in.

     T M 1 2.5 in. 1.25 in. 2    d c d 2 _{4.909 in}2 4 _{1.9175 in}4 ₂ _{3.835 in}4 4       A c I c J I For a semicircle, 2 3 1.3021 in3 3   Q c (a) At point H, _H 0 ◄ (9)(1.25) (1.5)(1.3021) 2.934 0.407 3.34 ksi 3.835 (1.9175)(2.5)        H Tc VQ J It  _H 3.34 ksi ◄ (b) At point K, (13.5)(1.25) 8.80 ksi 1.9175       K Mc I  _K  8.80 ksi ◄ (9)(1.25) 2.93 ksi 3.835 _K Tc   J K 2.93 ksi ◄

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75 mm 45 mm 1500 N 1200 N 45 mm A B z x y 20 mm a b

PROBLEM 8.38

Two forces are applied to the pipe AB as shown. Knowing that the pipe has inner and outer diameters equal to 35 and 42 mm, respectively, determine the normal and shearing stresses at (a) point a, (b) point b.

SOLUTION









2 2 2 4 4 3 4 3 4 21 mm, 17.5 mm 423.33 mm 2 2 1 158.166 10 mm 79.083 10 mm 2 2 o i o i o i o i d d c c A c c J c c I J                

For semicircle with semicircular cutout,



3 3



3 3

2

2.6011 10 mm

3 o i

Q  c c  

At the section containing points a and b,

3 3 3 1500 N 1200 N 0 (45 10 )(1500) 67.5 N m (75 10 )(1200) 90 N m (90 10 )(1200) 108 N m z x z x P V V M M T                         (a) 1500 ₆ ( 90)(21 10 )₉3 423.33 10 79.083 10 x P M c A I              20.4   MPa  3 9 (108)(21 10 ) 0 158.166 10 x Tc V Q J It          14.34   MPa  (b) 1500 ₆ ( 67.5)(21 10 )₉3 423.33 10 79.083 10 z P M c A I              21.5    MPa  3 6 9 9 3 (108)(21 10 ) (1200)(2.6011 10 ) 158.166 10 (79.083 10 )(7 10 ) z V Q Tc J It                19.98   MPa 

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4 in. 6 in. 4 in. H y z K 150 lb 50 lb x 10 in. 150 lb 200 lb D

PROBLEM 8.39

Several forces are applied to the pipe assembly shown. Knowing that the pipe has inner and outer diameters equal to 1.61 in. and 1.90 in., respectively, determine the normal and shearing stresses at (a) point H, (b) point K.

SOLUTION Section properties: 2 2 2 4 4 4 150 lb (200 lb)(10 in.) 2000 lb in. (150 lb)(10 in.) 1500 lb in. (200 lb 50 lb)(10 in.) (150 lb)(4 in.) 900 lb in. 200 150 50 0 0 (0.95 0.805 ) 0.79946 in (0.95 0.805 ) 0.30989 in 4 z y z y P T M M V V A I J                          4 2I 0.61979 in  

(a) Point H: 150 lb ₂ (1500 lb in.)(0.95 in.)₄

0.79946 in 0.30989 in z H M c P A I      

187.6  psi 4593  psi 4.79 _H  ksi 

4 (2000 lb in.)(0.95 in.) 3065.6 psi 0.61979 in     H Tc J  3.07 _H  ksi  150 lb (900 lb in.)(0.95 in.) M c P 

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50 mm 225 mm 20 mm A H E D B z x y t ⫽ 8 mm 60⬚

PROBLEM 8.40

The steel pipe AB has a 100-mm outer diameter and an 8-mm wall thickness. Knowing that the tension in the cable is 40 kN, determine the normal and shearing stresses at point H.

SOLUTION

Vertical force:

40cos30 34.64 kN Horizontal force: 40sin 30 20 kN

Point H lies on neutral axis of bending.

Section properties: 2 2 3 2 3 2 1 100 mm, 50 mm, 42 mm, 2 ( ) 2.312 10 mm 2.312 10 m o o o i o o i d c d c c t A  c c              3 6 34.64 10 2.312 10 P A       _  14.98    MPa 

For thin pipe,

3 3 (2)(20 10 ) 2 2.314 10 V A     _  17.29   MPa 

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8 in. 2 in. 2 in. 6 kips 2.5 kips 6 kips H y z x B A

PROBLEM 8.41

Three forces are applied to a 4-in.-diameter plate that is attached to the solid 1.8-in. diameter shaft AB. At point H, determine (a) the principal stresses and principal planes, (b) the maximum shearing stress.

SOLUTION

At the section containing point H,

2 2 4 4 4 12 kips (compression) 2.5 kips (2)(2.5) 5 kip in. (8)(2.5) 20 kip in. 1 1.8 in. 0.9 in. 2 2.545 in 0.5153 in 4 2 1.0306 in                  P V T M d c d A c I c J I   For a semicircle, 3 3 2 0.486 in Q  c 

(57)

Use Mohr’s circle.

ave 2 2 1 ( 4.715) 2 2.3575 ksi 4.715 5.676 2 6.1461 ksi         _ _     R

(a) _a _ave R 3.79 ksi_a  

ave _b  R 8.50 ksi_b    (2)(5.676) tan 2 2.408 4.715 p    33.7_a    123.7 b    