Mechanics of Materials 7th Edition Beer Johnson Chapter 8

CHAPTER 8

P

PROBLEM 8.1

P

A

D B a

C 10 ft

a

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 |V |max  90 kips |M |max  (45)(10)  450 kip  in. For W10  39 rolled steel section, d  9.92 in.,

b f  7.99 in., t f  0.530 in.,

tw  0.315 in., I x  209 in 4 c

(a)

1 d  4.96 in. 2

S x  42.1 in 3

yb  c  t f  4.43 in.

m 

|M |max 450  Sx 42.1

b 

yb  4.43  m    (10.69)  9.55 ksi c  4.96 

 m  10.69 ksi 

A f  b f t f  4.2347 in 2 yf 

1 (c  yb )  4.695 in. 2

Qb  A f y f  19.8819 in 3

 xy 

|V |maxQb (45)(19.8819)   13.5898 ksi I xt w (209)(0.315) 2

R

b

(b)

 max 

(c)

Since  max >  all ( 18 ksi),

2

 b  2     xy  14.4043 ksi  2 

 max  19.18 ksi 

 R  19.18 ksi

W10  39 is not acceptable. 

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

For W10  39 rolled steel section, d  9.92 in.,

b f  7.99 in.,

tw  0.315 in.,

I x  209 in 4 ,

c (a)

1 d  4.96 in. 2

t f  0.530 in., S x  42.1 in 3

yb  c  t f  4.43 in.

m 

|M |max 450  Sx 42.1

b 

yb  4.43  m    (10.69)  9.55 ksi c  4.96 

 m  10.69 ksi 

A f  b f t f  4.2347 in 2 1 (c  yb )  4.695 in. 2 Qb  A f y f  19.8819 in 3 yf 

 xy 

|V |max Qb (22.5)(19.8819)   6.7949 ksi (209)(0.315) I x tw 2

  2  8.3049 ksi R   b    xy 2  

b

(b)

 max 

(c)

Since  max <  all ( 18 ksi),

2

 max  13.08 ksi 

 R  13.08 ksi

W10  39 is acceptable. 

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P C

A B 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 |V |max  700 kN  700  103 N |M |max  (700  103 )(2.5)  1.75  106 N  m For W920  449 rolled steel beam, d  947 mm,

b f  424 mm,

tw  24.0 mm,

I x  8780  106 mm 4 , S x  18,500  103 mm3

c

(a)

1 d  473.5 mm, 2

t f  42.7 mm,

yb  c  t f  430.8 mm

 m  94.595 MPa  m  94.6 MPa 

|M |max 1.75  106 m   Sx 18,500  106 yb 430.8 (94.595)  86.064 MPa m  473.5 c A f  b f tt  18.1048  103 mm 2

b 

1 (c  yb )  452.15 mm 2 Qb  A f y f  8186.1  103 mm3  8186.1  106 m3 yf 

 xy 

|V |max Qb (700  103 )(8186.1  106 )   27.194 MPa I x tw (8780  106 )(24.0  103 ) 2

2

   86.064  2    27.1942  50.904 MPa R   b    xy   2   2 

b

(b)

 max 

(c)

Since 94.6 MPa   all ( 100 MPa),

2

 max  93.9 MPa 

 R  93.9 MPa

W920  449 is acceptable. 

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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 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 |V |max  850 kN  850  103 N |M |max  (850  103 )(2.0)  1.70  106 N  m For W920  449 rolled steel section, d  947 mm,

b f  424 mm,

tw  24.0 mm,

I x  8780  106 mm 4 , S x  18,500  103 mm3

c

(a)

1 d  473.5 mm 2

t f  42.7 mm,

yb  c  t f  430.8 mm

 m  91.892 MPa  m  91.9 MPa 

|M |max 1.70  106 m   Sx 18,500  106

yb 430.8 (91.892)  83.605 MPa m  473.5 c A f  b f t f  18.1048  103 mm 2

b 

1 (c  yb )  452.15 mm 2 Qb  A f y f  8186.1  103 mm3  8186.1  106 m3 yf 

 xy 

|V |max Qb (850  103 )(8186.1  106 )   33.021 MPa I x tw (8780  106 )(24.0  103 ) 2

2

   83.605  2 2 R   b    xy     33.021  53.271 MPa 2 2    

b

(b)

 max 

(c)

Since 95.1 MPa <  all ( 100 MPa),

2

 max  95.1 MPa 

 R  95.1 MPa

W920  449 is acceptable. 

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PROBLEM 8.5

40 kN 2.2 kN/m

A

C B 4.5 m

2.7 m

(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  M C  0: 7.2RA  (2.2)(7.2)(3.6)  (40)(2.7)  0 RA  22.92 kN VA  RA  22.92 kN VB  22.92  (2.2)(4.5)  13.02 kN VB  13.02  40  26.98 kN VC  26.98  (2.2)(2.7)  32.92 kN MA  0 MB  0 

1 (22.92  13.02)(4.5)  80.865 kN  m 2

MC  0

S min 

M

max

 all



80.865  103  490  106 m3 165  106  490  103 mm3

Shape

S (103 mm3)

W360  39

578

W310  38.7

547

W250  44.8

531

W200  52

511



(a) 

d  310 mm tw  5.84 mm

t f  9.65 mm

Use W310  38.7. 



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PROBLEM 8.5 (Continued)

(b)

m 

MB 80.865  103   147.834  106 Pa 6 S 547  10

 m  147.8 MPa 

m 

V

max

dtw



32.92  103  18.1838  106 Pa (310  103 )(5.84  103 )

 m  18.18 MPa  1 d  155 mm yb  c  t f  155  9.65  145.35 mm 2 y  145.35  b  b m    (147.834)  138.630 MPa c  155  c

At point B,

w 

V (26.98  103 )   14.9028 MPa dtw (310  103 )(5.84  103 ) 2

R

 max 

 b  2    w   2 

b 2

(69.315) 2  (14.9028)2  70.899 MPa

 max  140.2 MPa 

 R  69.315  70.899

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PROBLEM 8.6

275 kN B

C

A

D 275 kN 3.6 m

1.5 m

1.5 m

(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 RB  504.17 kN  V

max

 275 kN

Smin 

M

max

 all

M 

RC  504.17  max

 412.5 kN  m

412.5  102  2578  106 m3 6 160  10  2578  103 mm3

Shape 

Sx (103 mm3)

W760  147

4410

W690  125

3490

W530  150

3720

W460  158

3340

W360  216

3800 (a)

d  678 mm

 

(b)



m 



M

m  V

max

Aw

c



max

S V

max

dtw

 

412.5  103  118.195  106 Pa  3490  106 275  103  34.667  106 Pa  (678  103 )(11.7  103 )

1 67.8 d   339 mm, t f  16.3 mm, 2 2

b 

t f  16.3 mm

Use W690  125. 

tw  11.7 mm 

 m  118.2 MPa   m  34.7 MPa 

yb  c  t f  339  16.3  322.7 mm 

yb  322.7  m    (118.195)  112.512 MPa c  339  2



R



 max 

 b  2    m   2 

b 2

(56.256) 2  (34.667)2  66.080 MPa 

 max  122.3 MPa 

 R  56.256  66.080 

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20 kips

20 kips

PROBLEM 8.7

2 kips/ft A

B

10 ft

C

30 ft

D

10 ft

(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 RA  50 kips 

V

max

 30 kips

S min 

M

max

 all

RD  50 kips 

M 

max

 200 kip  ft  2400 kip  in.

2400  100 in 3 24 Shape



S (in 3 )

W24  68

154

W21  62

127

W18  76

146

W16  77

134

W12  96

131

W10  112

126 Use W21  62. 

(a)

d  21.0 in. t f  0.615 in. tw  0.400 in.

(b)

M

m 

S V

m  c

b 



max

max

dtw



2400  18.8976 ksi 127

 m  18.9 ksi 

30  3.5714 ksi (21.0)(0.400)

1 21.0 d   10.50 in. 2 2

 m  3.57 ksi 

yb  c  t f  10.50  0.615  9.8850 in.

yb  9.8850  m    (18.8976)  17.7907 ksi c  10.50  2

 b  2    m  2  

R

 max 

b 2

(8.8954) 2  (3.5714) 2  9.5856 ksi

 R  8.8954  9.5856  18.4810 ksi

 max  18.48 ksi ◄

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PROBLEM 8.8

1.5 kips/ft

A

C B 12 ft

6 ft

(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  M B  0: 12 RA  (1.5)(18)(3)  0 RA  6.75 kips   M A  0: 12 RB  (1.5)(18)(9)  0 RB  20.25 kips 

V M

 11.25 kips

max

 27 kip  ft  324 kip  in.

max

Smin 

M

 all



(a)

max



324  13.5 in 3 24

Shape

S (in 3 )

W12  16

17.1

W10  15

13.8

W8  18

15.2

W6  20

13.4

Use W10  15.



d  10.0 in. t f  0.270 in. 



tw  0.230 in.

(b)

m 

M



max

S

324  23.478 ksi 13.8

 m  23.5 ksi  m 

V

max

dtw



11.25  4.8913 ksi (10.0)(0.230)

 m  4.89 ksi  1 10.0  c  d   5.00 in.  2 2

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PROBLEM 8.8 (Continued)

yb  c  t f  5.00  0.270  4.73 in.

b 

yb  4.73  m    (23.478)  22.210 ksi c  5.00  2

R

 max 

 b  2    m  2  

b 2

R

2

 22.210  2    (4.8913)  12.1345 ksi 2  

22.210  12.1345 2

 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:

RA  97.5 kN  RD  97.5 kN 

|V |max  97.5 kN |M |max  286 kN  m

For W530  92 rolled-steel section, d  533 mm,

b f  209 mm,

tw  10.2 mm,

c

I  554  106 mm 4 (a)

m 

t f  15.6 mm,

1 d  266.5 mm 2 S  2080  103 mm3

| M |max 286  103   137.5  106 Pa S 2080  106

 m  137.5 MPa  yb  c  t f  250.9 mm A f  b f t f  3260.4 mm 2 y 

1 (c  yb )  258.7 mm 2

Q  A f y  843.47  103 mm3 At midspan:

V 0

b 

b  0

yb 250.9 (137.5)  129.5 MPa m  c 266.5

 max  129.5 MPa 

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PROBLEM 8.9 (Continued)

At sections B and C: M 270  103   129.808 MPa S 2080  106 250.9 y (129.808)  122.209 MPa b  b m  c 266.5 VQ VA f y (82.5  103 )(3260.4  106 )(258.7  103 )   b  It Itw (554  106 )(10.2  103 )

m 

 12.3143 MPa 2

R

 max 

 b  2     b  62.333 MPa  2 

b 2

 max  123.4 MPa 

R

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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

M

max

 48 kN  m at section E, which lies 1.6 m to the right of B.

V 0 For W250  28.4 rolled-steel section, d  259 mm

b f  102 mm

t f  10.0 mm

I  40.1  106 mm 4

tw  6.35 mm

S  308  103 mm3 1 d  129.5 mm 2

c

(a)

M

m 

max

S



48  103 308  106

 m  155.8 MPa ◄ yb  c  t f  119.5 mm 1 b f t f  1020 mm 2 2

Af 

1 (c  yb )  124.5 mm 2

y 

Q  A f y  126.99  103 mm3  126.99  106 m3

At section E,

(b)

V 0

b 

yb  m  143.8 MPa c

b 

VQ 0 Itw

 max   b

M  M max

 max  143.8 MPa ◄

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PROBLEM 8.10 (Continued) At section C, V  40 kN

M  32 kN  m

b 

Myb (32  103 )(119.5  103 )   95.6 MPa I 40.0  106

b 

VQ (40  103 )(126.99  106 )   19.95 MPa Itw (40.1  106 )(6.35  103 ) 2

R

 max 

 b  2    b  2  

b 2

47.82  19.952  51.08 MPa

 R  99.6 MPa (less than value at section E)

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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

 all  24 ksi

From Prob. 5.75,

|M |max  54 kip  ft  648 kip  in. at C. At C,

|V |  18 kips

For S12  31.8 rolled-steel shape, d  12.0 in.,

b f  5.00 in.,

tw  0.350 in.

I z  217 in 4 ,

c

m 

t f  0.544 in., S z  36.2 in 3

1 d  6.00 in. 2 |M | 648   17.9006 ksi S z 36.2

(a)

 m  17.90 ksi 

yb  c  t f  5.456 in. yb  m  16.2776 ksi c A f  b f t f  2.72 in 2

b 

b 2

 8.1388 ksi

1 (c  yb )  5.728 in. 2 Q  A f y  15.5802 in 3 y

b 

VQ (18)(15.5802)   3.6925 ksi I z tw (217)(0.350) 2

  R   b    b2  8.13882  3.69252  8.9373 ksi  2 

 max 

b 2

 R  8.1388  8.9373

(b)

 max  17.08 ksi 

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PROBLEM 8.12 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. 5.76 and selected S15  42.9 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

 all  24 ksi

From Prob. 5.76,

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

At D,

For S15  42.9 shape, d  15.0 in.,

b f  5.50 in.

tw  0.411 in., c

m 

t f  0.622 in.,

I z  446 in 4 ,

S z  59.4 in 3

1 d  7.5 in. 2 |M | 1152   19.3939 ksi 59.4 S

(a)

 m  19.39 ksi 

yb  c  t f  6.878 in. yb  m  17.7855 ksi c A f  b f t f  3.421 in 2

b 

b 2

 8.8928 ksi

1 (c  yb )  7.189 in. 2 Q  A f y  24.594 in 3 y

b 



VQ (57.6)(24.594)   7.7281 ksi I z tw (446)(0.411) 2

  R   b    b2  8.89282  7.72812  11.7816 ksi   2 



 max 

b 2

 R  8.8928  11.7816 

(b)

 max  20.7 ksi 

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PROBLEM 8.13 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. 5.77 and selected S510  98.2 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

 all  160 MPa

From Problem 5.77, M

max

 256 kN  m at point B

V  360 kN at B

For S510  98.2 rolled-steel section, d  508 mm,

b f  159 mm,

I x  495  106 mm 4 ,

tw  12.8 mm, c (a)

t f  20.2 mm S x  1950  103 mm3

1 d  254 mm 2

m 

M

max

Sx



256  103  131.3 MPa 1950  106



yb  c  t f  233.8

b 

yb  m  120.9 MPa c

b 2

 60. 45 MPa

A f  b f t f  3212 mm 2 y 

1 (c  yb )  243.9 mm 2

Q  A f y  783.4  103 mm3

b 

VQ (360  103 )(783.4  106 )   44.5 MPa Itw (495  106 )(12.8  103 ) 2

R

(b)  max 

 b  2    b  2  

b 2

60.452  44.52  75.06 MPa

 R  60.45  75.06  135.5 MPa

 max  135.5 MPa ◄

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PROBLEM 8.14 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. 5.78 and selected S460  81.4 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 RA  65 kN 

Reactions:

RD  35 kN 

|V |max  65 kN |M |max  175 kN  m

For S460  81.4 rolled-steel section, d  457 mm, tw  11.7 mm, I  333  106 mm 4

(a)

m 

M

max

S

b f  152 mm, t f  17.6 mm, 1 c  d  228.5 mm 2 S  1460  103 mm3



175  103  119.863  106 Pa 1460  106

 m  119.9 MPa  (b)

yb  c  t f  210.9 mm A f  b f t f  2675.2 mm 2 y 

1 (c  yb )  219.7 mm 2

Q  A f y  587.74  103 mm3 At section C,

b 

yb 210.9 (119.863)  110.631 MPa m  c 228.5

b 

VQ VA f y (35  103 )(2675.2  106 )(219.7  103 )    5.2799 MPa It Itw (333  106 )(11.7  103 ) 2

R

 max 

 b  2     b  55.567 MPa 2  

b 2

 max  110.9 MPa 

R

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PROBLEM 8.14 (Continued)

At section B,

M 162.5  103   111.301 MPa S 1460  106 y 210.9 b  b m  (111.301)  102.728 MPa c 228.5 VQ VA f y (65  103 )(2675.2  106 )(219.7  103 ) b     9.8055 MPa It Itw (333  106 )(11.7  103 )

m 

2

R

 max 

 b  2     b  52.292 MPa  2 

b 2

 max  103.7 MPa 

R

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PROBLEM 8.15 Determine the smallest allowable diameter of the solid shaft ABCD, knowing that  all  60 MPa and that the radius of disk B is r  80 mm.

A r B

P

150 mm

C 150 mm D

T ⫽ 600 N · m

SOLUTION M axis  0: T  Pr  0 P  RA  RC 

T 600   7.5  103 N r 80  103

1 P 2

 3.75  103 N M B  (3.75  103 )(150  103 )  562.5 N  m Bending moment: (See sketch). Torque: (See sketch). Critical section lies at point B. M  562.5 N  m,

 J  c3  c 2 c3 

2





T  600 N  m

M2  T2

 all

M2  T2

 all



2





max

(562.5) 2  (600) 2 60  106

 8.726  106 m3 c  20.58  103 m

d  2c  41.2  103 m

d  41.2 mm 

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PROBLEM 8.16 Determine the smallest allowable diameter of the solid shaft ABCD, knowing that  all  60 MPa and that the radius of disk B is r  120 mm.

A r B

P

150 mm

C 150 mm D

T ⫽ 600 N · m

SOLUTION M AD  0,

T  Pr  0

RA  RC 

P

T 600   5  103 N 3 r 120  10

1 P 2

 2.5  103 N M B  (2.5  103 )(0.150  103 )  375 N  m Bending moment: (See sketch). Torque: (See sketch). Critical section lies at point B. M  375 N  m, J   c3  c 2

c3 

2





T  600 N  m M2  T2

M2  T2

 all

 all 

2



c  19.5807  103 m



max

3752  6002  7.5073  106 m3 60  106 d  2c  39.2  103 m  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:

 max 



c 2 M y  M z2 J 



1/2



 M y2  M z2  T 2



1/2 

 max

SOLUTION Maximum bending stress:

m 

|M | c  I

m 

Tc J

M y2  M z2 c I

Maximum torsional stress:

m

M y2  M z2 c



2

2I



c J

M y2  M z2

Using Mohr’s circle, 2



 max  

c2 T 2c 2 2 2 M M   y z J2 J2

 m  2    m   2 

R c J





M y2  M z2  T 2

m

R

2



c J

c 2 M y  M z2 J 

M y2  M z2 



1/2



c J

M y2  M z2  T 2

 M y2  M z2  T 2



1/2 





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PROBLEM 8.18

y

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.

A 60 mm Q

B 90 mm

100 mm C

4 kN

80 mm

140 mm

D z

x

SOLUTION My  0: 60  103 Q  (90  103 )(4  103 )  0 Q  6  103 N  6 kN 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.

About z axis:

Tmax  360 N  m

between B and C.

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PROBLEM 8.18 (Continued)

At B,  100  Mz    (315)  175 N  m  180  M x2  M z2  T 2  1752  412.52  3602  574.79 N  m

At C,  140  Mx    (412.5)  262.5 N  m  220  M x2  M z2  T 2  3152  262.52  3602  545.65 N  m

Largest value is 574.79 N  m.

 max 

max M x2  M z2  T 2 c J M x2

J max  c

 M z2  T 2

 max



574.79 60  106

 9.5798  106 m3  9.5798  103 mm3





 4 4   c4 J 2 co  ci   co3 1  i4 2  co c co

  3   1 4    co 1      2   2  

 1.47262co3 1.47262co3  9.5798  103 

d o  2co 

co  18.67 mm d o  37.3 mm 

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6 in.

A

8 in. B C

P1 3 in.

PROBLEM 8.19

P2

D

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.

10 in. 10 in.

SOLUTION Let P2 be in kips.

 M shaft  0:

6 P1  8P2  0

Torque over portion ABC:

P1 

4 P2 3

T  8P2

Bending in horizontal plane:

M Cy  10 

Bending in vertical plane:

M Bz  3P1  3

1 P2  5P2 2

4 P2 3

 4 P1

Critical point is just to the left of point C. T  8P2

M y  5P2

d  1.75 in.

c

M z  2 P2

1 d  0.875 in. 2



(0.875) 4  0.92077 in 4 2 c T 2  M y2  M z2  all  J 0.875 8 (8P2 ) 2  (5P2 ) 2  (2 P2 ) 2  9.164 P2 0.92077 J 

P2  0.873 kips

P2  873 lb 

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PROBLEM 8.20

y 7 in. 7 in.

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.

7 in. 4 in. A

7 in.

P

B

4 in. C

B

z

E

6 in.

D x

500 lb 500 lb

SOLUTION

Forces in vertical plane: M x  0: (4)(500)  6 P  (4)(500)  0 P  666.67 lb

Torques: AB : T  0 BC : T  (4)(500)  2000 lb  in. CD : T  4(500)  2000 lb  in. DE : T  0

Critical sections are on either side of disk C.

T  2000 lb  in. M z  3500 lb  in. M y  4667 lb  in.

 all 

c J

M y2  M z2  T 2

Forces in horizontal plane:

J  3  c c 2 

M y2  M z2  T 2

 all

4667 2  35002  20002 8  103  0.77083 in 3 

c  0.7888 in.

d  2c

d  1.578 in. 

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PROBLEM 8.21 90⬚ M H O

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,

T

H 

(a) V M

␤

K

K  T

(b)

M2 T2

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

O 90⬚

c J

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

2

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:

Q

For a semicircle,

2 3 c 3

For a circle cut across its diameter,

t  d  2c

For a circular section,

I

 xy Due to T:

1 J 2

 

2 3 VQ (V ) 3 c 2 Vc 2   1  It  2 J  (2c) 3 J

 xy 

Tc J

Since these shearing stresses have the same orientation,

 xy 

c2  Vc  T  J  3 

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PROBLEM 8.21 (Continued)

Bending stress at point K:

x 

Mu 2 Mu  I J

Where u is the distance between point K and the neutral axis,   u  c sin   c sin      c cos  2   2 Mc cos  x  J

Using Mohr’s circle, 2

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

Cross section 2



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6 in.

PROBLEM 8.22

P2

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.

8 in.

A

B C

P1 3 in.

D

10 in. 10 in.

SOLUTION From Prob. 8.19, shaft diameter  1.75 in. c J 

1 d  0.875 in. 2

 2

c 4  0.92077

Torque over portion ABC: T  (6)(1080)  (8)(810)  6480 lb  in.

(a)

Just to the left of point B:

H 

V  1080 lb

M  3240 lb  in.

  90

T  6480 lb  in.

c J

M2  T2 

0.875 (3240)2  (6480)2 0.92077

 H  6880 psi  K  

c (M cos  ) 2  J

 32 VC  T 

2



c 2  VC  T   J 3 

 0.875  2    (1080)(0.875)  6480   0.92077  3  

 K  6760 psi 

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PROBLEM 8.22 (Continued)

(b)

Just to the left of point C:

V 

(162)2  (405) 2  436.2 lb

  tan 1 M 



162  21.80 405

(1620) 2  (4050)2  4362 lb  in.

1620  21.80 4050   90  21.8  21.8  46.4

  tan 1







H 

0.875 (6480)2  (4362) 2  0.92077

 H  7420 psi 

2 2 VC  T    (436.2)(0.875)  6480  6734 lb  in.  3 3 M cos   4362 cos 46.4  3008 lb  in.

K 

0.875 (3008)2  (6734)2  0.92077

 K  7010 psi 

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PROBLEM 8.23 120 mm

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.

160 mm M A

120 mm C

F

D E 80 mm

B 60 mm

SOLUTION f  T  Gear C:

Gear D:

600 rpm  10 Hz 60 sec /min P 2 f



80  103  1273.24 N  m (2 )(10)

FC 

T rC

FC 

1273.24  15.9155  103 N 3 80  10

FD 

T rD

FD 

1273.24  21.221  103 N 60  103

Forces in vertical plane:  7  M Cz  (120  103 )  Fc   1336.90 N  m  10  M Dz 

120 M Cz  572.96 N  m 280

Forces in horizontal plane:  7  M D y  (120  103 )  FD   1782.56 N.m  10  120 M Cy  M Dy  763.95 N  m 280 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. 1271

PROBLEM 8.23 (Continued)

At C ,

M y2  M z2  T 2  1998.01 N  m

At D,

M y2  M z2  T 2  2264.3 N  m

 all 

c J





M y2  M z2  T 2

 J  c3  2 c



max

M y2  M z2  T 2

c  28.855  103 m

 all



max



2264.3  37.738  106 m3 60  106

d  2c  57.7  103 m

d  57.7 mm ◄

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PROBLEM 8.24 120 mm

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

160 mm M A

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.

120 mm C

F

D E 80 mm

B 60 mm

SOLUTION f  T  Gear C:

Gear D:

720 rpm  12 Hz 60 sec / min P 2 f



80  103  1061.03 N  m (2 )(12)

FC 

I rC

FC 

1061.03  13.2629  103 N 80  103

FD 

T rD

FD 

1061.03  17.6838  103 N 3 60  10

Forces in vertical plane:  7  M C z  (120  103 )  FC   1114.08 N  m 10   M Dz 

120 M Cz  477.46 N.m 280

Forces in horizontal plane:  7  M D y  (120  103 )  FD   1485.44 N  m  10  MC y 

120 M D y  636.62 N  m 280

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PROBLEM 8.24 (Continued)

At C ,

M y2  M z2  T 2  1665.01 N  m

At D,

M y2  M z2  T 2  1886.87 N  m

 all 

c J



M y2  M z2  T 2

 J  c3  2 c



M y2



M z2

 all 3

c  27.153  10 m



max

 T2



max



1886.87  31.448  106 m3 60  106

d  2c  54.3  103

d  54.3 mm ◄

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PROBLEM 8.25

8 in. 4 in. M

3.5 in. A

D

B

E F

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.

C 6 in.

SOLUTION 20 hp  (20)(6600)  132  103 in.  lb/s 240 240 rpm   4 Hz 60 (a)

Shaft ABC:

T

P 2 f



132  103  5252 lb  in. (2 )(4)

T 5252   875.4 lb rC 6

Gear C:

FCD 

Bending moment at B:

M B  (8)(875.4)  7003 lb  in.

 all 

c J

M2 T2

J  3  c  c 2

M2 T2

 all

(5252) 2  (7003)2   1.1671 in 3 7500

d  1.811 in. 

c  0.9057 in. d  2c

(b)

Shaft DEF: Bending moment at E:

T  rD FCD  (3.5)(875.4)  3064 lb  in. M E  (4)(875.4)  3502 lb  in.

 all 

c J

M2 T2

J  3  c  c 2

M2 T2

 all

c  0.7337 in. d  2c



(3502) 2  (3064)2  0.6204 in 3 7500 d  1.467 in. 

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PROBLEM 8.26

8 in. 4 in. M

3.5 in. A

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

D

B

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.

E F C 6 in.

SOLUTION 20 hp  (20)(6600)  132  103 in.  lb/s 360 360 rpm   6 Hz 60 (a)

Shaft ABC :

Gear C:

T FCD 

P 2 f



132  103  3501 lb  in. (2 )(6)

T 3501   583.6 lb rC 6

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

 all 

c J

M2 T2

J  3  c  c 2

M2 T2

 all

c  0.791 in.

(b) Shaft DEF: Bending moment at E:



46692  35012  0.77806 in 3 7500

d  1.582 in. 

d  2c

T  rD FCD  (3.5)(583.6)  2043 lb  in. M E  (4)(583.6)  2334 lb  in.

 all 

c J

M2 T2

J  3  c  c 2

M2 T2

c  0.6410 in.

 all



d  2c

23342  20433  0.41362 in 3 7500

d  1.282 in. 

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PROBLEM 8.27

100 mm

M

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.

C B C A

90 mm

D E

SOLUTION f  T

Gear A:

240 rpm  4 Hz 60 sec/ min P 2 f



10  103  397.89 N  m (2 )(4)

FrA  T  0 F

T 397.89   4421 N rA 90  103

Bending moment at B: M B  LAB F  (100  103 )(4421)  442.1 N  m

 all 

c J

M2 T2

J  3  c  c 2 c3 

2



M2 T2

 all

M2 T2

 all



c  18.479  103 m

(2) 442.12  397.892  6.3108  106 m3 6  (60  10 ) d  2c  37.0  103 m

d  37.0 mm 

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PROBLEM 8.28

100 mm

M

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.

C B C A

90 mm

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.

D E

SOLUTION f  T

Gear A:

240 rpm  4 Hz 60 sec/min P 2 f



10  103  397.89 N  m (2 )(4)

FrA  T  0 F

T 397.89   4421 N rA 90  103

Bending moment at B: M B  LAB F  (100  103 )(4421)  442.1 N  m

 all 

co J

M2 T2

co 

J  (co4  ci4 )   co 2 co ci4  co4 

1 d o  25  103 m 2

M2 T2

 all

2co M 2  T 2

 all

 (25  103 ) 4 

(2)(25  103 ) 442.12  397.892  (60  106 )

 390.625  109  157.772  109  232.85  109

ci  21.967  103 m

di  2ci  43.93  103 m

d  43.9 mm 

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PROBLEM 8.29

4 in.

M

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.

6 in. F 8 in.

A BC

3 in.

6 in.

H

C

D

G 4 in.

E 4 in.

SOLUTION 60 hp  (60)(6600)  396  103 in.  lb/ sec 600 rpm f   10 Hz 60 sec/min Torque on gear B:

TB 

P 2 f



396  103  6302.5 lb  in. 2 (10)

Torques on gears C and D: 40 TB  4201.7 lb  in. 60 20 TD  TB  2100.8 lb  in. 60 TC 

Shaft torques: AB : BC : CD : DE :

TAB TBC TCD TDE

0  6302.5 lb  in.  2100.8 lb  in. 0

Gear forces: TB 6302.5   2100.8 lb 3 rB T 4201.7 FC  C   1050.4 lb 4 rC T 2100.8 FD  D   525.2 lb 4 rD FB 

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PROBLEM 8.29 (Continued) Forces in horizontal plane:

Forces in vertical plane:

At B  ,

M z2  M y2  T 2  2450.92  6477.62  6302.52  9364 lb  in.

At C  ,

M z2  M y2  T 2  6127.32  3589.22  6302.52  9495 lb  in. (maximum)

 all 

c J



M z2  M y2  T 2

J  3  c  c 2



M z2



M y2



max

T2



 all

9495  1.1868 in 3 8  103 c  0.911 in. d  2c 

d  1.822 in. 

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PROBLEM 8.30

4 in.

M

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

6 in. F 8 in.

A

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.

BC C

3 in.

6 in.

H

D

G 4 in.

E 4 in.

SOLUTION 60 hp  (60)(6600)  396  103 in.  lb/ sec 600 rpm f   10 Hz 60 sec/min Torque on gear B:

TB 

P 2 f



396  103  6302.5 lb  in. 2 (10)

Torques on gears C and D: 30 TB  3151.3 lb  in. 60 30 TD  TB  3151.3 lb  in. 60 TC 

Shaft torques: AB : TAB  0 BC : TBC  6302.5 lb  in. CD : TCD  3151.3 lb  in. DE : TDE  0

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PROBLEM 8.30 (Continued)

Gear forces:

Forces in vertical plane: T 6302.5 FB  B   2100.8 lb rB 3

At B  ,

FC 

TC 3151.3   787.8 lb rC 4

FD 

TD 3151.3   787.8 lb rD 4

M z2  M y2  T 2  1838.22  6214.92  6302.52

Forces in horizontal plane:

 9040.2 lb  in. (maximum)

At C  ,

M z2  M y2  T 2  4595.52  2932.42  6302.52  8333.0 lb  in.

 all 

c J



M z2  M y2  T 2

J  3  c  c 2 



M z2



M y2



max

T2



 all

9040.3  1.1300 in 3 8  103

c  0.8960 in.

d  2c

d  1.792 in. 

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PROBLEM 8.31

12 in. A 1.8 in.

b a

c

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.

1.2 kips

d 0.5 in.

1.2 kips

6 in.

e f

0.5 in.

1.0 in. B 3.5 in. 1.0 in.

SOLUTION Let  be the slope angle of line AB. tan  

6 12

  26.565

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. A  (1.0)2  1.0 in 2

Section properties:

I 

1 (1.0)(1.0)3  0.083333 in 4 12

c  0.5 in. (a)

 

Point a:

P Mx 0.53666 (1.93196)(0.5)    1.0 0.083333 A I

  11.06 ksi ◄

  0◄ (b)

Point b:

 

P 0.53666  A 1.0

  0.537 ksi ◄

Q  (0.5)(1.0)(0.25)  0.125 in 3

  (c)

Point c:  

VQ (1.07331)(0.125)  It (0.083333)(1.0)

  1.610 ksi ◄

P Mx 0.53660 (1.93196)(0.5)    1.0 0.083333 A I

  12.13 ksi ◄   0 ◄

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PROBLEM 8.32

12 in. A 1.8 in.

b a

c

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.

1.2 kips

d 0.5 in.

1.2 kips

6 in.

e 0.5 in.

1.0 in. B

f

3.5 in. 1.0 in.

SOLUTION Let  be the slope angle of line AB. tan  

6 12

  26.535

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 I 

1 (1.0)(1.0)3  0.083333 in 4 12

c  0.5 in. (a)

Point d:

 

P My 1.07331 (1.87831)(0.5)    1.0 0.083333 A I

  12.34 ksi ◄ 0 ◄ (b)

Point e:

 

P 1.07331  A 1.0

  1.073 ksi ◄

Q  (0.5)(1.0)(0.25)  0.125 in 3

(c)

Point f:

 

VQ (0.53666)(0.125)  It (0.083333)(1.0)

  0.805 ksi ◄

 

P My 1.07331 (1.87831)(0.5)    1.0 0.083333 A I

  10.20 ksi ◄ 0 ◄

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100 mm

a

PROBLEM 8.33

150 mm

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.

c

100 mm b

D

0.75 m

a c

A b

200 mm

B

E

14 kN 0.9 m 0.3 m 0.6 m

SOLUTION DB 

0.752  1.82

 1.95 m Vertical component of TDB :

 0.75    (10.4)  4 kN  1.95 

Horizontal components of TDB :

 1.8    10.4   9.6 kN  1.95 

At section containing points a, b, and c, P  9.6 kN

V  14  4  10 kN

M  (1.5)(4)  (0.6)(14)  2.4 kN  m Section properties: A  (0.150)(0.200)  0.030 m 2 I 

1 (0.150)(0.200)3  100  106 m 2 12

c  0.100 m At point a,

x  

P Mc 9.6  103 (2.4  103 )(0.100)     2.08 MPa A I 0.030 100  106

 x  2.08 MPa ◄  xy  0 ◄

At point b,

x  

P Mc 9.6  103 (2.4  103 )(0.100)     2.72 MPa A I 0.030 100  106

 x  2.72 MPa ◄  xy  0 ◄

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PROBLEM 8.33 (Continued)

At point c,

x  

P 9.6  103   0.320 MPa A 0.030

◄

Q  (150)(100)(50)  750  103 mm3  750  106 m3

 xy  

VQ It



(10  103 )(750  106 )  0.500 MPa (100  106 )(0.150)

◄



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PROBLEM 8.34

60 mm 9 kN

A

30⬚ G H

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.

60 mm

K

12 mm

12 mm 40 mm

B

SOLUTION Fx  0: Bx  0 MA  0: By (120 sin 30 )  9(60 sin 30)  0 By  4.5 kN At the section containing points H and K, P  4.5 cos 30  3.897 kN V  4.5 sin 30  2.25 kN M  (4.5  103 )(40  103 sin 30)  90 N  m A  10  24  240 mm 2  240  106 m 2 1 I  (10)(24)3  11.52  103 mm 4 12  11.52  109 m 4

(a)

At point H,

 xy  (b)

At point K,

P 3.897  103  A 240  106

  16.24 MPa 

3 V 3 2.25  103  2 A 2 240  106

  14.06 MPa 

x  

x  

P Mc 3.897  103 (90)(12  103 )    A I 240  106 11.52  109

  110.0 MPa    0 

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PROBLEM 8.35 A

60 mm 9 kN

30⬚ G H 12 mm 40 mm

60 mm

K

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.

12 mm B

SOLUTION M B  0: (120 cos 30) RA  (60sin 30)(9)  0 RA  2.598 kN

Fy  0: By  9  0

By  9 kN 

Fx  0: 2.598  Bx  0 Bx  2.598 kN  At the section containing points H and K, P  9 cos 30  2.598 sin 30  9.093 kN V  9 sin 30  2.598 cos 30  2.25 kN M  (9  103 )(40  103 sin 30)  (2.598  103 )(40  103 cos 30)  90 N  m A  10  240  240 mm 2  240  106 m 2 I

(a)

1 (10)(24)3  11.52  103 mm 4  11.52  109 m 4 12

At point H, P 9.093  103  A 240  106

  37.9 MPa 

3 V 3 2.25  103  2 A 2 240  106

  14.06 MPa 

x    xy  (b)

At point K, P Mc  A I 9.093  103 (90)(12  103 )   240  106 11.52  109

x  

  131.6 MPa    0 

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PROBLEM 8.36

A 60 mm 9 kN

G 30⬚

H

12 mm 40 mm

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.

60 mm

K

12 mm B

SOLUTION M B  0: (9)(60sin 30)  120 RA  0 RA  2.25 kN Fx  0: 2.25cos 30  Bx  0 Bx  1.9486 kN 

Fy  0: 2.25sin 30  9  By  0 By  7.875 kN  At the section containing points H and K, P  7.875cos 30  1.9486sin 30  7.794 kN V  7.875sin 30  1.9486 cos 30  2.25 kN M  (7.875  103 )(40  103 sin 30)  (1.9486  103 )(40  103 cos 30)  90 N  m A  10  24  240 mm 2  240  106 m 2 I

(a)

At point H,

At point K,

P 7.794  103  A 240  106

 x  32.5 MPa 

3 V 3 2.25  103  2 A 2 240  106

 xy  14.06 MPa 

x    xy 

(b)

1 (10)(24)3  11.52  103 mm 4  11.52  109 m 4 12

x  

P Mc 7.794  103 (90)(12  103 )    A I 240  106 11.52  109

 x  126.2 MPa   xy  0 

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PROBLEM 8.37

9 kip · in. C

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.

1.5 kips

H

K

9 in.

SOLUTION Diameter  2.5 in. At the section containing points H and K, P0

V  1.5 kips

T  9 kip  in.

d  2.5 in.

M  (1.5)(9)  13.5 kip  in.

c

1 d  1.25 in. 2

A   c 2  4.909 in 2 Q

For a semicircle, (a)

 4

c 4  1.9175 in 4

J  2I  3.835 in 4

2 3 c  1.3021 in 3 3

H  0 ◄

At point H,

H  (b)

I 

At point K,

Tc VQ (9)(1.25) (1.5)(1.3021)     2.934  0.407  3.34 ksi J It 3.835 (1.9175)(2.5)

K   K 

Mc (13.5)(1.25)   8.80 ksi I 1.9175

Tc (9)(1.25)   2.93 ksi J 3.835

 H  3.34 ksi ◄  K   8.80 ksi ◄  K  2.93 ksi ◄

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PROBLEM 8.38

y

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.

45 mm 45 mm

A

1500 N 1200 N

a

b 75 mm

B

z

20 mm x

SOLUTION co  J 

do  21 mm, 2

 2

c

4 o

 ci4







di  17.5 mm A   co2  ci2  423.33 mm 2 2 1  158.166  103 mm 4 I  J  79.083  103 mm 4 2 ci 

For semicircle with semicircular cutout, Q





2 3 co  ci3  2.6011  103 mm3 3

At the section containing points a and b, P  1500 N

Vz  1200 N

Vx  0

M z  (45  103 )(1500)  67.5 N  m M x  (75  103 )(1200)  90 N  m T  (90  103 )(1200)  108 N  m (a)

(b)

 

P M xc 1500 (90)(21  103 )    A I 423.33  106 79.083  109

  20.4 MPa 

 

Tc VxQ (108)(21  103 )   0 J It 158.166  109

  14.34 MPa 

 

P M zc 1500 (67.5)(21  103 )    A I 423.33  106 79.083  109

 

(1200)(2.6011  106 ) Tc Vz Q (108)(21  103 )    J It 158.166  109 (79.083  109 )(7  103 )

  21.5 MPa    19.98 MPa 

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PROBLEM 8.39

y 200 lb

150 lb

D H K z

10 in.

4 in. 4 in.

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.

150 lb 6 in.

50 lb x

SOLUTION Section properties: P  150 lb T  (200 lb)(10 in.)  2000 lb  in. M z  (150 lb)(10 in.)  1500 lb  in. M y  (200 lb  50 lb)(10 in.)  (150 lb)(4 in.)  900 lb  in. Vz  200  150  50  0 Vy  0 A   (0.952  0.8052 )  0.79946 in 2

(a)



(0.954  0.8054 )  0.30989 in 4 4 J  2 I  0.61979 in 4 P M c 150 lb (1500 lb  in.)(0.95 in.) Point H: H   z   2 A I 0.79946 in 0.30989 in 4 I

 H  4.79 ksi 

 187.6 psi  4593 psi

(b)

Point K:

H 

Tc (2000 lb  in.)(0.95 in.)   3065.6 psi J 0.61979 in 4

K 

P M yc 150 lb (900 lb  in.)(0.95 in.)    2 A I 0.79946 in 0.30989 in 4

 H  3.07 ksi 

 187.6 psi  2759 psi

 K  2.57 ksi 

Tc  same as for  H J

 K  3.07 ksi 

K 

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PROBLEM 8.40

y 50 mm t ⫽ 8 mm

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.

20 mm

A D

225 mm

H 60⬚

E x

B

z

SOLUTION

Vertical force: 40cos 30  34.64 kN Horizontal force: 40sin 30  20 kN

Point H lies on neutral axis of bending. Section properties: d o  100 mm,

co 

1 d o  50 mm, 2

ci  co  t  42 mm,

A   (co2  ci2 )  2.312  103 mm 2  2.312  103 m 2

  For thin pipe,

P 34.64  103  A 2.312  106

 2

  14.98 MPa 

V (2)(20  103 )  A 2.314  103

  17.29 MPa 

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PROBLEM 8.41

y 2 in.

6 kips

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.

2 in. 6 kips 2.5 kips

A

8 in.

H

B z

x

SOLUTION At the section containing point H, P  12 kips (compression) V  2.5 kips T  (2)(2.5)  5 kip  in. M  (8)(2.5)  20 kip  in. d  1.8 in. c 

1 d  0.9 in. 2

A   c 2  2.545 in 2 I 

 4

c 4  0.5153 in 4

J  2 I  1.0306 in 4 For a semicircle, 2 3 c  0.486 in 3 3 Point H lies on neutral axis of bending. P 12 H    4.715 ksi A 2.545 Tc VQ (5)(0.9) (2.5)(0.486) H     J It 1.0306 (0.5153)(1.8) Q

 5.676 ksi 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. 1294

PROBLEM 8.41 (Continued)

Use Mohr’s circle. 1 (4.715) 2  2.3575 ksi

 ave 

2

R

 4.715  2    5.676  2 

 6.1461 ksi

(a)

 a   ave  R

 a  3.79 ksi 

 b   ave  R

 b  8.50 ksi 

tan 2 p 

(2)(5.676)  2.408 4.715

 a  33.7  b  123.7 

(b)

 max  R

 max  6.15 ksi 

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PROBLEM 8.42

y

3 kN

B

C

H

D

120 mm

x

9 kN

A

The steel pipe AB has a 72-mm outer diameter and a 5-mm wall thickness. Knowing that the arm CDE is rigidly attached to the pipe, determine the principal stresses, principal planes, and the maximum shearing stress at point H.

150 mm z

120 mm

E

SOLUTION Replace the forces at C and E by an equivalent force-couple system at D. FD  9  3  6 kN TD  (9  103 )(120  103 )  (3  103 )(120  103 )  1440 N  m At the section containing point H, P0 V  6 kN T  1440 N  m M  (6  103 )(150  103 )  900 N  m 1 Section properties: d o  72 mm co  d o  36 mm ci  co  t  31 mm 2 A   (co2  ci2 )  1.0524  103 mm 2  1.0524  103 m 2

I 



4

(co4  ci4 )  593.84  103 mm 4  593.84  109 m 4

J  2 I  1.1877  106 m 4 2 3 (co  ci3 )  11.243  103 mm3  11.243  106 m3 3 At point H, point H lies on the neutral axis of bending.  H  0. Q

For half-pipe,

H 

Tc VQ (1440)(36  103 ) (6  103 )(11.243  106 )     55.0 MPa J It 1.1877  106 (593.84  109 )(10  103 )

Use Mohr’s circle.

c  0 R  55.0 MPa

a  c  R

 a  55.0 MPa ◄

b  c  R

 b  55.0 MPa ◄

 a  45, b  45  max  R

 max  R

◄

 max  55.0 MPa ◄

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PROBLEM 8.43

y B

A 13-kN force is applied as shown to the 60-mm-diameter cast-iron post ABD. At point H, determine (a) the principal stresses and principal planes, (b) the maximum shearing stress.

D

13 kN 300 mm H 100 mm A z

150 mm

E 125 mm x

SOLUTION DE  1252  3002  325 mm Fx  0

At point D,

 300  Fy     (13)  12 kN  325   125  Fz     (13)  5 kN  300  Moment of equivalent force-couple system at C, the centroid of the section containing point H:    i j k     M  0.150 0.200 0  1.00i  0.75 j  1.8k kN  m 0 12 5 Section properties: d  60 mm c 

1 d  30 mm 2

A   c 2  2.8274  103 mm 2 I 

 4

c 4  636.17  103 mm 4

J  2 I  1.2723  106 mm 4 For a semicircle,

Q

2 3 c  18.00  103 mm3 3

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PROBLEM 8.43 (Continued)

At point H,

H   H 

P Mc 12  103 (1.8  103 )(30  103 )     89.13 MPa A I 2.8274  103 636.17  109

Tc VQ (0.75  103 )(30  103 ) (5  103 )(18.00  106 )     20.04 MPa J It 1.2723  106 (636.17  109 )(60  103 )

(a)

 ave 

H 2

 44.565 MPa 2

R

 tan 2 p 

H  2     H  48.863 MPa 2  

 a   ave  R

 a  4.30 MPa 

 b   ave  R

 b  93.4 MPa 

2 H

H

 0.4497

 a  12.1, b  102.1   max  R  48.9 MPa 

(b)

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PROBLEM 8.44

y 1 in. 2 in.

P A

60°

D E

H

8 in.

A vertical force P of magnitude 60 lb is applied to the crank at point A. Knowing that the shaft BDE has a diameter of 0.75 in., determine the principal stresses and the maximum shearing stress at point H located at the top of the shaft, 2 in. to the right of support D.

z B 5 in.

x

SOLUTION Force-couple system at the centroid of the section containing point H: Fx  0,

Vy  0.06 kips,

Vz  0

M z  (5  2  1)(0.06)  0.24 kip  in. M x  (8sin 60)(0.06)  0.41569 kip  in. d  0.75 in. I 

 4

c4 

c

 4

1 d  0.375 in. 2

(0.375) 4  15.5316  103 in 4

J  2 I  31.063  103 in 4 At point H, Mzy (0.24)(0.375)   5.7946 ksi Iz 15.5316  103

H   H 

Tc (0.41569)(0.375)   5.0183 ksi J 31.063  103

 ave 

Use Mohr’s circle.

1  H  2.8973 ksi 2

2

R

H  2     H  5.7946 ksi  2 

 a   ave  R

 max  8.69 ksi 

 b   ave  R

 min  2.90 ksi 

tan 2 p 

2 H

H

 a  30.0

 max  R



2(5.0183)  1.7321 5.7946

b  120.0

 max  5.79 ksi 

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PROBLEM 8.45

50 kips 0.9 in.

2 kips C

0.9 in.

2.4 in. 2 in.

Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

6 kips h ⫽ 10.5 in.

1.2 in. 1.2 in.

a

b c

4.8 in. 1.8 in.

SOLUTION Calculate forces and couples at section containing points a, b, and c.

Forces

Couples

h  10.5 in. P  50 kips Vx  6 kips Vz  2 kips M z  (10.5  2)(6)  51 kip  in. M x  (10.5)(2)  21 kip  in. Section properties.

A  (1.8)(4.8)  8.64 in 2 1 (4.8)(1.8)3  2.3328 in 4 12 1 I z  (1.8)(4.8)3  6.5888 in 4 12 Ix 

Stresses.

 

P Mzx Mxz   A Iz Ix



Vx Q Izt

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PROBLEM 8.45 (Continued)

(a)

Point a:

x  0, z  0.9 in., Q  (1.8)(2.4)(1.2)  5.184 in 3

   (b)

Point b:

(6)(5.184) (16.5888)(1.8)

 Point c:

  2.31 ksi    1.042 ksi 

x  1.2 in., z  0.9 in., Q  (1.8)(1.2)(1.8)  3.888 in 3

 

(c)

50 (21)(0.9) 0 8.64 2.3328

50 (51)(1.2) (21)(0.9)   8.64 16.5888 2.3328

(6)(3.888)  0.781 ksi (16.5888)(1.8)

  6.00 ksi    0.781 ksi 

x  2.4 in., z  0.9 in., Q  0

 

50 (51)(2.4) (21)(0.9)   8.64 16.5888 2.3328

  9.69 ksi    0 

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PROBLEM 8.46

50 kips 0.9 in.

2 kips C

0.9 in.

2.4 in. 2 in.

PROBLEM 8.45 Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

6 kips h ⫽ 10.5 in.

Solve Prob. 8.45, assuming that h  12 in.

1.2 in. 1.2 in.

a

b c

4.8 in. 1.8 in.

SOLUTION Calculate forces and couples at section containing points a, b, and c.

Forces

Couples

h  12 in. P  50 kips Vx  6 kips Vz  2 kips M z  (12  2)(6)  60 kip  in. M x  (12)(2)  24 kip  in. Section properties.

A  (1.8)(4.8)  8.64 in 2 1 (4.8)(1.8)3  2.3328 in 4 12 1 I z  (1.8)(4.8)3  6.5888 in 4 12 Ix 

Stresses.

 

P Mzx Mxz   A Iz Ix



Vx Q Izt

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PROBLEM 8.46 (Continued)

(a)

Point a:

x  0, z  0.9 in., Q  (1.8)(2.4)(1.2)  5.184 in 3

   (b)

Point b:

(6)(5.184) (16.5888)(1.8)

 Point c:

  3.47 ksi    1.042 ksi 

x  1.2 in., z  0.9 in., Q  (1.8)(1.2)(1.8)  3.888 in 3

 

(c)

50 (24)(0.9) 0 8.64 2.3328

50 (60)(1.2) (24)(0.9)   8.64 16.5888 2.3328

(6)(3.888) (16.5888)(1.8)

  7.81 ksi    0.781 ksi 

x  2.4 in., z  0.9 in., Q  0

 

50 (60)(2.4) (24)(0.9)   8.64 16.5888 2.3328

  12.15 ksi    0 

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PROBLEM 8.47

60 mm

Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

24 mm a

c

b

15 mm

180 mm 40 mm

750 N

32 mm

16 mm

30 mm 500 N C 10 kN

SOLUTION A  (60)(32)  1920 mm 2  1920  106 m 2 1 (60)(32)3  163.84  103 mm 4 12  163.84  109 m 4 1 I y  (32)(60)3 12  579  103 mm 4 Iz 

 576  109 m 4 At the section containing points a, b, and c, P  10 kN Vy  750 N, Vz  500 N M z  (180  103 )(750)  135 N  m

Side View

M y  (220  103 )(500)  110 N  m T 0



P Mz y Myz   A Iz Iy



Vz Q I yt

Top View

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PROBLEM 8.47 (Continued)

(a)

(b)

(c)

Point a:

Point b:

Point c:

y  16 mm, z  0, Q  Az  (32)(30)(15)  14.4  103 mm3



10  103 (135)(16  103 )  0 1920  106 163.84  109

  18.39 MPa 



(500)(14.4  106 ) (576  109 )(32  103 )

  0.391 MPa 

y  16 mm, z  15 mm, Q  Az  (32)(15)(22.5)  10.8  103 mm3



10  103 (135)(16  103 ) (110)(15  103 )   1920  106 163.84  109 576  109

  21.3 MPa 



(500)(10.8  106 ) (576  109 )(32  103 )

  0.293 MPa 

y  16 mm, z  30 mm, Q  0



10  103 (135)(16  103 ) (110)(30  106 )   1920  106 163.84  109 576  109

  24.1 MPa   0 

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PROBLEM 8.48

60 mm

Solve Prob. 8.47, assuming that the 750-N force is directed vertically upward.

24 mm a

b

c 15 mm

180 mm 40 mm

750 N

32 mm

16 mm

30 mm

PROBLEM 8.47 Three forces are applied to the bar shown. Determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

500 N C 10 kN

SOLUTION A  (60)(32)  1920 mm 2  1920  106 m 2 1 (60)(32)3 12  163.84  103 mm 4

Iz 

 163.84  109 m 4 1 (32)(60)3 12  576  103 mm 4

Iy 

 576  109 m 4

At the section containing points a, b, and c, P  10 kN

T 0

Vy  750 N Vz  500 N M z  (180  103 )(750)  135 N  m

Side View

M y  (220  103 )(500)  110 N  m

 

P M z y M yz   A Iz Iy

 

Vz Q I yt

Top View

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PROBLEM 8.48 (Continued)

(a)

(b)

(c)

Point a:

Point b:

Point c:

y  16 mm, z  0, Q  Az  (32)(30)(15)  14.4  103 mm3



10  103 (135)(16  103 )  0 1920  106 163.84  109

  7.98 MPa 



(500)(14.4  106 ) (163.84  109 )(32  103 )

  0.391 MPa 

y  16 mm, z  15 mm, Q  Az  (32)(15)(22.5)  10.8  103 mm3



10  103 (135)(16  103 ) (110)(15  103 )   1920  106 163.84  109 576  109

  5.11 MPa 



(500)(10.8  106 ) (163.84  109 )(32  109 )

  0.293 MPa 

y  16 mm, z  30 mm, Q  0



10  103 (135)(16  103 ) (110)(30  103 )   1920  106 163.84  109 576  109

  2.25 MPa   0 

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PROBLEM 8.49

y

Two forces are applied to the small post BD as shown. Knowing that the vertical portion of the post has a cross section of 1.5  2.4 in., determine the principal stresses, principal planes, and maximum shearing stress at point H.

B

6000 lb 500 lb 1.5 in.

2.4 in.

4 in. H D

1 in. z

6 in.

3.25 in. x 1.75 in.

SOLUTION Components of 500-lb force: (500)(1.75)  140 lb 6.25 (500)(6)  480 lb Fy   6.25 Fx 

Moment arm of 500-lb force:    r  3.25i  (6  1) j Moment of 500-lb force:    i j k   M  3.25 5 0  2260k lb  in. 140 480 0 P  480 lb

At the section containing point H, Vz  6000 lb,

M z  2260 lb  in.,

M x  (4)(6000)  24,000 lb  in.

1 (2.4)(1.5)3  0.675 in 4 12 P M zx 480 (2260)(0.75)      2644 psi 3.6 0.675 A Iz

A  (1.5)(2.4)  3.6 in 2

H

Vx  140 lb

H 

Iz 

3 Vz 3 6000   2500 psi 2 A 2 3.6

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PROBLEM 8.49 (Continued)

Use Mohr’s circle.

 ave  

2644  1322 psi 2 2

 2644  2    (2500)  2828 psi 2  

R

 a   ave  R

 a  1506 psi 

 b   ave  R

 b  4150 psi 

tan 2 p 

2 H

H



(2)(2500)  1.891 2644

 a  31.1,  max  R

b  121.1 

 max  2830 psi 

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PROBLEM 8.50

y

Solve Prob. 8.49, assuming that the magnitude of the 6000-lb force is reduced to 1500 lb.

B

PROBLEM 8.49 Two forces are applied to the small post BD as shown. Knowing that the vertical portion of the post has a cross section of 1.5  2.4 in., determine the principal stresses, principal planes, and maximum shearing stress at point H.

6000 lb 500 lb 1.5 in.

2.4 in.

4 in. H D

1 in. z

6 in.

3.25 in. x 1.75 in.

SOLUTION Components of 500-lb force: (500)(1.75)  140 lb 6.25 (500)(6)  480 lb Fy   6.25 Fx 

Moment arm of 500-lb force:    r  3.25i  (6  1) j    i j k   M  3.25 5 0  2260k lb  in. 140 480 0 P  480 lb

At the section containing point H, Vz  1500 lb,

M z  2260 lb  in.,

M x  (4)(1500)  6000 lb  in.

1 (2.4)(1.5)3  0.675 in 4 12 P M zx 480 (2260)(0.75)      2644 psi 3.6 0.675 A Iz

A  (1.5)(2.4)  3.6 in 2

H

Vx  140 lb

H 

Iz 

3 Vz 3 1500   625 psi 2 A 2 3.6

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PROBLEM 8.50 (Continued)

Use Mohr’s circle.

 ave 

1  H  1322 psi 2 2

 2644  2    (625)  1462.30 psi  2 

R

 a   ave  R

 a  140.3 psi 

 b   ave  R

 b  2780 psi 

tan 2 p 

2 H

H



(2)(625)  0.4728 2644

 a  12.6  max  R

b  102.6   max  1462 psi 

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PROBLEM 8.51

y 50 mm

Three forces are applied to the machine component ABD as shown. Knowing that the cross section containing point H is a 20  40-mm rectangle, determine the principal stresses and the maximum shearing stress at point H.

150 mm A H

40 mm

0.5 kN

z 20 mm

D

B 3 kN 160 mm

x

2.5 kN

SOLUTION Equivalent force-couple system at section containing point H: Fx  3 kN, Fy  0.5 kN, Fz  2.5 kN M x  0, M y  (0.150)(2500)  375 N  m M z  (0.150)(500)  75 N  m

A  (20)(40)  800 mm 2  800  106 m 2 Iz 

1 (40)(20)3 12

 26.667  103 mm 4  26.667  109 m 4

H 

P Mzy  A Iz

(75)(10  103 ) 3000  800  106 26.667  109  24.375 MPa 

3 | Vz | 3 2500   2 A 2 800  106  4.6875 MPa

H 

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PROBLEM 8.51 (Continued)

Use Mohr’s circle. 1 H 2  12.1875 MPa

 ave 

2

R

 24.375  2    (4.6875)  2 

 13.0579 MPa

 a   ave  R

 a  25.2 MPa 

 b   ave  R

 b  0.87 MPa 

tan 2 p 

2 H

H



(2)(4.6875)  0.3846 24.375

 a  10.5, b  100.5

 max  R

 max  13.06 MPa 

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PROBLEM 8.52

y 50 mm

Solve Prob. 8.51, assuming that the magnitude of the 2.5-kN force is increased to 10 kN.

150 mm A

40 mm

H

0.5 kN

z B

20 mm

3 kN 160 mm

D

x

2.5 kN

PROBLEM 8.51 Three forces are applied to the machine component ABD as shown. Knowing that the cross section containing point H is a 20  40-mm rectangle, determine the principal stresses and the maximum shearing stress at point H.

SOLUTION Equivalent force-couple system at section containing point H: Fx   3 kN,

Fy   0.5 kN, Fz   10 kN

M x  0, M y  (0.150)(10,000)  1500 N  m

M z   (0.150)(500)  75 N  m A  (20)(40)  800 mm 2  800  106 m 2 Iz 

1 (40)(20)3  26.667  103 mm 4  26.667  109 m 4 12

H 

P Mzy 3000 (75)(10  103 )     24.375 MPa A Iz 800  106 26.667  109

H 

3 Vz 3 10,000    18.75 MPa 2 A 2 800  106

1 2

 c   H  12.1875 MPa 2

 24.375  2 R    (18.75)  22.363 MPa  2 

a  c  R

 a  34.6 MPa ◄

b  c  R

 b   10.18 MPa ◄

tan 2 p 

2 H

H



(2)(18.75)  1.5385 24.375

 a  28.5, b  118.5  max  R

 max  22.4 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. 1314

PROBLEM 8.53 a

b

d

y

e 60 mm 30 mm 60 mm

400 mm 75 mm

x

C

150 mm

9 kN

Three steel plates, each 13 mm thick, are welded together to form a cantilever beam. For the loading shown, determine the normal and shearing stresses at points a and b.

t ⫽ 13 mm

C

13 kN

SOLUTION Equivalent force-couple system at section containing points a and b: Fx  9 kN, Fy  13 kN, Fz  0 M x  (0.400)(13  103 )  5200 N  m M y  (0.400)(9  103 )  3600 N  m Mz  0 A  (2)(150)(13)  (13)(75  26)  4537 mm 2  4537  106 m 2 1  1 I x  2  (150)(13)3  (150)(13)(37.5  6.5) 2   (13)(75  26)3 12   12  3.9303  106 mm 4  3.9303  106 m 4 1 1 I y  2  (13)(150)3  (75  26)(13)3 12 12 6 4  7.3215  10 mm  7.3215  106 mm 4

For point a, Qx  0, Qy  0 For point b, A*  (60)(13)  780 mm 2 x  45 mm y  31 mm Qx  A* y  24.18  103 mm3  24.18  106 m3 Qy  A* x  35.1  103 mm3  35.1  106 m3

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PROBLEM 8.53 (Continued)

Direction of shearing stress for horizontal and for vertical components of shear: At point a,

 

Mx y Myx  Ix Iy (5200)(37.5  103 ) (3600)(75  103 )  3.9303  106 7.3215  106

  86.5 MPa    0 

At point b,  









Mx y Myx   Ix Iy (5200)(37.5  103 ) (3600)(15  103 )   3.9303  106 7.3215  106 |Vx ||Qy | I yt



  57.0 MPa 

|Vy ||Qx | I xt

(9  103 )(35.1  106 ) (13  103 )(24.18  106 )   (7.3215  106 )(13  103 ) (3.9303  106 )(13  103 )







 3.32 MPa  6.15 MPa 

  9.47 MPa 

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PROBLEM 8.54 a

b

d

y

e 60 mm 30 mm 60 mm

400 mm 75 mm

x

C

150 mm

9 kN

Three steel plates, each 13 mm thick, are welded together to form a cantilever beam. For the loading shown, determine the normal and shearing stresses at points d and e.

t ⫽ 13 mm

C

13 kN

SOLUTION Equivalent force-couple system at section containing points d and e. Fx  9 kN, Fy  13 kN, Fz  0 M x  (0.400)(13  103 )  5200 N  m M y  (0.400)(9  103 )  3600 N  m Mz  0 A  (2)(150)(13)  (13)(75  26)  4537 mm 2  4537  106 m 2 1  1 I x  2  (150)(13)3  (150)(13)(37.5  6.5)2   (13)(75  26)3 12   12  3.9303  106 mm 4  3.9303  106 m 4 1  1 I y  2  (13)(150)3   (75  26)(13)3 12  12  7.3215  106 mm 4  7.3215  106 m 4

For point d, A*  (60)(13)  780 mm 2 x  45 mm

y  31 mm

Qx  A y  24.18  103 mm3  24.18  106 m3 *

Qy  A* x  35.1  103 mm3 = 35.1  106 m3

For point e,

Qx  0, Qy  0

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PROBLEM 8.54 (Continued)

At point d,

  

M x y M yx  Ix Iy (5200)(37.5  10 3 ) (3600)(15  103 )  3.9303  106 7.3215  106

  42.2 MPa 

Due to Vx :



|Vx ||Qy | I yt

(9000)(35.1  106 ) (7.3215  106 )(13  103 )



 3.32 MPa 

Due to Vy :



|Vy ||Qx | I xt



(13,000)(24.18  106 )  6.15 MPa  (3.9303  106 )(13  103 )

  2.83 MPa 

By superposition, the net value is At point e,



M x y M y x (5200)(37.5  103 ) (3600)(75  103 )    Ix Iy 3.9303  106 7.3215  106

  12.74 MPa    0 



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PROBLEM 8.55

y 75 mm a a x P2 P1

b 1.2 m

Two forces P1 and P2 are applied as shown in directions perpendicular to the longitudinal axis of a W310  60 beam. Knowing that P1  25 kN and P2  24 kN, determine the principal stresses and the maximum shearing stress at point a.

b W310 ⫻ 60

0.6 m

SOLUTION At the section containing points a and b, M x  (1.8)(25)  45 kN  m M y  (1.2)(24)  28.8 kN  m Vx  24 kN

Vy  25 kN

For W310  60 rolled-steel section, d  302 mm,

b f  203 mm,

t f  13.1 mm,

I x  128  106 mm 4  128  106 m 4 , x

Normal stress at point a:

y 

z 

bf 2

tw  7.49 mm

I y  18.4  106 mm 4  18.4  106 m 4

 75  26.5 mm

1 d  151 mm 2

M x y M y x (45  103 )(151  103 ) (28.8  103 )(26.5  103 )    Ix Iy 128  106 18.4  106

 53.086 MPa  41.478 MPa  11.608 MPa Shearing stress at point a:

 xz  

 Vx A x Vy A y  I yt f I xt f

A  (75  103 )(13.1  103 )  982.5  106 m 2 x  y 

 xz  

bf 2



75  64 mm 2

d tf   144.45 mm 2 2

(24  103 )(982.5  106 )(64  103 ) (25  103 )(982.5  106 )(144.45  103 )  (18.4  106 )(13.1  103 ) (128  106 )(13.1  103 )

 6.2609 MPa  2.1160 MPa  4.1449 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. 1319

PROBLEM 8.55 (Continued)

 ave 

11.608  5.804 MPa 2 2

R

 11.608  2    (4.1449)  7.1321 MPa  2 

 max   ave  R

 max  12.94 MPa 



 min   ave  R 

 min  1.328 MPa 



 max  R 

  max  7.13 MPa 

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PROBLEM 8.56

y 75 mm a a x P2 P1

b 1.2 m

Two forces P1 and P2 are applied as shown in directions perpendicular to the longitudinal axis of a W310  60 beam. Knowing that P1  25 kN and P2  24 kN, determine the principal stresses and the maximum shearing stress at point b.

b W310 ⫻ 60

0.6 m

SOLUTION At the section containing points a and b, M x  (1.8)(25)  45 kN  m M y  (1.2)(24)  28.8 kN  m Vx  24 kN,

Vy  25 kN

For W310  60 rolled-steel section, d  302 mm,

b f  203 mm,

t f  13.1 mm,

I x  128  106 mm 4  128  106 m 4 ,

x  0,

Normal stress at point b:

z 

tw  7.49 mm

I y  18.4  106 mm 4  18.4  106 m 4

1 y   d  t f  137.9 mm 2

M x y M y x (45  103 )(137.9  103 )   0 Ix Iy 128  106

 48.480 MPa Shearing stress at point b:

 yz  

Vy A y I xt w

A  A f  b f t f  2659.3 mm 2  2659.3  106 m 2 x  0,

 yz  

1 1 y   d  t f  144.45 mm 2 2

(25  103 )(2659.3  106 )(144.45  103 )  10.0169 MPa (128  106 )(7.49  103 )

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PROBLEM 8.56 (Continued)

 ave  

48.480  24.240 MPa 2 2

R

 48.48  2    (10.0169)  26.228 MPa  2 

 max   ave  R

 max  1.988 MPa 



 min   ave  R 

 min  50.5 MPa 



 max  R 

 max  26.2 MPa 

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20 kips 4 in.

PROBLEM 8.57

W8 3 28 y

1.6 kips 1.6 kips

5 kips

b

Four forces are applied to a W8  28 rolled-steel beam as shown. Determine the principal stresses and maximum shearing stress at point a. x

20 in. a 3 in. b a

SOLUTION Calculate forces and couples at section containing point of interest.

P  20 kips

Vx  3.2 kips

Vy  5 kips

M x  (20)(5)  (4.03)(20)  180.6 kip  in. M y  (20  4)(3.2)  76.8 kip  in. Section properties:

Point a:

A  8.24 in 4

d  8.06 in.

tw  0.285 in.

I x  98.0 in 4

6.54  3  0.27 in. 2 M x P M y a    x a  y a A Ix Iy xa  

b f  6.54 in.

t f  0.465 in.

I y  21.7 in 4 ya  

8.06  4.03 in. 2

20 (180.6)(4.03) (76.8)(0.27)   8.24 98.0 21.7   2.4272  7.4267  0.95558  4.0439 ksi 

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PROBLEM 8.57 (Continued) Shearing stress at point a due to Vx :

Shearing stress at point a due to Vy :

A  (3)(0.465)  1.395 in 2

A  (3)(0.465)  1.395 in 2 x 

6.54 3   1.77 in. 2 2

y 

Q   Ay  5.2975 in 3

Q   Ax  2.4692 in 3

 xz 

Vx Q (3.2)(2.4692)   0.78306 ksi I yt (21.7)(0.465)

Combined shearing stress:

8.06 0.465   3.7975 in. 2 2

 xz 

V yQ I xt



(5)(5.2975)  0.58125 ksi (98.0)(0.465)

 a  0.78306  0.58125  0.20181 ksi  ave 

4.0439  0  2.0220 ksi 2 2

R

 4.0439  0  2    (0.20181)  2.0320 ksi 2  

 max   ave  R

 max  4.05 ksi ◄

 min   ave  R

 min   0.010 ksi ◄

 max  R

 max  2.03 ksi ◄

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20 kips 4 in.

PROBLEM 8.58

W8 3 28 y

1.6 kips 1.6 kips

5 kips

Four forces are applied to a W8  28 rolled-steel beam as shown. Determine the principal stresses and maximum shearing stress at point b. x

b

20 in. a 3 in. b a

SOLUTION Calculate forces and couples at section containing point of interest.

P  20 kips

Vx  3.2 kips

Vy  5 kips

M x  (20)(5)  (4.03)(20)  180.6 kip  in. M y  (20  4)(3.2)  76.8 kip  in. Section properties:

Point b:

A  8.24 in 4

d  8.06 in.

tw  0.285 in.

I x  98.0 in 4

xb  0

b   

b f  6.54 in.

t f  0.465 in.

I y  21.7 in 4

yb  0 P M x yb M y xb   A Ix Iy 20  0  0  2.4272 ksi 8.24

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PROBLEM 8.58 (Continued)

Shearing stress at point b due to Vx:

 xz  0

Shearing stress at point b due to Vy: A(in 2 )

y (in.)

Ay (in 3 )



3.0411

3.7975

11.5486



1.01603

1.7825



1.81107 13.3597

Q  Ay  13.3597 in 3

b   ave 

V yQ I xt w



(5)(13.3597)  2.3916 ksi (98.0)(0.285)

2.4272  0  1.2136 ksi 2 2

R

 2.4272  0  2    (2.3916) 2  

 2.6819 ksi

 max   ave  R

 max  1.468 ksi ◄

 min   ave  R

 min  3.90 ksi ◄

 max  R

 max  2.68 ksi ◄

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PROBLEM 8.59

B a

A force P is applied to a cantilever beam by means of a cable attached to a bolt located at the center of the free end of the beam. Knowing that P acts in a direction perpendicular to the longitudinal axis of the beam, determine (a) the normal stress at point a in terms of P, b, h, l, and  , (b) the values of  for which the normal stress at a is zero.

b

A C

h l

␤ P

SOLUTION 1 3 1 3 bh I y  hb 12 12 M (h/2) M y (b/2)   x  Ix Iy

Ix 

6M x 6M y  bh 2 hb 2 P  P sin  i  P cos  j r  l k 

M  r  P  l k  ( P sin  i  P cos  j )  Pl cos  i  Pl sin  j M x  Pl cos 

M y  Pl sin 

6 Pl cos  6 Pl sin  6 Pl  cos  sin      2 2 bh  h b  bh hb

(a)

 

(b)

 0

cos  sin   0 h b

tan  



b h b

  tan 1    h



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PROBLEM 8.60 l ⫽ 1.25 m

a

A vertical force P is applied at the center of the free end of cantilever beam AB. (a) If the beam is installed with the web vertical (  0) and with its longitudinal axis AB horizontal, determine the magnitude of the force P for which the normal stress at point a is 120 MPa. (b) Solve part a, assuming that the beam is installed with   3.

B

A W250 ⫻ 44.8 P

␤

SOLUTION For W250  44.8 rolled-steel section, S x  531  103 mm3  531  106 m3 S y  94.2  103 mm3  94.2  106 m3 At the section containing point a, M x  Pl cos  ,

M y  Pl sin 

Stress at a:

 

Allowable load.

(a)

  0:

Pall 

Pall 

Mx My Pl cos  Pl sin     Sx Sy Sx Sy

 all  cos   l  S x

120  106 1.25

sin     S y 

1

1    531  106  0   

1

 51.0  103 N Pall  51.0 kN 

(b)

  3:

Pall 

120  106  cos 3 sin 3    6 1.25  531  10 94.2  106 

1

 39.4  103 N Pall  39.4 kN 

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PROBLEM 8.61* B d

A 5-kN force P is applied to a wire that is wrapped around bar AB as shown. Knowing that the cross section of the bar is a square of side d  40 mm, determine the principal stresses and the maximum shearing stress at point a.

a

d 2

A

P

SOLUTION Bending:

Point a lies on the neutral axis.

 0 T c1ab 2

Torsion:

 

and

c1  0.208 for a square section.

Since

T 

where a  b  d

Pd P P , T   2.404 2 . 2 2 0.416d d

Transverse shear: V  P 1 2 d 2 t d

A

V 

1 4 d 12 1 y  d 4

I 

Q  Ay 

1 3 d 8

VQ P  1.5 2 It d

By superposition,

   T  V  3.904  

P d2

(3.904)(5  103 ) (40  103 )2

 12.2  106 Pa  12.2 MPa

 max  12.2 MPa 

By Mohr’s circle,

 min  12.2 MPa   max  12.2 MPa 



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PROBLEM 8.62* 3 in.

Knowing that the structural tube shown has a uniform wall thickness of 0.3 in., determine the principal stresses, principal planes, and maximum shearing stress at (a) point H, (b) point K.

H 6 in. K 4 in. 2 in. 10 in.

0.15 in. 9 kips

SOLUTION At the section containing points H and K, V  9 kips M  (9)(10)  90 kip  in. T  (9)(3  0.15)  25.65 kip  in. Torsion:

Ꮽ  (5.7)(3.7)  21.09 in 2

 

T 25.65   2.027 ksi 2t Ꮽ (2)(0.3)(21.09)

Transverse shear: QH  0 QK  (3)(2)(1)  (2.7)(1.7)(0.85)  2.0985 in 3 1 1 (6)(4)3  (5.4)(3.4)3  14.3132 in 4 12 12 (9)(2.0985) VQK 0   4.398 ksi K  (14.3132)(0.3) It

I 

H

Bending:

H 

Mc (90)(2)   12.576 ksi, I 14.3132

K  0

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PROBLEM 8.62* (Continued)

(a)

Point H:

 ave 

12.576  6.288 ksi 2 2

 12.576  2    (2.027)  6.607 ksi  2 

R



 max   ave  R

 max  12.90 ksi 

 min   ave  R 

 min  0.32 ksi 

tan 2 p 



2



 p  8.9, 81.1 

 0.3224 

 max  R  6.61 ksi (b)

Point K:

 0

 max  6.61 ksi 

  2.027  4.398  6.425 ksi

 max  6.43 ksi   min  6.43 ksi 



   45   max  6.43 ksi 



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PROBLEM 8.63*

3 in.

The structural tube shown has a uniform wall thickness of 0.3 in. Knowing that the 15-kip load is applied 0.15 in. above the base of the tube, determine the shearing stress at (a) point a, (b) point b.

a 1.5 in.

b

2 in.

A

4 in.

10 in.

15 kips

SOLUTION Calculate forces and couples at section containing points a and b. Vx  15 kips M z  (2  0.15)(15)  27.75 kip  in. M y  (10)(15)  150 kip  in.

Shearing stresses due to torque T  M z : Ꮽ  [3  (2)(0.15)][4  (2)(0.15)]  9.99 in 2

q

27.75 Mz   1.3889 kip/in. 2Ꮽ (2)(9.99)

At point a,

t  0.3 in.  a 

q 1.3889   4.630 ksi  t 0.3

At point b,

t  0.3 in.  b 

q 1.3889   4.630 ksi  t 0.3

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PROBLEM 8.63* (Continued)

Shearing stresses due to Vx : At point a,

A(in 2 )

x (in.)

Ax (in 3 )



0.45

0.75

0.3375



1.02

1.35

1.377



0.45

0.75

0.3375

Part



2.052 Q  Ax  2.052 in 3 t  (2)(0.3)  0.6 in. 1 1 (4)(3)3  (3.4)(2.4)3  5.0832 12 12 (15)(2.052) VQ   x   10.092 ksi (5.0832)(0.6) I yt

Iy 

At point b,

b  0

Combined shearing stresses. (a)

At point a,

 a  4.630  10.092   5.46 ksi 

 a  5.46 ksi 

(b)

At point b,

 b  4.630   0  4.63 ksi 

 b  4.63 ksi 

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PROBLEM 8.64*

3 in.

For the tube and loading of Prob. 8.63, determine the principal stresses and the maximum shearing stress at point b.

a 1.5 in.

b

2 in.

A

15 kips

4 in.

PROBLEM 8.63* The structural tube shown has a uniform wall thickness of 0.3 in. Knowing that the 15-kip load is applied 0.15 in. above the base of the tube, determine the shearing stress at (a) point a, (b) point b.

10 in.

SOLUTION Calculate forces and couples at section containing point b. Vx  15 kips M z  (2  0.15)(15)  27.75 kip  in. M y  (10)(15)  150 kip  in.

Couples

Forces

1 1 (4)(3)3  (3.4)(2.4)3  5.0832 in 4 12 12 M x (150)(1.5) b   y b    44.26 ksi 5.0832 Iy Iy 

Shearing stress at point b due to torque M z: Ꮽ  [3  (2)(0.15)][4  (2) (0.15)]  9.99 in 2

q

27.75 Mz   1.38889 kip/in. 2Ꮽ (2)(9.99)

 

q 1.38889   4.630 ksi 0.3 t

Shearing at point b due to Vx :

 0

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PROBLEM 8.64* (Continued)

Calculation of principal stresses and maximum shearing stress.

 ave 

44.26  0  22.13 ksi 2 2

R



 44.26  0  2    (4.630)  22.61 ksi 2  

 max   ave  R

 max  0.48 ksi 

 min   ave  R 

 min  44.7 ksi 

 max  R

 max  22.6 ksi 

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12.5 kips

2 kips/ft B

C

A

D

9 ft

3 ft

3 ft

PROBLEM 8.65 (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 M D  0:  15RA  (10.5)(9)(2)  (3)(12.5)  0 RA  15.1 kips 

S min 

M

max

 all



For W14  22, (b)

Point E:



RD  15.4 kips 

(57.003  12 kip  in.)  28.502 in 3 24 ksi Shape

S (in3)

W16  26

38.4

W14  22

29.0

W12  26

33.4

W10  30

32.4

W8  35

31.2

Aweb  dtw  (13.7)(0.230)  3.151 in 2

ME (57.003)(12)   23.587 ksi S 29.0 1 c  d  6.85 in. yb  c  t f  6.515 in. 2 y  6.515  b  b m    (23.587)  22.433 ksi c  6.85 

m 

Since  m  0, Point C:

 max 

 m  23.6 ksi 

 max   b  22.4 ksi  m m

b 

(a) Use W14  22. 

M (46.2)(12)  C   19.1172 ksi 29.0 S V 15.4    4.8873 ksi Aweb 3.151

yb  6.515  m    (19.1172)  18.1823 ksi c  6.85 

b

 m  19.12 ksi  m  4.89 ksi  2

R

 b  2     m  10.3216 ksi 2  

 max  19.41 ksi 

R 2 For  max , point B controls;

 max  22.4 ksi 

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PROBLEM 8.66

200 mm

500 N

Neglecting the effect of fillets and of stress concentrations, determine the smallest permissible diameters of the solid rods BC and CD. Use  all  60 MPa.

180 mm 160 mm 1250 N

A

D C B

SOLUTION

 all  60  106 Pa J  3  c  c 2 c3 

2

M2 T2

 all

M2 T2



 all

d  2c

Bending moments and torques. Just to the left of C:

M  (500)(0.16)  80 N  m T  (500)(0.18)  90 N  m M 2  T 2  120.416 N  m

Just to the left of D:

T  90 N  m M  (500)(0.36)  (1250)(0.2)  430 N  m M 2  T 2  439.32 N  m

Smallest permissible diameter d BC . c3 

(2)(120.416)  1.27765  106 m3  (60  106 )

c  0.01085 m  10.85 mm d BC  21.7 mm  Smallest permissible diameter dCD . c3 

(2)(439.32)  4.6613  106 m3  (60  106 )

c  0.01670 m  16.7 mm

dCD  33.4 mm 

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PROBLEM 8.67

200 mm

500 N

Knowing that rods BC and CD are of diameter 24 mm and 36 mm, respectively, determine the maximum shearing stress in each rod. Neglect the effect of fillets and of stress concentrations.

180 mm 160 mm 1250 N

A

D C B

SOLUTION Over BC:

c

1 d  12 mm  0.012 m 2

Over CD:

c

1 d  18 mm  0.018 m 2



M 2 T2c J

2



M2 T2 c3

Bending moments and torques. Just to the left of C: M  (500)(0.16)  80 N  m T  (500)(0.18)  90 N  m M 2  T 2  120.416 N  m

Just to the left of D: T  90 N  m M  (500)(0.36)  (1250)(0.2)  430 N  m M 2  T 2  439.32 N  m

Maximum shearing stress in portion BC.

 max 

(2)(120.416)  44.36  106 Pa 3  (0.012)

 max  44.4 MPa 

Maximum shearing stress in portion CD. 

 max 

(2)(439.32)  47.96  106 Pa   (0.018)3

 max  48.0 MPa 

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PROBLEM 8.68

150 mm F

The solid shaft AB rotates at 450 rpm and transmits 20 kW from the motor M to machine tools connected to gears F and G. Knowing that  all  55 MPa and assuming that 8 kW is taken off at gear F and 12 kW is taken off at gear G, determine the smallest permissible diameter of shaft AB.

225 mm

A 225 mm 150 mm

60 mm M

100 mm

D

60 mm

E G

B

SOLUTION f 

450  7.5 Hz 60

Torque applied at D: TD 

P 2 f



20  103 (2 )(7.5)

 424.41 N  m Torques on gears C and E: 8 TD  169.76 N  m 20 12 TE  TD  254.65 N  m 20 TC 

Forces on gears: FD 

TD 424.41   4244 N rD 100  103

FC 

TC 169.76   2829 N rC 60  103

FE 

TE 254.65   4244 N rE 60  103

Forces in horizontal plane:

Torques in various parts: AC :

T 0

CD :

T  169.76 N  m

DE :

T  254.65 N  m

EB :

T 0

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PROBLEM 8.68 (Continued) Critical point lies just to the right of D. T  254.65 N  m M y  1007.9 N  m



Forces in vertical plane:

M z  318.3 N  m

M y2  M z2  T 2



max

 1087.2 N  m

 all 

c J



M y2  M z2  T 2

J   c3  c 2 



M y2



M z2

 all



max

 T2



max

1087.2 55  106

 19.767  103 m3 c  23.26  103 m d  2c  46.5  103 m

d  46.5 mm 

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8 in.

8 in.

6 kips 35⬚ A 8 in.

B 1.5 in. 1.5 in.

a

d

b

e

c

f

PROBLEM 8.69 A 6-kip force is applied to the machine element AB as shown. Knowing that the uniform thickness of the element is 0.8 in., determine the normal and shearing stresses at (a) point a, (b) point b, (c) point c.

SOLUTION Thickness  0.8 in. At the section containing points a, b, and c, P  6 cos35  4.9149 kips

V  6sin 35  3.4415 kips

M  (6sin 35)(16)  (6 cos35)(8)  15.744 kip  in. A  (0.8)(3.0)  2.4 in 2 I 

(a)

1 (0.8)(3.0)3  1.80 in 4 12

At point a,

x 

P Mc 4.9149 (15.744)(1.5)    A I 2.4 1.80

 x  11.07 ksi   xy  0 



(b)

At point b,

x 

P 4.9149  A 2.4

 x  2.05 ksi 





 xy 

3V 3 3.4415    2A 2 2.4

 xy  2.15 ksi 

(c)

At point c,

x 

P Mc 4.9149 (15.744)(1.5)    A I 2.4 1.80

 x  15.17 ksi   xy  0 



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PROBLEM 8.70 l

A thin strap is wrapped around a solid rod of radius c  20 mm as shown. Knowing that l  100 mm and F  5 kN, determine the normal and shearing stresses at (a) point H, (b) point K.

H K c

F

SOLUTION

At the section containing points H and K, T  Fc J

Point H:

Point K:

 2

c4

M  Fl I

Mc Flc   c4 I 4



4 Fl  c2



Tc Fc 2   c4 J 4



2F  c2

Due to torque:

Combined:

4

c4



 0

Point K lies on the neutral axis.

Due to shear:



V F

For a semicircle,



Tc 2F  J  c2

Q

2 3 c , t  d  2c 3



F 23 c3 VQ 4F   4  It c (2c) 3 c 2 4



2F 4F  2 3 c 2 c



10 F 3 c 2

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PROBLEM 8.70 (Continued)

F  5 kN  5  103 N, l  100 mm  0.100 m

Data:

c  20 mm  0.020 m (a)

(b)

Point H:



(4)(5  103 )(0.100)  (0.020)3

  79.6 MPa 



(2)(5  103 )  (0.020)2

  7.96 MPa   0 

Point K:



(10)(5  103 ) 3 (0.020)2

  13.26 MPa 

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P

PROBLEM 8.71

P R

R

A close-coiled spring is made of a circular wire of radius r that is formed into a helix of radius R. Determine the maximum shearing stress produced by the two equal and opposite forces P and P. (Hint: First determine the shear V and the torque T in a transverse cross section.)

T r

V

P'

SOLUTION  Fy  0:

P V  0

 M C  0:

T  PR  0

V  P

T  PR

Shearing stress due to T.

T 

Tc 2T 2 PR   3 J c  r3

Q

2 3 r , t  d  2r 3

Shearing stress due to V. For semicircle, For solid circular section,

I 

V By superposition,

1  J  r3 2 4

 

V 23 r 3 VQ 4V 4P    4   2 It r (2r ) 3 r 3 r 2 4

 max   T   V

 max  P(2R  4r/3)/ r 3 

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PROBLEM 8.72

y 1 in.

H

K

x

Forces are applied at points A and B of the solid cast-iron bracket shown. Knowing that the bracket has a diameter of 0.8 in., determine the principal stresses and the maximum shearing stress at (a) point H, (b) point K.

2500 lb B z A

2.5 in.

3.5 in.

600 lb

SOLUTION At the section containing points H and K, P  2500 lb (compression) Vy  600 lb Vx  0 M x  (3.5  1)(600)  1500 lb  in. My  0 M z  (2.5)(600)  1500 lb  in.

Forces

Couples 1 d  0.4 in. 2 A   c 2  0.50265 in 2 c



c 4  20.106  103 in 4 4 J  2 I  40.212  103 in 4 I

For semicircle, 2 3 c 3  42.667  103 in 3

Q

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PROBLEM 8.72 (Continued)

(a)

At point H, P Mc 2500 (1500)(0.4)    A I 0.50265 20.106  103  24.87  103 psi

H 

Tc (1500)(0.4)   14.92  103 psi 3 J 40.212  10 24.87   12.435 ksi 2

H   ave

2

 24.87  2 R    (14.92)  19.423 ksi  2 

 max   ave  R

 max  31.9 ksi 

 min   ave  R

 min  6.99 ksi 

1 2

 max  19.42 ksi 

 max  ( max   min ) (b)

At point K, P 2500   4.974  103 psi A 0.50265 Tc VQ K   J It (1500)(0.4) (600)(42.667  103 )   40.212  103 (20.106  103 )(0.8)

K 

 16.512  103 psi

 ave  

4.974  2.487 ksi 2 2

 4.974  R    (16.512) 2  16.698 ksi 2  

 max   ave  R

 max  14.21 ksi 

 min   ave  R

 min  19.18 ksi 

1 2

 max  ( max   min )

 max  16.70 ksi 

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PROBLEM 8.73

3 kips K

30⬚

2.5 in.

A 5 in.

2 in.

B

Knowing that the bracket AB has a uniform thickness of 85 in., determine (a) the principal planes and principal stresses at point K, (b) the maximum shearing stress at point K.

SOLUTION Resolve the 3-kip force F at point A into x and y components. Fx   F cos 30  (3) cos 30  2.598 kips Fy  F sin 30  (3)sin 30  1.5 kips At the section containing points H and K, P   Fx  2.598 kips, V  Fy  1.5 kips M  (5)(1.5)  7.5 kip  in.

Section properties.

t 

5 in.  0.625 in., 8

1 (0.625)(2.5)3  0.8138 in 4 , 12 At point k, y  0.75 in. I 

5 A    (2.5)  1.5625 in 2 , 8 c  1.25 in.

Q  (0.625)(0.50)(1.00)  0.3125 in 3 Stresses at point K. P My 2.598 (7.5)(0.75)       5.249 ksi A I 1.5625 0.8138 VQ (1.5)(0.3125)     0.9216 ksi It (0.8138)(0.625)

 x  5.249 ksi,

 y  0,

 xy  0.9216 ksi

Mohr’s circle. X : ( x ,  xy )  (5.249 ksi, 0.9216 ksi) Y : ( y ,  xy )  (0, 0.9216 ksi) C: ( ave , 0)  (2.6245 ksi, 0) x  y  2.6245 ksi 2 (2.6245)2  (0.9216) 2  2.7816 ksi 0.9216 tan 2 p   0.35112 2 p  19.35 2.6245 R

(a)

 max   ave  R  2.6245  2.7816

(b)

 min   ave  R  2.6245  2.7816

 max  0.157 ksi at 80.3



 min  5.41 ksi at 9.7 

 max  R  2.7816 ksi

 max  2.78 ksi 

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PROBLEM 8.74

y 120 kN 75 mm 75 mm

50 mm 50 mm

50 kN

C

For the post and loading shown, determine the principal stresses, principal planes, and maximum shearing stress at point H.

30⬚ 375 mm

H

K

z

x

SOLUTION Components of force at point C: Fx  50 cos 30  43.301 kN Fz  50 sin 30  25 kN Fy  120 kN

Section forces and couples at the section containing points H and K: P  120 kN (compression) Vx  43.301 kN Vz  25 kN M x  (25)(0.375)  9.375 kN  m My  0 M z  (43.301)(0.375)  16.238 kN  m

A  (100)(150)  15  103 mm 2  15  103 m 2 1 I x  (150)(100)3  12.5  106 mm 4  12.5  106 m 4 12

Couples

Forces

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PROBLEM 8.74 (Continued)

Stresses at point H:

H   H 

(120  103 ) (9.375  103 )(50  103 ) P Mxz     29.5 MPa A Ix 15  103 12.5  106

3 Vx 3 43.301  103   4.33 MPa 2 A 2 15  103

Use Mohr’s circle.

1 2

 ave   H  14.75 MPa 2

 29.5  2 R    4.33  15.37 MPa  2 

 a   ave  R 

 b   ave  R  tan 2 p 

 a  30.1 MPa   b  0.62 MPa 

2 H  0.2936  H

 a  8.2  max  R

b  81.8 

 max  15.37 MPa 

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PROBLEM 8.75

6 in.

3 in. 600 lb

1500 lb

600 lb

5 in.

Knowing that the structural tube shown has a uniform wall thickness of 0.25 in., determine the normal and shearing stresses at the three points indicated.

1500 lb

2.75 in. 3 in. 0.25 in. a

20 in.

b c

SOLUTION

Bending moment

Shear

bo  6 in. bi  bo  2t  5.5 in. ho  3 in. hi  ho  2t  2.5 in. 1 (bo ho3  bi hi3 )  6.3385 in 4 Ix  12 1 (hobo3  hibi3 )  19.3385 in 4 Iz  12

Normal stresses.  At a :  M x M z   z  x At b : Iz Ix   At c : 

(60)(3) (30)(1.5)  19.3385 6.3385 (60)(2.75) (30)(1.5)  19.3385 6.3385 (60)(0) (30)(1.5)  19.3385 6.3385

 a  16.41 ksi   b  15.63 ksi   c  7.10 ksi 

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PROBLEM 8.75 (Continued)

Shearing stresses.

 a  0 

Point a is an outside corner. At point b, Qzb  (1.5)(0.25)(2.875)  1.0781 in 3

 b, v x 

VxQz (3)(1.0781)   0.66899 ksi I zt (19.3385)(0.25)

At point c,  2.75  3 Qzc  Qzb  (2.75)(0.25)    2.0234 in  2  (3)(2.0234) VQ  c,v x  x z   1.256 ksi (19.3385)(0.25) I zt

At point b, Qxb  (2.75)(0.25)(1.375)  0.9453 in 3

 b, v z 

Vz Q y I xt



(1.2)(0.9453)  0.71585 ksi (6.3385)(0.25)

At point c,

(symmetry axis)

 c,vz  0 Net shearing stress at points b and c:

 b  0.71585  0.66899  c  1.256

 b  0.0469 ksi   c  1.256 ksi 

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PROBLEM 8.76

B

The cantilever beam AB will be installed so that the 60-mm side forms an angle  between 0 and 90° with the vertical. Knowing that the 600-N vertical force is applied at the center of the free end of the beam, determine the normal stress at point a when (a)   0, (b)   90. (c) Also, determine the value of  for which the normal stress at point a is a maximum and the corresponding value of that stress.

a 300 mm

b

40 mm A C 60 mm

␤

600 N

SOLUTION Sx 

1 (40)(60) 2  24  103 mm3 6

 24  106 m3 Sy 

1 (60)(40)2  16  103 mm3 6

 16  103 m3 M  Pl  (600)(300  103 )  180 N  m M x  M cos   180 cos  M y  M sin   180 sin 

a 

M x M y 180 cos  180 sin     Sx Sy 24  106 16  106

3    (7.5  106 )  cos   sin   Pa 2   3    7.5  cos   sin   MPa 2  

(a)

  0.

 a  7.50 MPa

(b)

  90.

 a  11.25 MPa

(c)

Maximum.



  11.25 MPa 

d a 3    7.5   sin   cos    0 d 2   sin  



  7.50 MPa 

3 cos  2 

tan  

3 2

  56.3  

3

 a  7.5  cos 56.3  sin 56.3   2  

  13.52 MPa 

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PROBLEM 8.C1 Let us assume that the shear V and the bending moment M have been determined in a given section of a rolled-steel beam. Write a computer program to calculate in that section, from the data available in Appendix C, (a) the maximum normal stress  m , (b) the principal stress  max at the junction of a flange and the web. Use this program to solve parts a and b of the following problems: (1) Prob. 8.1 (Use V ⫽ 45 kips and M  450 kip  in.) (2) Prob. 8.2 (Use V ⫽ 22.5 kips and M  450 kip  in.) (3) Prob. 8.3 (Use V ⫽ 700 kN and M  1750 kN  m.) (4) Prob. 8.4 (Use V ⫽ 850 kN and M  1700 kN  m.)

SOLUTION We enter the given values of V and M and obtain from Appendix C the values of d , b f , t f , t w , I , and S for the given WF shape. d , yb  c  t f 2 M 1 y  y  c  tf , a  , b  a  b  S 2  c  VQ Q  bf t f y, b  It w c

We compute

1 From Mohr’s circle,  max   b  R 2 1

1



2

 max   b    b    b2 2 2  Program Outputs Problem 8.1 Given Data: V  45 kips, d  9.92 in., t f  0.530 in.,

M  450 kip  in. b f  7.985 in.

Problem 8.2 Given Data: V  22.5 kips, d  9.92 in.,

M  450 kip  in. b f  7.985 in.

tw  0.315 in.

t f  0.530 in.,

tw  0.315 in.

I  209.00 in.4 , S  42.10 in.4 Answers: (a)  A  10,688.84 psi (b)

 m  19,169.08 psi

I  209.00 in.4 , S  42.10 in.4 Answers: (a)  A  10,688.84 psi (b)

 m  13,073.82 psi

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PROBLEM 8.C1 (Continued) Program Outputs (Continued) Problem 8.3 Given Data: V  700 kN,

M  1750 kN  m

Problem 8.4 Given Data: V  850 kN,

M  1700 kN  m

d  9.30 mm, b f  423 mm

d  930 mm, b f  423 mm

t f  43 mm,

t f  43 mm,

tw  24 mm

tw  24 mm

I  8470 (106 mm 4 )

I  8470 (106 mm 4 )

S  18,200 (103 mm3 ) Answers:

S  18,200 (103 mm3 ) Answers:

(a)

 A  96.15 MPa

(a)

 A  93.40 MPa

(b)

 m  95.39 MPa

(b)

 m  96.56 MPa



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PROBLEM 8.C2 P A

c

K y

b

A cantilever beam AB with a rectangular cross section of width b and depth 2c supports a single concentrated load P at its end A. Write a computer program to calculate, for any values of x/c and y/c, (a) the ratios  max / m and  min / m , where  max and  min are the principal stresses at Point K(x, y) and  m the maximum normal stress in the same transverse section, (b) the angle  p that the principal planes at K form with a transverse and a horizontal plane through K. Use this program to check the values shown in Fig. 8.8 and to verify that  max exceeds  m if x 艋 0.544c, as indicated in the second footnote on Page 560.

B

␴min

␴max

␪p c

x

SOLUTION  y Since the distribution of the normal stresses is linear, we have    m   c MC Pxc m   where I I



We use Equation (8.4), Page 498:

3 P y2  1  2  2 A c 



y 3 I 1 c    m 2 A xc

Dividing (3) by (2),

(2) (3)

2

3  I 121 b(2c) 1 1 1 or, since   c2 ,  m 2 A b (2c) 3 x y Letting X  and Y  , Equations (1) and (4) yield c c    mY

  m

(1)

 y c

2

(4)

x c

1 Y 2 2X

Using Mohr’s circle, we calculate 2 1Y 2 1 1  R       2   m Y 2   2 2   X  max 1  YR 2 m

 min 1  Y R 2 m

  

2



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PROBLEM 8.C2 (Continued)

tan 2 P 

 

2

P 



1 Y 2

1Y 2 XY



2 X ( 2y )

1Y 2 1 tan 1  2  XY

  



Note: For y  0, the angle  P is

, which is opposite to what was arbitrarily assumed in Figure P8.C2.

Program Outputs x For  2, c

For

x  8, c

x c

 min m

 max m



y c

 min m

1.0 0.8 0.6 0.4 0.2 0.0 –0.2 –0.4 –0.6 –0.8 –1.0

0.000 –0.010 –0.040 –0.090 –0.160 –0.250 –0.360 –0.490 –0.640 –0.810 –1.000

1.000 0.810 0.640 0.490 0.360 0.250 0.160 0.090 0.040 0.010 0.000

0.00 6.34 14.04 23.20 33.69 45.00 –33.69 –23.20 –14.04 –6.34 –0.00

1.0 0.8 0.6 0.4 0.2 0.0 –0.2 –0.4 –0.6 –0.8 –1.0

0.000 –0.001 –0.003 –0.007 –0.017 –0.062 –0.217 –0.407 –0.603 –0.801 –1.000

To check that  max   m if x 艋 0.544c, we run the program for that

 max m

exceeds 1 for several values of For

x  0.544, c

y c

x c

 max m 1.000 0.801 0.603 0.407 0.217 0.063 0.017 0.007 0.003 0.001 0.000

 0.544 and for

x c

 0.00 1.61 3.80 7.35 15.48 45.00 –15.48 –7.35 –3.80 –1.61 –0.00

 0.545 and observe

in the first case, but does not exceed 1 in the second case.

y c

 min m

 max m

0.30 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40

–0.700 –0.690 –0.680 –0.670 –0.660 –0.650 –0.640 –0.630 –0.619 –0.608 –0.598

0.9997 1.0001 1.0004 1.0005 1.0005 1.0003 1.0000 0.9996 0.9990 0.9983 0.9975

 39.92 39.72 39.51 39.30 39.09 38.88 38.66 38.44 38.21 37.98 37.74

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PROBLEM 8.C2 (Continued) Program Outputs (Continued) For

x  0.545, c

y c 0.30 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40

 min m –0.698 –0.689 –0.679 –0.669 –0.659 –0.649 –0.639 –0.628 –0.618 –0.607 –0.596

 max m 0.9982 0.9986 0.9989 0.9990 0.9990 0.9988 0.9986 0.9982 0.9976 0.9970 0.9962

 39.91 39.71 39.50 39.29 39.08 38.87 38.65 38.42 38.20 37.96 37.73



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PROBLEM 8.C3 Disks D1 , D2 , . . . , Dn are attached as shown in Fig. 8.C3 to the solid shaft AB of length L, uniform diameter d, and allowable shearing stress  all . Forces P1, P2, . . . , Pn of known magnitude (except for one of them) are applied to the disks, either at the top or bottom of its vertical diameter, or at the left or right end of its horizontal diameter. Denoting by ri the radius of disk Di and by ci its distance from the support at A, write a computer program to calculate (a) the magnitude of the unknown force Pi, (b) the smallest permissible value of the diameter d of shaft AB. Use this program to solve Prob. 8.18.

y

ci

L

P1

A Pn

ri z D1

B

D2 P2

Di Pi

x

Dn

SOLUTION 1. Determine the unknown force Pi by equating to zero the sum of their torques Ti about the x axis. 2. Determine the components ( Fy )i and ( Fz )i of all forces. 3. Determine the components Ay and Az of reaction at A by summing moments about axes Bz  // z and

By  // y : 1 ( Fy )i ( L  ci ) L 1 Az   ( Fz )i ( L  ci ) L

M z   0: Ay L  ( Fy )i ( L  ci )  0,

Ay  

M y  0: Az L  ( Fz )i ( L  ci )  0,

4. Determine ( M y )i , ( M z )i and torque Ti just to the left of disk Di : ( M y )i  Az ci 

 ( F ) c  c 1 z k

i

k

k

( M z )i   Ay ci 

 ( F ) c  c  y k

i

k

1

k

Ti 

 T c  c  k

i

k

0

k

where < > indicates a singularity function. 5. The minimum diameter d required to the left of Di is obtained by first computing ( J/c)i from Equation (8.7).

 M y i   M z i2  Ti2 2

J c   i

 all

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PROBLEM 8.C3 (Continued)

6. Recalling that J  12  c 4 and, thus, that

 Jc i  12  ci3 , we have ci  2  Jc i

1/3

and di  4  Jc 

1/ 3 i



This is the required diameter just to the left of disk Di . 7. The required diameter just to the right of disk Di is obtained by replacing Ti with Ti 1 in the above computation. 8. The smallest permissible value of the diameter of the shaft is the largest of the values obtained for Di .

Program Output Problem 8.19 Length of shaft  28 in.

 (ksi)  8 For Disk 1, Force  0.500 kips Radius of disk  4.0 in. Distance from A  7.0 in. For Disk 2, Force  0.000 kips Radius of disk  6.0 in. Distance from A  14.0 in. For Disk 3, Force  0.500 kips Radius of disk  4.0 in. Distance from A  21.0 in. Unknown force  –0.667 kips AY  0.500 kips,

AZ  0.333 kips

BY  0.500 kips,

BZ  0.333 kips

Just to the left of Disk 1: MY  2.3333 kip  in. MZ  –3.5000 kip  in. T  0.0000 kip  in. Diameter must be at least 1.389 in. Just to the right of Disk 1: T  2.00 kip  in. Diameter must be at least 1.437 in.

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PROBLEM 8.C3 (Continued) Program Output (Continued) Just to the left of Disk 2: MY  4.6667 kip  in. MZ  –3.5000 kip  in. T  2.0000 kip  in. 

Diameter must be at least 1.578 in. Just to the right of Disk 2: T  –2.00 kip  in. Diameter must be at least 1.578 in. Just to the left of Disk 3: MY  2.3333 kip  in. MZ  –3.5000 kip  in. T  –2.0000 kip  in. Diameter must be at least 1.437 in. Just to the right of Disk 3: T  0.00 kip  in. Diameter must be at least 1.389 in.

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PROBLEM 8.C4 The solid shaft AB of length L, uniform diameter d, and allowable shearing stress  all rotates at a given speed expressed in rpm (Fig. 8.C4). Gears G1, G2, . . . , Gn are attached to the shaft and each of these gears meshes with another gear (not shown), either at the top or bottom of its vertical diameter, or at the left or right end of its horizontal diameter. One of these gears is connected to a motor and the rest of them to various machine tools. Denoting by ri the radius of disk Gi, by ci its distance from the support at A, and by Pi the power transmitted to that gear (sign) or taken of that gear (sign), write a computer program to calculate the smallest permissible value of the diameter d of shaft AB. Use this program to solve Probs 8.27 and 8.68.

y

L

ci A

ri z G1 G2

B Gi

x

Gn

SOLUTION 1. Enter w in rpm and determine frequency f  w/60. 2. For each gear, determine the torque Ti  Pi /2 f , where Pi is the power input () or output (–) at the gear. 3. For each gear, determine the force Fi  Ti /ri exerted on the gear and its components ( Fy )i and ( Fz )i . 4. Determine the components Ay and Az of reaction at A by summing moments about axes Bz  // z and By  // y : 1 ( Fy )i ( L  ci ) L 1 Az   ( Fz )i ( L  ci ) L

M z   0: Ay L  ( Fy )i ( L  ci )  0, Ay   M y  0: Az L  ( Fz )i ( L  ci )  0,

5. Determine ( M y )i ,( M z )i , and torque Ti just to the left of gear Gi :

 ( F ) c  c  (M )   A c   ( F ) c  c  T   T c  c 

( M y )i  Az ci 

z k

i

k

1

k

z i

y i

y k

i

k

1

k

i

k

i

k

0

k

where   indicates a singularity function. 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. 1361

PROBLEM 8.C4 (Continued) 6. The minimum diameter d required to the left of Gi is obtained by first computing ( J/c)i from Equation (8.7).

 M y i   M z i2  Ti2 2

J c   i

7. Recalling that J  12  c 4 and, thus, that we have ci  2  Jc 

1/3 i

and di  4  Jc 

 all

 Jc i  12  ci3 ,

1/3



i

This is the required diameter just to the left of gear Gi . 8. The required diameter just to the right of gear Gi is obtained by replacing Ti with Ti 1 in the above computation. 9. The smallest permissible value of the diameter of the shaft is the largest of the values obtained for di .

Program Outputs Problem 8.25

Problem 8.27

  450 rpm

  600 rpm

Number of Gears: 3

Number of Gears: 3

Length of shaft  750 mm

Length of shaft  24 in.

  55 MPa

  8 ksi

For Gear 1,

For Gear 1,

Power input  –8.00 kW

Power input  60.00 hp

Radius of gear  60 mm

Radius of gear  3.00 in.

Distance from A in mm  150

Distance from A in inches  4.0

For Gear 2,

FY  0

Power input  20.00 kW

Fz  2.100845

Radius of gear  100 mm

For Gear 2,

Distance from A in mm  375

Power input  –40.00 hp

For Gear 3,

Radius of gear  4.00 in.

Power input  –12.00 kW Radius of gear  60 mm

Distance from A in inches  10.0 FY  1.050423

Distance from A in mm  600

FZ  0

AY  –0.849 kN, AZ  4.386

For Gear 3,

BY  –3.395 kN, BZ  2.688

Power input  –20.00 hp

Just to the left of Gear 1:

Radius of gear  4.00 in.

MY  657.84 Nm MZ  127.32 Nm T  0.00 Nm

Distance from A in inches  18.0 FY  0 FZ  –0.5252113

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PROBLEM 8.C4 (Continued) AY  –0.6127 kips, AZ  –1.6194 kips

Diameter must be at least 39.59 mm. Just to the right of Gear 1: T  –169.77 N  m

BY  –0.4377 kips, BZ  0.438 kips Just to the left of Gear 1:

Diameter must be at least 40.00 mm.

MY  –6.478 kip  in.

Just to the left of Gear 2:

MZ  2.451 kip  in.

MY  1007.98 N  m

T  0.000 kip  in.

MZ  318.31 N  m

Diameter must be at least 1.640 in.

T  –169.77 N  m

Just to the right of Gear 1:

Diameter must be at least 46.28 mm.

T  6.3025 kip  in.

Just to the right of Gear 2:

Diameter must be at least 1.813 in.

T  254.65 N  m

Just to the left of Gear 2:

Diameter must be at least 46.52 mm.



MY  –3.589 kip  in.

Just to the left of Gear 3:

MZ  6.127 kip  in.

MY  403.19 N  m

T  6.303 kip  in.

MZ  509.30 N  m

Diameter must be at least 1.822 in.

T  254.65 N  m

Just to the right of Gear 2:

Diameter must be at least 40.13 mm.

T  2.1008 kip  in.

Just to the right of Gear 3:

Diameter must be at least 1.677 in.

T  0.00 N  m

Just to the left of Gear 3:

Diameter must be at least 39.18 mm.

MY  0.263 kip  in.



MZ  2.626 kip  in. T  2.101 kip  in. Diameter must be at least 1.290 in. Just to the right of Gear 3: T  0.000 kip  in. Diameter must be at least 1.189 in. 

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PROBLEM 8.C5

y

Write a computer program that can be used to calculate the normal and shearing stresses at points with given coordinates y and z located on the surface of a machine part having a rectangular cross section. The internal forces are known to be equivalent to the force-couple system shown. Write the program so that the loads and dimensions can be expressed in either SI or U.S. customary units. Use this program to solve (a) Prob. 8.45b, (b) Prob. 8.47a.

My

b

Vy h

C P Vz x

Mz z

SOLUTION Enter:

b and h

b3 h bh3 Iz  12 12 For point on surface, enter y and z. Note: y and z must satisfy one of following:

Program: A  bh

y2 

h2 4

Iy 

and

z2 

b2 4

(1)

b2 h2 and y 2  4 4 If either (1) and (2) are satisfied, compute z2 

or



(2)

P Myz Mz y   A Iy Iz

If z2  b2/4, then point is on vertical surface and h  h Qz  b   y   2   2

 2

 h2 y 2  1 y   b    2  2  8

v y Qz I zb

2

If y  h /4, the point is on horizontal surface, and  b2 z 2  b  b 1 Qy  h   z   z   h    2  2  2 2  8 Vz Qy  I yh Force-couple system:

P  50 kips Vz  6 kips Vy  2 kips M y  (6 kips) (8.5 in.)  51 kip  in. M z  (2 kips) (10.5 in.)  21 kip  in.

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PROBLEM 8.C5 (Continued)

Force-Couple at Centroid

Problem 8.45b P  50.00 kips M Y  51.00 kip  in. VY  2.00 kips

M Z  21.00 kip  in. VZ  6.00 kips

++++++++++++++++++++++++++ At point of coordinates: y  0.90 in. z  1.20 in.   6.004 ksi

  0.781 ksi Force-Couple System P  10 kN Vy  750 N Vz  500 N M y  (500 N)(220 mm)  110 N  m M z  (750 N)(180 mm)  135 N  m Force-Couple at Centroid

Problem 8.47a P  10000.00 N M Y  110.00 N  m VY  750.00 N

M Z  135.00 N  m VZ  500.00 N

++++++++++++++++++++++++++ At point of coordinates: y  16.00 mm x  0.00 mm   18.392 MPa

  0.391 MPa

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PROBLEM 8.C6 A 9 kN

d 120 mm

30⬚ H 12 mm 40 mm

K

Member AB has a rectangular cross section of 10  24 mm. For the loading shown, write a computer program that can be used to determine the normal and shearing stresses at Points H and K for values of d from 0 to 120 mm, using 15-mm increments. Use this program to solve Prob. 8.35.

12 mm B

SOLUTION Cross section: Enter A  (0.010 m)(0.024 m)  240  106 m 2 I  (0.010 m)(0.024 m)3 /12  138.24  109 m 4 R  0.5(0.029 m)  12  103 m

Compute reaction at A. M B  0: (9 kN)(120  d ) sin 30°  A (120) cos 30°  0 (120 mm  d ) A  (9 kN) tan 30 120 mm Free Body:

From A to section containing Points H and K,

Define:

If d  80 mm, Then STP  1 Else STP  0

Program force-couple system.

F   A sin 30  (9 kN) cos 30°(Step) V   A cos 30°  (9 kN) sin 30°(Step) M  A(80 mm) cos 30°  (9 kN)(80 mm  d )sin 30 (STP) At Point H,

 H   F/ A 3 2

 H  V/ A At Point K,

 K   F/A  Mc/I K  0

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PROBLEM 8.C6 (Continued)

Program Output Problem 8.35 Stresses in MPa

d (mm)

H

H

K

K

0.0

43.30

0.00

43.30

0.00

15.0

41.95

3.52

65.39

0.00

30.0

40.59

7.03

87.47

0.00

45.0

39.24

10.55

109.55

0.00

60.0

37.89

14.06

131.64

0.00

75.0

36.54

17.58

153.72

0.00

90.0

2.71

7.03

96.46

0.00

105.0

1.35

3.52

48.23

0.00

120.0

0.00

0.00

0.00

0.00



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PROBLEM 8.C7*

y x H

The structural tube shown has a uniform wall thickness of 0.3 in. A 9-kip force is applied at a bar (not shown) that is welded to the end of the tube. Write a computer program that can be used to determine, for any given value of c, the principal stresses, principal planes, and maximum shearing stress at Point H for values of d from 3 in. to 3 in., using one-inch increments. Use this program to solve Prob. 8.62a.

10 in. d 3 in. 3 in.

4 in. z

9 kips

c

SOLUTION Force-couple system. Enter:

V  9 kips  M x  (9 kips)(10 in.)  90 kips  in. T  (9 kips)c Area enclosed. a  (a  t )(b  t ) T 9c T   2ta 2ta  T  Shearing stress due to torsion b t  Q  dt    2 2 I  ab3 /12  (a  2t )(b  2t )3 /12 VQ V  It  V  Shearing stress due to V  total   T   Y

Bending:

H

b Mx   2  I 2

Principal stresses:

1   2  ave   H ; R   H    total 2 2    max   ave  R;  min   ave  R; 1 2

2

  total   H  2   total  ;  max     2    ave 

 P  tan 1 

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PROBLEM 8.C7* (Continued)

Rectangular tube of uniform thickness t  0.3 in. Outside dimensions. Horizontal width:

a  6 in.

Vertical depth:

b  4 in.

Vertical load:

P  9 kips;

Line of action at

x  c

Find normal and shearing stresses at point H.

( x  d , y  b/2)

Problem 8.62a c  2.85 in.

Program output for value of d in.



ksi

V

ksi

T

ksi

 Total ksi

 max ksi

 min ksi

 max ksi

P

Degrees

----------------------------------------------------------------------------------------------------------------------------12.58 14.65 8.36 3.00 3.49 2.03 5.52 2.08 18.49 2.00

12.58

2.33

2.03

4.35

13.94

1.36

7.65

16.00

1.00

12.58

1.16

2.03

3.19

13.34

0.76

7.05

12.78

 0.00

12.58

0.00

2.03

2.03

12.89

0.32

6.61

8.73 

1.00

12.58

1.16

2.03

0.86

12.63

0.06

6.35

3.89

2.00

12.58

2.33

2.03

0.30

12.58

0.01

6.30

1.36

3.00

12.58

3.49

2.03

1.46

12.74

0.17

6.46

6.46



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