Mom chap 8 - mechanics of materials 7th edition solition manual chapter 8 PDF

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CHAPTER 8 PROBLEM 8 Solve Prob. 8, assuming that P 22 kips and a 20 in. PROBLEM 8 A W10 39 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 j...

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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.,

bf  7.99 in., t f  0.530 in.,

t w  0.315 in., I x  209 in4 c 

(a)

1 d  4.96 in. 2

S x  42.1 in3

y b  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 in2 1 ( c  yb )  4.695 in. 2 Qb  A f y f  19.8819 in 3 yf 

 xy 

|V |maxQb (45)(19.8819)   13.5898 ksi I x tw (209)(0.315) 2

  R   b    xy2  14.4043 ksi  2 

b

(b)

 max 

(c)

Since max >  all (  18 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.,

t f  0.530 in.,

tw  0.315 in.,

I x  209 in ,

Sx  42.1 in3

4

1 c  d  4.96 in. 2 (a)

y b  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 

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

 xy 

|V | max Qb (22.5)(19.8819)   6.7949 ksi I xt w (209)(0.315) 2

  R   b    2xy  8.3049 ksi  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 10 3 )(2.5)  1.75  10 6 N  m For W920  449 rolled steel beam, d  947 mm,

b f  424 mm,

t w  24.0 mm,

I x  8780 106 mm4 , 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| 1.75 10 6 m  max   Sx 18,500 10 6 yb 430.8   (94.595)  86.064 MPa c m 473.5 A f  b f tt  18.1048 103 mm2

b 

1 (c  y b )  452.15 mm 2  Qb  A f y f  8186.1  10 3 mm 3  8186.1 10 6 m 3 yf 

xy 

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

2

   86.064  2 R   b    2xy     27.194  50.904 MPa 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,

Ix  8780 10 mm , Sx  18,500 103 mm3

1 c  d  473.5 mm 2 (a)

t f  42.7 mm, 6

4

yb  c  t f  430.8 mm

m  91.892 MPa  m  91.9 MPa 

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

yb

430.8 m  (91.892)  83.605 MPa c 473.5 Af  bf t f  18.1048 10 3 mm 2

b 

1 (c  yb )  452.15 mm 2  Q b  A f y f  8186.1  10 3 mm 3  8186.1 10 6 m 3 yf 

 xy 

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

2

   83.605  2 R   b   xy2     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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2.2 kN/m

PROBLEM 8.5

40 kN

C

A 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: 7.2R A  (2.2)(7.2)(3.6)  (40)(2.7)  0 RA  22.92 kN VA  RA  22.92 kN V B  22.92  (2.2)(4.5)  13.02 kN V B  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

Smin 

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 S 547  10 6

m  147.8 MPa 

m 

V

max

dt w



32.92  103 (310  103 )(5.84  103 )

 18.1838  106 Pa

m  18.18 MPa  1 d  155 mm y b  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   b    w2   2 

 max 

b 2

(69.315) 2  (14.9028) 2  70.899 MPa

 max  140.2 MPa 

 R  69.315  70.899

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275 kN

PROBLEM 8.6

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 R B  504.17 kN  V

max

 275 kN

S min 

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  V max Aw

c 



M

max

S V max dt w

 

412.5  103 3490  10 6

 118.195  106 Pa 

275  103 (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

 34.667  106 Pa 

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   b   m2   2 

 max 

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 R A  50 kips

V

max

 30 kips M

Smin 

max

 all

RD  50 kips

M 

max

 200 kip  ft  2400 kip  in.

2400  100 in 3 24 Shape



S (in 3)

W24  68 W21  62

154

W18  76 W16  77

146

W12  96 W10  112

131

127 134 126 Use W21  62. 

(a) d  21.0 in.

(b)

M

m  m 



max

S V

max

dt w



2400  18.8976 ksi 127

 m  18.9 ksi 

30  3.5714 ksi (21.0)(0.400)

1 21.0  10.50 in. c  d  2 2

b 

t f  0.615 in. tw  0.400 in.

 m  3.57 ksi 

y b  c  t f  10.50  0.615  9.8850 in.

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

  R   b    m2  (8.8954) 2  (3.5714) 2  9.5856 ksi  2 

 max 

b 2

 R  8.8954  9.5856  18.4810 ksi

 max  18.48 ksi ◄

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PROBLEM 8.8

1.5 kips/ft

C

A 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

max

M

 11.25 kips  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 W8  18

15.2

W6  20

13.4

13.8

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   c 

1 10.0 d   5.00 in. 2 2

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PROBLEM 8.8 (Continued)

y b  c  t f  5.00  0.270  4.73 in.

b 

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

2

   22.210  2 R   b    2m     (4.8913)  12.1345 ksi 2 2    

max 

b 2

R 

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:

R A  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,

bf  209 mm,

tw  10.2 mm,

c 

I  554 10 6 mm 4 (a)

m 

t f  15.6 mm,

1 d  266.5 mm 2 S  2080  10 3 mm3

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

 m  137.5 MPa  yb  c  t f  250.9 mm Af  bf tf  3260.4 mm2 y 

1 (c  yb )  258.7 mm 2

Q  A f y  843.47  103 mm3

At midspan:

V  0

b 

b  0

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

 max  129.5 MPa 

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PROBLEM 8.9 (Continued)

At sections B and C: M 270  10 3   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 It w (554  10 6)(10.2  10 3)

m 

 12.3143 MPa 2

  R   b    2b  62.333 MPa 2 

 max 

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, 

M

max

all

 160 MPa

 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

bf  102 mm

tw  6.35 mm

t f  10.0 mm

I  40.1  10 6 mm 4

S  308  10 3 mm 3 1 d  129.5 mm 2

c

(a)

M

m 

max

S
...