C H APTER 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
|M |max
 90kips
 (45)(10)  450kip in.
 39 rolled steel section,
For W10
d  9.92in.,
b f  7.99in.,
t f 0.530in.,
t w  0.315in.,
I x 209in
S x 42.1in
1
c d 
2
(a)
4
4.96in.y ct  b
m

b

||M
max
Sx
4.43in.
450
10.69 ksi
m
42.1


yb
 4.43 
m  
 (10.69)  9.55 ksi
c
 4.96 
A b t
f

f
3
 4.2347 in 2
f f
yf 
1
2
(c
 yb )  4.695in.
Qb  A f y f  19.8819 in 3
 xy

|V |maxQb
I xt w

(45)(19.8819)
(209)(0.315)
 13.5898 ksi
2
R 

b
 R  19.18 ksi
(b)
 max
(c)
Since  max >  all ( 18ksi),
2
 b 
2
 2    xy  14.4043 ksi
 
 max
10
39
W

 19.18 ksi

is not acceptable. 
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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1241
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  22.5kips
|M |max  (22.5)(20)  450 kip in.
For W10  39 rolled steel section,
d  9.92in.,
b f 7.99in.,
tw  0.315in.,
I x 209in ,
1
c d 
2
(a)

m

b
4.96in. y ct 


|M |max
Sx
yb

c

b
t f 0.530in.,
Sx 
42.1in
4
f
4.43in.
450
42.1
m
3
10.69 ksi

m


 4.43 

 (10.69)  9.55 ksi
 4.96 
A f  b f t f  4.2347 in 2
y f  1 (c  yb )  4.695in.
2
Qb  A f y f  19.8819 in 3

xy

|V |max Qb
I xtw

(22.5)(19.8819)
(209)(0.315)
 6.7949 ksi
2
 b 
2
   xy  8.3049 ksi
 2 
R 


b
 R  13.08 ksi
(b)

(c)
Since  max <  all (  18ksi),
max
2
PROPRIETARY MATERIAL.
Copyright

max
 13.08 ksi 
W10  39 is acceptable.
© 2015 McGraw-Hill Education.
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on a website, in whole or part.
1242
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
|M |max
3
 700kN 700 10 N
6
 (700
 103 )(2.5)
  1.75

10 N m
For W920  449 rolled steel beam,
d  947 mm,
b f  424mm,
t w  24.0 mm,  I x
1
c d 
2
(a)
6

473.5 mm,y ct 
m

|M |max
b

yb
Sx
m
c


t f 42.7 mm,
8780  10 mm ,
1.75  10
b
4
f
Sx
m
6
 94.595 MPa
 94.6 MPa
m
6
473.5
3
430.8 mm
18,5 00  10
430.8
3
18,500 10m m

(94.595)  86.064 MP a
A f  b f tt  18.1048 103 mm 2
y f  1 (c  yb )  452.15 mm
2
Qb  A f y f
 xy

8186.1 103 mm
 3 8186.1 106 m3
|V |max Qb
I x tw
2
(700  10 )(8186.1  10 )
3

6
(8780  10 )(24.0  10  )
6
3
 27.194 MPa
2
 b 
 86.064 
2
2
   xy   2   27.194  50.904 MPa


 2 
R 

b
 R  93.9 MPa
(b)
 max
(c)
Since 94.6 MPa
2
  all ( 100 MPa),
 max
920 449
W

 93.9 MPa

is acceptable. 
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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on a website, in whole or part.
1243
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  850kN 850 10 3N
|M |max  (850
 103 )(2.0)
  1.70

106 N m
For W920  449 rolled steel section,
d  947mm,
w
 24.0mm,
1
c d 
2
(a)
b f 424mm,

m

b
 I x 8780 10mm ,
473.5mmy ct 

|M |max

yb
Sx

c
t f 42.7
 mm,
6
m


b
4
f
S x 18,500 10 3mm
430.8mm

1.70  106
18,500  10
430.8
473.5
3
m

6
 91.892 MPa
m
 91.9 MPa 
(91.892)  83.605 MPa
A f  b f t f  18.1048  103 mm 2
yf 
1
2
(c  yb )  452.15mm
Qb  A f y f

xy

8186.1 103 mm3
|V |max Qb
I x tw
2


b

(c)
Since 95.1 MPa <
2
6
3
(8780  10 )(24.0  10 )
 83.605 
  
 2 
 33.021 MPa
2
2
33.021
 R  95.1 MPa
(b)
max
6
(850  10 )(8186.1  10 )
3

 b 
2
  xy
 2 
R 
8186.1 106 m3

all (

100 MPa),
PROPRIETARY MATERIAL.
Copyright
53.271 MPa
920 449
W

max
 95.1 MPa 
is acceptable. 
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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.
1244
PROBLEM 8.5
40 kN
2.2 kN/m
A
C
B
4.5m
2.7m
(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
  C 0: 7.2RA  (2.2)(7.2)(3.6)

RA  22.92 kN
VA  RA  22.92 kN
(40)(2.7)
0
VB  22.92  (2.2)(4.5)  13.02 kN

VB  13.02  40

VC  26.98
26.98kN

(2.2)(2.7)
32.92 kN
MA  0
1
0
MB 
 (22.92 13.02)(4.5)

2
80.865 kN m
MC  0
S min 
M
max
80.865  103
 all

165  106
490
106m
3
 490  103 mm 3
Shape
S (103 mm3)
W360  39
578
W310  38.7
547
W250  44.8
531
W200  52
511

(a)
d  310mm
Use W310  38.7. 
t f 9.65mm
tw  5.84 mm
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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on a website, in whole or part.
1245
PROBLEM 8.5 (Continued)
(b)
m
m
80.865  103
MB
S
 
V



547  106
147.834
106 Pa
32.92  10
 
max
3
(310  10 3 )(5.84  103 )
dtw
18.1838
1
155mm y ct   b   f 155 9.65
c d 
2
y
 145.35 
b  b  m  
 (147.834)  138.630 MPa
c
 155 
At point B,
w

 147.8 MPa

m
 18.18 MPa

 max
 140.2 MPa

106 Pa
145.35mm
V
(26.98  103 )

 14.9028 MPa
dtw
(310  10 3 )(5.84  10 3 )
 b 

 2 
2
R 
 max
m

 b  R
2
 w2
 (69.315)
 2

69.315
(14.9028) 2
70.899
70.899 MPa
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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on a website, in whole or part.
1246
PROBLEM 8.6
275 kN
B
C
A
D
275 kN
1.5 m
3.6 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
V
RB  504.17kN
RC 504.17
  
 275kN
M max 412.5kN m
max
S min 
M
412.5  10 2
max
 
10 6 m
2578
160  106
 all
3
 2578  103 mm 3
Sx (103 mm3)
Shape

W760  147
4410
W690  125
3490
W530  150
3720
W460  158
3340
W360  216
3800
(a)
d  678mm
(b)
m
m
  max
M

V
V
Aw
1
c d 
2
b
412.5  103
max


118.195
275  103
 
max
(678  103 )(11.7  103 )
dtw


3490  10 6
S
67.8

339
t mm,
2
f
mm,
  
tyc16.3
t f 16.3mm
106 Pa
m
106 Pa
34.667
b
Use W690  125. 
 mm
tw 11.7
f
339
m
16.3
 118.2 MPa

 34.7 MPa

322.7mm
yb
 322.7 
m  
 (118.195)  112.512 MP a
c
 339 
 b 
 2
 
R 
2
 m2
 (56.256)
 2
(34.667)2
66.080 MPa
b
 max

 2

R  56.256
66.080
 max
 122.3 MPa
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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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.
1247
PROBLEM 8.45
50 kips
0.9 in.
2 kips
C
0.9 in.
2.4 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.
2 in.
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
 2)(6)  51k ip in.
M z  (10.5
M x  (10.5)(2)  21ki p  in.
Section properties.
A  (1.8)(4.8)  8.64in 2
Ix 
Iz 
1
12
1
12
3
 2.3328 in 4
3
 6.5888in 4
(4.8)(1.8)
(1.8)(4.8)
Stresses.
  
P
A

Mzx
x
Iz
Ix
z

Vx Q
I zt
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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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.
1300
PROBLEM 8.45 (Continued)
(a)
Point a:
x  0, z
 n., Q
0.9i
(1.8)(2.4)(1.2) 5.184i n 3
50
 0
8.64
 

(b)
Point b:

x  1.2i n.,


(c)
Point c:
x 2.4 in.,
(6)(5.184)




042 ksi
(1.8)(1.2)(1.8)
3.888i n
50
(51)(1.2)
(21)(0.9)
8.64
16.5888
2.3328

(6)(3.888)
(16.5888)(1.8)
z 0.9 in.,

2.3328
(16.5888)(1.8)

z 0.9i n., Q


(21)(0.9)
 0.781ksi

 2.31ksi

 1.



3
6.00 ksi 
0.781 ksi 


Q 0

50
(51)(2.4)
(21)(0.9)
8.64
16.5888
2.3328

9.69 ksi 



0

PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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1301
PROBLEM 8.46
50 kips
0.9 in.
2 kips
C
0.9 in.
2.4 in.
Solve Prob. 8.45, assuming that h  12 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.
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
 2)(6)

 60k ip in.
M z  (12
M x  (12)(2)  24 kip  in.
Section properties.
A  (1.8)(4.8)  8.64in 2
Ix 
Iz 
1
12
1
12
3
 2.3328 in 4
3
 6.5888in 4
(4.8)(1.8)
(1.8)(4.8)
Stresses.
  
P
A

Mzx
x
Iz
Ix
z

Vx Q
I zt
PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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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.
1302
PROBLEM 8.46 (Continued)
(a)
Point a:
x  0, z
 n., Q
0.9i
(1.8)(2.4)(1.2) 5.184i n 3
 

(b)
Point b:

x  1.2i n.,

50
 0
8.64
(6)(5.184)
(16.5888)(1.8)

z 0.9i n., Q



(24)(0.9)
2.3328


042 ksi
(1.8)(1.2)(1.8)
3.888i n

50
(60)(1.2)
(24)(0.9)
16.5888
2.3328

 3.4

 1.

3
8.64
(6)(3.888)

7 ksi

81 ksi
 7.81ksi


 0.7


 12

(16.5888)(1.8)
(c)
Point c:
x 2.4 in.,
z 0.9 in.,


Q 0

50
(60)(2.4)
(24)(0.9)
8.64
16.5888
2.3328

.15 ksi

0

PROPRIETARY MATERIAL.Copyright © 2015 McGraw-Hill Education.
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1303
PROBLEM 8.47
60 mm
24 mm
a
Three forces are applied to the bar
shown. Determine the normal and
shearing stresses at (a) point a,
(b) point b, (c) point c.
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
Iz 
1
12
(60)(32)
3
 163.84  103 mm 4
 163.84  109 m 4
Iy 
1
12
(32)(60)3
 579  103 mm 4
 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)
 135N  m
Side View
3
M y  (220  10 )(500)
 110N  m
T 0


z
P Mz y

 y
A
Iz
Iy


VzQ
I yt
Top View
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1304
PROBLEM 8.47 (Continued)
(a)
Point a:
 Q
y  16 mm,  z 0,
3
3
Az  
(32)(30)(15) 14.4 10 mm

3
3
 10  10 6  (135)(16  109 )  0
1920  10
163.84  10


(500)(14.4  10 )
39 MPa

 18.


 0.39

6
(b)
Point b:
9
 z
y  16 mm,
3
 mm, Q
15
10  10
3
(c)
Point c:




1920  10


)
(135)(16
10
3
6

163.84  10
(500)(10.8  106 )
9
10  10
1920  10
6

(135)(16
10)
163.84  10
(110)( 15 10 )
3

576  10
9
.3 MPa
  21

 0.2


0
3

9
93 MPa
3
 mm, Q
30
3

  (32)(15)(22.5) 10.8 103 mm3
Az
(576  10  )(32  10 )
 z
y  16 mm,

1 MPa
(576  10 )(32  10 )
9
(110)( 30 10 )
6

576 10
9

 24.1MPa

0


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1305
PROBLEM 8.48
60 mm
24 mm
a
b
Solve Prob. 8.47, assuming that the
750-N force is directed vertically
upward.
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
Iz 
1
12
3
(60)(32)
 163.84  103 mm 4
 163.84  109 m 4
Iy 
1
12
3
(32)(60)
 576  103 mm 4
9
4
 576  10 m
At the section containing points a, b, and c,
P  10 kN
T 0
V y  750 N
Vz  500 N
M z  (180  103 )(750)
 135N  m
Side View
3
M y  (220  10 )(500)
 110N  m


z
P Mzy

 y
A
Iz
Iy


VzQ
I yt
Top View
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1306
PROBLEM 8.48 (Continued)
(a)
Point a:
 Q
y  16 mm,  z 0,
(135)(16  10 )
10  10
3

 1920  10


3
3
Az  
(32)(30)(15) 14.4 10 mm
3
6
 163.84  10
9
(500)(14.4  10  )
0
8 MPa

 7.9


 0.39


 5.11MPa


 0.2


 2.2

6
(b)
Point b:
9
 z
y  16 mm,



3
 mm, Q
15
10  10
3

1 MPa
(163.84  10 )(32  10 )
1920  10
  (32)(15)(22.5) 10.8 103 mm3
Az


)
(135)(16
10
3
6

163.84  10
9
(110)( 15 10 )
3

576  10
9
6
(c)
Point c:
(500)(10.8
10 ) 9
9 
(163.84  10 )(32  10 )
 z
y  16 mm,


10  10
 mm, Q
30
1920  10
6
0
3
3

(135)(16

10)
163.84  10
93 MPa
9
(110)( 30 10 )
3

576  10
9
5 MPa

0

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1307
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).
J
c 
 i
7. Recalling that J
we have ci
 12  c and, thus, that
4
1/3

2
y
i
2
  M z i  Ti2
 all
 Jc i  12  ci3 ,
1/3
 2  Jc i and di  4  Jc 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 d i .
Program Outputs
Problem 8.25

 450 rpm
Problem 8.27

 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
Distance from A in inches  10.0
Radius of gear  60 mm
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
Distance from A in inches  18.0
MZ  127.32 Nm
T  0.00 Nm
FY  0
FZ  –0.5252113
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1362
PROBLEM 8.C4 (Continued)
Diameter must be at least 39.59 mm.
AY  –0.6127 kips, AZ  –1.6194 kips
Just to the right of Gear 1:
BY  –0.4377 kips, BZ  0.438 kips
Just to the left of Gear 1:
T  –169.77 N  m
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
Diameter must be at least 46.52 mm.
Just to the left of Gear 2:

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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1363
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
b and h
Enter:
Program: A  bh
Iy 
bh3
12
bh
Iz 
3
12
For point on surface, enter y and z.
y and z must satisfy one of following:
Note:
or
y2 
h2
4
and
z2 
b2
4
(1)
z2 
b2
4
and
y2 
h2
4
(2)
If either (1) and (2) are satisfied, compute
P

2
If z
M yz
 b2/4, then point is on vertical surface and
h
Qz b y
2

2
If y
zy
 A  Iy  Iz

h

b
y 22


h1 y

 2

 8
2
2



v y Qz
I zb
2
 h /4, the point is on horizontal surface, and
 b b 
Qy h z 
z h
 2  22

Force-couple system:

b1 z

 2

 8
2
2



VzQ y
Iyh
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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1364
PROBLEM 8.C5 (Continued)
Problem 8.45b
Force-Couple at Centroid
P  50.00 kips
 in.   MZ
M Y  51.00kip
21.00kip in.
VY  2.00kips
6.00kips

VZ
++++++++++++++++++++++++++
At point of coordinates:
y  0.90 in.
z 1.20 in.
 6.004 ksi
  0.781 ksi

Force-Couple System
P  10kN  V y
N
750
Vz500N
M y  (500 N)(220 mm) 110 N m
M z  (750 N)(180 mm) 135 N m
Problem 8.47a
Force-Couple at Centroid
P  10000.00 N
M Y  110.00 N m   M Z
VY  750.00N
135.00N m
VZ  500.00N
++++++++++++++++++++++++++
At point of coordinates:
y 16.00 mm
x 0.00 mm
 18.392 MPa
  0.391 MPa

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1365
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 6m 2
I  (0.010 m)(0.024 m)3 /12  138.24  109 m 4
R  0.5(0.029m ) 12 10 3m
Compute reaction at A.
M B  0: (9 kN)(120  d) sin 30°
 A (120) cos 30°  0
(120mm  d)
tan 30 
A  (9kN)
120 mm
Free Body:
Define:
From A to section containing Points H and K,
If d
 80 mm, Then STP 1 Else STP 0
Program force-couple system.

F  A sin30
(9 kN) cos 30°(Step)
V   A co s 30°  (9 kN) sin 30°(Step)
 30° (9kN)(80

M  A(80 mm) cos
mm
d)s in3 0 (STP)
At Point H,
H
  F/A
H
 V/A
K
  F/A  Mc/I
K
0
3
2
At Point K,
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1366
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
45.0
40.59 7.03
39.24 10.55
87.47 

109.55
0.00
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
105.0
1.35
3.52
48.23
0.00
120.0
0.00
0.00
0.00
0.00

0.00
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1367
y
PROBLEM 8.C7*
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. 8.62 a.
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
  Shearing stress due to torsion
T
b t 
 
2 2
  (a  2t )(b 2t )/3 12
I  ab3 /12
Q  dt 
VQ
It
 V  Shearing stress due to V
 total   T   Y
Bending:
V

H

b
Mx  
2
I
2
Principal stresses:
1
 
2
  H ; R   H    total
2
 2 

 max 
R;
ave  R;  min
ave
 ave
2
P
2
 12 tan 1  total  ;  max   2H    total


 ave 
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.
1368
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 /2)
b
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
3.49
2.03
5.52
2.08
18.49
3.00
2.00
12.58
2.33
2.03
4.35
1.00
12.58
1.16
2.03
3.19
 0.00
12.58
0.00
2.03
2.03
12.89
0.326.61
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
13.94
4 13.3
0.76
1.36
7.65

7.05
16.00
12.78
8.73


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.
1369