Chapter 1 Problems Solutions

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Chapter 1 Problems
Solutions
d1
PROBLEM 1.1
d2
125 kN B
C
A
Two solid cylindrical rods AB and BC are
welded together at B and loaded as shown.
Knowing that d1  30 mm and d 2  50 mm,
find the average normal stress at the
midsection of (a) rod AB, (b) rod BC.
60 kN
125 kN
0.9 m
1.2 m
SOLUTION
(a)
Rod AB:
Force:
P  60  103 N tension
Area:
A
Normal stress:
(b)
 AB 

4
d12 

4
(30  103 ) 2  706.86  106 m 2
P
60  103

 84.882  106 Pa
A 706.86  106
 AB  84.9 MPa 
Rod BC:
Force:
P  60  103  (2)(125  103 )  190  103 N
Area:
A
Normal stress:
 BC 

4
d 22 

4
(50  103 )2  1.96350  103 m 2
P
190  103

 96.766  106 Pa
A 1.96350  103
 BC  96.8 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.
3
PROBLEM 1.7
0.4 m
C
0.25 m
0.2 m
B
Each of the four vertical links has an 8  36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter. Determine
the maximum value of the average normal stress in the links connecting
(a) points B and D, (b) points C and E.
E
20 kN
D
A
SOLUTION
Use bar ABC as a free body.
M C  0 :
(0.040) FBD  (0.025  0.040)(20  103 )  0
FBD  32.5  103 N
Link BD is in tension.
3
M B  0 :  (0.040) FCE  (0.025)(20  10 )  0
FCE  12.5  103 N
Link CE is in compression.
Net area of one link for tension  (0.008)(0.036  0.016)  160  106 m 2
For two parallel links,
(a)
 BD 
A net  320  106 m 2
FBD
32.5  103

 101.563  106
6
Anet
320  10
 BD  101.6 MPa 
Area for one link in compression  (0.008)(0.036)  288  106 m 2
For two parallel links,
(b)
 CE 
A  576  106 m 2
FCE
12.5  103

 21.701  106
6
A
576  10
 CE  21.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
9
4 kips
6
C
in.
u
B
1
308
Link BD consists of a single bar 1 in. wide and
1
in. thick. Knowing that each pin has a 83 -in.
2
diameter, determine the maximum value of the
average normal stress in link BD if (a)  = 0,
(b)  = 90.
.
2 in
A
PROBLEM 1.10
D
SOLUTION
Use bar ABC as a free body.
(a)
  0.
M A  0: (18 sin 30)(4)  (12 cos30) FBD  0
FBD  3.4641 kips (tension)
Area for tension loading:
Stress:
(b)
3  1 

A  (b  d )t  1     0.31250 in 2
8  2 

F
3.4641 kips
  BD 
A
0.31250 in 2
  11.09 ksi 
  90.
M A  0:  (18 cos30)(4)  (12 cos 30) FBD  0
FBD  6 kips i.e. compression.
Area for compression loading:
Stress:
1
A  bt  (1)    0.5 in 2
2
F
6 kips
  BD 
A
0.5 in 2
  12.00 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.
12
PROBLEM 1.21
P 5 40 kN
120 mm
b
A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete
footing, (b) the size of the footing for which the average bearing
stress in the soil is 145 kPa.
100 mm
b
SOLUTION
(a)
Bearing stress on concrete footing.
P  40 kN  40  103 N
A  (100)(120)  12  103 mm 2  12  103 m 2
 
(b)
P
40  103

 3.3333  106 Pa
A 12  103
Footing area. P  40  103 N
 
P
A
3.33 MPa 
  145 kPa  45  103 Pa
A
P


40  103
 0.27586 m 2
3
145  10
Since the area is square, A  b 2
b
A 
0.27586  0.525 m
b  525 mm 
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.
23
PROBLEM 1.45
Three 34 -in.-diameter steel bolts are to be used to attach the steel plate shown to
a wooden beam. Knowing that the plate will support a load P = 24 kips and that
the ultimate shearing stress for the steel used is 52 ksi, determine the factor of
safety for this design.
P
SOLUTION
For each bolt,
A

4
d2 
 3
2
2
   0.44179 in
44
PU  A U  (0.44179)(52)
 22.973 kips
For the three bolts,
PU  (3)(22.973)  68.919 kips
Factor of safety:
F. S . 
PU
68.919

24
P
F. S .  2.87 
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.
51
PROBLEM 1.46
Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing that the plate will support a load P = 28 kips, that the ultimate shearing
stress for the steel used is 52 ksi, and that a factor of safety of 3.25 is desired,
determine the required diameter of the bolts.
P
SOLUTION
For each bolt,
Required:
P
24
 8 kips
3
PU  ( F. S.) P  (3.25)(8.0)  26.0 kips
U 
d 
PU
P
4P
  U 2  U2
A
d
d
4
4 PU
 U

(4)(26.0)
 0.79789 in.
 (52)
d  0.798 in. 
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.
52
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