PROBLEM 1.4
Two solid cylindrical rods AB and BC are welded together at B and loaded
as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average
normal stress at the midsection of (a) rod AB, (b) rod BC.
SOLUTION
(a)
Rod AB
P = 40 + 30 = 70 kN = 70 × 103 N
A=
σ AB =
(b)
π
4
d12 =
π
4
(50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10 −3 m 2
P
70 × 103
=
= 35.7 × 106 Pa
A 1.9635 × 10−3
σ AB = 35.7 MPa 
Rod BC
P = 30 kN = 30 × 103 N
A=
σ BC =
π
4
d 22 =
π
4
(30)2 = 706.86 mm 2 = 706.86 × 10 −6 m 2
P
30 × 103
=
= 42.4 × 106 Pa
A 706.86 × 10−6
σ BC = 42.4 MPa 
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6
BeerMOM_ISM_C01.indd 6
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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.
M B  0 :  (0.040) FCE  (0.025)(20  103 )  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
A
576  106
 CE  21.7 MPa 
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99
BeerMOM_ISM_C01.indd 9
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PROBLEM 1.17
When the force P reached 8 kN, the wooden specimen shown failed in
shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.
SOLUTION
Area being sheared:
A = 90 mm × 15 mm = 1350 mm 2 = 1350 × 10−6 m 2
Force:
P = 8 × 103 N
Shearing stress:
τ =
P
8 × 103
−
= 5.93 × 106 Pa
A 1350 × 10−6
τ = 5.93 MPa 
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19
BeerMOM_ISM_C01.indd 19
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PROBLEM 1.25
P
A
16 mm
750 mm
750 mm
50 mm
B
Knowing that   40° and P  9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress in
the pin is not to exceed 120 MPa, (b) the corresponding average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
C
12 mm
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle:
P
FAB
FAC


sin 20 sin110 sin 50
P sin110
FAB 
sin 20
(9)sin110

 24.727 kN
sin 20
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28
28
BeerMOM_ISM_C01.indd 28
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PROBLEM 1.25 (Continued)
(a)
Allowable pin diameter.
 
d2 
FAB
F
2FAB
 AB 2 
where FAB  24.727  103 N
2 AP
d2
24d
2 FAB


(2)(24.727  103 )
 131.181  106 m 2
 (120  106 )
d  11.4534  103 m
(b)
11.45 mm 
Bearing stress in AB at A.
Ab  td  (0.016)(11.4534  103 )  183.254  106 m 2
b 
(c)
FAB
24.727  103

 134.933  106 Pa
6
Ab
183.254  10
134.9 MPa 
Bearing stress in support brackets at B.
A  td  (0.012)(11.4534  103 )  137.441  106 m 2
b 
1
2
FAB
A

(0.5)(24.727  103 )
 89.955  106 Pa
6
137.441  10
90.0 MPa 
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29
29
BeerMOM_ISM_C01.indd 29
12/12/2014 6:00:52 PM
P'
150 mm
45
45
P
75 mm
PROBLEM 1.30
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 620 kPa, determine
(a) the largest load P that can be safely applied, (b) the corresponding
tensile stress in the splice.
SOLUTION
  90  45  45
A0  (150)(75)  11.25  103 mm 2  11.25  103 m 2
  620 kPa  620  103 Pa
P sin 2
 
2 A0
(a)
P
2 A0
(2)(11.25  103 )(620  103 )

sin2
sin 90
 13.95  103 N
(b)
 
P  13.95 kN 
P cos 2 
(13.95  103 )(cos 45)2

A0
11.25  103
 620  103 Pa
  620 kPa 
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34
34
BeerMOM_ISM_C01.indd 34
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A
w
908
B
PROBLEM 1.38
Link BC is 6 mm thick, has a width w  25 mm, and is made of a steel with
a 480-MPa ultimate strength in tension. What was the safety factor used if the
structure shown was designed to support a 16-kN load P?
480 mm
C
D
P
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
M A  0:
(480) FBC  (600) P  0
FBC 
Ultimate load for member BC:
600
(600)(16  103 )
 20  103 N
P
480
480
FU   U A
FU  (480  106 )(0.006)(0.025)  72  103 N
Factor of safety:
F.S. 
FU
72  103

FBC
20  103
F.S.  3.60 
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42
42
BeerMOM_ISM_C01.indd 42
12/12/2014 6:00:56 PM
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 110-kN load, that the ultimate shearing
stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired,
determine the required diameter of the bolts.
SOLUTION
For each bolt,
Required:
P=
110
= 36.667 kN
3
PU = (F. S.) P = (3.35)(36.667) = 122.83 kN
τU =
d =
PU
P
4P
= π U 2 = U2
A
πd
d
4
4PU
πτU
=
(4)(122.83 × 103 )
= 20.8 × 10−3 m
6
π (360 × 10 )
d = 20.8 mm 
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52
BeerMOM_ISM_C01.indd 52
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PROBLEM 1.55
Top view
200 mm
180 mm
12 mm
In the structure shown, an 8-mm-diameter pin
is used at A, and 12-mm-diameter pins are
used at B and D. Knowing that the ultimate
shearing stress is 100 MPa at all connections
and that the ultimate normal stress is 250 MPa
in each of the two links joining B and D,
determine the allowable load P if an overall
factor of safety of 3.0 is desired.
8 mm
A
B
C
B
A
C
B
20 mm
P
8 mm
8 mm
D
D
12 mm
Front view
Side view
SOLUTION
Statics: Use ABC as free body.
M B  0 : 0.20FA  0.18P  0
M A  0 : 0.20FBD  0.38P  0
Based on double shear in pin A, A 
FA 
2 U A


4
10
FA
9
10
P
FBD
19
P
d2 

4
(0.008)2  50.266  106 m 2
(2)(100  106 )(50.266  106 )
 3.351  103 N
3.0
F .S .
10
P
FA  3.72  103 N
9
Based on double shear in pins at B and D, A 
FBD 
2 U A


4
d2 

4
(0.012) 2  113.10  106 m 2
(2)(100  106 )(113.10  106 )
 7.54  103 N
3.0
F .S.
10
P
FBD  3.97  103 N
19
Based on compression in links BD, for one link, A  (0.020)(0.008)  160  106 m 2
2 U A (2)(250  106 )(160  106 )

 26.7  103 N
F .S .
3.0
10
P
FBD  14.04  103 N
19
Allowable value of P is smallest,  P  3.72  103 N
FBD 
P  3.72 kN 
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61
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