Strength of Material Fourth Edition By Andrew Pytel, Ferdinand L.Singer Chapter1-5

302 Pages·2009·32.78 MB·English

Pytel and Singer
Solution to Problems in Strength of Materials 4th Edition
Authors: Andrew Pytel and Ferdinand L. Singer
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Chapter 1 - Simple Stresses
Simple Stresses
1. Normal Stress
2. Shear Stress
3. Bearing Stress
4. Thin-walled Pressure Vessel
Normal Stresses
Stress is defined as the strength of a material per unit area of unit strength. It is the force on a member
divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa.
P
=
A
where P is the applied normal load in Newton and A is the area in mm2. The maximum stress in tension or
compression occurs over a section normal to the load.
Normal stress is either tensile stress or compressive stress. Members subject to pure tension (or tensile
force) is under tensile stress, while compression members (members subject to compressive force) are
under compressive stress.
Compressive force will tend to shorten the member. Tension force on the other hand will tend to lengthen
the member.
Solution to Problem 104 Normal Stress
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the
2
outside diameter of the tube if the stress is limited to 120 MN/m .
Solution 104
P = A
P = A
where:
P = 400kN = 400000N
= 120MPa
A = 1 D2–1 (1002)
4 4
A = 1 (D2–10000)
4
thus,
400000 = 120[ 1 (D2(cid:237)10000)]
4
400000 = 30 D2(cid:237)300000
400000 + 300000
D2 =
30
D = 119 35mm
answer
Solution to Problem 105 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Weight of bar = 800 kg
Maximum allowable stress for bronze = 90 MPa
Maximum allowable stress for steel = 120 MPa
Required: Smallest area of bronze and steel cables
Solution 105
By symmetry:
P = P = 1(7848)
br st 2
P = 3924N
br
P = 3924N
st
For bronze cable:
P = A
br br br
3924 = 90A
br
A = 43 6mm2
answer
br
For steel cable:
P = A
st st st
3924 = 120A
st
2
A = 32 7mm
answer
st
Solution to Problem 106 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Diameter of cable = 0.6 inch
Weight of bar = 6000 lb
Required: Stress in the cable
Solution 106
M = 0
C
5T + 10 3 T = 5(6000)
34
T = 2957 13lb
T = A
2957 13 = 1 (0 62)
4
= 10458 72psi
answer
Solution to Problem 107 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Axial load P = 3000 lb
2
Cross-sectional area of the rod = 0.5 in
Required: Stress in steel, aluminum, and bronze sections
Solution 107
For steel:
A = P
st st st
(0 5) = 12
st
= 24ksi
answer
st
For aluminum:
A = P
al al al
(0 5) = 12
al
= 24ksi
answer
al
For bronze:
A = P
br br br
(0 5) = 9
br
= 18ksi
answer
br
Solution to Problem 108 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Maximum allowable stress for steel = 140 MPa
Maximum allowable stress for aluminum = 90 MPa
Maximum allowable stress for bronze = 100 MPa
Required: Maximum safe value of axial load P
Solution 108
For bronze:
A = 2P
br br
100(200) = 2P
P = 10000N
For aluminum:
A = P
al al
90(400) = P
P = 36000N
For Steel:
A = 5P
st st
P = 14000N
P P = 10000N = 10kN
For safe , use answer
Solution to Problem 109 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Maximum allowable stress of the wire = 30 ksi
2
Cross-sectional area of wire AB = 0.4 in
2
Cross-sectional area of wire AC = 0.5 in
Required: Largest weight W
Solution 109
For wire AB: By sine law (from the force polygon):
T W
AB
=
sin40 sin40
T = 0 6527W
AB
A = 0 6527W
AB AB
30(0 4) = 0 6527W
W = 18 4kips
For wire AC:
T W
AC
=
sin60 sin80
T = 0 8794W
AC
T = A
AC AC AC
0 8794W = 30(0 5)
W = 17 1kips
W = 17 1kips
Safe load answer
Solution to Problem 110 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Size of steel bearing plate = 12-inches square
Size of concrete footing = 12-inches square
Size of wooden post = 8-inches diameter
Maximum allowable stress for wood = 1800 psi
Maximum allowable stress for concrete = 650 psi
Required: Maximum safe value of load P
Solution 110
For wood:
P = A
w w w
P = 1800[ 1 (82)]
w 4
P = 90477 9lb
w
From FBD of Wood:
P = P = 90477 9lb
w
For concrete:
P = A
c c c
P = 650(122)
c
P = 93600lb
c
From FBD of Concrete:
P = P = 93600lb
c
P = 90478lb
Safe load answer
Solution to Problem 111 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
2
Cross-sectional area of each member = 1.8 in
Required: Stresses in members CE, DE, and DF
Solution 111
From the FBD of the truss:
M = 0
A
24R = 16(30)
F
R = 20k
F
At joint F:
F = 0
V
3DF = 20
5
DF = 331k (Compression)
3
At joint D: (by symmetry)
BD = DF = 331k (Compression)
3
(cid:520)FV = 0
DE = 3BD + 3DF
5 5
DE = 3(331) + 3(331)
5 3 5 3
DE = 40k (Tension)
At joint E:
F = 0
V
3CE + 30 = 40
5
CE = 162k (Tension)
3
Stresses:
Stress = Force/Area
162
3
= = 9 26ksi (Tension)
answer
CE 1 8
40
= = 22 22ksi (Tension)
answer
DE 1 8
331
3
= = 18 52ksi (Compression)
answer
DF 1 8
Solution to Problem 112 Normal Stress
Strength of Materials 4th Edition by Pytel and Singer
Given:
Maximum allowable stress in tension = 20 ksi
Maximum allowable stress in compression = 14 ksi
Required: Cross-sectional areas of members AG, BC, and CE
Solution 112
F = 0
V
R = 40 + 25 = 65k
AV
= 0
AV
18R = 8(25) + 4(40)
D
R = 20k
D
F = 0
H
R = RD = 20k
AH
Check:
M = 0
D
12R = 18(R ) + 4(25) + 8(40)
AV AH
12(65) = 18(20) + 4(25) + 8(40)
780 ft kip = 780 ft kip
(OK!)
For member AG (At joint A):
F = 0
V
3 AB = 65
13
AB = 78 12k
F = 0
H
AG + 20 = 2 AB
13
AG = 20 33k Tension
AG = A
tension AG
20 33 = 20A
AG
2
A = 1 17in
answer
AG
For member BC (At section through MN):
M = 0
F
6( 2 BC) = 12(20)
13
BC = 72 11k
Compression
BC = A
compression BC
72 11 = 14A
BC
A = 5 15in2
answer
BC
For member CE (At joint D):
F = 0
H
2 CD = 20
13
CD = 36 06k
F = 0
V
DE = 3 CD = 3 (36 06) = 30k
13 13
At joint E: