MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-1a-1
PROBLEM 4-1a
Statement:
A differential stress element has a set of applied stresses on it as indicated in each row of Table
P4-1. For row a, draw the stress element showing the applied stresses, find the principal stresses
and maximum shear stress using Mohr's circle diagram, and draw the rotated stress element
showing the principal stresses.
Given:
σx := 1000
σy := 0
σz := 0
τxy := 500
τyz := 0
τzx := 0
Solution:
See Figure 4-1a and Mathcad file P0401a.
500
y
1. Draw the stress element, indicating the x and y axes.
1000
x
2. Draw the Mohr's circle axes, indicating the τ and σ
axes with CW up and CCW down.
3. Plot the positive x-face point, which is (+1000, -500),
and label it with an "x."
FIGURE 4-1aA
4. Plot the positive y-face point, which is (0, +500), and
label it with a "y."
Stress Element for Problem 4-1a
5. Draw a straight line from point x to point y. Using the point where this line intersects the σ-axis as the center of
the Mohr circle, draw a circle that goes through points x and y.
6. The center of the circle will be at
σc :=
σx + σy
σc = 500
2
7. The circle will intersect the σ-axis at two of the principal stresses. In this case, we see that one is positive and
the other is negative so they will be σ1 and σ3. The third principal stress is σ2 = 0.
2
8. Calculate the radius of the circle
 σx − σy 
2

 + τxy
 2 
R :=
R = 707.1
τ CW
τ CW
τ1-3
τ 1-2
500
500
y
-500
500
σ3
1000
0
1500
σ1
τ2-3
-500
σ
σ3
500
σ2
0
1000
1500
σ1
σ
2φ
500
x
τ CCW
500
τ CCW
FIGURE 4-1aB
2D and 3D Mohr's Circle Diagrams for Problem 4-1a
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P0401a.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
9. Calculate the principal stresses
4-1a-2
σ1 := σc + R
σ1 = 1207
σ3 := σc − R
σ3 = −207
σ2 := 0
10. Draw the three Mohr's circles to represent the complete 3D stress state.
y
11. Calculate the principal shear stresses
τ12 := 0.5⋅ ( σ1 − σ2)
τ12 = 603.6
τ23 := 0.5⋅ ( σ2 − σ3)
τ23 = 103.6
τ13 := 0.5⋅ ( σ1 − σ3)
τ13 = 707.1
207
1207
22.5°
x
The maximum principal stress is always τ13.
12. Determine the orientation of the principal normal
stress (σ1) with respect to the x-axis. From the 2D
Mohr's circle diagram, we see that the angle 2φ from x
to σ1 is CCW and is given by
 σx − σc 
ϕ := ⋅ acos

2
 R 
1
ϕ = 22.5 deg
FIGURE 4-1aC
Rotated Stress Element for Problem 4-1a
13. Draw the rotated 2D stress element showing the two nonzero principal stresses.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-3-1
PROBLEM 4-3
Statement:
For the bicycle pedal-arm assembly in Figure 4-1 with rider-applied force of 1500 N at the pedal,
determine the maximum principal stress in the pedal arm if its cross-section is 15 mm in dia. The
pedal attaches to the pedal arm with a 12-mm screw thread. What is the stress in the pedal screw?
Given:
Distances (see figure)
Rider-applied force
a  170  mm
Frider  1.5 kN
Pedal arm diameter
d pa  15 mm
Screw thread diameter
d sc  12 mm
b  60 mm
z
Solution:
See Figure 4-3 and Mathcad file P0403.
a
1. From the FBD in Figure 4-3A (and on
the solution for Problem 3-3), we see that
the force from the rider is reacted in the
pedal arm internally by a moment, a
torque, and a vertical shear force. There
are two points at section C (Figure 4-3B)
that we should investigate, one at z = 0.5
d pa (point A), and one at y = 0.5 d pa (point
B).
Frider a  Mc = 0
 M x:
Frider b  Tc = 0
Mc
b
Arm
y
Fc
Pedal
x
2. Refering to the FBD resulting from
taking a section through the arm at C, the
maximum bending moment Mc is found by
summing moments about the y-axis, and
the maximum torque Tc is found by
summing moments about the x-axis.
 M y:
Tc
C
Frider
FIGURE 4-3A
Free Body Diagram for Problem 4-3
z
Section C
A
Maximum bending moment:
Mc  Frider a
Mc  255  N  m
Maximum torque:
Tc  Frider b
Fc  Frider
3.
x
Tc  90 N  m
Vertical shear:
B
Arm
y
FIGURE 4-3B
Fc  1.500  kN
Points A and B at Section C
Determine the stress components at point A where we have the effects of bending and torsion, but where the
transverse shear due to bending is zero because A is at the outer fiber. Looking down the z-axis at a stress
element on the surface at A,
Distance to neutral axis
cpa  0.5 d pa
cpa  7.5 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4
π d pa
Moment of inertia of
pedal-arm
Ipa 
Bending stress
(x-direction)
σx 
Stress in y-direction
σy  0  MPa
Torsional stress
due to Tc
τxy 
Principal stresses at A,
equation (4.6a)
Mc cpa
Tc cpa
τxy  135.8  MPa
2  Ipa
σx  σy
2
σx  σy
2
CW
2
 σx  σy 
2
 
  τxy
2


2
 σx  σy 
2
 
  τxy
 2 
σ2A  0  MPa
σ3A  23 MPa
Determine the stress components at point B where we have the effects of transverse shear and torsion, but
where the bending stress is zero because B is on the neutral plane. Looking down the y-axis at a stress
element at B,
2
π d pa
Cross-section area
of pedal-arm
Apa 
Torsional stress
due to Tc and shear
stress due to Fc
τzx 
Normal stresses
σx  0  MPa
Principal stresses at B
σ1B  124  MPa
Apa  176.7  mm
4
4 Fc

 τxy
3 Apa
σ1B 
σ3B 
5.
4
σx  769.6  MPa
Ipa
σ1A 
σ1A  793  MPa
3
Ipa  2.485  10  mm
64
σ3A 
4.
4-3-2
τzx  124.5  MPa
CW
σz  0  MPa
σx  σz
2
σx  σz
2
2
 σx  σz 
2
 
  τzx
2


2
 σx  σz 
2
 
  τzx
 2 
σ2B  0  MPa
The maximum principal stress is at point A and is
2
σ3B  124  MPa
σ1A  793  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
4-3-3
Determine the stress in the pedal screw.
Bending moment
Msc  Frider b
Msc  90 N  m
Distance to neutral axis
csc  0.5 d sc
csc  6  mm
Moment of inertia of
pedal screw
Isc 
Bending stress
(y-direction)
σy 
Stress in z-direction
σz  0  MPa
Torsional stress
τxy  0  MPa
4
π d sc
3
Isc  1.018  10  mm
64
Msc csc
Isc
4
σy  530.5  MPa
Since there is no shear stress present at the top of the screw where the bending stress is a maximum, the
maximum principal stress in the pedal screw is
σ1  σy
σ1  530.5  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-33a-1
PROBLEM 4-33a
Statement:
For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the
bending stress at point A and the shear stress due to transverse loading at point B. Also the
torsional shear stress at both points. Then determine the principal stresses at points A and B.
Given:
Tube length
L  100  mm
F
y
Arm length
a  400  mm
Arm thickness
t  10 mm
Arm depth
h  20 mm
Applied force
F  50 N
Tube OD
OD  20 mm
A
B
T
T
x
M
L
R
Tube ID
ID  14 mm
Modulus of elasticity
E  207  GPa
FIGURE 4-33
Free Body Diagram of Tube for Problem 4-33
Solution:
See Figure 4-33 and Mathcad file P0433a.
1. Determine the bending stress at point A. From the FBD of the tube in Figure 4-33 we see that
Reaction force
R  F
R  50.0 N
Reaction moment
M  F  L
M  5.00 N  m
Distance from NA
to outside of tube
ct  0.5 OD
ct  10.0 mm
Moment of inertia
It 
Bending stress
at point A
σxA 

64
π
4
 OD  ID
4

M  ct
It  5968 mm
4
σxA  8.38 MPa
It
2. Determine the shear stress due to transverse loading at B.
Cross-section area
A 
π
4

2
 OD  ID
Maximum shear
V  R
Maximum shear stress
(Equation 4.15d)
τVmax  2 
2

V
A  160.2  mm
2
τVmax  0.624  MPa
A
3. Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above
Torque on tube
T  F  a
Polar moment of
inertia
J 
Maximum torsional
stress at surface
τTmax 
T  20.0 N  m

32
π
4
 OD  ID
T  ct
J
4

J  11936  mm
4
τTmax  16.76  MPa
4. Determine the principal stress at point A.
Stress components
σxA  8.378  MPa
σzA  0  MPa
τxz  τTmax
τxz  16.76  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-33a-2
Principal stresses
σ1 
σxA  σzA
2
2
 σxA  σzA 
2
 
  τxz
2


σ1  21.46  MPa
2
 σxA  σzA 
2
 
  τxz
2


σ3  13.08  MPa
σ2  0  MPa
σ3 
τ13 
σxA  σzA
2
σ1  σ3
τ13  17.27  MPa
2
5. Determine the principal stress at point B.
Stress components
σxB  0  MPa
σyB  0  MPa
τxy  τTmax  τVmax
τxy  16.13  MPa
Principal stresses
σ1 
σxB  σyB
2
2
 σxB  σyB 
2
 
  τxy
2


σ1  16.13  MPa
2
 σxB  σyB 
2
 
  τxy
2


σ3  16.13  MPa
σ2  0  MPa
σ3 
τ13 
σxB  σyB
2
σ1  σ3
2
τ13  16.13  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-46-1
PROBLEM 4-46
Statement:
The V-links of Figure P4-17 are rotated by the crank arm through a shaft that is 60 mm dia by
3.23 m long. Determine the maximum torque applied to this shaft during motion of the
V-linkage and find the maximum stress and deflection for the shaft. See Problem 4-43 for more
information.
Given:
Paper roll dimensions
OD  900  mm
ID  220  mm
Shaft dims
d  60 mm
Lshaft  3230 mm
Lroll  3230 mm
3
ρ  984  kg m
Roll density
G  79 GPa
Modulus of rigidity
Assumptions: The greatest torque will occur when the link is horizontal and the paper roll is located as shown
in Figure P4-17 or Figure 4-46.
Solution:
See Figure 4-46 and Mathcad file P0446.
y
1. Determine the weight of the roll on the V-arms.

4
π
W 
2
2

 OD  ID  Lroll  ρ  g
W  18.64  kN
2. Summing moments about the shaft center,
T 
OD
2
W
T  8.390  kN  m
3. Calculate the polar moment of inertia.
J 
π d
4
W
6
J  1.272  10  mm
32
4
T
Ry
4. The maximum torsional stress will be at the outside
diameter of the shaft. The radius of the OD is,
r 
d
r  30 mm
2
60-mm-dia shaft
450.0
FIGURE 4-46
Free Body Diagram used in Problem 4-46
5. Determine the maximum torsional stress using equation (4.23b).
τmax 
Tr
J
τmax  197.8  MPa
6. Use equation (4.24) to determine the angular shaft deflection.
θ 
T  Lshaft
J G
θ  15.447 deg
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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Pearson Education, Inc., Upper Saddle River, NJ 07458.