ME 360 – H07
Name _________________________
1) Norton text problem 7-2.
Solution attached
2) Norton text problem 7-4 (refers to Norton text problem 3-4).. Use cup radius = -1.10 inches
for concave surface.
Solution attached
ME 360 – H07
Name _________________________
3) Norton text problem 7-10 (refers to Norton text problem 4-10). Use fulcrum length across the
full width of the diving board.
Solution attached
4) Norton text problem 7-16.
Solution attached
MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-2-1
PROBLEM 7-2
Statement:
Given:
Estimate the dry coefficient of friction between the two pieces in Problem 7-1 if their S ut = 600
MPa.
Length of block
L  5  cm
Yield strength
S y  400  MPa
Width of block
w  3  cm
Ultimate strength
S ut  600  MPa
Normal force
F  400  N
Assumptions: The compressive yield strength is the same as the tensile yield strength. Then, S yc  S y.
Solution:
See Mathcad file P0702.
Using equations 7.3 and 7.4, the coefficient of friction is estimated as
S us  0.80 S ut
μ 
S us
3  S yc
S us  480 MPa
μ  0.40
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-4-1
PROBLEM 7-4
Statement:
For the trailer hitch from Problem 3-4 on p. 169, determine the contact stresses in the ball and ball
cup. Assume that the ball is 2-in dia and the ill-fitting cup that surrounds the it is an internal
spherical surface 10% larger in diameter than the ball.
F
Given:
Solution:
Ball diameter
Cup diameter
Pull force
d  2  in
D  2.2 in
Fpull  4.905  kN
Tongue weight
Wtong  0.981  kN
Poisson's ratio
ν  0.28
Modulus of elasticity
E  30.0 10  psi
6
See Figure 7-4 and Mathcad file P0704.
Total force
2
F 
Fpull  Wtong
2
FIGURE 7-4
Diagram Showing Contact Force
for Problem 7-4
F  1125 lbf
Ball radius
R1  0.5 d
R1  1.000 in
Cup radius
R2  0.5 D
R2  1.100 in
Geometry constant
B 
1
2
 
1

 R1


R2 
1
1
B  0.045 in
2
Material constants
m1 
1ν
8 1
m1  3.072  10
E
psi
m2  m1
1
Contact patch
radius
a 
Contact area
A  π a
Average pressure
p avg 
F
p max 
3
Maximum pressure
 3 m1  m2 
F
 
B
8

3
2
2
A  0.022 in
p avg  52.1 ksi
A
2
a  0.0829 in
 p avg
p max  78.1 ksi
Stresses
Axial
σzmax  p max
In-plane
σxmax  
1  2 ν
2
σzmax  78.1 ksi
 p max
σxmax  60.9 ksi
σymax  σxmax
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
P0704.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Max shear stress
τyzmax 
p max
2
 

7-4-2
1  2 ν
2

2
9
 ( 1  ν)  2  ( 1  ν)

τyzmax  26.4 ksi
Depth at max
shear stress
zτmax  a 
2  2 ν
7  2 ν
zτmax  0.05228 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-10-1
PROBLEM 7-10
Statement:
An overhung diving board is shown in Figure P7-4a. A 100-kg person is standing on the free end.
The board sits on a fulcrum that has a cylindrical contact surface of 5-mm radius. What is the size
of the contact patch between the board and the fulcrum if the board material is fiberglass with E =
10.3 GPa and n = 0.3?
Given:
Fulcrum radius
R1  5  mm
Board curvature
R2  ∞ mm
Mass of person
Material properties:
Aluminum fulcrum
M  100  kg
2000 = L
R1
ν1  0.34
E1  71.7 GPa
Fiberglass board
R2
ν2  0.30
E2  10.3 GPa
700 = a
Board dimensions:
Width (Prob 4-10)
w  305  mm
Distance to right support a'  0.7 m
Contact length
L  2  m
Solution:
P
FIGURE 7-10
Free Body Diagram for Problem 7-10
See Figure 7-10 and Mathcad file P0710.
Weight of person
P  M  g
P  0.981 kN
Summing moments about the support on the left end of the board,
Fulcrum reaction
Geometry constant
F  P
B 
1
2
L
F  2.802 kN
a'
 
P L  F  b = 0
1
 R1

1


B  0.100 mm
R2 
1
2
Material constants
m1 
1  ν1
5
m1  1.233  10
E1
2
m2 
1  ν2
5
m2  8.835  10
E2
1
MPa
1
MPa
1
Contact patch
half-width
Contact patch
width
 2 m1  m2 F 
a   
 
B
w
π
a2  2  a
2
a  0.0767 mm
a2  0.153 mm
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P0710.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-16-1
PROBLEM 7-16
Statement:
Determine the size of the contact patch and the maximum contact stresses for a 20-mm-dia steel
ball rolled against a flat aluminum plate with 1 kN.
Given:
Ball radius
R1  10 mm
Plate curvature
R2  ∞ mm
Load
F  1  kN
Material properties
Steel ball
ν1  0.28
E1  206.8  GPa
ν2  0.34
Aluminum plate
E2  71.7 GPa
Solution:
See Mathcad file P0716.
1. Calculate geometry and material constants, contact patch dimension, and pressures.
Geometry constant
B 
1
2
 
1
 R1



R2 
1
B  0.05 mm
1
2
Material constants
m1 
1  ν1
6
m1  4.456  10
E1
2
m2 
1  ν2
1
MPa
5
m2  1.233  10
E2
1
MPa
1
Contact patch
radius
 3 m1  m2 
a   
F
B
8

Contact area
A  π a
Average pressure
p avg 
F
p max 
3
Maximum pressure
3
2
A  0.789 mm
2
p avg  1267 MPa
A
2
a  0.501 mm
 p avg
p max  1900 MPa
2. Determine the stresses in the ball at the surface
Axial
σzmax  p max
In-plane
σxmax1  
σzmax  1900 MPa
1  2  ν1
2
 p max
σxmax1  1482 MPa
σymax1  σxmax1
3. Determine the stresses in the ball below the surface
Max shear
stress
τyzmax1 
 1  2 ν1 2

   1  ν1  2   1  ν1
9
2
2 

p max

τyzmax1  641.5 MPa
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
P0716.xmcd
mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
Depth at max
shear stress
zτmax1  a 
2  2  ν1
7  2  ν1
7-16-2
zτmax1  0.316 mm
4. Determine the stresses in the plate at the surface
Axial
σzmax  p max
In-plane
σxmax2  
σzmax  1900 MPa
1  2  ν2
2
 p max
σxmax2  1596 MPa
σymax2  σxmax2
5. Determine the stresses in the plate below the surface
Max shear
stress
τyzmax2 
 1  2 ν2 2

   1  ν2  2   1  ν2
9
2
2 

p max

τyzmax2  615.2 MPa
Depth at max
shear stress
zτmax2  a 
2  2  ν2
7  2  ν2
zτmax2  0.326 mm
© 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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P0716.xmcd
mechanical,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.