assignment #3 solution - spring 2014
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8/11/2019 Assignment #3 Solution - Spring 2014
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MAE 311: Machines and Mechanisms I Spring 2014
Assignment #3 Due Date: 3/7/2014
Copyright 2014 – Phillip M. Cormier and Jobaidur Khan
Problem 1 (25 Points)
The steel eyebolt shown in Figure 1 is loaded with a force of 75 pounds. The bolt is formed from round
stock with a diameter of 0.25 inches, and has in inner bend radius of 0.5 inches in the eye and at the end
of the shank. Estimate the stress at the inner and outer surfaces of section B, where = 30°.
Figure 1: Problem 1 Diagram
Sol. d = 0.25 in, r o = 0.75 on, r
i = 0.5 in,
From table 3-4, for circular cross-section, we get, R = 0.125 in, r c = 0.5 + 0.125 = 0.625 in
in618686.0
125.0625.0625.02
125.0
r r r 2
r r
22
2
22
cc
2
n
in006314.0618686.0625.0r r e nc
in118686.05.0618686.0r r c ini
in131314.0618686.075.0r r c ioo
Cross-section area,2
2
in049087.04
dA
The angle of the line of radius centers is,
30
2
1sin
R dR
2dR sin 11
Moment, in.lbf 44.2330Sin225.05.075Sin2dR FM
kpsi7.18 psi716,185.0006314.0049087.0
118686.044.23
049087.0
30Sin75
Aer
Mc
A
FSin
i
ii
psik 5.12 psi478,1275.0006314.0049087.0
131314.044.23
049087.0
30Sin75
Aer
Mc
A
FSin
o
oo
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MAE 311: Machines and Mechanisms I Spring 2014
Assignment #3 Due Date: 3/7/2014
Copyright 2014 – Phillip M. Cormier and Jobaidur Khan
Problem 2 (25 Points)
The latch spring shown in Figure 2 supports a load of 3 pounds. Given the noted dimensions, determine
the following:
a) Determine the stresses at the top and bottom of the beam, just before the bend. b) Using curved beam theory, determine the inner and outer stresses at the bend.
c) Compare the stresses between the curved portion and bent portion to determine an effectivestress concentration factor.
Figure 2: Problem 2 Diagram
Sol. (a)
kpsi02.8 psi8021
12in1094.0in75.0
in1094.05.0in4lb3
I
Mc3
(b) From table 3-4, for rectangular cross-section,
we get, r i = 0.125 in, r o = r i + h = 0.125 + 0.1094 = 0.2344 in
in174006.0
125.02344.0ln
1094.0
r r ln
hr
io
n
in005694.0174006.01797.0r r e nc
in049006.0125.0174006.0r r c ini
in060394.0174006.02344.0r r c ioo
Cross-section area,2in08205.01094.075.0 bhA
M = – 3 lb
4 in = – 12 lb.in
The negative sign is due to convention, now the stresses,
kpsi1.10 psi070,10125.0005694.008205.0
049006.012
Aer
Mc
i
ii
kpsi62.6 psi618,62344.0005694.008205.0
060394.012
Aer
Mc
o
oo
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MAE 311: Machines and Mechanisms I Spring 2014
Assignment #3 Due Date: 3/7/2014
Copyright 2014 – Phillip M. Cormier and Jobaidur Khan
(c) Stress concentrations,
26.102.8
1.10K i
i
825.0
02.8
62.6K o
o
Problem 3 (50 Points Total)
Compute the factor of safety for points A and B using both the distortion energy theory and max shear
stress theory. The bar is made from AISI 1006 cold-drawn steel, and is loaded with force F = 550 N, P
= 4000 N, and T = 25 N-m. Determine the required diameter change to bring the most restrictive caseup to a factor of safety of 3.
Figure 3: Problem 3 Diagram
Sol. Using Table A-20, yp = 280 MPaLet’s analyze the point B,
x from bending
xz
y
T
x from P
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MAE 311: Machines and Mechanisms I Spring 2014
Assignment #3 Due Date: 3/7/2014
Copyright 2014 – Phillip M. Cormier and Jobaidur Khan
For point B the bending stress from F will cause a normal stress in the x direction,
MPa Pam
m N
d
M
I
c M F F bend x 16610166
015.0
1.05503232 6
33
For point B the axial stress (tension) from P will cause a normal stress in the x direction,
MPa Pa
m
N
d
P
A
P axial 6.22106.22
015.0
400044 6
22
So, total axial stress at B, x = x,bend + axial = 188.6 MPa
Shear stress for torsion,
MPa7.37m015.0
m. N2516
d
T16
J
Tr 33T
So, total shear stress, MPa7.37Txy
So, according to DE, 4.17.3736.188280
32222 n
nn xy x
yp
So, according to MSST, 38.17.3746.188280
42222 n
nn xy x
yp
Let’s analyze the point A,
For point A the axial stress (tension) from P will cause a normal stress in the x direction,
MPa Pa
m
N
d
P
A
P axial 6.22106.22
015.0
400044 6
22
x from P
xz
y
T
F
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MAE 311: Machines and Mechanisms I Spring 2014
Assignment #3 Due Date: 3/7/2014
Copyright 2014 – Phillip M. Cormier and Jobaidur Khan
Shear stress for torsion,
MPa7.37m015.0
m. N2516
d
T16
J
Tr 33T
Shear stress from F,
MPam
N
d
F
A
F F 15.4
015.03
55016
3
16
3
422
So, total shear stress, MPa F T xz 55.3315.47.37
So, according to DE, 27.455.3536.22280
32222 n
nn xz x
yp
So, according to MSST, 75.355.3546.22280
42222 n
nn xz x
yp
This is obvious from the previous calculation that MSST in point B gives most restrictive case. So, the
design calculation becomes,
333
FF bendx
d
2.560
d
m1.0 N55032
d
M32
I
cM
For point B the axial stress (tension) from P will cause a normal stress in the x direction,
222
5093400044
d d
N
d
P
A
P axial
So, total axial stress at B, 23 50932.560d d
axial bend x x
Shear stress for torsion,333T
d
3.127
d
m. N2516
d
T16
J
Tr
So, total shear stress,3Txy
d
3.127
So, according to MSST,
mmmd
d d
d d d n xy x
yp
2002.0
03.127450932.560910280
3.1274
50932.560
3
10280 4
22236
2
3
2
23
622
The diameter needs to be increased by 33%.
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