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ME 452: Machine Design II
Solution to Homework Set 11
Problem 1.
Calculate the axial load that will stress an SAE 3/8” UNF Grade 5 bolt to its proof strength assuming:
a.
Shear failure of the nut threads when one thread takes the entire load.
b.
Shear failure of the nut threads when all the threads equally share the load.
c.
Shear failure of the bolt threads when one thread takes the entire load.
d.
Shear failure of the bolt threads when all of the threads equally share the load.
e.
Tensile failure of the bolt on area A
t
.
Solution
The given bolt is specified to be an SAE 3/8” UNF grade 5 bolt.
From Table 89, see page 418, for SAE grade number 5 and bolt diameter range 0.25  1.0 in:
The minimum proof strength = 85 kpsi, the minimum yield strength = 92 kpsi, and the minimum tensile
strength = 120 kpsi.
(a)
If one thread of the nut takes all the load, the shear area can be written as
s
o
Ad
w
p
π
=
The constant
o
w
is an area reduction factor that defines the percentage of the pitch occupied by metal at
the major diameter.
For UNF threads, the value of
o
w
can be taken to be 0.88 (see class notes).
From Table (82), page 399, for 3/8” UNF threads, the number of threads per inch is given as
24 threads/in
N
=
Therefore, the pitch of the thread, see page 396, can be written as
1
p
N
=
Therefore, the pitch of the thread is given as
1
0.042 in/thread
24
p
==
Therefore, the shear area for the nut when one thread takes all the load is given as
2
3
0.88 0.042
0.0435 in
8
s
A
=××
×
=
The axial load required for shearing can be written as
max
s
FA
τ
=
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where
max
τ
is the shear strength of the material.
The shear strength of the material can be approximated
using the vonMises criterian to be 57.7% of its proof strength.
Therefore, the shear strength of the
material can be estimated as
max
0.577
p
S
=
Therefore, the shear strength of the material is given as
max
0.577 85
49.05 kpsi
=
×=
Therefore, the axial load required for shearing if one thread of the nut takes all the load is given as
3
49.05 10
0.0435
2134 lb
F
=×
×
=
(b)
When all the nut threads equally share the load, the axial load required for shear failure can be
written as
= the axial load for one thread
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 Staff
 Machine Design, Shear, Stress

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