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Standard XII
Physics
Newton's Second Law: Rotatory Version
A motor drivi...
Question
A motor driving a solid circular steel shaft transmits 40 kW of power at 500 rpm. If the diameter of the shaft is 40 mm, the maximum shear stress in the shaft is
MPa.
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Solution
T
=
P
×
60
×
10
6
2
π
N
=
40
×
60
×
10
6
2
π
(
500
)
= 763943.7268 N-mm
τ
m
a
x
=
16
T
π
d
3
=
16
×
763943.7268
π
(
40
)
3
= 60.792 MPa
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