A shaft is transmitting 100 kW at 160 rpm. Find suitable diameter for the shaft if maximum
torque transmitted exceeds the mean by 25%.Given maximum shear stress 70 MPa.
Answers
Answer:
The diameter of the shaft is 80.5 mm.
Explanation:
Torsion equation is applied for the diameter of the solid shaft.
Step1
Given:
Power of the shaft is 100 kw.
Revolution per minute is 160 RPM.
Allowable shear stress is 70 Mpa.
Maximum torque is 20% more than the mean torque.
Step2
Mean torque is calculated as follows:
P=T\omegaP=Tω
P=T(\frac{2\pi N}{60})P=T(
60
2πN
)
100\times 1000=T(\frac{2\pi 160}{60})100×1000=T(
60
2π160
)
T=5968.31 N-m
Step3
Maximum torque is calculated as follows:
T_{max}=(1+\frac{20}{100})TT
max
=(1+
100
20
)T
T_{max}=1.2TT
max
=1.2T
T_{max}=1.2\times 5968.31T
max
=1.2×5968.31
T_{max}=7161.97 N-m
Step4
Apply torsional equation for diameter of shaft as follows:
\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}τ
max
=
16
πd
3
T
max
70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}70×10
6
=
16
πd
3
7161.97
d^{3}=5.211\times 10^{-4}d
3
=5.211×10
−4
d=0.0805 m
or,
d=80.5 mm
Thus, the diameter of the shaft is 80.5 mm.
Explanation:
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