Respuesta :
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:
[tex]P=T\omega[/tex]
[tex]P=T(\frac{2\pi N}{60})[/tex]
[tex]100\times 1000=T(\frac{2\pi 160}{60})[/tex]
T=5968.31 N-m
Step3
Maximum torque is calculated as follows:
[tex]T_{max}=(1+\frac{20}{100})T[/tex]
[tex]T_{max}=1.2T[/tex]
[tex]T_{max}=1.2\times 5968.31[/tex]
T_{max}=7161.97 N-m
Step4
Apply torsional equation for diameter of shaft as follows:
[tex]\tau _{max}=\frac{T_{max}}{\frac{\pi d^{3}}{16}}[/tex]
[tex]70\times 10^{6}=\frac{7161.97}{\frac{\pi d^{3}}{16}}[/tex]
[tex]d^{3}=5.211\times 10^{-4}[/tex]
d=0.0805 m
or,
d=80.5 mm
Thus, the diameter of the shaft is 80.5 mm.
The suitable diameter of the shaft will be "80.5 mm".
Torque:
According to the question,
- Power = 100 kw
- Revolution = 160 RPM
- Stress = 70 Mpa
Mean torque be:
→ [tex]P = T \omega[/tex]
[tex]= T\times \frac{2 \pi}{60}[/tex]
[tex]100\times 1000 = T(\frac{2 \pi 160}{60} )[/tex]
[tex]T = 5968.31 \ N-m[/tex]
hence,
The maximum torque:
→ [tex]T_{max} = (1+\frac{20}{100} )T[/tex]
[tex]= 1.2 \ T[/tex]
[tex]= 1.2\times 5968.31[/tex]
[tex]= 7161.97 \ N-m[/tex]
By using torsional equation,
→ [tex]T_{max} = \frac{T_{max}}{\frac{\pi d^3}{16} }[/tex]
[tex]70\times 10^6 = \frac{7161.97}{\frac{\pi d^3}{16} }[/tex]
Diameter be:
→ [tex]d^3 = 5.211\times 10^{-4}[/tex]
[tex]d = 0.0805 \ m \ or,\ 80.5 \ mm[/tex]
Thus the answer above is correct.
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