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A shaft is to transmit 3.5 kW power while rotating at 350 rpm. If the shaft is made of plain carbon steel with 100 MPa yield strength, calculate: a) Torque being transmitted by the shaft b) Diameter of the shaft

Answer :

Answer:

a)T=95.5414 N.m

b)d=21.35 mm

Explanation:

Given that

P=3.5 KW

Speed N=350 RPM

Yield strength= 100 MPa

So Shear strength = 0.5 x 100 =50 MPa

We know that

[tex]P=\dfrac{2\pi NT}{60}[/tex]

Where N is the speed ,T is the torque and P is the power.

Now by putting the values

[tex]P=\dfrac{2\pi NT}{60}[/tex]

[tex]3500=\dfrac{2\pi 350T}{60}[/tex]

T=95.5414 N.m

T=95,541.4 N.mm

We also know that

[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]

d is the diameter of shaft

[tex]Shear\ strength=\dfrac{16T}{\pi d^3}[/tex]

[tex]50=\dfrac{16\times 95541.4}{\pi d^3}[/tex]

d=21.35 mm

Torque,T=95.5414 N.m

Diameter ,d=21.35 mm

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