Question

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 1

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Answer 2

a)T=95.5414 N.m

b)d=21.35 mm

Step-by-step 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

`P=(2\pi NT)/(60)`

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

Now by putting the values

`P=(2\pi NT)/(60)`

`3500=(2\pi 350T)/(60)`

T=95.5414 N.m

T=95,541.4 N.mm

We also know that

`Shear\ strength=(16T)/(\pi d^3)`

d is the diameter of shaft

`Shear\ strength=(16T)/(\pi d^3)`

`50=(16* 95541.4)/(\pi d^3)`

d=21.35 mm

Torque,T=95.5414 N.m

Diameter ,d=21.35 mm