Determine the diameter of a solid shaft, which will transmit 300 kW at 250 rpm and the working conditions to be satisfied are:

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written 7.1 years ago by

tarashankarsharma • 550

modified 6.1 years ago by

sanketshingote • 100

The twist should not be more than 1^0 in a shaft of length 2m and The maximum shear stress should not exceed 30 N/mm^2 Take modulus of rigidity =1x10^5 N/mm^2

strength of materials

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written 7.1 years ago by

tarashankarsharma • 550

part 1 of 2 part 2 of 2

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written 6.1 years ago by

teamques10 ★ 70k

• modified 6.1 years ago

Data: (P) Power = 300 KW = 300 \times 10^3 W, (d) Diameter = ?

N = 250 rpm.

(L) Length = $2 \ m = 2 \ \times 10^3 \ mm$

$(\theta)$ Angle of twist = 1° = $ 1 \times \frac{\pi}{180} \ (radians)$

$(\zeta)$ maximum shear stress = …

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