Question

A solid shaft
120mm dia.
is required to transmit 200 kw at
100 rpm if the angle
angle of
twist not to be exceed 2°
find the length of the shaft ?
​

Answer 1

Answer:

The length of the shaft L = 5.57 meters (approx).

Explanation:

Given data:

Diameter (d) = 120 mm

Power (P) = 200 kW

Rotational speed (N) = 100 rpm

Angle of twist (ϕ) = 2°

Shear stress (τ) = ?

Modulus of rigidity (G) = 80 GPa = 80 x 10³ N/mm²

To Find :

The length of the shaft.

Solution:

To calculate the length of the shaft, we need to use the formula for torsional stress and then solve for the length.

Formula to calculate shear stress is:

$\tau = \frac{ 16T}{ \pi d^3}$

where,

$\tau = \frac{(P \times 60) }{ (2\pi \times N)}$ [Torque formula]

$\tau = \frac{ (200 \times 60) }{(2\pi \times 100)}$

= 19.0982 kNm

Substituting the values, we get

$\tau = \frac{(1619.098210^3)}{(\pi \times (120)^{3})}$

= 5.079 N/mm²

Now, we can use the formula for the angle of twist to find the length of the shaft.

$\phi = \frac{ (TL)}{(GJ)}$

where,

L = length of the shaft

$J = \frac{( \pi d^4)}{32}$

Substituting the values, we get

$2\degree = \frac{((19.0982 \times 10^3) \times L) }{ (80 \times 10^3 \times ( \frac{\pi}{32}) \times (120)^4)}$

$\frac{2\degree \times (80 \times 10^3 \times ( \frac{\pi}{32}) \times (120)^4)}{ (19.0982 \times 10^3) }= L$

On solving this equation, we get the length of the shaft L = 5.57 meters (approx).

Therefore, the length of the solid shaft required to transmit 200 kW at 100 rpm with an angle of twist not exceeding 2° is approximately 5.57 meters.

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