A solid shaft is to transmit 300 kW at 120 rpm. If the shear stress is not to exceed 100 MPa, find the shaft diameter. What percent saving in weight would be obtained if this shaft was replaced by a hollow one whose internal diameter equals 0.6 of the external diameter; the length, material and

Question

A solid shaft is to transmit 300 kW at 120 rpm. If the shear stress is not to exceed 100 MPa, find the shaft diameter. What percent saving in weight would be obtained if this shaft was replaced by a hollow one whose internal diameter equals 0.6 of the external diameter; the length, material and maximum allowable shear stress being same?

Accepted Answer

We know that the power transmitted is given by

[latex] P=\frac{2 \pi N T}{60} [/latex]

Here power, P = 300 kW and N = 120 rpm. Substituting these values, we get

[latex] T=\frac{300 imes 10^3 imes 60}{2 \pi(120)}=23873.24 Nm [/latex]

For the solid shaft, we know that

[latex] au_{\max }=\frac{16 T}{\pi d^3} [/latex]

or [latex] d=\sqrt[3]{\frac{16 T}{\pi au_{\max }}} [/latex]

Now putting T = 23873.24 N m and [latex] au_{\max }=100 MPa [/latex], we get d = 106.73 mm as the shaft diameter.
For the hollow shaft, similarly, we get

[latex] au_{\max }=\frac{16 T d^2}{\pi\left(d_2^4-d_1^4 ight)} [/latex]

where [latex] d_1 [/latex] is the internal diameter and [latex] d_2 [/latex] is the external diameter. Putting T = 23873.24 Nm and [latex] d_1=0.6 d_2 [/latex], we get

[latex] au_{\max }=100=\frac{16 imes 23873.24 imes d_2}{\pi\left[d_2^4-(0.6)^4 d_2^4 ight]} [/latex]

or [latex] d_2=\sqrt[3]{\frac{16 imes 23873.24 imes 10^3}{\pi\left(1-0.6^4 ight) imes 100}}=111.79 mm [/latex]

The internal diameter [latex] d_1 [/latex] is 0.6 × 111.79 = 67.07 mm. Thus, clearly for same material and length of the two shafts, we have percent saving in weight as

[latex] \frac{A_{ ext {solid }}-A_{ ext {hollow }}}{A_{ ext {solid }}} imes 100 \% [/latex]

where A is the area of shaft. Substituting the values, we get

[latex] \frac{(106.73)^2-\left(111.79^2-67.07^2 ight)}{(106.73)^2} imes 100 \%=29.78 \% [/latex]

Thus, the percentage saving in weight is 29.78%.

Question 2.10: A solid shaft is to transmit 300 kW at 120 rpm. If the shear...

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