Question

The diameter of shaft A is thrice that of the diameter of shaft B. Power transmitted by shaft A when compared to shaft B will be 
1. 3 times
2. 9 times
3. 27 times
4. 81 times

Answer 1

Correct Answer - Option 3 : 27 times

Concept:

The power transmitting capacity of the shaft is given by

\(P = \frac{{2 \times \pi \times N \times T}}{{60}}\)

And from the torsion equation:

\(\frac{T}{J} = \frac{{{\tau _{max}}}}{r} = \frac{{G\theta }}{l}\)

∴ \(T = {\tau _{max}} \times \frac{J}{r}\)

\(J = \frac{\pi }{{32}} \times {D^4}\)

∴ \(T = \frac{{{\tau _{max}}\; \times \;\pi \; \times {D^3}\;}}{{16}}\)

As Power ∝ T, and T ∝ D³

So, for a given material and rpm

Power transmitted capacity ∝ (Diameter) ³

Calculation:

Given:

D_A = 3 D_B

\(\frac{{{\rm{Power\;transmitted\;by\;shaft\;A}}}}{{{\rm{Power\;transmitted\;by\;shaft\;B}}}} = {\left( {\frac{{{{\rm{d}}_{\rm{A}}}}}{{{{\rm{d}}_{\rm{B}}}}}} \right)^3} = {\left( {\frac{{3{{\rm{d}}_{\rm{B}}}}}{{{{\rm{d}}_{\rm{B}}}}}} \right)^3} = {\left( 3 \right)^3} = 27\)

Power transmitted by shaft A when compared to shaft B is 27 times.