Question

Two heavy rotating masses are connected by shafts of lengths l₁, l₂ and l₃, and the corresponding diameters are d₁, d₂ and d₃, this system is reduced to a torsionally equivalent length of the shafts is
1. \({{\rm{l}}_1} + {{\rm{l}}_2}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_2}}}} \right)^4} + {{\rm{l}}_3}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_3}}}} \right)^4}\)
2. \({{\rm{l}}_1} + {{\rm{l}}_2}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_2}}}} \right)^3} + {{\rm{l}}_3}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_3}}}} \right)^3}\)
3. \(\frac{{{{\rm{l}}_1} + {{\rm{l}}_2} + {{\rm{l}}_3}}}{3}\)
4. l₁ + l₂ + l₃

Answer 1

Correct Answer - Option 1 : \({{\rm{l}}_1} + {{\rm{l}}_2}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_2}}}} \right)^4} + {{\rm{l}}_3}{\left( {\frac{{{{\rm{d}}_1}}}{{{{\rm{d}}_3}}}} \right)^4}\)

Concept:

The shafts of lengths l₁, l_2, and l₃ are connected in series and are replaced by a shaft of the equivalent of the length of the same material.

Stiffness of shaft, k = \(\frac{{{\rm{GJ}}}}{{\rm{L}}}\)

Since shafts are in series, thus equivalent stiffness

\(\frac{1}{{{{\rm{k}}_{{\rm{eq}}}}}} = {\rm{\;}}\frac{1}{{{{\rm{k}}_1}}} + {\rm{\;}}\frac{1}{{{{\rm{k}}_2}}} + \frac{1}{{{{\rm{k}}_3}}}\)

\(\frac{{\rm{L}}}{{{\rm{GJ}}}} = {\rm{\;}}\frac{{{{\rm{l}}_1}}}{{{\rm{G}}{{\rm{J}}_1}}} + {\rm{\;}}\frac{{{{\rm{l}}_2}}}{{{\rm{G}}{{\rm{J}}_2}}} + \frac{{{{\rm{l}}_3}}}{{{\rm{G}}{{\rm{J}}_3}}}\)

\(\frac{L}{{{\rm{d}}_1^4}} = {\rm{\;}}\frac{{{{\rm{l}}_1}}}{{{\rm{d}}_1^4}} + {\rm{\;}}\frac{{{{\rm{l}}_2}}}{{{\rm{d}}_2^4}} + \frac{{{{\rm{l}}_3}}}{{{\rm{d}}_3^4}}\)

\({\rm{L}} = {\rm{\;}}\frac{{{{\rm{l}}_1}{\rm{d}}_1^4}}{{{\rm{d}}_1^4}} + {\rm{\;}}\frac{{{{\rm{l}}_2}{\rm{d}}_1^4}}{{{\rm{d}}_2^4}} + \frac{{{{\rm{l}}_3}{\rm{d}}_1^4}}{{{\rm{d}}_3^4}}\)

\({\bf{L}} = {{\bf{l}}_1} + \;{{\bf{l}}_2}{\left( {\frac{{{{\bf{d}}_1}}}{{{{\bf{d}}_2}}}} \right)^4} + {{\bf{l}}_3}{\left( {\frac{{{{\bf{d}}_1}}}{{{{\bf{d}}_3}}}} \right)^4}\)