Question

In a simply supported shaft carrying a uniformly distributed mass, the maximum deflection at the midspan is:
1. \(\frac{{5mg{l^2}}}{{384EI}}\)
2. \(\frac{{5mg{l^4}}}{{384EI}}\)
3. \(\frac{{mg{l^4}}}{{384EI}}\)
4. \(\frac{{3mg{l^2}}}{{384EI}}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{5mg{l^4}}}{{384EI}}\)

Concept:

The maximum deflection (at centre) of a simply supported beam with uniformly distributed load (UDL) is given by

\(\delta = \frac{{5w{L^4}}}{{384EI}}\)

where w is the weight per unit length,

As we know, 

w = mg

∴ Maximum deflection is:

\(\delta = \frac{{5mg{L^4}}}{{384EI}}\)

where m is mass per unit length.

Here m is not the total mass of the beam, it is mass per unit length.