Question

A fixed ended beam is subjected to a load W at $\frac{1}{3}rd$ span as shown in figure. The collapse load is:

• A. $\frac{11.66M_P}{L}$
• B. $\frac{15M_P}{L}$
• C. $\frac{16.5M_P}{L}$
• D. $\frac{18M_P}{L}$

Answer 1

The correct option is B $\frac{15M_P}{L}$

No. of independent mechanisms = Possible number of plastic hinges - Degree of redundancy
= 4 -2 = 2
$\because$ Degree of redundancy = 2
$\therefore$ Number of plastic hinges required for collapse = 2 + 1 = 3
Case - I : Two plastic hinge at supports, and one plastic hinge at the point where cross - section changes. The plastic hinge will form at B in the limb BC and its value will be $M_P$

External work done = Internal work done
$\Rightarrow W\times \frac{L}{3}\theta =2M_P\theta +M_P(\theta +\theta )+M_P\theta$
$\Rightarrow W\times \frac{L}{3}\theta =5M_P\theta$
$\Rightarrow W=15\frac{M_P}{L}$
Case - II : Two plastic hinge at the supports and one below the concentrated load.

$\Delta =\frac{L}{3}{\theta }_1=\frac{2}{3}L\theta \Rightarrow {\theta }_1=2\theta$
External work done = Internal work done
$\Rightarrow W\times \frac{L}{3}{\theta }_1=2M_P{\theta }_1+2M_P(\theta +{\theta }_1)+M_P\theta$
$\Rightarrow \frac{2}{3}WL\theta =2M_P\times \left(2\theta \right)+2M_P(\theta +2\theta )+M_P\theta$
$\Rightarrow \frac{2}{3}WL\theta =11M_P\theta$
$\Rightarrow W=16.5\frac{M_P}{L}$
So, collapse load = $15\frac{M_P}{L}$