Question

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

Answer 1

No. of independent mechanisms
$\text{= Possible number of plastic hinges - Degree of redundancy}$
$=4-2=2$

$\because \text{Degree of redundancy}=2$
$\therefore \text{Number of plastic hinges required for collapse}=2+1=3$

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 $M_P$


$\text{External work done = Internal work done}$

$\Rightarrow W\times \frac{L}{3}\theta =2M_P\theta +M_P(\theta +\theta )+M_P\theta$

$\Rightarrow W\times \frac{L}{3}\theta =5M_P\theta$

$\Rightarrow W=15\frac{M_P}{L}$

Case - II :
Two plastic hinge at the supports and one below the concentrated load.


$(\because \frac{L}{3}{\theta }_1=\frac{2}{3}L\theta \Rightarrow {\theta }_1=2\theta )$

$\text{External work done = Internal work done}$

$\Rightarrow W\times \frac{L}{3}{\theta }_1=2M_P{\theta }_1+2M_P(\theta +{\theta }_1)+M_P\theta$

$\Rightarrow \frac{2}{3}WL\theta =2M_P\times \left(2\theta \right)+2M_P(\theta +2\theta )+M_P\theta$

$\Rightarrow \frac{2}{3}WL\theta =11M_P\theta$

$\Rightarrow W=16.5\frac{M_P}{L}$

$\text{So, collapse load =}15\frac{M_P}{L}$