Question

A propped cantilever of uniform 'Mp' is loaded as shown in figure below. Collapse load for the beam will be _______.

Answer 1

No.of reactions 'r' = 3
No of equilibrium equation 'S' =2
Static indeterminacy
$D_s=D_{se}+D_{si}$
= 3 - 2 = 1

No. of Possible hinges 'N' = 3
[at A,B, & at distance L/2 from end A]

No. of plastic hinges required to form a mechanism
$n=D_s+1=2$

No. of independent mechanism
$I=N-D_s$
$=3-2=1$

[Two independent beam mechanism]

1st beam mechanism:


External work done
$W_e=Load\times \text{Displacement under the load}$

$W_e=+W_c{\delta }_1-\frac{W_c}{2}\frac{L}{4}\theta =W_{c}L\theta \left[\begin{array}{c}\frac{1}{2}-\frac{1}{8}\end{array}\right]=\frac{3W_{c}L\theta }{8}$

Internal workdone
$w_i=Moment\times Rotation=3M_P\theta$

External work done = Internal work done

$\frac{3W_{c}L\theta }{8}=3M_p\theta$

$W_c=\frac{8M_p}{L}$

2nd beam mechanism:


$w_e=\frac{W_c}{2}\times \frac{L\theta }{4}=\frac{W_{c}L\theta }{8}$

$W_i=M_P\theta$

$W_e=W_i$

$\frac{W_{c}L\theta }{8}=M_P\theta$

$W_c=\frac{8M_p}{L}$

True collapse load is
$\frac{8M_P}{L}\left[Least\right]$