Subject: Structure Analysis 2

Topic: Determination of Collaps Load for Beams

Difficulty: Hard

No of plastic hinge = N = 6.

1) Let beam AB collapse:

Using virtual work principle:

EWD = IWD

$30*\gamma = M_PO+ M_P(O + \alpha) + M_P\alpha$

$30*2O = M_PO + M_P(O + \frac{O}{2}) + M_P(\frac{O}{2})$

$60O = M_PO(1+1+\frac{1}{2} + \frac{1}{2})$

$M_P = 20 KN.m$

2) Let Span BC collapse:

Using virtual work principle:

EWD = IWD

$20*(\frac{1}{2}*3*\gamma) = M_PO + M_P2O + M_PO$

$20*(0.5*3*\gamma) = 4M_PO$

$O=\frac{\gamma}{1.5}$

$\gamma = 1.5O$

$20*(0.5*3*1.5O) = 4M_PO$

$M_P = \frac{45}{4} = 11.25$

$M_P = 11.25 KN.m$

3) Let Span CD collapse:

Using Virtual work principle.

$M_P = \frac{WL^2}{11.656}$

$M_P = \frac{10*4^2}{11.656}$

$M_P = 13.72 KN.m$

Failure occurs in Span where $M_P$ is highest i.e.Span AB

$M_P = 20 KN.m$