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

$240*\gamma = M_PO+ M_P2O + M_PO$

$240*4O = M_PO + M_P2O + M_PO$

$O = \frac{\gamma}{4} \gamma=4O$ (Since O is very small)

$M_P = 240 KN.m$

2) Let Span BC collapse:

Using virtual work principle:

EWD = IWD

$Σ(W.\gamma) = Σ(M_P.O)$

$60\gamma = M_PO + M_P(O+ \alpha) + M_P(\alpha)$

${\alpha = \frac{O}{2}}$

$60.2O = M_PO + M_P(O + \frac{O}{2}) + M_P(\frac{O}{2})$

$60.2O = M_PO[1+1+\frac{1}{2}+\frac{1}{2}]$

$M_P = \frac{120}{3} = 40$

$M_P = 40 KN.m$

3) Let Span CD collapse:

Using Virtual work principle, Standard case.

$M_P = \frac{WL^2}{11.656}$ W=Collapse load

$M_P = \frac{120*6^2}{11.656}$

$M_P = 370.62 KN.m$

From 1,2 &3,

Failure occurs in Span where M_P is highest i.e.Span CD

$M_P = 370.62 KN.m$