Subject: Structure Analysis 2

Topic: Determination of Collaps Load for Beams

Difficulty: Hard

No of plastic hinge (P.H)= N = 6.

1) Let beam AB collapse:

Using virtual work principle:

EWD = IWD

$40*\gamma = M_P\theta+ M_P2\theta + M_P\theta$

$\theta=\frac{\gamma}{2} \gamma=2\theta$

$40.2\theta = 4 M_P\theta$

$M_P = 20 KN.m$

2) Let Span BC collapse:

Using virtual work principle:

EWD = IWD

$20*(\gamma) = M_P\theta + M_P2(\theta + \alpha) + M_P\alpha$

$\theta = \frac{\gamma}{1} \gamma = \theta$

$\theta = 3\alpha \alpha =\frac{\theta}{3}$

$20.\theta=M_P\theta + M_P(\theta + \frac{\theta}{3}) + M_P(\frac{\theta}{3})$

$M_P = \frac{20}{2.67}$

$M_P =7.49 KN.m$

3) Let Span CD collapse:

Using Virtual work principle, Standard Case.

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

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

$M_P = 370.62 KN.m$

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

$M_P = 370.62KN.m$