Subject: Structure Analysis 2

Topic: Determination of Collaps Load for Beams

Difficulty: Hard

1) Collapse of the Span AP :

With plastic hinges at A,B and at the centre of the Span.

Using Virtual work principle

EWD = IWD

$M_P = \frac{400}{6} = 66.67 KN.m$

$M_P = 66.67 Kn.m$

3) Collapse of the span CD:

with plastic hinges at C,D and at the centre of the Span.

Using virtual work done:

FWD = IWD

$100* \gamma = 1.5 M_P\theta + 1.5 M_P\theta + 1.5 M_P\theta + M_P\theta$

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

$100*3\theta = M_P\theta(1.5 + 1.5 + 1.5 + 1)$

$300\theta = 5.5M_P\theta$

$M_P=\frac{300}{5.5} = 54.55 KN.m$

$M_P = 54.55 KN.m$

$(50*\gamma) + (80*\gamma) + (50*\gamma) = 2M_P\theta + 2M_P\theta + 2M_P\theta + 1.5M_P\theta $

$M_P = \frac{520}{7.5} = 69.33$

$M_P = 69.33 Kn,m$

2) Collapse of the span BC

with the plastic hinges at B,C and at the centre of the span.

using virtual work done

EWD = IWD

$25*\frac{1}{2}*8*\gamma = 1.5M_P\theta +1.5M_P\theta + 1.5M_P\theta + 1.5M_P\theta$

$\theta=\frac{\gamma}{4} \gamma=4\theta$

4) Collapse of the span CD

with the plastic hinges at D and understand the load at c.

EWD = IWD

$(45*\gamma) + (45*\gamma) =M_P\theta +M_P\theta +M_P\alpha $

$270\theta = M_P\theta (1+1+2)$

$M_P = \frac{270}{4} = 65.7 Kn.m$

$M_P = 65.7 KN.m$

From all the above possible consideration, the collapse Mechanism for the span AB requires the greatest value of $M_P$

$M_P = 69.33 KN.m$