Question

The collapse load for a continuous beam, shown in figure of uniform cross-section will be ______.

Answer 1

Span AB

$W_u\times \frac{L}{2}\theta =M_P(\theta +\theta )+M_P\theta$

$W_u=\frac{6M_P}{L}$

Span BC


$\Delta =\frac{L}{3}{\theta }_1=\frac{2L}{3}\theta \Rightarrow {\theta }_1=2\theta$

$W_u\frac{2}{3}L\theta =M_P\theta +M_P(\theta +{\theta }_1)$

$W_u\times \frac{2}{3}L\theta =4M_P\theta$

$W_u=\frac{6M_P}{L}$