A fixed end beam is subjected to a concentrated load (W) as shown in the figure. The beam has two different segments having different plastic moments capacities ( $M_{p}, 3 M_{p}$ ) as shown.

The minimum value of load (W) at which the beam would collapse (ultimate load) is ______.

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Solution

First mechanism:

$3 M_{p} \times θ + M_{p} θ + 2 M_{p} ϕ = W \times \frac{2 L}{3} \times ϕ . . . . (1)$

$\frac{4 L}{3} \times θ = \frac{2 L}{3} \times ϕ \Rightarrow ϕ = 2 θ$

Putting
$ϕ = 2 θ in equation (1), we get$

$8 M_{p} θ = W \times \frac{2 L}{3} \times 2 θ$

$\Rightarrow W = \frac{6 M_{p}}{L} . . . . (2)$

Second mechanism

$3 M_{p} \times θ + 2 M_{p} θ + M_{p} θ = W \times \frac{2 L}{3} \times θ$

$\Rightarrow 6 M_{p} = W \times \frac{2 L}{3}$

$\Rightarrow W = \frac{9 M_{p}}{L} . . . . (3)$

$∴ Collapse load (W) = m i n {\frac{6 M_{p}}{L}, \frac{9 M_{p}}{L}} = {\frac{6 M_{p}}{L}}$

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