A fixed end beam is subjected to a load W at 1/3rd span from the left support as shown in figure. The collapse load of the beam is ______.

\frac{16 M_{p}}{L}

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\frac{15 M_{p}}{L}

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\frac{15.5 M_{p}}{L}

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\frac{16.5 M_{p}}{L}

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Solution

The correct option is B $\frac{15 M_{p}}{L}$
No. of plastic hinges formed at collapse = r + 1 = 3

Case I:

$\frac{L}{3} \times θ = \frac{2 L}{3} \times α$

$θ = 2 α$
$- 2 M_{p} θ - 2 M_{p} θ - 2 M_{p} α - M_{p} α + W \times \frac{L}{3} θ = 0$

$\Rightarrow - 4 M_{p} θ - M_{p} θ - \frac{M_{p} θ}{2} = - \frac{W L}{3} θ$

$\Rightarrow \frac{11}{2} M_{p} θ = \frac{W L}{3} θ$

$\Rightarrow W = \frac{16.5}{L} M_{p}$

Case II:

$- 2 M_{p} θ - M_{p} θ - M_{p} θ - M_{p} θ + \frac{W L}{3} \times θ = 0$

$W = \frac{15 M_{p}}{L}$

Minimum collapse load, $W = M i n i m u m o f (\frac{16.5 M_{p}}{L}$ , $\frac{15 M_{p}}{L}$ )

Hence, Minimum collapse load,
$W = \frac{15 M_{p}}{L}$

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