The ultimate collapse load $(P)$ in terms of plastic moment $M_{\mathrm{p}}$ by kinematic approach for a propped
cantilever of length $L$ with $P$ acting at its mid-span as shown in the figure, would be
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$\begin{array}{llll}\text { (A) } P=\frac{2 M_{p}}{L} & \text { (B) } P=\frac{4 M_{\mathrm{p}}}{L} & \text { (C) } P=\frac{6 M_{\mathrm{p}}}{L} & \text { (D) } P=\frac{8 M_{\mathrm{p}}}{L}\end{array}$
cantilever of length $L$ with $P$ acting at its mid-span as shown in the figure, would be
$\begin{array}{llll}\text { (A) } P=\frac{2 M_{p}}{L} & \text { (B) } P=\frac{4 M_{\mathrm{p}}}{L} & \text { (C) } P=\frac{6 M_{\mathrm{p}}}{L} & \text { (D) } P=\frac{8 M_{\mathrm{p}}}{L}\end{array}$
Question ID - 155963 | SaraNextGen Top Answer The ultimate collapse load $(P)$ in terms of plastic moment $M_{\mathrm{p}}$ by kinematic approach for a propped
cantilever of length $L$ with $P$ acting at its mid-span as shown in the figure, would be
$\begin{array}{llll}\text { (A) } P=\frac{2 M_{p}}{L} & \text { (B) } P=\frac{4 M_{\mathrm{p}}}{L} & \text { (C) } P=\frac{6 M_{\mathrm{p}}}{L} & \text { (D) } P=\frac{8 M_{\mathrm{p}}}{L}\end{array}$
The ultimate collapse load $(P)$ in terms of plastic moment $M_{\mathrm{p}}$ by kinematic approach for a propped
cantilever of length $L$ with $P$ acting at its mid-span as shown in the figure, would be
$\begin{array}{llll}\text { (A) } P=\frac{2 M_{p}}{L} & \text { (B) } P=\frac{4 M_{\mathrm{p}}}{L} & \text { (C) } P=\frac{6 M_{\mathrm{p}}}{L} & \text { (D) } P=\frac{8 M_{\mathrm{p}}}{L}\end{array}$