Determine the ultimate load for a simply supported, rectangular section beam, breadth [latex]b[/latex], depth [latex]d[/latex], having a span [latex]L[/latex] and subjected to a uniformly distributed load of intensity [latex]w[/latex].

The maximum bending moment occurs at mid-span and is equal to [latex]w L^{2} / 8[/latex] (see Section 3.4). The plastic hinge therefore forms at mid-span when this bending moment is equal to [latex]M_{\mathrm{P}}[/latex], the corresponding ultimate load intensity being [latex]w_{\mathrm{U}}[/latex]. Thus [latex]\frac{w_{\mathrm{U}} L^{2}}{8}=M_{\mathrm{P}}[/latex] (i) From Ex. 18.1 , Eq. (ii) [latex]M_{\mathrm{P}}=\sigma_{\mathrm{Y}} \frac{b d^{2}}{4}[/latex] so that [latex]w_{\mathrm{U}}=\frac{8 M_{\mathrm{P}}}{L^{2}}=\frac{2 \sigma_{\mathrm{Y}} b d^{2}}{L^{2}}[/latex] where [latex]\sigma_{\mathrm{Y}}[/latex] is the yield stress of the material of the beam.

Question 18.4: Determine the ultimate load for a simply supported, rectangu...

Structural and Stress Analysis

Determine the ultimate load for a simply supported, rectangular section beam, breadth $b$ , depth $d$ , having a span $L$ and subjected to a uniformly distributed load of intensity $w$ .

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The maximum bending moment occurs at mid-span and is equal to $w L^{2} / 8$ (see Section 3.4). The plastic hinge therefore forms at mid-span when this bending moment is equal to $M_{\mathrm{P}}$ , the corresponding ultimate load intensity being $w_{\mathrm{U}}$ . Thus

$\frac{w_{\mathrm{U}} L^{2}}{8}=M_{\mathrm{P}}$ (i)

From Ex. 18.1, Eq. (ii)

$M_{\mathrm{P}}=\sigma_{\mathrm{Y}} \frac{b d^{2}}{4}$

so that

$w_{\mathrm{U}}=\frac{8 M_{\mathrm{P}}}{L^{2}}=\frac{2 \sigma_{\mathrm{Y}} b d^{2}}{L^{2}}$

where $\sigma_{\mathrm{Y}}$ is the yield stress of the material of the beam.