A beam carrying a uniform load is simply supported with the supports set back a distance a from the ends as shown in the figure

Question

A beam carrying a uniform load is simply supported with the supports set back a distance [latex]a [/latex]from the ends as shown in the figure. The bending moment at [latex]x [/latex]can be found from summing moments to zero at section [latex]x [/latex]:

[latex]\sum M=M+\frac{1}{2} w(a+x)^{2}-\frac{1}{2} w l x=0 [/latex]

or

[latex]M=\frac{w}{2}\left[l x-(a+x)^{2}\right] [/latex]

where [latex]w [/latex]is the loading intensity in [latex]\mathrm{lbf} / \mathrm{in} [/latex]. The designer wishes to minimize the necessary weight of the supporting beam by choosing a setback resulting in the smallest possible maximum bending stress

(a) If the beam is configured with [latex]a=2.25 [/latex]in, [latex]l=10 [/latex]in, and [latex]w=100 \mathrm{lbf} / [/latex]in, find the magnitude of the severest bending moment in the beam.

( [latex]b [/latex]) Since the configuration in part ( [latex]a [/latex]) is not optimal, find the optimal setback [latex]a [/latex]that will result in the lightest-weight beam.

Accepted Answer

(a) Moment at center,

[latex]\begin{aligned}&x_{c}=\frac{(l-2 a)}{2} \&M_{c}=\frac{w}{2}\left[\frac{l}{2}(l-2 a)-\left(\frac{l}{2} ight)^{2} ight]=\frac{w l}{2}\left(\frac{l}{4}-a ight)\end{aligned} [/latex]

At reaction, [latex]\left|M_{r} ight|=w \mathrm{a}^{2} / 2 [/latex]

[latex]a=2.25, l=10 \mathrm{in}, w=100 \mathrm{lbf} / \mathrm{in} [/latex]

[latex]\begin{aligned}&M_{c}=\frac{100(10)}{2}\left(\frac{10}{4}-2.25 ight)=125 \mathrm{lbf} \cdot \mathrm{in} \&\left|M_{r} ight|=\frac{100\left(2.25^{2} ight)}{2}=253 \mathrm{lbf} \cdot \mathrm{in} \quad ext { }\end{aligned} [/latex]

(b) Optimal occurs when [latex]M_{c}=\left|M_{r} ight|\ [/latex]

[latex]\frac{w l}{2}\left(\frac{l}{4}-a ight)=\frac{w a^{2}}{2} \Rightarrow a^{2}+a l-0.25 l^{2}=0 [/latex]

Taking the positive root

[latex]a=\frac{1}{2}\left[-l+\sqrt{l^{2}+4\left(0.25 l^{2} ight)} ight]=\frac{l}{2}(\sqrt{2}-1)=0.207 l [/latex]

for [latex]l=10 \mathrm{in}, w=100 \mathrm{lbf}, a=0.207(10)=2.07 \mathrm{in} [/latex]

[latex]M_{\min }=(100 / 2) 2 \cdot 07^{2}=214 \mathrm{lbf} \cdot \mathrm{in} [/latex]

Question 3.14: A beam carrying a uniform load is simply supported with the...

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