The intensity of loading in a simply supported beam of 8 m span varies gradually from 2 KN/m at one end to 6 k N/m at the other end. Draw the shear force and bending moment diagram.

Question

The intensity of loading in a simply supported beam of 8 m span varies gradually from 2 KN/m at one end to 6 k N/m at the other end. Draw the shear force and bending moment diagram. [KU, Jan 2010]

Accepted Answer

The loading diagram is shown in Fig. 14 (a). The intensity of loading at a distance x from A is

[latex] an heta=\frac{6-2}{8}=\frac{4}{8}=\frac{1}{2}[/latex]

[latex]w=\left\lgroup 2+\frac{x}{2} ight group \mathrm{\ kN} / \mathrm{m}[/latex]

Taking moments about B, we have

[latex]\begin{aligned}8 R_A &=2 imes 8 imes 4+\frac{1}{2} imes 8 imes 4 imes \frac{8}{3} \&=64+42.67=106.67 \R_A &=13.33 \mathrm{\ kN} \R_B &=\frac{1}{2}(2+6) imes 8-13.33=18.67 \mathrm{\ kN}\end{aligned}[/latex]

S.F.D.

[latex]\begin{aligned}F_x &=R_A-\frac{1}{2}\left\lgroup2+2+\frac{x}{2} ight group imes x \&=13.33-\left\lgroup2+\frac{x}{4} ight group x=0 \F_A &=13.33 \mathrm{\ kN} ext { and } F_B=-18.67 \mathrm{\ kN} \F_x &=0 ext { when } 13.33-\left\lgroup 2+\frac{x}{2} ight group x=0\end{aligned}[/latex]

or

[latex]\begin{aligned}\frac{x^2}{2}+2 x-13.33 &=0 \x^2+4 x-26.66 &=0 \x &=\frac{1}{2}[-4 \pm \sqrt{16+4 imes 26.66}] \x &=3.537 \mathrm{~m}\end{aligned}[/latex]

The S.F.D. is shown in Fig. 14 (b).

B.M.D.

[latex]\begin{aligned}M_x &=\frac{40}{3} imes x-2 imes x imes \frac{x}{2}-\frac{1}{2} imes x imes \frac{x}{2} imes \frac{x}{3} \&=\frac{40}{3} x-x^2-\frac{x^3}{12} \M_A &=0, M_B=0\end{aligned}[/latex]

At x = 3.537 m, [latex]M_{\max }=\frac{40}{3} imes 3.537-(3.537)^2-\frac{1}{12}(3.537)^3=30.962 \mathrm{\ kN.m} [/latex]

Question A2.16: The intensity of loading in a simply supported beam of 8 m s...

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