A uniform beam is simply supported over a span of 6 m. It carries a trapezoidally distributed load with intensity varying from 30 kN/m at the left-hand support to 90 kN/m at the right-hand support. Find the equation of the deflection curve and hence the deflection at the mid-span point. The second moment of area of the cross-section of the beam is $120\times10^{6}\operatorname{mm}^{4}$ and Young’s modulus E=206,000 N/mm².

Answer: 41 mm (downward)

Question

A uniform beam is simply supported over a span of 6 m. It carries a trapezoidally distributed load with intensity varying from 30 kN/m at the left-hand support to 90 kN/m at the right-hand support. Find the equation of the deflection curve and hence the deflection at the mid-span point. The second moment of area of the cross-section of the beam is $120\times10^{6}\operatorname{mm}^{4}$ and Young’s modulus E=206,000 N/mm².

Answer: 41 mm (downward)

Accepted Answer

The beam is as shown in Fig. S.16.9.
Taking moments about B,

[latex]R_{\mathrm{A}} imes6-{\frac{30 imes6^{2}}{2}}-{\frac{60}{2}} imes6 imes{\frac{6}{3}}=0[/latex]

which gives

[latex]R_{\mathrm{A}}=150\,\mathrm{kN}[/latex]

The bending moment at any section a distance z from A is then

[latex]M=-150z+\frac{30z^{2}}{2}+\left(90-30 ight)\left(\frac{z}{6} ight)\left(\frac{z}{2} ight)\left(\frac{z}{3} ight)[/latex]

i.e.,

[latex]M=-150z+15z^{2}+{\frac{5z^{3}}{3}}[/latex]

Substituting in the second of Eqs (16.31),

[latex]u^{\prime\prime}=-{\frac{M_{y}}{E I_{y y}}},\quad ν^{\prime\prime}=-{\frac{M_{x}}{E I_{x x}}}[/latex] (16.31)

[latex]E I{\biggl(}{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}{\biggr)}=150z-15z^{2}-{\frac{5z^{3}}{3}}[/latex]

[latex]E I{\biggl(}{\frac{\mathrm{d}v}{\mathrm{d}z}}{\biggr)}=75z^{2}-5z^{3}-{\frac{5z^{4}}{12}}+C_{1}[/latex]

[latex]E I v=25z^{3}-{\frac{5z^{4}}{4}}-{\frac{z^{5}}{12}}+C_{1}z+C_{2}[/latex]

When z = 0, υ=0 so that [latex]C_{2}=0,[/latex] and when z = 6m, υ=0. Then

[latex]0=25 imes6^{3}-{\frac{5 imes6^{4}}{4}}-{\frac{6^{5}}{12}}+6C_{1}[/latex]

from which

[latex]C_{1}=-522[/latex]

and the deflected shape of the beam is given by

[latex]E I v=25z^{3}-{\frac{5z^{4}}{4}}-{\frac{z^{5}}{12}}-522z[/latex]

The deflection at the mid-span point is then given by

[latex]E I v_{m id-s p a n}=25 imes3^{3}-{\frac{5 imes3^{4}}{4}}-{\frac{3^{5}}{12}}-522 imes3=-1012.5\mathrm{kNm}^{3}[/latex]

Therefore,

[latex]v_{\mathrm{mid}-\mathrm{span}}={\frac{-1012.5 imes10^{12}}{120 imes10^{6} imes206000}}=-41.0\mathrm{mm}\ \ (\mathrm{downwards})[/latex]

Question 16.9: A uniform beam is simply supported over a span of 6 m. It ca......

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