Question 16.9: A uniform beam is simply supported over a span of 6 m. It ca......
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)
The beam is as shown in Fig. S.16.9.
Taking moments about B,
which gives
R_{\mathrm{A}}=150\,\mathrm{kN}The bending moment at any section a distance z from A is then
M=-150z+\frac{30z^{2}}{2}+\left(90-30\right)\left(\frac{z}{6}\right)\left(\frac{z}{2}\right)\left(\frac{z}{3}\right)i.e.,
M=-150z+15z^{2}+{\frac{5z^{3}}{3}}Substituting in the second of Eqs (16.31),
u^{\prime\prime}=-{\frac{M_{y}}{E I_{y y}}},\quad ν^{\prime\prime}=-{\frac{M_{x}}{E I_{x x}}} (16.31)
E I{\biggl(}{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}{\biggr)}=150z-15z^{2}-{\frac{5z^{3}}{3}}
E I{\biggl(}{\frac{\mathrm{d}v}{\mathrm{d}z}}{\biggr)}=75z^{2}-5z^{3}-{\frac{5z^{4}}{12}}+C_{1}
E I v=25z^{3}-{\frac{5z^{4}}{4}}-{\frac{z^{5}}{12}}+C_{1}z+C_{2}
When z = 0, υ=0 so that C_{2}=0, and when z = 6m, υ=0. Then
0=25\times6^{3}-{\frac{5\times6^{4}}{4}}-{\frac{6^{5}}{12}}+6C_{1}from which
C_{1}=-522and the deflected shape of the beam is given by
E I v=25z^{3}-{\frac{5z^{4}}{4}}-{\frac{z^{5}}{12}}-522zThe deflection at the mid-span point is then given by
E I v_{m id-s p a n}=25\times3^{3}-{\frac{5\times3^{4}}{4}}-{\frac{3^{5}}{12}}-522\times3=-1012.5\mathrm{kNm}^{3}Therefore,
v_{\mathrm{mid}-\mathrm{span}}={\frac{-1012.5\times10^{12}}{120\times10^{6}\times206000}}=-41.0\mathrm{mm}\ \ (\mathrm{downwards})