Question A2.13: A 6 m long cantilever carries loads of 2 kN at 2 m and 5 m r...
A 6 m long cantilever carries loads of 2 kN at 2 m and 5 m respectively from the fixed end and a UDL of 10 kN over its entire length. Draw shear force and bending moment diagrams. [KU, June 2010]
The loding diagram is shown in Fig. 13 (a).
UDL, \quad w=\frac{10}{6}=\frac{5}{3} \mathrm{\ kN} / \mathrm{m}
Span AC :
\begin{aligned}&F_x=-w x=-\frac{5}{3} x \\&F_A=0, F_C=-\frac{5}{3} \times 1=-\frac{5}{3} \mathrm{\ kN}\end{aligned}
Span CD :
\begin{aligned}&F_x=-\left[\frac{5}{3} x+3\right], F_c=-\left[\frac{5}{3}+3\right]=-4.67 \mathrm{\ kN} \\&F_D=-\left[\frac{5}{3} \times 4+3\right]=-9.67 \mathrm{\ kN}\end{aligned}
Span DB :
\begin{aligned}&F_x=–\left[\frac{5}{3} x+3+2\right]=-\left[\frac{5}{3} x+5\right] \\&F_D=-\left[\frac{5}{3} \times 4+5\right]=-11.67 \\&F_B=-\left[\frac{5}{3} \times 6+5\right]=-15 \mathrm{\ kN}\end{aligned}
The S.F.D. is shown in Fig. 13 (b).
B.M.D.
Span AC :
\begin{aligned}&M_x=-w \frac{x^2}{2}=-\frac{5}{3} \times \frac{x^2}{2}=\frac{5}{6} x^2=0.833 x^2 \\&M_A=0, M_C=-\frac{5}{6} \times 1=-0.833 \mathrm{\ kN} . \mathrm{m}\end{aligned}
Span CD :
\begin{aligned}&M_x=-\left[\frac{w x^2}{2}+3(x-1)\right]=-\left[\frac{5}{6} x^2+3(x-1)\right] \\&M_D=-\left[\frac{5}{6} \times 4^2+3(4-1)\right]=-[13.33+9]=-22.33 \mathrm{\ kN} \cdot \mathrm{m}\end{aligned}
Span DB :
\begin{aligned}&M_x=-\left[\frac{w x^2}{2}+3(x-1)+2(x-4)\right]=-\left[\frac{5}{6} x^2+3(x-1)+2(x-4)\right] \\&M_B=-\left[\frac{5}{6} \times 6^2+3(6-1)+2(6-4)\right]=[30+15+4]=-49 \mathrm{\ kN} . \mathrm{m}\end{aligned}
The B.M.D. is shown in Fig. 13 (c).
