Question A2.12: Draw the SF and BM diagrams for a cantilever loaded as shown...
Draw the SF and BM diagrams for a cantilever loaded as shown in Fig. 11. [KU, June 2009]
The loading diagram is shown in Fig. 12 (a).
S.F.D.
Span AC :
\begin{aligned}&F_x=-1 \mathrm{\ kN} \\&F_A=F_C=-1 \mathrm{\ kN}\end{aligned}
Span CD:
\begin{aligned}&F_x=-[1+2(x-1)] \\&F_D=-[1+2(3-1)]=-5 \mathrm{\ kN}\end{aligned}
Span DE:
\begin{aligned}F_x &=-[1+2 \times 2]=-5 \mathrm{\ kN} \\D_E &=-5 \mathrm{\ kN}\end{aligned}
Span EB :
\begin{aligned}&F_x=-[1+2 \times 2+1]=-6 \mathrm{\ kN} \\&F_B=-6 \mathrm{\ kN}\end{aligned}
The S.F.D. is shown in Fig. 12 (b).
B.M.D.
Span AC:
\begin{aligned}&M_x=-1 \times x=-1 x \\&M_A=0, M_C=-1 \times 1=-1 \mathrm{\ kN} . \mathrm{m}\end{aligned}
Span CD:
\begin{aligned}&M_x=-\left[1 \times x+2(x-1) \frac{(x-1)}{2}\right]=-\left[x+(x-1)^2\right] \\&M_D=-\left[3+(3-1)^2\right]=-7 \text { kN.m }\end{aligned}
Span DE:
\begin{aligned}&M_x=-[1 \times x+2 \times 2(x-2)]=-[x+4(x-2)] \\&M_E=-[4+4(4-2)]=-12 \text { kN.m }\end{aligned}
Span EB :
\begin{aligned}&M_x=-[1 \times x+2 \times 2(x-2)+1(x-4)]=-[x+4(x-2)+(x-4)] \\&M_B=-[5+4(5-2)+(5-4)]=-[5+12+1]=-18 \mathrm{\ kN} . \mathrm{m}\end{aligned}
The B.M.D. is shown in Fig. 12 (c).
