Question A2.17: For the beam shown in Fig. 15, draw SFD and BMD.
For the beam shown in Fig. 15, draw SFD and BMD. [KU, Dec. 2009]
The loading diagram is shown in fig. 16 (a).
R_B+R_C=4+4+4 \times 6=32 \mathrm{\ kN}
Taking moments about C, we have
\begin{aligned}6 R_B &=4 \times 8-4 \times 2+4 \times 6 \times 3=96 \\R_B &=16 \mathrm{\ kN} \\R_C &=32-16=16 \mathrm{\ kN}\end{aligned}
S.F.D.
Span AB :
F_A=F_B=-4 \mathrm{\ kN}
Span BC:
Let x = distance from B.
\begin{aligned}&F_x=-4+R_B-4 x=-4+16-4 x=12-4 x \\&F_B=12-0=12 \mathrm{\ kN} \\&F_C=12-4 \times 6=-12 \mathrm{\ kN}\end{aligned}
Span CD:
\begin{aligned}&F_D=4 \mathrm{\ kN} \\&F_C=4-16=-12 \mathrm{\ kN}\end{aligned}
The S.F.D. is shown in Fig. 16 (b).
B.M.D.
Span AB :
M_A=0, M_B=-4 \times 2=-8 \mathrm{\ kN} \cdot \mathrm{m}
Span BC:
M_x=-4(2+x)+16 \times 6-2 \times 6^2=-32+96-72=-8 \mathrm{\ kN} \cdot \mathrm{m}
At x = 3 m from B
M = –4 × 5 + 4 8 – 18 = 10 kN.m
For \quad M_x=0,2 x^2-12 x+8=0 \text { or } x^2-6 x+4=0
x=\frac{1}{2}(6 \pm \sqrt{36-16})=\frac{1}{2}(6 \pm 4.47)=0.764 \mathrm{~m} \text { and } 5.236 \mathrm{~m}
Span CD:
M_D=0, M_C=-4 \times 2=-8 \mathrm{\ kN} . \mathrm{m}
The B.M.D. is shown in Fig. 16 (c).
