Question 9.18: Draw S.F.D. and B.M.D. of an overhanging beam as shown in Fi...
Draw S.F.D. and B.M.D. of an overhanging beam as shown in Fig. 9.26.
\Sigma Y =0, R_{A} + R_{B}=150 + 50
\quad= 200 kN ….. (1)
\sum M_{A}=0,
R_{B} \times 3+50 \times 2=10+150 \times 4
\quad= 610
R_{B}= 170 kN
R_{A} = 30 kN
Shear Force Diagram:
(S F)_{X_{1} X_{1}^{1}}=+150 k N
(S F)_{C}=(S F)_{B}=+150 kN
(S F)_{X_{2} X_{2}^{1}}=+150 k N-R_{B}=150-170
\quad=-20 kN \text { i.e., }(S F)_{B}=(S F)_{F}=-20 kN
(S F)_{X_{3} X_{3}^{1}}=150-170
\quad= −20 kN
(S F)_{E}=(S F)_{A}=-20 kN
(S F)_{X_{4} X_{4}^{1}}=150-170-30
\quad= −50 kN
(S F)_{A}=(S F)_{D}=-50 kN
Bending Moment Diagram:
(B M)_{X_{1} X_{1}^{1}}=-150 \cdot x_{1}
(B M)_{C}=0
(B M)_{B}=−150 kNm
(B M)_{X_{2} X_{2}^{1}}=-150 \cdot x_{2}+170\left(x_{2}-1\right)
(B M)_{B}=−150 kNm
(B M)_{E}=−150 \times 2.5 + 170 (2.5 − 1)
\quad= −120 kNm
(B M)_{X_{3} X_{3}^{1}}=-150 \cdot x_{3}+170\left(x_{3}-1\right)-10
(B M)_{E}=−150 \times 2.5 + 170 (2.5 − 1) – 10
\quad= −130 kNm
(B M)_{A}=−150 \times 4 + 170 (4 − 1) – 10
\quad= −100 kNm
(B M)_{X_{4} X_{4}^{1}}=-150 \cdot x_{4}+170\left(x_{4}-1\right)-10+30\left(x_{4}-4\right)
(B M)_{A}=−150 \times 4 + 170x (4 − 1) – 10
\quad= −100 kNm
(B M)_{D}=−150 \times 6 + 170 (6 − 1) – 10 + 30 (6 −4)
= 0