Question 9.14: Vertical forces 20 kN, 40 kN and uniformly distributed load ...
Vertical forces 20 kN, 40 kN and uniformly distributed load of 20 kN/m is acting in 3 m length as shown in Fig. 9.22. Find the resultant force of the system and draw S.F.D and B.M.D.
(U.P.T.U., Ist Sem, 2001−2002)
\Sigma Y=0, R_{c}=20+40+(20 \times 3)
= 120 kN
\Sigma M_{c}=0,-20 \times 3-40 \times 2-(20 \times 3) \times 1.5+M_{c}=0
\Sigma M_{c}=230 kNm
Shear Force Diagram:
Consider sections X_{1} X_{1}^{1} and X_{2} X_{2}^{1} from left side i.e., at x_{1} and x_{2}, respectively. Thus sign convention will get changed for shear force diagram.
(S F)_{X_{1} X_{1}^{1}}=-20-20 x_{1}
(S F)_{A}=−20 kN, (S F)_{B} =−20–2 \times 1=−40 kN
(S F)_{X_{2} X_{2}^{1}}=-20-40-20 x_{2}
(S F)_{B} =−20 – 40 – 20=−80 kN
(S F)_{C} =−20 – 40 – 20 \times 3
= −120 kN
Bending Moment Diagram:
(B M)_{X_{1} X_{1}^{1}}=-20 \cdot x_{1}-20 \cdot x_{1} \cdot \frac{x_{1}}{2}
(B M)_{A}=0,(B M)_{B}=-20-\frac{20 \times 1 \times 1}{2}
= −30 kNm
(B M)_{X_{2} X_{2}^{1}}=-20 \cdot x_{2}-40\left(x_{2}-1\right)-20 \cdot x_{2} \cdot \frac{x_{2}}{2}
(B M)_{B}=-20 \times 1-0-\frac{20}{2}
= −30 kNm
(B M)_{C}=-20 \times 3-40(3-1)-20 \cdot 3 \cdot \frac{3}{2}
= −60 – 80 – 90
= −230 kNm