Question 9.12: Draw S.F.D and B.M.D for cantilever beam subjected to U.D.L ...

To maintain the beam in equilibrium, wall will offer an upward force R_{A} and moment in anticlockwise direction as shown in Fig. 9.20 at point A.
Shear Force Diagram:

(S . F .)_{x_{1} x_{1}^{1}}=+3 \cdot x_{1}

(S F)_{D}=0,(S F)_{C}=3 \times 4=+12  kN

(S F)_{x_{2} x_{2}^{1}}=+3 \times 4=+12  kN

(S F)_{C}=(S F)_{B}=+12  kN

(S F)_{x_{3} x_{3}^{1}}=+12+10=22  kN

(S F)_{B}=(S F)_{A}=+22  kN

Bending Moment Diagram:

(B M)_{x_{1} x_{1}^{1}}=-3 x_{1} \times x_{1} / 2

(B M)_{D}=0,(B M)_{C}=-3 \times 4 \times \frac{4}{2}=-24  kNm

(B M)_{x_{2} x_{2}^{1}}=-(3 \times 4) \times\left(x_{2}-2\right)

(B M)_{D}= −12 (4−2)

= −24 kNm

(B M)_{B}= −12(5−2)= −36  kNm

(B M)_{x_{3} x_{3}^{1}}=-(3 \times 4)\left(x_{3}-2\right)-10\left(x_{3}-5\right)

(B M)_{B}=-3 \times 4(5-2)-0=-36  kNm

(B M)_{A}=−3 \times 4 (6−2)–10 (6−5)=−58  kNm

For support reaction and moment,

\Sigma Y =0, R_{A}=10+(3 \times 4)=22  kN

\Sigma M _{ A }=0,- M _{ A }+10 \times 1+(3 \times 4) \times 4=0

M _{ A } = 58  kNm

Note: Here full load of U.D.L (3 × 4) is acting at mid i.e. at 2 m from D thus from X_{2} X_{2}^{1} thedistance of load will be (x_{2}− 2).