Question 9.15: Determine reaction and moment offered by wall on beam and dr...
Determine reaction and moment offered by wall on beam and draw S.F.D. and B.M.D. for cantilever beam as shown in Fig. 9.23.
\Sigma Y =0, R_{A}=9+(5 \times 3)+16
= 40 kN
\sum M_{A}=0, \quad-M_{A}+9 \times 1+6+(5 \times 3)\left(3+\frac{3}{2}\right)+16 \times 6=0
M_{A} = 178.5 kNm
Shear Force Diagram:
(S F)_{X_{1} x_{1}^{1}}=16+5 \cdot x_{1}
(S F)_{E}=16 kN ,(S F)_{D}=16+5 \times 3
= 31 kN
(S F)_{X_{2} X_{2}^{1}}=16+(5 \times 3)=31 kN
(S F)_{D}=(S F)_{C}=31 kN
(couple or moment has no effect in shear force diagram)
(S F)_{X_{3} X_{3}^{1}}=16+(5 \times 3)=31 kN
(S F)_{C}=(S F)_{B}=31 kN
(S F)_{X_{4} X_{4}^{1}}=16+(5 \times 3) +9=40 kN
(S F)_{B}=(S F)_{A}=40 kN
Bending Moment Diagram:
(B M)_{X_{1} X_{1}^{1}}=-16 x_{1}-5 x_{1} \times \frac{x_{1}}{2}
(B M)_{E}=0
(B M)_{D}=-16 \times 3-5 \times 3 \times \frac{3}{2}=-70.5 kNm
(B M)_{X_{2} X_{2}^{1}}=-16 x_{2}-(5 \times 3)\left(x_{2}-1.5\right)
(B M)_{D}=-16 \times 3-(5 \times 3)(3-1.5)=-70.5 kNm
(B M)_{C}=-16 \times 4-(5 \times 3)(4-1.5)-6=-101.5 kNm
(B M)_{X_{3} X_{3}^{1}}=-16 \cdot x_{3}-(5 \times 3)\left(x_{3}-1.5\right)-6
(B M)_{C}=-16 \times 4-(5 \times 3)(4-1.5)-6=-95.5 kNm
(B M)_{B}=-16 \times 5-(5 \times 3)(5-1.5)-6=-138.5 kNm
(B M)_{X_{4} X_{4}^{1}}=-16 \cdot x_{4}-(5 \times 3)\left(x_{4}-1.5\right)-6-9\left(x_{4}-5\right)
(B M)_{B}=-16 \times 5-(5 \times 3)(5-1.5)-6=-138.5 kNm
(B M)_{A}=-16 \times 6-(5 \times 3)(6-1.5)-6-9(6-5)
= −178.5 kNm