Question 9.16: Draw S.F.D. and B.M.D. of overhanging beam as shown in the F...
Draw S.F.D. and B.M.D. of overhanging beam as shown in the Fig. 9.24.
\Sigma Y =0, R_{B} + R_{C}= 6 + (15 \times 2)
=36
\sum M_{B}=0,
R_{C} \times 4+6 \times 2+10=30 \times 5
R_{C}= 32 kN
R_{B} = 4 kN
Shear Force Diagram:
(S F)_{X_{1} X_{1}^{1}}=+15 \times x_{1}
(S F)_{D}=0
(S F)_{C}=15 \times 2 = 30 kN
(S F) x_{2} x_{2}^{1}=(15 \times 2)-R_{C}
= 30 – 32 = −2 kN
(S F)_{X_{2} X_{2}^{1}}=-2 kN \quad(S F)_{C}=(S F)_{E}=-2 kN
(S F)_{X_{3} X_{3}^{1}}=(15 \times 2)-R_{C}
(S F)_{E}=(S F)_{B}=-2 kN
(S F)_{X_{4} X_{4}^{1}}=(15 \times 2)-R_{C}-R_{B}
= 30 – 32 – 4
= −6 kN
( SF )_{ B }=( SF )_{ A }=-6 kN
Bending Moment Diagram:
(B M)_{X_{1} X_{1}^{1}}=-15 x_{1} \times \frac{x_{1}}{2}
(B M)_{D}=0
(B M)_{C}=-15 \times 2 \times \frac{\not 2}{\not 2}
= −30 kNm
(B M)_{X_{2} X_{2}^{1}}=-(15 \times 2)\left(x_{2}-1\right)+R_{C}\left(x_{2}-2\right)
=-30\left( x _{2}-1\right)+32\left( x _{2}-2\right)
(B M)_{C}=-30(2-1)+0=-30 kNm
(B M)_{E}=-30(4-1)+32(4-2)
= −26 kNm
(B M)_{X_{3} X_{3}^{1}}=-30\left(x_{3}-1\right)+32\left(x_{3}-2\right)+10
(B M)_{E}=-30(4-1)+32(4-2)+10
= −16 kNm
(B M)_{B}=-30(6-1)+32(6-2)+10
(B M)_{B}=−12 kNm
(B M)_{X_{4} X_{4}^{1}}=-30\left(x_{4}-1\right)+32\left(x_{4}-2\right)+10+4\left(x_{4}-6\right)
(B M)_{B}=−12 kNm
(B M)_{A}=-30(8-1)+32(8-2)+10+4(8-6)
= 0