Question 9.19: Draw SFD and BMD for an overhanging beam as shown in Fig. 9....
Draw SFD and BMD for an overhanging beam as shown in Fig. 9.27.
hTe figure can be modified about point B where Couple = 25 × 2 = 50 kNm (anticlockwise) act.
\Sigma Y =0, R_{A} + R_{C}=(5 \times 2 2) + (60 \times 2 1)
\quad= 70 kN ….. (1)
\Sigma M _{ A }=0, R_{C} \times 4+50+(5 \times 2) \times 1=60 \times 1 \times(4+0.5)
4 R_{C}+60=270
R_{C}= 52.50 kN
R_{A} = 17.50 kN
Shear Force Diagram:
(S F)_{X_{1} X_{1}^{1}}=+60 \cdot x_{1}
(S F)_{D}=0,(S F)_{C}=+60 kN
(S F)_{X_{2} X_{2}^{1}}=+60 \times 1-R_{C}
= 60 – 52.50
= 7.5 kN
(S F)_{C}=(S F)_{B}=+7.5 kN
(S F)_{B}=7.5 kN.
(S F)_{X_{3} X_{3}^{1}}=60-52.5
= 7.5 kN
(S F)_{B}=(S F)_{A}=7.5 kN
(S F)_{X_{4} X_{4}^{1}}=+60-52.50-17.5+5\left(x_{4}-5\right)
=-10+5\left(x_{4}-5\right)
(S F)_{A}=−10 + 5 (5 − 5)
= −10 kN
(S F)_{E}=−10 + 5 (7 − 5)
= 0
Bending Moment Diagram:
(B M)_{X_{1} X_{1}^{1}}=-60 x_{1} \times \frac{x_{1}}{2}
(B M)_{D}=0
(B M)_{C}=-60 \times \frac{1 \times 1}{2}=-30 kNm
(B M)_{X_{2} X_{2}^{1}}=-60\left(x_{2}-0.5\right)+52.50\left(x_{2}-1\right)
(B M)_{C}=-60(1-0.5)+0=-30 kNm
(B M)_{B}=−60 (3 – 0.5) + 52.50 (3 − 1)
= −45 kNm
(B M)_{X_{3} X_{3}^{1}}=-60\left(x_{3}-0.5\right)+52.50\left(x_{3}-1\right)+50
(B M)_{B}=−60 (3 – 0.5) + 52.5 (3 − 1) + 50
= +5 kNm
(B M)_{A}=−60 (5 – 0.5) + 52.5 (5 − 1) + 50
= −10 kNm
(B M)_{X_{4} X_{4}^{1}}=-60\left(x_{4}-0.5\right)+52.5\left(x_{4}-1\right)+50+17.50\left(x_{4}-5\right)-5\left(x_{4}-5\right)\left(x_{4}-5\right) / 2
(B M)_{A}=−10 kNm
(B M)_{E}=0
To determine point of Contraflexure put (\text { B.M. })_{ X _{3} X _{3}^{1}}=0 and x _{3} = a
(B M)_{F}=0=-60(a-0.5)+52.5(a-1)+50
0 = −60·a + 30 + 52.5a – 52.5 + 50
a = 3.67 m