Question 9.17: Find the value of x and draw S.F.D. and B.M.D. for the beam ...

\sum M_{A}=0,

R_{B} \times(2+x)=1000 \times(3+x)+(2000 \times 2)(x+1)

4000 (2 + x) = 3000 + 1000x + 4000x + 4000

8000 + 4000x = 7000 + 5000x

\quad1000 = 1000x

x = 1 m

Shear Force Diagram:

(S F)_{X_{1} X_{1}^{1}}=+1000  N

(S F)_{D}=(S F)_{B}=+1000  N

(S F)_{x_{2} x_{2}^{1}}=+1000-4000+2000\left(x_{2}-1\right)

(S F)_{B}=−3000  N

(S F)_{C}= 1000 – 4000 + 2000 (3 − 1)

\quad= 1000 N

(S D)_{x_{3} x_{3}^{1}}=+1000-4000+(2000 \times 2)

= +1000 N

(S F)_{C}=(S F)_{A}=+1000  N

Note in portion BC, (S.F.) changes its sign and let it is zero at point E which is at distance ‘a’ from point D. Using equation (S F)_{X_{2} X_{2}^{1}} ,

(S F)_{E}=1000 – 4000 + 2000 (a − 1)

0 = −3000 + 2000 (a − 1)
a = 2.5 m

Bending Moment Diagram:

(B M)_{x_{1} x_{1}^{1}}=-1000 \cdot x_{1}

(B M)_{D}=0,(B M)_{B}=-1000  Nm

(B M)_{x_{2} x_{2}^{1}}=-1000 \cdot x_{2}+4000\left(x_{2}-1\right)-2000\left(x_{2}-1\right) \frac{\left(x_{2}-1\right)}{2}

(B M)_{B}=-1000  Nm

(B M)_{C}=-1000 \times 3+4000(3-1)-2000 \frac{(3-1)^{2}}{2}

= +1000 Nm

(B M)_{E}=-1000 \times 2.5+4000(2.5-1)-2000 \frac{(2.5-1)}{2}

= +1250 Nm

(B M)_{X_{3} x_{3}^{1}}=-1000 \cdot x_{3}+4000\left(x_{3}-1\right)-(2000 \times 2)\left(x_{3}-2\right)

(B M)_{C}=−1000 × 3 + 4000 (3 – 1) – 4000 (3 − 2)

= 1000 Nm

(B M)_{A}=−1000 × 4000 (4 − 1) – 4000 (4 − 2)

= 0

To determine Point of Contraflexure,

Put (B M)_{X_{2} X_{2}^{1}}=0

-1000 \cdot x_{2}+4000\left(x_{2}-1\right)-2000 \frac{\left(x_{2}-1\right)^{2}}{2}=0

-x_{2}+4\left(x_{2}-1\right)-\frac{2\left(x_{2}-1\right)^{2}}{2}=0

-x_{2}+4 x_{2}-4-\left(x_{2}^{2}+1-2 x_{2}\right)=0

+3 x_{2}-4-x_{2}^{2}-1+2 x_{2}=0

-x_{2}^{2}+5 x_{2}-5=0

x_{2}^{2}-5 x_{2}-5=0

x_{2}=\frac{+5 \pm \sqrt{(+5)^{2}-4 \times 1 \times 5}}{2},

x_{2}=\frac{5 \pm \sqrt{5}}{2}

x_{2}=3.62 m and 1.38 m where first value does not lie for x_{2} as shown in BMD, thus
x_{2}= 1.38 m