Question 9.10: Figure 9.18 shows a beam pivoted at A and simple supported a...

\Sigma Y =0, R_{A}+R_{B}= Area of triangle

=\frac{1}{2} \times 3 \times 12

= 18                                                                            ….. (1)

\Sigma M _{ A }=0,

R_{B} \times 3=18 \times \frac{2}{3} \times 3

R_{B} = 12 kN
R_{A} = 6 kN

In this question we have to consider section X_{1} X_{1}^{1} from leſt side A at distance x_{1}. The reason is if section XX is considered from right side the remaining area will be trapezium whose centroid would be difficult to consider.

Shear Force Diagram:

(S \cdot F \cdot)_{X_{1} X_{1}^{1}}=+R_{A}- Load of UVL upto x_{1} from A

= 6 – Area of triangle up to x_{1}

=\left(6-\frac{1}{2} \times x_{1} \times h\right)

Now height, h of UVL at x_{1} distance can be determined by three ways:
(i) By law of similar triangles: \frac{h}{x_{1}}=\frac{12}{3} \text { i.e., } h=4 x_{1}

(ii) By trigonometry: if ∠CAB = α

then \tan \alpha =\frac{h}{x_{1}} and \tan \alpha =\frac{12}{3}

\text { i.e., } \frac{h}{x_{1}}=\frac{12}{3} \text { i.e., } h=4 x_{1}

(iii) By method of proportionality:
The intensity of load at 3 m from A to B = 12 kN
thus the intensity of load at x_{1} distance, i.e., h=\left(\frac{12 \times x_{1}}{3}\right)
Substituting value of h in equation (1),

(S F)_{X_{1} X_{1}^{1}}=6-2 x_{1}^{2}

(S F)_{A}=+6  kN

(S F)_{B}=6-\frac{1}{2} \times 3 \times 4 \times 3=-12  kN

Let S.F. changes its sign at distance ‘a’ from the end A,

(S F)_{C}=6-\frac{1}{2} \times a \times 4 \times a

0 = 6 – 2a²

a = 1.732 m

Bending Moment Diagram:

(B M)_{X_{1} X_{1}^{1}}=+R_{A} \times x_{1}-\left(\frac{1}{2} \times x_{1} \times 4 x_{1}\right) \times \frac{x_{1}}{3}

=\left(6 \cdot x_{1}-\frac{2 \cdot x_{1}^{3}}{3}\right)

( BM )_{ A }=0

(B M)_{B}=6 \times 3-\frac{2}{3}(3)^{3}

= 0

(B M)_{C}=6 \times 1.732-\frac{2(1.732)^{3}}{3}

= 6.93 kNm