A simply supported beam of length $3 L$ is subjected to the loading shown in the figure.
It is given that P=1 N, L=1 m and Young's modulus E = 200 GPa. The cross-section is a square with dimension 10mm x 10 mm. The bending stress (in Pa) at the point A located at the top surface of the beam at the distance of 1.5 L from the left end is

0

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Solution

The correct option is A 0
$\sum F_{v} = 0$
$R_{1} + R_{2}$ =0
$\sum M_{1}$ =0
$3 R_{2} L - 2 P L + P L = 0$
$R_{2} = \frac{P}{3}; R_{1} = \frac{- P}{3}$
B.M at A $M_{A} = (R_{1} \times 1.5 L) + (P \times 0.5 L)$
$= (\frac{- P}{3} \times 1.5 L) + 0.5 P L = 0$
$∴$ B.M. at A=0
$∴$ Bending stress = 0 Pa

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The correct option is A 0∑Fv=0 R1+R2 =0 ∑M1 =0 3R2L−2PL+PL=0 R2=P3;R1=−P3 B.M at A MA=(R1×1.5L)+(P×0.5L) =(−P3×1.5L)+0.5PL=0 ∴ B.M. at A=0 ∴ Bending stress = 0 Pa