Question

A simply supported beam 6m long is carrying a uniformly distributed load of 5kN/m over a length of 3m from the right end.Draw the S.F and B.M diagrams for the beam and also calculate the maximum B.M on the section​

Answer 1

Answer:

In case of simply supported beam, bending moment will be zero at supports. And it will be maximum where shear force is zero. Bending moment at point B = M(B) = R1 x Distance of R1 from point B.

Answer 2

Given:

l=6m

W=5kN/m

a=3m

$\begin{aligned}R_{B} \times 6 &=(5 \times 3) \times 4.5=67.5 \\R_{B} &=\frac{67.5}{6}=11.25 \mathrm{kN} \\R_{A} &=(5 \times 3)-11.25=3.75 \mathrm{kN}\end{aligned}$

Share force

$\begin{array}{l}F_{A}=+R_{A}=+3.75 \mathrm{kN} \\F_{C}=+3.75 \mathrm{kN} \\F_{B}=+3.75-(5 \times 3)=-11.25 \mathrm{kN}\end{array}$

S.F and B.M diagrams are in attachment