Question

Find support reactions at A and B and then calculate the axial force N, shear force V, and bending moment M at mid-span of AB. Let L 5 4 m, q0 5 160 N/m, P 5 200 N, and M0 5 ? 380 N m.

Answer 1

Answer:

$A=198N$,  $B=-38.33N$, $N=0$, $V=118N$, $M=-37.33\ Nm$

Explanation:

It is given that $L=4m, \ q_{0}=160N/m,\ P=200N,\ M_{0}=380 N-m$.

Thus taking equilibrium in x-direction (horizontal)

∑$F_{x}=0$ ⇒ $B_{x}=-\frac{3}{5}* P=-\frac{3}{5} *200 N=-120N$

Taking Equilibrium moments for point A.

Giving ∑$M_{A} =0$

Thus

$B_{y}*L+\frac{4}{5}*P*(L+\frac{L}{2})=M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3}$

So it can be written as

$B_{y}=\frac{1}{L}*(M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3} -\frac{4}{5} *P*(L +\frac{L}{2}))$

$B_{y}=\frac{1}{4}*(380+\frac{1}{2}*160*4*\frac{4}{3} -\frac{4}{5} *200*(4 +\frac{4}{2}))$

$B_{y}=-38.33N$

Now taking Equilibrium in y-direction (Vertical).

∑$F_{y} =0$

thus it becomes as

$A_{y}+B_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L$

it can be written as

$A_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L-B_{y}$

$A_{y}=-\frac{4}{5}*200+\frac{1}{2} *160*4-(-38.33)$

$A_{y}=198N$

Now, As for equilibrium in x=direction

∑$F_{x}=0$ ⇒ $N=0$

And for equilibrium in y- direction

∑$F_{y} =0$

$A_{y}=V+\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}$

it can be written as

$V=A_{y}-\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}$

$V=198-\frac{1}{2}*\frac{160}{2} *\frac{4}{2}$

$V=118N$

Now taking equilibrium of moments about A,

∑$M_{A} =0$

$M_{0}+M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}$

it can be written as

$M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}-M_{0}$

$M=\frac{1}{2}*\frac{160*4}{4} *\frac{2*4}{3*2}+118*\frac{4}{2}-380$

$M=-37.33\ Nm$