Answer :
Answer:
[tex]A=198N[/tex], [tex]B=-38.33N[/tex], [tex]N=0[/tex], [tex]V=118N[/tex], [tex]M=-37.33\ Nm[/tex]
Explanation:
It is given that [tex]L=4m, \ q_{0}=160N/m,\ P=200N,\ M_{0}=380 N-m[/tex].
Thus taking equilibrium in x-direction (horizontal)
∑[tex]F_{x}=0[/tex] ⇒ [tex]B_{x}=-\frac{3}{5}* P=-\frac{3}{5} *200 N=-120N[/tex]
Taking Equilibrium moments for point A.
Giving ∑[tex]M_{A} =0[/tex]
Thus
[tex]B_{y}*L+\frac{4}{5}*P*(L+\frac{L}{2})=M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3}[/tex]
So it can be written as
[tex]B_{y}=\frac{1}{L}*(M_{0}+\frac{1}{2}*q_{0}*L*\frac{L}{3} -\frac{4}{5} *P*(L +\frac{L}{2}))[/tex]
[tex]B_{y}=\frac{1}{4}*(380+\frac{1}{2}*160*4*\frac{4}{3} -\frac{4}{5} *200*(4 +\frac{4}{2}))[/tex]
[tex]B_{y}=-38.33N[/tex]
Now taking Equilibrium in y-direction (Vertical).
∑[tex]F_{y} =0[/tex]
thus it becomes as
[tex]A_{y}+B_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L[/tex]
it can be written as
[tex]A_{y}=-\frac{4}{5}*P+\frac{1}{2} *q_{0}*L-B_{y}[/tex]
[tex]A_{y}=-\frac{4}{5}*200+\frac{1}{2} *160*4-(-38.33)[/tex]
[tex]A_{y}=198N[/tex]
Now, As for equilibrium in x=direction
∑[tex]F_{x}=0[/tex] ⇒ [tex]N=0[/tex]
And for equilibrium in y- direction
∑[tex]F_{y} =0[/tex]
[tex]A_{y}=V+\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}[/tex]
it can be written as
[tex]V=A_{y}-\frac{1}{2}*\frac{q_{0} }{2} *\frac{L}{2}[/tex]
[tex]V=198-\frac{1}{2}*\frac{160}{2} *\frac{4}{2}[/tex]
[tex]V=118N[/tex]
Now taking equilibrium of moments about A,
∑[tex]M_{A} =0[/tex]
[tex]M_{0}+M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}[/tex]
it can be written as
[tex]M=\frac{1}{2}*\frac{q_{0}*L }{4} *\frac{2*L}{3*2}+V*\frac{L}{2}-M_{0}[/tex]
[tex]M=\frac{1}{2}*\frac{160*4}{4} *\frac{2*4}{3*2}+118*\frac{4}{2}-380[/tex]
[tex]M=-37.33\ Nm[/tex]
