- #1
Dell
- 590
- 0
find the deflection of the beam at point B
i drew the free body diagram of the beam and found that
Ay=0
Ax=qL
Cy=qL
Cx=qL
M(x)= (qL)*x - (qx)*x/2 +(qL)*<x-L/2>
M(x)=(qL)x -(q/2)x2 +(qL)<x-L/2>
EIΦ(x)=(qL/2)x2 -(q/6)x3 +(qL/2)<x-L/2>2 + C1
EIY(x)=(qL/6)x3 -(q/24)x4 +(qL/6)<x-L/2>3 + C1*x + C2
EIY(0)=0= 0+ C2
===>c2=0
but i need another condition to find C1, i think i need to use the deflection of the bar DC somehow, but I am not quite sure how, i tried using triangles and saying that the deflection of point C is the same as the deflection of the bar DC*sin(45) but that didnt work
i drew the free body diagram of the beam and found that
Ay=0
Ax=qL
Cy=qL
Cx=qL
M(x)= (qL)*x - (qx)*x/2 +(qL)*<x-L/2>
M(x)=(qL)x -(q/2)x2 +(qL)<x-L/2>
EIΦ(x)=(qL/2)x2 -(q/6)x3 +(qL/2)<x-L/2>2 + C1
EIY(x)=(qL/6)x3 -(q/24)x4 +(qL/6)<x-L/2>3 + C1*x + C2
EIY(0)=0= 0+ C2
===>c2=0
but i need another condition to find C1, i think i need to use the deflection of the bar DC somehow, but I am not quite sure how, i tried using triangles and saying that the deflection of point C is the same as the deflection of the bar DC*sin(45) but that didnt work