How to find Ay and Ma reactions in a structural analysis problem?

[ride5150]

to figure out the Ay and Ma reactions, i tried to find the effect of the 9kn/m force. i did this by integrating:

integral from 0 to 8: 9x

i found this to be 288kn

for the moment caused by the 9kn/m force:

integral from 0 to 8: 9(x^2)

i found this to be 1,535kn*m clockwise

i then added this to the moment about A caused by the 100kn force in the positive x direction:

(100kn)(12m) + 1,536kn * m= 2,736 kn*m

so my answer for the first part would be:

reactions:

Ax= -100kn
Ay= 288kn (in positive y direction)
Ma= 2,736 kn * m counterclockwise.

am i correct?

 

[ride5150]

no one?

 

[SteamKing]

I think you need to look at the distributed load more carefully.

The equivalent force is 9 * 8 / 2 = 36 kN, not 288 kN

Similarly, the moment of this load appears to be wrong.

 

[ride5150]

I think you need to look at the distributed load more carefully.

The equivalent force is 9 * 8 / 2 = 36 kN, not 288 kN

Similarly, the moment of this load appears to be wrong.

right, its 36kn directly inline with A, but wouldn't the distributed load have an effect on Ay as well?

also, since its a distributed load, wouldn't i have to find the moment caused by it through integration?

 

[SteamKing]

You can, but your integration set up is flawed. The distributed load from B to C can be replaced by a force and a couple acting at B. Similarly, the 100 kN load at C can be shifted to B without changing the reactions at A.

 

[pongo38]

Why would the vertical distributed load cause a horizontal reaction at A? When you integrate let x=0 at C and x=8 at B. It's easier that way.

 

[Sohan]

Seems like you are complicating the problem by bringing integration into the picture.
Firstly you can can find the resultant of the distributed load by doing as follow:
Resultant:0.5*9*8=36kN(basically the area of the distributed load i e a triangle in this case)

Now to find the location where this resultant acts, you simply need to find the centroid of the triangle from side that is denoted by 9kN. it's given by h/3 i.e 8/3=2.67m from the left.
now use the equilibrium equation of statics
ƩFy=0
Ay-36=0 ie Ay=36kN
ƩFx=0
Ax+100=0 ie Ax=-100kN(the negative sign indicated the reaction is towards the left and not towards the right if u had imagined it to be so)
ƩMA=0(moment about A)
M+36*2.67+100(12)=0 ie M=-1296.12kN-m(negative sign indicates a anti clockwise moment ,which makes intuitive sense since clockwise moments give a anti clockwise reaction at the support A)
Hence u got all the reactions .