# Reaction at a point

## Homework Statement

http://imgur.com/AjUdF

Segments AB and BC of the beam ABC are pin connected a small distance to the right of joint B. Axial loads act at A and at the midspan of AB. A concentrated moment is applied at joint B.
Find the reaction at A, B, and C.

ƩF

ƩM= (Fi)(di)

## The Attempt at a Solution

I split the bar at joint B and drew two separate free-body diagrams. For the right side of the bar, I have Cy and Cx at C, Bx from the pin, and the moment about B (100 ft-lb).

I solved:

ƩMb=10Cy-100
Cy=10

From there I took the moment of the entire bar about point A, and found By=-20 lb, which I don't think is correct.

I think I'm mostly confused because there are no downward y forces in this problem, and I didn't think rollers could ever have a downward force.

Thanks!

## Answers and Replies

PhanthomJay
Science Advisor
Homework Helper
Gold Member
1. Homework Statement [/b]

http://imgur.com/AjUdF

Segments AB and BC of the beam ABC are pin connected a small distance to the right of joint B. Axial loads act at A and at the midspan of AB. A concentrated moment is applied at joint B.
Find the reaction at A, B, and C.

ƩF

ƩM= (Fi)(di)

## The Attempt at a Solution

I split the bar at joint B and drew two separate free-body diagrams. For the right side of the bar, I have Cy and Cx at C, Bx from the pin, and the moment about B (100 ft-lb).

I solved:

ƩMb=10Cy-100
Cy=10

From there I took the moment of the entire bar about point A, and found By=-20 lb, which I don't think is correct.

I think I'm mostly confused because there are no downward y forces in this problem, and I didn't think rollers could ever have a downward force.

Thanks!
In this problem, the roller supports are free to slide in the x direction, and thus cannot support any load in that direction, but they are quite capable of supporting vertical forces in the y direction. The applied moment is applied at the support joint B, not at the pin to the right of B. When you look at a FBD of the right section from the pin to C, you have at most just x and y forces at the pin and at C, no applied moment . By using the equilibrium equations, you can find Cy and the vert pin force rather easily.

Then look at the left section from A to the pin. Use the equilibrium equations again on this section to solve for Ay , By, and the horiz force at the pin. Then back to the right section for Cx.

Sorry, but how would I find any forces on the right side? Wouldn't my equilibrium equations just be:

Fy= By+Cy=0
Mb=(10)(Cy)=0

PhanthomJay
Science Advisor
Homework Helper
Gold Member
Sorry, but how would I find any forces on the right side? Wouldn't my equilibrium equations just be:

Fy= By+Cy=0
Mb=(10)(Cy)=0
Yes, more or less, but remember you are not looking at the joint B at the support in this FBD, you are looking at the pin just to the right of B, call it B'. So your equations should be
Sum of forces in y direction = B'y + Cy = 0, and
Sum of moments about B' = 10Cy = 0.

from this last equation, Cy = ???
And then from the first, B'y = ???
You still have B'x and Cx forces to contend with.

So they're both just equal to zero?

PhanthomJay
Science Advisor
Homework Helper
Gold Member
So they're both just equal to zero?
Yes....