Reaction Forces at A, B, & C in Beam ABC

[giacomh]

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.

Homework Equations

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!

 

[PhanthomJay]

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.

Homework Equations

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.

 

[giacomh]

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]

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.

 

[giacomh]

So they're both just equal to zero?

 

[PhanthomJay]

So they're both just equal to zero?

Yes...