# Homework Help: Working out support reaction of a frame - Statics

1. Oct 7, 2012

### Kasthuri

1. The problem statement, all variables and given/known data

For the frame shown (see attached picture) , work out the horizontal reaction at B (ie find Bx)

2. Relevant equations

EQUILIBRIUM EQUATIONS:

sum of moment = 0 (CCW is positive)
sum of vertical forces = 0 (up is positive)
sum of horizontal forces = 0 (left to right is positive)

3. The attempt at a solution

I thought this was easy but for some reason everytime I equate the forces, Bx gets cancelled out.

[sum of moments at A = 0];
-38(0.6) + By(0.3) = 0
By = 76 kN

[sum of moments at B = 0];
-Ay(0.3) + 38(0.6) = 0
Ay = 76 kN

[sum of moments at D = 0];
Ax(0.6) + Bx(0.6) + By(0.3) = 0 (***)

[sum of moments at C = 0];
(-38)(0.2) + Ax(0.4) + Bx(0.4) = 0

so, Ax = 19 - Bx

substitute Ax = 19 - Bx in (***) :
(19 - Bx)(0.6) + Bx(0.6) + By(0.3) = 0

but By = 76 kN so:

(19 - Bx)(0.6) + Bx(0.6) + 76(0.3) = 0

BUT Bx just cancels out??

I have no clue how to approach this question any other way
Any help/guidance would be great!
Thanks!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Oct 7, 2012

### Kasthuri

I got it :)
I realised that when taking the moment about C, the force Bx does not need to included.
So Bx = -57 kN

3. Oct 8, 2012

### PhanthomJay

NO. By produces a moment also.

What you mean to say is that since member BC is a 2-force member with the force BC directed along the axis, then the force BC produces no moment about C.

4. Oct 13, 2012

### Kasthuri

Thanks for that PhanthomJay! :) My exam is very very soon and I'm glad you cleared that up for me.