- #1
Kasthuri
- 17
- 0
Homework Statement
For the frame shown (see attached picture) , work out the horizontal reaction at B (ie find Bx)
Homework 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)
The Attempt at a Solution
I thought this was easy but for some reason everytime I equate the forces, Bx gets canceled 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!