Working out support reaction of a frame - Statics

[Kasthuri]

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!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Kasthuri]

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

 

[PhanthomJay]

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

NO. By produces a moment also.

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

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.

 

[Kasthuri]

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

 

[rezeew]

Hello,

Thank you for reaching out for assistance with your homework problem. I understand the importance of approaching problems in a systematic and logical manner. Let's break down the problem and go through it step by step.

First, let's review the given information. The problem asks you to find the horizontal reaction at point B of a frame, as shown in the attached picture. The frame is a static system, meaning that all forces are balanced and it is not moving. This is a key piece of information that we will use to solve the problem.

Next, let's review the equations provided. The equilibrium equations state that the sum of forces and moments in a static system must equal zero. This means that the forces acting on the frame must be balanced and there must be no net rotation. These equations will be our main tools in solving the problem.

Now, let's take a closer look at your attempt at a solution. You correctly used the equilibrium equations to find the vertical reactions at points A and B. However, when you tried to use the moments equation to find the horizontal reaction at point B, you ran into a problem. This is because you were using the same equation twice, which is not necessary. Instead, we can use the moments equation to solve for the horizontal reaction at point A, and then use that result to find the horizontal reaction at point B.

So, let's start by using the moments equation at point A, as you did in your attempt at a solution. This equation is given by:

sum of moments at A = 0

We can plug in the known values from the problem:

-38(0.6) + By(0.3) = 0

This gives us:

By = 76 kN

Next, we can use this result to find the horizontal reaction at point A. We can use the sum of forces equation at point A, which is given by:

sum of horizontal forces = 0

We can plug in the known values from the problem:

Bx + Ax = 0

Since we know the value of By, we can substitute it in for Ax:

Bx + 76 = 0

This gives us:

Bx = -76 kN

Now, we can use this result to find the horizontal reaction at point B. We can use the sum of moments equation at point B, which is given by:

sum of moments at B = 0

We