Calculating a force on a member (Statics)

[bardia sepehrnia]

I isolated the member ABC and drew the free body diagram:

α is then calculated using inverse tan: Tan^-1=(6.25+15)/50=23.03
Then force of member BD on the joint can be found by sum of all moments around point A.

Then Ax is calculated which is equal to BD×Cos(α)=235.2×Cos(23.03) Ax=216.48
Then I draw free body diagram of the block with joint A:

Then P=A_x=216.48.
I cannot to the second part of this question and calculate the force on the member BC!?
I can calculate Ay in first free body diagram however would that at all help me find the force on Member BC? Is the resultant force of vectors Ay and Ax the force acting on member AB?

 

[CivilSigma]

To find the force in member BC, try cutting the member AC just after point B by a bit.

Then for equilibrium to take place, there will be a new internal force introduced when you "cut". This force is the internal force of member BC. Take moments about point A and then solve for this force.

 

[bardia sepehrnia]

To find the force in member BC, try cutting the member AC just after point B by a bit.

Then for equilibrium to take place, there will be a new internal force introduced when you "cut". This force is the internal force of member BC. Take moments about point A and then solve for this force.

I don't understand. Can you elaborate? Am I not supposed to firs calculate the force on AB member or is that irrelevant?

 

[CivilSigma]

It's irrelevant to find force in member AB.

You need to "cut" the rod somewhere between point B and C if you want to find force BC.

We know there is a force in member BC. Imagine you cut the member somewhere from in between point B and C - call it point W.

Now you have two rods: A-B-W and W-C. Since the original rod was under equilibrium, the internal force resulting from "this cut" in member A-B-W and W-C must be equal in magnitude and opposite in direction. This internal force also represents the axial force in member BC.


Here are a few links that may be helpful :

http://www2.hawaii.edu/~takebaya/lessons/method-of-sections.pdf

 

[PhanthomJay]

I really think that there is a typo in the problem, that it means to ask for the force in BD rather than the force in BC, which is only a part of the member ABC. BC has both axial and shear forces, as well as a varying bending moment. But anyway, as suggested, cut ABC just to the right of B, then look at the FBD of BC to find the force, or axial and shear components of that force.