# Drawing the Forces for this Structure?

1. Feb 25, 2012

### mneox

The question is to draw the shear and moment diagram for this frame structure.

I think I can do the diagram part alright, but I'm confused as to how I can draw the FBDs for this structure. What I'm thinking is to split into two members, AB and BC.

For member AB, I know that there's two reaction forces at A, and (I think) also two reaction forces at B? Like so:

I'm pretty sure that the FBD for member AB I got right. However, I'm unsure of how to draw the FBD for member BC. How am I supposed to orient the reaction forces at C? I tried looking for examples in my textbook similar to this case but couldn't find any. Here's my attempt:

Is this right? I tried solve for my unknown forces from these two FBDs and while member AB seemed fine, I swear I am doing something wrong for member BC. The answer key listed the max shear force to be 11.7, which I achieved in member AB. But when I solved for shear forces in member BC, I got a shear force exceeding that of 11.7, so that means I must have done something wrong. I have no idea where though.

2. Feb 25, 2012

### PhanthomJay

Your FBDs are good for calculating the reaction and joint loads. Be sure in the diagonal member to convert them into loads perpendicular (shear)and parallel (axial) to the member. Also be sure when drawing the shear and moment diagrams for the diagonal member that you get rid of that resultant force from the triangularly distributed load and show the actual distributed load instead.

3. Feb 25, 2012

### mneox

Ah yeah earlier I had thought about converting the loads on the diagonal member into perpendicular and parallel loads, but I was not sure if I was approaching it correctly. Maybe it's just late and my brain is half dead, but I can't seem to think of how to do it. Would I use trig? Cause that's what I did the first time around and got answers that seemed off.

And I knew to not use the resultant force in the diagram, but thanks for the reminder anyways!

4. Feb 25, 2012

### PhanthomJay

You need geometry and trig and a good sketch. For forces Bx and By, for example, First resolve Bx into its vector components parallel and perpendicular to incline. The parallel comp is Bx cos 45 and the perp comp is Bx sin 45. Then do the same for By, where its parallel and perp comps are found in the same manner. Then add up the perp comps of Bx and By to get the force at B perp to the diagonal member. Watch plus and minus signs

As an example, suppose Bx was 10 and By was 20. The shear force at B perp to the diagonal member would be 7.07 + 14.14 = 21.21, but watch plus and minus signs.

5. Feb 26, 2012

### mneox

Thanks Jay. I successfully got the answer.