A-Frame, method of joints.

[takio]

Hello

The two A-frames stands side by side and linked by the top cross beam and middle cross beam.
and the middle bar has a rod sticking out at the center. Force is applied at a distance D from the middle bar.

I wanted to calculate the internal forces in the beams AC, CB, and AB.
I think I can calculate forces using 1)+2), using truss analysis (method of joints). Am I on right track?

For 1)
F= W/2

For 2)
M=WD,, But I have also M=W/2*L, and theses two has different directions. This gets me confused.

Would you give me some ideas on the calculations?

Thank you

 

[rb1957]

solve FD as a free body first.

note that your triangle trusses are not true trusses. why?
hint ... members in a true truss carry axial loads only.

another day in paradise, or is paradise one day closer ?

 

[rb1957]

"M=W/2*L" ... there's a lot of short hand going on here. You're dealing with two different moments ...

First "L" is the length of the beam DF and "l" shown on your sketch is L/2
then the moment in DF is W/2*L/2

if l is the distance to the load, then what is the length of DF ? 2l ??
then the moment in DF (due the direct force W) is W/2*l

But the offset moment (due to the load W being offset from DF) is Wd
This will get reacted at the two ends ... if DF is 2l long then the reaction at D will be Wd/2

another day in paradise, or is paradise one day closer ?

 

[takio]

Hello rb1957

Thank you so much for your comments !
I am sorry that I didn't specify the other length. I edited it.
and, I am s

Ideal truss
- Pin-connected : no moments.
- Internal force in members : only compression or tension, no shear
- Assumptions : frictionless joints, nodes are only at the ends of members, loads are applied at nodes.

I think that's why it not a true truss, and why I was wondering how I can calculate this kind of problems.
and I also found some examples that loads are not applied at nodes. links below.

[URL unfurl="true"]https://engineeringlibrary.org/reference/formulas-for-simple-frames-air-force-stress-manual[/url]
see Table 5-5
Rectangular Forces and Moments on Triangular Frames

[URL unfurl="true"]https://www.youtube.com/watch?v=rp4kBguEZIA[/url]

Thank you

 

[rb1957]

'k

so you've correctly modified the load, W, in it's offset position, d, to be a direct load W and a moment Wd onto the beam.
So you can treat these two as individual loads, and superimpose the results.

1) the direct load W ...
the reactions are W/2 as the load is applied mid-span.
the moment this creates in DF, W/2*l, doesn't impact the the triangular trusses (does it ?)
So DF applies a load W/2 to the triangular truss. How can you modify this load to make it easier to work with in the truss ?
You can "just as easily" solve CB as a free body. Your immediate problem is going to be moment effectiveness of the joints at B and C !
You'll also see immediately another difficulty ... I'll wait until you notice this !

You can easily solve the triangle as a free body and so get the reactions at A and B.
This may help solve CB, depending on the joint moment effectivity ...
This will, when you look at AC, show you some part of the solution, or more likely just give you some bread crumbs leading you into the analysis trap !?

2) then same for the moment ...

another day in paradise, or is paradise one day closer ?

 

[takio]

Hello

Thank you for your comment.
I have calculated the frame, using the moment and direct load.
I hope this makes sense.

Thank you

 

[rb1957]

stop.

the first thing to do is the calculate the external reactions (at A and B) before you try to figure out the internal reactions.

the 2nd thing would be to make components out to the load. IF (and I say IF) the joints are pinned then there is only 1 statically determinate solution.

Your sum Mb is wrong ... Fac has a moment arm about b.

another day in paradise, or is paradise one day closer ?