Solving Portal Frame Using Slope Deflection Equations

can't read your script. How are you accounting for the redundancy (in your hand calc) ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Is this any better to view?

clearer (easier to read) but I can't read (understand) python ... so let someone else have a go.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Ah right, haha, no worries then, rb1957. Hopefully someone else can help.

looking at your joint equilibrium ... M12+M22 = 0 ... without knowing what these terms are (???) but looking at the pic above, should it be M12-M22 = 0 ?

does M12 = M22 ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

What I learned at Uni is the internal equilibrium equations at the free joints (where these joints have non zero displacements/rotations), the sum of the moments equals zero.

And yes, M12 = M22. You can see this in the BeamGuru diagram. It's just BeamGuru calculated M12 and M22 as -145.322 kNm each; I'm getting 35.277 kNm.

on a quick cursory review it looks like you are missing the psi components of the joints to account for the fact that this frame can experience sway:

if M12 = M22 then M12+M22 .NE. 0. Your equilibrium equations are driving M12 = -M22, no?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Let's assume the frame is a three hinged arch...hinge at 1,3 and 5 (big red dot). The structure is statically determinate and it is easy to calculate the moment at 2 to be 228.5 (I took the load to be gravity load, not sloping as shown on the OP sketch).



The so called "hand" calculation provided is not easily understood, at least not by me. The slope deflection method would not be my choice, but it is a perfectly sound method. Since the structure and the loading are symmetrical, consider half the structure (from point 1 to 3). Rotation is zero at points 1 and 3. Horizontal deflection is zero at point 3.

Calculate rotation and deflection at point 3 with only the fixed moment at point 1. Apply horizontal force H and moment M at point 3, then solve to bring theta 3 and delta 3 to zero.

Another method, if you just want to know the answer, is to consult the book by Kleinlogel, which considers several different kinds of load on a gable frame.

Thanks, BAretired. I have however solved the portal frame with pinned supports using the unit load/virtual work method. I was very happy, but then when I added a lateral load (to simulate wind) I quickly realised that the frame could not take it, and there was massive horizontal deflections - because the pinned supports don't resist the rotation. Similarly, once I solved this problem I was going to add a lateral load.

I think Celt83 pointed out the problem. I was not including the psi terms from the slope deflection equations. I have only ever analysed structures where psi is zero. I had a look into it yesterday, but I think I might as well stick to using the direct stiffness method, as its quite complicated to find psi for this particular structure.

A three hinged arch is more flexible than a rigid gable frame, but the deflections are not "massive". Not many engineers would consider the columns fixed at the base. They are not hinged either, so perhaps a rotational spring would be ideal.

For purposes of a homework assignment, a fixed end can be assumed. I would agree that the stiffness method is best in a work environment, preferably with the aid of a computer. To perform a true hand calculation for this frame is onerous and likely to result in arithmetic errors.