Second order effects

[user]

How can I activate second order (P-delta) effects in SeismoStruct analysis?

[seismosoft]

There is no need for such an activation option since Geometric Nonlinearity is an intrinsic part of the formulation of all SeismoStruct elements and analysis types (with the obvious exception of Eigenvalue analysis). Hence, second order effects are always fully accounted for.

SeismoSoft Support

[gmenichini]

Dear SeismoSoft Support
I have to analyze the second order effect in a prefabricated single-storey building, so in order to know how SeismoStruct take in account this effect, I started with a simple structure.
I designed a cantilever beam with eccentric compression load at the top, how you can see in the figure below:
http://it.tinypic.com/r/augp5t/8
The cantilever is constituted by a elastic frame element, with 5m of height and a square section with side of 0,1m, then i discretized the beam by subdividing it in 10 elements
For modeling the eccentricity I used a rigid link between the node where the load is applied and the node of beam’s top.

I applied a compression load of 140 KN and checked the “geometric nonlinearity” options; when I run analysis obtained the follow x displacement on top 0,6304m.
I tried to compare this result whit a hand calculated one found by solving the displacement’s equilibrium differential equation written in deformed configuration.

Well, the result that I found with the hand-calc is: x displacement on top equal to 0,7240m

Why the difference is so marked? Maybe I did something wrong in modeling?

[huffte]

Hi gmenichini.

The first thing which comes to my mind is to:

(a) make certain the the value for E used in your hand calculations exactly matches the value used in your elastic material in the SeismoStruct model.

(b) run the SeismoStruct model with geometric nonlinearity unchecked to evaluate the amplification effect of geometric nonlinearity on displacement.

(c) make a hand calculation for what the displacement should be ignoring geometric nonlinearity.

These steps should help pinpoint the problem, whether it is in the model or in the hand calculations.

If these don't reveal any resolution, you may certainly e-mail your model to either myself or the SeismoSoft support team.

Best of luck gmenichini.

[gmenichini]

Hi huffte and thanks for the answer
a) I checked the elastic modulus E both in SeismoStruct's elastic material and in my hand calculations and I used the same value 2,0E+8 kPa

b) I ran the Seismostruct model with geometric nonlinearity unchecked and I found that the x displacement at the top is 0,1050m instead of 0,6304m with geometric nonlinearity activated.

c) I made an hand calculation without geometric nonlinearity through the known relation for a cantilever with moment on top:
x disp= (M*H^2)/2*EJ
where:
M= is the moment on top equal to axial load*eccentricity (0,1m)
H =is the height of cantilever
J=is the moment of inertia

For the cantilever beam in question the result is x disp=0,1050m exactly equal to SeismoStruct result.

[huffte]

OK. So, let's summarize:

SeismoStruct nonlinear gives a top displacement of 0.6304 meters.
SeismoStruct linear gives a top displacement of 0.1050 meters.
Hand calculations linear gives a top displacement of 0.1050 meters.
Hand calculations nonlinear gives a top displacement of 0.7240 meters.

May I ask what closed form solution for nonlinear hand calculations you used?

I used another analysis program and came up with a displacement of 0.7555 meters for the nonlinear case, and 0.1050 meters for the linear case. But I am not sure how that program is treating geometric nonlinearity.

I also used a code-based amplification factor calculated as 1/(1-P/Pcr) = 1/(1-140/164.493) = 6.716, which gives a displacement of 0.105(6.716) = 0.7052 meters, which is certainly not exact. My calculation of PCr is not exact either. I get 0.6996 meters top displacement with my Seismo model, which uses the applied load and moment with no rigid link.

I haven't seen the model, but it is possible that you may have introduced some artificial stiffness at the top of the element with the rigid link. Why not just apply a load of 140 kN simultaneously with a moment of 140X0.1 = 14 kN-m at the top and do away with the rigid link?

It is also possible that the hand calculations for the nonlinear case are not exact - that is why I asked for the reference.

I'll be curious to learn if that was the problem gmenichini.

[gmenichini]

Hi huffte
Well, I try to explain the closed form solution that I used for my non linear hand claculation
With reference to the follow figure
http://it.tinypic.com/r/jqjqjk/8
we can write the external and internal moment’s equilibrium equation:
Internal moment: Mi (z)=EJ*v’’(z)
External moment: Me(z)=P(e0+a-v(z))
So we have:
EJ*v’’(z)= P(e0+a-v(z))
and then
v’’(z)+B^2*v(z)=B^2*(e0+a)
where the parameter B^2 is:
B^2=P/EJ
With the boundary conditions
v(0)=0
v’(0)=0
v(L)=a
we can integrate the differential equation and the solution is:
v(z)=[e0*(1-cos(Bz))]/cos(BL)
and so the displacement on top is
v(L)= [e0*(1-cos(BL))]/cos(BL)

The reference text of this solution is “Migliacci A., Mola F., Progetto agli stati limite delle strutture in C.A., Milano, Masson Italia Editore, 1985”

I followed your advice and I modified my SeimoStruct model by removing the rigid link, and I applied a axial load (140 kN) and a moment (14 kNm) and I found 0,6877m of x displacement on top, the difference from your result is may be given by the specific weight of material that I have considered equal to 0.

Also I have noticed that this problem is more evident if I increase the axial load.
By considering the model without rigid link, I have:

P=100 kN M=10 kNm
top x disp.:
hand calc: 0,1948m
sesimo calc: 0,1939m

P=120 kN M=12 kNm
top x disp.:
hand calc: 0,3402m
sesimo calc: 0,3364m

P=150 kN M=15 kNm
top x disp.:
hand calc: 1,3137m
sesimo calc: 1,0961 m

[huffte]

Yes. I see your point gmenichini. I'll also not in particular, that the solution seems to converge as the number of subdivisions in to elements increases. If I run with a single element I get something like .35m and my .699m was obtained with 16 elements.

[gmenichini]

Thank you huffte for your help.

I took my cantilever beam, and subdivided it in 25 elements then I did some tests:

P=100 kN M=10 kNm
top x disp.:
hand calc: 0,1948m
sesimo calc: 0,1944m

P=120 kN M=12 kNm
top x disp.:
hand calc: 0,3402m
sesimo calc: 0,3380m

P=140 kN M=14 kNm
top x disp.:
hand calc: 0,7240m
sesimo calc: 0,6937m

P=150 kN M=15 kNm
top x disp.:
hand calc: 1,3137m
sesimo calc: 1,1085 m

The solution of Seismostruct seems to diverge from hand-calc one with the increasing of axial load, is this may be caused by the fact that the hand-calc consider only “small” deformation instead SeismoStruct consider large deformation?

[huffte]

That's a valuable observation gmenichini. It would appear that SeismoStruct does require a finer element discretization when the axial load in members is large enough to warrant 2nd-order effects.

The hand calculation solution appears to be in order, so it is difficult to pinpoint the reason behind the divergence between hand calculation and SeismoStruct as the load gets closer and closer to a critical, buckling value.

It's an interesting problem and I'll see if I can come up with any new insights.

Best of luck gmenichini.