How does WALLAP calculate the new displacement with the changes of wall stiffness? 1

[IGEngSoft]

Hi, I just have one question to ask regarding the use of WALLAP for a deep excavation problem. WALLAP allows the users to reduce the wall stiffness in order to model the long-term stiffness of the wall at the end of the excavation. I am just curious to know how WALLAP carries out the analysis with the new wall stiffness. Since all forces are in equilibrium following the previous stage analysis - out-of-balance force is zero, I would expect that just changing the wall stiffness would not lead to any more displacements in the solutions. But WALLAP does gives large additional displacement based on the updated wall stiffness. I searched through WALLAP's manual but it doesn't say how this is done within the program. I suspect the program just recalculate the whole thing again from the beginning with the new stiffness but would like to know whether there are other reasons for that.

Thanks a lot!

 

[DrWALLAP]

You are right to point out that the forces are all in equilibrium but when the wall becomes less stiff there is no longer strain compatibility. WALLAP uses an "initial stress" technique to compensate for the incompatibility. It does not start from scratch. BTW - although not mentioned in the User Guide, relaxation only works using the 2D-FE option not Subgrade Reaction (sorry for that)

 

[IGEngSoft]

DrWALLAP,

Thanks for your reply. I would appreciate if you could explain a bit more on this "Initial stress" technique to compensate for the incompatibility. Does this approach re-evaluate the initial stresses at both sides of the wall based on the specific excavation level (at the stage where wall stiffness is reduced)and then solve the equilibrium equation again to get the results? I just try to figure out how this is done within the program.

 

[DrWALLAP]

Imagine a load hanging from a wire. The load is in equilibrium with the force in the wire. Think of the wire as made of two wires each of half the cross-sectional area of the original wire. Nothing changed. Half the load is taken by each wire. What happens if we reduce the modulus of the wires by 50% i.e. we halve the modulus. It is equivalent to removing one of the two wires. So how would we model the removal of one wire? All the load will now have to be taken by the remaining wire. i.e. the remaining wire is subject to an additional load equal to that which was previously supported by the removed wire. That load is your "initial stress". In the case of the wall, we look at the bending stress supported by the part of the wall which was "removed" and apply those moments as loads.