Plate Maximum Axial Force changes with Sectiion Modulus

Question

Hi all, 
 
 I am using an elastoplastic plate (in plaxis 3D) and trying to control the axial and flexural capacity of the plate such that it approximates the behavior of a slurry wall. In the horizontal direction, due to the slurry wall joints, the flexural capacity is very limited while the axial force capacity is still decent. I tried the following: 
 - for vertical direction (direction 1), I calculate the yield stress and plastic section modulus as usual. 
 - for the horizontal direciton (direction 2), I calculate the yield stress (sigy,22) as usual, but I set the plastic section modulus (W22) to a very low number. 
 What i noticed that the maximum axial force in direction 2 (Np,2,eq) displayed to the right in the input window is very small (see attached capture), despite the high yield stress (sigy,22) that i provided. It turned out that its value is being affected by the value of W22. This is contradictory to the equations provided in the reference manual (Equations 383 and 358) which show that Np,2,eq is function of sigy,22 and d only. 
 Would someone please help with this? 
 Thank you

Vasileios Basas · Answer

Hi Omar, 
 Indeed, the manual is not clear on this topic. To compute the equivalent plastic axial force, N p,eq , the following mathematical operations are performed: 
 (a) Estimation of the plastic bending moment: 
 M p = W ∙ &sigma; y 
 (b) Assuming that bending moment is the most important mode of deformation, the plastic bending moment at a given gauss point, M p,gp , is considered equal to plastic bending moment specified in the input, i.e.: M p,gp = M p 
 (c) On other hand, taking into account that plate elements have rectangular cross-sections and assuming that the plastic moment can be calculated based on the elastic section modulus, the plastic moment at a given gauss point, M p,gp , should correspond to: M p,gp = W eq ∙ &sigma; y,eq,gp 
 where W eq is the equivalent (i.e. for a rectangular cross-section) elastic section modulus at a gauss point level, which, for a second-order Gauss-Legendre numerical integration, is given by: 
 W eq = I / y = (1.0 ∙ d 3 / 12) / (1 / &radic;3 ∙ d / 2) = &radic;3 ∙ d 2 / 6 
 (d) From the above equations, it is possible to estimate the equivalent yield stress at gauss point, &sigma; y,eq,gp , as: 
 &sigma; y,eq,gp = M p,gp / W eq = 6 ∙ M p / (&radic;3 ∙ d 2 ) 
 (e) The equivalent plastic axial force, Np,eq, can be, finally, estimated by: 
 N p,eq,gp = &sigma; y,eq,gp ∙ d = M p,gp / W eq ∙ d = 6 ∙ Mp ∙ d / (&radic;3 ∙ d 2 ) = 6 ∙ Mp / (&radic;3 ∙ d) 
 I hope this makes it clearer for you.