Mechanical Properties of a 2D Numerically Modeled Solid Lattice

I have deliberately allowed slider variable ranges that can result in a break up of the solid. It that happens, the learner just needs to refresh the page, which in Windows means hitting the F5 key. Here I compute the longitudinal and shear modulii of the 2D solid. For the longitudinal modulus (LM) I apply force to the right end and anchor the left end to cause about 1% compression and extension. Of course this is analogous to placing a weight of mass, M, on a vertical spring and letting it go. The weight exhibits vertical oscillations between the starting height and the displacement dy=Mg/k where k is the spring constant. To cause these oscillations to relax, I had to multiply the disc velocities by a drag coefficient which is adjustable by the learner. For the shear modulus (SM) I apply force to the top row and anchor the bottom row and cause about 0.1 radian of angle change. In this simulation the bonding is done with a Lennard-Jones (LJ) potential. LJ potentials have extremely large repelling forces when the particles get too close and longer range attractive forces when the particles are farther apart. The LJ potential is of the form V(r)=4VLJ[(sigma/r)^2b-(sigma/r)^b] where VLJ is the potential minimum, sigma is a length scaling parameter, r is the particle separation, and b=6 is an integer. The larger the value of b, the more abrupt is the force change with distance, r. The solid is initially set up so that all the particles are separated by a distance that causes the nearest neighbor (NN) forces to be zero. The force between a site and its second nearest neighbor (SNN) is not initially zero. An obvious result of that set up is that alternate rows of particles have to be staggered by 1/2 the spacing in the horizontal direction and the vertical spacing of rows is squareroot(3)/2 times the horizontal spacing of particles in any column so that the solid has 6 fold symmetry. SNN would be at just the squareroot(2) times the NN distance. With b=6 that would mean that the force due to the SNN would be 1/8 that of the NN. With the 6 fold symmetry SNN distance is twice NN distance so the SNN force is just 1/64 that of the NN force, so 6 fold is much more stable. It turns out that the square lattice tries to degenerate to the 6 fold symmetry anyway. The graphic shows the positions of the discs as they are being strained by the applied forces. The colors of the discs are coded in terms of the restoring forces that they feel and the code is shown by the color bar at the left. The longitudinal strain is just the displacement of the right hand column divided by the initial width of the array. The shear strain is the lean angle of the columns measured in radians. The longitudinal stress,sigmaL, is the total force applied to the right column divided by the column height. The longitudinal modulus of rigidity, E or LM, is the stress divided by the strain. The shear stress, sigmaS, is the total force applied to the top row divided by the width of the top row. The SM is the shear stress divided by the shear strain.
The SM calculation has difficulty for the 6 fold symmetry case because that symmetry exhibits very little NN distance change during shear and thus has very minimal restoring force. For that reason the SM is estimated from a combination of the LM and Poissons ratio, nu. During the LM strain, the length of the array is is increased and there is a corresponding decrease in the height of the array. These result in a change in total volume (Area in 2D) change. From the fractional change in volume and the longitudinal strain, the Poisson ratio is computed. The SM is just SM=LM/(2(1+nu)).
From the LM and SM and the mass density, the longitudinal and shear wave speeds, c, can be computed using the equation c=squareroot(LM/density). These agree quite well with those computed in a companion page named "Speed of Sound in a 2D Numerically Modeled Solid Lattice". Plots of the strain Vs time as well as the change of short range spacing are shown below the graphic.