Stress/displacement model change: static

These models (except for pmce_cps4_se1.inp, pmce_cps4r_se1.inp, pmce_cgax4rh_se1.inp, and pmce_c3d8r_se1.inp) include nonlinear geometric terms. This loading regime puts the model in uniaxial stress in the plane stress, axisymmetric, and three-dimensional models; in the plane strain model it is in a biaxial state of stress. Hence, the stress can be found by multiplying the strain by the elastic modulus or, in the case of plane strain, by $E^{\prime }=E/\left(1-{\nu }^2\right)$. Only the strain value will be listed.

General tests:

Step 1

Uniform axial strain should exist in this step for all tests. The value should be ln ($l/L_o$), where l = 4.9 and $L_o$ = 5.0. These values give ${ϵ}_{xx}$ = −2.0203 × 10⁻².

Step 2

The stress and strain in the elements that are not removed should become zero. The nodes on elements 1 and 6 should have $u_x$ = 0.0, and the nodes on elements 5 and 10 should have $u_x$ = −0.1.

Step 3

The displacement of the nodes in this step should have no effect on the results that were obtained in Step 2.

Step 4

For the plane strain, axisymmetric, and three-dimensional models there will be a state of uniform axial strain in this step. The magnitude will be ${ϵ}_{xx}$ = 4.0005 × 10⁻² (ln ($l/L_o$), where l = 5.1 and $L_o$ = 4.9).

For the plane stress and truss elements there is a change in thickness of the elements in Step 1. The thickness is not changed when elements are removed. Therefore, the elements added back into the model in this step will not have the same axial stiffness (and, hence, axial strain) as the elements that were not removed. The variation in ${ϵ}_{xx}$ is as follows: elements 1, 5, 6, and 10 have ${ϵ}_{xx}$ = 4.07 × 10⁻²; elements 2, 4, 7, and 9 have ${ϵ}_{xx}$ = 3.92 × 10⁻²; elements 3 and 8 have ${ϵ}_{xx}$ = 3.99 × 10⁻².

The axisymmetric models are loaded in the z-direction.

Specific tests:

The models that have the same loading as the general tests have the same analytical solution.

pmce_cps4_se1.inp

Because this is a test without NLGEOM, the strain is always based on the change in displacement divided by the original length. This produces ${ϵ}_{xx}$ = −2 × 10⁻² in Step 1 and 4 × 10⁻² in Step 4.

pmce_c3d8_se1.inp

There should be zero response in the model in Step 1 and Step 2. In Step 3 there should be thermal strains in the model equal to $\alpha \left(\theta -{\theta }^I\right)=80\alpha$ for the middle elements and $40\alpha$ for the other elements. (These thermal strains are the same value since the $\alpha$ value for the middle elements is one-half of that for the other elements.) There should be no elastic strain in the model and no stress.

pmce_cpe4_sp.inp and pmce_cpe4_sp1.inp

In Step 1 the model will yield uniformly. ${ϵ}_{xx}^{pl}$ = −1.879 × 10⁻⁴. In Step 4 only the middle elements of the model will yield. ${ϵ}_{xx}^{pl}$ will be approximately −1.76 × 10⁻⁴.

pmce_cpe4i_se1.inp

The strains in Steps 1 and 4 are ${ϵ}_{xx}$ = −2 × 10⁻⁴ and 4 × 10⁻⁴, respectively. This applies to both the rebar and the elements.

pmce_cpe8_se1.inp and pmce_cpe8_sh1.inp

In Step 1 there will be a gradient of ${ϵ}_{xx}$ in the model. In Step 2 elements 1 and 6 will be in tension and Elements 5 and 10 will be in compression. In Step 4 there will be a gradient of ${ϵ}_{xx}$ in the model.