Nonlinear finite elements/Homework 11/Solutions/Problem 3
Problem 3: Taylor impact test[edit | edit source]
- Cylindrical impact specimen strikes rigid target.
- Frictionless contact between cylinder and target.
- = 30 mm; = 6 mm.
- = 188 m/s; = 718 K.
- = 46 GPa; = 129 GPa; = 1.76e-5 /K.
- = 8930 kg/m; = 386 W/m-K; = 414 J/kg-K.
- Yield stress given by the Johnson-Cook model.
- = 90 MPa; = 292 MPa; = 0.025; = 0.31; = 1.09.
- = 1.0 /s; = 294 K; = 1356 K.
- Experimental deformed profile:
|Point||x (mm)||y (mm)||Point||x (mm)||y (mm)||Point||x (mm)||y (mm)|
Solution[edit | edit source]
Part 1[edit | edit source]
Plot of final deformed shape of specimen and comparison with experimental data.
This problem is solved using LS-DYNA. Due to symmetry, only a quarter of the cylinder is modelled. The plot of the final deformed shape of the specimen is shown in Fig 6.
Part 2[edit | edit source]
Comment of differences between simulation and experiment.
The error in the predicted final width of the base of the cylinder is less than 1%
However, the width of the bulge is underestimated by the simulation by almost 10% not predicting enough strain hardening at 718 K or that there is some volumetric locking during the simulation.
The final length of the cylinder is also underestimated by the simulation. This could because the Johnson-Cook model predicts a higher yield stress at 718 K than is observed in experiments. It could also be because our mesh is too coarse.
Overall, agreement with experiment is not very good and could be improved with improvements in the plasticity model and the finite element model.
Part 3[edit | edit source]
Plot of energy. Is energy conserved?
Though the total energy at the end of the simulation is less than the the energy at the beginning of the simulation, the variation in the energy is small. Hence, we can say that the energy is reasonably well conserved during the simulation.