Week 4- Rolling operation

Objective: Simulating the rolling process of a copper alloy sheet between two steel rollers.

Rolling is the process where a sheet is plastically deformed due to forces applied by two rollers and is a form of cold working process, i.e. plastic deformation is carried out below the recrystallization temperature of the sheet material.

Geometry:

Materials:
The sheet is made of copper alloy and an isotropic bilinear model is used to represent its strain hardening and non-linear behaviour beyond its yield point. Its linear elastic stiffness is 110 GPa, non-linear tangent stiffness is 1.15 GPa and yield strength is 280 MPa.

The rollers are made of structural steel which is assumed to be linear material.

Model Setup:
1). Contacts: Since the sheet has to slide between the rollers, frictional contacts are defined between the sheet and each roller with the coefficient of friction being 0.2. In these contacts, the workpiece acts as the target body and the roller behaves as the contact body. The formulation of the contacts is set to Augmented Lagrange to avoid convergence problems due to chattering of contact status, the contact stiffness defined as 0.1 times the program-controlled stiffness value and the interface treatment is set to Add offsets, ramped effects.

2). Analysis settings:
The simulation is carried out in 14 loadsteps (0-14 seconds) and for each loadstep, the automated time stepping is program controlled i.e. the gap between each substep where the solution attempts to converge is decided by the solver itself. To account for any geometric non-linearities which may arise, large deflections is turned on. In order to ensure that there are no convergence issues due to excess node deformations, stabilization also has been turned on with the energy dissipation factor being set to 0.1.

3). Joints:
The rollers need to rotate and at the same time exert sufficient force on the workpiece in order to plastically deform it. If we opt for revolute joints, we would have to add an additional load as a boundary condition in order to simulate the roller exerting force on the workpiece but if we use cylindrical joints, the rotation of the joint will also take into consideration the force which is being exerted over the workpiece. Thus, cylindrical joints are used for the roller joints.

4). Joint Loads:
The workpiece needs to traverse linearly between the rollers thus is given a displacement along the negative Y direction (of the global coordinate system). The total distance to be travelled by the workpiece is 90mm which is to be done in 14 steps. The displacement is given by:

The roller diameter is 50mm and its circumference is 157.08mm. For the workpiece to traverse a distance of 90mm, the roller would have to rotate by an angle of 206.26 °. This rotation is divided equally among the 14 loadsteps, with the roller having to rotate 14.73 ° each loadstep.

5). Mesh:
The mesh on the workpiece is refined by using body sizing and the refined mesh size is 3mm. It is important to note here that this mesh setting is still considerably course and may not provide accurate solutions, but for a finer mesh, we may have to increase the number of loadsteps since for finer workpiece mesh, travelling 90mm in 14 loadsteps leads to convergence issues which can be fixed by tavelling 90mm in 20-25 steps.

#Nodes = 4847
#Elements = 741

Results:

1). Equivalent stress in the whole setup:

2). Equivalent stress in the workpiece:

As seen from the results, the stresses generated in both the workpiece and the roller are abnormally high. While this solution may not the the accurate one since the mesh on both the workpiece and the roller is coarse but in order to reduce the stress concentrations, the workpiece thickness can be increased. Moreover, the roller stress is probably being overestimated since the structural steel being used for the rollers is a linear elastic material, and thus beyond the yield point, the stress will be defined according to the young's modulus instead of the tangent modulus (in the isotropic hardening bilinear model).

3). Equivalent plastic strain in the workpiece:

4). Z directional deformation of the workpiece:

Due to the plastic compressive forces being exerted on the sheet causes an expansion in the Z direction (poisson's ratio). After the 14 seconds (i.e. after the workpiece has travelled 90mm), the Z deformation is about 4mm on one side and 7mm on the opposite side.

5). Y directional deformation of the workpiece:

Here, only the tip of the workpiece has been considered for the deformation in Y axis. As can be seen, the tip has a deformation of 90mm, with the middle part experiencing a little extra deformation of about 0.503mm due to bulging out from the compressive forces being applied on the workpiece. This makes sense as we have made the workpiece travel 90mm.