RD-E: 2300 Brake

A brake system is simulated using a finite Lagrangian mesh element.

A frictional mechanism is studied, which consists of a brake system, defined by a disk pinched between two pads. The main aspects of the model are the initial rotary motion of the disk and the interface definition, between the disk and the pads. Carefully watch the accuracy of the simulation results compared to the analytical solution.

ex_23_brake
Figure 1.

Options and Keywords Used

  • Brick elements and HEPH formulation
  • Interface (/INTER/TYPE7) and friction
  • Boundary conditions (/BCS)

    For the disk's rigid bodies, all DOF, except the rotation around Y are fixed. For the pads' rigid bodies, all DOF; except translation around Y are fixed.

  • Rigid body (/RBODY)

    Two rigid bodies are created to put the disk into motion: the first (called RBODY1) contains all the nodes of the disk, except those in the disk's internal periphery, which are contained in the second rigid body (called RBODY2). Both rigid bodies are activated in the first step of computation; however, RBODY1 is deactivated in the D02 file.

    Two other rigid bodies are created to model the pads' faces where concentrated loads are applied.

    fig_23-2
    Figure 2. Rigid Bodies on the Disk (RBODY1 on the left and RBODY2 on the right)

    fig_23-3
    Figure 3. Rigid Body on a Pad
  • Initial velocities (/INIVEL)

    An initial rotational velocity ω 0 = 120 rad/s is applied to the disk's master nodes during the first computation phase.

  • Concentrated load (/CLOAD)

    Two concentrated opposite forces are applied to the rigid bodies' master nodes for the pads.

  • Skew frame (Skews)
  • Function (/FUNCT)

Input Files

The input files used in this example include:
Lagrangian formulation
<install_directory>/hwsolvers/demos/radioss/example/Lagrangian_formulation/BRAKE2*

Model Description

The purpose of this example is to highlight the capacity of Radioss to simulate frictional mechanisms. The braking system retained consists of a disk pinched in between two pads.

A disk with a hole in the center rotates at ω 0 = 120 rad/s around its axis. It is subjected to frictional contact using two small brake pads, placed on two faces.
Disk Description
Radius
100 mm
Width
50 mm
Thickness
5 mm
Mass
1 kg
Inertia
0.57x10-2 kg/m2 (about its free rotation axis)
Pads Description
Length
65 mm
Width
28 mm
Thickness
5 mm

A constant P = 300N pressure is applied on the back of each pad to push them against the disk. A Coulomb friction coefficient is assumed as being 0.15.

Units: m, s, kg, MPa

fig_23-1
Figure 4. Geometry of the Problem
The material used for the disk follows an isotropic elasto-plastic law (/MAT/LAW2) using the Johnson-Cook plasticity model, with the following characteristics:
Material Properties
Initial density
7800 kg/m3
Young's modulus
210000 [ MPa ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqGqFfpeea0xe9vq=Jb9 vqpeea0xd9q8qiYRWxGi6xij=hbba9q8aq0=yq=He9q8qiLsFr0=vr 0=vr0db8meaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWadaqaai Gac2eacaGGqbGaaiyyaaGaay5waiaaw2faaaaa@3BE6@
Poisson ratio
0.3
Yield stress
206 [ MPa ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqGqFfpeea0xe9vq=Jb9 vqpeea0xd9q8qiYRWxGi6xij=hbba9q8aq0=yq=He9q8qiLsFr0=vr 0=vr0db8meaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWadaqaai Gac2eacaGGqbGaaiyyaaGaay5waiaaw2faaaaa@3BE6@
Hardening parameter
450 [ MPa ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqGqFfpeea0xe9vq=Jb9 vqpeea0xd9q8qiYRWxGi6xij=hbba9q8aq0=yq=He9q8qiLsFr0=vr 0=vr0db8meaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWadaqaai Gac2eacaGGqbGaaiyyaaGaay5waiaaw2faaaaa@3BE6@
Hardening exponent
0.5
Maximum stress
340 [ MPa ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqGqFfpeea0xe9vq=Jb9 vqpeea0xd9q8qiYRWxGi6xij=hbba9q8aq0=yq=He9q8qiLsFr0=vr 0=vr0db8meaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWadaqaai Gac2eacaGGqbGaaiyyaaGaay5waiaaw2faaaaa@3BE6@
The material used for the pads follows a linear elastic law, with the following characteristics:
Material Properties
Initial density
7300 [ kg m 3 ]
Young's modulus
160000 [ MPa ] MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqGqFfpeea0xe9vq=Jb9 vqpeea0xd9q8qiYRWxGi6xij=hbba9q8aq0=yq=He9q8qiLsFr0=vr 0=vr0db8meaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWadaqaai Gac2eacaGGqbGaaiyyaaGaay5waiaaw2faaaaa@3BE6@
Poisson ratio
0.3

Model Method

The two parts are modeled using a regular mesh having 720 BRICK elements for the disk and 80 such elements for the pads. The HEPH formulation is used to describe the BRICK elements.

Two steps are necessary to compute the model: First, an initial velocity 0.03 ms is applied to the disk. In the second step, pressure is applied to the pads to push them onto the disk.

Results

Angular Velocity of the Disk

The normal contact force between the pads and the disk is FN = 600N. Then the tangential friction force on the surface of the disk is obtained at FT = 0.15 x FN = 90N. The torque around the axis of the disk is T = r x FT = 7.1 Nm, with r = 0.0789 m, which corresponds to the orthogonal projection on a radial axis with regard to the distance between the center of the disk and the point of the pad where the load is applied. This leads to an angular deceleration of α = T / I R = 1246 rad/s2.

The necessary time to stop the disk can be computed as: (1) t = ω 0 / α = 0.096   seconds
The simulation by Radioss using the explicit solver allows similar results to be obtained. Figure 5 shows the time history for angular velocity. The disk stops at t = 0.095 s, which corresponds to the analytical solution.

fig_23-4
Figure 5. Angular Velocity of the Disk

Disk Rotation

Figure 6 shows the total rotation of the disk, which rotates 5.65 rad before stopping.

fig_23-5
Figure 6. Total Rotation of the Disk

Reaction Forces

The reaction forces value in Figure 7 is about 90 N, which corresponds to the analytical value.

fig_23-6
Figure 7. Reaction Forces

Energies

The total energy remains constant during computation. After braking, the kinetics energy decreases smoothly while the contact energy increases. There is no hourglass energy as a HEPH solid element is used.

fig_23-7
Figure 8. Energies

Contact Forces

Figure 9 presents the tangential contact forces for three consecutive moments.

fig_23-8
Figure 9. Variation of X-component of Tangential Contact Forces

Conclusion

The accuracy of the results obtained, using the simulation and corresponding to the analytical solution, proves that Radioss is able to simulate mechanisms, such as braking systems.