# Linear elasticity modeling load using traction vs. mixed BC

In classical linear elasticity, when modeling a force/load boundary condition, it appears that we could either:

1. Use a pure Neumann boundary condition, where the 3 traction components are specified. In 3-D the 2 tangential traction components would be zero, with the non-zero component being the normal component.

2. Use a mixed boundary condition. Here, the 2 tangential traction components would be zero. The normal traction component is unknown. However, the normal displacement is now known, and the 2 tangential displacements are unknown.

Questions:

• Is it correct to say that 1 and 2 are both accurate methods of modeling this force boundary condition?

• Is there an actual analytical expression for converting the 1 to 2?

• If we know the normal traction component, could we actually uniquely convert this to a normal displacement value?

## 1 Answer

The two descriptions you use describe different physical situations.

In the first, you apply a given force (which you have assumed is normal to the surface, though it could of course also have a tangential component in general) and the displacement is then simply going to be determined by how the object to which the force is applied reacts to the force. An example is if you place a weight of known mass on top of a deformable object -- say, you park a truck on a bridge: it has a fixed weight, and the bridge deforms according to its elastic properties.

The second situation you describe is where you apply a certain displacement, regardless of how much force that requires. An example is if you push a steel tool into a soft object: the steel is going to hard enough to not deform noticeably so that you can achieve whatever deformation you desire regardless of the force necessary.

Since these are two entirely different physical situations, there is no way to convert one description into the other. In other words, there is no easy way (other than to solve the equations of elasticity) to compute the displacement you obtain for the given force in situation 1 above, or to compute the necessary force to obtain a given displacement in situation 2. In other words, the answers to your second and third question is "no, this is not possible without actually solving the equations of elasticity".

• I was reading this PPT presentation on how a 3pt bend test was modeled. There was a load in mid point, and 2 points on the bottom of the beam are supported. There was some device that measured the reaction force, $R_f$ at these 2 bottom supports. They modeled this experiment by applying a normal displacement BC at the mid point until the simulation generated the force at the 2 bottom supports that matched $R_f$. I was confused why they would do this instead of applying a force BC at the mid pt. I think they could have simply just applied a force BC where the force is equal to $2R_f$. – roulette01 Mar 16 '19 at 18:05