 represents a boundary load condition for PDEs with predicate pred indicating where it applies, with model variables vars and global parameters pars.

represents a boundary load condition with local parameters specified in pars[lkey].

# Details  • SolidBoundaryLoadValue specifies a boundary condition for SolidMechanicsPDEComponent and is used as part of the modeling equation:
• • SolidBoundaryLoadValue is typically used to apply a pressure or force at a boundary.
• • SolidBoundaryLoadValue forces the movement of a solid with dependent variables , and , independent variables and time variable .
• Stationary variables vars are vars={{u[x1,,xn],v[x1,,xn],},{x1,,xn}}.
• Time-dependent variables vars are vars={{u[t,x1,,xn],v[t,x1,,xn],},t,{x1,,xn}}.
• Frequency-dependent variables vars are vars={{u[x1,,xn],v[x1,,xn],},ω,{x1,,xn}}.
• Model parameters pars are specified as for SolidMechanicsPDEComponent.
• The following model parameters pars can be given:
•  parameter default symbol "Force" {0,0,0} , force in "Pressure" {0,0,0} , pressure in "Thickness" - , thickness in • Forces are automatically converted to a pressure.
• In two dimensions, the parameter "Thickness" is taken into account to convert the pressure from units to .
• A prescribed displacement condition boundary can be used with:
•  analysis type applicable Transient Yes Frequency Response Yes Eigenfrequency No Stationary Yes
• SolidDisplacementCondition evaluates to a NeumannValue.
• The boundary predicate pred can be specified as in NeumannValue.

# Examples

open allclose all

## Basic Examples(1)

Compute the deflection of a spoon held fixed at the end and with a force applied at the top. Set up variables and parameters:

Set up the PDE and the geometry:

Visualize the displacement:

## Scope(5)

Create a pressure boundary load condition:

Create a force boundary load condition:

Create a time-dependent boundary load condition:

Create a time-dependent parametric boundary load:

Compute the deflection of a beam held fixed at the left end and with a pressure applied in the negative axis at the right end:

Set up a geometry and the PDE:

Solve the equation with a pressure load:

Solve the equation with a force load:

Because the surface area that the pressure and force act on is 0.02 and since the force is equal to pressure per area , the displacements are the same:

## Possible Issues(1)

In the case where boundary load values have predicates that do not overlap with the region boundary, results may be inaccurate. Set up a PDE:

Set up a boundary load with a discontinuity at and at :

If the geometry does not reflect the discontinuity in the boundary condition, the result will be less accurate. The solution will be more accurate if the region reflects the discontinuities at and at . Set up two meshes:

Solve the equation with two different geometries:

Compare the direction displacement along and :