In the thread Stress calculations using finite elements User21 showed an example how to define a force over the entire body during FEM calculation as boundary condition. See the screenshot below from the corresponding position in this thread.

enter image description here

In the description of the definition of the boundary condition - force on the entire body - User21 has defined the differential equation system as follows.

$ps$ == $\{0, -9.8\}$

Which unit has this power? Is the unit $N/m^2$?

If $N/m^2$ is the correct unit, then I can understand how to calculate the normalized body force from the density and volume of the body and insert it into the right side of the differential equation.

If $N/m^2$ is not the right unit, then I have the following questions on you:

How is the density of the material or the mass considered here? Could you please show how to use this correctly in the equation?

In my case I have a centripetal acceleration due to rotation and the equation would look like this:

$ps$ == $\{\omega ^2 \cdot x, \omega ^2 \cdot y\}$

$\omega$ is the angular velocity of the body for which the deformations are to be calculated with FEM. And the expression

$\omega^2 \cdot r$

is the centripetal acceleration, where $r$ is the distance from the center of rotation.

However also here I have the problem, density resp. masses are not considered.

Does anybody have an answer to the question how to use the density and the mass correctly in the equation?

Many thanks in advance!

  • $\begingroup$ In solid mechanics, body forces like gravity or inertial forces (like those due to acceleration) are not boundary conditions. Instead, they appear in the actual PDE itself. Surface forces (also called traction forces) are boundary conditions, as are physical constraints such as zero displacement or slope. $\endgroup$ Jun 19, 2020 at 19:40
  • $\begingroup$ Many thanks for the important addition!!! $\endgroup$
    – ronin2222
    Jun 19, 2020 at 20:51
  • $\begingroup$ Typically, this would be a comment, not an answer. $\endgroup$
    – bbgodfrey
    Mar 12, 2021 at 15:12

1 Answer 1


The units are defined in $N/m³$.


Let's look at the left side of the equation.

    Div][({{0, -((Y \[Nu])/(1 - \[Nu]^2))}, {-((Y (1 - \[Nu]))/(
        2 (1 - \[Nu]^2))), 0}}.Inactive[Grad][v[x, y], {x, y}]), {x, 
    y}] + Inactive[
    Div][({{-(Y/(1 - \[Nu]^2)), 
       0}, {0, -((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2)))}}.Inactive[Grad][
      u[x, y], {x, y}]), {x, y}], 
 Inactive[Div][({{0, -((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2)))}, {-((
        Y \[Nu])/(1 - \[Nu]^2)), 0}}.Inactive[Grad][
      u[x, y], {x, y}]), {x, y}] + 
  Inactive[Div][({{-((Y (1 - \[Nu]))/(2 (1 - \[Nu]^2))), 
       0}, {0, -(Y/(1 - \[Nu]^2))}}.Inactive[Grad][
      v[x, y], {x, y}]), {x, y}]}

Let us consider only the first term of the entire equation

$\nabla_{\{x,y\}}\cdot \left(\left\{\left\{0,-\frac{Y \nu}{1-\nu^2}\right\},\left\{-\frac{Y(1-\nu)}{2 \left(1-\nu^2\right)},0\right\}\right\}.\nabla_{\{x,y\}}v[x,y]\right)$

Now we replace all symbols with the corresponding units

$\frac{1}{m} \cdot \left(\left\{\left\{0,\frac{N}{m^2}\right\},\left\{\frac{N}{m^2},0\right\}\right\} \cdot \frac{1}{m} \cdot m\right)$

After the exclusion remains


2.Definition of gravitational force

A beam with the length $L = 1 m$ and the height $H=2h=2*0,05 m$ is assumed. An acceleration constant $g=0.3 \frac{m}{s^2}$ is defined. We set the density to $\rho=3 \frac{kg}{m^3}$. The width $b$ of the beam for the 2D to $1 m$.

Let us calculate the volume force normalized to $1 m^3$ so in $\frac{N}{m^3}$.

$F_{norm} = \frac{V \rho g}{V} =\frac{2h b L \rho g}{2h b L}=\rho g=3 \frac{kg}{m^3} \cdot 0.3 \frac{m}{s^2} = 0.9\frac{N}{m^3}$.

Now we put this value in the right side of the equation. We choose the negative $y$-direction for the force and stretch the beam on the left side (Dirichlet boundary conditions).

$ps==\{0, -0.9 \}$

The complete description of the calculation of a bending beam can be found under the link: Stress calculations using finite elements.

After calculating the bending and the stresses, we can compare the results with the theory.

The deformed state deformed state

Bending line on the neutral fiber at $y=0$. Bending line

Normal stress \sigma_x along the $x$-axis at $y=+h$. normal stress +h

Normal stress $\sigma_x$ along the $x$-axis at $y=-h$. normal stress -h

Normal stress $\sigma_x$ at $x=\frac{L}{2}$ and and $-h<y<h$. enter image description here

Shear stress $\tau_{xy}$ along $x$-axis and $y=0$. enter image description here

Shear stress $\tau_{xy}$ at $x=\frac{L}{2}$ and and $-h<y<h$. enter image description here

3. Accelerazione centripeta The centripetal acceleration can thus be defined in the following way.

$ps==\rho \cdot \omega^2 \cdot \{x, y\}$

$\omega$ is an instantaneous angular velocity in $\frac{1}{s^2}$, $\{x,y\}$ in $m$ and $\rho$ in $\frac{kg}{m^3}$. The multiplication results in $\frac{N}{m^3}$.

The cetripetal force corresponds to a vector field which increases in amplitude uniformly in all directions from the coordinate origin.


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