I am a bit puzzled about the meaning of statically indeterminate in particular in terms of force distribution.

The following system is clearly statically indeterminate: There are four reaction forces acting on the body while only three equations for static equilibrium, two for forces, one for torque are available.

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My first question is, that would happen if I managed to build a system that stiff? What would the force distribution be like? What would a energy based solution method or e.g. a rigid body simulation tool output in such a case?

As far as I have understood, for statically indeterminate systems I would have to consider deformations that allow every part of the system to engage and have a well defined forces. Is this system statically determinate if I replace all rods with springs? My guess is that it is still statically indeterminate because the body is still rigid. How can a determine a feasible solution for this?

  • $\begingroup$ Consider the case if table with four legs and a heavy object placed somewhere on top. Clearly this is statically indeterminate as three forces would be needed to support the table, and also clearly the table still works with 4 legs. The answer is that in real life nothing is perfectly rigid, and slight deformations change the load distribution. $\endgroup$
    – JAlex
    Commented Jan 10, 2023 at 22:52

1 Answer 1


For statically indeterminate systems you need to consider both the material behavior and the kinematics of deformation in addition to the equations of equilibrium in order to determine the forces. So, when you ask about a "rigid body simulation tool", such a tool would have to use some deformable material model, e.g. linear elasticity, and then take a limit as the elastic modulus goes to infinity. A different material model, e.g. some nonlinear elastic model, would yield a different set of forces even in the limit that the model becomes rigid.

"Is this system statically determinate if I replace all rods with springs?"

No, but in such a case you would not have to worry about the internal deformation of the rectangular block which you could treat directly as rigid. It remains statically indeterminate because you would have to consider the rigid body rotation and translation of the block to relate the deformations of the springs to one another. The three equilibrium equations alone still cannot determine the four spring forces. You need kinematics of the motion and how that deforms the springs, and you need the "material behavior" of each of the springs (linear, non-linear, stiffness, ...).

To solve this system with springs attached to the rigid rectangle you have your 3 equilibrium equations for the 4 spring forces, $F_A$, $F_B$, $F_C$, and $F_D$. I assume you also have some external loading on the system, and that this is a known set of forces/moments/distributed loads.

Then you have the kinematics of the rectangle. Let's take our origin to be at the center of the rectangle. Let's also describe the rigid body motion of the rectangle by the displacements $u_0$ and $v_0$ of the center and its rotation $\theta_0$. So we have introduced 3 new unknown quantities. Right now we have our 4 unknown spring forces and our 3 unknown rigid body motions, but only 3 equilibrium equations.

Next, for each of the spring connection points we can find the displacements $(u_A,v_A)$, $(u_B,v_B)$, $(u_C,v_C)$, and $(u_D,v_D)$ in terms of $u_0$, $v_0$ and $\theta_0$. If the rotation angle is small then these equations are linear in $u_0$, $v_0$ and $\theta_0$, otherwise they are nonlinear and a bit more messy. For small rotations the equations for $A$ are, $u_A = u_0 - \theta_0 y_A$ and $v_A = v_0 + \theta_0 x_A$. The equations for the other springs are identical with $A$ replaced by the other points.

Now, once you have $(u_A,v_A)$, $(u_B,v_B)$, $(u_C,v_C)$, and $(u_D,v_D)$ you can calculate the change in length of each of the springs in terms of $u_0$, $v_0$ and $\theta_0$.

Finally, the change in length of each of the springs must be related to the force in that spring through its constitutive response, e.g. Hooke's law if it is a linear spring. This provides 4 spring equations that relate the spring forces $F_A$, $F_B$, $F_C$, and $F_D$ to the rigid body motions $u_0$, $v_0$ and $\theta_0$.

So you can now solve the 3 equilibrium and 4 spring equations for the 4 forces and 3 rigid body motions.


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