Textbook Calculation vs Real-world Intuition

This week I'll share a toy example that is similar to a question I have at work.

Robotics - What’s wrong with rigid body?

In robotics and many other engineering fields, we often make an assumption that objects, e.g., robot links and parts are rigid, i.e., they do not change shape or deform.

Here is a more detailed description: a rigid body in robotics is a solid object whose shape and size remain constant regardless of the forces acting upon it. Unlike flexible or deformable objects, all points within a rigid body maintain fixed distances from each other during motion. In robotic systems, individual links or segments of a robot arm are typically modeled as rigid bodies, connected by joints. This simplification allows engineers to use fundamental principles of mechanics, such as Newton's laws and Euler's equations, to analyze and predict the robot's motion, dynamics, and behavior.

So what’s wrong? An interesting and counterintuitive number came out when I do the following calculation about reaction force. In this example, I want to understand “How the axial torque of an open-chain robotic arm can generate a reaction force?”

An example

Here is the problem, as shown in the following picture, assuming the arm is massless, with a mass m attached at the end. The arm is held in all 6 degree-of-freedom (DOFs), a torque is applied in the z-direction along the frame. What are the reaction forces at the frame?

Force balance equations

In each of the 6 DOFs:

Kinematics

The arm can only rotate about z, therefore:

\(\ddot{\theta}_y = \ddot{\theta}_x = \ddot{z} = \ddot{x} = 0\)

Rotation and Motion Relationship

The rotation about z causes a motion in y:

\(\ddot{\theta}_z l_x = \ddot{y}\)

Torque Analysis

\(M_z = m l_x^2 \ddot{\theta}_z = m l_x^2 \frac{\ddot{y}}{l_x} = m l_x \ddot{y}, \forall M_z \neq 0\)

With the force balance equations on y axis, we have:

\(M_z = m l_x \frac{F_y}{m} = l_x F_y. \text{ We get, }F_y = \frac{M_z}{l_x}\)

Note that the smaller lx is, the larger the forces are which are generated by applying a torque in Mz. Did you expect to see a large Fy?

The problem

Calculating reaction force is something important to mechanical design. Often times, engineers will use software such as Ansys for numerical simulation the does not treat the robot as rigid body. This is then we notice a big different in results between hand calculation with rigid body assumption and simulation solution based on finite element analysis (FEA). Essentially, the large reaction force in Fy simply disappeared in Ansys. Or maybe it is not there from the very beginning? I’m currently still trying to understand concretely why there is such difference! Will share more when I find the answer!