I am currently creating a MATLAB script to determine the internal forces present in a uniform beam as it is accelerated by a torsional spring. I was able to derive the equation for the axial force present in the beam, but I am not sure how to determine the shear force of bending moment at any time. My setup is approximated by this image:

eRotating cantilever beam

My intuition tells me that there should be a nonlinear shear force across the radius of the beam, with the largest shear occurring near the rotational axis. I believe the bending moment should follow a similar distribution.

I was able to simulate the dynamics of the system, so I know the angular velocity, acceleration, and spring force at any point, but I am not sure how to translate this to the shear and moment. I attempted an integral of the force required to accelerate each point of the beam, but I don't believe that my result is accurate. If I could see the derivation process to determine the shear and moment distributions, that would be invaluable. Thank you!


1 Answer 1


If the beam is rigid (deformation is small and negligible) and is rigidly attached to the cylinder, the rotation does not affect the responses of shear and moment, as the center of mass/force remains constant. Therefore, the same as a cantilever beam, the shear is linear throughout the beam; the moment is zero at the tip and is the maximum at the support.

  • $\begingroup$ Hmm. I am still having trouble convincing myself of this. Intuitively, I would imagine that the shear would also increase from tip to pivot, as the parts of the beam close to the pivot have to exert a tangential force to accelerate all of the mass above it. I am not sure why this intuition fails. $\endgroup$
    – DrCobras28
    Nov 9, 2021 at 15:56
  • $\begingroup$ @DrCobras28 The inertia force acting radially produces a pulling force on the support cylinder (flying outward horizontally), the only potential to cause the horizontal shear is torsion, however, as the beam rotates at the same rate as the shaft, there is no force differential but the friction/cohesion between the particles because lacking restrain. $\endgroup$
    – r13
    Nov 9, 2021 at 16:19

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