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From what I understand, shear deformations are neglected in this course?

Hi,

Yes indeed - thatâ€™s correct.

SeÃ¡n

Thanks for your answer. Do you cover this in any of your other courses? Or will it be added in the future?

No I donâ€™t have anything on this and Iâ€™m afraid I donâ€™t have any immediate plans to add anything - the pipeline of production is already pretty full for the next 8-12 months. Just not enough time to fit everyoneâ€™s requests in unfortunately.

S

In minute 5:39, why did you choose different polynomials for displacements and rotations?. For the first you used grade 3 polynomial, and for the second one grade 2 polynomial ?

When you say the interpolation function is for linear displacements, what do you mean?. If th equation is grade 3 polynomial how could the displacemnts be linear?

Hi @edwinhenaove, the fact that the rotations are a lower order polynomial is simply a result of the fact that the rotations are obtained as the first derivative of the displacements. By definition, the rotation at a point is the slope of a tangent to the curve passing through that point - we obtain this slope as the first derivative, which results in a lower order polynomial.

When I used the term linear, I meant translational, rather than rotational displacements.