Whether you use closed-form or numerical integration, the deflection at the free end of a laterally loaded linear-elastic, prismatic column is known to be . This result is easily verified in OpenSees with either an elasticBeamColumn element or a material nonlinear element with elastic sections.

Now suppose we philosophized a bit then posited the following modeling approach…

Instead of a single element for that column, let’s use a series of zero length rotational springs connected by flexurally rigid segments. The rotational springs are collocated with the integration points from whatever quadrature rule you like, e.g., three-point Gauss-Legendre or four-point Gauss-Lobatto as shown below, or any other integration option.

My question is, what stiffness should be assigned to each rotational spring in order to compute the correct free end deflection, , using a springs and rigid segments model?

A. EI B. EI/L C. 3EI/L D. EI/w_{i} E. EI*w_{i} F. None of the above

In options D and E, w_{i} is the integration weight on the column domain [0,L] associated with quadrature point/spring location x_{i}, meaning the stiffness is not the same for each spring.

Post your answer with an explanation in the comments section below. I will explain the correct answer in a follow up post. Responses are due by May 31, 2022.

While you can develop a model in OpenSees or use good ‘ol pencil and paper to figure out the correct answer, I don’t recommend you use this modeling approach anywhere else. As usual, complicated is not better.

I have been involved in the development, maintenance, and growth of OpenSees since its early days. I am interested in learning Python and improving my academic writing.
View all posts by Michael H. Scott

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4 thoughts on “A Complicated Equivalent”

I choose F, if the spring stiffness is constant, as x increases, the spring angle will become smaller and smaller, which does not conform to the cubic deflection curve. If the spring stiffness decreases linearly with the length of L, then all springs have the same angle of rotation, and the resulting parabola is close to a circle, which also does not conform to a cubic parabola. Therefore, the stiffness of the spring should decrease second-order with the length of L.

First, I give my answer to the question as F (None of the above)

I solved the problem with the “virtual work” method and verified the solutions with the OpenSees models. I share only the “three-point Gauss-Legendre” model script. You can check it from:

My comment on this question:
The stiffness value of the rotational springs (RS) does not depend on the flexural stiffness of the cantilever column, but it depends on the geometry of the system and the location of loading. You can obtain a unique solution if you assign the same stiffness to all RSs in the system. However, if you choose to assign different stiffness values for each RS, you can get an infinite number of solutions for the stiffness of RSs which produce the same lateral displacement value as the single element system gives.

I agree with your comment. You can assign different stiffness values to the springs and find any one of infinite solutions. Thanks for the input and for sharing your model!

I choose F, if the spring stiffness is constant, as x increases, the spring angle will become smaller and smaller, which does not conform to the cubic deflection curve. If the spring stiffness decreases linearly with the length of L, then all springs have the same angle of rotation, and the resulting parabola is close to a circle, which also does not conform to a cubic parabola. Therefore, the stiffness of the spring should decrease second-order with the length of L.

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Greetings,

First, I give my answer to the question as F (None of the above)

I solved the problem with the “virtual work” method and verified the solutions with the OpenSees models. I share only the “three-point Gauss-Legendre” model script. You can check it from:

https://github.com/hakki-deniz/Equivalent-System

My comment on this question:

The stiffness value of the rotational springs (RS) does not depend on the flexural stiffness of the cantilever column, but it depends on the geometry of the system and the location of loading. You can obtain a unique solution if you assign the same stiffness to all RSs in the system. However, if you choose to assign different stiffness values for each RS, you can get an infinite number of solutions for the stiffness of RSs which produce the same lateral displacement value as the single element system gives.

Great question from Mr. Scott.

Best regards.

LikeLiked by 1 person

I agree with your comment. You can assign different stiffness values to the springs and find any one of infinite solutions. Thanks for the input and for sharing your model!

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