Dear @karamba3d -team,
I am connecting several plate strips with line hinges in between and I want to define the rotational stiffness of the line hinges to the same stiffness as the one of the plate, i.e. the plate’s bending stiffness. I thought it would be as simple as I*E, but I get strange results. Could you maybe shed some light on this, that would be great.
Thanks,
David
Dear @DavAndLar,
E*I/L gets you the bending stiffness of a plate strip or beam. A hinge has no geometric extension perpendicular to its axis.
If it is your aim to avoid any kinks at the joints, you have to choose a very large rotational stiffness. The value need not be too large however, since this may lead to an ill-conditioned stiffness matrix and thus inaccurate results.
– Clemens
Thank you Clemens @karamba3d for your answer.
E*I/L gets you the bending stiffness of a plate strip or beam
Yes your are absolutely right, my fault.
If it is your aim to avoid any kinks at the joints, you have to choose a very large rotational stiffness.
I want to assign a procentage of the exact bending stiffness of the plate. The end goal is to simulate a continous timber plate strip with cuts with different depth in it. So e.g. I want to assign 12,5% of the exact bending stiffness in one hinge, where I have a h/2 deep cut. As a starting point however, I wanted to validate my simulation by assigning the full bending stiffness of the plate strip, wouldn’t that also result in a “kink-free” result?
best,
david
As a basic test case, here I’m calculating the deflection of a cantilever. In the top case I have a homogenous cantilever strip without hinges, and in the lower one I have multiple shorter strips coupled with hinges. If I calculate the rotational stiffness as E*I/L I get a value of 0,0305 [kNm/m/rad] however, I need to assign a value larger than 1000 [kNm/m/rad] to achieve the same results as without hinges.
The lower strip in the figure is the deflection of a strip with hinges with a rotational stiffness of 16,4 [kNm/m/rad], which is a random number chosen to paint the picture.
I am arguing that I should get the exact same deflection with a proparly assigned spring stiffness, and not just a spring stiffness big enough, isn’t that the case? Could you please help clarify this @karamba3d
Hello @DavAndLar,
could you post the GH definition? It would then be easier to discuss the matter.
I am not quite sure whether I understand you correctly.
– Clemens
Yes of course! Attached you’ll find the example file @karamba3d
Line Hinge Benchmark.gh (47.4 KB)
Hello @DavAndLar,
you compare a cantilever with hinges with a cantilever without hinges.
In order to make the impact of a hinge disappear it needs to be infinitely stiff, otherwise there will always be a resulting kink which increases the displacement as compared to the model without hinge.
– Clemens
But the deflection of the tip at x=L is the accumulated angle change on two neighbouring elements. If I have at some points a spring with infinite stiffness, I claim that there will be no angle change at those points (i.e. the two neighbouring elements remain parallell). Shouldn’t that result in less deflection, compared to an analysis without springs?
You are right in that the resultant displacement at the tip is the result of the accumulated deformation of the elements.
A line hinge has zero length perpendicular to its axis though. It is not like a normal element with corresponding. When a line hinge deforms in bending the result is a kink. When a beam or shell deforms in bending the result is curved line or surface.
Both sides of a line hinge are at the same position, so one needs to make the hinge infinitely stiff to guarantee a continuous deformation distribution.
– Clemens
Well, if you zoom in on a deformed beam or a shell it is really a polygon where each element remain straight, i.e. the deformation and angle change is happening in the nodes.
But maybe I should think of it in another way. That the spring is causing “additional rotation” and that the elements are pinned to the spring. Meaning of the spring is weak, we have a lot of additional rotation and if the spring is infinite stiff, it’s as if the two neighbouring elements would’ve been directly pinned to each other. Does that make sence?
Yes.
– Clemens
