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# how to applying Moment load?

#### ping

##### New member
when I analyse some shaft of solid, and I want to applying moment load to this shaft, there are warning __the moment component of a specified load will have no effect because the surface to which it is applied is not associated with any shells in the summary of run. I can't understard why and what can I do!

Ping,

I see you have asked this question several times. I will try to help.

Have you modelled your shaft with solid elements, shell elements, or beam elements?

You need to be aware that solid elements do not have any rotations defined at the nodes - just the actual translations. Accordingly, you can not apply moments directly to a solid element model - you need to come up with a system of node loads or surface pressures that equate to the desired torque. Then, when the run is complete, you will only be able to get the node translations, not the rotations directly. But you can still calculate the twist rotation, by getting the surface translations, and dividing by the radius.

If you have modelled with shell elements, only five degrees of freedom are defined - the so-called drilling freedom (i.e. rotation normal to the surface of the shell) is undefined, so you can't apply a moment which acts about the normal to the shell - again, you have to model your load using some other system of node loads or element pressures etc.

If you model a shaft with beam elements, all translations and rotations are defined, AND you can apply twist lods directly, AND you can get the resulting twist rotations directly.

Hope this helps.

JHardy

Thank you very very much!

I thought no one would answer my question, so I haven't been to this site a bit longer time. When I saw your words,I surprised pleasantly! thank you again!!

Yes, I had modelled my shaft with solid, then I got the warning

above.

Now I'm modelling my shaft with beam element, and try to solve the simplest case of shaft with only torsion. there are two supports of bearing on two ends of this shaft and two moment load(twist) which are same quantity and reverse direction. SoI think that there are two points constraints with all translation and rotation constrain fixed except X rotation because Y Direction of Vector in WCS is 0, 1,0.But some error happened, I can't finish 'run'.Why????

I try many times and using several methods,but I can't.

Ping,

Try fixing your shaft torsionally at one end, instead of applying an equal and oposite torsion to to both ends. I suspect that Pro/M is probably calculating a zero stiffness term for the torsional degree of freedom for the global stiffness matrix of the shaft, and crashes out as a result. The reason is that the whole model is unconstrained with respect to torsion.

Rember that computers run with a finite number of decimal places of precision, so even though you have applied two equal and opposite torsions, there could be some very small calculated net torsion on the shaft. However, the shaft has zero stiffness, and a very small (but finite) torsion divided by zero (or very, very small) stiffness would still give an infinite (or at least, very large) torsional rotation.

Alternatively, if you are uncomfortable with fixing one end of the shaft, try applying a very small torsional spring stiffness to some point of the shaft, and keeping the two equal and opposite torques. Check the torsional rotations, and the reaction at the soft spring - if you have done everything right, the model should run OK, the twist rotations should be OK, and it should give a negligible reaction at the soft spring.

Hope this helps!

JHardy

I have tried fixing one point as constraint's point instead of applying load with an equal and opposite torsion. And I was success to run and got correct result which is the same as the result from my pen!

This success is not only for the simplest example like I said above, but also for more complicated case where there are several bending moments and torsions and I have gotten the correct 'shear and bending moment diagrams'

Pro/M is great!!!

I'll try applying a very small torsional spring stiffness to some point of shaft until success. Before I can do that, may be I need to get more knowledge of torsional spring.

Appreciate all of the words of your!

I have another question.

I don't know what some item are meaning in Summary of Run Box

Such as Ixx;Ixy; Ixz; Iyy; Iyz; Izz below 'Mass Moment of Inertia about the Center of Mass'.I don't think they are some property of the section of my model.Becaose I can get Iyy=Ixx=0.785398 which is Moment of enertia of circle section with R=1, but I get Iyy=Ixx=7.00801e-6 in summary of run when I analyse one simple straight shatf with this section and length 15.

What's it Ixx;Ixy; Ixz; Iyy; Iyz; Izz below 'Mass Moment of Inertia about the Center of Mass'

Ping,

The moments of inertia which are being reported are the MASS moments of inertia of the model, NOT the AREA moments of inertia of the beam sections (second moment of area).

For some reason I have never understood, both the area moments and the mass moments are commonly given the same name (Ixx etc), even though they have quite different meanings and units - Length^4 for area moments, Mass*Length^2 for mass moments.

Hope this helps.

JHardy

This couple days is so busy that I haven't any time to go to Web sites.May be I have to leave studing Pro/M for some while becouse I have to do some project.Perhaps it is a little bit pity for me now.

Your help made me pass some bar of studing Pro/M in self-study and know more knowledge about principle of Pro/M or FEA although the question was very simple.So I want to say 'thank you' again!!!

I hope I can help some one on pro/m one day.

So far I need to get more knowledge about what kind of calculative result need 'Mass Moments of Inertia',for stress? strain?.... I only know 'Area section moment of inertia' is the important parameter for stress strain or deformation in material mechanics.May be I need some book in library.

Best regard!

fhashemi

Thank you fhashemi! I'll try your method.

Ping,

Mass moments of inertia are important attributes of any solid object in dynamic analysis - they provide the necessary information regarding the inertia of the object with respect to rotation about any of its axes.

In the same way that the basic equation of dynamics is:

Force = Mass x Acceleration

it is also the case that:

Torque = Mass Moment of Inertia x Angular Acceleration

That is, the smaller the mass moment of inertia of an object, the faster it will speed up when subjected to a torque about an axis.

It sounds like you could use a good reference book on dynamics, finite element analysis etc. One of the best books I have come across is Building Better Products with Finite Element Analysis by Adams and Askenazi (1999, Onword Press, Santa Fe). It is a terrific guide to practical use of finite elements generally, and is particularly relevant to Pro/Mechanica.

Hope this helps.

fhashemi

Thanks - I am just starting to come to grips with Pro/M SE - your tip is clearly the best way to apply a torque to a solid model.

Thanks again.

JHardy:

Thanks!

I just tried your method of adding torsional spring in some point of beam(shaft).I think it’s great way to apply torque to beam.

I can’t get correct deformation of rotation of shaft and it’s difficult to decide the appropriate torsional stiffness for this spring.

In the real engineering project, which method do you use.

Thanks- you remind me the the mass of inertia.

Fhashemi

I tried your method, that is great.

But I encountered some problem when I use this method to analyze the simplest 3D shaft with only torque

I applied the constrain on the one of the end of shaft and applied the moment(torque) on the other end. The process was all right, and I can run successfully,But when I displaied the Results of deformation, I found the result perhaps is’t the situation that should happend in real world.

The result, I think, should be that the radius of two end become larger together and the shaft torsion.

But the only end become larger at the direction of radius and the other end that I applied constrain wasn’t deformed at all.

Then I tried different constrains to the fixed end in roder to get the result I want to get,but I got either same situation as above or run failed.

I do’t know where is wrong, My thought? My method to apply comstrain?or some other thing?

Ping,

It's a bit hard to know where to start - are you SURE you know what answer you really expect?

I don't really understand the behaviour you are describing that you say you expect to see - the only deformation for a pure torque on the end of a shaft should be pure torsional rotation - no radial deformations etc.

I suggest you start by trying to analyse a very simple model, with a classical solution that you can easily check. For example, analyse a circular shaft, 10 mm diameter, one metre long, with a torque of 1 Newton.metre on one end, fixed at the other. Check the resulting deflections, stresses and strains against the classical solution - this should help you to see where you are going wrong.

When you are confident with your ability to solve the standard classical problems, you can attack your real world problems.

Please post the results you get if you are still having trouble.

Hope this helps.

JHardy:

I think it's difficult to express what I want to say,

and the easy way is illustration.But I can't upload picture to this forum.

Would you like to tell me your Email, then I can paste the picture to explain

My Email:[email protected]

Thank you!

Sorry guys. I was away for a while and didn't have chance to get back to you.

Ping, it's very easy when using a tool like Pro/MECHANICA to just fix all degrees of freedom when applying constraints. I see this all the time. Many times in real life models there are situations where we have fixed boundary conditions but the part has the ability to expand or contract at the boundary due to Poison's effect. In an FEA tool when you apply fully fixed constraints, you are disallowing Poison's effect at the boundary. This can cause unrealistic deformations as well as stress concentrations. So with that in mind, take a look at your situation and see if you need to modify your constraints. Instead of applying one fully fixed constraint to a surface, consider applying several partially fixed constraints to points or edges. You want the combination of the constraints to fully fix your model, but allow for Poison's effect. Take care when applying constraints to points an edges. Create them such that there would not be any reactions in directions you don't want.

You had also sent me an e-mail asking me about hand meshing. In order to hand mesh, you need Pro/MECHANICA Independent mode (outside of Pro/ENGINEER). Most people don't have the license for the independent mode. It's extra.

The FEM mode that you used, is not Pro/MECHANICA itself. With the FEM mode Pro/ENGINEER uses functionality from Pro/MECHANICA to set up FEA models for exporting to third party tools such as ANSYS, NASTRAN, etc.

Ping,

I think you will find that the apparent growth outwards in the radial direction is an “optical illusion” arising from viewing an exaggerated plot of the results. Check the actual calculated deflections at each node point – that is, look closely at the reported values of deflection in the radial and tangential directions. I think you will find that the deflection is all tangential, and that the radial deformation is zero, corresponding to pure twist, as expected.

What you are apparently seeing arises because you have an exaggerated scale of deflection. The program plots the position of each node and element at a position calculated as being:

Scale factor * Actual Deformation

from its original position.

Now, when you calculate the new coordinates of a node by moving it in a straight line, direction tangential to the original curve, but with a magnitude many times its actual displacement, you will find that it appears to move out radially, as well as circumferentially.

That is, it’s just a plotting glitch, and it is common to most FEA packages in my experience.

Hope this helps.

I am trying to do this same thing, my search lead me here and I appreciate the information. I have an additional question: When constraining the one end of he shaft, I fix the X,Y and Z directions, do I also fix all of the rotational constraints or do I allow one for rotation?

stressrise,

If you are modelling with solid elements, you don't need to apply any rotational constraints, because the solid element formulation only deals with translations of nodes, and the rotations are not defined.

If you are modelling with plate or beam idealisations, you will need to apply rotational constraints as well, as appropriate.

Hope this helps.