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- Re: Frame analysis UDL and Point Loads

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1584 Views, 4 Replies

10-10-2012 02:43 AM

Having trouble with frame analysis and agreeing with calculations (which admittedly might be flawed).

I understand that in a normal part to apply a load to a face is to make it a UDL (uniformly distributed load) by default. To apply a point load, I was told to create a small sketch and use it to split a face to create, in effect, a small face on which to apply a load, effectively making it a point load.

Now in the frame analysis environment, there is clearly an option for 'force' and one for 'continuous force' which the description clearly describes as a uniformly distributed load. However, when I run the simulation (on a 2m long 100mm x 50mm x 4mm Rectangular hollow beam) with a normal load in the middle of the beam (1m offset); the maximum moment comes out at 500000 Nmm or 500 Nm. By my reckoning the moment should be 1000 Nm as the basic formula for a moment from a point load is M = F*D (Moment = Force * Distance). However, to get 500 Nm, it looks like Inventor is using M = (F*x^2)/2 where x is the midspan of the beam (in this case 1m). Therefore my question is...is a normal load in frame analysis the same as a UDL? If so..how to I make it a point load?

Screenshot is attached. There are also some errors stating instability of type 2 at various nodes, no idea why as I've used both Pinned and Floating constraints at either end of the beam.

Sam

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10-12-2012 12:50 AM in reply to:
samstreet101

Any suggestions on this at all?

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10-12-2012 01:40 AM in reply to:
samstreet101

I may be wrong here, but I don't think that a moment force reaction is what you're after judging by your screenshot.

The scenario you present is a classic simply supported beam with a centrally located downward force, in which case:

A shear stress calc would be F/2A with A being the external cross section area minus the internal CSA.

I assume also that you're applying 250N of force? My shear stress comes to 110 kN/m^2.

Bending stress using FL/2Z = 6.76 MN/m^2

Deflection I work out to be 0.04mm using 1/48 * FL^3/EI where E=205GPa and I = 46.2 cm^4

Not sure if this helps, but I can't help think that a moment isn't required?

Please click "Accept as Solution" if this response answers your question.

-------------------------------------------------------------------------------------

Autodesk Inventor Professional 2012

Windows 7 Professional, 64-bit (Service Pack 1)

Intel(R) Core(TM i7-2600 CPU, Quadro 2000 graphics card, 8G RAM

-------------------------------------------------------------------------------------

Autodesk Inventor Professional 2012

Windows 7 Professional, 64-bit (Service Pack 1)

Intel(R) Core(TM i7-2600 CPU, Quadro 2000 graphics card, 8G RAM

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10-17-2012 04:45 AM in reply to:
PhilSaw

Hi Phil,

Thanks for the reply. To clarify, I was using the general bending equation (M/I = sigma/y = E/r where M=moment, I=second moment of area, sigma=stress, y=distance to neutral axis, E=young's modulus, and r=bend radius) I'm trying to reconcile a classroom problem with something modelled in Inventor. I know the glyph looks like a point load applied in the centre but then so does a normal load applied to a face in a normal stress analysis environment, however, when you do this, you're actually applying a uniform load to the selected face. I'm trying to work out whether the same thing is happening here. You're right when you say I don't actually need to know the moment, it's simply the fact that Inventor said the moment in the centre was 500Nm as opposed to 1000Nm, made me think that Inventor was treating the load as a UDL rather than a point load.

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10-22-2012 05:24 PM in reply to:
samstreet101

Hi Sam,

I believe the 500 N*m moment load max at the center of the beam is correct. Here's why:

0) Solve for the reaction forces R_y. They are +500N at each support (1/2 of the concentrated load P=-1000N at the center of the beam, using the normal sign convention)

1) Draw a free-body diagram for the left side of the beam, but "cutting" the beam in half.

2) Sum the moments about the center of the beam (M_c) taking counter-clockwise as the positive direction / normal sign convention:

- Sum of moments about c = 0 = M_c - (P)(0m) - (R_y)(1m)

3) Solving for M_c we get:

- M_c = R_y * 1m = (500 N) (1m) = 500 N*m <= (agrees with computed value in Frame Analysis)

For your second question about why we get instabilities messages from the solver using floating or pinned constraints: A simply supported beam has one end fixed, and one end floating / frictionless. We'll get different results if we have say both ends fixed. The beam wont have an angle of deflection at a fixed end. Also, if both ends are pinned, they are fixed in the translational DOF. So the beam ends cannot move towards each other when flexed, so the beam is kind of streching as well as bending. Also, with basic beam theory, we mainly deal with 2D. Frame Analysis is 3D, so for one end pinned and the other as a 'floating / roller' we still have a DOF for the floating constraint going in the 3rd orthogonal direction (into the plane). This is why an 'instablility' messge originates. Depending on the loading of the model, these can sometimes be ignored and regarded as zero.

There is a 'Being Inventive' blog post about applying loads and constraints to a simply supported beam in Stress Analysis (being mindful about fixed versus floading, and UDL versus concentrated load): Here

Apologies for the late reply. Please let us know if you have additional questions, comments or suggestions.

Best regards, -Hugh

Simulation QA Engineer

DLS

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