# Finding the thickness of a plate with a slot, after finding K (SCF)

1. Dec 21, 2011

### john237084

1. The problem statement, all variables and given/known data
Hi all, I was wondering if anyone can help with this?

A uniform Steel strut (material Modulus of Elasticity 210 GN/m2, Poisson’s Ratio 0.3) of length 400mm and width 80 mm is restrained as appropriate at one end and under an axial tensile load of 20 KN applied to the other end face. It has been initially designed with a thickness of 5mm. This gives a hand calculated stress value of 50 MN/ m2, which is a quarter of the material’s Yield Stress, of 200 MN/m2, giving a factor of safety of four.

The strut now requires a round ended slot in the centre, right through, of Length L and width W. Determine the minimum thickness of strut of the same length, width and material, capable of withstanding the same tensile force with the same factor of safety. Assume Von Mises failure criterion apply, i.e. the maximum Von Mises (unaveraged) stress should not exceed 50 MN/m2. Your final thickness must be an integer millimetre value to use standard size material.

2. Relevant equations
Kt = σmax/σnom
ratio for Kt = hole diameter/plate width
σnom = P/A, A = smallest cross sectional area at the section with the highest stress.
σmax = Kt*(P/A)
A = t*40 t = thickness.

3. The attempt at a solution
Petersons chart 4.1 shows d/W to be 0.5 = Kt of 2.16. For a finite plate with a central hole in uniaxial stress.

σnom = σmax/Kt = 23.148 Mn/m2, through trial and error 21.6mm x 40mm = 864mm2 = 0.000864m2.(Which is obviously a plate thickness of 22mm for the integer value)

σnom = 20*10^3/864*10^-6 = 23.148148 Mn/m2, to prove this I put it back into σmax = Kt*(P/A) and it = 2.16*(20*10^3/864*10^-6) = 50MN/m2

So I modelled this thickness (21.6mm) to see if it gave a max stress of 50 Mn/m2 and it didn't. So I modelled some smaller thicknesses and found through modelling that the FEA results showed 16.5mm to be 49.9 MN/m2 max stress?

Is this because you don't take the shear stress into account and the software does?

So if you could help I'd really appreciate it. Am I using the wrong K? I looked for a formulae for a slot but the only one I could find is for a vertical slot, not a horizontal slot?

Also am I right to assume that von-mises failure criterion, in simple tension, is the same as the yield stress of 200 MN/m2?

Sorry if this seems long winded and a bit hard to understand, but I'm new to this stuff and quite long in the tooth. I really apreciate you taking the time to look, I'm not looking for someone to answer the whole thing, but I would like to know that I'm not wasting my time.

Thanks in advance.

Last edited: Dec 21, 2011
2. Dec 22, 2011

### nvn

john237084: What are the values for W, L, and d, or did I miss them?

By the way, MN/m^2 is called MPa. Always use the correct, special name for a unit. E.g., 50 MPa, not 50 MN/m^2. You need a caret (^) symbol before your exponents. E.g., 864 mm^2, not 864 mm2. Also, always leave a space between a numeric value and its following unit symbol. E.g., 21.6 mm, not 21.6mm. The unit symbol for kilonewton is spelled kN, not KN. Uppercase K means kelvin. Always use correct capitalization of units.

3. Dec 22, 2011

### john237084

I apologise for my ignorance. Like I said I'm very new to this.

Thankyou for also pointing out my glaringly massive mistake in missing out the dimensions

The slot is 85 mm x 40 mm, which makes the radius of the rounded end 20 mm.

Thanks again.

4. Dec 23, 2011

### nvn

john237084: Is L = 85 mm the distance from radius centre point to radius centre point of the slot? Or is L = 85 mm the distance from peak to peak of the slot rounded ends? What is the distance between the two radius centre points of the slot?

Is this a school assignment or test question?

Your current FEA result looks (perhaps) slightly inaccurate. Were your finite elements too big? What was your approximate element size adjacent to the full rounds of the slot? Was it 1.0 mm, 2.0 mm, or what?

Did you have element stress averaging turned on or off? The given problem statement strangely seems to claim element averaging should be turned off (?), although this makes only a slight difference.
No, that is not the reason. The von Mises stress is almost identical to axial stress in this particular problem.
Yes. You cannot use a round hole. A round hole is completely different from a slot. (Stress concentration factors are influenced by a lot of phenomena, and a slot is a different scenario than a round hole.)
That is correct. Neither Peterson nor Roark seem to contain a longitudinal slot. Therefore, it currently appears your scenario is unavailable. Are you required to also look up Kt in a book? Or are you only required to use FEA?
Yes.

5. Dec 28, 2011

### john237084

Hi, sorry I've took so long to reply.

Yes the 85 mm is the overall length of the slot, which would make the distance between the radius centre points 45 mm.

This is an assignment, the first one I've been given and I was struggling to find K. I was required to analytically find the axial loading, stress concentrations and the Von-Mises failure criterion. So, if Von-Mises is the same as the yield, then that's not a problem. Without the thickness, found with K, I couldn't work out the axial stresses at the critically stressed area.

The finite element size was 5 mm (any smaller and the P.C gave up the ghost) for the overall strut, with a local element density of 25 around the slot.

We were told to use the unaveraged stresses with the FEA.

I guess if there is no way I can find a scenario for K, then I'm stuffed. It's funny because the lecturer told us to use a hole and I remember thinking, if that was the case, then why give us all different lengths of slot?

Thanks.

6. Dec 28, 2011

### nvn

john237084: Perhaps the lecturer wanted you to discover how different a slot is from a round hole (?).

5 mm mesh sounds too big. I think 0.5 mm mesh adjacent to the (side of the) round would be good. Or 1 mm mesh. If you do this, I think you might find your strut thickness in post 1 is slightly inaccurate, by perhaps more than 1 mm (?). I think you could reduce your model disk/memory size by using a quarter model for this analysis, instead of a full model. Regardless, if you obtain a different strut thickness than what is currently listed in post 1, please let us know.

What type of finite element are you using? And how many nodes does each element have?

7. Dec 7, 2013

### Darklordz28

Final thickness required for 3D FEA

Hi there
Well, seems this thread has to be alive again!.
I think we both from same uni john237084 !! I got the same assessment as yours (since 2011) with small changes in dimensions, but the software PLM-UGS upgraded to NX 8.5.
Here it is:
A uniform Steel strut (material Modulus of Elasticity 200 GN/m2, Poisson’s Ratio 0.3) of length 0.5 m and width 100 mm is restrained as appropriate at one end and under an axial tensile load of 25 kN applied to the other end face.
It has been initially designed with a thickness of 5mm. This gives a hand calculated stress value of 50 MN/ m2, which is a quarter of the material’s Yield Stress, of 200 MN/m2, giving a factor of safety of four.
The strut now requires a round ended slot in the centre, right through, of Length 55 mm and width 10 mm. Determine a thickness of strut of the same length, width and material, capable of withstanding the same tensile force with the same factor of safety. Assume Von Mises failure criterion apply, i.e. the maximum Von Mises (unaveraged) stress should not exceed 50 MN/m2. Your final thickness must be an integer millimetre value to use standard size material.

I tackled the problem with same method finding the KC value first then find the possible thickness to apply 3-D FEA using the software however, it's only for circular hole not a slot!

Please give me a hint how to resolve this … your help would be much appreciated.

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