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

• john237084
In summary: I couldn't find one for a horizontal slot, just a vertical slot.Yes, you are correct to assume that von-mises failure criterion, in simple tension, is the same as the yield stress of 200 MN/m2.
john237084

## Homework Statement

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.

## Homework 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.

## 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 modeled this thickness (21.6mm) to see if it gave a max stress of 50 Mn/m2 and it didn't. So I modeled 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.

Last edited:
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.

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.

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.
john237084 said:
The FEA results showed 16.5 mm to be 49.9 MPa max [von Mises] stress? Is this because you don't take the shear stress into account and the software does?

No, that is not the reason. The von Mises stress is almost identical to axial stress in this particular problem.
john237084 said:
Am I using the wrong Kt?

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.)
john237084 said:
I looked for a formula for a slot, but the only one I could find is for a [transverse] slot, not a [longitudinal] slot?

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?
john237084 said:
Am I right to assume that the von Mises failure criterion, in simple tension, is the same as the yield stress of 200 MPa?

Yes.

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.

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?

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.

#### Attachments

• Capture.JPG
6.7 KB · Views: 669

## 1. How do I find the thickness of a plate with a slot?

The thickness of a plate with a slot can be found by using the stress concentration factor (K) and the dimensions of the slot.

## 2. What is stress concentration factor (K)?

Stress concentration factor (K) is a dimensionless factor that represents the ratio of the highest stress in a material to the nominal stress. It is used to determine the effect of stress concentration caused by geometric irregularities, such as a slot, on a material.

## 3. What is the formula for finding the thickness of a plate with a slot using K?

The formula for finding the thickness of a plate with a slot is t = K*d, where t is the thickness of the plate and d is the depth of the slot. This formula assumes that the plate is made of a homogeneous material with a uniform thickness.

## 4. How do I determine the value of K for a specific material?

The value of K for a specific material can be determined by consulting a handbook or reference table for stress concentration factors. These values are typically provided for different material types and slot dimensions.

## 5. Can the thickness of a plate with a slot be found using K for any type of material?

No, the value of K is specific to the material and geometry of the plate and slot. It is important to use the correct value of K for the specific material and slot dimensions to ensure accurate results.

• Engineering and Comp Sci Homework Help
Replies
11
Views
4K
• Engineering and Comp Sci Homework Help
Replies
2
Views
2K
• Engineering and Comp Sci Homework Help
Replies
9
Views
6K
• Mechanical Engineering
Replies
12
Views
8K
• Engineering and Comp Sci Homework Help
Replies
2
Views
6K
• Mechanical Engineering
Replies
5
Views
10K