ASME Sec 8 UHA 44

[Nashanas]

Hello,

I have a seemingly dumb question. Does anyone know if the formulas given in the UHA 44 to calculate % strain give values in % or should be multplied with 100 to get the percentage?

e.g If I have a 5mm thick plate with 420mm span, and I want to find the strain. Using formula for cylinders I get 0.034, which to me seems very less, I think that it needs to be multiplied by 100 to be expressed in %.

 

[Geoff13]

I believe the Table UG-79.1 formulas are in % strain. Not clear what your Rf value is so can't double-check your numbers. A question on Code rules would be better posted in: forum794

However, if you've just got a flat plate spanning between two supports and sagging under self-weight plus some loading : Calculate your mid-span stress, and use your material E value to convert stress to strain.

 

[Nashanas]

Thank you for response. In future I will post in the mentioned forum. But now for the sake of discussion, the formula that I am using for finding R is:

R = (a^2 * b^2)/2a

here a = deflection = 3
b = half of the span = 210mm
so R = 7351.5mm

Using this value of R in the strain forumula for cylinders I got the value 0.034.

Now, for curiousity I want to know the value of delfection at which the strain will reach 15%. If the above value 0.034 is percentage then I am unable to find the 15% value, because even at deflection a = 210mm (if a = b then R = a = B), whcih would mean that the plate has been bent to form a "U", the strain is only 1.19%. However if these values should be multiplied by 100 then it becomes 119% which makes sense, and also the above value 0.034 becomes 3.4%.

 

[Geoff13]

With such a large radius on such a thin plate the 0.034% strain makes perfect sense.

Since 15% strain is a huge value you can only get it in your thin plate by going to a very small radius. By substituting 15% into the UG-79.1 formula, starting from a flat plate, you can solve that Rf would be about 17 mm.

Looking at it another way, for 15% strain you want the circumference at the inside radius to be 15% smaller than at the centreline radius. That means 0.5t must be 15% of Rf. In your case 0.5t is 2.5 mm so Rf is again about 17 mm. This logic is the basis of the UG-79.1 formula, so of course it gives the same answer.

I suspect secondary effects are relevant at this sort of strain, and this simple geometric logic would not longer work.

 

[HTURKAK]

The R value 7351 mm is correct..

the strain value 0.034 is also correct...



0.015=50*5/R R=1667 mm


(1667-def)**2+ 210**2=1667**2 say def =d

so d**2-3334*d+210**2=0 solve this quadratic eq. to obtain d ..

good luck..