Solving Stress Concentrator Homework: Plate with Hole in Center

[steve2510]

Homework Statement

A plate with a hole in the centre is loaded in tension It has a height of 86mm (W), a thickness of 12 mm, a hole diameter of 33mm and a yield stress of 346MPa. The tensile load is 117kN.
A) Determine the stress on the top edge of the hole
B)Determine the stress on the top edge of the plate above the hole
C)Determine the maximum moment that the plate can withstand based on the stress at the top edge of the hole
D)Determine the maximum moment that the plate can withstand based on the stress at the top edge of the plate above the hole

Homework Equations

$\frac{\sigma s}{\sigma ref}$=K_s
σref = $\frac{P}{Wt}$

The Attempt at a Solution

So for part A and B i have a chart i can use to find the stress concentrator factor, i get $\frac{d}{W}$=0.384 which leads to to values for K1 the inside of the hole as 3.65 and K2 the top of the plate 0.85( From My chart) I then used σref = $\frac{P}{Wt}$ to find my reference stress.
σref = $\frac{117x10^3}{0.086*0.012}$ = 113.37x10^6 Pascals
Then using
$\frac{\sigma s}{\sigma ref}$=K_s >> σs = K*σref
For the top of the hole, σs = 3.65 * 113.37*10$^{6}$ = 413.8*10$^{6}$ Pa
For the Top of the plate above the hole, 0.85*113.37*10$^{6}$=96.36*10$^{6}$Pa

I believe i am correct up to this point. For part C and D i don't really know exactly what its asking for, i have another chart which has stress concentrator values for a plate in bending or i have the bending equation σ=-$\frac{My}{I}$
I'm using the chart, but I'm not quite sure if its correct.

Using the other chart i find that the stress concentrators are:
K2 = 1 ( top of the plate)
K1 = 0.75 ( Bottom of the plate)
This chart has 2 equations attached with it :
$\frac{\sigma s}{\sigma ref}$=K_s & $\frac{6M}{W^{2}t}$ = $\sigma$ref
So the Reference stress is biggest when K is its smallest value, therefore K1 is used. So for part C, based off the stress at the top of the hole σref = $\frac{413.8*10^6}{0.75}$
$\sigma$ref = 551.73*10^6 Pa
$\frac{6M}{W^2 * t}$ = $\sigma$ref
M = $\frac{551.73*10^6}{0.086^2 * 0.012}$ = 8161Nm
This answer isn't correct according to my sheet. I'm not sure what I'm doing wrong, for part D i would just do the same just with different stress value. Maybe I'm missing something as i haven't used the yield stress presented to me in the question.

Any help would be much appreciated.

 

[PhanthomJay]

Check stress concentration factors in bending, and correct your equation for ref stress you used 6M/W^2t instead of 6M/Wt^2. I once at the University had to solve for stress concentration factor at a hole using finite elements and partial differential equations that would challenge even the most skilled of the calculus gurus. And it was done by hand without any fancy computers. And I all the better for it I guess, having survived that awesome task. All bragging rights reserved.

 

[osman04]

I think it is better to use the yield stress instead of the stress that you had found in the first part.
Try fs = 346MPa
at the top of the hole σref = 346/0.75= 461.3 MPa
σ_ref=6M/(w^2 t)
from this equation M = 6.8235496 KN.m
same for the top plate
K2 = 1
σref = 346 MPa
so M = 5.118 KN.m
I think it is better to use the yield stress because you have been asked to determine the maximum moment and that occurs when the stress equal to the yield stress before failure could happen
I hope you got the answer now
cheers

 

[nvn]

steve2510: You failed to attach the diagrams that go with the question, and you failed to attach your stress concentration charts. Therefore, we have no way of knowing whether you are referring to in-plane or out-of-plane bending. And you did not state whether you are referring to in-plane or out-of-plane bending. Besides (even if we assume in-plane bending), we have no way of knowing exactly how your charts are defined. Therefore, we do not really know what you are talking about, nor whether or not you used the charts correctly. Furthermore, for a plate with a hole subjected to axial tension, sigma_nom (sigma_ref) is usually defined as P/[t*(W - d)], not P/(W*t). Or, for in-plane bending, sigma_nom (sigma_ref) is usually defined as perhaps something like 6*M*W/[t*(W^3 - d^3)], not 6*M/(t*W^2). That is how most charts are defined; but it completely depends on your particular charts, which we would need to see, in case you have nonstandard charts.

Also, I am not too familiar with K values less than 1.00. Therefore, some of your numbers do not seem to make sense yet. But like I said, we would need to see how your charts are defined, and what you are talking about, in order to check your work.

 

[rezeew]

I would like to commend you for your thorough attempt at solving this problem. It is clear that you have a good understanding of stress concentration and its effect on a plate with a hole. However, there are a few points that I would like to clarify and suggest for improvement in your solution.

Firstly, when determining the stress at the top edge of the hole, you have correctly used the stress concentration factor, but you have not accounted for the yield stress of the material. In this case, the stress at the top edge of the hole should be calculated as follows:

σs = K1 * yield stress = 0.75 * 346 MPa = 259.5 MPa

Similarly, for the stress at the top edge of the plate above the hole, the correct calculation would be:

σs = K2 * yield stress = 1 * 346 MPa = 346 MPa

Moving on to parts C and D, it seems that you have used the incorrect equation for calculating the maximum moment that the plate can withstand. The equation you have used, σ = -\frac{My}{I}, is for calculating the stress due to bending, not the maximum moment. The correct equation to use for this problem would be:

Mmax = \frac{\sigma ref * W^2 * t}{6}

Based on the corrected values for the stress at the top edge of the hole and the plate, the maximum moments would be:

For part C: Mmax = \frac{259.5 MPa * (0.086 m)^2 * 0.012 m}{6} = 371.2 Nm

For part D: Mmax = \frac{346 MPa * (0.086 m)^2 * 0.012 m}{6} = 494.4 Nm

I would also like to point out that the yield stress provided in the question may not be relevant for determining the maximum moment that the plate can withstand, as it is only applicable for the stress at the top edge of the hole or the plate. The maximum moment that a plate can withstand depends on its overall strength and stiffness, which may vary depending on the material and other factors.

In conclusion, your approach to solving this problem was commendable, but I would suggest double-checking your equations and considering the yield stress in your calculations. Keep up the good work and good luck with your future problem-solving