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steve2510
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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
[itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks
σref = [itex]\frac{P}{Wt}[/itex]
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 [itex]\frac{d}{W}[/itex]=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 = [itex]\frac{P}{Wt}[/itex] to find my reference stress.
σref = [itex]\frac{117x10^3}{0.086*0.012}[/itex] = 113.37x10^6 Pascals
Then using
[itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks >> σs = K*σref
For the top of the hole, σs = 3.65 * 113.37*10[itex]^{6}[/itex] = 413.8*10[itex]^{6}[/itex] Pa
For the Top of the plate above the hole, 0.85*113.37*10[itex]^{6}[/itex]=96.36*10[itex]^{6}[/itex]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=-[itex]\frac{My}{I}[/itex]
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 :
[itex]\frac{\sigma s}{\sigma ref}[/itex]=Ks & [itex]\frac{6M}{W^{2}t}[/itex] = [itex]\sigma[/itex]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 = [itex]\frac{413.8*10^6}{0.75}[/itex]
[itex]\sigma[/itex]ref = 551.73*10^6 Pa
[itex]\frac{6M}{W^2 * t}[/itex] = [itex]\sigma[/itex]ref
M = [itex]\frac{551.73*10^6}{0.086^2 * 0.012}[/itex] = 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.
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