# Timeline for How can I determine the required thickness of these plate materials?

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Feb 3 '18 at 20:34 comment I haven't got any of these parts yet, I am designing the device. Kamran thank you very much, that is very kind but I don't need any free work done. What I am looking for is a technique to validate my design on my own. I think if someone can validate the units in your formulas then the problem will be solved.
Feb 3 '18 at 16:31 comment @DouglasHeld, I will calculate it, just give me the dimensions of copper plate, bolt centers and the dimension of your specimen. We are off skiing at Utah. give me a few days. Thanks. off the top of my head your plate is already strong enough, but I check. I N = 0.22481 lbs.
Feb 3 '18 at 9:18 comment I don't think it would make sense to compare Sigma, a force per CUBIC length, against tensile strength, a force per SQUARE length (PSI). So, I guess there is an error.
Feb 3 '18 at 9:12 comment OK, I added an image to the question to represent our lever distance l. Let's say it is 1" and plate width is 12" and thickness is 1/2". My moment M is 7680N * 1in * (1lb/9.8N) * (1ft/12in) = 65lb-ft. Then the slab bending moment I is 1ft * (0.5in * (1ft/12in) )^3 / 12 = 6.0X10(-6) ft^4. (What is a foot^4??) Then next the stress sigma, considering the second moment y is half the thickness; y = 1/4". Sigma = 65 lb-ft * 1/4in * (1ft/12in) / 6.0E10-6 ft^4 = 225694 lb per cubic foot.
Feb 2 '18 at 23:48 comment I picked your number but I did not convert N to lbs, you'd need to do it. X is multiplication sign and lower case l means the vertical distance of the bolt center to the edge of specimen. That is the moment your copper plate needs to support.
Feb 2 '18 at 23:32 comment And you get 7680 how? And what does x represent?
Feb 2 '18 at 3:32 comment @DouglasHeld, let's just plug in the numbers using English units. Copper yield tensile strength =4830 psi, say we set factor of safety at 40% thus our limit is say 2000 psi allowable. Then say the thickness of your plate, h is 1/8" so you get 2000= 7680x l (say the distance from bolt cen to edge of specimen in inches).1/16"/I. and you calculate I as mentioned in my answer. If your answer is bigger than 2000 you need thicker copper, if smaller you consider thinner plate and check it again!
S Feb 2 '18 at 2:34 history
added formatting and punctuation in attempt to understand the message
Feb 2 '18 at 1:01 comment My apologies for using my phone and not having my reading glasses. Basically we are measuring straight from the center of the bolt to a dash line continueation of edge of your specimen. The y is not the young modules, it is the heigth of the point in cross section of your plate under consideration for stress and because this is bigest at the top and bottom surfaces of your plate, we use height/2, h/2. I is called second moment of inertia.
Feb 2 '18 at 0:49 comment kamran, I cannot visualize this measurement $l$ thought I have re-read several times. Do you mean to measure diagonally from the bolt center to the fulcrum point of the bend? Similar to my red line? And can you explain where you get 0.707? And the $y$ in the stress calculation -- is this the Young's Modulus of the material?
Feb 2 '18 at 0:41 review
S Feb 2 '18 at 2:34
Feb 1 '18 at 23:28 history answered