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- Thread starter Suitengu
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PhanthomJay

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PhanthomJay

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btw, what would di cross section be? the 8x36mm area or the 36x0.2*10^3 mm area?

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PhanthomJay

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Factor of Safety = Ultimate stress / Allowable stress

A.S. = 400/3

= 133.3 MPa

Stress = Force/Cross-sectional Area

133.3 = F/(6(18-10))

I subtracted the bolt's diameter from width of BD.

F = 6398.4 N

Sum of the moments about C (anticlockwise +ve) = P(120+160) - 6398.4(120) = 0

P = 2.742kN (which is so wrong)

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PhanthomJay

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A couple of points here. First, you didn't check the shear stress in the pins. That might limit the load to less than what you calculated based on the allowable tensile stress in the link. For example, if the link can withstand 6398N, the shear stress in the 10mm diameter pin is F/A = 6398/(pi)(.0005)^2 = 81.5MPa which exceeds the allowable 50MPa shear stress in the pins. You've got to check that 6mm dia pin also (the one in double shear) to see if that may govern.

Factor of Safety = Ultimate stress / Allowable stress

A.S. = 400/3

= 133.3 MPa

Stress = Force/Cross-sectional Area

133.3 = F/(6(18-10))

I subtracted the bolt's diameter from width of BD.

F = 6398.4 N

Sum of the moments about C (anticlockwise +ve) = P(120+160) - 6398.4(120) = 0

P = 2.742kN (which is so wrong)

Secondly, I don't know if you are using a steel code for the net area calculation in the ternsion member. A 10mm pin has a slightly higher hole size, such that this fact combined with a code adder my be reduce the effective width of the 18mm link down to say 5 or 6mm instead of the 8mm you used. It might make a difference as to what governs, but if the problem does not state a hole size and does not reference a Code, then I guess its OK to assume just a 10mm hole for a 10mm pin, which may not be realistic.

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Like how do you test for pins in double shear? Sorry for asking these questions but I am also without a textbook and I am just going off of what little I got in notes. I plan to get one this week however so this wont happen again. But in your shoes, walk me through the first step you would do? Also I subtracted that 10mm bolt hole diameter from the link's cross-sectional area. How is that justified as I dont see it affecting the area in any way and if so why didnt we subtract two bolt holes although that would be more than possible to do?

Oh you also made a mistake, for you to get MPa you should have left the 5mm as it is. You converted it to meters and N per meter squared is not MPa

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PhanthomJay

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Well just forget it and your calc for net area is correct.Hmmm...I am at a lost then mon on what to try out. As this is my first week in the class I am not very certain about the steps to follow and so on and so forth. Also that last part you said came as a shock to me as its the first I am hearing of this.

If you look at your sketch, can you visualize that the 10mm pins are in single shear because there is only one shear plane across the pin, whereas at the 6 mm pin, there are 2 shear planes? This means that when analyzing a bolt or pin for double shear, you can double the area of the bolt when calculating the average shear stress in that bolt.Like how do you test for pins in double shear? Sorry for asking these questions but I am also without a textbook and I am just going off of what little I got in notes. I plan to get one this week however so this wont happen again. But in your shoes, walk me through the first step you would do?

I don't know what you mean. Without the hole, the area of the link is 18*6 = 108mm^2. With the hole, the net area is (18-10)*6 = 48mm^2Also I subtracted that 10mm bolt hole diameter from the link's cross-sectional area. How is that justified as I dont see it affecting the area in any way

If you and I were to pull on a bar at each end, with you pulling on one end and me pulling on the other end, with each of us pulling with a force of 100N, the bar sees 100N of force, and the tensile stress is uniform throughout. Similarly, the end of each link with the hole sees the same force, and the reduction in area at the hole is just for the one hole, not two holes. If there were 2 bolt holes side by side at one end, then you'd have to subtract them both from the width to get the net width, but this is not the case.and if so why didnt we subtract two bolt holes although that would be more than possible to do?

Well now being from the States I'm not into this metric and Pascal stuff, and I slipped a decimal point as a typo when converting 5mm to meters, but the result is correctOh you also made a mistake, for you to get MPa you should have left the 5mm as it is. You converted it to meters and N per meter squared is not MPa

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Okay so since we are sure that F cannot be 6398 N, how does one go back and calculate a different force there. I'm still not sure where to go from here. Oh and to check the 6mm diamter bolt, I would:

Allowable shear stress = Force/(2*A)

Force = (50*2(1/4)(6)^2*pi)

= 2,827N

But how does that help me and where would I go from there?

I think you may need to just walk me through this problem from start to finish as I am more confused than ever although I now understand why the bolt was subtracted in finding the net area. Sorry for the stress mon. No pun intended.

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PhanthomJay

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Your use of the word 'mon' should have been an easy giveaway of your Jamaican roots, but I tend to be naive in these matters, and thought it was a misspelling.I am not from the states either mon. Jamaica is where I hail from so I see what your saying. And now I finally understand the concept on why they subtract the bolt hole.

Okay so since we are sure that F cannot be 6398 N, how does one go back and calculate a different force there. I'm still not sure where to go from here. Oh and to check the 6mm diamter bolt, I would:

Allowable shear stress = Force/(2*A)

Force = (50*2(1/4)(6)^2*pi)

= 2,827N

This is good. Now proceed as you did in the beginning when you did the tensile link force calc and backtracked in the statics by taking moments to solve for what the applied P force would be if link tension controlled. If you have a 2827N force at C, what is P? If its less than the value you arrived at for the link tension case, then the pin force at at C controls over it. BUT, you still need to look at the case where the 10mm pin shear stresses may govern the value of P. The lowest P wins.

No stress, man.I think you may need to just walk me through this problem from start to finish as I am more confused than ever although I now understand why the bolt was subtracted in finding the net area. Sorry for the stress mon. No pun intended.

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lol. Alright I did it also for the shear stress in the 10mm diameter pins and then use it also to find P and found that it produced the lowest value for P so yes it does win. I got it to be 1,683 kN. Thank you for your help mon. I got it in the nick of time as well as I almost gave up.

P's were; 2,742N, 2,120N and 1683N

Btw, why wouldnt the P obtained for checking the pin at C be used when its the biggest P and I guess it wouldnt break at that value?

P's were; 2,742N, 2,120N and 1683N

Btw, why wouldnt the P obtained for checking the pin at C be used when its the biggest P and I guess it wouldnt break at that value?

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PhanthomJay

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You are welcome, nice work.lol. Alright I did it also for the shear stress in the 10mm diameter pins and then use it also to find P and found that it produced the lowest value for P so yes it does win. I got it to be 1,683 kN. Thank you for your help mon. I got it in the nick of time as well as I almost gave up.

P's were; 4,487 N, 2827N and 1683N

Thank you again

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Btw, why wouldnt the P obtained for checking the pin at C be used when its the biggest P?

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PhanthomJay

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I didn't notice you were editing while I was posting.

Btw, why wouldnt the P obtained for checking the pin at C be used when its the biggest P?

I checked all your numbers, and you have now correctly identified the 'P' loadings. P would be 2742N if you were just looking at the link max allowable tensile stress and paying no heed to the pins; P would be 2120N if you were just looking at the pin at C max allowable stress, paying no heed to the other pin or the link; and P would be 1683N if you were just looking at the 10mm pins and paying no attention to the 6mm pin or the link. But now you must compare all three values and pick the lowest. If you choose P as 2742N, the link holds up fine, but the pins get overstressed; if you choose 2120N for the P load, then the pin at C and the link are fine, but the 10 mm pins are overstressed. If you choose P as 1683N, then all pins and the link are within their allowable stresses. That's why you pick the lowest P as your answer. OK?

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