Took a picture from the book and attatched it.
The Attempt at a Solution
Dont I only need either the U normal stress or the U shear stress? Not even sure where to start. Thats just my problem.
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.I tried doing it like this but it seems I am still not following the correct steps to get the solution as mine does not add up.
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)
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
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)
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.
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
I didn't notice you were editing while I was posting.Didnt even notice that you were posting while I was editing.
Btw, why wouldnt the P obtained for checking the pin at C be used when its the biggest P?