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Calculating the diameter of a pin in double shearby rad10k
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#1
Jan1011, 06:36 AM

P: 61

1. The problem statement, all variables and given/known data
Calculate the diameter of a meduim carbon steel pin in double shear subjected to a suddenly applied shearing force of 150000N? 2. Relevant equations A=p/t p=300000 ( double shear) t=580 ( ultime shear strength) 3. The attempt at a solution Double shear 150000 x 2 = p A = p/t 30000/580 = 517.24138 square root = 22mm Answer 22mm Is this correct ? if not could someone please help me where i am going wrong thanks 


#2
Jan1011, 01:14 PM

P: 137

Two things: Since it's doubleshear, you divide the shearing force by two, not multiply. Also, you seem to have gotten the equation for area of a circle incorrect  it's not d^{2}. It's (pi*d^{2})/4.



#3
Jan1011, 03:39 PM

P: 61

so the force should be 150000 / 2 = 75000 but then it is suddenly applied so 75000* 2
the force is then 150000N ? 


#4
Jan1011, 04:37 PM

P: 696

Calculating the diameter of a pin in double shear
What are your units of t?
And are you using a shear strength for gradually applied loads, rather than the more appropriate value for suddenly applied loads? 


#5
Jan1011, 05:02 PM

P: 61

Hi, t is the ultimate shear strength of medium carbon steel taken from a table in my course work book. It is 580. Thanks



#6
Jan1111, 08:40 AM

P: 137

No. Since it's a doubleshear pin  the force is equally spread across two shear points. Thus, the force at either individual point is 75000 N. Use that force, the shear strength, and the equation for area of a circle (since it's asking for diameter, we assume the pin's crosssection is a circle) that has diameter in it. Another way to think of it is that your crosssectional area is doubled, since it's doubleshear, and you'd use the force of 150000 N in the equation. Either way, you'll come out with the same answer. Were this pin in singleshear, the force would simply be 150000 N and 1x the circular crosssectional area. Also, as pongo mentioned  what are your units for t? It would seem to be MPa, which is N/mm^{2}  but that would seem a little high for medium. But, if that number (in MPa) is what your book states, then go with it. One last thing  your original post states "subjected to a suddenly applied shearing force of 150000N." Are you certain that that's how it is worded? Or does the problem refer to the 150000 N as an "applied" force? I would think they'd use "applied," but maybe they're trying to fool you. 


#7
Jan1111, 03:19 PM

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P: 6,036

'Suddenly' applied I think implies impact loading , resulting in forces that are approximately twice as high as static "slowly" applied forces. Each case would have its own safety or load factor.



#8
Jan1111, 06:14 PM

P: 61

Yes the wording in the book is " Suddenly applied force" of 150000N. The book also states as phantomjay says "a suddenly applied force is doubled " that was my reason for suggesting that I divide the double shear in half then double it back again for the "suddenly applied force" giving me 150000N again..... also
the units of t are in N/mm2 and are for ultimate strength in shear for medium carbon steel. Having said all this I still seem to be confuse about the formula for finding the diameter of the pin. Thanks 


#9
Jan1211, 12:01 AM

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rad10k: I agree with PhanthomJay and you. You were already doing it correctly when you doubled the suddenly applied load. And N/mm^2 are called MPa. Also, 580 MPa indeed sounds plausible for shear ultimate strength of medium carbon steel. This could be, e.g., steel UNS G10600 (SAE 1060, AISI 1060), quenched and tempered, or something like that. Think ISO property class 10.9 bolts , called "medium carbon steel, quenched and tempered," which are somewhat stronger than your steel. Because you know the formula for area of a circle (post 2), and because you know the formula for shear stress, tau, plug in all the values you know, and then solve for diameter, d. Give it a try again, and show your work, and PhanthomJay or someone will check your math.
You almost did it correctly in post 1, except, among other mistakes discussed previously, you also lost track of the correct number of zeros. That is why the international standard says the following.



#10
Jan1211, 07:42 AM

P: 61

Thanks for the advice :o)
Ok here is a new attempt at a solution: Original force 150 000N Suddenly Applied = 150 000 * 2 = 300 000N Double Shear 300 000 / 2 = 150 000N P = 150 000N t = 580 N/mm2 ( I have table titled "Shearing and Torsional Strength of Materials" for medium carbon steel I have the choice of " Yield Stress = 280 N/mm2" or " Ultimate Shear Stress = 580 N/mm2" I have chosen to use ultimate shear stress ???) Formula : p/t /2 / (0.25*pi) = d 150 000N / 580 = 258.62 258.620 / 2 = 129.31 0.25 * 3.14 = 0.785 129.31 / 0.785 = 164.72 Square root of 164.72 = d d = 13 mm ( Rouded up ) Any Better ?? thanks 


#11
Jan1211, 09:10 AM

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P: 2,121

rad10k: Close, but not quite. Hint 1: You already divided your applied load by 2 for double shear, so why did you divide by 2 again? Try again.
Also, remember what I wrote in post 9. Always leave a space between a numeric value and its following unit symbol. E.g., 150 000 N, not 150 000N. N/mm^2 is called MPa. Always use the correct, special name for a unit. E.g., 580 MPa, not 580 N/mm^2. Also, do not round so much. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. 


#12
Jan1211, 09:11 AM

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rad10k....Does the problem mention a safety factor under impact load?



#13
Jan1211, 10:20 AM

P: 61

No does not mention and saftey factor or load factors
ok so... Question ; Calculate the diameter of a medium carbon steel pin in double shear subjected to a suddenly applied shearing force of 150 000 N ? Solution : 150 000 N * 2 For Suddenly Applied = 300 000 N 300 000 N / 2 For Double Shear = 150 000 N P = 150 000 N t = 580 Mpa p/t / (0.25*pi) 150 000 N / 580 Mpa / 0.785 . square root = 18 mm Thanks for everyones patience :o) 


#14
Jan1211, 11:37 AM

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rad10k: Your answer is correct. By the way, always use correct capitalization of unit symbols. E.g., MPa, not Mpa. See NIST for the correct spelling of any unit symbol.



#15
Jan1211, 11:49 AM

P: 61

excellent Thanks everyone and thanks nvn for the advice on correct way of displaying units :o) I tdoes make it much easier to read.



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