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Online site for hardware strength
Does anyone know of an online site that has strength data for standard bolts and fastener hardware of different grades?
For instance I want to figure out the shear and yield strength of a 1/2" carriage bolt. I think even the low grade is strong enough for what I want, but that's just my gut, I'd rather find an authoritative source.
I've done a bunch of google searching but can't find a thing, still I can't believe that it's not out there somewhere.
Thanks!


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Try the link below. It shows common standard and metric grade markings as well as PSI ratings.
Link:


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psimonson,
Thanks! That helps, now what I need is a way to turn that into a safe working load for a given size fastener. I looked up a formula in an old 20th edition "Machinery's Handbook" c1975, and found a formula that said:
W = St(0.55dd  0.25d)
Where:
W = Working strength or permissible load in pounds (after allowance is made for initial load due to tightening [not sure how to account for this though ts])
St = Allowable working stress in tension, in psi
d = Nominal outside diameter of bolt in inches
[I used dd above to mean "d squared"  ts]
then I plugged the values from the Textron site you pointed me to, and made up a spreadsheet (I am happy to eamil it to anyone, but I don't know how to post it here other than sticking it on a web site) but got some strange results. The working strength for 1/2" and above all came out to be believable results, but for all grades and all calculations, the working load values for sizes below 1/2" all came out with negative numbers...????
I can see why this happens when I study the formula, but I don't understand how this can be? Maybe there is a typo in this 20th edition "Machinery's Manual"? Does someone else have a different formula?
I would like to believe the working load numbers for 1/2" bolts, even #2 grade came out to 688, 713, and 925, for "proof load", "yield strength", and "tensile strength" respectively. These seem quite reasonable, but I don't trust these numbers when the values for smaller bolts are "negative" strengths. Even if it were supposed to be "absolute value", the numbers get larger negatives as the bolt size gets smaller, so that can't be it either. That would say 0.25" is stronger than 0.375".
Any experts out here?


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Lets take this a few steps at a time. You do not need to deduct the pretension load from the allowable tension a bolt can be designed for. Although this seems a little strange at first, the pretension load is actually reduced by the working tension load. I'm sure that could start a debate, but that's the way it works.
When calculating the allowable shear load for a bolt, you need to know whether the threads will be in the shear plane of the connection. If the threads are included in the shear plane, you need to reduce the area for the thread depth. This is normally taken at about 70% of the gross shear area if the threads are included the shear plane.
Allowable stresses for tension and shear are calculated as follows:
Fu=Ultimate Stress or Minimum Specified Yield Stress(p/si)
Ft=Allowable tensile stress = .33Fu
Fv=Allowable shear stress=.22Fu
For a 3/4" diameter Grade 8 bolt:
Area(gross)=(3.1417 x .375 x .375 )=.44(in)2
T=Allowable tensile force=(.44)x(.33)x150,000=21,780lbs
V=Allowable shear force=(.44)x(.22)x150,000=14,500lbs
or:
T=Area x Ft Reduce the area for threads by .7
V=area x Fv Reduce the area for threads included by .7
Reduce these value by .7 if the threads are going to be in the shear plane of the bolt, and for the tensile strength of the threaded potion of the fastener.
You can perform the same calculations for any fastener, for any size, as long as you know Fu ( Minimum Specified Yield Stress ) for the base material. The chart referenced in this thread has this information in the first column to the right of the material description.
The above allowable working stresses are from the American Institute for Steel Construction, Allowable Stress Design, Ninth Edition. Hope this helps.


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Beagle,
THANK YOU! That looks like exactly the kind of information I was looking for. We are just sitting down to dinner now so I will look at this and do some calculations later. I was real diappointed in that machinery's Manual. I think that perhaps they left a square out of the second "d" term, but I will go through what you've given me later and see what I get.
Thanks very much!


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Beagle,
Well on further reflection I have a few questions/observations. First, your comments indicate that:
Fu=Ultimate Stress or Minimum Specified Yield Stress(p/si)
Yet in your example for a 3/4" bolt, you used 150kpsi which was given as the tensile strength of a grade 8 not the yield strength. The yield strength is given as 130kpsi. Should we actually use that value?
Second, you calculated tensile without taking into account the reduced area for thread root, yet I believe tensile strength of a threaded bolt has to always be based on the area of the thread root. Do you agree?
I realize that 70% is probably a pretty good estimate of thread root cross section area, but realizing that this will vary with diameter and thread pitch, I found these values in a table:
diam. TPI shank thread
area area
0.25 20 0.049 0.026 53%
0.375 16 0.110 0.068 62%
0.5 13 0.196 0.126 64%
0.625 11 0.307 0.202 66%
0.75 10 0.442 0.302 68%
0.875 9 0.601 0.419 70%
1 8 0.785 0.551 70%
For a 3/4" bolt using 130kpsi as Fu (yield strength) and the area values above and your constants for tensile and shear:
Tensile Shank Thread
Strength Shear Shear
(Thread) Strength Strength
12956 12635 8637
Do we agree so far?
Finally as I was trying to figure out what was wrong with the formula from my "Machinery's Manual", it occurred to me to ask,
"Why can't I just take the tensile strength of the bolt grade and multiply it by the cross sectional area of the bolt at the thread root to get tensile strength?"
(yes I talk to myself sometimes...
But this value for the 3/4" grade 8 bolt would obviously be substantially higher strength than the formula you gave me suggests, 45300 pounds.
I can see why a constant factor is necessary for calculating shear from yield, but why can't you take tensile strength in psi and multiply by the square area to get the tensile strength of the bolt?
In any case, I am getting closer.
Finally, I must admit to being very diappointed in that "Machinery's Handbook". I always kind of thought it was pretty reliable, I suppose there must be some kind of typo in the formula. I tried squaring the second "d" term, and at least got all positive values, but they still look about 2x too large. Maybe someone has a later version and can tell us if that formula ever got corrected? Look in the index under "Bolts, Working strength of bolts", pg 1168 to 1170 in the 20th edition.
Again Beagle, thanks!


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A couple of clarifications:
Fu is an ultimate stress value, so you use the minimum specified yield stress, not the tensile stress to calculate it.
You do reduce the area of the fastener for the root diameter at the thread to calculate allowable tensile load through the thread. I thought I indicated that, sorry if I didn't.


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For your machinery's handbook equation, do some algebra first.
W = St x d x (.55d  .25)
Well, that doesn't work either. The problem you have is once the size gets less than .455", the number goes negative. There should be a formula for smaller sizes somewhere. I don't have my handbook with me so I can't look at it. But now you have made me curious...


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Sorry I don't have access to the handbook. I've been schooled on Allowable Stress Design. Calculated Allowable Stress x the net/gross area of the fastener ( pi x r x r).
The "Factors of Safety" are built into the Allowable Stress reductions.


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Sorry for the confusion, I used the correct value, but typed the wrong property description.
Fu=Minimum Specified Tensile Stress. For the grade 8 bolt, the value is 150ksi, or 150,000psi. Thanks for paying attention. Peer reviews are invaluable.


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