Bolt thread loading

[johnfin1]

I was looking at this chart and I dont understand why the numbers are different. Says the working load for a 1/4 unf bolt is 160 lbs (tensile) yet underneath on the lower chart is says 2700 for the same bolt.

http://dodgeram.org/tech/specs/bolts/SAE_bolt_strength.html

 

[ESPcomposites]

For the upper chart, it looks like they are assuming a 6ksi strength (working strength). They called it load, but they meant strength (i.e. its in psi).

For the lower chart, they are using the tensile strength of the material (as opposed to a design strength). Obviously, it would be much higher.

Brian

http://www.espcomposites.com

 

[johnfin1]

The first chart says (lbs) for load as in you can lift something with a threaded bolt into the load, straight lift using a 1/4 bolt and the load weight of 160lbs. So if the the thread shear calcs say are at 6000 psi, what does that equate to in a dead lift. I am confused here.

 

[SnTMan]

The upper chart says "Safe Working Load", the lower chart says "Breaking Strength", approx a 17:1 FOS.

This chart is worth what it cost you

Regards,

Mike

 

[johnfin1]

Mike, I did some threaded bolt calcs using the shear area of the bolt and such and came up with around 6k psi for the threads to shear. Are you saying that I can use around 17:1 to figure out what the bolt could safely lift or is there a more acurate way to figure that out?

 

[rb1957]

why wouldn't you use the chart labelled "safe working load" for the load the bolt can safely lift ?

 

[johnfin1]

First of all its only in grade 2 UNF and second it would be nice to learn something and see how those numbers are calculated. Kind of a cross check too, dont believe every chart I read on the internet.

 

[ESPcomposites]

I am pretty sure it is what I already said in the second post. But to clarify:

First table:

- P/A (root area) = 6,000 gives you the first load = 160. This is just a "design stress" of 6ksi that they chose.

Second table:

- P/ A (root area) = Ftu gives you another load. At about Ftu = 75ksi, this gives about 2700 lb.

So it looks pretty straight forward to me.

Brian

http://www.espcomposites.com

 

[johnfin1]

I can see what the chart says, how do you get 160 lbs from a shear strength of 6000ksi.

 

[SnTMan]

johnfin1, get yourself a Machinery's Handbook. Contained within is nearly everything you need to know about calculating bolt / thread strengths, including root areas, likely tensile strengths, thread engagements, etc. It is a pretty extensive subject to cover in a message board.

To crunch some numbers from your chart:

Working Load Chart: 160 lb / 6000 psi = 0.02667 in^2 root area for 1/4 UNF. I get at least 0.0326 in^2 from said Handbook

Breaking Strength Chart: 2700 lb / 0.02667 in^2 = 101,000 psi. I doubt I would trust this for a GR 2 "iron" bolt.

2700 / 160 = about 17 (factor of safety, if you believe 2700).

All tensile, all assuming enough thread engagement to fully develop the strength of the bolt. Which is pretty much normal practice.

Nothing against Geno, but I'm sure there is much better info available on the 'net.

Regards,

Mike

 

[ESPcomposites]

agree with SnTMan, have a look at a machinery handbook.

John, I think your problem is that you are trying to use the shear strength to calculate a tension failure mode, which does not make sense. To calculate the tension LOAD capacity, there are two failure modes. One for tension failure, using the root area, and relates to the loads you are asking about (160 vs 2700). The other failure mode, due to shear of the threads is a separate matter. The actual capability is the minimum of either.

Either way, it does roughly add up. Your question asked why are they so grossly different (160 vs 2700). That should be clear now. As stated previously it is simply the fact that one calc is basing it on a design stress and the other a material strength. Nothing more to it really.

Brian

http://www.espcomposites.com

 

[johnfin1]

Ahh, the math. I understand now. Got a copy of the book at the library, 23rd addition, 2 feet thick. I calculated 8700 not 6000 for a 1/4-20 bolt, goes up higher for fine thread, not sure why, thought course had a higher rating. Found the following formula and put it into mathcad. See attachment.

 

[johnfin1]

If I use this calculator and the machinerys hand book the 1/4-20 is only good for 114lbs

http://www.fastenal.com/web/services.ex?action=FEDS&calculator=Load

Their(Fastenal) cross section area (Ats) differs from the Machinerys handbook (tensile strength area).

....""Thread Stength
Two fundamentals must be considered when designing a threaded connection

1.Ensure that the threaded fasteners were manufactured to a current ASTM, ANSI, DIN, ISO or other recognized standard.
2.Ensure that the design promotes bolts to break in tension prior to the female and/or male threads stripping. A broken bolt is an obvious failure. However, when the threads strip prior to the bolt breaking, the failure may go unnoticed until after the fastener is put in service.
Internal Thread Strength Formula
F = Su * Ats
Su = shear strength of the nut or tapped material
Ats = cross-sectional area through which the shear occurs

The strength of bolts loaded in tension can be easily determined by the ultimate tensile strength. To determine the amount of force required to break a bolt, multiply its ultimate tensile strength by its tensile stress area, As.

Determining the strength of the threads is more complicated. Since the male threads pull past the female threads, or vice-versa, the threads fail in shear and not in tension. Therefore, the stripping strength of an assembly depends on the shear strength of the nut and bolt materials.

Formula for Ats (when shear occurs at the roots of the thread)
Ats = ? n Le Dsmin[1/(2n) + 0.57735 (Dsmin - Enmax)]
Dsmin = min major dia. of external threads Enmax = max pitch dia. of internal threads n = thread per inch Le = length of thread engagement

 

[drawoh]

johnfin1,

My college text book on machine design was Design of Machine Elements, by Virgil M. Faires.

Faires discusses the problem of a mechanic with a standard set of tools, tightening your bolt whatever way he darn well pleases. He comes up with the following equation...

Fe = Sy As^3/2/6

For a 1/4-28UNF bolt, stress area As=.0364in². My textbook shows a proof stress for a grade[ ]2 bolt of 55000psi.

Fe = 55000lb/in²[×](.0364in²)^3/2 = 52lb

Note how the unit balance does not work. This is an arbitrary analysis of an uncontrolled assembly procedure. If you specify torque or some other assembly procedure, you can manage higher loads.

It would be nice to know Geno's Garage did their calculations.

JHG

 

[johnfin1]

Exactly what I have been saying, the tensile strength x the tensile = pull out strength which does not come close to the chart.

This formula from Fastenal Corp is confusing and seems way off. The shear area (Ats)seems like it is 10x off

Internal Thread Strength Formula
F = Su * Ats
Su = shear strength of the nut or tapped material
Ats = cross-sectional area through which the shear occurs

Ats = pi n Le Dsmin[1/(2n) + 0.57735 (Dsmin – Enmax)]
Dsmin = min major dia.
of external threads
Enmax = max pitch
dia. of internal threads
n = thread per
inch
Le = length of engagement

 

[SnTMan]

johnfin1, you may notice two areas in the MH, Tables 3. The first "Area of Minor Diameter" is based on the diameter at the thread root. Some classes of work require the use of this smaller area. The second "Tensile Stress Area" is based on pitch diameter(s), per the equation given.

Of course the different areas will give different strengths.

Per the Tables 3 the fine threads have larger areas. We can easily see this by looking at any two bolts, the threads are not as "tall" on the fine thread.

The differing and less than obvious assumptions and level of quality from sources such as Fastenal, Genos', etc. is why Codes and standrds were developed.

Regards,

Mike

 

[johnfin1]

Drawoh, what is the 3/2 exponent in that formula. Cant seem to get 52lbs out of that.

 

[rb1957]

55000*(0.0364^1.5)/6 = 64 "lbs" ... no?

 

[johnfin1]

I take it the /6 is the safety factor, right?

 

[johnfin1]

How come when the bolt size goes down the clamping force goes up??

http://www.engineersedge.com/calculators/torque_calc.htm