Shear capacity of Bolts calculations 1

[wontweldcantweld]

Hi there,

I’m looking for help with some shear strength calcs for bolted joints as this isn’t my field of expertise;

I have x 6 M16 BSEN ISO 3506 Class 70 Bolts, each with a shear capacity (Psb) of 48,700N
By my working out, the Shear area (As) per bolt should be 2.0106 x 10 ^-4 m^2
And by using psb (Shear Strength) = Psb / As,
We get 48,700n / 2.0106 x 10 ^-4 m^2, and so psb = 242216253.9 n/m^2
Now, because there’s 6 bolts in total, do I simply just multiply it by six?
Meaning that 1453297523 n/m^2 is the shear strength of the joint?

To me, it seems like very basic maths but I'm being told its wrong and that the shear capacity given as 48,700N is left on its own and not multiplied by the number of bolts, as it is (meant) to be given as the shear capacity of the entire bolted joint in the first place!

I think the problem may lie with my understanding of the term 'shear capacity', as I've researched it and it seems there's very little on the subject and in many cases it refers me back to the shear strength - can anyone explain the relationship between these two and point me in the right direction?

Cheers

 

[hokie66]

I don't want to check your arithmetic with all those long numbers. But assuming the bolt shear capacity is 48.7 kN as you say, the shear capacity of 6 bolts would be 6 x 48.7 kN = 292 kN. Now, a bolt group might have less shear capacity than that due to other factors.

Your calculation to get stress in the strange units of n/m^2, then multiplying that by the number of bolts, makes no sense.

 

[OldBldgGuy]

CISC or AISC manuals give very clear direction on bolted connections.

 

[racookpe1978]

Do they teach a different "significant digits" formula for use in the metric system? 8<)

Seriously. Use 2 decimal places with scientific notation if you must.
See for bolt stress limits. Newton/m^2 ????

http://www.eng-tips.com/viewthread.cfm?qid=403942

(Nbr of bolts) x shear force limit per bolt .
BUT! Verify you are using the shear limit for each bolt based on the shank diameter or the minimum thread diameter. The difference will be based on whether or not you can guarantee that the bolts will never face a shear force across the threaded length, or if they canpotential ever face the stress stress across their threads.

Second.

It is NOT clear when you are working with (nor when you are trying to look up ?) the total shear limit of the bolt itself (Newtons, in metric), or the shear stress per unit area of the bolt material itself (Newton/area) .

 

[BrianE22]

Are you sure your bolts are the only material resisting shear? In most bolted connections the mating parts react shear via friction on the faces of the parts.

 

[hokie66]

BrianE22,
That is incorrect. What you are referring to are connections which are supposed to not slip at all, and those are not "most bolted connections". Friction grip connections have lower rated capacities than bearing connections.

 

[wontweldcantweld]

Morning all,

hokie66 & racookpe1978 I never mentioned shear stress in any of my calcs, however, this is exactly what I'm trying to determine - Not a value for the shear stress, but what actually is the relationship between the shear capacity and shear strength, as now I've been told of another engineer that these are essentially the same yet I've been given values for both? where does the shear stress come into this

racookpe1978 I also just copy & pasted the values from a separate document, I get what you meant about 2d.p sorry about that

I also should have included that it should be taken that the bolts area is taken from its shank diameter and that the bolts are the only material resisting shear, so no friction on the faces of the parts

Thanks for your input, like I said this isn't my field of expertise

 

[andriver]

Not a value for the shear stress, but what actually is the relationship between the shear capacity and shear strength, as now I've been told of another engineer that these are essentially the same yet I've been given values for both? where does the shear stress come into this

Not really sure what you mean...shear capacity and shear strength are one in the same

If you multiply your allowable shear stress by your area you get your capacity (shear strength). Divide your applied shear by your shear area to get your actual shear stress in the bolt.

You can choose to compare allowable stress to computed stress, or allowable strength to your computed strength. Same difference.

 

[racookpe1978]

Expressing andriver a little bit differently:

Nbr of bolts x maximum rated capacity/bolt in shear. This is a "catalog" value, accurate if your catalog is correct and you actually use the material from that catalog as it is designed.

Or

Total area of the material being sheared x shear stress/area. This is a theoretical value, accurate if your areas and your theoretical values are correct, and if all of your assumptions are correct.

 

[Tmoose]

Hi wontweld,

What is your bolted joint configuration?
Assuring all 6 bolts are loaded at the same time is a big task.
It is Perhaps more likely that only 1 or 2 will really end up with all the load, until various component yield.

 

[RolMec]

Hello wwcw,

pls. allow me to dig a bit into your input..
- following the standard you invoke you need to specify the stainless material being A (austenitic) or C (martensitic)
- for grade 70 there's A - material with fu = 700 N/mm² , fy = 450 N/mm² and C - material with fu = 700 N/mm² , fy = 410 N/mm²
Conclusion 1: These parameters show that we talk very ductile mech. properties.
- however, you still need to define whether a) your load is predominantly static or pulsating or alternating and b) how many shear planes are involved and c) which level of safety you require..
Assumption: 1 shear plane, static load, pure shear: With 16 mm shank dia. and 48.7 kN shear capac. the only sensible (to me) math. setting would be to assume you have C - type material and the allowed shear load of one bolt was derived by dividing (fy*A) by sqrt(3) to get an allowable shear load. But this arrives at 47.6 kN and thus doesn't fit the given 48.7 kN with ;-) 2 digit accuracy. It's rather something like the best guess I could make.
Conclusion 2: Whatever the factor of division, it can imo suit as factor of safety only if the system does have but 1 shear plane and the load is static AND all concerned parties agreed upon that. I've seen fos of 2 for shear of very ductile materials, but also much less (1.1 was the [well checked & proven in tests] extreme).

As what concerns the other issue, I don't see why at #6 bolts the loadbearing capac. of the joint should not be = no. of bolts * loadbearing capac. of the single bolt. However, a sketch o.e. would be quite helpful to ;-) assess on that. There looms the question of the thickness of the members, design of the pattern, ...

Best of success, & Regards
R.

 

[Sengland]

Is your loading on the bolts concentric or eccentric? If it's concentric then you can assume the bolts carry the load evenly, if not you need to translate the force using either the elastic method or instantaneous centre of rotation (ICR) method, both of which end up with some bolts carrying more load than others.