neverenuff

My question relates to shear loading on UCA and LCA bolts. I was always taught that in double shear the shank of the bolt must pass through both sides of the bracket. The shear loading is reduced if the threaded portion of the bolt does not pass all the way through. My question is what practical experience says about this issue? are you guys concerned if the threaded portion of the bolt is in the shear area ? if using an oversize bolt then maybe the additional strength (larger diameter) will overcome the reduced shear loading of the threads? I'm getting ready to install a B.Lee's 3-link and found the grade 8 bolts they sent do not have enough shank to pass thru the hiem joint much less all the way through the brackets... trying to figure out what gives, seems to me these should be longer bolts

McCaffrey

Are you saying the threads should end half way through welded-on bracket? Seems like that would be optimal but fairly difficult to get right since you only have about 1/4" accuracy to play with.

neverenuff

I'm saying that optimal, the bolt shank should pass through both sides of bracket so the threads do not start until after that. The shear load against the shank is greater(stronger) than against the threaded portion (smaller diameter, weaker)... my question is does that really matter? as a general rule, are you guys concerned if threads are in the shear area?

cdn_xj

You are correct: the shear load capacity of a bolt is less at the threads due to the minor diameter. But here are some things to consider:

The rule of thumb regarding the amount of threading on standard SAE bolts is (2x diameter)+ 1/4" for any length under 6" and (2x diameter)+ 1/2" for any length over 6".

This can be challenging for manufacturing if you require a specific length of bearing surface. Either you wind up with threads in a shear plane or a bolt that is excessively long.

Further compounding the matter is the fact that if the shoulder of the bolt is too long, you will now have to start adding washers in order to prevent the nut from bottoming out on the shoulder before proper clamping force is applied. (Incidentally, i should point out that adding washers in this scenario does nothing to diminish the integrity of the assembly, it's just not a very elegant solution.)

In other words, it's REALLY difficult to get bolts that are EXACTLY the right length short of designing around the amount of shoulder that a bolt has (impractical) or having bolts custom made (expensive).

Now, for the more empirically minded:

Minimum tensile and yield strengths of bolts are required by the applicable ASTM Specifications. In the case of Grade 5 and Grade 8 fasteners, the MINIMUM tensilse strengths are:
Grade 5 120,000 psi
Grade 8 150,000 psi

BUT, shear strengths are neither published nor required by ASTM specs. That having been said, the Industrial Fastener Institute has determined that a shear strength of 60% of the MTS is a reasonable expectation and assumption to make for design purposes.

SO what does that mean for Grade 5 and Grade 8 bolts?

Well,
Grade 5 120,000 psi x .60 = 72,000 psi
Grade 8 150,000 psi x .60 = 90,000 psi

But, due to varying fastener diameters, we need to take the calculated minimum shear strength and factor in the the diameter via the cross sectional area of the bolt in order to find out the specific shear strength of a specific diameter of bolt.

There are 3 ways to calculate this: using the bolt's major diameter (diameter of the shoulder) or the minor diameter (diameter at the bottom of the threads), or you can use the Tensile Stress Area which is in between the two.

SO, for the sake of arguement let's say that you are using a 1/2"-13 NC bolt, and we'll calculate the shear strength based on the minor diameter of the bolt. The minor sectional area can be looked up on a table. In this case it is 0.1257".

Grade 5 72,000 psi x 0.1257 = 9,050.4 lbs
Grade 8 90,000 psi x 0.1257 = 11,313 lbs

Providing the math has been done correctly and sufficiently sized and graded bolts for the given application are used, it is very difficult to simply shear a bolt; assuming that the bolt assembly has been properly torqued and has been maintained so that bending moments are eliminated (ie. kept tight) you would need to apply at LEAST 9,000 lbs of force to shear a Grade 5 bolt and at least 11,000 lbs to shear a 1/2" NC Grade 8 bolt.

So, to make a long story short, if a reasonably sized bolt is used, I wouldn't worry about the threads being in the shear plane.

neverenuff

CDN - easy reading great understanding. Thanks for the explanation it does give a sense of well being. I work in aviation were specific NAS bolts are used in shear applications but are all shank with minimal thread. I believe my LCA bolts will be 1/2" grade 8 at the axle and 5/8" grade 8 at the crossmember, should be sufficent.

cdn_xj

Glad that I could be of help.

Something that I failed to mention is that since we are talking about a double shear application we would need to take that 9,000 lbs and 11,000 lbs and double it.

So, in order to completely shear the control arm bolt (assume a 1/2"-NC grade 8 bolt) to the point that the end of the arm falls freely, you would need to apply somewhere north of 22,000 lbs. of force.

And if you manage to do that while driving I'd suspect your problems are far greater than whether your control arm bolts will hold.

And before anyone else brings it up, yes, I know control arm bolts have failed. But WHY did they fail? Were they properly torqued and maintained assemblies? Or were those failed assemblies loose, allowing free play and bending moments to exist, which would weaken the bolt through repeated cyclic flexing of the bolt as the loads shift back and forth?

Outside of straight shearing forces, a body needs to bend before it can break. So as long as you ensure that no undesireable bending forces exist, you are only dealing with shearing forces which, as we've demonstrated above, needs to be EXTREMELY high in our scenario.