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I need to determine the shear stress of a bolt in single shear and a bolt in double shear, is the equation
Shear stress = force / cross sectional area
The bolt in question is a m10 bolt, class not known yet although will be 8.8 or 12.9, force applied on threaded section so overall diameter of bolt assumed to be 8.5mm
for single shear divide by two?
How would I go about determining the safety factor?
BTW Im putting a 400N load throught this bolt in single shear, its not constant its that max possible load, any input?
Thanks 

rb1957 (Aerospace) 
18 Jan 08 7:46 
wait up ... you're shearing over the threads ?
conventionally, with a bolt shank in double shear, your effective area is twice, but there is a knockdown if the middle layer (the distance between these two shear faces) is small ... MILHDBK5 has factors (table 8.1.2.1(b)) ... give the geometry and i'll look it up for you.
but i don't like shearing over the threads 

400 N is a ridiculously small force going through an M10 screw with property class 8.8 or 12.9. This calculation is irrelevant. Regards,
Cory
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distance between the two shears is 10mm,
need to do the calc just to confirm to myself that its suitable
thanks 

rb1957 (Aerospace) 
18 Jan 08 11:56 
with the middle layer being about a diameter, the bolt would be fully effective in double shear.
400N is a very small load ... i not that towards the end of your post you start mentioning "single shear" ... i assume the bolt is in double shear 2*400N (still a small load for a 3/16" bolt. i would knockdown the allowable to account for the threads (some sort of stress conc.). 

Gymmeh (Mechanical) 
18 Jan 08 13:59 
My book shows M10 Class 5.8 to have a proof strength of 380MPa... 

how do you determine the shear stess of a bolt? I have determined the shear stress for the situation so I now need to know if the fastener will take it. thanks 

Correction to my previous post. The material properties are material strengths, not stresses. I apologize for the misnomers. Ted 

I hate to add this, but then again maybe I don't. If the assembly is not very tight and the clearances not close then there may be added bending stresses to consider. Of course if it is tight enough and depending upon the arrangement, then there may be no shear stress because friction is transferring the load. My point is that when someone asks such a basic question, oh so basic, then we should be careful that answers that seem to be correct may be misleading because the person that asked it was not competent to ask it properly. Paul www.ostand.com 

right single shear situation
method
determine the force acting on the bolt, divide this by the area of the bolt to get the shear stress that will be acting in the bolt.
now determine a bolt class to use 8.8 12.9 etc, take the tensile yield stress, now 60% of this is the shear yield stress, divide this by the bolt diameter used above and you get the value the bolt will shear at.
is that correct? how do i go about determining the safety factor?


wahoo88 (Mechanical) 
20 Jan 08 17:24 
Oakley,
To determine the safety factor, divide the yield or ultimate strength of the bolt (in MPA, or psi), by the calculated stress in the bolt (in the same units). So for your situation, assuming single shear, threads not in shear, m10, class 8.8 bolt:
.6(660 MPa) SFy =  = 77.6 400 N / pi(5 mm)^2
.6(830 MPa) SFu =  = 97.8 400 N / pi(5 mm)^2
Andy


rb1957 (Aerospace) 
21 Jan 08 6:42 
wahoo88 ... in his 1st post he says, "threads in shear". but then i was confused with him mentioning "double shear".
i think proinwv2 brings up some good points. 

Gymmeh (Mechanical) 
21 Jan 08 9:23 
I recommend getting your hands on or have your company buy you  Mechanical Engineering Design by Shingley.
As wahoo points out, you have nice safety factors over 75 with the design.
I am with proinwv, I really want to help but... For a basic bolt analysis this seems well over designed, is there more to this situation that requires extra strength?, Is fatigue a major consideration? ... Knowing the application or governing regulations really can affect the problem and solution.
Best Regards


6060842 (Mechanical) 
22 Jan 08 13:43 
Hi
I always use
As = P/A = P /(Pi*Db^2/4)
As = Average Shear Db = Diameter of bolt (Minor Dia) P = Load
A good reference on bolted connections is to go with the procedures defined in VDI2230  Systematic calculation of high duty bolted joints. This is a document drawn up by the German Society of engineers  Not free but well worth a gander
Tom 

Shear is not distributed uniformly across a crosssection when a beam (a bolt in this case) is loaded in bending.
I would follow Gymmeh's advice to purchase Shigley's book on mechanical engineering (Shigley & Mischke, published by McGrawHill).
Here's where the fundamentals come into play. The reason that we generally don't have to apply Mohr's circle or diddle around with Von Mise's equation when we're dealing with combined bending and shear in this particular beam is that shear is a maximum in the center while bending stresses are at a maximum at the outside edges of the beam. Check both.
However, shear stresses in beams distribute themselves in a manner that is not evenly loaded like tensile stresses applied axially. The maximum shear stress for a round beam is found along a plane perpendicular to the direction of loading, passing through the cg of the beam and is as follows:
tau(max) = 4V/3A (for a round beam)
Also, remember that your shear yield strength (as wahoo88 pointed out) is not the same as your tensile yield strength. I calculate shear strength to be 0.577*Sy. But even that is open to debate. 

dimjim (Mechanical) 
12 Jun 08 18:23 
I recommend getting your hands on or have your company buy you  Mechanical Engineering Design by Shingley.
Shingley should be Shigley. I now that is what you meant.


Gymmeh (Mechanical) 
17 Jun 08 11:45 
Yes that is correct. sry my bad. 



