Bolt Tearout
Bolt Tearout
(OP)
Hello everyone,
I am designing a bracket to put a bolted link in double shear. I want to make sure the bracket thickness is great enough such that the bolt shears before the hole in the bracket elongates. It seems I remember that a good estimate of bearing surface to calculate stress is 1/3 of the circumfrence x the plate thickness. This seems reasonable, but the numbers I am getting do not... Heres what I am using:
- 5/8" Grade 8 bolt in double shear (assumed shear yield strength is 2/3 yield so, 100,000 psi)
- Steel plate with a yield of 50,000psi
- Want bolt to shear before hole elongates
- Assume plate bearing area is 1/3 circ.*thickness (t)
Force to shear bolt =
= (area of bolt) * (yield strength) * (2 b/c double shear)
= (pi/4*.625^2) * (100,000) * 2
= 61,360 lbs
Required plate thickness (t) calc:
F= (bearing area) * (2 b/c double shear) * (allowable stress)
61,360=(pi*.625*1/3*t)* (2) * (50,000)
t=.94"
a 1" thick plate seems mighty excessive. What am I doing wrong?
I am designing a bracket to put a bolted link in double shear. I want to make sure the bracket thickness is great enough such that the bolt shears before the hole in the bracket elongates. It seems I remember that a good estimate of bearing surface to calculate stress is 1/3 of the circumfrence x the plate thickness. This seems reasonable, but the numbers I am getting do not... Heres what I am using:
- 5/8" Grade 8 bolt in double shear (assumed shear yield strength is 2/3 yield so, 100,000 psi)
- Steel plate with a yield of 50,000psi
- Want bolt to shear before hole elongates
- Assume plate bearing area is 1/3 circ.*thickness (t)
Force to shear bolt =
= (area of bolt) * (yield strength) * (2 b/c double shear)
= (pi/4*.625^2) * (100,000) * 2
= 61,360 lbs
Required plate thickness (t) calc:
F= (bearing area) * (2 b/c double shear) * (allowable stress)
61,360=(pi*.625*1/3*t)* (2) * (50,000)
t=.94"
a 1" thick plate seems mighty excessive. What am I doing wrong?





RE: Bolt Tearout
i think you're combining a bearing area with a shear stress ... i think you're allowable stress should be the bearing allowable, something in excess of 100 ksi. oh, and your use of dia/3 is conservative, most would use the full dia.
second, this is a bearing calc, not shear tear-out. shear tear-out refers to the "lug" around the hole shearing. a simple way to approximate this area is to take a 45deg line from the bolt center; the area is thickness * the lengnth between the edge of the hole and the edge of the part * 2 (as there are two faces (+45 and -45). the allowable for this is the shear strength of the lug material.
RE: Bolt Tearout
<tg>
RE: Bolt Tearout
RE: Bolt Tearout
rb1957, you mentioned a allowable bearing stress. This is the first I have heard of this. Can you please elaborate. As for using the full circumfrence, I do not agree. At most, I should use 1/2 of it since that is all that is capable of transmitting load to the plate (the back half is pulling away from the back side of the hole, thus, no force is transmitted)
Also, the bolt is not spinning (or should), the bushing should rotate inside of its outer sleeve.
Thanks for the help so far!
RE: Bolt Tearout
http://
http://www.boltcouncil.org/guide1.htm
Another good source is Handbook of Bolts and Bolted Joints by Bickford and Nassar:
http://w
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Bolt Tearout
The bearing area is diameter*thickness. the allowable is Fbr (usually expressed for e/D=2 or 1.5). this is convered in most strcutures books (like Bruhn or Niu), but being a car guy you may not recognise these texts !
good luck
RE: Bolt Tearout
RE: Bolt Tearout
1/3 * circum = .654
Dia = .625
Thus, using the 1/3 circum you have a lower bearing stress.
RE: Bolt Tearout
Sorry, I misread your post. You stated to use the diameter, I misread and thought you said the full circ. As Kwan pointed out, they are almost the same.
Either way though, what allowable stress should I use? I thought I should use the yield stress of the material, but it sounds as if this may not be the case. You mentioned e/D=2. What is e? You are right, I do not recognize those texts :)
RE: Bolt Tearout
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Bolt Tearout
still, blakrapter, what flavour of steel are you using ?
RE: Bolt Tearout
Sort of switching gears here, but what determines the bearing strength? Typically, yield and uts are determined by tensile loads on a round specimen. I suppose that strength goes up significantly when you place a compressive load it it because it is compressing the lattice of the material?
RE: Bolt Tearout
RE: Bolt Tearout
but it is good design practice to make the joint critical in shear
RE: Bolt Tearout
RE: Bolt Tearout
You should review ASTM E 238 Standard Test Method for Pin-Type Bearing Test of Metallic Materials, available at:
http://ww
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
RE: Bolt Tearout