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DBCox (Automotive)
10 Jul 06 14:02
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?

rb1957 (Aerospace)
10 Jul 06 14:17
a couple of things ...

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.
telecomguy (Mechanical)
10 Jul 06 16:40
I too was expecting to see some dimensions related to the hole geometry relative to the part edge so I was assuming I misunderstood the question.

<tg>
JStephen (Mechanical)
10 Jul 06 17:24
You might check AISC-ASD for bolt spacing and bearing requirements- keeping in mind it's intended for mainly static applications (IE, bolt not sliding back and forth in hole).
DBCox (Automotive)
10 Jul 06 18:00
Thank you for the replies everyone.  Let me clarify what the setup is.  I have a pivoting suspension linkage that pivots about the bolt axis on a bushing.  The bushings are flanged and pressed into a housing, which is welded perpendicular to the main linkage.  There is one on each end, thus it pivots on both ends.  The bracket will wrap around such that the bolt goes through the bracket on both sides of the bushing, thus the double shear I talked about.  The force will be acting parallel to the linkage (which is also parallel to the plates). I probably mispoke when I said tearout, because the elongation of holes is only the begining of tearout.  As for the distance from the center of the bolt to the edge of the bracket, it will be ~1"

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!
Helpful Member!  CoryPad (Materials)
10 Jul 06 18:12
Here are two design guides for bolted joints that refer to bearing/tear-out calculations:

http://gltrs.grc.nasa.gov/cgi-bin/GLTRS/browse.pl?1990/RP-1228.html

http://www.boltcouncil.org/guide1.htm

Another good source is Handbook of Bolts and Bolted Joints by Bickford and Nassar:

http://www.crcpress.com/shopping_cart/products/product_contents.asp?id=&parent_id=&amp;sku=DK5574&pc=

Regards,

Cory

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rb1957 (Aerospace)
11 Jul 06 8:33
elongation of the hole is caused by a force (in the direction of elongation).  the pin (transferring the force) is harder than the plate material and so the pin bears against the plate (and elongates the hole).

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
swearingen (Civil/Environmental)
11 Jul 06 10:11
blakrapter, you and rb1957 are saying almost the same thing about the bearing area.  You say 1/3 of the CIRCUMFERENCE x t, he's saying the full DIAMETER x t.  Note the difference...
Kwan (Aerospace)
11 Jul 06 12:18
You would typically use Dia * t for bearing area.  blakrapter is using 1/3*circum*t.  Well, this would be okay if you wanted to ensure a bearing failure.  However, the problem states that you want a bolt shear failure.  You better use Dia * t.  Why?

1/3 * circum = .654
Dia  = .625  

Thus, using the 1/3 circum you have a lower bearing stress.
DBCox (Automotive)
11 Jul 06 15:01
rb1957,

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 :)
CoryPad (Materials)
11 Jul 06 17:03
The stress to use is the bearing yield stress or the bearing ultimate stress.  MIL-HDBK-5 has values for various alloys.  In a pinch, you could use the ultimate tensile stress as an estimate for bearing yield stress.

Regards,

Cory

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rb1957 (Aerospace)
12 Jul 06 9:39
agreed, I didn't read the post carefully enough (seems more work than necessary to calculate the circumference (using the diameter) and then apply only 1/3 of it, when you want the diameter to start with), and as it's been pointed out, it is slightly unconservative.

still, blakrapter, what flavour of steel are you using ?
DBCox (Automotive)
12 Jul 06 11:41
We will be using a mild steel.  Probably 1020 normalized or something close in strength properties.

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?
Kwan (Aerospace)
12 Jul 06 12:38
I looked up 1025 in Mil-hdbk-5 and it shows 90 ksi for Fbru (ultimate bearing strength).  The ultimate tensile strength is 55 ksi. So I wouldn't use tensile.  You could double tensile and thus, ensure you are shear critical.
rb1957 (Aerospace)
12 Jul 06 13:19
don't quite see the point ... if Fbr is 90 ksi (from a reference that can be checked), why use 100 ? ... that's a bit like rounding pi to 3.

but it is good design practice to make the joint critical in shear
Kwan (Aerospace)
12 Jul 06 13:25
Actually I would use a slightly higher bearing allowable to ensure that you are shear critical.  But, I would design the joint to be bearing critical.  This way you won't have a sudden failure.  If you stay bearing critical, the joint would stay together (hole would elongate).  Also remember that bearing critical is better for joints if you have multiple fasteners.  This way as the joint yields, the load redistributes evenly to the full joint (rather than a sudden "unzipping" of the joint).
CoryPad (Materials)
12 Jul 06 13:49
blakrapter,

You should review ASTM E 238 Standard Test Method for Pin-Type Bearing Test of Metallic Materials, available at:

http://www.astm.org/cgi-bin/SoftCart.exe/STORE/filtrexx40.cgi?U+mystore+ojlk9492+-L+BEARING:STRENGTH+/usr6/htdocs/astm.org/DATABASE.CART/REDLINE_PAGES/E238.htm

Regards,

Cory

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EdDanzer (Mechanical)
21 Jul 06 0:26
I think you may need to consider bending of the bolt creating point loading on the edge of the threaded hole. With the plate being of a lower hardness than the bolt you will see the hole egg shape prior to the bolt shearing.

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