How Bolt Table Was Derived

[Debra555]

I read the article:

http://www.deckfailure.com/JLCDeckLedgers.html

I wonder if someone might kindly help me see how this was derived.

I wonder if you might illustrate the calculations for the table DETAIL 1 fastener schedule?
Also a typical 1/2 dia bolt load for SPF is 180 lbs. Can someone show how this is obtained.

Any references on line to *sample problems* that illustrate these specific topics?

 

[woodman88]

The NDS gives all the information for this. You can go to

http://www.awc.org/standards/nds.html

to check it out. Tech Note 12 (A free download I have attached) gives a quick rundown of the 1997 calculations.

Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.

 

[Debra555]

I wonder if you might illustrate the calculations for the table DETAIL 1 fastener schedule?
Also a typical 1/2 dia bolt load for SPF is 180 lbs. Can someone show how this is obtained.

Thank you for the reference ........ Can someone answer the specific question........ Thankyou for the help

 

[PSlem]

Formula is 180 lb/bolt/(1/2 deck span x (40PSF+10PSF)) x 12in/ft.

 

[Debra555]

How was the 180 lbs/bolt derived for SPF? Can you illustrate how to get this value when using a 1/2 inch dia bolt?

Thanks

 

[BAretired]

Oops PSlem, not quite right.
If a bolt can resist 180# in shear and is spaced at 'S' feet, then it can resist 180/S pounds per foot.

The floor reaction per lineal foot is w*L/2 where w is the load per square foot on the floor.

Equating the two, 180/S = wL/2
Or S = 180*2/(wL)
Example:
For L = 12' and w = 40+10 = 50 psf,
S = 360/(50*12) = 0.6' = 7.2" which is the value in the table.

BA

 

[PSlem]

=180/(12/2*(40+10))*12 =7.2" So same answer.

 

[BAretired]

Yes, you are correct, PSlem. I did not see your division sign following the word "bolt". Sorry. Put it down to old age.

BA

 

[Debra555]

How do we determine the ALLOWABLE bolt load for SPF using a 1/2 inch diameter bolt?
Also: Is this allowable the MAX SAFE LOAD that can be placed on this bolt? Please explain in detail.
Please illustrate the development of the allowable bolt equation.
Thanks

 

[BAretired]

Are you talking about a bolt or a lag screw? Detail 1 illustrates a lag screw. In Canada, we would find factored values for lag screws or bolts from the Wood Design Manual. For a 1/2" dia. lag screw, the minimum length of penetration in the main member is 64 mm (2.5") so Detail 1 would not be acceptable with only 1.5" penetration.

I am not familiar with the NDS standard but here is a link to an article which discusses lag screws.

BA

 

[Debra555]

You are saying the penetration is in the band joist? Correct.
Can you illustrate how they obtained 180 lbs for the allowable bolt load.
Also what SF would be considered acceptable for Ledger bolt connections?

I look forward to an illustration of how the 180 lbs per bolt was developed. Give a sample calc.
Also what would be considered a MAX for a 1/2 in lag screw/bolt (typical for a ledger connection).
Thanks

 

[BAretired]

Why don't you read my last post?

BA

 

[BAretired]

And also, Debra, why don't you do a little research of your own? Use Mr. Google.

BA

 

[steele6707]

Using the 2005 NDS, Table 11J, 1.5” side member thickness, and ½” diameter lag screw, condition Zperpendicular (side & main loaded perpendicular to the grain), G=.42 (SPF) you get a value of 170 pounds. This value is based on a penetration of 8D (4”) which the example does not have. The penetration is 1.5”. The minimum penetration required is 4D (or 2”). If you used the reduction factor anyway you would get 170 * (p/8D or 1.5/4) = 63#. This assumes a Cd of 1.0 (10 years) and a Cm of 1.0 (Moisture content modifier). There are no provisions for main member penetration less than 4D and the number is based on main and side member having the same specific gravity. The example uses a southern pine side member which has a higher specific gravity, but I have not seen anything that discusses main and side members of different specific gravities.
You could go to a 3/8” lag screw and get a 4D penetration. Your value would become 110 * (1.5/3) = 55#. However, the different specific gravity issue is not solved.
I do not do this calc very often so anyone may provide any corrections needed. You could also go into yield limit equations Table 11.3.1A and calculate values. I believe Table 11J provides the least value of the yield limit equations.

 

[Debra555]

Thank you BAretired. I do use Mr. Google. I don't do these calcs often....sorry for asking for help.

Using the 2005 NDS, Table 11J, 1.5” side member thickness, and ½” diameter lag screw, condition Zperpendicular (side & main loaded perpendicular to the grain), G=.42 (SPF) you get a value of 170 pounds.

Can you show me this table. Thanks
Also, I would like to know how to determine this value without the table. How is it calculated? Thanks for your help.

This value is based on a penetration of 8D (4”) which the example does not have. The penetration is 1.5”. The minimum penetration required is 4D (or 2”). If you used the reduction factor anyway you would get 170 * (p/8D or 1.5/4) = 63#. This assumes a Cd of 1.0 (10 years) and a Cm of 1.0 (Moisture content modifier). There are no provisions for main member penetration less than 4D and the number is based on main and side member having the same specific gravity. The example uses a southern pine side member which has a higher specific gravity, but I have not seen anything that discusses main and side members of different specific gravities.

I have noticed other tables (ie:ledger to band joist) shows spacing at 16" OC and greater (ie: - in other words greater than what this example shows from deck failure example) for 1/2 " lag bolts for SPF. This gives a value GREATER than the 180 lbs per bolt (which is suppose to be the MAX). Why is this? (assuming of course the restrictions given)??

You could go to a 3/8” lag screw and get a 4D penetration. Your value would become 110 * (1.5/3) = 55#. However, the different specific gravity issue is not solved.

Yes, I see. I would be greatful to calculate this 110 lbs by hand. Is there a formula for this?

I do not do this calc very often so anyone may provide any corrections needed. You could also go into yield limit equations Table 11.3.1A and calculate values. I believe Table 11J provides the least value of the yield limit equations.

Yes, your help is appreciated. I'm sure others will see this and correct it if wrong.

 

[bastianguy]

They need a bigger bolt for a bigger project in my opinion. I know some experts in the field will say a 1/2 bolt is perfect but I disagree.

 

[SlideRuleEra]

Debra - You mention that you are "Using the 2005 NDS, Table 11J..."

Note that the article you are reading

http://www.deckfailure.com/JLCDeckLedgers.html

"...was published in August of 2003..."

Suggest that you look for your answers in NDS 2001, not NDS 2005.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[BAretired]

Detail F1 in the original link is not permitted in CSA O86-01 as the lag screw has less than the minimum penetration in the main member. CSA O86-01 requires a minimum of 5d or 2.5" and for that, there is a reduction factor of 0.625. Detail 1 shows only 1.5" penetration and is not permitted, so there is no recognized formula to justify the value of 180# used for one lag screw.

In CSA O86, there is a factor Jpl (factor for reduced penetration) which varies from 0.625 to 1 for penetration of 5d to 8d respectively and linearly in between). If you extrapolated to 3d or 1.5" penetration, you might rationalize a Jpl value of .375, but it would not meet the code and in my opinion, should not be used.

Detail 1 is simply not acceptable by CSA O86-01 and, from the above posts it is not acceptable by NDS either.



BA

 

[OldPaperMaker]

Debra555,
I think that Woodman88's attachment shows an example of the calculations that are used to develop the values in tables 11A to R. They use the Yield Limit Equations that are found in section 11.3 of the NDS.

For more detailed examples of the calculations see the book "Design of Wood Structures ASD/LRFD" by Donald E. Breyer, et at, sixth edition. The book has about 20 pages of calculations about lag bolt capacities, pages 13.55 to 13.73.

 

[Debra555]

Debra555,
I think that Woodman88's attachment shows an example of the calculations that are used to develop the values in tables 11A to R. They use the Yield Limit Equations that are found in section 11.3 of the NDS.

YES, you are correct, but calculations for MODE II are not there. I believe this will give the LOWEST value of Z. Is this correct?

For more detailed examples of the calculations see the book "Design of Wood Structures ASD/LRFD" by Donald E. Breyer, et at, sixth edition. The book has about 20 pages of calculations about lag bolt capacities, pages 13.55 to 13.73.

I don't have that text......Is it possible to scan a typical calc for MODE II (or provide here?). I'm wanting to verify how the 180 lbs was developed. Also, would this be a MAX load (if the result for Z was based on the adequate penetration depth)?