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?

 

[Debra555]

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.

If this was permitted, what equation would be used to verify the Bolt load? Can you help with a sample calc.?

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.

Thanks. Can you post this particular clause referenced above.

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

Yes, understood. For decks, what would be the MAX acceptable bolt load? (if using a) Lag screws, or b) Lag bolts)

 

[Debra555]

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.

Yes, understood. The point I'm trying to find out is HOW the MAX Bolt load is calculated? So, if this is different from NDS 2001, then that is the one I would like. Are they different? If so, how do these equations differ. Thanks for the input. I look forward to the help given. You guys are great. Thanks...

 

[BAretired]

I do not know the formulas in NDS. I have perused the method in CSA O86-01 but it is too involved to attempt to include in a post. Furthermore, I do not know whether or not it agrees with NDS.

For what it is worth, I am providing this link to one page of the Selection Tables from the Canadian Design Manual 2001.

Please note that these are factored loads, consistent with the Canadian standard. For a 1/2" lag screw the Q'r³ value is 1.15 kN. That would be for full penetration of 8d = 4". This would correspond to a working load of 1.15/1.45 = 0.793 kN or about 178#.

For 2.5" penetration, the allowable load would be 112#. A penetration below 2.5" is outside of the range of applicability of the table.

BA

 

[BAretired]

Sorry, my mistake, the value in the table for 64mm (2.5") penetration is 1.15kN. The reduction factor has already been applied. I don't think the values should be extrapolated to a penetration of 1.5".

BA

 

[woodman88]

Per the IBC/IRC and the NDS lateral design values for bolts, lag screws, wood screws, nails, etc. You need to take the least value for four differance yield modes (for main member and side member where required). See the pdf I attached above for these equations.

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 do not know the formulas in NDS. I have perused the method in CSA O86-01 but it is too involved to attempt to include in a post. Furthermore, I do not know whether or not it agrees with NDS.

For what it is worth, I am providing this link to one page of the Selection Tables from the Canadian Design Manual 2001.

Please note that these are factored loads, consistent with the Canadian standard. For a 1/2" lag screw the Q'r3 value is 1.15 kN. That would be for full penetration of 8d = 4". This would correspond to a working load of 1.15/1.45 = 0.793 kN or about 178#.

YES this is helpful. Thankyou.... Would this be a MAX for this 1/2 in dia lag screw?? (ie: Code MINIMUM)

For 2.5" penetration, the allowable load would be 112#. A penetration below 2.5" is outside of the range of applicability of the table.

 

[Debra555]

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.

So for 1/2" in dia lag bolts (Using SPF for ledger), what are ALL components and STANDARDS we must consider? Can you specify.
Is this MAX bolt load, when usin NDS considered a Code minimum? Thank you.

 

[Debra555]

Per the IBC/IRC and the NDS lateral design values for bolts, lag screws, wood screws, nails, etc. You need to take the least value for four differance yield modes (for main member and side member where required). See the pdf I attached above for these equations.

YES....Understood. The equation for MODE II is not listed. Can you specify this equation.
Also: Can one attempt the same calc assuming a simpler method ..... as an example that would match that of MODE II (using a SF instead)?

 

[Debra555]

Please note that these are factored loads, consistent with the Canadian standard. For a 1/2" lag screw the Q'r3 value is 1.15 kN. That would be for full penetration of 8d = 4". This would correspond to a working load of 1.15/1.45 = 0.793 kN or about 178#.

Sorry, just check table: What is formula? What is the 1.45?
Bolt load = Q'r^3 / 1.45 correct? Thanks

 

[woodman88]

Per the NDS a 0.5" lag screw can not be used in a 1.5" main member. Per what others state, this is also true for the Canada standards/codes. I would say that the table you are looking at does not meet the current engineering standards.
Per the attached pds, Table 1 on page 6 gives the equation for mode II. Along with the equations for A, B and C for the equations. Then you have the other tables values and equations to do.
There are difference methods of calculating the end values. The NDS ones puts most of the table equations together. But this does not make the equations easier.

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.

 

[BAretired]

I do not know the formulas in NDS. I have perused the method in CSA O86-01 but it is too involved to attempt to include in a post. Furthermore, I do not know whether or not it agrees with NDS.

For what it is worth, I am providing this link to one page of the Selection Tables from the Canadian Design Manual 2001.

Please note that these are factored loads, consistent with the Canadian standard. For a 1/2" lag screw the Q'r³ value is 1.15 kN. That would be for full penetration of 8d = 4". This would correspond to a working load of 1.15/1.45 = 0.793 kN or about 178#.

My error. The value of 1.15kN is for a penetration of 64mm (2.5"). For 102mm (4") penetration, the table shows two values, 2.31 and 1.94kN.

YES this is helpful. Thankyou.... Would this be a MAX for this 1/2 in dia lag screw?? (ie: Code MINIMUM)

The factored values shown in the table are dependent on the direction of grain. The maximum factored load for a 1/2" lag screw parallel to the grain with 4" pen. would be 2.96kN or about 666#. In your case, however, the load is perpendicular to the grain so 1.94kN (436#)is the maximum factored load for 4" penetration. It is 1.15kN (260#) for 2.5" penetration. No value is given for 1.5" pen.

For 2.5" penetration, the allowable load would be 112#.

My error. The factored load for 64mm (2.5") pen. is 1.15kN = 259# which corresponds to a working stress load of 178# using a load factor of 1.45.

Assuming LL + DL = 40 + 10, Factored load = 1.5(40) + 1.25(10) = 72.5 = 1.45(40 + 10).

BA

 

[BAretired]

Debra555,

To tackle the problem using equations, you are faced with a complicated study of the code. To do that, you will need to acquire a copy of the code which governs in your area. But for your enlightenment, here are a couple of pages from the Canadian Wood Design Manual 2001. See link below.

BA

 

[BAretired]

I seriously question whether the code people are on the right track. I suspect they are not.

BA

 

[Debra555]

I seriously question whether the code people are on the right track. I suspect they are not.

In your opinion, based on the above a) would you use 1/2 in dia lag screws for decks? Please explain. b) What would you say the the MAX bolt load would be at 5d penetration for a SPF ledger? c) I noticed you are assuming dia of 1/2 in bolt is 0.5. Is it not 0.371 inches (root dia)? Explain..

Given that deck failures are generally due to improper ledger connection and that in most cases these are built by the home owner, or poorly informed contractor without a permit, what would you suggest for bolting the ledger + pattern of bolting? Please explain....

 

[nac521]

Attached is a mode II calculation (which controls in this case) per NDS 2005 that results in a 180 lb load. However, as mentioned previously, the 4d minimum penetration is not met if you use the nominal diameter. An argument can be made that the code doesn't specifically state to use the nominal diameter (NDS 11.1.3.6) and since the root diameter (.371") is used to calculate capacity (NDS 11.3.6), can it be used to calculate the minimum penetration requirement?? (.371x4 < 1.5") Additionally, the 2001 NDS deleted the used of the p/4D penetration reduction factor for lag screws since all six yield limit equations are now used (instead of only 3 equations in the 1997 edition). The p/4D reduction is still used if you design per the tables because the tables are based on an 8D minimum main member thickness. This may result in conservative results because not all controlling modes are based on the side member, but all table values still get reduced.

 

[Debra555]

Attached is a mode II calculation (which controls in this case) per NDS 2005 that results in a 180 lb load.

Yes....Thanks. Can you illustrate the source of this equation? Attach?
Also does this result give the same one as the equation in Technical Report 12 - illustrated above?

However, as mentioned previously, the 4d minimum penetration is not met if you use the nominal diameter. An argument can be made that the code doesn't specifically state to use the nominal diameter (NDS 11.1.3.6) and since the root diameter (.371") is used to calculate capacity (NDS 11.3.6), can it be used to calculate the minimum penetration requirement?? (.371x4 < 1.5")

What do you think? Can it be used tominimum penetration requirement??

Additionally, the 2001 NDS deleted the used of the p/4D penetration reduction factor for lag screws since all six yield limit equations are now used (instead of only 3 equations in the 1997 edition).

When used for deck ledger attachments, I assume Mode II will always govern. Correct?? If not, what other Mode might govern, in your opinion?

The p/4D reduction is still used if you design per the tables because the tables are based on an 8D minimum main member thickness.

Since the tables are based on 8D, would it not be better to use the calcs for Mode II @ 4D min penetration. If the peneterataion is < 4d, can one use p/4D to adjust value? Would this be a representation of a RESONABLE Max bolt load? Would this be considered as a Code minimum?

This may result in conservative results because not all controlling modes are based on the side member, but all table values still get reduced.