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1/4" Lag Screw Lateral Capacity Calculation

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npsSE

Structural
Aug 19, 2022
2
Reference: thread507-436818

I get the same result as OP for 1/4" lag screw capacity. The discrepancy (with NDS) appears to be how Fe is calculated. Since Dr = 0.173 < .25, Fe=16600G^1.84=4637 psi (see footnote 2 of NDS Table 12.3.3), for G=0.50.

It appears that 2018 NDS Table 12J and the online Connection Calculator use the nominal D of 0.25 to calculate Fe=5600 psi.

As other replies point out, instinctively, a 1/4" lag screw should have greater capacity than a 10d common wire nail, so using the higher NDS table value may be appropriate.
 
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The value in the NDS table is correct. For Table 12.3.3, you should be using D instead of Dr. Section 12.3.7 only tells you to use Dr with Tables 12.3.1A and 12.3.1B. Using D = 1/4" results in Fem = 4465 psi.

I took a look at the calculations in the thread that you linked to. RWW0002 correctly identified the issue with OP's calculations. The formula for Z,IV includes an Rd factor in the denominator which OP had set equal to 3.2*Kθ. This was an error due to Footnote 1 of Table 12.3.1B in the 2018 NDS. Rd should instead be equal to Kd*Kθ or 2.23*Kθ. This explains the difference in the results since 105 lbs * 3.2/2.23 = 150 lbs.

Aside from the Rd factor which shows up in all of the Z formulas, OP's calculations are correct, so refer to them for any other discrepancies you may have in your calculations.



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Footnote 2 of Table 12J states that tabulated values are based on reduced body diameter. I assume that should also include Fe calculation. For a 1/4" lag, Dr= 0.173", and G=0.5, Fe=16600G^1.84=4637psi, Fyb=70,000psi. I get Z(IV)= 140 lbs. Table 12J lists 150 lbs for ts=1.5". (I must have transposed 140 and 104 in my head and thought my result was the same as OP.) If I use Fe=5600psi and D=Dr=0.173" I get Z(IV)=153 lbs. So (why) does table 12J use the full body diameter for Fe calculation?
 
npsSE, I am the OP from the linked thread and the author of the calculation sheet in the linked thread. I believe the answer to your question is the same as the answer to my question in the linked thread. The answer was provided by RWW0002. The answer is that, in the calculation of Z(IV), Dr^2 should be divided by 2.2*K_theta instead of 3.2*K_theta.
 
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