Prequalified joints (again)

Date 01-08-2009 18:57

Hotwork,
Look for the little note at the bottom of the Groove Preparation column for this joint detail. It says S2 = T1 - (S1 + f). So S2 = 3.5" - (1.625 + 0.25). Note that 0.25" is the minimum allowed "f" dimension, and therefore S1 is 1.625". Therefore your answer is 1.625" as the formula will be 1.625" = 3.5" - (1.625" + 0.25").
Mankenberg

By BryonLewis (****)

Date 01-08-2009 19:04 Edited 01-08-2009 20:06

I got:

S2 = T1 - ( S1 + f )
S2 = 3.5 - (2.125 + .25)
S2 = 3.5 - 1.875
S2 = 1.625 or 1 5/8

But with those numbers: S1 = 2.125 S2 = 1.625 and f=.25 Then with S1+S2+f=T1 2.125 + 1.625 + .25 = 4"
Damnit that's not right either.

By hotwork (*)

Date 01-08-2009 19:40

This may be incorrect but using that formula I keep comimg up with 1 1/8 as my anwser If T1 = 3.50 and S1= 2.125 + .25 and that is 2.375 so 3.50- 2.375 is 1.125 or am I messed up.

By kipman

Date 01-08-2009 19:42

Bryon,
S1 should be 1.625 (3.5 - 0.25 / 2).
Mankenberg

By ctacker (****)

Date 01-08-2009 19:49

isn't that formula for plates greater than 6-1/4" and less than or = to 2", otherwise you should use the table. at least thats how I read it too.

By BryonLewis (****)

Date 01-08-2009 19:57 Edited 01-08-2009 20:00

I would think by the nature of the grooves that S1 would equal S2. But by my understanding of the chart and my math says that they are not going to be equal. I am confused by the chart for SAW only. If T1 is 3" to 3 1/2" then S1 = 2 1/8. That is what I plugged into the equation.

By BryonLewis (****)

Date 01-08-2009 20:24

hotwork,
I consulted Excel and it appears that you are the winner. Here we go again.

T1=3.5
S1=According to the chart is 2.125
f=.25
S2=T1-(S1+f)
S2=3.5-(2.125+.25)
S2=3.5-2.375
S2=1.125
My mistake was in the -(2.125+.25) part of the equation. It was an algebraic brain fart.
S1+S2+f=T1
2.125 + 1.125 + .25 = 3.5
Because of the Note h stating: Double groove-welds may have grooves of unequaled depth........
It all makes sense now. Remember those Damned notes.

By hogan

Date 01-08-2009 19:00

All information needed is provided. Apply the variables to the equation provided. Using the 1/4 as the min for f will allow for the max to be calculated for the S1 and S2, as the question asks.