Blodgetts "Design of Welded Structures" question

Both AWWA D-100 and API 650 have design guidelines for tank stiffeners.

>Does he mean 4C
I have the same book and the same "error". I think 4C is correct.

>How do I know what the I for the stiffener should be?
I think the assumption is that the I of the stiffener is large compared to the I of the tank, as long as you use a "respectable" size section, relative to the tank, hence it probably doesn't matter what the I of the stiffener is.

-- drej --

Thanks Drej, That is what I thought too.

Going to the next example, "Considering Plate Section as a Beam", Blodgett calculates the maximum deflection on page 6.5-5 except for the (12) in the numerator and the (.5)^3 in the denominator of the calculation.

Are these two values part of the I=bh^3/12? If so, should the I not be the total I for both the plate and the stiffener?

I think he also makes another error in the equation. The constant .0625 should be .00652 per the equation 3Da on page 8.1-8.

The I is the moment of inertia for the vertical tank plate only....you are caculating the deflection of a single plate spanning from the tank bottom to the tank top...no stiffener is involved in the deflection between the bottom and stiffener...just the plate....and yes, the 12 and the .5^3 is for the I (bh^3/12 with the b disappearing since it equals 1).

In these examples, the stiffener determination is not shown. What you have to do is determine the reaction at the top of the plate (the little arrow reaction symbol shown in condition 4C, Table 1). This would be a uniform load of X lbs/ft. This then would be a simple beam design spanning from tank wall to tank wall over the 120".

There seems to be a factor of 10 error here - note that his (Blodgett's) beam formula for condition 3Da on page 8.1-7 shows [Δ] max = .01304 WL^3 / EI which agrees with the beam formulae in AISC's Steel Manual. This equates to the factor .0652 when you replace W with pL/2...
So I think that his formula on page 8.1-8 is incorrect and should be .0652 - not .00652.

JAE,
"So I think that his formula on page 8.1-8 is incorrect and should be .0652 - not .00652."

I thought that it should have been .00652 because he has the same equation for ?(max) on page 6.4-2, figure 3, condition B and on page 8.1-8.

Also, in the formula for condition 3Da on page 8.1-7, the ? max = .01304WL^3/EI. This equates to the same as the formula on page 8.1-8... Unless I am missing something, I think the equation on 8.1-8 is correct and there is a typo in the example on page 6.5-5 in the ? max problem.

Do you agree with what I have pasted below?

Thanks.

Well, I just located the errata for the AISC manual and yes....the number SHOULD be 0.01304 WL^3/EI and NOT 0.1304.

So the correct number is 0.00652.

Your hand calc is what I originally did - but with the erroneous 0.1304 number from the 2rd Edition of the AISC LRFD Manual...which was wrong and corrected by errata in the third edition LRFD.

JAE,
"In these examples, the stiffener determination is not shown. What you have to do is determine the reaction at the top of the plate (the little arrow reaction symbol shown in condition 4C, Table 1). This would be a uniform load of X lbs/ft. This then would be a simple beam design spanning from tank wall to tank wall over the 120". "

Just so that I understand what you are saying to do.. On page 8.1-8, formula 3Da, I should solve for R1. This uniform load could then be used in Equation 3B on page 8.1-7 to solve for M(max) and delta max. This will tell me what the I should be for the stiffener around the top.

Thanks for your help.

Yes - correct on the R1 usage.

If your stiffener is continuous at the corners, then deflections would be dramatically less than that given by 3B. The stiffener would need to be analyzed as a horizontal frame. For a square tank (in plan), you could use the formula in 4Ba on sheet 8.1-13.

On page 8.1-2, Beam #5, there is a 5Da and 5Db shown in the chart. I can find 5B, 5C, and 5E on page 8.1-18.

Can anyone tell me where to find 5Da or 5Db?