Thank you for responses. I was surprised how gray this subject is. We proposed to use bearing capacity based on the larger embedment if eccentricity (M/P) exceeds 0.1 of footing width, which ensures that linear bearing stress on this side is at least 4 times larger than on the side of shallow...
Initially posted this question here: https://www.eng-tips.com/viewthread.cfm?qid=519497 Hope to attract more attention on this forum. Appreciate responses!
Yakpol.
I am designing the arch culverts supported on the spread footing. My question is: can I use larger bearing resistance when maximum stress occurs under the heel of the footing?
Tradionally, the bearing resistance is calculated using footing embedment depth (Df per AASHTO) inside the culvert (at...
You may come up with a long cell formula for linear interpolation. Excel still does not have a good built-in function to do this simple task. Or you can use VBA function presented below.
Public Function INTER1(Xval As Double, x As Range, Y As Range)
Dim Nrow%
Nrow =...
HTURKAK,
ACI does not require to increase development length for 2-bar bundles. So, the difference in bond with a single bar must be small and safe to ignore.
Curios about your opinion about ACI/AASHTO or any other code provisions about bar spacing. E.g. would it be equal to reinforce with #8 spaced at 4" vs bundled (2-#8) spaced at 8" ?
The bond area of 2-bar bundle is the same as for 2 single bars. Would it matter? Older ACI/AASHTO provisions were...
Thank you for replies. I will try to change properties of the file. Doubt it's going to help. The file I am using has plenty of macros and they run fine. But UDF does not. Wonder if there's a way to register, so it may act as native function.
I was surprised to find that UDF do not work in the latest MS office worksheet. UDF name appears on the list after typing "=" sign, but then it becomes unrecognized (#NAME?).
If anybody encountered it and managed to solve? Thanks!
One more thing. Disfor() requires these two functions to be placed in ThisWorkbook module, they help to keep screen and print updated.
Private Sub Workbook_Activate()
Application.CalculateFull
End Sub
Private Sub Workbook_BeforePrint(Cancel As Boolean)
Application.CalculateFull
End Sub...
The numbers displayed have the similar format to referenced in the formula. In addition, the second argument (optional) limits precision. Like disfor(A1, 3)
Skip, thanks for showing this technique. Implementing it I came up with this array formula. It looks weird, but works:
{=LEFT(A1,LEN(A1)-SUM(ISNUMBER(NUMBERVALUE(RIGHT(A1, ROW(INDIRECT("1:" & LEN(A1))))))*1))}
