ATomanovich (User)
Frog Prince (Platinum Boarder)
Posts: 181

update to v1.1. 9 Years, 3 Months ago

Karma: 53

It was brought to my attention in the "Shear" worksheet of the "RECTBEAM (31805).xls" workbook that the logic for determining the minimum shear reinforcing was incorrect. Prior to the ACI 31802 Code the minimum shear reinforcing was determined as Av = 50*b*s/fy. In ACI 31802 and ACI 31805 a new formula Av = (0.75*SQRT(f'c)*b*s/fy (Eqn. 1113) was introduced, but the value of Av was not to be less than 50*b*s/fy. I had both formulas in the worksheet, but I used the "MIN" command, when I should have used the "MAX" command in the comparison of the values of the 2 formulas, with the intent that the minimum value would never be less than 50*b*s/fy. As it turns out, ACI Eqn. 1113 only controls the minimum value when the concrete strength, f'c, is greater than 4,444 psi. Thus, revising the "MIN" to a "MAX" corrected the situation.
This same correction was also made in the shear calculation portions of the "Complete Analysis" and "Torsion" worksheets as well. This workbook is now version 1.1.



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ATomanovich (User)
Frog Prince (Platinum Boarder)
Posts: 181

Update to v1.2 8 Years, 5 Months ago

Karma: 53

I'm kind of embarrassed to admit that I obviously "slept through" first the ACI 31802 Code and then the ACI 31805 Code as well in that I was not aware that the capacity reduction factor, "phi", for flexural tension is not always to be taken = 0.90, as it had been up through the ACI 31899 Code. It has to be determined, and the range for the "phi" factor is between 0.90 (tension controlled) and 0.65 (compression controlled).
For any of the rest of you who might not have been aware of this change, starting with ACI 31802 based on the actual reinforcing that is used in the cross section, one has to first calculate the strain in the tension reinforcing, then calculate the "phi" factor, and finally calculate the ultimate moment capacity for the cross section. What this means is that one can no longer directly solve for the required tension reinforcing for a given ultimate design moment, as was always done in the past when "phi" was = 0.90. Thus, you'll see in the workbook that one of the changes that I made was to eliminate the worksheet that directly solved for the required tension reinforcing. However, in the first 2 worksheets that deal with flexure, off of the calculation page to the right, the program calculates the required reinforcing for the given moment (assuming "phi" = 0.90) to give the user a hint of the amount of required reinforcing to input into the calculation.
Another change in the recent codes is that the maximum flexural reinforcing ratio is no longer controlled by 3/4 of the balanced reinforcing ratio, rho(max) = 0.75*rho(bal). It is now controlled by limiting the strain in the tension reinforcing to a minimum value of 0.004. When the calculated strain is >= 0.005, then it is a "tension controlled" section and the "phi" factor = 0.90.
I apologize for this oversight on my part, but I'm consoled to some degree to have learned that I was far from being the only one who had missed this change in the Codes.



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ATomanovich (User)
Frog Prince (Platinum Boarder)
Posts: 181

Update to V1.2 5 Years, 12 Months ago

Karma: 53

It was brought to my attention in the "RECTBEAM (31805)" workbook in the "Biaxial" worksheet there was a typo error in the text display of the formula for pure axial capacity for an unreinforced section (Cell L48). The formula displayed was showing a "phi" factor of 0.70, when it should have been showing 0.65. The actual capacity calculation in Cell K48, as well as the comment box in Cell K48 were both correctly using/showing "phi" = 0.65.
I have corrected this typo error, and the version of this workbook is now 1.2.



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ATomanovich (User)
Frog Prince (Platinum Boarder)
Posts: 181

RECTBEAM (31805).xls update to version 1.3 4 Years, 9 Months ago

Karma: 53

In the "Complete Analysis", "Shear", and "Torsion" worksheets of the "RECTBEAM (31805).xls" spreadsheet workbook it was brought to my attention that the program was not limiting the ultimate shear strength provided by the shear reinforcing, "(phi)Vs" to the Code allowable maximum, (phi)Vs(max) = 8*(0.75)*SQRT(f'c)*b*d. Actually, I had the verbiage on this in the appropriate comment boxes, but for some reason I did not have the program logic set up to limit the value. I also added a visible note which appears in bold, red text whenever the value of (phi)Vs is controlled by the allowable maximum value.
Also, in the "Biaxial" worksheet and under the heading of "Tie Min. Size & Max. Spac:", the program was supposed to be only referencing the input cells for the longitudinal bar sizes in Cells D18 and D20. However, it was also inadvertently referencing Cell D19, which pertains to the total number of side reinforcing bars.
I have made the necessary corrections to the above mentioned items.
While I was at it, I decided to add a new worksheet to this workbook. The "Slenderness (Sidesway)" worksheet can be used to analysis the slenderness effects for a single, unbraced column subject to sidesway. This worksheet is very well suited for the quick slenderness analysis of a pier of a foundation, where there is either no supporting soil around the pier or where the soil is considered inadequate to provide continuous lateral support for the pier.
This workbook is now version 1.3.



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ATomanovich (User)
Frog Prince (Platinum Boarder)
Posts: 181

RECTBEAM (31805).xls update to V1.4 3 Years, 1 Month ago

Karma: 53

In both the "Complete Analysis" and "Flexure" worksheets, I discovered an incorrect cell reference that in some rare cases for doubly reinforced beams would result in the depth of the assumed compression block (a) and the depth to the neutral axis (c) to be displayed with zero values. The subsequent calculations were not effected, as it was merely a display error. These "rare cases" occur when the depth to the compression reinforcing is greater than or equal to the depth to the neutral axis (c). For this condition, the intended compression reinforcing actually is not in compression, and the program simply ignores this reinforcing and calculates the capacity as a singly reinforced section.
While I was at it, I also "cleaned up" some display issues in a few of the illustrations where some input values were being displayed.



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