Reinforcing requirements based on % of concrete volume 1

[Bedrock1977]

I am working on a budgetary number for a parking ramp. The large footing pads do not have an actual rebar design yet, but there is a note stating that the reinforcing requirements shall be 1% of the concrete volume.

The sizes are as follows:

10-0 x 10-0 x 28" (8.641975 CY or 233.3333 CF)
12-0 x 12-0 x 32" (14.22222 CY or 384 CF)
14-0 x 14-0 x 36" (21.77778 CY or 588 CF)
20-0 x 20-0 x 42" (51.85185 CY or 1400 CF)
25-0 x 25-0 x 48" (92.59259 CY or 2500 CF)
20-0 x 31-0 x 48" (91.85185 CY or 2480 CF)
31-0 x 20-0 x 48" (91.85185 CY or 2480 CF)
34-0 x 20-0 x 48" (100.7407 CY or 2720 CF)

How do you go about calculating what the rebar would be? In my experience in jobs of this size, the footing pads have a double layer of rebar (one top and bottom).

If you need any further information from me, please let me know.

Regards,
Chris

 

[hokie66]

1% of concrete volume (for steel volume) seems simple to me. Whether the estimate proves correct or not will be known once the design is done.

 

[frostryan]

Dont know if this will help but I came across a conversion factor by weight for unknown rebar.
Multiply your CF x 165= x tons/2000 lbs per ton = tons of concrete w/ rebar.
Your example:
233.33 x 165 = 38499.45 / 2000 = 19.249 tons

Let me know if this does anything for you.

 

[hokie66]

frostryan's conversion factor is well off the mark. 1% of 233.33 CF is 2.33 CF of steel, which weighs about 1140 pounds. I have no idea if that is what the footing requires, but that is what the note says.

 

[cvg]

generally require area of steel as a percentage of the area of the structural element. not sure how volume of steel entered into the picture.

What I have done previously is take a fully designed, similar structure, calculate the weight of steel for that structure as a percentage of the concrete volume and use that number for estimating. Say 140 lbs steel per cubic yard which will give you a fairly heavily reinforced structure such as a box culvert

 

[hokie66]

cvg,
The 1% volume would give 132 pounds/CY, so not far from your estimating value.

 

[Illbay]

frostryan's formula essentially says that rebar is 4.4% by volume (or given a prismatic section, this is the steel area as a percent of gross area of section).

If you subscribe to a unit-weight-of-reinforced-concrete formulation such as frostryan's, a bunch of algebra yields the following:

Vs ≡ (w-wc)/(ws-wc)*V [EQUATION 1]

Where V = Total volume of reinforced concrete element
Vs = Total volume of steel in the element volume
w = unit weight of total volume (= 165 lb/ft3 according to frostryan's formula)
wc = unit weight of concrete (= 150 lb/ft3 - a typical value)
ws = unit weight of steel (= 490 lb/ft3)

This seems pretty steep to me, but ymmv. It would probably be useful to run a bunch of typical concrete elements that you use in your particular application and see what you come up with.

If you calculate the weight of your steel for a given concrete volume, you can then calculate "w" as follows:

w = Ws/(ws*V)*(ws-wc)+wc [EQUATION 2]

Where Ws = Total weight of reinforcement for a given total volume V.

You'd then have a handy-dandy value for "w" (again, in frostryan's example w = 165 lb/ft3) to plug into EQUATION 1, to estimate your steel weight from a given volume of concrete. I suggest this calculation is idiosyncratic to the typical types of concrete elements you deal with, and your particular detailing preferences. But if you have a large enough "database" against which to apply EQUATION 2, then EQUATION 1 would be a very handy formula. I'd suggest having a value of "w" for beams of s certain size, another for columns (possibly by type), another for slabs, and so on.

OH...simply multiply the result of EQUATION 1 by ws (= 490 lb/ft^3, e.g.) to obtain the WEIGHT per unit volume, since you typically want that value rather than just the volume:

Ws = ws*(w-wc)/(ws-wc)*V [EQUATION 1a]

That variable "w" is the linchpin for the whole procedure.

"No one is completely useless. He can always serve as a bad example." --My Dad ca. 1975

 

[Illbay]

hokie66 "The 1% volume would give 132 pounds/CY, so not far from your estimating value"​

If you assume 1% by volume (or area of prismatic section - which incidentally is the minimum reinforcement required for a column by ACI 318), then:

w = 0.01(ws-wc)+wc = 0.01*(490 lb/ft3 - 150 lb/ft3) + 150 lb/ft3

= 153.4 lb/ft3

(Compare with the value of 165 given in frostryan's formula).

"No one is completely useless. He can always serve as a bad example." --My Dad ca. 1975

 

[rapt]

I cannot believe that this is causing so much angst.

If the concrete volume is 233.33ft3 and steel weighs 490lbs/ft3, then at 1% reinforcing by volume

steel reqd = 1 / 100 *233.33 * 490 = 1143lbs.

What has concrete density got to do with it?

 

[Ron]

rapt...we can overthink anything! hokie66's first answer was the right one and nothing necessary beyond!