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Design of Reinforced Concrete Beam with Minimum Steel as per ACI 318-08
This example explores the minimum steel requirements of ACI.

Given:

• f’c = 6,000 psi
• fy = 60 ksi

Required:

• Determine if this section can carry a factored moment, Mu, of 40 kip*ft, while satisfying all ACI requirements

Assumptions:

1. Plain sections remain plain (ACI 318-08 section 10.2.2)
2. Maximum concrete strain at extreme compression fiber = 0.003 (ACI section 10.2.3)
3. Tensile strength of concrete is neglected (10.2.5)
4. Compression steel is neglected in this calculation.

Let’s start by constructing the stress and strain diagrams:

• Next, we’ll calculate d, the depth from the extreme compression fiber to the center of reinforcement in the tensile zone.
d = h – Clear Spacing – dstirrup – dreinforcement /2
d = 24” – 1.5” – 0.5” - 0.5”/2
d = 21.75”

Whitney stress block
Next, we want to use equilibrium to solve for a, the depth of the Whitney stress block.
From the rules of equilibrium we know that C must equal T
C = T
C = 0.85 x f’c x b x a
• Defined in ACI section 10.2.7.1
• b = width of compression zone
• a = depth of Whitney stress block
C = 0.85 x 6000psi x 12” x a = 61,200 lb/in x a
T = fs x As
• fs = stress in the steel (we make the assumption that the steel yields, and will later confirm if it does).
• As = area of tensile steel
T = 60000psi x (3 x 0.2 in2) = 36,000 lb

Solve for a:
• 61,200 lb/in x a = 36,000 lb
a = 0.588”

strain diagram
Add caption
Now that we know the depth of the stress block, we can calculate c, the depth to the neutral axis.
From ACI 318 section 10.2.7.1 – a = β1 x c β1 is a factor that relates the depth of the Whitney stress block to the depth of the neutral axis based on the concrete strength. It is defined in 10.2.7.3
β1 = 0.65 ≤ 0.85 - ((f’c – 4000psi)/1000)) x 0.05 ≤ 0.85
β1 = 0.85 – ((6000psi – 4000psi)/1000) x 0.05 = 0.75
c = a / β1 = 0.588”/0.75
c = 0.784”


With c, we can calculate the strain in the extreme tensile steel using similar triangles. With this strain calculated, we can check our assumption that the steel yields, and determine if the section is tension controlled.
c/εc = d/(εc + εt)
εt = (d x εc)/c – εc = (21.75” x 0.003)/0.784” – 0.003
εt = 0.080


Determine if the section is tension controlled:

– Per ACI section 10.3.4 a beam is considered tension controlled if the strain in the extreme tension steel is greater than 0.005.
– The calculated steel strain in our section is 0.080 which is greater than 0.005 therefore this beam section is tension controlled.
Φ = 0.90
• Determine the strain at which the steel yields and check our assumption that the steel in fact yielded:
E = fy/εy
• E = Young’s modulus which is generally accepted to be 29,000 ksi for steel
• fy = steel yield stress
• εy = yield strain
εy = 60ksi / 29,000 ksi = 0.00207
0.080 is greater than 0.00207 therefore our assumption is correct and the steel yields prior to failure.

Next, let’s determine if the beam section satisfies the minimum steel requirements of ACI:


• Per ACI section 10.5.1, the minimum steel requirement is:
As, min = ((3 x square root(f’c))/fy) x bw x d ≥ (200/fy) x bw x d
• As a side note, the 200/fy minimum controls when f’c is less than 4,500 psi
As, min = ((3 x squareroot(6,000 psi))/60,000psi) x 12” x 21.75”
As,min = 1.01 sq.in
As = 3 x 0.2sq.in = 0.6sq.in < 1.01sq.in therefore we do not satisfy the minimum steel requirements of ACI
Per ACI 318 Section 10.5.3 - The above mentioned provision does not need to be satisfied if, at every section, As provided is at least one-third greater than that required by analysis.We need to calculate the required amount of steel for this section to carry a 40 kip*ft moment and determine if the amount of steel provided is at least 1/3rd greater.

Let’s calculate the area of steel required to carry a moment of 40 kip*ft:

Using force equilibrium, let’s find the relationship of As to a:
• 0.85 x f’c x b x a = As x fy → 0.85 x 6,000psi x 12in x a = As x 60,000psi → a = 0.98/in x As
Calculate the moment about the center of the compressive force, and set it equal to the factored moment:
ΦMn = Φ x T x (d – a/2)
• Φ = 0.9 as we previously calculated
• T = Asfy
• ΦMn = Mu
40kip*ft x 12 kip*in/kip*ft = 0.9 x As x 60 ksi (21.75in - a / 2)
Substitute a = 0.98/in x As into the above equation and rearrange it a bit:
26.26(kip/in3) x As2 - 1174.5(kip/in) x As + 480(kip*in) = 0
What we have is a quadratic equation to solve!

• Quadratic formula:
(-b ± √(b2 - 4ac))/2a
As = (-(-1174.5) - √((-1174.5)2 - 4 x 26.46 x 480))/(2 x 26.46)
As = 0.41in2
• Determine if we satisfy the minimum steel requirements per ACI 318 section 10.5.3
1 1/3 x As = 1.333 x 0.41in2 = 0.55 in2
We have 0.6 in2 which is greater than 0.55 in2 therefore the section satisfies all ACI requirements for an applied factored moment of 40 kip*ft.








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