004 LEER Over Reinforced Concrete Beam Design

8/10/2019 004 LEER Over Reinforced Concrete Beam Design

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Over-Reinforced Concrete Beam Design ACI 318-0

Note: Over-reinforced typically means the tensile steel doesfor this example, over-reinforced means the section

controlled” per ACI standards even though the stee

Given:

• f’c = 6,000 psi

• f y = 60 ksi

Required:

• Determine the nominal moment capacity, !Mn

Assumptions:

1. Plain sections remain plain (ACI 318-08 section 10.2.2

2. Maximum concrete strain at extreme compression fibsection 10.2.3)

3. Tensile strength of concrete is neglected (10.2.5)

4. Compression steel is neglected in this calculation.

Friday, August 10, 12



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 - Clear Spacing Between Bars/2

– d = 24” – 1.5” – 0.5” - 1” - 1”/2

– d = 20.5”





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 = f s x As

• f s = stress in the steel (we make the assumption that the steel yconfirm if it does).

• As = area of tensile steel

T = 60000psi x (8 x 0.79 in2) = 379,200 lb





Solve for a:

• 61,200 lb/in x a = 379,200 lb

• a = 6.196”

With c, we can calculate the strain in the extreme tensile steel using similar trWith this strain calculated, we can check our assumption that the steel yieldsdetermine if the section is tension controlled.

First, let’s calculate dt, the depth from the extreme compression fiber to the etensile steel

dt = 24” - 1.5” - 0.5” - 1”/2 = 21.5”

c/!c = dt/(!c + !t)

!t = (dt x !c)/c – !c = (21.5” x 0.003)/8.261” – 0.003

"t = 0.0048

Now that we know the depth of the stress block, we can calculate c, the depthneutral axis.

From ACI 318 section 10.2.7.1 – a = "1 x c

"1 is a factor that relates the depth of the Whitney stress block to the depth oneutral axis based on the concrete strength. It is defined in 10.2.7.3

"1 = 0.65 # 0.85 - ((f’c – 4000psi)/1000)) x 0.05 # 0.85

"1 = 0.85 – ((6000psi – 4000psi)/1000) x 0.05 = 0.75

c = a / "1 = 6.196”/0.75

c = 8.261”





• 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.

– The calculated steel strain in our section is 0.0048 which is less than 0.005 therefore this beam section is not tension con

– Per ACI section 10.3.5 !t at nominal strength for flexural members shall not be less than 0.004 (0.0048 is acceptable)

• The strain for our section is between the limit for compression controlled (.002 per ACI section 10.3.3) and the limit for tensionsections, 0.005. Therefore, we must interpolate to determine our $ factor in accordance with ACI 9.3.2.2

– (0.9 - 0.65)/(0.005 - 0.002) = ($ - 0.65)/(0.0048 - 0.002)

– !

= 0.88

• Determine the strain at which the steel yields and check our assumption that the steel in fact yielded:

– E = f y/!y

• E = Young’s modulus which is generally accepted to be 29,000 ksi for steel

• f y = steel yield stress

• !y = yield strain

– !y = 60ksi / 29,000 ksi = 0.00207


Determine if the section is tension controlled:




When we checked if the section was tensioned controlled, we check the strain at the extreme layer of steel. We must calculate t

centroid of the tensile steel to determine if it yields:

• !s = (d x !c)/c – !c = (20.5” x 0.003)/8.261” – 0.003

• "t = 0.0044

• 0.0044 is greater than 0.00207 therefore our assumptions are 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))/f y) x bw x d % (200/f y) x bw x d

• As a side note, the 200/f y minimum controls when f’c is less than 4,500 psi

– As, min = ((3 x squareroot(6,000 psi))/60,000psi) x 12” x 20.5”

– As,min = 0.95 in2

– As = 8 x 0.79 in2 = 6.32 in2 > 0.95 in2 therefore we satisfy the minimum steel requirements of ACI





Finally, let’s calculate the nominal moment capacity of the section:

Using moment equilibrium, we can calculate the moment capacity by taking the moment about the center of the tensile force, or compressive force.

Calculate the moment about the center of the compressive force

$Mn = $ x T x (d – a/2)

• $ = 0.88 as we previously calculated

• T = Asf y

$Mn = 0.88 x 6.32 in2 x 60 ksi (20.5in - 6.196in / 2)

!Mn = 5,807 kip*in = 483.9 kip*ft





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004 LEER Over Reinforced Concrete Beam Design