Over-Reinforced Concrete Beam Design ACI

Transcription

1 Note: Over-reinforced typically means the tensile steel does not yield at failure. for this example, over-reinforced means the section is not tension controlled per ACI standards even though the steel does yield Given: f c = 6,000 psi fy = 60 ksi Required: Determine the nominal moment capacity, ΦMn Assumptions: 1. Plain sections remain plain (ACI section ) 2. Maximum concrete strain at extreme compression fiber = (ACI section ) 3. Tensile strength of concrete is neglected (10.2.5) 4. Compression steel is neglected in this calculation.

2 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 = /2 d = 20.5

3 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 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 (8 x 0.79 in 2 ) = 379,200 lb

4 Solve for a: 61,200 lb/in x a = 379,200 lb a = Now that we know the depth of the stress block, we can calculate c, the depth to the neutral axis. From ACI 318 section 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 β1 = ((f c 4000psi)/1000)) x β1 = 0.85 ((6000psi 4000psi)/1000) x 0.05 = 0.75 c = a / β1 = /0.75 c = 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. First, let s calculate dt, the depth from the extreme compression fiber to the extreme tensile steel dt = /2 = 21.5 c/ε c = dt/(ε c + ε t ) ε t = (dt x ε c )/c ε c = (21.5 x 0.003)/ ε t =

5 Determine if the section is tension controlled: Determine if the section is tension controlled: Per ACI section a beam is considered tension controlled if the strain in the extreme tension steel is greater than The calculated steel strain in our section is which is less than therefore this beam section is not tension controlled. Per ACI section εt at nominal strength for flexural members shall not be less than ( is acceptable) The strain for our section is between the limit for compression controlled (.002 per ACI section ) and the limit for tensioned controlled sections, Therefore, we must interpolate to determine our Φ factor in accordance with ACI ( )/( ) = (Φ )/( ) Φ = 0.88 Determine the strain at which the steel yields and check our assumption that the steel in fact yielded: E = f y /ε y E = Young s modulus which is generally accepted to be 29,000 ksi for steel f y = steel yield stress ε y = yield strain ε y = 60ksi / 29,000 ksi =

6 When we checked if the section was tensioned controlled, we check the strain at the extreme layer of steel. We must calculate the strain at the centroid of the tensile steel to determine if it yields: ε s = (d x ε c )/c ε c = (20.5 x 0.003)/ ε t = is greater than 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 , the minimum steel requirement is: A s, min = ((3 x square root(f c))/f y ) x b w x d (200/fy) x b w x d As a side note, the 200/f y minimum controls when f c is less than 4,500 psi A s, min = ((3 x squareroot(6,000 psi))/60,000psi) x 12 x 20.5 A s,min = 0.95 in 2 A s = 8 x 0.79 in 2 = 6.32 in 2 > 0.95 in 2 therefore we satisfy the minimum steel requirements of ACI

7 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 the center of the compressive force. Calculate the moment about the center of the compressive force: ΦM n = Φ x T x (d a/2) Φ = 0.88 as we previously calculated T = A s f y ΦM n = 0.88 x 6.32 in 2 x 60 ksi (20.5in in / 2) ΦM n = 5,807 kip*in = kip*ft

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