Contents ACI 318 08 Example 001
User Manual: ACI 318-08 Example 001
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Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 ACI 318-08 Example 001 SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE EXAMPLE DESCRIPTION The flexural and shear design of a rectangular concrete beam is calculated in this example. A simply supported beam is subjected to an ultimate uniform load of 9.736 k/ft. This example is tested using the ACI 318-08 concrete design code. The flexural and shear reinforcing computed is compared with independent hand calculated results. GEOMETRY, PROPERTIES AND LOADING CL 10" A 13.5" 2.5" A Section A-A 10' = 120" Material Properties E= 3600 k/in2 ν= 0.2 G= 1500 k/in2 Section Properties d = 13.5 in b = 10.0 in I = 3,413 in4 Design Properties f’c = 4 k/in2 fy = 60 k/in2 TECHNICAL FEATURES TESTED Calculation of Flexural reinforcement, As Enforcement of Minimum tension reinforcement, As,min Calculation of Shear reinforcement, Av Enforcement of Minimum shear reinforcing, Av,min ACI 318-08 Example 001 - 1 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 RESULTS COMPARISON Independent results are hand calculated based on the equivalent rectangular stress distribution described in Example 6.1 in Notes on ACI 318-08 Building Code. Output Parameter ETABS Independent Percent Difference Design Moment, Mu (k-in) 1460.4 1460.4 0.00% Tension Reinf, As (in2) 2.37 2.37 0.00% Design Shear Force, Vu 37.73 37.73 0.00% Shear Reinf, Av/s (in2/in) 0.041 0.041 0.00% COMPUTER FILE: ACI 318-08 Ex001 CONCLUSION The computed results show an exact match for the flexural and the shear reinforcing. ACI 318-08 Example 001 - 2 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 HAND CALCULATION Flexural Design The following quantities are computed for all the load combinations: ϕ = 0.9, Ag = 160 sq-in As,min = = 200 bw d = 0.450 sq-in (Govern) fy 3 f c' fy bw d = 0.427 sq-in f c′ − 4000 0.85 = 1000 0.85 − 0.05 β1 = 0.003 = d 5.0625 in 0.003 + 0.005 = cmax amax = β1cmax = 4.303 in Combo1 wu = (1.2wd + 1.6wl) = 9.736 k/ft Mu = ∙ wu l 2 = 9.736 102/8 = 121.7 k-ft = 1460.4 k-in 8 The depth of the compression block is given by: a = d − d2 − 2Mu 0.85 f c'ϕb = 4.183 in (a < amax) The area of tensile steel reinforcement is given by: Mu 1460.4 = a 0.9 • 60 • (13.5 − 4.183 / 2 ) ϕ fy d − 2 As = As = 2.37 sq-in ACI 318-08 Example 001 - 3 Software Verification PROGRAM NAME: REVISION NO.: ETABS 0 Shear Design The following quantities are computed for all of the load combinations: ϕ = 0.75 f c′ : Check the limit of f c′ = 63.246 psi < 100 psi The concrete shear capacity is given by: ϕ Vc = ϕ2 f c′ bd = 12.807 k The maximum shear that can be carried by reinforcement is given by: ϕ Vs = ϕ8 f c′ bd = 51.229 k The following limits are required in the determination of the reinforcement: (ϕ Vc/2) = 6.4035 k ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for any load combination is calculated as follows: If Vu ≤ ϕ (Vc/2), Av = 0, s else if ϕ (Vc/2) < Vu ≤ ϕ Vmax Av (V − φVc ) Av = u ≥ φ f ys d s s min where: b Av w = max 50 s min f yt bw , f yt 3 4 f c′ else if Vu > ϕ Vmax, a failure condition is declared. ACI 318-08 Example 001 - 4 Software Verification ETABS 0 PROGRAM NAME: REVISION NO.: Combo1 ∙ Vu = 9.736 (5-13.5/12) = 37.727 k φ (Vc = / 2 ) 6.4035 k ≤= Vu 37.727 k ≤ φ V = 64.036 k max 10 10 3 Av , 4, 000 = max 50 s min 60, 000 60, 000 4 in 2 Av = max = 0.0083, 0.0079 0.0083 { } in s min Av = s (Vu − φVc ) in 2 in 2 = 0.041 = 0.492 φ f ys d in ft ACI 318-08 Example 001 - 5
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