Is 456 00 rc bm 001

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Software Verification PROGRAM NAME: REVISION NO.: SAFE EXAMPLE IS 456-00 RC-BM-001 Flexural and Shear Beam Design PROBLEM DESCRIPTION The purpose of this example is to verify slab flexural design in SAFE The load level is adjusted for the case corresponding to the following conditions: ƒ The stress-block extends below the flange but remains within the balanced condition permitted by IS 456-2000 ƒ The average shear stress in the beam is below the maximum shear stress allowed by IS 456-2000, requiring design shear reinforcement A simple-span, 6-m-long, 300-mm-wide, and 500-mm-deep T beam with a flange 100 mm thickness and 600 mm wide is modeled using SAFE The beam is shown in Figure The computational model uses a finite element mesh of frame elements, automatically generated by SAFE The maximum element size has been specified to be 200 mm The beam is supported by columns without rotational stiffnesses and with very large vertical stiffness (1 × 1020 kN/m) The beam is loaded with symmetric third-point loading One dead load case (DL20) and one live load case (LL80) with only symmetric third-point loads of magnitudes 20, and 80 kN, respectively, are defined in the model One load combinations (COMB80) is defined using the IS 456-2000 load combination factors of 1.5 for dead loads and 1.5 for live loads The model is analyzed for both of these load cases and the load combinations The beam moment and shear force are computed analytically The total factored moment and shear force are compared with the SAFE results The moment and shear force are identical After completing the analysis, design is performed using the IS 456-2000 code in SAFE and also by hand computation Table shows the comparison of the design longitudinal reinforcements Table shows the comparison of the design shear reinforcements EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE 600 mm 75 mm 100 mm 500 mm 75 mm 300 mm Beam Section 2000 mm 2000 mm 2000 mm Shear Force Bending Moment Figure The Model Beam for Flexural and Shear Design EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: GEOMETRY, PROPERTIES AND LOADING Clear span, Overall depth, Flange Thickness, Width of web, Width of flange, Depth of tensile reinf., Effective depth, Depth of comp reinf., l h ds bw bf dc d d' = = = = = = = = 6000 500 100 300 600 75 425 75 Concrete strength, Yield strength of steel, Concrete unit weight, Modulus of elasticity, Modulus of elasticity, Poisson’s ratio, f'c fy wc Ec Es v = = = = = = 30 460 25x105 2x108 0.2 Dead load, Live load, Pd Pl = = 20 80 SAFE mm mm mm mm mm mm mm mm MPa MPa kN/m3 MPa MPa kN kN TECHNICAL FEATURES OF SAFE TESTED ¾ Calculation of flexural and shear reinforcement ¾ Application of minimum flexural and shear reinforcement RESULTS COMPARISON Table shows the comparison of SAFE total factored moments in the design strip with the moments obtained by the analytical method They match exactly for this problem Table also shows the comparison of design reinforcements Table Comparison of Moments and Flexural Reinforcements Reinforcement Area (sq-cm) Method Moment (kN-m) SAFE 312 21.13 Calculated 312 21.13 As+ A +s ,min = 235.6 sq-mm EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE Table Comparison of Shear Reinforcements Reinforcement Area, Av s (sq-cm/m) Shear Force (kN) SAFE Calculated 156 7.76 7.73 COMPUTER FILE: IS 456-00 RC-BM-001.FDB CONCLUSION The SAFE results show an exact comparison with the independent results EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE HAND CALCULATION Flexural Design The following quantities are computed for all the load combinations: γm, steel = 1.15 γm, concrete = 1.50 α = 0.36 β = 0.42 As ,min ≥ 0.85 bd = 235.6 sq-mm fy COMB80 P = (1.4Pd + 1.6Pt) =156 kN M* = N *l = 312 kN-m xu ,max d ⎧0.53 ⎪ ⎪0.53 − 0.05 f y − 250 ⎪ 165 =⎨ ⎪0.48 − 0.02 f y − 415 ⎪ 85 ⎪ ⎩0.46 if f y ≤ 250 MPa if 250 < f y ≤ 415 MPa if 415 < f y ≤ 500 MPa if f y ≥ 500 MPa xu ,max = 0.4666 d The normalized design moment, m, is given by m= Mu b f d 2α f ck M = 312x106/(600 • 4252 • 0.36 • 30) = 0.26656 ⎛ Df ⎜ ⎝ d ⎞ ⎟ = 100/425 = 0.23529 ⎠ EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE xu − − β m = 0.305848 > ⎛ D f ⎞ = ⎜ ⎟ d 2β ⎝ d ⎠ γ f = 0.15 xu + 0.65 D f if D f > 0.2 d = 84.49781 γf ⎞ ⎛ M f = 0.45 f ck (b f − bw )γ f ⎜⎜ d − ⎟⎟ = 130.98359 kN-m ⎠ ⎝ Mw = Mu − Mf = 181.0164 kN-m Mw,single = αfckbwd2 m= x u,max ⎡ x u,max ⎤ ⎥ = 233.233 < Mw ⎢1 − β d ⎣ d ⎦ Mw = 0.309310 b f d 2α f ck xu − − β m = 0.36538 = d 2β Mf Mw = 2113 sq-mm + As = ( f y γ s )(d − 0.5 y f ) ( f y γ s ) z Shear Design τv = Vu = 1.2235 bd τmax = 3.5 for M30 concrete k = 1.0 δ = if Pu ≤ , Under Tension 100 As 100 As = 0.15 as 0.15 ≤ ≤3 bd bd ⎛ fck ⎞ ⎜ ⎟ ⎝ 25 ⎠ = 1.0466 τ c = 0.29 From Table 19 of IS 456:2000 code τcd = kδτc = 0.29 τcd +0.4 = 0.69 EXAMPLE IS 456-00 RC-BM-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE The required shear reinforcement is calculated as follows: If τcd + 0.4 < τv ≤ τc,max Asv (τ v − τ cd ) b ≥ = 7.73 sq-cm/m sv 0.87 f y EXAMPLE IS 456-00 RC-BM-001 -