Eurocode 2 04 rc sl 001

Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001 Eurocode 2 04 rc sl 001

Software Verification PROGRAM NAME: REVISION NO.: SAFE EXAMPLE Eurocode 2-04 RC-SL-001 Slab Flexural Design PROBLEM DESCRIPTION The purpose of this example is to verify slab flexural design in SAFE A one-way, simple-span slab supported by walls on two opposite edges is modeled using SAFE The slab is 150 mm thick and spans meters between walls To ensure one-way action, Poisson’s ratio is taken to be zero The computational model uses a finite element mesh, automatically generated by SAFE The maximum element size is specified as 1.0 meter The slab is modeled using thin plate elements The walls are modeled as line supports without rotational stiffnesses and with very large vertical stiffness (1 × 1015 kN/m) To obtain factored moments and flexural reinforcement in a design strip, one onemeter wide strip is defined in the X-direction on the slab, as shown in Figure Simply supported edge at wall m span Simply supported edge at wall Free edge m design strip Y X Free edge Figure Plan View of One-Way Slab One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly distributed surface loads of magnitudes and kN/m2, respectively, are defined in the model A load combination (COMB5kPa) is defined using the Eurocode 204 load combination factors, 1.35 for dead loads and 1.5 for live loads The model is analyzed for both load cases and the load combination The slab moment on a strip of unit width is computed analytically The total factored strip moments are compared with the SAFE results These moments are identical After completing the analysis, design is performed using the Eurocode 2-04 code by SAFE and also by hand computation Table shows the comparison of the design reinforcements computed by the two methods EXAMPLE Eurocode 2-04 RC-SL-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE GEOMETRY, PROPERTIES AND LOADING Thickness Depth of tensile reinf Effective depth Clear span T, h dc d ln, l1 = = = = 150 25 125 4000 mm mm mm mm Concrete strength Yield strength of steel Concrete unit weight Modulus of elasticity Modulus of elasticity Poisson’s ratio fc fsy wc Ec Es ν = = = = = = 30 460 25000 2x106 MPa MPa N/m3 MPa MPa Dead load Live load wd wl = = 4.0 5.0 kPa kPa TECHNICAL FEATURES OF SAFE TESTED ¾ Calculation of flexural reinforcement ¾ Application of minimum flexural reinforcement RESULTS COMPARISON Table shows the comparison of the SAFE total factored moments in the design strip the moments obtained by the hand computation method Table also shows the comparison of the design reinforcements Table Comparison of Design Moments and Reinforcements Load Level Reinforcement Area (sq-cm) Method Moment (kN-m) SAFE 25.797 5.401 Calculated 25.800 5.393 As+ Medium A +s ,min = 204.642 sq-mm EXAMPLE Eurocode 2-04 RC-SL-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE COMPUTER FILE: Eurocode 2-04 RC-SL-001.FDB CONCLUSION The SAFE results show a very close comparison with the independent results EXAMPLE Eurocode 2-04 RC-SL-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE HAND CALCULATION The following quantities are computed for the load combination: γm, steel = 1.15 γm, concrete = 1.50 η = 1.0 for fck ≤ 50 MPa λ = 0.8 for fck ≤ 50 MPa b = 1000 mm For the load combination, w and M are calculated as follows: w = (1.35wd + 1.5wt) b M= wl12 As ,min ⎧ 0.0013bw d ⎪ = max ⎨ fctm 26 bd ⎪ f yk ⎩ = 204.642 sq-mm COMB100 wd = 4.0 kPa wt = 5.0 kPa w = 12.9 kN/m M = 25.8 kN-m The depth of the compression block is given by: m= M bd 2ηf cd = 0.08256 ⎛x⎞ ⎡ λ⎛ x⎞ ⎤ mlim = λ ⎜ ⎟ ⎢1 − ⎜ ⎟ ⎥ = 0.294 ⎝ d ⎠ lim ⎣ ⎝ d ⎠ lim ⎦ δ − k1 ⎛x⎞ for fck ≤ 50 MPa = 1.8182 ⎜ ⎟ = k2 ⎝ d ⎠ lim EXAMPLE Eurocode 2-04 RC-SL-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE For reinforcement with fyk ≤ 500 MPa, the following values are used: k1 = 0.44 k2 = k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25 δ is assumed to be ω = − − 2m = 0.08682 ⎛ ηf bd ⎞ As = ω ⎜ cd ⎟ = 539.264 sq-mm > As,min ⎜ f ⎟ ⎝ yd ⎠ As = 5.393 sq-cm EXAMPLE Eurocode 2-04 RC-SL-001 -