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Aci 318 08 rc sl 001

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Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001 Aci 318 08 rc sl 001

Software Verification PROGRAM NAME: REVISION NO.: SAFE EXAMPLE ACI 318-08 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 inches thick and spans 12 feet between walls To ensure one-way action, Poisson’s ratio is taken to be zero 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 × 1020 k-in) The computational model uses a finite element mesh, automatically generated by SAFE The maximum element size is specified to be 24 inches To obtain factored moments and flexural reinforcement in a design strip, one onefoot-wide strip is defined in the X-direction on the slab, as shown in Figure Simply supported edge at wall 12 ft span Simply supported edge at wall Free edge 1ft design strip Y X Free edge Figure Plan View of One-Way Slab One dead load case (DL80) and one live load case (LL100) with uniformly distributed surface loads of magnitudes 80 and 100 psf, respectively, are defined in the model A load combination (COMB100) is defined using the ACI 318-08 load combination factors, 1.2 for dead loads and 1.6 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 After completing the analysis, design is performed in accordance with ACI 318-08 using SAFE and also by hand computation Table shows the comparison of the moments and design reinforcements computed using the two methods EXAMPLE ACI 318-08 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 = = = = 144 Concrete strength Yield strength of steel Concrete unit weight Modulus of elasticity Modulus of elasticity Poisson’s ratio fc fy wc Ec Es ν = = = = = = 4,000 60,000 3,600 29,000 Dead load Live load wd wl = = in in in in psi psi pcf ksi ksi 80 psf 100 psf 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 with 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-in) Method Moment (k-in) SAFE 55.220 0.212 Calculated 55.296 0.211 As+ Medium A +s ,min = 0.1296 sq-in COMPUTER FILE: ACI 318-08 RC-SL-001.FDB CONCLUSION The SAFE results show a very close comparison with the independent results EXAMPLE ACI 318-08 RC-SL-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE HAND CALCULATION The following quantities are computed for the load combination: ϕ = 0.9 b = 12 in As,min = 0.0018bh = 0.1296 sq-in ⎛ f c′ − 4000 ⎞ ⎟ = 0.85 ⎝ 1000 ⎠ β1 = 0.85 − 0.05 ⎜ cmax = 0.003 d = 1.875 in 0.003 + 0.005 amax = β1cmax = 1.59375 in For the load combination, w and Mu are calculated as follows: w = (1.2wd + 1.6wt) b / 144 wl12 Mu = As = min[As,min, (4/3) As,required] = min[0.1296, (4/3)2.11] = 0.1296 sq-in COMB100 wd = 80 psf wt = 100 psf w = 21.33 lb/in Mu = 55.296 k-in The depth of the compression block is given by: a = d − d2 − Mu 0.85 f c'ϕ b = 0.3108 in < amax The area of tensile steel reinforcement is then given by: As = Mu = 0.2114 sq-in > As,min a⎞ ⎛ ϕ fy ⎜ d − ⎟ 2⎠ ⎝ As = 0.2114 sq-in EXAMPLE ACI 318-08 RC-SL-001 -

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