As 3600 01 rc pn 001

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Software Verification PROGRAM NAME: REVISION NO.: SAFE EXAMPLE AS 3600-01 RC-PN-001 Slab Punching Shear Design PROBLEM DESCRIPTION The purpose of this example is to verify slab punching shear design in SAFE The numerical example is a flat slab that has three 8-m spans in each direction, as shown in Figure 0.3 m A B 8m C 8m D 0.3 m 8m 0.6 m 0.25 m thick flat slab 8m Columns are 0.3 m x 0.9 m with long side parallel to the Y-axis, typical 8m Concrete Properties Unit weight = 24 kN/m3 f'c = 30 N/mm2 8m Y X Loading DL = Self weight + 1.0 kN/m2 LL = 4.0 kN/m2 0.6 m Figure 1: Flat Slab for Numerical Example The slab overhangs beyond the face of the column by 0.15 m along each side of the structure The columns are typically 0.3 m x 0.9 m, with the long side parallel to the Y-axis The slab is typically 0.25-m thick Thick plate properties are used for the slab EXAMPLE AS 3600-01 RC-PN-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE The concrete has a unit weight of 24 kN/m3 and a f'c of 30 N/mm2 The dead load consists of the self weight of the structure plus an additional kN/m2 The live load is kN/m2 TECHNICAL FEATURES OF SAFE TESTED ¾ Calculation of punching shear capacity, shear stress, and D/C ratio RESULTS COMPARISON Table shows the comparison of the punching shear capacity, shear stress ratio, and D/C ratio obtained from SAFE with the punching shear capacity, shear stress ratio and D/C ratio obtained by the analytical method They match exactly for this problem Table Comparison of Design Results for Punching Shear at Grid Point B-2 Shear Stress Shear Capacity D/C ratio Method (N/mm2) (N/mm2) SAFE 1.803 1.086 1.660 Calculated 1.803 1.086 1.660 COMPUTER FILE: AS 3600-01 RC-PN-001.FDB CONCLUSION The SAFE results show an exact comparison with the independent results EXAMPLE AS 3600-01 RC-PN-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE HAND CALCULATION Hand Calculation For Interior Column Using SAFE Method d = [(250 − 26) + (250 − 38)] / = 218 mm Refer to Figure b0 = 518+ 1118 + 1118 + 518 = 3272 mm Note: All dimensions in millimeters 518 Y 109 150 150 A Column Critical section for punching shear shown dashed 109 B 109 Side Side Side 450 X Center of column is point (x1, y1) Set this equal to (0,0) 1118 450 109 Side D C Figure 2: Interior Column, Grid B-2 in SAFE Model γ VX =1− γ VY =1− ⎛ ⎞ 900 1+ ⎜ ⎟ ⎝ ⎠ 300 ⎛ ⎞ 300 1+ ⎜ ⎟ ⎝ ⎠ 900 = 0.536 = 0.278 The coordinates of the center of the column (x1, y1) are taken as (0, 0) EXAMPLE AS 3600-01 RC-PN-001 - Software Verification SAFE PROGRAM NAME: REVISION NO.: The following table is used for calculating the centroid of the critical section for punching shear Side 1, Side 2, Side 3, and Side refer to the sides of the critical section for punching shear as identified in Figure Item x2 y2 L d Ld Ldx2 Ldy2 Side −259 1118 218 243724 −63124516 x3 = ∑ Ldx y3 = ∑ Ldy Ld Side 559 518 218 112924 63124516 = = mm 713296 = = mm 713296 Ld Side 259 1118 218 243724 63124516 Side −559 518 218 112924 −63124516 Sum N.A N.A b0 = 3272 N.A 713296 0 The following table is used to calculate IXX, IYY and IXY The values for IXX, IYY and IXY are given in the "Sum" column Item L d x2 - x3 y2 - y3 Parallel to Equations IXX IYY IXY Side 1118 218 −259 Y-Axis 5b, 6b, 2.64E+10 1.63E+10 Side 518 218 559 X-axis 5a, 6a, 3.53E+10 2.97E+09 From the SAFE output at grid line B-2: VU = 1125.59 kN MUX = −28514.675 kN-mm MUY = 14231.447 kN-mm EXAMPLE AS 3600-01 RC-PN-001 - Side 1118 218 259 Y-Axis 5b, 6b, 2.64E+10 1.63E+10 Side 518 218 −559 X-axis 5a, 6a, 3.53E+10 2.97E+09 Sum N.A N.A N.A N.A N.A N.A 1.23E+11 3.86E+10 Software Verification PROGRAM NAME: REVISION NO.: SAFE At the point labeled A in Figure 2, x4 = −259 and y4 = 559, thus: vU = 1125.591 • 103 28514.675 • 103 [3.86 • 1010 (559 − 0) − (0)(−259 − 0)] − + 3272 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 14231.447 • 103 [1.23 • 1011 (−259 − 0) − (0)(559 − 0)] (1.23 • 1011 )(3.86 • 1010 ) − (0) vU = 1.5780 − 0.1293 − 0.0954 = 1.353 N/mm2 at point A At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus: 1125.59 • 103 28514.675 • 103 [3.86 • 1010 (559 − 0) − (0)(259 − 0)] vU = − + 3272 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 14231.447 • 103 [1.23 • 1011 (259 − 0) − (0)(559 − 0)] (1.23 • 1011 )(3.86 • 1010 ) − (0) vU = 1.5780 − 0.1293 + 0.0954 = 1.544 N/mm2 at point B At the point labeled C in Figure 2, x4 = 259 and y4 = −559, thus: vU = 1125.59 • 103 28514.675 • 103 [3.86 • 1010 (−559 − 0) − (0)(259 − 0)] − + 3272 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 14231.447 • 103 [1.23 • 1011 (259 − 0) − (0)(−559 − 0)] (1.23 • 1011 )(3.86 • 1010 ) − (0) vU = 1.5780 + 0.1293 + 0.0954 = 1.803 N/mm2 at point C At the point labeled D in Figure 2, x4 = −259 and y4 = −559, thus: vU = 1125.591 • 103 28514.675 • 103 [3.86 • 1010 (−559 − 0) − (0)(−259 − 0)] − + 3272 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 14231.447 •103 [1.23 •1011 ( −259 − 0) − (0)(−559 − 0)] (1.23 • 1011 )(3.86 • 1010 ) − (0) vU = 1.5780 + 0.1293 − 0.0954 = 1.544 N/mm2 at point D Point C has the largest absolute value of vu; thus vmax = 1.8143 N/mm2 EXAMPLE AS 3600-01 RC-PN-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE The shear capacity is calculated based on the smallest of AS 3600-01 equation 11-35, with the dom and u terms removed to convert force to stress ⎧ ⎛ ⎪0.17 ⎜ + βh ϕ fcv = ⎨ ⎝ ⎪ ⎩0.34ϕ f ′c ⎞ ⎟ϕ ⎠ f ′c = 1.803N/mm2 in accordance with AS 9.2.3(a) AS 9.2.3(a) yields the smallest value of ϕ fcv = 1.086 N/mm2, and thus this is the shear capacity Shear Ratio = EXAMPLE AS 3600-01 RC-PN-001 - vU 1.803 = = 1.660 1.086 ϕ f cv