Hong kong cop 04 rc pn 001

Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001 Hong kong cop 04 rc pn 001

Software Verification PROGRAM NAME: REVISION NO.: SAFE EXAMPLE Hong Kong CP-04 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/m 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 The concrete has a unit weight of 24 kN/m3 and a fcu 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 EXAMPLE Hong Kong CP-04 RC-PN-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE 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 B-2 Method Shear Stress Shear Capacity (N/mm2) (N/mm2) D/C ratio SAFE 1.122 0.415 2.704 Calculated 1.122 0.415 2.704 COMPUTER FILE: HONG KONG CP-04 RC-PN-001.FDB CONCLUSION The SAFE results show an exact comparison with the independent results EXAMPLE Hong Kong CP-04 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 = 954+ 1554 + 954 + 1554 = 5016 mm Note: All dimensions in millimeters 954 Y Critical section for punching shear shown dashed 327 150 150 327 A Column B Side Side Side 327 450 X Center of column is point (x1, y1) Set this equal to (0,0) 1554 450 327 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 Hong Kong CP-04 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 -477 1554 218 338772 -161594244 x3 = ∑ Ldx y3 = ∑ Ldy Ld Ld Side 777 954 218 207972 161594244 = = mm 1093488 = = mm 1093488 Side 477 1554 218 338772 161594244 Side -777 954 218 207972 -161594244 Sum N.A N.A b0 = 5016 N.A 1093488 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 1554 218 -477 Y-Axis 5b, 6b, 6.95E+10 7.71E+10 Side 954 218 777 X-axis 5a, 6a, 1.26E+11 1.66E+10 From the SAFE output at Grid B-2: VU = 1125.59 kN MUX = −26492.633 kN-mm MUY = 15643.779 kN-mm EXAMPLE Hong Kong CP-04 RC-PN-001 - Side 1554 218 477 Y-Axis 5b, 6b, 6.95E+10 7.71E+10 Side 954 218 -777 X-axis 5a, 6a, 1.26E+11 1.66E+10 Sum N.A N.A N.A N.A N.A N.A 3.90E+11 1.87E+11 Software Verification PROGRAM NAME: REVISION NO.: SAFE At the point labeled A in Figure 2, x4 = −477 and y4 = 777, thus: vU = 1125.591 • 103 26492.633 • 103 [1.87 • 1011 (777 − 0) − (0)(−477 − 0)] − + 5016 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 1564.779 • 103 [3.90 • 1011 (−477 − 0) − (0)(777 − 0)] (3.90 • 1011 )(1.87 • 1011 ) − (0) vU = 1.0294 − 0.0528 − 0.0398 = 0.9368 N/mm2 at point A At the point labeled B in Figure 2, x4 = 477 and y4 = 777, thus: 1125.591 • 103 26492.633 • 103 [1.87 • 1011 (777 − 0) − (0)(477 − 0)] vU = − + 5016 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 1564.779 • 103 [3.90 • 1011 (477 − 0) − (0)(777 − 0)] (3.90 • 1011 )(1.87 • 1011 ) − (0) vU = 1.0294 − 0.0528 − 0.0398 =1.0164 N/mm2 at point B At the point labeled C in Figure 2, x4 = 477 and y4 = −777, thus: vU = 1125.591 • 103 26492.633 • 103 [1.87 • 1011 (−777 − 0) − (0)(477 − 0)] − + 5016 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 1564.779 • 103 [3.90 • 1011 (477 − 0) − (0)(−777 − 0)] (3.90 • 1011 )(1.87 • 1011 ) − (0) vU = 1.0294 − 0.0528 − 0.0398 = 1.1219 N/mm2 at point C At the point labeled D in Figure 2, x4 = −477 and y4 = −777, thus: vU = 1125.591 • 103 26492.633 • 103 [1.87 • 1011 (−777 − 0) − (0)(−477 − 0)] − + 5016 • 218 (1.23 • 1011 )(3.86 • 1010 ) − (0) 1564.779 • 103 [3.90 • 1011 (−477 − 0) − (0)(−777 − 0)] (3.90 • 1011 )(1.87 • 1011 ) − (0) vU = 1.0294 − 0.0528 − 0.0398 = 1.0164 N/mm2 at point D Point C has the largest absolute value of vu, thus vmax = 1.122 N/mm2 EXAMPLE Hong Kong CP-04 RC-PN-001 - Software Verification PROGRAM NAME: REVISION NO.: SAFE The shear capacity is calculated based on the minimum of the following three limits: 1.5 M y 1.5 M ⎛ x + V =V⎜ f + eff ⎜ Vx Vy ⎝ v= Veff ud ⎞ ⎟ = ⎟ ⎠ = 0.415 N/mm2 (CP 3.7.6.2, 3.7.6.3) (CP 3.7.7.3) CP 3.7.7.3 yields the value of v = 0.415 N/mm2, and thus this is the shear capacity Shear Ratio = EXAMPLE Hong Kong CP-04 RC-PN-001 - vU 1.122 = = 2.704 v 0.415