The design formulae and data provided in this Annex are for education, training and assessment purposes only. They are based on the Hong Kong Code of Practice for Structural Use of Concrete 2013 (HKCP-2013). Unless otherwise specified, their application is subject to the following conditions: Grade of concrete not higher than 45. The moment redistribution is not more than 10%, i.e. β b < 0.9. The structural element is subjected to gravitational loads, i.e. dead and imposed loads, only. Mild Steel High Tensile Steel Grade 250 500B or 500C Specified characteristic strength, f y (N/mm 2 ) 250 500 Appearance Plain Ribbed Notation R T Table 1.1 Properties of Reinforcement Bars Preferred nominal size (mm) Nominal cross-sectional area (mm 2 ) 10 78.5 12 113 16 201 20 314 25 491 32 804 40 1257 Table 1.2 Sizes of Reinforcement Bars (Extracted from Table 2 of CS2:2012) Materials Density (kn/m 3 ) Reinforced concrete 24.5 Cement mortar 23 Natural stone (granite) 29 Soil 20 Table 1.4 - Examples of Density of Material (Extracted from Appendix A of the Code of Practice for Dead and Imposed oads 2011) Material/design consideration Values of m for US Reinforcement 1.15 Concrete in flexure or axial load Concrete in shear strength without shear reinforcement 1.50 1.25 Bond strength 1.40 Others (e.g. bearing stress) >= 1.50 Table 1.3 Values of Partial Safety Factors for Material Strength ( m ) for US (Extracted from Table 2.2 of HKCP 2013) Usage q k (kpa) Domestic 2.0 Offices for general use 3.0 Department stores, shops, etc. 5.0 Table 1.5 - Examples of Imposed oad (Extracted from Table 3.2 of the Code of Practice for Dead and Imposed oads 2011) Dead oad Imposed oad Adverse effect 1.4 1.6 Beneficial effect 1.0 0 Table 1.6 - Values of Partial Safety Factors for oad ( f ) for US (Extracted from Table 2.1 of HKCP 2013) IVE Construction Page 1
h S w a 1 = S w/2 (if S w < h) BEAM Clear Span, n Effective Span, S w a 2 = h/2 (if h < S w) SUPPORT 1 SUPPORT 2 Elevation Figure 1.5 - Effective Span β b = moment at the section after redistribution moment at the section before redistribution [1.1] imit to neutral axis x 0.5d [2.1] The K-value K = M/(bd 2 f cu ) [2.2] β b = 1.0 K' = 0.156 [2.7] β b = 0.8 K' = 0.132 Singly-reinforced section, K < K The lever arm z = [0.5 + (0.25 K/0.9) 0.5 ] d [2.3] If K 0.0428 z = 0.95 d [2.6] A s = M / (0.87 f y z) [2.4] Check depth of neutral axis for flanged section (d z) / 0.45 h f [2.11] Doubly-reinforced section, K > K z = 0.775 d [2.5] Check d' / x < 0.38 [2.10] A' s = (K K') f cu bd 2 [2.8] 0.87 f y (d - d') A s = K' f cu bd 2 + A' s [2.9] 0.87 f y z K 0.043 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.156 z/d 0.950 0.941 0.928 0.915 0.901 0.887 0.873 0.857 0.842 0.825 0.807 0.789 0.775 Table 2.2 ever Arm Factor IVE Construction Page 2
Notes: (a) The length of the cantilever, l3, should be less than half the adjacent span. (b) The ratio of the adjacent spans should lie between 2/3 and 1.5. (c) For simply-supported beam, l pi =. Figure 2.11 Definition of l pi for Calculation of Effective Flange Width (Figure 5.1 of HKCP-2013) c/c distance of adjacent slabs Figure 2.12 Effective Flange Width (Figure 5.2 of HKCP-2013) Elements Minimum (%) Maximum (%) Beam Flexural tension steel Rectangular section 0.13 Flanged section (b w /b < 0.4) 0.18 4 Flexural compression steel Rectangular section 0.20 Column 0.80 6 Wall Vertical bars 0.40 4 Horizontal bars 0.25 Table 2.1 Minimum and Maximum Percentage of Reinforcement (Grade 500) IVE Construction Page 3
Shear Stress v = V b v d b v = b for rectangular beam section b v = b w for flanged beam section [3.1] where v c = 0.79 ( 100 A s 400 1 ) 1/3 ( ) 1/4 b v d d m [3.2] ( ( ( 100 A s b v d 400 d 400 d m ) should not be taken as greater than 3. (A s is the steel area of longitudinal tension steel.) ) 1/4 should not be taken as less than 0.67 for members without shear reinforcement. ) 1/4 should not be taken as less than 1 for members with minimum shear reinforcement. 1.25, partial factor of safety for shear strength of concrete Table 3.1 Values of Design Concrete Shear Stress, v c (Extracted from Table 6.3 of HKCP-2013) If f cu > 25 MPa, the value of v c has to be multiplied by (f cu / 25) 1/3. A sv / s v b v (v v c ) / (0.87 f yv ) [3.3] A sv / s v 0.4 b v / (0.87 f yv ) [3.4] V n = (v c + 0.4) b v d [3.5] Max. Allowable Shear Stress = 0.8 f cu [3.6] Max s v 0.75d [3.7] Spacing of inks in (mm) ink Size 80 100 125 150 175 200 225 250 275 300 325 8 1.257 1.005 0.804 0.670 0.574 0.503 0.447 0.402 0.366 0.335 0.309 10 1.964 1.571 1.257 1.047 0.898 0.785 0.698 0.628 0.571 0.524 0.483 12 2.827 2.262 1.810 1.508 1.293 1.131 1.005 0.905 0.823 0.754 0.696 16 5.027 4.021 3.217 2.681 2.298 2.011 1.787 1.608 1.462 1.340 1.237 Table 3.23 A sv / s v Values for Single ink (2 legs) (in mm 2 per mm) IVE Construction Page 4
Table 3.4 Basic /d Ratio for R. C. Section (Table 7.3 of HKCP-2013) Table 3.5 Modification Factor for Tension Reinforcement (Table 7.4 of HKCP-2013) IVE Construction Page 5
Table 3.6 Modification Factor for Compression Reinforcement (Table 7.5 of HKCP-2013) At outer support Near middle of end span At first interior support At middle of interior span At interior supports Moment 0 0.09F -0.11F 0.07F -0.08F Shear 0.45F - 0.6F - 0.55F Notes: No redistribution of the moments calculated from this table should be made. Characteristic imposed load may not exceed the characteristic dead load. oad should be substantially uniformly distributed over three or more spans. Variation in span length should not exceed 15% of the longest. Table 4.1 Force Coefficients for Continuous Beams with Approximately Equal Span under udl (Table 6.1 of HKCP-2013) At outer support (simply supported) Near middle of end span At first interior support At middle of interior span At interior supports Moment 0 0.086F -0.086F 0.063F -0.063F Shear 0.4F - 0.6F - 0.5F Notes: 1. Area of each bay exceeds 30 m 2. 2. Characteristic imposed load does not exceed 5kPa. 3. The ratio of characteristic imposed load to the characteristic dead load does not exceed 1.25. 4. An allowance of 20% redistribution of the moments at the supports has been made. 5. oad should be substantially uniformly distributed over three or more spans. Table 5.1 Force Coefficients for One-way Slabs with Approximately Equal Span under udl (Extracted from Table 6.4 of HKCP-2013) IVE Construction Page 6
Bar Size Bar Spacing in mm 100 125 150 175 200 225 250 275 300 350 8 503 402 335 287 251 223 201 183 168 144 10 785 628 524 449 393 349 314 286 262 224 12 1131 905 754 646 565 503 452 411 377 323 16 2011 1608 1340 1149 1005 894 804 731 670 574 Table 5.2 Steel Area in mm 2 per m Width Shear and moment for simply-supported beam symmetrically loaded with two partial udl loads, w 1 and w 2, as shown in the figure on the right: w w 1 2 w 1 1 2 1 Shears at the supports V = w 1 1 + 0.5 w 2 2 [5.1] Mid-span moment M = 0.5w 1 1 2 + 0.5w 2 2 ( 1 + 0.25 2 ) [5.2] Effective height of a column: e = β o [6.1] where, o = The clear height between end restraints The value of β is given in the following table: End Condition at the Top End Condition at the Bottom 1 2 3 1 0.75 0.80 0.90 2 0.80 0.85 0.95 3 0.90 0.95 1.00 Condition 1. The end of the column is connected monolithically to beams on either side which are at least as deep as the overall dimension of the column in the plane considered. Where the column is designed to a foundation structure, this should be of a form specifically designed to carry the moment. Condition 2. The end of the column is connected monolithically to beams on either side which are shallower than the overall dimension of the column in the plane considered. Condition 3. The end of the column is connected to members which, while not specifically designed to provide restraint to rotation of the column will, nevertheless, provide some nominal restraint. Table 6.1 Values of β for Braced Columns (Table 6.11 of HKCP-2013) K cu K K = Stiffness of the member = 4E(bh 3 /12)/ (for rectangular beam with fixed end) 1.4G k + 1.6Q k K b1 1.0G k K b2 M es = The unbalance fixed end moment (FEM) of beams = (FEM of b 1 ) (FEM of b 2 ) FEM = (1/12)w 2 (for beam under udl) M cl = Moment to the lower column K cl = M es K cl / (K cu + K lu + 0.5K b1 + 0.5K b2 ) M cu = Moment to the upper column = M es K cu / (K cu + K lu + 0.5K b1 + 0.5K b2 ) Figure 6.2 Simplified Sub-frame Analysis for Determination of Design Moment for Column IVE Construction Page 7
Column under Axial Force N = 0.4f cu A c + 0.75f y A sc [6.3] Biaxial Bending N = 0.35f cu A c + 0.67f y A sc [6.4] For M x /h' M y /b', M x ' = M x + β(h'/b')m y [6.5] For M x /h' < M y /b', M y ' = M y + β(b'/h')m x [6.6] N/(bhf cu ) 0 0.1 0.2 0.3 0.4 0.5 0.6 β 1.00 0.88 0.77 0.65 0.53 0.42 0.30 Table 6.2 Values of the Coefficient β for Biaxial Bending (Table 6.14 of HKCP-2013) IVE Construction Page 8
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(1) Section across column face design for bending Pad Footing e a c f (2) Perimeter of the column check v max (3) Section at 1d from the column from design for shear 1.5d Column h g (4) Perimeter at 1.5d from the column face check punching shear b d d PAN Figure 7.3 Critical Sections for R C Design of Square Footing Check distribution of rebars in square footing: 1.5(c+3d) [7.1] No. of Piles Configuration Tensile Force of the Bottom Reinforcement 0.2ϕ 2 T = F c /(4d) ϕ 3 T = F c /(9d) a v Critical section for shear check 4 T = F c /(8d) Figure 7.5 Critical Section for Shear Check in Pile Cap (Plan) (Adapted from Figure 6.19 of HKCP-2013) Table 7.1 Tensile Force for the Reinforcement in Simple Pile Cap IVE Construction Page 10