Annex - R C Design Formulae and Data (2014-08-01)tycnw01.vtc.edu.hk/cbe2022/Annex - R C Design...

Annex - R C Design Formulae and Data

IVE Construction Aug 2014

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. βb < 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, fy (N/mm2) 250 500

Appearance Plain Ribbed

Notation R T

Table 1.1 – Properties of Reinforcement Bars

Preferred nominal size (mm)

Nominal cross-sectional area

(mm2)

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)

Material/design consideration

Values of m for ULS

Reinforcement 1.15

Concrete in flexure or axial load

1.50

Concrete in shear strength without shear reinforcement

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 ULS (Extracted from Table 2.2 of HKCP – 2013)

Usage qk (kPa)

Domestic 2.0

Offices for general use 3.0

Department stores, shops, etc. 5.0

Table 1.5 - Examples of Imposed Load

(Extracted from Table 3.2 of the Code of Practice for Dead and Imposed Loads – 2011)

Materials Density (kN/m3)

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 Loads – 2011)

Dead Load Imposed Load

Adverse effect 1.4 1.6

Beneficial effect 1.0 0

Table 1.6 - Values of Partial Safety Factors

for Load (f) for ULS (Extracted from Table 2.1 of HKCP – 2013)



Elevation

h

Sw Sw

Clear Span, Ln

Effective Span, La1 = Sw/2 (if Sw < h)

a2 = h/2 (if h < Sw)

BEAM

SUPPORT 1 SUPPORT 2

Figure 1.5 - Effective Span

Limit to neutral axis x ≤ 0.5d [2.1]

The K-value K = M/(bd2fcu) [2.2]

βb = 1.0 K' = 0.156 [2.7]

βb = 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]

As = M / (0.87 fy z) [2.4]

Check depth of neutral axis for flanged section

(d – z) / 0.45 ≤ hf [2.11]

Doubly-reinforced section, K > K’

z = 0.775 d [2.5]

Check d' / x < 0.38 [2.10]

A's= (K – K') fcu bd2

[2.8] 0.87 fy (d - d')

As = K' fcu bd2

+ A's [2.9] 0.87 fy z

βb = moment at the section after redistribution

[1.1]moment at the section before redistribution

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 – Lever Arm Factor



Figure 2.11 – Definition of lpi for Calculation of Effective Flange Width

(Figure 5.1 of HKCP-2013)

Figure 2.12 – Effective Flange Width (Figure 5.2 of HKCP-2013)

Elements Minimum (%) Maximum (%)

Beam

4

Flexural tension steel

Rectangular section 0.13

Flanged section (bw/b < 0.4) 0.18

Flexural compression steel

Rectangular section 0.20

Column 0.80 6

Wall

4 Vertical bars 0.40

Horizontal bars 0.25

Table 2.1 – Minimum and Maximum Percentage of Reinforcement (Grade 500)

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, lpi = L.

c/c distance of adjacent slabs



Shear Stress v = V bv = b for rectangular beam section

bv = bw for flanged beam section [3.1]

bvd

vc = 0.79 (

100 As )1/3 (

400 )1/4

1 [3.2]

bvd d m where

( 100 As

) should not be taken as greater than 3. (As is the steel area of longitudinal tension steel.) bvd

( 400

)1/4 should not be taken as less than 0.67 for members without shear reinforcement. d

( 400

)1/4 should not be taken as less than 1 for members with minimum shear reinforcement. d

m 1.25, partial factor of safety for shear strength of concrete

Table 3.1 – Values of Design Concrete Shear Stress, vc (Extracted from Table 6.3 of HKCP-2013)

If fcu > 25 MPa, the value of vc has to be multiplied by (fcu / 25)1/3.

Asv / sv ≥ bv(v – vc) / (0.87 fyv) [3.3]

Asv / sv ≥ 0.4 bv / (0.87 fyv) [3.4]

Vn = (vc + 0.4) bvd [3.5]

Max. Allowable Shear Stress = 0.8 √ fcu [3.6]

Max sv 0.75d [3.7]

Link Size Spacing of Links in (mm)

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 – Asv / sv Values for Single Link (2 legs) (in mm2 per mm)



Table 3.4 – Basic L/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)



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 0.09FL -0.11FL 0.07FL -0.08FL

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.

Load 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 0.086FL -0.086FL 0.063FL -0.063FL

Shear 0.4F - 0.6F - 0.5F

Notes:

1. Area of each bay exceeds 30 m2.

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. Load 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)



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 mm2 per m Width

Shear and moment for simply-supported beam symmetrically loaded with two partial udl loads, w1 and w2, as shown in the figure on the right:

Shears at the supports V = w1L1 + 0.5 w2L2 [5.1]

Mid-span moment M = 0.5w1L12 + 0.5w2L2(L1 + 0.25L2) [5.2]

Effective height of a column: Le = βLo [6.1] where, Lo = The clear height between end restraints

The value of β is given in the following table:

End Condition at the

Bottom 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.

1 2 3

End

C

ondi

tion

at

the

Top

1 0.75 0.80 0.90

2 0.80 0.85 0.95

3 0.90 0.95 1.00

Table 6.1 – Values of β for Braced Columns (Table 6.11 of HKCP-2013)

K = Stiffness of the member

K = 4E(bh3/12)/L (for rectangular beam with fixed end)

Mes = The unbalance fixed end moment (FEM) of beams

= (FEM of b1) – (FEM of b2)

FEM = (1/12)wL2 (for beam under udl)

Mcl = Moment to the lower column

= Mes Kcl / (Kcu + Klu + 0.5Kb1 + 0.5Kb2)

Mcu = Moment to the upper column

= Mes Kcu / (Kcu + Klu + 0.5Kb1 + 0.5Kb2)

Figure 6.2 – Simplified Sub-frame Analysis for Determination of Design Moment for Column

1.0Gk 1.4G

k + 1.6Qk

Kcu

Kcl

Kb1

Kb2

w1 w1w

2

L1 L1

L2



Column under Axial Force

N = 0.4fcu Ac + 0.75fyAsc [6.3]

N = 0.35fcu Ac + 0.67fyAsc [6.4]

Biaxial Bending

For Mx/h' ≥ My/b', Mx' = Mx + β(h'/b')My [6.5] For Mx/h' < My/b', My' = My + β(b'/h')Mx [6.6]

N/(bhfcu) 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)



Figure 7.3 – Critical Sections for R C Design of Square Footing

Pad Footing

Column

(2) Perimeter of the column – check vmax

(1) Section across column face – design for bending

(3) Section at 1d from the column from – design for shear

(4) Perimeter at 1.5d from the column face – check punching shear

PLAN

L

a

b

c

d

e f

gh

d

1.5d

Check distribution of rebars in square footing: L ≤1.5(c+3d) [7.1]

No. of Piles

Configuration Tensile Force of the

Bottom Reinforcement

2

T = FcL/(4d)

3

T = FcL/(9d)

4

T = FcL/(8d)

Table 7.1 – Tensile Force for the Reinforcement in

Simple Pile Cap

L

L

L

L

L

L

ϕ

av

0.2ϕ

Critical section for shear check

Figure 7.5 – Critical Section for Shear Check in Pile Cap (Plan)

(Adapted from Figure 6.19 of HKCP-2013)

Annex - R C Design Formulae and Data (2014-08-01)tycnw01.vtc.edu.hk/cbe2022/Annex - R C Design...

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