High Rise Part B to D

CVEN90024 DESIGN OF HIGH-RISE STRUCTURES

MENGMENG DONG 324294

5/24/2011

CVEN90024 Design of High-rise Structures Mengmeng DONG 324292

1

Table of Contents INTRODUCTION ................................................................................................................................................................................................................. 2

PART B-CORE SYSTEM AND LATERAL LOAD DISTRIBUTION ............................................................................................................................................ 3

PART C-WIND LOAD AND SEISMIC LOAD ............................................................................................................................................................ 12

THE LATERAL ALONG-WIND LOAD............................................................................................................................................................................ 12

THE LATERAL NORTH SOUTH DESIGN EARTHQUAKE LOAD ...................................................................................................................................... 23

PART D-SPACEGASS MODEL AND ANALYSIS .................................................................................................................................................................. 27

CONCLUSION ..................................................................................................................................................................................................... 32


2

Introduction

Project 1 is about design a high-rise office building by analysis the core and floor system, lateral load distribution, wind and seismic load to determine how the tall building behaviors under each cases.(e.g. the deflection of typical members, the force carried out by six different cells, the overturning bending moment under wind and seismic load)

The perimeter of the building is as defined by floor system developed for Part A of this project. The floors crossing section of the tall building is 42*42m which has a 19*19m core areas and has 42 floors in total. Project B is only consider one load condition being the wind from the south and the wind load for each of the orthogonal directions can be based on the project area of building on to a surface perpendicular to the wind direction. For this load case determine manually the percentage of wind load resisted by each of six core cells by hand calibration.

Part C is determine the lateral along-wind loads for the north south direction only up to the height of the building by using dynamic method and finds the base-overturning moment. In addition, also determine the lateral north south design earthquake load up the height of building and compare which of the two loads is dictates.

Part D is about developing a simple SpaceGass model using a column element for the two conditions. Carry out a lateral load analysis for the north south direction with the appropriate eccentricity by applying the derived along-wind case from the Part C. First model will have just six column elements going up the height of the building. The second model will have six columns linked by header beams. By using SpaceGass to analysis how the building behaviors under two different models e.g. deflection at the top of the building and the forces at the base of the building.


3

Part B-Core System and Lateral Load Distribution

Step 1: Find the Centroid of cells 1-6 and the second moment of area Ixx and Iyy about the centroidal axis

The core layout is given below in Figure 1

Figure 1: Core layout

The centroid of cell 1 is calculated in following steps

The cell 1 is divided into 4 segments as shown in Figure 2

Taking the data from Figure 2, the calculations are tabulated as follows:


4

Figure 2: Cell 1 configuration

Segment b(m) d(m) A(m2) xi(m) yi(m) xiA(m3) yiA(m3) -xi(m) -yi(m) Ixx(m4) Iyy(m

4)

1 0.400 8.000 3.200 0.200 4.000 0.640 12.800 1.322 0.140 17.130 5.636

2 2.600 0.400 1.040 1.700 7.850 1.768 8.164 -0.178 -3.710 14.327 0.619

3 0.300 8.000 2.400 3.150 4.000 7.560 9.600 -1.628 0.140 12.847 6.378

4 2.600 0.300 0.780 1.700 0.200 1.326 0.156 -0.178 3.940 12.115 0.464

∑ 7.420 ∑ 11.294 ∑ 30.720 ∑ ∑ Note: xi and yi is the center of each segment

Thus, the centroid of cell 1 in x-direction ∑

∑

The centroid of cell 1 in y-direction ∑

∑

-xi is the x-axis distance between centroid of cell 1 and center point of the segment 1 While -yi is the y-axis distance between centroid of cell 1 and center point of the segment 1

Second moment of area Ixx and Iyy are calculated using formulas below:

Ixx =

Iyy =

E.g. For Segment 1, Ixx =

Iyy =

Hence ∑ m4


5

∑ m4

Similar to cell 1 calculation, the centroid and Ixx and Iyy about the centroidal axis of cell 2 to cell 6’s calculation is showing below.



4)

1 0.300 8.000 2.400 6.250 4.000 15.000 9.600 -3.250 -0.248 12.800 25.368

2 6.200 0.400 2.480 9.500 7.850 23.560 19.468 0.000 3.602 32.216 7.944

3 0.300 8.000 2.400 12.750 4.000 30.600 9.600 3.250 -0.248 12.800 25.368

4 6.200 0.300 1.860 9.500 0.200 17.670 0.372 0.000 -4.048 30.487 5.958

5 0.150 7.300 1.095 9.500 4.050 10.403 4.435 0.000 -0.198 4.863 0.002

∑ 10.235 ∑ 97.233 ∑ 43.475 ∑ 93.166 ∑ 64.641

the centroid of cell 2 in x-direction ∑

∑


∑

∑ m4

∑ m4


6



4)

1 0.300 8.000 2.400 15.850 4.000 38.040 9.600 -1.628 0.128 12.839 6.378

2 2.600 0.300 0.780 17.300 7.850 13.494 6.123 -0.178 3.978 12.348 0.464

3 0.400 8.000 3.200 18.800 4.000 60.160 12.800 1.322 0.128 17.119 5.636

4 2.600 0.400 1.040 17.300 0.200 17.992 0.208 -0.178 -3.672 14.038 0.619

∑ 7.420 ∑ 129.686 ∑ 28.731 ∑ 56.344 ∑ 13.097

The centroid of cell 3 in x-direction ∑

∑


∑

∑ 56.344m4 ∑ 13.097m4


7



4)

1 0.400 8.000 3.200 0.200 15.000 0.640 48.000 1.322 0.128 17.119 5.636

2 2.600 0.400 1.040 1.700 18.800 1.768 19.552 -0.178 -3.672 14.038 0.619

3 0.300 8.000 2.400 3.150 15.000 7.560 36.000 -1.628 0.128 12.839 6.378

4 2.600 0.300 0.780 1.700 11.150 1.326 8.697 -0.178 3.978 12.348 0.464

∑ 7.420 ∑ 11.294 ∑ 112.249 ∑ 56.344 ∑ 13.097

the centroid of cell 4 in x-direction ∑

∑


∑

∑ 56.344m4

∑ 13.097m4


8



4)

1 0.300 8.000 2.400 6.250 15.000 15.000 36.000 3.250 0.216 12.800 25.368

2 6.200 0.400 2.480 9.500 18.800 23.560 46.624 0.000 -3.584 31.893 7.944

3 0.300 8.000 2.400 12.750 15.000 30.600 36.000 -3.250 0.216 12.800 25.368

4 6.200 0.300 1.860 9.500 11.150 17.670 20.739 0.000 4.066 30.760 5.958

5 0.150 7.300 1.095 9.500 14.950 10.403 16.370 0.000 0.266 4.863 0.002

∑ 10.235 ∑ 97.233 ∑ 155.733 ∑ 93.116 ∑ 64.641


∑


∑

∑ 93.116m4

∑ 64.641m4


9



4)

1 0.300 8.000 2.400 15.850 15.000 38.040 36.000 1.628 -0.140 12.847 6.378

2 2.600 0.300 0.780 17.300 18.800 13.494 14.664 0.178 -3.940 12.115 0.464

3 0.400 8.000 3.200 18.800 15.000 60.160 48.000 -1.322 -0.140 17.130 5.636

4 2.600 0.400 1.040 17.300 11.150 17.992 11.596 0.178 3.710 14.327 0.619

∑ 7.420 ∑ 129.686 ∑ 110.260 ∑ 56.419 ∑ 13.097


∑


∑

∑ 56.419m4 ∑ 13.097m4


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Step 2: find the center of stiffness of the core

The center stiffness of the core is found using:

∑

∑

∑

∑

Note: 1.Element stiffness is proportional to the second moment of area if the elements that is found the first step:

, where E is the modulus of elasticity and H, which represents the height of the building are constant for all elements.

2. Kxxj and Kyyj is the lateral stiffness of each core in the x and y direction respectively. Ixxi and Iyyj can be used as K is proportional to I.

Hence, ∑

∑

∑

∑

9.5m

∑

∑

∑

∑

Look at Figure 1 that given core and arrangement and sizing of cells, all dimensions and sizes are symmetrical about two axes. Consequently, the center of stiffness of the core will be around the geometrical center of the core. The center stiffness of the core calculated above (9.5, 9.665) is really close to the geometrical center of the core (9.5, 9.5).


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Step 3: find the center of area of the wall perpendicular to the N-S wind.

For this project consider the load wind from the south only. The point of application of the load shall be offset from the centroid of the projected surface by 5%. This refers to the projected surface which designed in Part A.

Part A the width of the projected surface is 41.5m

Hence, the wind force is applied 41.5*0.05=2.075m toward the east from the center of the core

Step 4: find the element forces in the north direction by using the equation

∑

∑

Kxxj and Kyyj is the lateral stiffness of ach core in the x and y direction which can be taken as Ixxj and Iyyj

eccentricity ex is the distance from the center of stiffness to the load applied which equal 2.075m

xj and yj are the distance in the x and y direction, respectively, from the center of stiffness and the centroid of each core

In this case, Vy can be assumed to be 1, given that the proportion of the load distributed by each is element is to be calculated

The first term of the equation represents the translational force and the second term is the rotational force due to the eccentricity of the load

Table below sum up all the values need to calculate the Fyj

Cell Centroid X (m)

Centroid Y (m)

xj= (m)

yj= (m)

Ixxj=Kxxj

(m4) Iyyj=Kyyj

(m4) Kxxj*xj

2 Kyyj*yj2 Fyj(%)

1 1.522 4.140 7.978 5.525 56.419 13.097 3590.905 399.758 18.4

2 9.500 4.248 0.000 5.417 93.166 64.641 0.000 1896.967 22.6

3 17.478 3.872 -7.978 5.793 56.344 13.097 3586.128 439.491 9.00

4 1.522 15.128 7.978 -5.463 56.344 13.097 3586.128 390.876 18.4

5 9.500 15.216 0.000 -5.551 93.116 64.641 0.000 1991.712 22.6

6 17.478 14.860 -7.978 -5.195 56.419 13.097 3590.905 353.458 9.00

∑ 411.809 ∑ 181.669 ∑ 14354.067 ∑

5472.263 ∑ 100

E.g. the 1 cell Fyi =

∑

∑


12

Part C –Wind load and seismic load

Design wind loads for a tall building in Melbourne area (AS1170.2)

Relevant information follows:

Location and Terrain: Terrain category 3 zone with no shielding in Melbourne (Region A5)

Topography: assume ground to be flat

Dimensions: Average roof height 168m (Number of floors=42 and height of each floor is 4m)

Horizontal directions: 42m*42m square section

Building orientation: Major axis North-South

In order to calculate the wind load, calculate the wind pressure according to AS1170.2, Cl.2.4.1 as follows:

2.4.1 Design wind pressure


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Step 1: Calculate the design wind speed

Note: The orientation of the building major axis is specified for this project:

North-South Direction

Next, determine Vsit, β according to Cl. 2.2:


14

Next, each of VR, Md, Mz,cat , Ms and Mt are defined from the relevant sections as pointed.

Region wind speed (VR) Cl. 3.2

According to the Building Code of Australia (BCA), the structure should be treated as Level 3. Hence take average recurrence interval, R, for loading and overall structural response equal to 1000 years.

Note: -V1000 is a common return period for buildings

-1000 year return period: 1/1000 probabilities that the wind speed is exceeded

From Table 3.1 in AS/NZS1170.2, V1000 = 46 m/s (Region A)

Wind direction multiplier (Md) Cl. 3.3

Md = 1 for Region A5 with direction of wind is N (Table 3.2)

Terrain/Height Multiplier (Mz, cat) Cl. 4.2

Terrain category is Category 3; this corresponds to Cl. 4.2.1(c). This actual value of Mz,cat is obtained from Table 4.1(A):

Mz, cat =1.22

Note: Mz, cat value is for the top floor and needs to be obtained by interpolation at other levels where Vsit, B is being calculated.

Shielding multiplier (Ms) Cl. 4.3

Shielding may be provided by upwind buildings or other surrounding structures. In this exercise, there are no other buildings of greater height in any direction –therefore, the shielding multiplier is taken as 1.

Ms = 1.0

Topography multiplier (Mt) Cl. 4.4 (b)


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Elsewhere (i.e., Australia), the larger value of the following:

(i) Mt= Mh-defined by Cl. 4.4.1

(ii) Mt= Mlee-defined by Cl. 4.4.2

Hill shape multiplier (Mh)

The hill shape multiplier (Mh) is to be assessed for each cardinal direction. This takes into account the most adverse topographic cross-section that occurs within the range of directions within 22.5°on either side of the cardinal direction. The value is determined by the Cl. 4.4.2(a):

(a) For H/ (2Lu) < 0.05, Mh = 1.0 Cl. 4.4.4 (a)

Where

H= height of the hill, ridge or escarpment

Lu= horizontal distance upwind from the crest of the hill, ridge or escarpment to a level half the height below the crest

Ground is flat, hence H/ (2Lu) =0

Mh =1.0

Topography multiplier (Mlee)

The lee (effect) multiplier (Mlee) shall be evaluated for New Zealand sites in the lee zones. For all other sites, the lee multiplier shall be 1.0.

Mlee =1.0

Hence, Mt =1.0

Step 2: Calculate the Aerodynamic shape factor, Cfig

Values of Cfig are used in determining the pressures applied to each surface. For calculating pressures, the sign of Cfig indicates the direction of the pressure on the surface or element, positive values indicating pressure acting towards the surface and negative values indicating pressure acting away from the surface (less than ambient pressure, i.e. suction).


16

The aerodynamic shape factor (Cfig) is to be determined for specific surfaces or parts of surfaces as follows:

Cfig, i= Cp, i Kc, i, for internal pressures Cl. 5.2(1)

Cfig,e = Cp,eKaKc,eKlKp, for external pressures Cl. 5.2(2)

Cfig = Frictional drag forces, ignored since d/h < 4.read, Cl. 5.5

Where:

Cp, e= external pressure coefficient

Cp, i= internal pressure coefficient

Ka= area reduction factor

Kc, e= combination factor applied to external pressures

Kc, i= combination factor applied to internal pressures

Kl= local pressure factor

Kp= porous cladding reduction factor

Aerodynamic shape factor for internal pressure (Cp, i) Table. 5.1(A)

Assume the building is sealed:


17

Note: When calculating the base overturning moments Cp,i will cancel out as the above Figure points out. Cp,i is a useful parameter when estimating the external pressure on the surfaces.

Combination factor applied to external and internal pressure (Kc,e and Kc,i)

Kc,e = 0.9

Kc,i = 1.0

Cfig,i= Cp,iKc,i=0*1 = 0 OR Cfig,i= -0.2*1 = -0.2

Aerodynamic shape factors for external pressure (Cfig,e ) Table. 5.(A)

Cfig,e = Cp,eKaKc,eKlKp, for external pressures


18

Cp,e=0.8

Cp, e =-0.5

Cp, e =-0.65

Note: when calculating the along wind base overturning moment, the forces on the side walls are not included.


19

Cp, e =-1.3

Note: When calculating the along wind base overturning moment, the forces on the roofs are not included.

Area reduction factor for roofs and side walls (Ka) Cl. 5.4.2

Ka = 0.8

Local pressure factor for cladding (Kl) Cl. 5.4.4

The local pressure factor (Kl) shall be taken as 1.0 in all cases except when determining the wind forces applied to cladding, their fixings, the members that

directly support the cladding, and the immediate fixings of these members.

Kl = 1.0


20

Permeable cladding reduction factor (Kp) Cl. 5.4.5

The permeable cladding reduction factor (Kp) shall be taken as 1.0 except that where an external surface consists of permeable cladding

Kp = 1.0

Windward wall Cfig,e = 0.8*0.8*0.9*1.0*1.0 = 0.576

Leeward wall Cfig,e = -0.5*0.8*0.9*1.0*1.0 = -0.36

Note: Cfig,e and Cfig,e are parameters used to calculate the pressure at different levels on the Windward and the leeward walls

Step 3: Calculation for Dynamic response factor (Cdyn)

√

Cl. 6.2.2

Turbulence intensity z=h, Ih = 0.1134 (Table 6.1 by interpolation)

Background factor,

√

(Equation 6.2(2))

For b=42m, s=0 (for base bending moment),

√

Cl. 6.2.2

Hs =1.0

na = 46 / H =46/168 =0.27


21

√ √

Size reduction factor,

[

][

]

For b=42m

[

][

]

[ ][ ]

Reduced frequency,

ξ(ration of structural damping to critical): take as 0.03 for concrete structures

√

Cl. 6.2.2

For b=42m

√

√

Step 4：Calcualtion of base moment

Calculations were carried out by spreadsheet. Summaries of the results are given in the following tables.

Height of sector(m) Mz,cat 3 Vdes,θ(m/s) Windward qz*Cfig(kPa)

Leeward qh*cfig (kPa)

Windward qz*Cfig*Cdyn*A(kN)

Leedward qz*Cfig*Cdyn*A(kN)

Moment Contribution (MNm)


22

166 1.220 56.102 1.071 -0.680 172.869 -109.758 46.916

162 1.217 55.991 1.067 -0.680 172.189 -109.758 45.676

158 1.215 55.881 1.062 -0.680 171.511 -109.758 44.441

154 1.212 55.770 1.058 -0.680 170.834 -109.758 43.211

150 1.210 55.660 1.054 -0.680 170.158 -109.758 41.987

146 1.206 55.476 1.047 -0.680 169.035 -109.758 40.704

142 1.202 55.292 1.040 -0.680 167.916 -109.758 39.430

138 1.198 55.108 1.033 -0.680 166.800 -109.758 38.165

134 1.194 54.924 1.026 -0.680 165.688 -109.758 36.910

130 1.190 54.740 1.019 -0.680 164.580 -109.758 35.664

126 1.186 54.556 1.013 -0.680 163.475 -109.758 34.427

122 1.182 54.372 1.006 -0.680 162.374 -109.758 33.200

118 1.178 54.188 0.999 -0.680 161.277 -109.758 31.982

114 1.174 54.004 0.992 -0.680 160.184 -109.758 30.773

110 1.170 53.820 0.985 -0.680 159.094 -109.758 29.574

106 1.166 53.636 0.979 -0.680 158.008 -109.758 28.383

102 1.162 53.452 0.972 -0.680 156.926 -109.758 27.202

98 1.157 53.213 0.963 -0.680 155.525 -109.758 25.998

94 1.150 52.918 0.953 -0.680 153.809 -109.758 24.775

90 1.144 52.624 0.942 -0.680 152.102 -109.758 23.567

86 1.138 52.330 0.932 -0.680 150.405 -109.758 22.374

82 1.131 52.035 0.921 -0.680 148.717 -109.758 21.195

78 1.125 51.741 0.911 -0.680 147.039 -109.758 20.030

74 1.078 49.588 0.837 -0.680 135.058 -109.758 18.116

70 1.078 49.588 0.837 -0.680 135.058 -109.758 17.137

66 1.102 50.692 0.874 -0.680 141.139 -109.758 16.559

62 1.094 50.324 0.862 -0.680 139.097 -109.758 15.429

58 1.086 49.956 0.849 -0.680 137.070 -109.758 14.316

54 1.078 49.588 0.837 -0.680 135.058 -109.758 13.220

50 1.070 49.220 0.824 -0.680 133.061 -109.758 12.141

46 1.058 48.668 0.806 -0.680 130.093 -109.758 11.033

42 1.046 48.116 0.788 -0.680 127.159 -109.758 9.951


23

38 1.032 47.472 0.767 -0.680 123.778 -109.758 8.874

34 1.016 46.736 0.743 -0.680 119.969 -109.758 7.811

30 1.000 46.000 0.720 -0.680 116.220 -109.758 6.779

26 0.976 44.896 0.686 -0.680 110.709 -109.758 5.732

22 0.952 43.792 0.652 -0.680 105.331 -109.758 4.732

18 0.920 42.320 0.609 -0.680 98.369 -109.758 3.746

14 0.878 40.388 0.555 -0.680 89.593 -109.758 2.791

10 0.830 38.180 0.496 -0.680 80.064 -109.758 1.898

6 0.830 38.180 0.496 -0.680 80.064 -109.758 1.139

2 0.830 38.180 0.496 -0.680 80.064 -109.758 0.380

∑ 938.370

Note:

qz=0.6* Vdes,θ2

A = 42 * 4 = 168m2

Totoal along-wind base bending moment = 938.370 MNm for a 168m tall building

Design the earthquake load for a tall building in Melbourne area (AS1170.4)

Find Base shear

Design for 500 years return period

The annual probability of exceedance factor Kp = 1.0 Table 3.1

C = 0.88ZS/T≤2.94Z For T≤1.5 sec


24

C = 1.32ZS/T2 For T≥1.5 sec

T(sec) = 1.25(0.05)H0.75 for other structures (ultimate)

H=168m

T(sec) =1.25(0.05)1680.75 = 2.916 secs

Hazard factor (Z) = 0.08 for Melbourne Table 3.2

S = 1 for rock sites Class Be

C = 1.32ZS/T2 as T>1.5 sec

=1.32*0.08*1.0/2.9162

=0.0124

Density of building ρ = 0.35 tonne/m3

W =ρ *A*H*g

=0.25*42*42*168*9.81

=726800kN

Rf = 2.6 assume limited ductile shear wall

Vertival distribution of VB

the building will be divided into 42 nodes according to the number of floors auusing uniform distribution of mass

∑ for T>2.5 sec


25

Calculationd were carried out by spreadsheet. Summaries of the results are given in the table following tables.

Stastic Analysis

Nodes mi(kg) hi(m) mihi2(kgm2) Fi (kN) Bending moment(MNm)

0 1618.349 168 45676271.560 179.758 30.199

1 2157.798 164 58036139.450 228.400 37.458

2 2157.798 160 55239633.028 217.394 34.783

3 2157.798 156 52512176.147 206.660 32.239

4 2157.798 152 49853768.807 196.198 29.822

5 2157.798 148 47264411.009 186.008 27.529

6 2157.798 144 44744102.752 176.089 25.357

7 2157.798 140 42292844.037 166.442 23.302

8 2157.798 136 39910634.862 157.067 21.361

9 2157.798 132 37597475.229 147.964 19.531

10 2157.798 128 35353365.138 139.132 17.809

11 2157.798 124 33178304.587 130.572 16.191

12 2157.798 120 31072293.578 122.284 14.674

13 2157.798 116 29035332.110 114.268 13.255

14 2157.798 112 27067420.183 106.523 11.931

15 3056.881 108 35655456.881 140.321 15.155

16 2157.798 104 23338744.954 91.849 9.552

17 2157.798 100 21577981.651 84.920 8.492

18 2157.798 96 19886267.890 78.262 7.513

19 2157.798 92 18263603.670 71.876 6.613

20 2157.798 88 16709988.991 65.762 5.787

21 2157.798 84 15225423.853 59.919 5.033

22 2157.798 80 13809908.257 54.349 4.348

23 2157.798 76 12463442.202 49.050 3.728

24 2157.798 72 11186025.688 44.022 3.170

25 2157.798 68 9977658.716 39.267 2.670


26

26 2157.798 64 8838341.284 34.783 2.226

27 2157.798 60 7768073.394 30.571 1.834

28 2157.798 56 6766855.046 26.631 1.491

29 2157.798 52 5834686.239 22.962 1.194

30 3056.881 48 7043053.211 27.718 1.330

31 2157.798 44 4177497.248 16.440 0.723

32 2157.798 40 3452477.064 13.587 0.543

33 2157.798 36 2796506.422 11.006 0.396

34 2157.798 32 2209585.321 8.696 0.278

35 2157.798 28 1691713.761 6.658 0.186

36 2157.798 24 1242891.743 4.891 0.117

37 2157.798 20 863119.266 3.397 0.068

38 2157.798 16 552396.330 2.174 0.035

39 2157.798 12 310722.936 1.223 0.015

40 3056.881 8 195640.367 0.770 0.006

41 2157.798 4 34524.771 0.136 0.001

42 2157.798 0 0.000 0.000 0.000

∑ 880706759.633 ∑ 437.947

Note:

A=42*42=1764.000m2(the cross sectional area of the floor plan)

VB=3466.000kN

G+ΨQ(typical floors)=12.000kPa

G+ΨQ(Plant-room levels)=17.000kPa

G+ΨQ(top floor)=9.000kPa

The total base overturning moment for earthquke load =437.947 MNm for a 168m tall building.

Note:

The total base overturning moment for earthquke load =437.947 MNm while the totoal along-wind base bending moment = 938.370MNm for a 168m tall

building. The along-wind base bending moment is obviously two times larger than the overturning moment. Hence the along-wind loads is dicates.


27

Part D- SpaceGass Model and Analysis

The 3-D Models output from SpassGass are showing in Figure 8 and 9

Figure 8: 3-D view of Model A-without header beams

Figure 9: 3-D view of Model B-with header beams


28

Result：

Deflection at the top of the building for Model A and Model B

The deflection at the top of the building is 850mm and 300mm respectively to Model A and Model B. It’s obviously that Model B gives a much smaller deflection compare to Model A. Model A just thin slab acting between walls and due to the slim nature of the slab, there will be a low flexural stiffness, the extend of coupling action is limiting. However in Model B by introducing beams between the cantilever elements this flexural stiffness can be greatly enhanced hence increasing the coupling action and hence improving the overall lateral resistance. Figure 10 graphically explain the above reason behind the header beam function.

Note: the deflection at the top of the building is hand calculated by using the formulas below:

Where W is the force action on the top of the building, l is the height of the building, E is Young’s Modulus of the concrete.

Figure 10: Schematic of a coupled core system


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Force at the base of the building for Model A and Model B The moments for X and Y direction in Model A for each of the six columns are tabulate in the Table below.

Model A

Column

X–Axis Force

(kN)

Y-Axis Force

(kN)

Z-Axis Force

(kN)

X-Axis Moment

(kNm)

Y-Axis Moment

(kNm)

Z-Axis Moment

(kNm)

Total base Moment Percentage (%)

Compare to Part B Force Percentage (%)

2 117.2 0.0 1133.9 171879.6 -22.6 -7625.9 18.3 18.4

3 556.0 0.0 3015.3 212736.2 -23.3 -39330.5 22.65 22.6

4 128.1 0.0 2495.5 85000.6 -22.6 -8604.5 9.05 9

5 -117.2 0.0 1133.9 171879.6 -22.6 7625.9 18.3 18.4

6 -556.0 0.0 3015.3 212736.2 -23.3 39330.5 22.65 22.6

7 -128.1 0.0 2495.5 85000.6 -22.6 8604.5 9.05 9

Load 0.0 0.0 -13289.4 0.0 0.0 0.0

Total Reaction 0.0 0.0 13289.4 939232.7 -137.1 0.0 100 100

Note:

All the numbers are taking from the SpaceGass report. The sum of overturning base moment in Model A is 939.2327MNm while the sum of overturning base moment finds in Part C is equal to

938.370MNm. Those two values are quite similar (almost the same).

For the total base moment action on each column the percentages are also close to the Force percentages which find in Part B of the Project.

From above statement, its can see the Model A is analysis is quite successful.

The moments for X and Y direction in Model B for each of the six columns are tabulate in the Table below.


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Model B

Column

X–Axis Force

(kN)

Y-Axis Force

(kN)

Z-Axis Force

(kN)

X-Axis Moment

(kNm)

Y-Axis Moment

(kNm)

Z-Axis Moment

(kNm)

Coupling forces Moment (kNm)

Total core Moment Percentage (%)

Coupling forces Percentage (%)

2 93.8 17398.9 1187.7 75021.101 -7.269 -2415.5 97917.05 10.31 7.99

3 558.3 12199.1 2945.6 92853.985 -7.499 -13139.0 119882.2 12.76 9.89

4 162.4 17802.8 2541.4 37100.599 -7.269 -3031.8 47899.96 5.10 3.95

5 -93.8 -17398.9 1187.7 75021.101 -7.269 2415.5 97917.05 10.31 7.99

6 -559.3 -12199.1 2945.6 92853.985 -7.499 13139.0 119882.2 12.76 9.89

7 -164.2 -17802.8 2541.4 37100.599 -7.269 3031.8 47899.96 5.10 3.95

Load 0.0 0.0 -13289.4 0.0 0.0 0.0 0.0

Total Reaction 0.0 0.0 13289.4 409951.370 -44.10 0.0 529281.330 56.34 43.66

Note:

Due to header beams at Model B, the coupling effect will decrease the overturning moment at the base but activate the push and pull action on the cores.

So in the Table above seeing decreasing values of overall moment compare to Model A, but there are values in forces in y-direction compare to Model A.

The base moment and coupling forces percentages are all in term the total overturning moment. The core moment takes approximately 43.66% of the overturning moment while the coupling forces takes about 56.34% of the overturning moment.

By add the sum of overturning moment to the moment due to these coupling forces, get the sum of overturning moment that is equal 937.370MNm which is approximately equal to the overturning moment in part C (938.370MNm)

In additional, depends entirely on the number of header beams that chose, the more header beams, the greater the coupling action. Because for Model B assumes that every floor has header beams, therefore, the percentages of the core moment are smaller than the percentages of the coupling forces.


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Conclusion

Finding in Part B: The percentages of wind load resisted by six core cells in Part B is respectively to 18.4, 22.6, 9, 18.4, 22.6, and 9

Findng in Part C: The total base overturning moment for earthquke load =437.947 MNm while the totoal along-wind base bending moment = 938.370MNm for a 168m tall building. The along-wind loads is the one that dictates.

Find in Part D: For SpaceGass Model A, the overturning moment at base is all taking by the core cell and the values is 939.370MNm which is nearly the same compare to Part C value. The deflection at the top of the building is 850mm. For SpaceGass Model B, the overturning moment at base is combining from the moment caused by core cell and coupling forces. The core cell and the coupling forces take approximately 43.66% and 56.34% respectively of the total overturning base moment. The deflection at the top of the building is 300mm.

The header beams in the Model B is increasing the coupling action and hence improving the overall lateral resistance. Therefore the Model B is a better model to be used.

Documents

High Rise Part B to D