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31 0

14

. .

334 , 43 , 60067 , 2656

( 47) 1 2700 ( 47) 1 2702 ( 4 ) 24 373 ( 4 ) 24 3

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S. K. Ghosh Associates Inc.

www.skghoshassociates.com

- 1 -

REINFORCED CONCRETE DESIGN IN COMPLIANCE WITHACI 318M-08: DESIGN FOR WIND

Please stay tuned. We will be starting at8:00 am UAE Time.

You will be able to listen to the seminarusing computer speakers

If you are encountering technical difficulties, please call 00 1 847 991 2700 Visit us at: www.skghoshassociates.com

SKGA Web Seminarin cooperation with and under sponsorship of

the Department of Municipal Affairs, Emirate of Abu Dhabi, UAE

- 2 -

REINFORCED CONCRETE DESIGN INCOMPLIANCE WITH ACI 318M-08

A web seminar series in cooperation with and under sponsorship ofthe Department of Municipal Affairs, Emirate of Abu Dhabi, UAE.

Part 14: BUILDING DESIGN EXAMPLE DESIGN FOR WIND


Palatine, IL and Aliso Viejo, CA


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- 3 -

- 4 -

Example Office Building

Part of Chapter 2 updated to ACI 318-08 andconverted to metric units.

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- 5 -


Plan view of the example building

7900 7900 7900 7900 7900 7900 7900

6 7 0 0

6 7 0 0

6 7 0 0

Unit: mm

- 6 -


Elevation view of the example building

1 1 @

3 7 0 0 = 4 0

, 7 0 0

4 9 0 0

Unit: mm

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- 7 -

Example Office Building Design of Main Wind-Force-Resisting System (MWFRS)

- 8 -

Design Data

Design Data

Building location: Abu Dhabi,

Latitude : 24 28 N

Longitude : 54 22 E

Material properties

Concrete: f c = 40 MPa, w c = 2400 kg/m 3

Reinforcement: f y = 420 MPa

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- 9 -

Design Data

Design Data

Service loads

Live loads: roof = 1.0 kN/m 2

floors = 2.5 kN/m 2 plus

1.0 kN/m 2 for partitions

SIDL: roof = 2.0 kN/m 2 + 900 kN penthouse

floors = 2.0 kN/m 2

- 10 -

Design Data

Design Data

Wind design data

Basic wind speed V = 38 m/sec

(Figure 8, Vickery report)

Exposure B (IBC 1609.4, ASCE 6.5.6.3)

For Occupancy Category II, I = 1.0

(IBC Table 1604.5, ASCE Table 6-1)

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- 11 -

Design Data

Design Data

Member dimensions

Slab: 200 mm

Beams: 550 550 mm

Interior columns: 650 650 mm

Edge columns: 600 600 mm

Wall thickness: 300 mm

- 12 -

Wind Load Determination

Step 1: Basic wind speed, V, and winddirectionality factor, K d

V = 38 m/sec

Kd = 0.85 for main wind-force-resisting systems ofbuildings per ASCE Table 6-4

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- 13 -


ASCE Table 6-4 Wind directionality factor, k d

- 14 -


Step 2: Importance factor, I

I = 1.0 per IBC Table 1604.5, ASCE Table 6-1 forOccupancy Category II

Occupancy

Category

Non-Hurricane Prone Regionsand Hurricane Prone Regions

with V = 38 45 m/sec

Hurricane ProneRegions with V > 45

m/sec

I 0.87 0.77

II 1.00 1.00

III 1.15 1.15

IV 1.15 1.15

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- 15 -


Step 3: Exposure category and velocity pressureexposure coefficient, K zValues of K z are to be determined from ASCE Table 6-3.In lieu of linear interpolation, K z may be calculated atany height z m above ground level by the followingequations:

- 16 -


= 3-second gust speed power law exponent

from ASCE Table 6-2

= 7.0 for Exposure B

Zg = nominal height of the atmospheric

boundary layer from ASCE Table 6-2

= 365.76 m for Exposure B

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- 17 -


ASCE Table 6-2 Terrain Exposure Constant

Exposure zg (m) b b c l (m) zmin(m)*

B 7.0 365.76 1/7 0.84 1/4.0 0.45 0.30 97.54 1/3.0 9.14

C 9.5 274.32 1/9.5 1.00 1/6.5 0.65 0.20 152.4 1/5.0 4.57

D 11.5 213.36 1/11.5 1.07 1/9.0 0.80 0.15 198.12 1/8.0 2.13

^ ^

*zmin = minimum height used to ensure that the equivalent height z is greater of 0.6h or z min .

For buildings with h zmin , z shall be taken as z min.

- 18 -


Table: Velocity pressure exposure coefficient K z

Level Height above ground level, z (m) K z12 45.6 1.109

11 41.9 1.082

10 38.2 1.0549 34.5 1.024

8 30.8 0.991

7 27.1 0.955

6 23.4 0.916

5 19.7 0.872

4 16.0 0.822

3 12.3 0.762

2 8.6 0.688

1 4.9 0.586

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- 19 -


Step 4: Topographic factor, K zt

Assuming the example building is situated on levelground and not on a hill, ridge, or escarpment, K zt isequal to 1.

- 20 -


ASCE Figure 6-4 Topographic factor, K zt

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- 21 -


ASCE Figure 6-4 Topographic factor, K zt

- 22 -


Step 5: Gust effect factors, G and G f

Gust effect factor depends on whether a building isrigid or flexible (ASCE 6.5.8). A rigid building has afundamental natural frequency n 1 greater than orequal to 1 Hz, while a flexible building has afundamental natural frequency less than 1 Hz(ASCE 6.2).

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- 23 -


N-S direction:

Ta = 0.0488 (45.6) 0.75 = 0.856 sec

(ASCE Eq. 12.8-7)

n1= 1/T a = 1.17 Hz > 1.0 Hz

Thus, the example building is rigid in N-S directionand G = 0.85

- 24 -


E-W direction:

Ta

= 0.0466 (45.6) 0.9 = 1.45 sec

n1 = 1/T a = 0.7 Hz

Thus, the example building is flexible in E-Wdirection.

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- 25 -


ASCE Eq. 6-8

where,

- 26 -


intensity of turbulence at height

(ASCE Eq. 6-5 and Table 6-2 for Exposure B)

ASCE Eq. 6-9

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- 27 -


= 0.6h zmin= 0.6 45.6 = 27.36 m > z min = 9.14 m

(Table 6-2 for Exposure B)

background response

ASCE Eq. 6-6

- 28 -


integral length scale of turbulence at equivalent height


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- 29 -


The resonant response factor R is computed from:

ASCE Eq. 6-10

damping ratio (assumed to be 0.01)

- 30 -


ASCE Eq. 6-11

N1 = reduced frequencyASCE Eq. 6-12

mean hourly wind speed at height


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- 31 -


ASCE Eq. 6-13a

ASCE Eq. 6-13a

- 32 -


ASCE Eq. 6-13a

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- 33 -


Step 6: Enclosure classification

It is assumed in this example that the building isenclosed per IBC 1609.2, ASCE 6.5.9.

- 34 -


Step 7: Internal pressure coefficient, GC pi

Internal pressure coefficients are to be determinedfrom ASCE Figure 6-5, based on building enclosureclassification.

For an enclosed building, GC pi = 0.18.

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- 35 -


ASCE Figure 6-5 Internal pressure coefficient, GC pi

- 36 -


Step 8: External pressure coefficients, C pN-S direction (ASCE Figure 6-6):

Windward wall: C p = 0.8

Leeward wall (L/B = 20.7/55.9 = 0.37): C p = -0.5

Side wall: C p = - 0.7

Roof (h/L = 45.6/20.7 = 2.2):

Cp = -1.3 over entire roof (20.7m < h/2 = 22.8m).

May be reduced to 0.80 -1.3 = -1.04 for area

greater than 92.9 m 2 per ASCE Figure 6-6.

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- 39 -


ASCE Figure 6-6 C p for MWFRS: Walls

- 40 -


ASCE Figure 6-6 C p for MWFRS: Roofs

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- 41 -


- 42 -


Step 9: Velocity pressure, q z

The velocity pressure at height z is determined byEq. 6-15 in ASCE 6.5.10:

qz = 0.613 K z Kzt Kd V2 I

where all terms have been defined previously.

ASCE Eq. 6-15

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- 43 -


Velocity Pressure q z (V = 38 m/sec)Level Height above ground level, z (m) K z qz (N/m 2)

12 45.6 1.109 834

11 41.9 1.082 814

10 38.2 1.054 793

9 34.5 1.024 770

8 30.8 0.991 746

7 27.1 0.955 719

6 23.4 0.916 689

5 19.7 0.872 656

4 16.0 0.822 618

3 12.3 0.762 574

2 8.6 0.688 518

1 4.9 0.586 441

where q = q z for windward walls at height z above groundq = q h for leeward walls, side walls, and roof, evaluated at height hq i = q h for all walls and roofs of enclosed buildings

- 44 -


Step 10: Design wind pressure, p

For rigid buildings of all heights, design windpressures on the main wind-force-resisting systemare calculated by Eq. 6-17:

p = q GC p q i (GC pi) ASCE Eq. 6-17

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- 45 -


Table: Design Wind Pressure in N-S Direction (V = 38 m/sec)Location Level Height above

ground level,

z(m)

External Pressure Internal Pressure

q

(N/m 2)

G C p qGC p(N/m 2)

q i(N/m 2)

GC pi q iGC pi(N/m 2)

Windward 12 45.6 834 0.85 0.8 567 834 0.18 150

11 41.9 814 0.85 0.8 554 834 0.18 150

10 38.2 793 0.85 0.8 539 834 0.18 150

9 34.5 770 0.85 0.8 524 834 0.18 150

8 30.8 746 0.85 0.8 507 834 0.18 150

7 27.1 719 0.85 0.8 489 834 0.18 150

6 23.4 689 0.85 0.8 469 834 0.18 150

5 19.7 656 0.85 0.8 446 834 0.18 150

4 16.0 618 0.85 0.8 420 834 0.18 150

3 12.3 574 0.85 0.8 390 834 0.18 1502 8.6 518 0.85 0.8 352 834 0.18 150

1 4.9 441 0.85 0.8 300 834 0.18 150

Leeward --- All 834 0.85 -0.5 -354 834 0.18 150

Side --- All 834 0.85 -0.7 -496 834 0.18 150

Roof --- 45.6 834 0.85 -1.04 -737 834 0.18 150

- 46 -


Table: Design Wind Forces in N-S Direction (V = 38 m/sec)

Level Heightabove

groundlevel, z (m)

TributaryHeight

(m)

Windward Leeward TotalDesign

Wind Force(kN)

External DesignWind Pressure,

q zGC p (N/m 2)

DesignWind Force,

P* (kN)


q hGC p (N/m 2)

Design WindForce, P*

(kN)

12 45.6 1.9 567 60.2 -354 -37.6 97.9

11 41.9 3.7 554 114.5 -354 -73.3 187.8

10 38.2 3.7 539 111.5 -354 -73.3 184.8

9 34.5 3.7 524 108.3 -354 -73.3 181.6

8 30.8 3.7 507 104.9 -354 -73.3 178.2

7 27.1 3.7 489 101.1 -354 -73.3 174.4

6 23.4 3.7 469 96.9 -354 -73.3 170.3

5 19.7 3.7 446 92.3 -354 -73.3 165.6

4 16.0 3.7 420 87.0 -354 -73.3 160.3

3 12.3 3.7 390 80.7 -354 -73.3 154.0

2 8.6 3.7 352 72.8 -354 -73.3 146.1

1 4.9 4.3 300 72.1 -354 -85.2 157.3

*P = qGC p Tributary height 55.9 m 1958.3

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- 47 -


For flexible buildings, ASCE Eq. 6-19 is to be used:

p = q G fCp q i (GC pi) ASCE Eq. 6-19

- 48 -

Wind Load DeterminationTable: Design Wind Pressure in E-W Direction (V = 38 m/sec)

Location Level Height aboveground

level,z(m)

External Pressure Internal Pressure

q(N/m 2)

G C p qGC p(N/m 2)

q i(N/m 2)

GC pi q iGC pi(N/m 2)

Windward 12 45.6 834 0.89 0.8 594 834 0.18 150

11 41.9 814 0.89 0.8 580 834 0.18 150

10 38.2 793 0.89 0.8 565 834 0.18 150

9 34.5 770 0.89 0.8 548 834 0.18 150

8 30.8 746 0.89 0.8 531 834 0.18 150

7 27.1 719 0.89 0.8 512 834 0.18 150

6 23.4 689 0.89 0.8 491 834 0.18 150

5 19.7 656 0.89 0.8 467 834 0.18 150

4 16.0 618 0.89 0.8 440 834 0.18 150

3 12.3 574 0.89 0.8 408 834 0.18 150

2 8.6 518 0.89 0.8 369 834 0.18 150

1 4.9 441 0.89 0.8 314 834 0.18 150

Leeward --- All 834 0.89 -0.27 -200 834 0.18 150

Side --- All 834 0.89 -0.7 -520 834 0.18 150

Roof --- 45.6 * 834 0.89 -1.00 -742 834 0.18 150

--- 45.6 834 0.89 -0.78 -576 834 0.18 150

--- 45.6 834 0.89 -0.62 -460 834 0.18 150

* from windward edge to 22.8 m, from 22.8 m to 45.6 m, from 45.6 m to 55.9 m

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- 49 -


Table: Design Wind Forces in E-W Direction (V = 38 m/sec)

Level

Heightabove

groundlevel, z

(m)

TributaryHeight

(m)

Windward Leeward TotalDesignWind

Force (kN)External DesignWind Pressure,

q zG fCp (N/m 2)

DesignWind

Force, P*(kN)


q hG fCp (N/m 2)

DesignWind Force,

P* (kN)

12 45.6 1.9 594 23.4 -200 -7.9 31.2

11 41.9 3.7 580 44.4 -200 -15.3 59.7

10 38.2 3.7 565 43.2 -200 -15.3 58.6

9 34.5 3.7 548 42.0 -200 -15.3 57.3

8 30.8 3.7 531 40.7 -200 -15.3 56.0

7 27.1 3.7 512 39.2 -200 -15.3 54.5

6 23.4 3.7 491 37.6 -200 -15.3 52.9

5 19.7 3.7 467 35.8 -200 -15.3 51.1

4 16.0 3.7 440 33.7 -200 -15.3 49.1

3 12.3 3.7 408 31.3 -200 -15.3 46.6

2 8.6 3.7 369 28.2 -200 -15.3 43.6

1 4.9 4.3 314 27.9 -200 -17.8 45.8

606.6*P = qGC p Tributary height 20.7 m

- 50 -

Wind Load Analysis

Method of analysis

3-D analysis

Rigid diaphragms, rigid-end offsets

Stiffness properties

Beams : EI eff = 0.7 EI gColumns : EI eff = 1.0 EI gShear walls : EI eff = 1.0 EI g

P-delta effects

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- 51 -

Wind Load Analysis

Load combination

1.4D (ACI Eq. 9-1)

1.2D + 1.6L (ACI Eq. 9-2)

1.2D + 1.6L + 0.8W (ACI Eq. 9-3)

1.2D + 0.5L + 1.6W (ACI Eq. 9-4)

0.9D + 1.6W (ACI Eq. 9-6)

- 52 -

Beam Design

7900 7900 7900 7900 7900 7900 7900

6 7 0 0

6 7 0 0

6 7 0 0

Beam C4 C5

Unit: mmBeam dimension: 550 mm 550 mm

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- 53 -

Beam Design

Table: Summary of Design Bending Moments and Shear Forces for BeamC4-C5 at the Second Floor Level

Load Case Location Bending Moment(kN-m)

Shear Force (kN)

Dead (D) Support -195 138Midspan 110

Live (L) Support -69 48Midspan 40

Wind(W)

Case 1 (E-W) Support 35 10

Case 2 (E-W) support 26 7Case 3 Support 26 7Case 4 support 21 6

- 54 -

Beam Design

Table: Summary of Design Bending Moments and Shear Forces for BeamC4-C5 at the Second Floor Level

Load combination Location Bending Moment (kN-m) Shear Force (kN)

1.4D Support -273 194

Midspan 154

1.2D + 1.6L Support -345 242Midspan 196

1.2D + 1.6L 0.8W Support -373 250

-317 234

Midspan 196

1.2D + 0.5L 1.6W Support -325 205

-213 174

Midspan 152

0.9D 1.6W Support -231 140

-120 109

Midspan 99

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- 55 -

Beam Design

Using No. 25 reinforcement and 10-mm stirrups:d = 550 40 10 25/2 = 487.5 mm

Min. As = (0.25 400.5

550 487.5)/(420) = 1009 mm2

= 1.4b wd/fy = 894 mm 2 (ACI 10.5.1)

Max. A s = (0.85 1fcbwd/fy) (0.003/0.007)= 7070 mm 2 (ACI 10.3.5)

Location Mu(kN-m)As

(mm 2) Reinf. Mn

(kN-m)Support -373 2116 5-No.25 429Midspan 196 1112 3-No.25 264

- 56 -

Beam Design

Shear design

Tributary area was calculated following ACI 13.6.8(Factored shear in slab system with beams).

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- 57 -

Beam Design

6.7 m

C

54

7.9 m

6.7 m

3.35 m 3.35 m1.2 m

3.35 m

Beam tributary area

- 58 -

Beam Design

Total trapezoidal area tributary to beam =2[1/2 (1.2 + 7.9) 3.35] = 30.5 m 2

Dead load = (0.2 2400 9.81/1000 30.5)+(0.35 0.55 2400 9.81/1000 7.25)+(2 30.5) =237.5 kN

wD = 237.5 / 7.25 = 32.8 kN/m

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- 59 -

Beam Design

Reduced floor live load =

= 2.08 kN/m 2

Live load = (2.08 + 1.0) 30.5 = 94 kNwL = 94 / 7.25 = 13.0 kN/m

- 60 -

Beam Design

Vu = 250 kN at support (from the load combinationtable for 1.2D + 1.6L +0.8W)

wu = 1.2w D + 1.6 w L = 60.2 kN/m

Vu = 250 60.2 0.488 = 220.6 kN at critical section

Vc = 0.17 (40) 0.5 550 487.5/1000 = 288 kN

Vu (= 220.6 kN) Vc (216 kN) [ = 0.75 ]

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- 61 -

Beam Design

Provide minimum shear reinforcement (ACI 11.4.6.3)

When using No.10 (A s =79 mm 2) for stirrups,

s = (2 79 420)/(0.062 40 0.5 550) = 308 mm= (2 79 420)/(0.35 550) = 345 mm

- 62 -

Beam Design

Maximum spacing:

d/2 = 510/2 = 255 mm (governs)

600 mm

Provide No.10 @ 250 mm at both ends of the beam.Stirrup can be discontinued at section where V u Vc /2. This occurs at 2.35 m from the face of thesupport.

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- 63 -

Beam Design

Reinforcing bar cutoff point

Cutoff 3 of the 5-No.25 top bars away from support

Mn (2-No.25) = 177 kN-m

Try load combination: 1.2D + 1.6L + 0.8W

RR = (60.1 7.25 2 /2 + 373 317) / 7.25 = 225.6 kN

60.1 (x2 /2) 225.6x + 373 177 = 0

Cutoff point x = 1.0 m from face of support

- 64 -

Beam Design

w u = 1.2w D + 1.6w L = 60.1 kN/m

M-=373 kN-mM

-

=317 kN-m

7.25 m

210.1 kN 225.6 kN

RRRLx

Top reinforcing bar cutoff point

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- 67 -

Beam Design

Governing Cutoff point x = 1.0 m from face of support

ACI 12.10.3

3-No.25 must extend a distance

d = 487.5 mm (governs) or

12d b = 12 x 25 = 300 mm

beyond the distance x = 1.0 mThus, total required bar length = 1.0 + 0.49 = 1.49 m; use

1.5 m

- 68 -

Beam Design

ACI 12.10.4

3-No.25 must extend a full development length l dbeyond the face of the support

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- 69 -

Beam Design

where,

t = reinforcement location factor = 1.3 for top bars

e = coating factor = 1.0 for uncoated bars

s = reinforcement size factor = 1.0 for No.25 bars

= lightweight aggregate factor = 1.0

cb = spacing or cover dimension

(governs)

- 70 -

Beam Design

(cb + Ktr)/d b = (53 + 0)/25 = 2.12 < 2.5

l d = 0.92 < 1.5 m

Retain cutoff length of 1.5 m

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- 71 -

Beam Design

ACI 12.10.5

Flexural reinforcement terminated in a tension zone:

Point of inflection is 4.6 m from face of support >1.5 m

Check ACI 12.10.5.1:

If Vu (cutoff point) 2Vn /3, tension reinforcementcan be terminated there

- 72 -

Beam Design

Vn = (Vc + Vs) = (Vc + Avfytd/s )

= 0.75 [288 + (157 420 487.5)/(1000x250)]

= 312 kN

2Vn /3 = 208 kN

Vu @ 1.5 m from face of support = 225.6 (60.1 1.5)= 135 kN < 208 kN

Therefore, 3-No.25 bars can be terminated at 1.5 mfrom face of support

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- 73 -

Beam Design

Structural integrity reinforcement (ACI 7.13.2.5)

Other than perimeter beams, when closed stirrupsare not provided

At least one-quarter of positive reinforcementrequired at midspan, but not less than two bars,shall be continuous or spliced with Class Btension splices over the supports

- 74 -

Beam Design

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- 75 -

Column Design

7900 7900 7900 7900 7900 7900 7900

6 7 0 0

6 7 0 0

6 7 0 0

Column C4

Unit: mmColumn dimension: 650 mm 650 mm

- 76 -

Column Design

Table: Summary of Design Axial force, Bending Moments, and ShearForces for Column C4 Supporting the Second Floor

Load Case AxialForces (kN)

BendingMoment (kN-m)

Shear Force (kN)

Dead (D) 5701 0 0Live (L) 1100 0 0

Wind(W)

Case 1 0 35 21Case 2 0 26 15Case 3 9 26 15Case 4 6 21 12

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- 77 -

Column Design

Table: Summary of Design Axial force, Bending Moments, and ShearForces for Column C4 Supporting the Second Floor



Shear Force (kN)

1.4D 7981 0 01.2D + 1.6L 8601 0 0

1.2D + 1.6L + 0.8W(Wind Case 1)

8601 28 17

1.2D + 0.5L + 1.6W

(Wind Case 1)

7391 56 34

0.9D - 1.6W(Wind Case 1)

5131 -56 -34

- 78 -

Column Design

Design for axial force and bending

650 650 mm column with 12 - No.28 bars

g = 1.75 % > 1.0 % O.K.

< 8.0 % O.K.

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- 79 -

Column Design

- 80 -

Column Design

Shear design

where N u = 5131 kN is the smallest axial forcecorresponding to the largest shear force on thesection and d = 650 40 10 28/2 = 586 mm

ACI Eq. (11-4)

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- 81 -

Column Design

Vu (= 34 kN) < Vc /2 = (0.75 765)/2 = 287 kN

Transverse reinforcement requirements must satisfyACI 7.10.5, with No.10 ties, the vertical spacing ofthe ties must not exceed the least of the following:

16(longitudinal bar diameter) = 16 28 = 448 mm (governs)

48(tie bar diameter) = 48 10 = 480 mmLeast column dimension = 650 mm

- 82 -

Column Design

Use No.10 ties @ 400 mm with the first tie locatedvertically not more than 400/2 = 200 mm above thetop of the slab and not more than 75 mm below thelowest horizontal reinforcement in the beams (ACI7.10.5.4 and 7.10.5.5)

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- 83 -

Column Design

Splice length of longitudinal reinforcement

ACI 12.17.2.1

From the P-M interaction diagram, it can be seen thatthe column reinforcement is in compression for allload combinations. As a result, lap splice length isdetermined in accordance with ACI 12.16.1.

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Column Design


ACI 12.16.1

For f y = 420 MPa and f c = 40 MPa,

l dc = 0.071f ydb = 0.071x420x28 = 835 mm

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Column Design


ACI 12.17.2.4

In both directions of the cross-section:

0.0015hs = 0.0015x650x400 = 390 mm 2

Area of ties provided in both direction:

Av = 3-legged No.10 = 3x79 = 237 mm 2 < 0.0015hsThus, lap splice length cannot be reduced by a factor

of 0.83.

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Column Design

Provide 850 mm splice length

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- 87 -

Column Design

- 88 -

Shear Wall Design

7900 7900 7900 7900 7900 7900 7900

6 7 0 0

6 7 0 0

6 7 0 0

Shear Wall

Unit: mm

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Shear Wall Design

Table: Summary of Design Axial forces, Bending Moments, and ShearForces at the Base of the Shear Wall on Line 7

Load Case AxialForce (kN)


Shear Force(kN)

Dead (D) 15,885 0 0Live (L) 2539 0 0

Wind(W)

Case 1 0 13,376 897Case 2 0 14,211 955

Case 3 45 10,034 673Case 4 34 11,021 741

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Shear Wall Design

Table: Summary of Design Axial forces, Bending Moments, and ShearForces at the Base of the Shearwall on Line 7



Shear Force(kN)

1.4D 22,239 0 01.2D + 1.6L 23,124 0 0

1.2D + 1.6L + 0.8W(Wind Case 2)

23,124 11,369 764

1.2D + 0.5L + 1.6W(Wind Case 2)

20,332 22,738 1528

0.9D - 1.6W(Wind Case 2)

14,297 -22,738 -1528

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Shear Wall Design

Shear design

The shear strength of the concrete is determined inaccordance with ACI 11.9.5 for walls subjected toaxial compression:

Vc = 0.17 (40) 0.5 300 5880/1000 = 1897 kN

where, d = 0.8 l w = 0.8x7350 = 5880 mm

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Shear Wall Design

Vc = 0.75x1897 = 1423 kN < V u = 1528 kN

Horizontal shear reinforcement shall be provided inaccordance with ACI 11.9.9.

Because the wall is more than 250 mm thick, twolayers of reinforcement is necessary in accordancewith ACI 14.3.4.

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Shear Wall Design

According to ACI 11.9.9.3, spacing of horizontalreinforcement shall not exceed

l w /5 = 7350/5 = 1470 mm,

3h = 3 300 = 900 mm, or

450 mm. (governs)

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Shear Wall Design

According to ACI 11.9.9.2, ratio of horizontal shearreinforcement shall not be less than 0.0025.

With 2 layers of No.14 bars, spacing required:

s = (2 154) / (300 0.0025) = 411 mm

Provide 2-No. 14 horizontal bars @ 400 mm( t = 0.00257).

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Shear Wall Design

By ACI 11.9.3, shear strength V n at any horizontalsection must be less than or equal to0.83 (40) 0.5 300 5880 1/1000 = 9260 kN

Vn = Vc + Vs= 1897 + (2 154 420 5880)/(400 1000)

= 3799 kN < 9260 kN O.K.

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Shear Wall Design

The ratio of vertical shear reinforcement area to grossconcrete area of horizontal section shall not be lessthan 0.0025 nor the value obtained by Eq. (11-30)

(ACI 11.9.9.4):

l = 0.0025 + 0.5(2.5-h w / l w)( t 0.0025)

= 0.0025 + 0.5 (2.5-45.6/7.35)(0.00257-0.0025)

= 0.00237

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Shear Wall Design

According to ACI 11.9.9.5, spacing of verticalreinforcement shall not exceed

l w /5 = 7450/5 = 1490 mm

3h = 3 300 = 900 mm or

450 mm

Provide 2-No. 14 vertical bars @ 400 mm

( l = 0.00257)

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Shear Wall Design

The provided vertical and horizontal reinforcement

satisfy the requirements of ACI 14.3.2 and 14.3.3 forminimum ratio of vertical and horizontal

reinforcement to gross concrete area, respectively,and ACI 14.3.5 for maximum bar spacing.

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- 99 -

Shear Wall Design

In accordance with ACI 14.4, the shear wall isdesigned for combined flexure and axial load as acompression member following the provisions ofACI 10.2, 10.3, 10.10, 10.11, 10.14, 14.2, and 14.3.

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Shear Wall Design

The wall is reinforced with 12-No. 28 bars in the 650

650 mm columns at both ends of the wall and 2-No.14 vertical bars @ 400 mm in the web.

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Shear Wall Design

- 102 -

Shear Wall Design

Development length of horizontal reinforcement

where, t = reinforcement location factor = 1.0 e = coating factor = 1.0 for uncoated bars s = reinforcement size factor = 0.8 for No.14 bar = lightweight aggregate factor = 1.0

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- 103 -

Shear Wall Design

cb = spacing or cover dimension

(cb + Ktr)/d b = (27 + 0)/14 = 1.93 < 2.5

(governs)

- 104 -

Shear Wall Design

Required l d (= 347 mm) can be accommodated withinthe 650 mm dimension of columns, so that hooksare not needed at the ends of the bars.

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Shear Wall Design

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If you are encountering technical difficulties, please call 00 1 847 991 2700

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- 107 -

Thank You!!

For more informationwww.skghoshassociates.com

Chicago Main Office334 East Colfax Street, Unit EPalatine, IL 60067

Phone: (847) 991-2700Fax: (847) 991-2702Email: [email protected]

Southern California Office43 Vantis DriveAliso Viejo, CA 92656

Phone: (949) 249-3739Fax: (949) 249-3989Email: [email protected]

1001011 - design-wind

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