Design of Shearwall-Frame Under Lateral Loads

ECCS 203-2009-Design Aids Shear wall -Frame example

By: Prof. Abdel Hamid Zaghw

1

EGYPTIAN CODE LATERAL FORCES

Simplified Modal Response Spectrum Method (Code item 8-7-3-2-2 Equations 8-16)

The design base shear is defined in the Egyptian Code by the following equation:

g

TSWF d

Ib

)( 1 (1)

where

bF is the design base shear

W is the weight of the building

I is the importance factor

is the correction factor for short period structures

1T is the fundamental period of the structure

g

TSd )( 1 is the acceleration response spectrum specified by the Egyptian Code

Building Weight W The weight of the building, W, is defined in code item 8-7-1-7 as

D.L. for residential buildings.

D.L.+0.5 L.L. for common buildings, malls, schools

D.L.+ L.L. for silos, tanks, stores, libraries, garages

Importance Factor γI The importance factor, I , is defined in code item 8-7-6 by the following table

Type of building Factor “γI”

I. Emergency facilities: hospitals,

fire stations, power plants, etc. 1.40

II. High occupancy buildings:

schools, assembly halls, etc. 1.20

III. Ordinary buildings. 1.00

IV. Buildings of minor importance

for public safety. 0.80

Correction factor λ The correction factor for short period structures, , is defined in code item

8-7-3-2-2 by the following Table



2

Case Factor “λ”

I. T1 < 2TC and number of

stories are more than 2

stories.

0.85

II. T1 > 2TC 1.00

Fundamental Period T1 The fundamental period of the structure, 1T , is defined in the Egyptian Code by the

following equation:

T1 = Ct H 3/4

sec.

Ct = 0.085 for space steel Frames

= 0.075 for space R.C. Frames

= 0.05 for other structures

and H is the total building height in meters

Acceleration Response Spectrum

The acceleration response spectrum,g

TSd )( 1 , is specified by the Egyptian Code

and defined by the following equations:

0.1

5.20.1

g

)( : 0 11

1RT

TS

g

aTSTT

B

gdB

R

S

g

aTSTTT

gdCB

5.2

g

)( : 1

1

g

a

T

T

R

S

g

aTSTTT

gcgd

DC 2.05.2g

)( :

1

1

1

g

a

T

TT

R

S

g

aTSTT

gDcgdD 2.05.2

g

)( : sec 4

2

1

1

1

The values of TB, TC, TD, and S are obtained from either spectrum

Type 1 or Spectrum Type 2

Spectrum Type 1, used all over Egypt except in the Mediterranean areas

Type 1 response spectrum

Subsoil

Class

S TB TC TD

A 1.0 0.05 0.25 1.2

B 1.35 0.05 0.25 1.2

C 1.5 0.10 0.25 1.2

D 1.8 0.10 0.30 1.2



3

Spectrum Type 2, used in the Mediterranean areas only.

Type 2 response spectrum

Subsoil

Class

S TB TC TD

A 1.0 0.15 0.40 2.0

B 1.20 0.15 0.50 2.0

C 1.15 0.20 0.60 2.0

D 1.35 0.20 0.80 2.0

Identification of ground types The following Table describes ground types A, B, C, and D

Ground

Type

Description of stratigraphic profile

A Rock or other rock-like soil,

including at most 3m of weaker

material at the surface.

B Deposits of very dense sand, gravel,

or

very stiff clay, at least 15 meters in

thickness.

C Deep deposits of dense or medium

dense sand, gravel, or stiff clay with

thickness extending at least 15

meters in thickness.

D

Deposits of loose-to-medium

cohesionless soil, or of

predominantly soft-to-firm cohesive

soil extending for 15 meters in

thickness.



4

R Factor

Structural system R

1.Bearing walls;

i) Shear walls-R.C.

ii) Shear walls- reinforced masonry

iii) Shear walls- unreinforced

masonry

4.5

3.5

2.0

2.Ordinary frames

i) Shear walls-R.C.

ii) Shear walls- reinforced masonry

5.0

4.5

3.Moment resisting frames; R.C or Steel.

i) With adequate ductility.

i) With limited ductility.

7.0

5.0

4.Dual systems, moment frames &walls

i) With adequate ductility.

i) With limited ductility.

6.0

5.0

5. Other Structures.

i) Water Tanks (framed)

ii) Network Towers

iii) Minaret, chimneys, silos.

2.0

3.0

3.5

Damping factor

Structural Type

Steel with welded

connections

1.20

Steel with bolted

connections

1.05

Reinforced Concrete 1.00

Presetressed Concrete 1.05

Shear wall buildings 0.95



5

Seismic Zones in Egypt

Earthquake zone Coefficient ag/g

Zone 1 0.1

Zone 2 0.125

Zone 3 0.15

Zone 4 0.20

Zone 5 0.25

Distribution of Earthquake Lateral Loads

The storey shear value, jF , at level j is related to the total base shear value bF and

the story weight wj and story drift ui according to the following equation:

ii

jjbj

uw

uwFF

For story drift ui proportional with story height zi the storey shear value, jF , at

level j is calculated according to the following equation:

ii

jjbj

zw

zwFF



6

Example

For the 12-storey frame-shearwall office building shown below is located in Cairo.

The soil type is composed of deposits of loose-to-medium cohesionless soil. The

columns and shearwalls have constant cross section throughout the height of the

building. The beams and slabs also have the same dimensions at all floor levels.

Other pertinent design data are as follows:

Service Loads Live load: 3 kN/m

2

Additional superimposed dead load (flooring): 2.0 kN/m2

Beam and Column Sections

Slab thickness is 0.2 m Shear wall thickness is 0.35 m

External columns are 0.65m x 0.65m Internal columns are 0.75m x 0.75m

All beams are 0.5m x 0.6m

Material Properties

Concrete: fcu = 30 N/ mm2 Steel: fy = 400 N/ mm

2

Wall Density = 18 KN/m3

Consider the building’s structural system as a duel system with frames having

limited ductility in the transverse direction and as moment resisting frames of

limited ductility in the longitudinal direction.

A

B

C

D

21 3 4 5 6 7 8

7 x 8m = 56m

3 x

7m

= 2

1m

Y

X

Building Plan

Calculate the following:

i) Seismic base shear according to the Egyptian Code.

ii) Storey shear in both transverse and longitudinal directions.

iii) Design the beam along axis 4 in the sixth floor.

iv) Design the exterior column A 3 and the interior column B 3 in the fourth

floor.

v) Design the beam column connections for the above columns.

vi) Design shearwall at the foundation level



7

12

11

10

9

8

7

6

5

4

3

2

1

11 x

3.6

m =

39.6

m4.4

7 x 8 m = 56 m

Longitudinal Section

Solution:

i) Seismic base shear according to the Egyptian Code:

Step (1): Building Weight On the basis of the given data and the dimensions described above weights of

ground, typical, and roof floor (in KN) were estimated as follows:

Ground Typical Roof

Live Load (KN)

LL on Slab 3,528 3,528 3,528

Dead Load (KN)

Slab

(O.W.+Flooring) 8,232 8,232 8,232

Beams 1,742 1,742 1,742

Columns 1,520 1,368 684

Shear Wall 445 400 200

Partitions 2,552 2,252 1,126

Total DL (KN) 14,490 13,993 11,983

DL + 0.5 LL (KN) 16,254 15,757 13,747

The total weight of the building, W, is 16,254+ 10 x15, 757 +13,747 =187,571 kN

Note: 1) Building weight (W) is taken as DL+0.5LL (office building) code item 8-7-1-7

2) Columns, shear walls, and partitions loads are calculated from the center line

of the floor height below to the center line of the floor height above.



8

Step (2): Seismic Base Shear Calculations

From the soil description it is classified as Type D. Since the building is located

in Cairo, the values of the various constants TB, TC, TD, and S are calculated

from Spectrum Type 1 as shown in the following table.

Subsoil

Class

S TB TC TD

D 1.8 0.10 0.30 1.2

The base shear on the building according to the Egyptian Code of Practice is equal

to

g

TSWF d

Ib

)( 1

Base Shear Calculation in the Transverse Direction

The weight of the building, W = 187,571kN

The importance factor, I = 1.2 (Office Building)

The fundamental period of the structure, 1T is equal to

T1 = Ct H 3/4

Note that this building is frame-shearwall in the transverse direction (Ct = 0.05)

T1 = 0.05 x (44)3/4

= 0.854 sec

The correction factor is 1.0 since T1 > 2 TC

Since Cairo is located in zone 3, 15.0g

ag

According to the building description the damping coefficient is equal to 1.0

and the R factor is 5.0 in both transverse and longitudinal directions.

Since TC < T1 < TD the acceleration response spectrum g

TSd )( 1 is calculated from

the following equation:

g

a

T

T

R

S

g

aTS gcgd2.0

5.2g

)(

1

1

)15.0(2.0854.0

30.0

5

0.18.1)15.0(5.2

g

)( 1

TSd

03.004741.0g

)( 1

TSd 04741.0g

)( 1

TSd



9

Substituting back in the base shear equation

g

TSWF d

Ib

)( 1

672,1004741.0571,1870.12.1 transbF kN

Base Shear Calculation in the Longitudinal Direction

The weight of the building, W = 187,571kN

The importance factor, I = 1.2

The fundamental period of the structure, 1T is equal to

T1 = Ct H 3/4

Note that this building is RC frame in the longitudinal direction (Ct = 0.075).

T1 = 0.075 x (44)3/4

= 1.28 sec

The correction factor is 1.0 since T1 > 2 TC

Since Cairo is located in zone 3, 15.0g

ag

According to the building description the damping coefficient is equal to 1.0

and the R factor is 5.0 in both transverse and longitudinal directions.

Since TD < T1 the acceleration response spectrum g

TSd )( 1 is calculated from the

following equation:

g

a

T

TT

R

S

g

aTS gDcgd 2.05.2g

)(

2

1

1

)15.0(2.028.1

2.130.0

5

0.18.1)15.0(5.2

g

)(

2

1

TSd

03.00297.0g

)( 1

TSd 03.0g

)( 1

TSd

Substituting back in the base shear equation

g

TSWF d

Ib

)( 1

753,603.0571,1870.12.1 bF kN



10

ii) Storey shear in both transverse and longitudinal directions.

The storey shear value, jF , at level j is related to the total base shear value bF and

the story weight wj and storey height zj according to the following equation:

ii

jjbj

zw

zwFF

Design seismic forces in transverse (short) direction

Floor Level

Height (hx) Story Weight(Wx) Wx*hx

Lateral Force(Fx)

Story Shear(∑Fx)

m KN KN*m KN KN

12(roof) 44 13,747 604,887 1438 1438

11 40.4 15,757 636,597 1513 2951

10 36.8 15,757 579,870 1378 4329

9 33.2 15,757 523,144 1244 5573

8 29.6 15,757 466,418 1109 6682

7 26 15,757 409,691 974 7656

6 22.4 15,757 352,965 839 8495

5 18.8 15,757 296,238 704 9199

4 15.2 15,757 239,512 569 9768

3 11.6 15,757 182,785 434 10202

2 8 15,757 126,059 300 10502

1 4.4 16,254 71,518 170 10672

∑= 187,571 4,489,683 10,672

Design seismic forces in longitudinal (long) direction

Floor Level

Height (hx) Story Weight(Wx) Wx*hx

Lateral Force(Fx)

Story Shear(∑Fx)

m KN KN*m KN KN

12(roof) 44 13,747 604,887 910 910

11 40.4 15,757 636,597 957 1,867

10 36.8 15,757 579,870 872 2,739

9 33.2 15,757 523,144 787 3,526

8 29.6 15,757 466,418 702 4,228

7 26 15,757 409,691 616 4,844

6 22.4 15,757 352,965 531 5,375

5 18.8 15,757 296,238 446 5,820

4 15.2 15,757 239,512 360 6,181

3 11.6 15,757 182,785 275 6,456

2 8 15,757 126,059 190 6,645

1 4.4 16,254 71,518 108 6,753

∑= 187,575 4,489,683 6,753

The following figures show the output of program BaseShear. The story forces

calculated by the program are almost identical to the results presented in the above

tables.



11

Response Spectrum Based on the Eqyption Code of Practice (2008)

(Program "BaseShear" created by: Prof. Abdel Hamid Zaghw - Cairo University - Egypt)

Project Name:

Location:

Response Spectrum Data

Response Spectrum Type: 1 Soil Subclass Type: D

Damping Coefficient : 1 R Factor: 5

ag/g: 0.15

Egyptian Code Design Spectrum

0

0.05

0.1

0.15

0.2

0.25

0.3

0 0.5 1 1.5 2 2.5 3 3.5 4

Period (T)

Sa /

g

Output of Program "BaseShear" for the response spectrum curve for the above

building location



12

Lateral Loads Based on the Eqyption Code of Practice (2008)


Base Shear Computations

Number of Stories: 12 Daming Coefficient: 1.2

Total Weight: 187571 Total Height: 44

Ct for Period Computaions: 0.05 Building Period: 0.8542

Sa/g: 0.04741 Base Shear : 10,672

Floor Weight Floor Height (m) Floor Shear

13747 1437.796

3.6

15757 1513.184

3.6

15757 1378.346

3.6

15757 1243.507

3.6

15757 1108.669

3.6

15757 973.8311

3.6

15757 838.993

3.6

15757 704.1548

3.6

15757 569.3167

3.6

15757 434.4785

3.6

15757 299.6404

3.6

16254 170.0003

4.4

Output of Program "BaseShear" for the lateral loads in the transverse

direction for the above building



13

Lateral Loads Based on the Eqyption Code of Practice (2008)


Base Shear Computations

Number of Stories: 12 Daming Coefficient: 1.2

Total Weight: 187571 Total Height: 44

Ct for Period Computaions: 0.075 Building Period: 1.2813

Sa/g: 0.03000 Base Shear : 6,753

Floor Weight Floor Height (m) Floor Shear

13747 909.7522

3.6

15757 957.4529

3.6

15757 872.1353

3.6

15757 786.8177

3.6

15757 701.5001

3.6

15757 616.1826

3.6

15757 530.865

3.6

15757 445.5474

3.6

15757 360.2298

3.6

15757 274.9122

3.6

15757 189.5946

3.6

16254 107.5661

4.4

Output of Program "BaseShear" for the lateral loads in the longitudinal

direction for the above building



14

A three dimensional model of the building was created using SAP2000. The

beams, columns, and shear wall boundary elements were modeled by frame

elements and the slab and shear wall were modeled by shell element. The lateral

loads calculated in the previous section were applied as distributed loads at each

floor in both the transverse and longitudinal directions.

Three dimensional representation of the building



15

iii) Design the beam along axis 4 in the sixth floor

The bending moment for the beam was calculated for the following load cases

1) 1.4 DL + 1.6 LL

2) 1.12 DL + (0.5) LL + S

3) 1.12 DL + (0.5) LL – S

4) 0.9 DL + S

5) 0.9 DL – S

7 7 7

A B C D 6th

Floor

Axis 4

Load Case

Beam AB Beam BC

A Mid

Span

B B Mid

Span

C

1.4 DL + 1.6 LL -351 150 -183 -258 130 -258

1.12 DL + (0.5) LL + S -460 142 109 -437 132 90

1.12 DL + (0.5) LL – S 0 120 -359 91 120 -437

0.9 DL + S -389 125 145 -385 120 143

0.9 DL – S 71 100 -323 142 120 -385

From the above table it is shown that each section in the beam should be designed

to withstand the following positive and negative moments:

Moment Type

Beam AB Beam BC

A Mid Span B B Mid

Span

C

Negative -460 --- -359 -437 --- -437 Positive 71 150 145 142 132 143



16



17

iv) Design the exterior column A 3 and the interior column B 3 in the

fourth floor

7 7 7

A B C D

4th

Floor

Axis 3

3rd

Floor

Axis 3

x x

SE C (x-x)

Local Axis

Design Loads for Exterior Column A3

Load Case

Axial

Load

(kN)

Moment kN.m

Top Bottom

My

(kN.m)

Mz

(kN.m)

My

(kN.m)

Mz

(kN.m)

1.4 DL + 1.6 LL -5728.447 221.0517 16.5191 -220.8357 -15.0685

1.12 DL + (0.5) LL + SX + .3 SY -4291.876 200.8478 -627.097 -209.4897 554.0159

1.12 DL + (0.5) LL – SX +. 3 SY -4110.134 157.7823 645.0652 -164.4847 -570.371

1.12 DL + (0.5) LL + SX –. 3 SY -3845.386 132.1003 -622.628 -125.2837 549.906

1.12 DL + (0.5) LL – SX – .3 SY -3142.737 8.8295 654.7479 17.9618 -579.2757

1.12 DL + (0.5) LL + .3 SX + SY -4749.172 265.9802 -187.0541 -291.9784 165.2753

1.12 DL + (0.5) LL – .3 SX + SY -4694.649 253.0605 194.5946 -278.4769 -172.0407

1.12 DL + (0.5) LL + .3 SX – SY -3845.386 132.1003 -622.628 -125.2837 549.906

1.12 DL + (0.5) LL – .3 SX – SY -3206.347 23.9024 209.4912 2.21 -185.7403

0.9 DL + SX + .3 SY -3198.89 155.9451 -630.3292 -164.6579 556.9645

0.9 DL – SX +. 3 SY -3017.148 112.8796 641.833 -119.6529 -567.4223

0.9 DL + SX –. 3 SY -2752.4 87.1977 -625.8602 -80.4518 552.8547

0.9 DL – SX – .3 SY -2570.657 44.1322 646.302 -35.4468 -571.5322

0.9 DL + .3 SX + SY -3656.186 221.0775 -190.2862 -247.1466 168.224

0.9 DL – .3 SX + SY -3601.663 208.1579 191.3624 -233.6451 -169.0921

0.9 DL + .3 SX – SY -2167.884 -8.0806 -175.3896 33.5404 154.5244

0.9 DL – .3 SX – SY -2113.361 -21.0002 206.259 47.0419 -182.7916



18

Design of the top fiber for Exterior Column A 3

Load Case Axial(kN)

My(kN.m)

Mz(kN.m)

ty(cm)

tz(cm) m

1.4 DL + 1.6 LL -5728.447 221.0517 16.5191 65 65 0.013535

1.12 DL + (0.5) LL + SX + .3 SY -4291.876 200.8478 -627.097 65 65 0.01908

1.12 DL + (0.5) LL – SX +. 3 SY -4110.134 157.7823 645.0652 65 65 0.017587

1.12 DL + (0.5) LL + SX –. 3 SY -3845.386 132.1003 -622.628 65 65 0.014835

1.12 DL + (0.5) LL – SX – .3 SY -3142.737 8.8295 654.7479 65 65 0.009623

1.12 DL + (0.5) LL + .3 SX + SY -4749.172 265.9802 -187.0541 65 65 0.010522

1.12 DL + (0.5) LL – .3 SX + SY -4694.649 253.0605 194.5946 65 65 0.009873

1.12 DL + (0.5) LL + .3 SX – SY -3845.386 132.1003 -622.628 65 65 0.014835

1.12 DL + (0.5) LL – .3 SX – SY -3206.347 23.9024 209.4912 65 65 0.006

0.9 DL + SX + .3 SY -3198.89 155.9451 -630.3292 65 65 0.012532

0.9 DL – SX +. 3 SY -3017.148 112.8796 641.833 65 65 0.011096

0.9 DL + SX –. 3 SY -2752.4 87.1977 -625.8602 65 65 0.00895

0.9 DL – SX – .3 SY -2570.657 44.1322 646.302 65 65 0.008271

0.9 DL + .3 SX + SY -3656.186 221.0775 -190.2862 65 65 0.006

0.9 DL – .3 SX + SY -3601.663 208.1579 191.3624 65 65 0.006

0.9 DL + .3 SX – SY -2167.884 -8.0806 -175.3896 65 65 0.006

0.9 DL – .3 SX – SY -2113.361 -21.0002 206.259 65 65 0.006

Design of the bottom fiber for Exterior Column A 3

Load Case Axial(kN)

My(kN.m)

Mz(kN.m)

ty(cm)

tz(cm) m

1.4 DL + 1.6 LL -5728.447 -220.8357 -15.0685 65 65 0.013502

1.12 DL + (0.5) LL + SX + .3 SY -4291.876 -209.4897 554.0159 65 65 0.017053

1.12 DL + (0.5) LL – SX +. 3 SY -4110.134 -164.4847 -570.371 65 65 0.015461

1.12 DL + (0.5) LL + SX –. 3 SY -3845.386 -125.2837 549.906 65 65 0.012393

1.12 DL + (0.5) LL – SX – .3 SY -3142.737 17.9618 -579.2757 65 65 0.007267

1.12 DL + (0.5) LL + .3 SX + SY -4749.172 -291.9784 165.2753 65 65 0.010832

1.12 DL + (0.5) LL – .3 SX + SY -4694.649 -278.4769 -172.0407 65 65 0.010148

1.12 DL + (0.5) LL + .3 SX – SY -3845.386 -125.2837 549.906 65 65 0.012393

1.12 DL + (0.5) LL – .3 SX – SY -3206.347 2.21 -185.7403 65 65 0.006

0.9 DL + SX + .3 SY -3198.89 -164.6579 556.9645 65 65 0.010492

0.9 DL – SX +. 3 SY -3017.148 -119.6529 -567.4223 65 65 0.008941

0.9 DL + SX –. 3 SY -2752.4 -80.4518 552.8547 65 65 0.006718

0.9 DL – SX – .3 SY -2570.657 -35.4468 -571.5322 65 65 0.006

0.9 DL + .3 SX + SY -3656.186 -247.1466 168.224 65 65 0.006

0.9 DL – .3 SX + SY -3601.663 -233.6451 -169.0921 65 65 0.006

0.9 DL + .3 SX – SY -2167.884 33.5404 154.5244 65 65 0.006

0.9 DL – .3 SX – SY -2113.361 47.0419 -182.7916 65 65 0.006

Hence m =0.01908

As= m.b.t =0.01908x650x650=8061mm2

Use 20Φ25 (Asc=9817 mm2)

(Note that the bold case is the most critical case)



19



20

Design Loads for Interior Column B 3

Load Case

Axial

Load

(kN)

Moment kN.m

Top Bottom

My

(kN.m)

Mz

(kN.m)

My

(kN.m)

Mz

(kN.m)

1.4 DL + 1.6 LL -9094.576 9.5047 96.4193 -6.2987 -89.3567

1.12 DL + (0.5) LL + SX + .3 SY -6378.243 115.548 -780.3507 -128.9465 645.4878

1.12 DL + (0.5) LL – SX +. 3 SY -6235.882 29.6687 885.4035 -39.125 -742.0216

1.12 DL + (0.5) LL + SX –. 3 SY -6198.075 -19.1351 -754.5385 32.6839 620.754

1.12 DL + (0.5) LL – SX – .3 SY -5845.518 -262.1448 941.3298 311.0744 -795.6114

1.12 DL + (0.5) LL + .3 SX + SY -6538.613 242.6206 -227.4509 -286.0778 188.7155

1.12 DL + (0.5) LL – .3 SX + SY -6495.905 216.8568 272.2753 -259.1314 -227.5373

1.12 DL + (0.5) LL + .3 SX – SY -6198.075 -19.1351 -754.5385 32.6839 620.754

1.12 DL + (0.5) LL – .3 SX – SY -5895.344 -232.087 358.3159 279.6368 -309.9831

0.9 DL + SX + .3 SY -4611.713 113.3097 -799.2318 -127.3772 662.9877

0.9 DL – SX +. 3 SY -4469.352 27.4303 866.5223 -37.5557 -724.5216

0.9 DL + SX –. 3 SY -4431.545 -21.3734 -773.4197 34.2532 638.254

0.9 DL – SX – .3 SY -4289.184 -107.2528 892.3345 124.0748 -749.2554

0.9 DL + .3 SX + SY -4772.083 240.3822 -246.3321 -284.5086 206.2155

0.9 DL – .3 SX + SY -4729.375 214.6184 253.3942 -257.5621 -210.0373

0.9 DL + .3 SX – SY -4171.523 -208.5615 -160.2915 254.2596 123.7697

0.9 DL – .3 SX – SY -4128.814 -234.3253 339.4347 281.2061 -292.4831

Design of the top fiber for Interior Column B 3

Load Case Axial(kN)

My(kN.m)

Mz(kN.m)

ty(cm)

tz(cm) m

1.4 DL + 1.6 LL -9094.576 9.5047 96.4193 75 75 0.022785

1.12 DL + (0.5) LL + Sx + .3 Sy -6378.243 115.548 -780.3507 75 75 0.015877

1.12 DL + (0.5) LL – Sx +. 3 Sy -6235.882 29.6687 885.4035 75 75 0.016006

1.12 DL + (0.5) LL + Sx –. 3 Sy -6198.075 -19.1351 -754.5385 75 75 0.013002

1.12 DL + (0.5) LL – Sx – .3 Sy -5845.518 -262.1448 941.3298 75 75 0.018588

1.12 DL + (0.5) LL + .3 Sx + Sy -6538.613 242.6206 -227.4509 75 75 0.007637

1.12 DL + (0.5) LL – .3 Sx + Sy -6495.905 216.8568 272.2753 75 75 0.008036

1.12 DL + (0.5) LL + .3 Sx – Sy -6198.075 -19.1351 -754.5385 75 75 0.013002

1.12 DL + (0.5) LL – .3 Sx – Sy -5895.344 -232.087 358.3159 75 75 0.006423

0.9 DL + Sx + .3 Sy -4611.713 113.3097 -799.2318 75 75 0.008263

0.9 DL – Sx +. 3 Sy -4469.352 27.4303 866.5223 75 75 0.00784

0.9 DL + Sx –. 3 Sy -4431.545 -21.3734 -773.4197 75 75 0.006

0.9 DL – Sx – .3 Sy -4289.184 -107.2528 892.3345 75 75 0.009059

0.9 DL + .3 Sx + Sy -4772.083 240.3822 -246.3321 75 75 0.006

0.9 DL – .3 Sx + Sy -4729.375 214.6184 253.3942 75 75 0.006

0.9 DL + .3 Sx – Sy -4171.523 -208.5615 -160.2915 75 75 0.006

0.9 DL – .3 Sx – Sy -4128.814 -234.3253 339.4347 75 75 0.006



21

Design of the bottom fiber for Interior Column B 3

Load Case Axial(kN)

My(kN.m)

Mz(kN.m)

ty(cm)

tz(cm) m

1.4 DL + 1.6 LL -9094.576 -6.2987 -89.3567 75 75 0.022785

1.12 DL + (0.5) LL + Sx + .3 Sy -6378.243 -128.9465 645.4878 75 75 0.013235

1.12 DL + (0.5) LL – Sx +. 3 Sy -6235.882 -39.125 -742.0216 75 75 0.01326

1.12 DL + (0.5) LL + Sx –. 3 Sy -6198.075 32.6839 620.754 75 75 0.010378

1.12 DL + (0.5) LL – Sx – .3 Sy -5845.518 311.0744 -795.6114 75 75 0.016376

1.12 DL + (0.5) LL + .3 Sx + Sy -6538.613 -286.0778 188.7155 75 75 0.007945

1.12 DL + (0.5) LL – .3 Sx + Sy -6495.905 -259.1314 -227.5373 75 75 0.007929

1.12 DL + (0.5) LL + .3 Sx – Sy -6198.075 32.6839 620.754 75 75 0.010378

1.12 DL + (0.5) LL – .3 Sx – Sy -5895.344 279.6368 -309.9831 75 75 0.00625

0.9 DL + Sx + .3 Sy -4611.713 -127.3772 662.9877 75 75 0.006

0.9 DL – Sx +. 3 Sy -4469.352 -37.5557 -724.5216 75 75 0.006

0.9 DL + Sx –. 3 Sy -4431.545 34.2532 638.254 75 75 0.006

0.9 DL – Sx – .3 Sy -4289.184 124.0748 -749.2554 75 75 0.00625

0.9 DL + .3 Sx + Sy -4772.083 -284.5086 206.2155 75 75 0.006

0.9 DL – .3 Sx + Sy -4729.375 -257.5621 -210.0373 75 75 0.006

0.9 DL + .3 Sx – Sy -4171.523 254.2596 123.7697 75 75 0.006

0.9 DL – .3 Sx – Sy -4128.814 281.2061 -292.4831 75 75 0.006

Hence m =0.022785

As= m.b.t =0.022785x750x750=12816mm2

Use 28Φ25 (Asc=13720 mm2)

(Note that the bold case is the most critical case)



22



23

v) Design beam column connections for the above columns

a- Design of exterior beam column connection

s

Material Properties

fy = 400 N/mm2

fcu = 30 N/mm2

Beam Data

Top reinforcement Astop = 8φ22 = 3040 mm2

Depth t = 600 mm

Width b = 500mm

Cover = 50 mm

Column Data

Height = 3.6 m

Depth t = 650 mm

Width b = 650 mm

Cover = 25 mm

Axial Load (1.4 DL + 1.6LL) = 5728 kN. (4th

floor)


floor)



24

Step 1 Check of strong column and weak beam

The strong column weak beam requirement is guaranteed by the satisfaction of the

following equation:

∑Mc ≥ ∑1.2Mg…………………… (Code 6-61)

where:

∑Mg is the summation of the maximum moment capacities that can be resisted by

the beams on the right and the left of the column

∑Mc is the ultimate moment capacity that can be resisted by the column

corresponding to the maximum ultimate axial force applied on the column

The top beam reinforcement calculated in section (iii) of this example is 8Φ22

(As=3041mm2). Using the design aid curves one finds that mkNM g .490

The external column reinforcement is 20Φ25 (As = 9817mm2)

Pucol. above the beam of the 4th

floor = 5091 kN Using the Interaction Diagrams with 2

y N/mm 400f & 0.9 one finds that kN.m 741topMc

Pucol. below the beam of the 4th


y N/mm 400f & 0.9 one finds that kN.m 765bottomMc

Hence ∑Mc =Mctop +Mcbottom =741+576=1317 kN.m

........Ok...................................................................... 1.2 2.69490

1317

g

c

M

M

Step 2 Calculate the Beam Ultimate Capacity with 1.25 fy

Tension in top steel T top = 7.132115.11000

40025.1304025.1

s

y

stop

fA

kN

Compression in bottom fibers C1 = aaabf

c

cu 7.61000/5005.1

3067.067.0

kN

Since T = C Therefore 3.1977.6

7.1321a mm

Lever arm 35.451502

3.197600cover

2

atyct mm

The beam ultimate moment capacity Mpr = 6.5961000

35.4517.1321 ctyT kN.m

Step 3 Check the horizontal shear

An estimate of the horizontal shear from the column, Qucol can be obtained by assuming

that the beams in the adjoining floors are also deformed so that plastic hinges form at

their junctions with the column. By farther assuming that the end moment in the beams

are resisted equally by the columns above and below the joint, one obtains for the

horizontal shear at the column ends ״ the column shear ״.



25

Qucol = heightstorey

)beam(prM =

6.3

6.596=165.7

This shear force must be resisted by concrete of column

23

.

. N/mm 0.392650650

107.165

..

x

x

db

Quq

colcol

colu

c

cucuc

fq

24.0 ……………………………………………………...(Code 4-18)

c = 949.1650650

10572807.0107.01

3

x

x

Ac

Pu……………………….(Code 4-19)

u

c

cucuc qmmNx

fq 2/09.2

5.1

30949.124.024.0

…………… Ok.



26

Step 4 check of ultimate shear force Qju

The net shear at section x-x of the joint is

Qju = T1 - Qucol =1321.7-165.7=1156 kN ………………………...(Code 6-56-b)

The ultimate shear force Qju must satisfy this equation

Qju ≤ Kj.Aj c

cuf

…………………………………………………...(Code 6-65-a)

C2

X1

X2

connection depth = C2

effective connection width

= the smallest of 1- ( b+C2)

2- (b+2x)effective area (Aj)

X

b

Where Aj is the effective cross sectional area through the beam column connection zone

It is the minimum of = tcol. (bbeam+2X) = 422500 mm2

X=75mm …(Code Figure 6-31)

= tcol. (bbeam+tcol) = 747500 mm2 …………………. Ok.

Hence Aj = 422500 mm2

Kj is a factor depends on type of connection (exterior , Interior or other )

Hence Kj = 1.2 ………………………………………………………( Code Table 6-15)

Kj.Aj c

cuf

= 1.2x 422500

5.1

30x10

-3 = 2267.37 kN > Qju ……………….Ok



27

Step 5 Design of Column stirrups inside the connection

Ast is the maximum of Ast1 =

1

)/(

)/(.313.0 1

k

g

syst

ccu

A

A

f

fys

………...(Code 6-57-a)

Ast2 =

)/(

)/(.1.0 1

syst

ccu

f

fys

………………………(Code 6-57-b)

Where Ag is the gross area of section = tcol.x bcol. = 650x650 = 422500 mm2

Ak

is the sectional area inside the exterior stirrups perimeter = 360000 mm

2

y1 is distance between stirrups axes perpendicular to the assumed direction

y1 = 600 mm

s is the spacing between stirrups it is assume it =100 mm

Hence Ast1=

1

360000

422500

)15.1/400(

)5.1/30(600100313.0

x=187 mm

2

Ast2=

)15.1/400(

)5.1/30(6001001.0

x=345 mm

2……………………………..Ok

Ast=345 mm2

Assume number of branches= 4

Astof one branch =

4

345=86.3 mm

2

useΦ12@100 mm

Note that: the column stirrups must continue through the beam column connection zone

and extend to 1/6 the clear height from beam face in both top & bottom ends



28

b- Design of interior beam column connection

s

s

Material Properties

fy = 400 N/mm2

fcu = 30 N/mm2

Beam Data

Top reinforcement Astop = 8φ22 = 3040 mm2

Bottom reinforcement Asbottom =5φ16=1005 mm2

Depth t = 600 mm

Width b = 500mm

Cover = 50 mm

Column Data

Height = 3.6 m

Depth t = 750 mm

Width b = 750 mm

Cover = 25 mm


floor)


floor)



29

Step 1 Check of strong column and weak beam

The strong column weak beam requirement is guaranteed by the satisfaction of the

following equation:

∑Mc ≥ ∑1.2Mg…………………… (Code 6-61)

where:

∑Mg is the summation of the maximum moment capacities that can be resisted by

the beams on the right and the left of the column

∑Mc is the ultimate moment capacity that can be resisted by the column

corresponding to the maximum ultimate axial force applied on the column

The top beam reinforcement calculated in section (iii) of this example is 8Φ22

(As=3041mm2). Using the design aid curves one finds that mkNM top .490

The bottom beam reinforcement calculated in section (iii) of this example is 5φ16

(As=1005 mm2). Using the design aid curves one finds that mkNMbottom .181

∑Mg = 671 kN.m

The internal column reinforcement is 28Φ25 (As = 13744mm2)

Pucol. before the beam of the 4th


y N/mm 400f & 0.9 one finds that kN.m 597topMc

Pucol. after the beam of the 4th


y N/mm 400f & 0.9 one finds that kN.m 965bottomMc

Hence ∑Mc =Mctop +Mcbottom =759+569 = 1328 kN.m

........Ok...................................................................... 1.2 98.1671

1328

g

c

M

M

Step 2 Calculate the Beam Ultimate Capacities

Tension in top steel T top = 7.132115.11000

40025.1304025.1

s

y

stop

fA

kN

Corresponding Compression in bottom fibers C1 =

aaabf

c

cu 7.61000/5005.1

3067.067.0

kN


7.1321a mm

Lever arm 35.451502

3.197600cover

21

atyct mm

The beam ultimate moment capacity Mpr1 = 6.5961000

35.4517.132111 ctyT kN.m



30

Tension in bottom steel T botoom= 9.43615.11000

40025.1100525.1

s

y

sbottom

fA

kN

Corresponding Compression in top fibers C2 =

aaabf

c

cu 7.61000/5005.1

3067.067.0

kN


9.436a mm

Lever arm 4.517502

2.65600cover

22

a

tyct mm

The beam ultimate moment capacity Mpr2 = 2261000

4.5179.436

22 ctyT kN.m

2

21 prpr MMMu = 411.3 kN.m

Step 3 Check the horizontal shear

By farther assuming that the end moment in the beams are resisted equally by the

columns above and below the joint, one obtains for the horizontal shear at the column

ends the column shear Qucol = heightstorey

)( 21 prpr MM =

6.3

2266.596 =228.5 kN

This shear force must be resisted by concrete of column

23

.

. N/mm 406.0750750

105.228

..

x

x

db

Quq

colcol

colu

c

cu

cuc

fq

24.0 …………………………………………………...(Code 4-18)

c = 1244.1)750750

109095(07.01)(07.01

3

x

x

Ac

Pu……………………...(Code 4-19)

u

c

cu

cuc qmmNxf

q 2/21.15.1

301244.124.024.0

…………… Ok.



31

Step 4 check of ultimate shear force Qju

The net shear at section x-x of the joint is

Qju = T1+C2-Qucol =1321.7+436.9-228.5=1530.1 kN ……………..(Code 6-56-b)

The ultimate shear force Qju must satisfy this equation

Qju ≤ Kj.Aj c

cuf

…………………………………………………...(Code 6-65-a)

Where Aj is the effective cross sectional area through the beam column connection zone

It is the minimum of = tcol. (bbeam+2X) = 562500 mm2

X=125mm …(Code Figure 6-31)

= tcol. (bbeam+tcol) = 937500 mm2 …………………. Ok.

Hence Aj = 562500 mm2

Kj is a factor depends on type of connection (exterior , Interior or other )

Hence Kj = 1.6 ………………………………………………………( Code Table 6-15)

Kj.Aj c

cuf

= 1.6x 562500

5.1

30x10

-3 = 4024.92 kN > Qju ……………….Ok



32

Step 5 Design of Column stirrups inside the connection

Ast is the maximum of Ast1=

1

)/(

)/(.313.0 1

k

g

syst

ccu

A

A

f

fys

………...(Code 6-57-a)

Ast2=

)/(

)/(.1.0 1

syst

ccu

f

fys

………………………(Code 6-57-b)

Where Ag is the gross area of section = tcol.x bcol. = 750x750 = 562500 mm2

Ak

is the sectional area inside the exterior stirrups perimeter = 490000 mm

2

y1 is distance between stirrups axes perpendicular to the assumed direction

y1 = 700 mm

s is the spacing between stirrups it is assume it =100 mm

Hence Ast1=

1

490000

562500

)15.1/400(

)5.1/30(700100313.0

x=186 mm

2

Ast2=

)15.1/400(

)5.1/30(7001001.0

x=403 mm

2……………………………..Ok

Ast=403 mm2

Assume number of branches= 4

Astof one branch =

4

403=100.6 mm

2

useΦ12@100 mm

Note that: the column stirrups must continue through the beam column connection zone

and extend to 1/6 the clear height from beam face in both top & bottom ends



33

vi) Design of shear wall at foundation level

Steps for checking the shear wall sections for Normal force and Moment

For a given section geometry and eccentricity (e):

- Assume a value for the distance (c) down to the neutral axis. This value is a

measure of the compression block depth (a) since a = 0.8c .

- Using the assumed value of (c), calculate the strains e's and es at the compression

and tension steel, respectively.

- From the strains, calculate the stresses f's and fs in the compression and tension

steel, respectively.

- Calculate the compression and tension forces on steel (Fs or Fs\ ) by multiplying

the steel stress with the corresponding steel area

Fs =As. fs ………………...(Code item 4-2-1)

Fs\ =As. f's ………………...(Code item 4-2-1)

- Calculate the compression forces on the concrete by the following equation

C = baf

c

cu .67.0

…………...(Code item 4-2-1)

- Calculate the ultimate axial load ∑Nures using the equilibrium equations (by taking

the summation of all forces).

- If ∑Nures Nuapplied applied on the cross section then a new value for (c) is

assumed the above steps are repeated.

- If ∑Nures = Nuapplied applied on the cross section then calculate the corresponding

moment of resistance ∑Mures is calculated (by taking the summation of moments

about the center of gravity of the shear wall for all forces.

- If ∑Mures ≥ Muapplied applied on the cross section then the cross section of shear

wall is assumed to be safe if not the concrete dimensions or reinforcement should

be modified.

- Note that: If it is not possible to satisfy the equation ∑Nures= Nuapplied, it means that

the assumed section is inadequate to carry the applied normal forces. In this case a

new trial section should be assumed and the above procedure is repeated.

- The above steps are demonstrated in the following pages.

http://localhost/atutor/index.php?cid=229



34



35

The following calculations describes one of the trails for the most critical case of loading on

the shear wall (1.12 DL + (0.5) LL + .3 SX + SY)

Input data Output data

C (NA depth) = 4483 mm e(at fy/ s )= 0.001739

fcu = 30 N/mm2 Nures = -23533.73 kN

fy = 400 N/mm2 Mures = 87236.38 kN.m

L boundary element = 1000 mm

B boundary element = 800 mm Mures > Muapp safe

bweb = 350 mm

L Shear Wall = 7000 Mm

Ltotal = 8000 mm

Nuapplied= -23533 kN

Muapplied= 71253 kN.m

Es = 200 KN/mm2

Reinforcement Contribution

X(mm) #of bars/row Φ As(mm2) eCalc es Fs(kN) Mures (kN.m)

50 5 22 1900.31 -0.0029665 -0.001739 -660.98 2610.85

150 2 22 760.12 -0.0028996 -0.001739 -264.39 1017.90

250 2 22 760.12 -0.0028327 -0.001739 -264.39 991.46

350 2 22 760.12 -0.0027658 -0.001739 -264.39 965.02

450 2 22 760.12 -0.0026989 -0.001739 -264.39 938.59

550 2 22 760.12 -0.0026319 -0.001739 -264.39 912.15

650 2 22 760.12 -0.002565 -0.001739 -264.39 885.71

750 2 22 760.12 -0.0024981 -0.001739 -264.39 859.27

850 2 22 760.12 -0.0024312 -0.001739 -264.39 832.83

950 5 22 1900.31 -0.0023643 -0.001739 -660.98 2015.98

1000 2 16 402.05 -0.0023308 -0.001739 -139.84 419.53

1200 2 16 402.05 -0.002197 -0.001739 -139.84 391.56

1400 2 16 402.05 -0.0020631 -0.001739 -139.84 363.59

1600 2 16 402.05 -0.0019293 -0.001739 -139.84 335.62

1800 2 16 402.05 -0.0017954 -0.001739 -139.84 307.65

2000 2 16 402.05 -0.0016616 -0.001662 -133.61 267.22

2200 2 16 402.05 -0.0015278 -0.001528 -122.85 221.13

2400 2 16 402.05 -0.0013939 -0.001394 -112.09 179.34

2600 2 16 402.05 -0.0012601 -0.00126 -101.32 141.85

2800 2 16 402.05 -0.0011263 -0.001126 -90.56 108.67

3000 2 16 402.05 -0.0009924 -0.000992 -79.80 79.80

3200 2 16 402.05 -0.0008586 -0.000859 -69.04 55.23

3400 2 16 402.05 -0.0007247 -0.000725 -58.28 34.97

3600 2 16 402.05 -0.0005909 -0.000591 -47.51 19.01

3800 2 16 402.05 -0.0004571 -0.000457 -36.75 7.35

4000 2 16 402.05 -0.0003232 -0.000323 -25.99 0.00

4200 2 16 402.05 -0.0001894 -0.000189 -15.23 -3.05

4400 2 16 402.05 -5.554E-05 -5.55E-05 -4.47 -1.79

4600 2 16 402.05 7.83E-05 7.83E-05 6.30 3.78

4800 2 16 402.05 0.0002121 0.0002121 17.06 13.65

5000 2 16 402.05 0.000346 0.000346 27.82 27.82



36

5200 2 16 402.05 0.0004798 0.0004798 38.58 46.30

5400 2 16 402.05 0.0006137 0.0006137 49.34 69.08

5600 2 16 402.05 0.0007475 0.0007475 60.11 96.17

5800 2 16 402.05 0.0008813 0.0008813 70.87 127.56

6000 2 16 402.05 0.0010152 0.0010152 81.63 163.26

6200 2 16 402.05 0.001149 0.001149 92.39 203.26

6400 2 16 402.05 0.0012828 0.0012828 103.15 247.57

6600 2 16 402.05 0.0014167 0.0014167 113.92 296.18

6800 2 16 402.05 0.0015505 0.0015505 124.68 349.10

7000 2 16 402.05 0.0016844 0.0016844 135.44 406.32

7050 5 22 1900.31 0.0017178 0.0017178 652.88 1991.28

7150 2 22 760.12 0.0017847 0.0017391 264.39 832.83

7250 2 22 760.12 0.0018517 0.0017391 264.39 859.27

7350 2 22 760.12 0.0019186 0.0017391 264.39 885.71

7450 2 22 760.12 0.0019855 0.0017391 264.39 912.15

7550 2 22 760.12 0.0020524 0.0017391 264.39 938.59

7650 2 22 760.12 0.0021193 0.0017391 264.39 965.02

7750 2 22 760.12 0.0021863 0.0017391 264.39 991.46

7850 2 22 760.12 0.0022532 0.0017391 264.39 1017.90

7950 5 22 1900.31 0.0023201 0.0017391 660.98 2610.85

Total -683.53 29012.53

Concrete Contribution

abf

Cc

cu

67.0

Section a b Yct Nu conc. Mu conc.

Boundary

element

1000 800 3500 -10720 37520

Web 4483x0.8-1000

= 2586.4

350 1706.8 -12130.2 20703.8

Total -22850.2 58223.8

uresN 73.235332.2285053.683 kN

uresM 33.872368.5822353.29012 kNm



37

The following Interaction Diagram shows the capacity of the shear wall. The

diagram also shows all design loading combinations upon which the wall is

subjected to. Since all loading combinations fall inside the interaction diagram

the shear is capable of resisting all the applied loading combinations.

Design strengh Interaction diagram for

shearwall section

0, 34061

21327, 23592

21177, 17096 71231, 16994

71253.5,23533

0

10000

20000

30000

40000

50000

60000

0 20000 40000 60000 80000 100000

Mu (kN.m)

Nu

(kN

)



38

Design of Shearwall-Frame Under Lateral Loads

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Transcript of Design of Shearwall-Frame Under Lateral Loads