13 - (Seismic) Lateral Loads Effects (Eng. a. Farghal)

Introduction. 2Page

Table of Contents.

Equivalent static load method. 12Page

Lateral Loads( ) Seismic

Steps of Calculating Seismic Load. 25PageWind Loads. 37PageSystems resisting lateral loads. 48PageDesign of Shear wall. 57PageDrift of structures due to seismic loads 66PageExamples. 70Page

By: Eng. Ahmed Farghal.

Page No.

CDownloading or printing of these notes is allowed for personal use only.Commercial use of these notes is not allowed.

. All copyrights reserved.Copyright

( )

&Wind

Lateral Loads

Wind loads.

Seismic loads.

1. Seismic Loads:

( )

Seismic sources

4- Failure of roof of large cave3- Volcanoes2- Movements of faults

1- Movements of tectonic plates

5- Mankind effect (explosion, fill and in-fill of dams ..... etc.)

6- Undefined reasons

2-

1-

Page No.



( )

Introduction.

Types of surface waves

Direction of propagation

Particle motion

Rayleigh waveLove wave

Direction of propagation

Particle motion

Love wave

Rayleigh wave

Classification of earthquakes

Focal depth > 300 km1- Deep focus earthquakes Epicentral

distance

Hypo

centr

al

distan

ce300 km > Focal depth > 70 km2- Intermediate focus earthquakes

Focal depth < 70 km3- Shallow focus earthquakes

Focusor Hypocenter

Epicenter

Focaldepth

Page No.



( )

Methods of measuring earthquakes magnitude

The Richter Scale

- The magnitude of most earthquakes is measured on richter scale.

- It was invented by Charles F. Richter in 1934.

- The richter magnitude is calculated from the amplitude of the

largest seismic wave recorded for the earthquake, no matter what

type of wave was the strongest.

The Mercalli Scale

- It is another way to measure the strength of an earthquake.

- It was invented by Giuseppe Mercalli in 1902.

- This scale uses the observations of the people who experienced

the earthquake to estimate its intensity.

Page No.



( )

3- Time History Analysis.

2- Multi-Modal Response Spectrum Method.

1- Equivalent Static Load (Simplified Modal Response Spectrum).

Methods of analysis of structures under seismic load

Page No.



( )

Fundamental period 1

T < TC1 4 T

Solid SlabSec 3

Sec 1Core

Sec 4Flat Slab

Sec 2 Shear wall

Sec 1

Sec 2

Sec 3

Sec 4

H

Uniform shape 3

Uniform statical system4

Page No.



( )

(L / L )x y

(e )o(x , y

e > 0.15 L

Page No.



( )

H> 0.15 H

13 LL

2L

1L

L1 L2-L1

< 0.20 L1 L3+L < 0.20

L

Page No.



( )

H< 0.15 H

13 LL

2L

1L

L L2-L < 0.30

L1 L3+L < 0.50

LL

L1 L2-L1

< 0.10

Page No.



( )

2- Multi-Modal Response Spectrum Method

Time (Sec)

Sd (T)

Response Spectrum Curve

Response Spectrum CurveDisplacement ( Rotation

3- Time History Analysis.

Page No.



( )

(Simplified Modal Response Spectrum Method)

Equivalent static load method:

Plan

Side ViewElevationL

L

B

F

F1 2

F

( More Critical )

B

XY

Z

F1

Critica

l More

b b

b

b2

X ,YManual

More Critical

EFx

EFy

XY

0.3

0.3

EE =T Fx +

EFx

EFy

0.3 EFy EE =T Fx + 0.3 EFy

Page No.



( )

Sd (T ) lwhere:

Fb = Ultimate base shear force

Fb = 1Wg

Sd =(T )1 (T )1Response Spectrum

g = Gravitational acceleration

Response Spectrum

Time (Sec)

Sd (T)

2.5 a Sg g1

a Sg g1

TB TC TD 4.0

Response Spectrum CurveDisplacement ( Rotation

Sd (T)1

T1

T

Response Spectrum

0 < T < TB

< T < TCTB

< T < TDTC

< T < 4.0 :TD

:

:

: Sd (T) = ag g1 S23 +

TTB

( 2.5 Rh - 23 )

Sd (T) = ag g1 S2.5 R

h

Sd (T) = ag g1 STT

C2.5 R

h > ag g10.20

Sd (T) = ag g1 STC2.5

Rh > ag g10.20

TD2

Curve

+

Page No.



( )

Sec

Zone Design acceleration g(a )

Zone 1Zone 2Zone 3Zone 4Zone 5AZone 5B

0.10

a g =

= Periodic time of different mode shapesT

B T, C T, D S, a, gT

where:

(Different mode shapes)

Page No.



( )

g0.125 g0.15 g0.20 g0.25 g0.30 g

Page No.



( )

2526

2728

2930

3132

3334

3536

37

2537

2223242526272829303132

22232425262728293031

32Z

one

150 100

125

150

200

250

Zon

e 2

Zon

e 3

Zon

e 4

Zon

e 5A

Zon

e 5B

P G A ( c m / s e c )

2

Oas

is

2627

2829

3031

3233

3435

36

WEST

DESE

RT

EAST

DESE

RT

Dak

hla

Oas

is

Kha

rga

RED S

EA

Bah

ariy

aO

asis

Siw

e

Asw

an

Idfu K

om O

mbo

Lux

or

Qen

a

Soha

g

Asy

ut

El M

inya

ME

DIT

ER

RIA

N S

EA

Qus

eir

Safa

ga

Hur

ghad

a

Ras

Muh

amm

ad

Mar

sa A

lam R

as B

anas

Hal

aib

SINAI

GULF

OF SU

EZ

GULF OF AUABA

Ben

i Sue

f

Al F

ayum

Giz

a

Cai

ro

Ism

ailia

El M

ansu

raT

antaD

aman

hur

Dam

ietta

Ala

mei

n

Ale

xand

ria

Mat

ruh

Salu

mPo

rt S

aid

El A

rish

Bah

ariy

a

Kos

etta

Subsoil ClassABCD

S TB TC TD1.00 0.05 0.25 1.201.35 0.05 0.25 1.201.50 0.10 0.25 1.201.80 0.10 0.30 1.20

For Type (1)

SubsoilClass Soil Description

A

B

C

D

TB =T, CTD =

elastic response spectrumspectrum

S = Soil Factor

B T, C T, D S,T

Rock

Dense

Soil

Soil

Meduim

Loose

Soil

TypeSoil

Page No.



( )

g1 = Importance factor

Type of Structures

I

II

III

IV

Importance

Category

1.40

1.20

1.00

0.80

ImportanceFactor (g )1

Subsoil ClassABCD

S TB TC TD1.00 0.15 0.40 2.001.20 0.15 0.50 2.001.25 0.20 0.60 2.001.35 0.20 0.80 2.00

For Type (2)

Page No.



( )

R = Response modification factor

R

:4.503.502.00

:5.004.504.50

:7.005.00

:6.005.00

3.003.50

:

Response modification factor5.00 (R)

Page No.



( )

Page No.



( )

R=

Res

pons

e m

odif

icat

ion

fact

or

R

:4.

503.

502.

00

:5.

004.

504.

50

:7.

005.

00

:6.

005.

00

3.00

3.50

:

Res

pons

e m

odif

icat

ion

fact

or5.

00(R

)

R

R =

4.5

R =

5.0

Fram

es w

ith B

raci

ngR

.C. S

hear

Wal

ls o

r C

ores

NO

R.C

. She

ar W

alls

Duc

tile

Fram

es

Non

Duc

tile

Fram

es

R =

7.0

R =

5.0

Fram

es w

ith B

raci

ng

R.C

. She

ar W

alls

or

Cor

es

or C

ores

or

Bra

cing

OR

+D

uctil

e Fr

ames

Non

Duc

tile

Fram

es

R =

6.0

R =

5.0

h = Damping factor corrected for horizontal response spectrum

Type of StructureSteel with Welded ConnectionsSteel with Bolted ConnectionsReinforced ConcretePrestressed Concrete

h

1.201.051.001.05

Reinforced Masonry Walls 0.95

Damping factor1.00 (h)

Prestressed Concrete

Page No.



( )

Ct HT1 =3/4

where:

Ct = Factor depend on structural system and material

Structural SystemSteel moment resisting frames 0.085

Ct

0.075

Reinforced concrete moment resisting frames (Space frames)

All other buildings

0.050

=T1 Fundamental period of the building

& framesCombinations of (cores or shear walls)

Non-ductile frames (flat slabs)Ductile frames (beams & columns)

Cores or Shear walls

where:

Fb = Ultimate base shear force

Sd =(T )1 (T )1Response Spectrum

g = Gravitational acceleration

H = Height of the building from foundation level H

Fb

Response Spectrum

Page No.



( )

Sd (T ) lFb = 1Wg+

Range T1

Frames + Shear wall + Shear wall Core 0.05C = )( 0.05C = )(

Non-ductile frames (Flat Slab)0.075C = )(

t t

t

Ductile frames (Beams + Slab)0.075C = )( t

Sd (T )1Response Spectrum

B T, C T, D S, a, gT

Response modification factor5.00 (R)

Damping factor1.00 (h)

Prestressed Concrete

1Importance factor (g )

Page No.



( )

T

0 < T < TB

< T < TCTB

< T < TDTC

< T < 4.0 :TD

:

:

: Sd (T) = ag g1 S23 +

TTB

( 2.5 Rh - 23 )

Sd (T) = ag g1 S2.5 R

h

Sd (T) = ag g1 STT

C2.5 R

h > ag g10.20


Rh > ag g10.20

TD2Sec

W =

l = Correction factor

T < TC1 2 l = 0.85If

T > TC1 2 l = 1.00If

w D.L. + a L.L.+

Total weight of the structure above foundation level

i

i S (w )

( i )w =i( ) Floor Area+=

ss- D.L. ( g ) & L.L. ( P ) are working loads ( working loads )

w g + a P+i ( ) Floor Area+= s s

NOTE

a0.25

0.50

1.00

a =

g ts = g + F.C.s c + wallsP L.L.s = If given in kN/m

2

+ O.W. of beams and columns+ O.W. of beams and columns

Page No.



( )

W =

NOTE

Page No.



( )

Steps of Calculating Seismic Load:

1 Calculate Ct HT1 =3/4

2

3 B T, C T, D S, a, gT1 ,g,h,RSd (T )1

Subsoil ClassABCD

S TB TC TD1.00 0.15 0.40 2.001.20 0.15 0.50 2.001.25 0.20 0.60 2.001.35 0.20 0.80 2.00

For Type (2)

Subsoil ClassABCD

S TB TC TD1.00 0.05 0.25 1.201.35 0.05 0.25 1.201.50 0.10 0.25 1.201.80 0.10 0.30 1.20

For Type (1)

Zone Design acceleration g(a )

Zone 1Zone 2Zone 3Zone 4Zone 5AZone 5B

0.10

Type of Structures

I

II

III

IV

Importance

Category

1.40

1.20

1.00

0.80

ImportanceFactor (g )1

1.00 (h)

5.00 (R)

Structural System Ct

0.075

Reinforced concrete moment resisting frames (Space frames)

All other buildings

0.050

& framesCombinations of (cores or shear walls)

Non-ductile frames (flat slabs)Ductile frames (beams & columns)

Cores or Shear wallsH

SubsoilClass Soil Description

A

B

C

D

Rock

Dense

Soil

Soil

Meduim

Loose

Soil

TypeSoil

Page No.



( )

Sd (T ) lFb = 1Wg+

g0.125 g0.15 g0.20 g0.25 g0.30 g

Range T1Sd (T )1

4

T < TC1 2l = 0.85T > TC1 2l = 1.00

i SW = (w )

ss- D.L. ( g ) & L.L. ( P ) are working loads ( working loads )

NOTE

a0.25

0.50

1.00

5 Get

6 Calculate W

Page No.



( )

T

0 < T < TB

< T < TCTB

< T < TDTC

< T < 4.0 :TD

:

:

: Sd (T) = ag g1 S23 +

TTB

( 2.5 Rh - 23 )

Sd (T) = ag g1 S2.5 R

h

Sd (T) = ag g1 STT

C2.5 R

h > ag g10.20


Rh > ag g10.20

TD2Sec

w D.L. + a L.L.+i




FbF3

F1

2F

F5F4

F1-3

F1-1

1-2F

F1-5F1-4

F2-3

F2-1

2-2F

F2-5F2-4

F3-3

F3-1

3-2F

F3-5F3-4

sw 1

sw 2

sw 3

sw 4

( i )

F

( i )

In Plan( )

F =ii = 1

i = nbiw iH

w Hi i

In Elevation ( )

Distribution of lateral force on each floor

+

where:F =i

H =i

ssD.L. ( g ) & L.L. ( P ) are working loads ( working loads )

H5

H4

H3

2H

H1

h 8

h 7

h 6

h 5

h 4

h 3

h 2

1h

F =b Ultimate base shear force

F4-3

F4-1

4-2F

F4-5F4-4

sw1 sw2

sw4sw3

Page No.



( )

Sd (T ) lFb = 1Wg+

w D.L. + a L.L.+i




Story ForcesTotal seismic load

iiS

i F = S F

12M = M + Q hbase U.L. 1

23M = M + Q h2 2

34M = M + Q h3 3

45M = M + Q h4 4

56M = M + Q h5 5

67M = M + Q h6 6

77M = M + Q h8 7

88M = Q h8

1Q = Q + F2 1

2Q = Q + F3 2

3Q = Q + F4 3

4Q = Q + F5 4

5Q = Q + F6 5

6Q = Q + F7 6

78Q = Q + F7

8Q = F 8

Overturning MomentShear Diagram

Load Diagram

7F

6F

5F

4F

3F

2F

F1h 1

2h

3h

4h

5h

6h

7h

8h

1H

H2

3H

4H

5H

F3

F1

2F

F7F6F5F4F

+

+

+

+

+

+

+

+

F8

112base U.L.M M + Q +h=F +H= 11F +H= + 22F +H + 33F +H + ...............

8F

b

b

Page No.



( )

i S F i S Q

Q = F

1

2

3

4

5

6

7

8

Q (kN)iiF (kN)H (m)Floor i

2H

H1

H3

H4

H5

H6

H7

H8

F1

2F

3F

4F5F

6F

7F8 8

787Q = Q + F

Q = Q + F6 7 6Q = Q + F5 6 5Q = Q + F4 5 4Q = Q + F3 4 3Q = Q + F2 3 2Q = Q + F1 2 1

iM (kN.m)

888

M7 8 7 7

6676

5565

4454

3343

2232

112base U.L.

w (kN)i No. i i

8w

w7

w6

w5

w4

w3

w2

w1

8 8

77

66

55

44

33

22

11w

( )

w & H i i

base shear (F )

-

S H i

+H

w +H

w +H

w +H

w +H

w +H

w +H

w +H

= M + Q +h

M M + Q +h

M M + Q +h

M M + Q +h

M M + Q +h

M M + Q +h

=

=

=

=

=

M M + Q +h=

M = Q +h

iiw +HS

where:Floor No. =

( i ) H =iH5

H4

H3

2H

H1

h 8

h 7

h 6

h 5

h 4

h 3

h 2

1h

iiw +H =

i = 1

i = niw iH

w Hi i

=i = 1

i = niw iH

w Hi i

iiw +H&iw

=i = 1

i = niw iH

w Hi = 1

i = niHHi

w +H

8F

b

i i

Page No.



( )

w D.L. + a L.L.+i




Q M - ii

i

23

. (w )

H

overturning (M ) -

< 1.5

B2

W overturningM

B

Check Overturning

( i )F =i

iiS

base U.L.M F +H= 11F +H= + 22F +H + 33F +H + ....................

Q =iM =i

iiSbase U.L.M F +H=overturningM =

stabilityM = W +

where:

W = Total weight of structure

Stability MomentOverturning Moment

=Factor Of Safety

=F.O.S.overturningMstabilityM

7F

6F

5F

4F

3F

2F

F1

F (M =base U.L. )b23+

H

F =b

F F =ii = 1

i = nbiw iHw Hi i

+

8F

Fb

1.40 1.40

base U.L.MoverturningM = 1.40

= Ultimate base shear force

Page No.



( )

Sd (T ) lFb = 1Wg+

mSliding Force =

Check Sliding

Resisting Force = +W

where:W = Total weight of structure

= Coefficient of frictionmResisting Force

Sliding Force=Factor Of Safety < 1.5

W

Fb

b

m +W

F1.40

NOTE

Ultimate loads

bF&base U.L.M1.40Working loads

Check sliding and Check overturning

base U.L.MoverturningM = 1.40

F bF

1.40=sliding

Page No.



( )

4

10 =

40

m

40 m

20

m

Plan

Elevation

+

F.C.

A ten floor hospital located in Cairo with dimensions (20 x 40 m). Height of each floor is 4.0 m. Soil below the building is very dense

2- Calculate the story shear and overturning moment at each floor

3- Find the bending moment and shearing forces acting at base

Given that :

level and draw its distribution on the height of the building.

level of the building and draw distribution of shear forces.

thickness equal 0.3 m. Due to Earthquake loads , it is required to :

Example.

sand and its coeff. of friction is 0.3. All floors are flat slab of average

4- Check The Stability of the building against silding and

2= 2.0 kN/mWalls 2= 3.0 kN/mL.L. 2= 3.0 kN/m

0.30

1- Calculate the ultimate base shear force .

overturning.

Page No.



( )

Solution

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4

CShear wall t = 0.05Table

0.05 40T1 =3/4

+ = 0.795 sec.

According to Soil type and building location

Zone (3)Cairo Map ag= 0.15Table g

Soil type (B)Very dense sand Table

SSoil type (B)Table

Response spectrum curve Type (1)

= 1.35T = 0.05T = 0.25T = 1.20

B

C

D

1- Simplified Modal Response Spectrum

Page (9)

Page (7)Page (8)

Page (13)

Page (9)

Total Height of building (H) = 40 m

Check T < TC1 4

T

< T < TDTC

Sd (T ) = ag g1 STT

C2.5 R

h > ag g10.201

1

1

Rh

=Table g1 1.40

==

5.001.00

Sd (T ) = 0.150.250.795

2.5 5.00

0.20

1 g +1.40 +1.35 + +1.00

= 0.0446 g

+ 0.15 g +1.40ag g10.20 = 0.042= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.50 sec

w D.L. + a L.L.+s =w t + a L.L.+s =

Table a 0.50=

( g + F.C. )s c + Walls

w 0.30 + 0.50 3.0+s = ( + 2.0 )+ 3.025+ 14.0= kN/m2

w 14.0Floor = 20+w 11200Total = 10+

40+ 11200= kN112000= kN

Sd (T ) lFb = 1Wg+

0.0446= g + 1.0112000

g

4995.2= kN

Page (14)

Page (10)

Page (15)

+

Fb

NOTEtav tav

Page No.



( )

47227.35

3632.87

10535.33

139865.60

119884.80

100267.29

81376.35

63575.27

32695.85

iM (kN.m)

20344.09

908.2240.010 908.22

817.4036.09 1725.62

4+

4036

8+

+ii Hw

Hiw ib i = n

i = 1

i =F F

m

90.82

181.64

272.47

363.29

454.11

544.93

635.75

726.57

220=

iHHi

i = n

i = 1ii HwHiw i

i = n

i = 1

=

No.

4995.20

4904.38

4722.74

4450.27

4086.98

3632.87

3087.94

32.0

28.0

24.0

20.0

16.0

12.0

4.0

8.0

iFloor H (m) F (kN)i iQ (kN)

8

7

6

5

4

3

2

1

2452.19

2- Distribution of lateral force on each floor

kN4995.2Shearing force at base =

12

48

24

1620

12+

28 32iH

36+ + +28+32 +20+24

)F (4995.2

ii = 1

i = 10

H = 40

1

=iF 22.705 iH

=i 220

16

Page No.



( )

Moment Diagram

3632.87

10535.33

81376.35

32695.85

20344.09

63575.27

47227.35

139865.60

119884.80

100267.29

33600=0.30

4- Check overturning

Safe

> 1.5= 9.4==Resisting ForceSliding ForceFactor Of Safety

336003568

112000+=

kN= 35681.404995.2=

3- Check sliding

kN.m99904=1.40 =139865.6

Mbase U.L.

1.40 =MoverturningMbase U.L.

1.40Fb

W+Resisting Force =

Sliding Force =

m

908.22

817.40

726.57

635.75

544.93

454.11

363.29

272.47

181.64

90.824995.20

4904.38

Safe

> 1.5= 11.2

1120000= kN.m

kN.m139865.6

4450.27

4086.98

3632.87

3087.94

2452.19

1725.62

Load Diagram Shear Diagram

=

4722.74

112000

=

=B2 + 220

=

=

W

Resisting MomentOver Turning Moment

Total +

Factor Of Safety

Resisting Moment

112000099904

908.22

Page No.



( )

Elevation

(0.00)

SuctionPressure

(Suction)

eC

= +

0.8

eC

= -

0.5

eC = - 0.8

eC = - 0.7

C =

- 0.

5e

C = - 0.7e

C = 0.8 + 0.5 = 1.3e

(Pressure)

eC

= +

0.8

2P e e

(0.00)

Plan

Wind load acting on the structure

2. Wind Loads.

qC k+ + kN/m

P where:

e

C e

k = Factor of exposure

Ground Roughness Length

Zone (A): Open exposure

Zone (B): Suburban exposure

Zone (C): City center exposure

Page No.



( )

=

=

=

where:

0 - 1010 - 2020 - 3030 - 5050 - 8080 - 120

Zone

Ground roughness

Height (m)length

k

120 - 160160 - 240

0.05 0.30 1.00

1.001.151.401.601.852.102.302.50 1.85 1.55

1.301.151.001.001.001.001.001.00

1.001.001.051.301.501.70

A B C

NOTE

q 2q 0.5 V C Cr t s kN/m2

V (m/sec)

(kN/m )2

(Open exposure or Suburban exposure or City center exposure)(Ground roughness length = 0.05 or 0.30 or 1.00)

k(Zone A)(Most critical case)

1000

(Most critical case)

Page No.



( )

=

V (m/sec)4239

36

33

30

r C t

C

1.00

1.201.401.601.80

t

1.801.00

1.80

:

NOTEC1.00 t

Page No.



( )

= kg/m 31.25= Factor of topography

C s

Turbulence

1.00:

(kN/m)

1.100.95

0.81

0.68

0.56

2

NOTEC

C1.00 s

&C s tV

2

q 0.5 V C Cr t s1000

=0.5

1000+1.25 +1.00 +1.00 2V

=2V6.25 +10

-4q

Page No.



( )

= Structural factor

=

q

.

Wind load (F) +area kN

e qC k+ +=e

+area

where:area = Area subjected to wind

kN

NOTE

Distribution of wind

2P( )H

2kN\m1P( )

kN\m2

2

2

kN\m3P( )

( ) 4P4 kN\m

= e 22 +CP q+k=

kkP1 qke 1C +

2=1.15

1=1.00+

=

k=

kP4P3

qq

C +k 4C 3 1.40=e 3k+

=1.60e 4

+

+

Zone A(Most critical case)

Page No.



( )

= P

2P e e qC k+ + kN/m=h

3h

2h

1h

Wind load distribution

(0.00)

Storey forces due to seismic load

-

-

-

-

-

-

Seismic loadsWind loads

Difference between wind & seismic loads:

F b+ )(+4 = h44P

h

kN

kN F

F2

1

2

1h

h4H

H3

H2

1H

kN

kN F

4F

3

h

3

4

b

H

e qC k+ += +areaF

F b+ )(+3 = h33PF b+ )(+2 = h22PF b+ )(+1 = h11P

iiS F +HTotal Moment at base =

11F +H= + 22F +H + 33F +H + ....................overturningM

Working loads


NOTE

-Ultimate loads -Working loads

Page No.



( )

S.L. L.L. D.L.

1.12 D.L. + L.L. + S.L.

0.8 [1.4 D.L. + 1.6 L.L. + 1.6 W.L.]

Factored loads of Ultimate Limit Design Method:

where:

a

W.L. = Seismic load = Live load= Dead load

= Wind load

Ultimate Load (U):

U =

(U)1.4 D.L. + 1.6 L.L.

a0.25

0.50

1.00

a =

Page No.



( )

6.0

Plan

6.0

4.0

Sec . Elevation

3.5

3.5

4.0

3.0

16

= 48

m

+

The shown figure of a store house which lies at Cairo. The building consist of a ground , mezzanine and 16 repeated floors and two

1- Calculate the wind load acting on the building .

Example.

basements. It is required to :

6.0

6.0

6.0

6.0

6.0 6.0 6.0 6.0

2- Check The Stability of the building against sliding and overturning.

F.C.

Given that :

2= 2.0 kN/mWalls 2= 3.0 kN/mL.L. 2= 3.0 kN/m

t = 0.20 msaverage

Page No.



( )

1.635

0.884

1.238

1.414

371.4

1.0170.68 == 1.01.3 +k +e + q +=P C

=eP2kN/m+k+e (q )C

+ kN+ = 371.4

Total wind force +

2084.4=

265.2 += 305.1

kN

+

+ kN

kN

101.017

0.884 =

=

265.2

305.1=

=

30

30

+

+10

1.635 + 30+6=

+ kN

kN

20

294.3

1.414

1.238

=

= 848.4

=

=

30

30

10

+

+

5k +

1

+4

2

k+3k+k

5P

PPPP

5F = h(+

+eC5 =

1

+e

e +

=4

=2

3

CC

= +eC

b)

1.85+ +0.68 =q = 1.3=0.68

0.68 =

=

=

1.601.3 +

1.40+1.3qq

++

=q +1.151.3 + 0.68 =

2kN/m

56 m

10 m

10 m

10 m

6 m

20 m

2

2

2

2

kN/m

kN/m

kN/m

kN/m

+4P4F = h(+ b)

+3P3F = h(+ b)

+2P2F = h(+ b)

+1P1F = h(+ b)

C =e 1.3=+0.8 0.5

Solution

2

q =0.5 V C Cr t s

10002= 33+ = 0.68 kN/m

2

Zone A ( More Critical )

305.1 kN

371.4 kN 2kN/m

2

21.017kN/m

0.884kN/m

kN/m

2

1.238

1.414kN/m

21.635

7 m

265.2 kN

848.4 kN

294.3 kN

848.4+ 294.3+

6.25 +10-4

Page No.



( )

79312.2

22+

=+60+ 294.3265.2 12 + 305.1 += +

H2++= HF F211 +

kN.m

47++ 848.432+ 371.4 +

H4+

H i+F H F3 3 4

Fi ++

56 m

10 m

10 m

10 m

6 m

20 m

32

22

12

305.1 kN

371.4 kN 2kN/m

47

2

21.017kN/m

0.884kN/m

kN/m

60

2

1.238

1.414kN/m

21.635

7 m

265.2 kN

848.4 kN

294.3 kN

H5+ F5 +

Check overturning

79312.23510000

Resisting Moment

Factor Of Safety

+Total

Over Turning MomentResisting Moment

W

=

= 2B

= 44.2= > 1.5Safe

=( Overturning Moment )Total Moment at base

w t + L.L.s = g + F.C.av c + Wallsw 0.20 + 3.0s = + 2.0 + 3.025+ 13.0= kN/m

2

w 13.0Floor = 30+w 11700Total = 20+

30+ 11700= kN234000= kN

Resisting Moment =302+234000 kN.m= 3510000

Page No.



( )

Check sliding

m

Sliding Force =

Resisting Force = +W

wind force 2084.4= kN

= +234000

2084.470200Factor Of Safety Sliding Force

Resisting Force= = 33.7= > 1.5

Safe

0.30 = 70200

Working loads


NOTE

Page No.



( )

Core

wallShear Frames

- Frames + Shear walls + Cores- Core + Shear walls- Frames + Core- Frames + Shear walls

Core (Tube)Shear walls

4- Combination between different systems

3- Frames

1- Shear walls or cores

Systems resisting lateral loads.

2- Coupled shear walls

Ductile frames (Beams + Columns)

Page No.



( )

XC.G.C.G.

x12x

C.M.

2A A 1

Columns - Core - Shear Walls

Columns - Core - Shear Walls

)(Slabs - Beams

(Stiffness( )

(

)

C.M.((C.R.) )

)

Center of Mass (C.M.) & Center of Rigidity (C.R.)

-

Center of mass (C.M.):

Center of rigidity (C.R.):

of ( F ).It is the center of gravity of area and it is the point of application

It is the point where the force ( F ) is resisted.

ww ++ 21

where:

iix = Distance between C.G. of A & datumX = Distance between C.M. & datum

X = iw ixA i

+

X = 11A +x + 22A +x1w + 2w +

1A + 2A1w + 2w +

iA +iw +

iiw = Weight of floor at this part of floor A

NOTE

C.M.

C.M.

b

b

Datum

C.M.

C.M.

Page No.



( )

(C.M.))(C.R.

Torsion ( ( C.R.) )

C.M.(

Shear wall (

)

)

(

() Core

C.R. )

(Torsion)

C.R. C.M.C.R. C.M.

Page No.



( )

FbFb

( Inertia ) Bending Moment Shear force

Case (b): Unsymmetrical shear walls in one direction (C.M. = C.R.)

Case (a): Symmetrical shear walls (C.M. = C.R.)

L

C.M.

4

swi = n

i = 1

i =M M+IiI

+F= iIF

i = 1

i = niI

To Get C.R.

C.M.C.R.

2x1

2x x1x

sw2sw1 2 1sw

IiC.R.xIX =1 i i

2x

Msw

C.R.

ex

1

C.R.

1 T

xx3

x

sw24

3swsw

1- Shear walls or cores

base&

where:

F = Force acting at the baseF = Force acting on the shear wall no. ( i )I = Moment of inertia of the shear wall no. ( i )

M = Bending Moment acting at the baseM = Bending Moment acting on the shear wall no. ( i )

ii

i

i

base

+F= 1IF1

1I + 2I2 2

+F= 2IF2

1I + 2I2 2

xC.M.

X =+ 22I +x

C.R.

+ 33I +x + 44I +x11I +x

+ 2I + 3I + 4I1I

C.R.X2 =e C.M.X

where:

Distance between C.M. & C.R.

i

b

b

b

b

b

Fb

Fb

Datum

I2I2I1 I1

I2I1 I3 I4

C.R. C.M.

C.R. C.M.

Lateral Force (F )

Page No.



( )

e =

( F )

e+

C.M.

x2

xx

Ii = n

i = 1

Fi =

1

M

iIi xi = n

T

x 2Ii = 1 ii

i+

IF +

+e 0.053 =e

sw2

M FT =

1sw

C.R.

e

MT

x4

sw3

3

sw4

* *

i

i x = (C.R.)

where:

= 1IF1

1I + 2I1I1 x

x 2I 11-

+ 3I + 4I + xI 22 xI 33 xI 44+ +

( )

( )2

( )2

( )2

( )++F

MT=2IF2

1I + 2I2I2 x

x 2I 11-

+ 3I + 4I + xI 22 xI 33 xI 44+ +( )2

( )2

( )2

( )++F

MT=3IF3

1I + 2I3I3 x

x2

I 11+

+ 3I + 4I + xI 22 xI 33 xI 44+ +( )2

( )2

( )2

( )++F

MT=4IF4

1I + 2I4I4 x

x2

I 11+

+ 3I + 4I + xI 22 xI 33 xI 44+ +( )2

( )2

( )2

( )++F

Ii = n

i = 1

i+

IF

i

MT+

iIi xi = n

x 2Ii = 1 ii

( )M T+

b

b

b

b

b

b

b

b

Fb

I2I1 I3 I4

Page No.



( )

L

.C.R. Zone ( 1 ) & Zone ( 2 ) -

(Shear wall) ( F = 1 kN ) -

.Zone ( 2 ) - ve

C.M. Zone ( 1 ) + ve -

I( )

42

y

F by

C.R.*

y

1

ye

C.R.x

+bx ye1x

Fbx

sw

F

2sw

sw4

3swxL

y3

ex

C.M.

C.R.x y

x

Ly

&

(exy) & )

y)(I

e( ))

(

x (I( )

Eccentricity

C.R.

NOTE

C.M.

sw2

1sw

C.R.

MTsw3

sw4

Zone ( 1 )Zone ( 2 )

Ii = n

i = 1

% of each Shear Wall = iIi x

i = n

x 2Ii = 1 ii

i+

I1 +

i ( )1 +( e+ * )

M = % of each Shear Wall M+i base

Case (c): Unsymmetrical shear walls in both direction

*+by xeF

C.M.x

C.M.y

b

Fb

(+ ve)(- ve)

Page No.



( )

((

C.M. 3e

))1(sw) &sw( 4 (Fby

)C.M.C.M.

((

Shear wallShear wall

(()

))

C.R.y

exy2

x

y

1

1x

4 y

))+

x

I iC.R.xIX =1 i i

- C.R.X2 =e C.M.X

=+ 44I +x11I +x+ 4I1I

e++e 0.05 L3 =e M FT =* *

xx xy by

))2(sw) &sw( 3 (Fbx

C.R.Y =1

- C.R.Y2 =e C.M.Y

e++e 0.05 L3 =e M FT =* *

yy yx bx

x

y

Ii = n

i = 1

Fy =iI x

i = n

i = 1

+I

F + MT+

i = n

x 2Ii = 1 ii

( ) = x2I 11 xI 44+( )

2

( )x x xi = n

y 2Ii = 1 ii

( ) = y2I

22yI 33+( )

2

( )y y x

ixi

xiby yx 2I ii ( )x[ + y

2I ii ( )y

xi

Ii = n

i = 1

Fx =iI y

i = n

i = 1

+I

F + MT+iyi

yibx xx

2I ii ( )x[ + y

2I ii ( )y

yi

1sw

2sw

sw4

3swxL

Ly

x

x x x

x x

I iyI i i =

+ 33I +y22I +y+ 3I2Iy

y y y

y y

Fby

Fbx

Page No.



( )

NOTE

Symmetrice = 0.05Lmin

( F )

e+

Ii = n

i = 1

Fi =iIi x

i = n

x 2Ii = 1 ii

i+

IF +

0.05 L=e M FT =* *

i

i x = (C.R.)

where:

( )MT+

bb

b

e minMore Critical

0.05 L= 1IF1 1

I1 xx

2I 11

++ xI 22( )

2

( )++F

0.05 L= 2IF21I + 2I

2I2 x+ ++F

b

b( )2

1I + 2I( )2 [ ]2

x 2I 11 + xI 22( )2

( )[ ]2

+F

+F

b

b

(

(

)

)

sw

C.R.

1x

sw2sw1 2 1sw

Fb

2x 2x

min e

1x

C.M.

I2I1 I2 I1

sw

C.R.

1x

sw2sw1 2 1sw

C.M.

2x 2x1x

min e

Fb

I2I1 I2 I1

= 0.05L = 0.05L

OR

D.L.T.L.

sw sw2sw1 2 1sw

I2I1 I2 I1

T.L.D.L.

sw sw2sw1 2 1sw

I2I1 I2 I1

C.M.T.L.C.M.T.L.

Page No.



( )

swsw

m= 0.8932

15

1sw =I +12

sw()( Inertia

I 2.4478

13.3 %

0.20 %

=

=

I

sw 2 =i = 1

i = 1

Ii = 6

I

sw 3 =i = 1

i = 6

I

2.4478 0.3255

2.4478 0.0052

0.3255

0.0052

36.50 %

2 ( 0.8932 + 0.3255 + 0.0052 )

=

)0.253

=

=

=2swI

4.0=3sw

I =

I

i = 6

i = 1

i = 61sw

I= 0.8932

12

(12+

m4= 2.4478

m4

4m

)3

2.510.0 m

3.5

21

sw

4.03sw

4.03

21sw sw

)(3.53

0.25 4

2.53.5

Illustrative Example

For the shown figure , if the thickness of all shear walls = 25 cm .It is required to :

1- Calculate the percentage of force ( P ) acting on each wall

Solution

+ )(2.53

0.25 Fb

NOTEShear wall

Fb

Page No.



( )

N ( ) due to own weight ,weight of walls and floors .

sw

Shear Wall

F=Fswisw +

2

i

i = 1

i = ni I

sw4

Plan

5sw

I

sw3

1sw

Elevation

sw6

2

Shear walls solved as a cantilever totally fixed in foundation and

subjected to moment ( ) due to lateral force and normal force M

Design of Shear wall.

isw

Page No.



( )

2F

F 6-2

1F

4 swF sw 36

F

F

5

3F

4

2

1

F

Fsw 5

sw

3

2

swsw

1

6

4

5

6

FF

F

F

F 5-2

F 4-2

F 3-2

F 2-2

F 1-2

iF

+ The Normal Force Due to Seismic or Wind LoadsO.W of shear wall

+ The part from the floor which it carry number of floors

=S.L.

load

M

N =

+1.12 N + N + N

0.8 [1.4 N + 1.6 N + 1.6 N ]

a

D.L. L.L. W.L.

D.L. L.L. S.L.

1.12 M + M + M

0.8 [1.4 M + 1.6 M + 1.6 M ]

a

D.L. L.L. W.L.

D.L. L.L. S.L.

a0.25

0.50

1.00

a =

Base Seismic U.L.M

Cases of Load Combinations

Case (1)

Case (2)

Case (3)

N =

M =

N =

M =

1.4 N + 1.6 N D.L. L.L.

1.4 M + 1.6 M D.L. L.L.

N =

M =

where:

F Hi= i% of each Shear Wall + % of each Shear Wall +

=W.L.M Base Wind WorkingM F Hi= i% of each Shear Wall + % of each Shear Wall +

D.L.+L.L.

D.L.+L.L.+W.L.

D.L.+L.L.+S.L.

Page No.



( )

NOTE

Wind&SeismicManual

L.L.&D.L. Moments

Ultimate M S.L.

Working M W.L.

SAPZeroW.L.N = ZeroS.L.N =and

Frames Flat Slabs

A = A

Get

s

) b

Using interaction diagram

10 cu=\ssA = A ( F

minsA

1000.6

cA=

=t

>> I The column considered as a fixed columnb c

K = 3hc12EI

For bottom story

Page No.



( )

e minMore Critical

( frames ) ( F )

Ki = n

i = 1

Fi =iKFi x

i = n

x 2Ki = 1 iFi

Fi+

KF +

Fi ( )M T+

b

b+

i

where: x = (C.R.) (frames)

0.05 L= 1KF1 1

K1 xx 2K 11

++ xK 22( )

2

( )++F

0.05 L= 2KF2 2

K2 x+ ++F

b

b

1K + 2K( )2 [ ]2+F

+F

b

b

(

(

)

)1K + 2K( )2 x

2K 11 + xK 22( )

2

( )[ ]2

NOTE

Symmetrice = 0.05 Lmin

e+0.05 L=e M FT =* *

b

Fb Fb

OR

T.L.D.L.C.M.T.L.C.M.T.L.

1x2x 2x

min e

1x

1F 1F2F 2F

C.R.

C.M.

K1 K2 K2 K1

= 0.05L

1F 1F2F 2F

K1 K2 K2 K1

D.L.T.L.

1F 1F2F 2F

K1 K2 K2 K1

1x2x 2x

1x

min e

1F 1F2F 2F

C.R.

C.M.

K1 K2 K2 K1

= 0.05L

K

Page No.



( )

+3Q

4Q

5Q

6Q

Q 7

8Q

1Q

2Q

Shear diagram

D 1

Load diagram

2D3D

4D5D

6D7D

8Dweb drift

Drift of structures due to seismic loads

Total Drift = Web Drift + Chord Drift

1- Web drift

Drift diagram

Q % of each framei

i

Total web drift for each frame ( ) =

where:

Web drift for each frame ( ) =

= Story drift of each frameQ = Shear force acting on the storyi

Importance of drift:Is to satisfy serviceability requirements.

The drift should be limited to fulfill the safety requirements fornon-structural elements.

................) - -

Shear

(

D

iSD

KFi

+Q % of each frameiSKFi

iD

Page No.



( )

% of each Frame =

Case (a): Web drift for symmetrical rigid frames

DDDD 1 2 2 1

(For Symmetrical Rigid Frames)Ki = ni = 1

Fi

KFi

% of each Frame = +1 + 1 +( e+ * )K

i = n

i = 1

iKFi xi = n

x2

Ki = 1 iFi

Fi

K Fi ( )+ For Unsymmetrical

Rigid Frames( )

+Q % of each framei

iWeb drift for each frame ( ) =D KFi

+Q i

= KFi

Ki = n

i = 1

Fi

K Fi Q i= i = ni = 1

K Fi

as K = where K Story stiffness=si = n

i = 1K Fi s

Q i

i

Total web drift ( ) =

Web drift ( ) =D

iSD

K sQ iS

K s

frames

2FF2 F11F

C.M.C.R.

Fb

Page No.



( )

y (x) = ( x H - )sh K

w x2

2

x

y(x)

w

KN

/m

H

wind load ( w ) ( uniform load )

( web drift )

2D3D

4Dstraight line

D 1

Case (b): Web drift for unsymmetrical rigid frames

frames

h

+Q % of each framei

i

Total web drift for each frame ( ) =

where:

Web drift for each frame ( ) =D

iSD

KFi

+Q % of each frameiSKFi

% of each Frame = +1 + 1 +( e+ * )K

i = n

i = 1

iKFi xi = n

x2

Ki = 1 iFi

Fi

KFi ( )+ For Unsymmetrical

rigid frames( )

2F

C.M.

M

C.R.

T

1F 3F 4F

FbNOTE

w

w

w

Page No.



( )

+++

H

Drift diagram

Chord drift

B

2- Chord drift

Chord drift ( ) = w He

4

8EI

M base

where:I = Composite moment of inertia of columns at C.G.e

I = 2 [ I + A ( L ) ] + 2 [ I + A ( L ) ]e

A1

1LL2

2A

I21I

1 1 1 2 2 22 2

C.G

Bending

Moment Diagram

H500 600

Serviceability Requirements

base get wFiw HM = % of each frame M =

2

2

web drift chord drift

cD

NOTE

D = D + D > DTotal web chord allowable =

w

KN

/me

e

ee

Page No.



( )

4

8=32

m

7 3 = 21 m+

+

7

3 =

21 m

Plan

Elevation

+

All columns = 25 80All beams = 25 60

For the shown figure of a residential building lies at Aswan . The building consist of ( 8 ) repeated floors . L. L. = 2.0 kN/m &

1- Calculate the story shear at each floor level and draw its

2- Find the bending moment and shearing forces acting at base

Given that :

distribution on the height of the building.

level of the R.C. columns and draw distribution of shear force

Due to Earthquake loads , it is required to :

Example.

2

F.C. = 1.5 kN/m and the soil is loose sand.2

3- Check Stability of the building against over turning

++

and bending moment.

Slab thickness = 160 mm

Page No.



( )

Solution

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4

CR.C. Frames t = 0.075

0.075 32T1 =3/4

+ = 1.01sec.


Zone (2)Aswan Fig. (8-1) ag= 0.125Table (8-2) g

Soil type (D)Loose sand

SSoil type (D)


= 1.80T = 0.10T = 0.30T = 1.20

B

C

D

Time (Sec)

Sd (T)

TB TC TD 4.0

Sd (T )1

T1


P. (2/5 code)P. (1/5 code)

P. (5/5 code)


Check T < TC1 4

T

< T < TDTC

Sd (T ) = ag g1 STT

C2.5 R

h > ag g10.201

1

1

Rh

=g1 1.00

==

5.001.00

Sd (T ) = 0.1250.301.01

2.5 5.00

0.20

1 g +1.00 +1.80 + +1.00

= 0.0334 g

+ 0.125 g +1.00ag g10.20 = 0.025= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.60 sec

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=


w 0.16 + 0.25 2.0+s = ( + 1.5 )25+ 6.0= kN/m2

w 6.0Floor = 21+

w 3596Total = 8+

21+

28768= kN

NOTE

ts

Eq. (8-13)P. (3/5 code)

Table (8-9)P. (4/5 code)

kN35960.25

slab+

beams+8++ )(+25 210.6+

columns25++4 (+ )160.25 +0.8


=

tav

P. (5/5 code)

Page No.



( )

Sd (T ) lFb = 1Wg+

0.0334= g + 1.028768

g

960.85= kN

+

Fb

4697.49

21779.27

17935.87

14199.23

10676.11

7473.28

2455.51

iM (kN.m)

854.09

4+8+

+ii Hw

Hiw ib i = n

i = 1

i =F F

m

26.69

53.38

80.07

106.76

133.45

160.14

186.83

213.52

144=

iHHi

i = n

i = 1ii HwHiw i

i = n

i = 1

=

No.

960.85

934.16

880.78

800.71

693.95

560.50

400.35

32.0

28.0

24.0

20.0

16.0

12.0

4.0

8.0

iFloor H (m) F (kN)i iQ (kN)

8

7

6

5

4

3

2

1

213.52


12

48

24

1620

12+

2832iH

+28+32 +20+24

)F (960.85

ii = 1

i = 8

H =

1

=iF 6.6726 iH

=i 144

16

Page No.



( )

Moment Diagram

854.09

2455.51

17935.87

7473.28

4697.49

14199.23

10676.11

21779.27

3- Check overturning

kN.m15556.62=1.40 =21779.27

Mbase U.L.


213.52

186.83

160.14

133.45

106.76

80.07

53.38

26.69

Safe

> 1.5= 19.4

302064= kN.m

kN.m21779.27

934.16

880.78

800.71

693.95

560.50

400.35

Load Diagram

kN960.85

Shear Diagram

=

Shearing force at base =

960.85

28768

=

=B2 + 221

=

=

W

Resisting MomentOver Turning Moment

Total +

Factor Of Safety

Resisting Moment

30206415556.62

213.52

kN.m21779.27Bending moment at base =

Page No.



( )

2.0=L.L 2kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

kN/m

Sec . Elev6.0

6.0

6.0

6.0

6.0

6.0

4.0=L.L Void

Plan6.0 6.06.0 6.0

8.0

8.0

10.0

=L.L

=L.L

=L.L

4.0=L.L

4.0=L.L

4.0

4.0

4.0

=L.L

=L.L

=L.L

4.0=L.L

4.0=L.L

3.0

8 =

24

m4.

04.

03.

53.

52

2

2

+

2

2

2

2

2

2

2

2

The shown figure of a building which lies at Cairo. The building consist of two basements, ground floor and mezzanine used as

1- Calculate the equivalent seismic load acting on the building .


Example.

hospital and 8 repeated floors used as residential building.Live load

2- Draw lateral load ,shear and over turning moment diagram over

is 215 mm and Floor cover + walls = 2.5 kN/m . The soil is weak.

the height of the structure.

and floor heights are shown in elevation, the average slab thickness2

kN/m10.0=L.L 2

Page No.



( )

Solution1- Simplified Modal Response Spectrum

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4


Zone (3)Cairo Fig. (8-1) ag= 0.15Table (8-2) g

Soil type (D)Weak soil

SSoil type (D)


= 1.80T = 0.10T = 0.30T = 1.20

B

C

D

P. (2/5 code)P. (1/5 code)



C t = 0.05

0.05 39T1 =3/4

+ = 0.78sec.

P. (5/5 code)


Check T < TC1 4

T ag g10.201

1

1

Rh

==

5.001.00

Eq. (8-13)P. (3/5 code)

Page No.



( )

=g1 1.00

Sd (T ) = 0.150.300.78

2.5 5.00

0.20

1 g +1.40 +1.80 + +1.00 = 0.0727 g

+ 0.15 g +1.40ag g10.20 = 0.042= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.60 sec

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=




P. (5/5 code)

=g1 1.40Table (8-9)

P. (4/5 code)

=g1 1.40

NOTE More critical g1

a 0.50=Table (8-7)P. (5/5 code)

30Area = 864+ +6.0-30 6.0 =2m

void

D.L. = 0.215( + 2.5)25+ 7.875= kN/m2t g + F.C.s c + Walls =

w D.L. + a L.L.+floor = ( ) + area

&

Floor 4

Floor 2

Floor 1

11

3

w 7.875 + 10.0+floor = ( ) + 8640.50

w 7.875 + 8.0+floor = ( ) + 8640.50

11124=10260=

w 7.875 + 4.0+floor = ( ) + 8640.25 7668=

Floor 12 w 7.875 + 2.0+floor = ( ) + 8640.25 7236=

WTotal = +111124( + +210260 + +87668 + +16912 )

= 100224

kN

kN

kN

kN

kN

Page No.



( )

NOTE

L.L.~ a

Sd (T ) lFb = 1Wg+

0.0727= g + 1.0100224

g

7286.28= kN

+

Fb

+ii Hw

Hiw ib i = n

i = 1

i =F F


12 39.0 7236 282204 993.30 993.30 2979.90

11 36.0 7668 276048 971.63 1964.93 8874.68

10 33.0 7668 253044 890.66 2855.59 17441.46

9 30.0 7668 230040 809.69 3665.28 28437.30

8 27.0 7668 207036 728.72 4394.01 41619.32

7 24.0 7668 184032 647.75 5041.76 56744.60

6 21.0 7668 161028 566.78 5608.54 73570.24

5 18.0 7668 138024 485.82 6094.36 91853.31

4 15.0 7668 115020 404.85 6499.21 117850.14

3 11.0 10260 112860 397.24 6896.45 145435.94

2 7.0 10260 71820 252.79 7149.24 170458.28

1 3.5 11124 38934 137.04 7286.28 195960.26

= 2070090

No.Floor

iM (kN.m)iH (m) F (kN)i iQ (kN)iw (kN) iw H i

ii Hwi = n

i = 1 iFi = n

i = 1

= 7286.28

Page No.



( )

Load Diagram

Moment Diagram

Shear Diagram

2

1

3

10

6

5

9

7

8

4

12

11

993.30

971.63

890.66809.69

728.72

647.75

566.78

485.82

404.85

397.24

252.79

137.04

993.30

1964.93

2855.59

3665.28

4394.015041.76

5608.54

6094.36

6499.21

6896.45

7149.24

7286.28

2979.90

8874.68

17441.46

28437.30

41619.32

56744.60

73570.24

91853.31

117850.14

145435.94

170458.28

195960.26

2

1

3

10

6

5

9

7

8

4

12

11

Page No.



( )

F.C.

12.0 4.06.0

6.06.0

1 2 3

Walls 1.5 kN/m = 2

6.0

6.06.0

4.0

6.0

4 5 6

2 = 1.5 kN/m

The shown figure of a student house which lies at Alexandria. The building consist of 15 repeated floors & Floor height = 3 m. The

shear walls 300 mm thickness .The soil is medium dense sand.

Example.

building consist of flat slab with average thickness 200 mm and

4- Design the cross-section at the base of the Shear wall on axis 3 and draw its details of reinforcement .

Given that :

L.L. 2 = 4.0 kN/m

Plan

1- Calculate the static wind load acting on the building in

2- Calculate the equivalent seismic load acting in Y- direction.

Y - direction and show its distribution over the height .

3- Draw lateral load ,shear and over turning moment diagram over the height of the structure due to the critical lateral load .

It is required to :

Y

Page No.



( )

46518.375

15+

1.053

1.474

1.685

442.20

1.211

=

0.81 == 1.001.3 +k +e + q +=P C

=eP2kN/m+k+e (q )C

+ kN+ = 442.20

Total wind force

315.90 5 + 363.30 += +

H2++= HF F211 +

+

751.86=

315.90 += 363.30

kN

+

+ kN

kN

101.211

1.053 =

=

315.90

363.30=

=

30

30

+

+10

+ kN151.685

1.474

= 758.25

=

=

30

30

10

+

1

+4

2

k+3k+k

PPP

1

+e

e +

=4

=2

3

CC

= +eC

=0.81

0.81 =

=

=

1.601.3 +

1.40+1.3qq

++

=q +1.151.3 + 0.81 =

kN.m

2kN/m

45 m10 m

37.5++ 758.2525+ 442.20 +

H4+

H i+F H F3 3 4

Fi ++

10 m

15 m

2

2

2

kN/m

kN/m

kN/m

+4P4F = h(+ b)

+3P3F = h(+ b)

+2P2F = h(+ b)

+1P1F = h(+ b)

C =e 1.3=+0.8 0.5

Solution

2

q =0.5 V C Cr t s

10002= 36+ = 0.81 kN/m

2

Zone A ( More Critical )

363.30 kN

442.20 kN 2kN/m

2

21.211kN/m

1.053kN/m

2

1.474

1.685kN/m

315.90 kN

758.25 kN

15

5758.25+

6.25 +10-4

=( Overturning Moment )Total Moment at base

VAlexandria = 36 m/sec

10 m

25

37.5

1- The static wind load

Page No.



( )

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4

C t = 0.05

0.05 45T1 =3/4

+ = 0.87sec.


Zone (2)Alexandria Fig. (8-1) ag= 0.125Table (8-2) g

Soil type (C)Medium dense sand

SSoil type (C)


= 1.25T = 0.20T = 0.60T = 2.00

B

C

D

< T < TDTC

Sd (T ) = ag g1 STT

C2.5 R

h > ag g10.201

1

1

Rh

=g1 1.00

==

5.001.00


P. (2/5 code)P. (1/5 code)

P. (5/5 code)


Check T < TC1 4

T

Sd (T ) = 0.1250.600.87

2.5 5.00

0.20

1 g +1.00 +1.25 + +1.00

= 0.0539 g

+ 0.125 g +1.00ag g10.20 = 0.025= g < Sd (T )1O.K.

T < TC1 2 l = 0.85= 1.20 sec

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=



P. (5/5 code)

w 0.20 + 0.25 4.0+s = ( + 1.5 )+ 1.525+ 9.0= kN/m2

w 9.0Floor = 12+w 3240Total = 15+

30+ 3240= kN48600= kN

Sd (T ) lFb = 1Wg+

0.0539= g +0.8548600

g

2226.61= kN

+

Fb

24+27+

+ii Hw

Hiw ib i = n

i = 1

i =F F

m360=

iHHi

i = n

i = 1ii HwHiw i

i = n

i = 1

=


30++42+45 +36+39ii = 1

i = 15

H = 33 21+ 18+ 15+ 12+ 9+ 6+3+

Page No.



( )

iM (kN.m) No. iFloor H (m) F (kN)i iQ (kN)

iH)F (2226.61 1

=iF 6.185 iH

=i 360

15 45.0 278.33 278.33 834.98

14 42.0 259.77 538.10 2449.27

13 39.0 241.22 779.31 4787.21

12 36.0 222.66 1001.97 7793.14

11 33.0 204.11 1206.08 11411.38

10 30.0 185.55 1391.63 15586.27

9 27.0 167.00 1558.63 20262.15

8 24.0 148.44 1707.07 25383.35

7 21.0 129.89 1836.95 30894.21

6 18.0 111.33 1948.28 36739.07

5 15.0 92.78 2041.06 42862.24

4 12.0 74.22 2115.28 49208.08

3 9.0 55.67 2170.94 55720.92

2 6.0 37.11 2208.05 62345.08

1 3.0 18.56 2226.61 69024.91

kN.m49303.51=1.40 =69024.91


Overturning Moment ( wind )

The case of seismic load is the critical one

46518.375= kN.mOverturning Moment ( Seismic ) 49303.51= kN.m

Page No.



( )

Load Diagram

Moment Diagram

Shear Diagram

21

3

10

65

9

78

4

1211

1314

15

21

3

10

65

9

78

4

1211

1314

15

278.33

259.77241.22

222.66204.11185.55167.00148.44

129.89111.33

92.7874.2255.6737.1118.56

278.33538.10

779.31

1001.97

1206.081391.63

1558.63

1707.07

1836.951948.28

2041.062115.28

2170.942208.05

2226.61

834.98

2449.27

4787.217793.14

11411.3815586.27

20262.1525383.35

30894.2136739.07

42862.2449208.08

55720.92

62345.08

69024.91

Page No.



( )

=+ =

i = 1

M

For Shear Walls on Axis

+

+

By Area Method get the Normal force

+=3Msw

8.0

N sg =D.L. Area+

+69024.91

4.0(+ 0.3 +)

No of floors

( +6.0

o.w

8.0

+

3

=

1.6

3)

12

I = +5.4(2

0.30

0.30

2 =

=3

+12

(6.0

)3

+ 4.0(

I

Ii = 4

sw

sw

m4=

4

swI 5

swI 4

==

=)

m5.40

1.60

14.0

4

m

4.0

6.0

8.0

3

kN.m8438.56

15+)3.0 +25

2sw

4- Design of shear wall 2 3 4 5 61

3sw

4sw

5sw

C.R.

C.M.

0.05 L=e min += 0.05 30 = 1.5 m

emin

1.6= +1.6 1 + ++ 5.42 ( )+1.6 3.0+

6.0 6.0 6.0 6.0 6.0

1.6 + 5.42 ( )+ 3.0+2

9.0+2

= 0.1223 =

Ii = n

i = 1


i = n

x 2Ii = 1 ii

i+

I1 +

i ( )1 +( e+ min)

+1.51

12.23 %

% of Sw3

% of Sw3

0.01223

base

=

0.20= ( + 1.5 )+ 1.525+

g ts = ( g + F.C. )s c + Walls

8.0= kN/m2

ND.L. = 7110 kN

N sP =L.L Area+ No of floors+

= +4.0 )( +6.0 8.0 15+ = 2880 kN

+N sg =D.L. Area+ No of floorso.w +

Page No.



( )

5 16 / m \5 10 / m

12 28

5 8 / m \

4.0

0.3

2812

M

cu

2

2

0.289+108683.2= =Nu+300

( F

10 - 4

100

+

+

= 0.9

t +25Fcu b

s \ =

2.0 25

Ac0.6100 =

0.6+ +

s

=As

min

A = A

=

2.0=

F

t b

= mm6000

= 7200mm

4000

Get

cu

300

( 300 +

+ 10 )

4000

-4

) 4000

+

3

8438.56= +10 =u 0.0703300+25 +4000

2tb 2

2812

6

x

1.12 N + N + NaD.L. L.L. S.L.

1.12 M + M + MaD.L. L.L. S.L.

N =

M =

ult

ult

= 1.12 + + 0.25 2880 + 0 = 8683.27110 + kN

= 8438.56+0 = kN.m8438.56

Page No.



( )

All columns are (300 800)

St. = 360/520

1F

7.0 7.0 7.0

4.0

8.0

8.0

2L.L. = 2.0 kN/m

8.0

6.0

4.0

4.0

4.0

4.0

8.0 8.0

The stiffness of exterior frames is twice the stiffness of interior

F2F3F4

1

frames

L.L. = 2.0 kN/m 2

2

2

L.L. = 7.0 kN/m 2

L.L. = 7.0 kN/m 2

F.C. 2 = 3.0 kN/m

For the given plan of a residential building, located in Hurghada (Seismic Zone 5A). The building consists of ground floor and

Example.

5 typical floors. The building is constructed on a weak soil.

Given that :

1- Calculate the shear base at ground floor level2- Draw the lateral load & shear distribution diagrams3- Calculate the seismic loads acting on the frame (F )

It is required to :

F 2 = 25cu N/mm

+ Partitions

+

t = 250sav mm

L.L. = 2.0 kN/m

L.L. = 2.0 kN/m

Plan Sec . Elev

Page No.



( )

Solution1- Simplified Modal Response Spectrum

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4


(Zone 5A)Hurghada ag= 0.25Table (8-2) g


SSoil type (D)


= 1.80T = 0.10T = 0.30T = 1.20

B

C

D

P. (2/5 code)



0.075 30T1 =3/4

+ = 0.96 sec.


Check T < TC1 4

T ag g10.201

1

1

Rh

==

5.001.00

Eq. (8-13)P. (3/5 code)

CR.C. Frames t = 0.075P. (5/5 code)Beams + Columns

Page No.



( )

=g1 1.00

Sd (T ) = 0.250.300.96

2.5 5.00

0.20

1 g +1.00 +1.80 + +1.00

= 0.0703 g

+ 0.25 g +1.00ag g10.20 = 0.050= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.60 sec

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=




P. (5/5 code)

16Area = 308+ +4.0-21 7.0 =2m

void

D.L. = 0.25( + 3.0)25+ 9.25= kN/m2t g + F.C.s c + Walls =

w D.L. + a L.L.+floor = ( ) + area

&

Floor 3

Floor 1

6

2 w 9.25 + 7.0+floor = ( ) + 3080.25 3388=

w 9.25 + 2.0+floor = ( ) + 3080.25 3003=

WTotal = +23388( + +410260 ) = 18788

kN

kN

kN

Sd (T ) lFb = 1Wg+

0.0703= g + 1.018788

g

1320.80= kN

+

Fb

Page No.



( )

+ii Hw

Hiw ib i = n

i = 1

i =F F


= 362824

No.Floor

iH (m) F (kN)i iQ (kN)iw (kN) iw H i

ii Hwi = 6

i = 1

6 30.0 3003 90090 327.96 327.96

5 26.0 3003 78078 284.23 612.19

4 22.0 3003 66066 240.50 852.69

3 18.0 3003 54054 196.77 1049.46

2 14.0 3388 47432 172.67 1222.13

1 8.0 3388 27104 98.67 1320.80

Load DiagramShear Diagram

21

3

654

327.96284.23240.50196.77172.67

98.67

327.96612.19

852.691049.46

1222.13

1320.80

3- Seismic loads acting on frame (F )1Plan is unsymmetric as C.M. = C.R.

- As the stiffness of exterior frames is twice the stiffness of interior

C.R.X =

- C.R. is at the center of the plan as K is symmetric for frames

( K = K & K = K )F1 F4 F2 F3 = 10.5 m221

frames ( K = K = 2 K = 2 K )F1 F4 F2 F3

Page No.



( )

12.0

4.0

XC.M.

14.07.0

C.M.

14.03.5

X = iw ixA i

+ iA +iw +C.M.

Datum

C.G.

C.G.A 1

2A

NOTEw

= ixA iiA +

iD.L. & L.L.

X = 7.0 +12.0C.M.+3.5 14.0 +16.0 +14.0+ = 11.14 m

- C.R.X=e C.M.X - 10.5=11.14 m= 0.64

+e 0.05 L=e * +0.64 0.05 = + 21.0 m1.69=

% of each Frame = +1 + 1 +( e+ *)K

i = n

i = 1

iKFi xi = n

x 2Ki = 1 iFi

Fi

K Fi ( )+

K = 2 K F1 F2

7.0 +12.0 14.0 +16.0+

i = n

i = 1KFi = KF1 KF2+ = =2 2 KF2 KF2+2 22+

K = K & K = K F1 F4 F2 F3

KF26

Page No.



( )

for F1F i

Zone (1)Zone (2)

8.0

8.0

7.07.07.0

10.53.5

TM

C.M.

0.64

C.R.

F1

= +1 + +10.5++2 ( )+ 10.5+

23.5+

2 +1.691% of F1 K F26K F1 K F1

K F1 K F2

= +1 + +10.5++2 ( )+ 10.5+

23.5+

2 +1.691K F26 K K F2K F22 K F2

2 F2

2

= 0.4096 = 40.96 %

frame zone (1)

= % F +1 F i = 0.4096 +F i

Load Diagram

134.33116.42

98.5180.6070.72

40.41

Page No.



( )

E6.06.0 6.06.0

A

B

C

D

21 3 4

6.0

5.0

5.0

5.0

5.0

65

F.C. Walls 1.5 kN/m = 22 = 1.5 kN/m

The figure shows a residential building which lies at Aswan. Itconsist of 15 repeated floors & Floor height = 3 m. The building

1- Calculate the equivalent static load acting on the building at

shear walls and cores 250 mm thickness .The soil is weak.

Example.

consist of prestressed slab with average thickness 200 mm and

3- Compute the torsion moment at the ground level.

Given that :

2- Determine the center of rigidity of the structure .

L.L. 2 = 6.0 kN/m

each floor and show its distribution over the height .

4- Calculate the percentage of lateral load acting on each shearwall and core.

It is required to :

Page No.



( )

Solution

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4

0.05 45T1 =3/4

+ = 0.87sec.




SSoil type (D)


= 1.80T = 0.10T = 0.30T = 1.20

B

C

D


P. (2/5 code)P. (1/5 code)


Check T < TC1 4

T ag g10.201

1

1

R

h

=

=

5.00

1.05

Eq. (8-13)P. (3/5 code)


Prestressed concrete

Page No.



( )

=g1 1.00

Sd (T ) = 0.1250.300.87

2.5 5.00

0.20

1 g +1.00 +1.80 + +1.05

= 0.0407 g

+ 0.125 g +1.00ag g10.20 = 0.025= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.60 sec


P. (5/5 code)

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=



w 9.5Floor = 20+w 5700Total = 15+

30+ 5700= kN85500= kN

Sd (T ) lFb = 1Wg+

0.0407= g + 1.085500

g

3479.85= kN

+

Fb

+ii Hw

Hiw ib i = n

i = 1

i =F F

iHHi

i = n

i = 1ii HwHiw i

i = n

i = 1

=


w 0.20 + 0.25 6.0+s = ( + 1.5 )25+ 9.5= kN/m2+ 1.5

Page No.



( )

15 45.0

14 42.0

13 39.0

12 36.0

11 33.0

10 30.0

9 27.0

8 24.0

7 21.0

6 18.0

5 15.0

4 12.0

3 9.0

2 6.0

1 3.0

24+27+

m360=

30++42+45 +36+39ii = 1

i = 15

H = 3321+ 18+ 15+ 12+ 9+ 6+

iH)F (3479.85 1

=iF 9.666 iH

=i 360

No. iFloor H (m) F (kN)i

3+

3- Center of rigidity and torsional moment

e6.06.0 6.06.0

a

b

c

d

21 3 4

6.0

5.0

5.0

5.0

5.0

65

C.M.

C.R.

For Shear walls

x x10.0

0.25

I 0.25 +10.0

= 20.83 m

=12

x

3

4

434.98

405.98

376.98

347.99

318.99

289.99

260.99

231.99

202.99

173.99

144.99

116.00

87.00

58.00

29.00

Load Diagram

434.98405.98376.98347.99318.99289.99260.99

231.99202.99173.99144.99116.00

87.0058.0029.00

Page No.



( )

x xy5.0

0.25

6.0

y 4.75 +0.25 +2 6.0 +0.25 +0.125+

= 1.66 m

= +2.6254.75 +0.25 +2 6.0 +0.25+

For Core

I 0.25 +4.75=12

x

3

+ 0.25 +4.75 + ( 2.625 - 1.66 )2

+2

6.0 +0.2512

3

+ 0.25 +6.0 +( 1.66 - 0.125)2

+ = 10.22 m4

X = iI ix+iIC.R.

X =

20.83 +0

C.M.

20.83 +6+ = 13.26 m= 20.83 +30+10.22 +21+20.83 +3 10.22+

L2

302

= =15.0 m

- C.R.X=e C.M.X - 13.26=15.0 m= 1.74

+e 0.05 L=e * +1.74 0.05 = +30.0 m3.24=

4- Percentage of lateral load carried by shear walls & coresI

i = n

i = 1


i = n

x 2Ii = 1 ii

i+

I1 +

i ( )1 +( e+ * )

e6.06.0 6.06.0

a

b

c

d

21 3 4

6.0

5.0

5.0

5.0

5.0

65

C.M.

C.R.

Zone (1)(+ ve)

Zone (2)(- ve)

Page No.



( )

20.66

For Shear Walls on Axis 1

= %

For Core

24.28

20.8320.83

For Shear Walls on Axis

2 =F 20.83- )3.24

311 +(+

2

+

=)

+2 2(+

213.26 + 16.747.26+

7.2620.83

%+ 10.22

2+7.74

38.73

For Shear Walls on Axis 6

= %

16.34= %

+ 10.22

20.8320.83

1 =F 20.83- )3.24

311 +(+

+ )+

2 2(+2

13.26 + 16.747.26+13.2620.83

+ 10.222

+7.74+ 10.22

20.8320.83

3 =F 20.83+ )3.24

311 +(+

+ )+

2 2(+2

13.26 + 16.747.26+16.7420.83

+ 10.222

+7.74+ 10.22

20.8310.22

4 =F 20.83+ )3.24

311 +(+

+ )+

2 2(+2

13.26 + 16.747.26+7.7410.22

+ 10.222

+7.74+ 10.22

Page No.



( )

Example.The shown figure of a residential building which lies at Cairo . Itconsist of a ground, mezzanine and 8 repeated floors and two

1- Calculate the story shear at each floor level .


basements. Live load and floor heights are shown in elevation.

2- Calculate the total lateral drift .

The soil below the building is weak.

24.0= kN/m

2

2

2.0

4.0

=

=

kN/m

kN/m

2

2

2

4.0

4.0

4.0

=

=

= kN/m

kN/m

kN/m

4.0= kN/m2

2

L.L

L.L

L.L

L.L

L.L

L.L

L.L

+5 6 m =30m

+

2

2

2

8.0

8.0

4.0

=

=

=

kN/m

kN/m

kN/m

2

210.0= kN/m

10.0= kN/m

L.L

L.L

L.L

L.L

L.L

+4 6

m =

24 m

3.0

12

= 36

m

4.0L.L = kN/m

3- Draw the distribution of web drift along the height of the building .

F.C.= mm160 2 = 1.5 kN/m

Given that :

Slab thickness

++700 )

2600 )

((

All Column are

All Beams are 22100=E N/mm

300300

Sec . Elev

Plan

Page No.



( )

Solution

Sd (T ) lFb = 1Wg+

Ct HT1 =3/4

CR.C. Frames t = 0.075

0.075 36T1 =3/4

+ = 1.10sec.



Soil type (D)Loose sand

SSoil type (D)


= 1.80T = 0.10T = 0.30T = 1.20

B

C

D


P. (2/5 code)P. (1/5 code)

P. (5/5 code)


Check T < TC1 4

T ag g10.201

1

1

Rh

==

5.001.00

Eq. (8-13)P. (3/5 code)

=g1 1.00Table (8-9)

P. (4/5 code)

Page No.



( )

Sd (T ) = 0.1250.301.10

2.5 5.00

0.20

1 g +1.00 +1.80 + +1.00

= 0.0307 g

+ 0.125 g +1.00ag g10.20 = 0.025= g < Sd (T )1O.K.

T > TC1 2 l = 1.00= 0.60 sec P. (5/5 code)

w D.L. + a L.L.+s =w t + a L.L.+s =

a 0.25=



Floor 1

w (Floor = 24+ 30+ 0.3slab

+beams

+

6++ )(+25 240.6+columns

25+ + 4.5+300.3 +0.7

w 0.16s = ( + 1.5 )25+ 5.5= kN/m2

0.25+ 10.0+

= 7116.75 kN

&Floor 2 3

Floor 4 11

Floor 12

5.5 )

w (Floor = 24+ 30+ 0.3+

+

6++ )(+25 240.6+25+ + 3.0+300.3 +0.7

0.25+ 8.0+= 6520.50 kN

5.5 )

w (Floor = 24+ 30+ 0.3+

+

6++ )(+25 240.6+25+ +3.0+300.3 +0.7

0.25+ 4.0+= 5800.5 kN

5.5 )

w (Floor = 24+ 30+ 0.3+

+

13 - (Seismic) Lateral Loads Effects (Eng. a. Farghal)

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Transcript of 13 - (Seismic) Lateral Loads Effects (Eng. a. Farghal)