Lecture Notes

LECTURE NOTES

Prepared by

Dr. Ashraf A El Damatty, Ph.D., P. Eng. Professor

Department of Civil & Environmental Engineering The University of Western Ontario

London, Ontario, Canada

Notes can not be copied or reproduced without approval of Dr. Ashraf El Damatty

Acknowledgement

The author would like to acknowledge the contribution of the following people in

preparing this set of notes.

Dr. Maged Youssef in developing the section describing the seismic requirements

of reinforced concrete structures, Mr. Mohamed Semelawy in preparing the

example for seismic design of RC moment resisting frames, Mr. Mahmoud Hassan

in preparing the example for seismic design of steel structures, Mr. Ahmed

Hamada in preparing the modeling section, and Mr. Mohamed Darwish and Mr.

Mohammad Siddique in typing the notes.

List of Proposed References

1) General, good coverage of code Title: Elements of Earthquake Engineering and Structural Dynamics By: Andre Filiatrault Publisher: Polytechnic international press ISBN: 2-553-01021-4 2) Structural Dynamics Author Chopra, Anil K. Title Dynamics of structures: theory and applications to earthquake engineering / Anil

K. Chopra. Publisher Englewood Cliffs, N.J. : Prentice Hall, 1995. 3) Steel Structures Author Bruneau, Michel, Ph.D. Title Ductile design of steel structures / Michel Bruneau, Chia-Ming Uang, Andrew

Whittaker. Publisher New York : McGraw-Hill, 1998. Author Canadian Institute of Steel Construction Title Handbook of Steel Construction Publisher Universal Offset Limited, 9th Edition, 2007. 4) Concrete and Masonry Structures Authors Thomas Paulay, and M. J. N. Priestley Title Seismic Design of Reinforced Concrete and Masonry Buildings/ Thomas Paulay,

and M. J. N. Priestley Publisher Willey- Interscience, 1992 Author Cement Association of Canada Title Concrete Design Handbook Publisher Ottawa, 3rd Edition, 2006

CHAPTER 1

EARTHQUAKE GROUND MOTIONS CHARACTERISTICS

Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-1

1- Causes and Effects of Earthquakes The earth is divided to six major tectonic plates; these are the Eurasian,

Pacific, American, African, Indian, and Antarctic. The vast majorities of

damaging earthquakes originate at, or adjacent to, the boundaries of tectonic

plates, due to relative deformations at the boundaries. Because of the nature

of the rough interface between adjacent plates, a stick-slip phenomenon

generally occurs, rather than smooth continuous relative deformation, or

creep. The relative deformation of the adjacent plates is resisted at the rough

interface by friction. When the induced stresses exceed the frictional

capacity of the interface, slip occurs releasing the elastic energy stored in the

rock in the form of shock waves propagating through the medium at the

ground-wave velocity.

The inertial response of structures to the ground accelerations resulting from

the energy released during fault slip is the primary interest to structural

engineers.

2- Seismic Waves

• The rupture point within the earth’s crust represents the source of

emission of energy. It is known as “hypocenter”, “focus”, or “source”

• The “epicenter” is the point on the earth’s surface immediately above

the hypocenter.

• The “Focal depth” is the depth of the hypocenter below the epicenter.

• The “Focal distance” is the distance from the hypocenter to a given

reference point.


The energy released by earthquakes is propagated by different types of

waves. The body waves, originate at the rupture zone and include P waves

(primary or dilatation waves) and “S” waves (secondary or shear waves).

• The “P waves” involve particles moving parallel to the direction of

propagation of the wave.

• The “S waves” involve particles moving perpendicular to the direction

of propagation of the wave.

“P” and “S” waves have different characteristic velocities Vp and Vs. The

velocities usually satisfy the following relation:

s

p

VV

≈√3

As a consequence, the time interval ∆T between the arrival of P and S waves

at a given site is thus proportional to the focal distance Xf, hence:

Xf = Vp ∆T/(√3-1)

Recording the P-S time interval at three or more sites enables the epicentral

position to be estimated as shown in Fig. 1

132

−

∆ pVT

X2

133

−

∆ pVT

131

−

∆ pVT

X3

X1

X1, X2, and X3 are recording stations

Fig. 1 Determination of Epicenter


3-Characteristics of Earthquakes

The effect of earthquakes on buildings depends on the seismic ground

motion that the structure experiences.

In general, there are six components of seismic ground motions; three

translation and three rotational components. Current strong motion

measuring instruments only record three translation components. There is

no reliable measurement on the rotational components at present.

Seismic ground motion is very complex. There is a continuous effort to

understand what are the relevant parameters that denote the damage

potential of such motions to buildings. Factors that affect the damage

potential are:

• Peak ground acceleration (PGA)

• Peak ground velocity (PGV)

• Peak ground displacement (PGD)

• Duration of Strong Shaking

• Frequency Content

The current opinion is that one needs at least two parameters to describe the

damage potential; one parameter denotes the intensity of shaking and

frequency content; and the other denotes the duration of shaking.

The combination of level of shaking and frequency content is reflected in a

single quantity called “Response Spectrum” which will be described later.


a) Magnitude

The accepted measure of magnitude is the Richter scale which is related to

the maximum deformation of the surface-wave motion recorded by a

standard Wood-Anderson seismograph located at a distance of 100 km from

the epicenter.

The accepted relationship between energy released E, and Richter magnitude

M is:

log10E = 11.4 + 1.5M

It is clear that an increase in one Richter unit leads to about 30 times

increase in the energy released.

(b) Intensity

Earthquake intensity is a subjective estimate of the effects of an earthquake

and is dependent on peak acceleration, velocity, duration, and frequency

content.

The most widely used scale is the modified Mercalli, MM, scale which was

refined by Richter in 1958.

The effective range is from MM2, which is felt by persons at rest on upper

floor of building to MM12 where damage is nearly total. A listing of the

complete scale is given below:


Masonry A : Good workmanship, mortar, and design; reinforced, especially laterally, and bound together by using steel, concrete, etc.; designed to resist lateral forces. Masonry B : Good workmanship and mortar; reinforced, but not designed in detail to resist lateral forces. Masonry C: Ordinary workmanship and mortar; no extreme weaknesses like failing to tie in at corners, but neither reinforced nor designed against horizontal forces. Masonry D: Weak materials, such as adobe; poor mortar; low standards of workmanship; weak horizontally. Modified Mercalli Intensity Scale of 1931 (abridged and rewritten by C. F. Richter)


4- Characteristics of Earthquake Accelerograms (a) Accelerograms Accelerographs record ground accelerations in optical or digital forms as

time history records. Integration of the records enables velocities and

displacements to be estimated. Three different time history accelerations are

shown in Fig. 2

Fig. 2 Time History Earthquake Records


(b) Vertical Acceleration

Vertical accelerations recorded by accelerographs are generally lower than

corresponding horizontal components and frequently are richer in low-period

components. It is often assumed that peak vertical accelerations are

approximately two-thirds of peak horizontal values.

However, there is an increasing evidence that peak horizontal and vertical

components are similar for near-field records.

(C) Influence of Soil Stiffness

It is generally accepted that soft soils modifies the characteristics of strong

ground motion transmitted to the surface from the underlying bed rock. Soft

soils usually result in amplification of long-period components and reduction

of the short-period ones.

Fig. 3 Comparison of lake bed (1-3) and rock (4-6) Accelerographs, Mexico City 1985, Paulay and Priestley (1992)


Fig. 3 shows accelerations recorded at adjacent sites on rock and on medium

depth lake deposits in the 1985 Mexico earthquake. The epicenter of this

earthquake was approximately 400 km from Mexico City and the peak

bedrock acceleration was about 0.05g. It can be noticed from Fig. 3 that this

acceleration was amplified about five times by the elastic characteristics of

the lake bed deposits. A modified ground motion, having predominant

energy in the period range of 2 to 5 sec was generated. As a consequence,

buildings with natural periods in this range were subjected to extremely

strong motions with many failures.

(d) Geographical Amplification

Geographic features may have a significant influence on local intensity of

ground motion. In particular, steep ridges may amplify the base rock

accelerations by resonance effects.

A structure built on top of ridge may thus be subjected to intensified

shaking.

Fig. 4

200 m 1000 m

50m No damage

Extensive damage Buildings abandoned


5- Attenuation Relationships

The attenuation relationship represents the reduction in peak ground

acceleration with distance from epicenter.

Three major factors contribute to the attenuation:

1) The energy released from an earthquake may consider to be radiated away

from the source as a combination of spherical and cylindrical waves. The

increase in surface area of the waves as they move away from the source

implies that accelerations will decrease with distance as the sum of a number

of terms proportional to Re-1/2, Re-1, Re

-2 and ln Re ; where Re is the distance

to the point source or cylindrical axis.

2) The energy is reduced due to material attenuation or damping of the

transmitting medium.

3) Attenuation may result from wave scattering at interfaces between

different layers of material.

For small-to-moderate earthquakes, the source could reasonably be

considered as a point source, and spherically radiating waves would

characterize attenuation.

For large earthquakes, with fault movement over several kilometers,

cylindrical waves might seem more appropriate. Hence attenuation

relationships for small and large earthquakes might be expected to be

different.

Typical attenuation relationships taken the form:

43

210 )( C

eMC CReCa +=


where: a0 is the peak ground acceleration

M is the Richter magnitude of the earthquake

Re is the epicentral distance

C1, C2, C3, and C4 are constants

Many relationships of the general form is given by above equations have

been proposed by various researchers.

6- Relation between Intensity and Ground Acceleration Fig. 5 shows the relationship between MM intensity and peak acceleration resulting from a number of studies.

Fig. 5 Relationship between Intensity and Peak Acceleration

(Paulay and Priestley (1992))

To some extent the scatter exhibited by Fig. 5 can be reduced by use of

effective ground acceleration (EPA) which is related to the peak response


acceleration of short period elastic oscillation rather than the actual

maximum ground acceleration.

Despite the wide scatter in Fig. 5, there is an approximate linear relationship

between the logarithm of PGA and MM intensity, I. An average relationship

may be written as:

(PGA)Ave = 10-2.4 + 0.34I m/sec

Meanwhile, a conservative estimate for (PGA) can be evaluated using the

relationship:

(PGA) = 10-1.95 + 0.32I m/sec


7- Return Periods: Probability of Occurrence To assess the seismic risk for a given site, it is necessary to know not

only the characteristics of strong ground motion that are feasible for the

site, but also frequency with which such events are expected. It is

common to express this by the return period of an earthquake of given

magnitude, which is the average recurrence interval for earthquakes of

equal or longer magnitude.

Generally, larger earthquakes occur less frequently than small ones.

The probability of occurrence (inverse of the return period) of

earthquakes of different magnitude M can be expressed by the following

distribution:

λ(M) = α Ve-βM

λ(M) is the probability of an earthquake of magnitude M or greater

occurring in a given volume V per unit volume.

α and β are constants related to the location of a given volume V.

Fig. 6 shows data for different tectonic zones compared with predictions

of the above equation.


Fig. 6 Magnitude-Probability Relationship (Paulay and Priestley (1992))

8- Design intensity The intensity of ground motion adopted for design depends on the

seismicity of the area and also on the level of structural response

contemplated and the acceptable risk associated with that response.

Three levels of response are usually identified:

1) Serviceability limit state where building operations are not disrupted

by the level of ground shaking.

2) A damage control limit state where repairable damage to the building

may occur.

3) A survival limit state under an extreme event earthquake, where severe

and possible irrepairable damage might occur but collapse and loss of life

are avoided.

CHAPTER 2

RESPONSE OF A SINGLE DEGREE OF FREEDOM SYSTEM


2-1 Equation of motion of a SDOF Basic components of a dynamic problem:

1. Mass (m = W/g) 2. Applied load P(t) 3. Inertia force = FI 4. Damping force = FD 5. Elastic force = FS

Single Degree of Freedom Model:

Using D’Alembert’s principle; a dynamic system can be considered to be in equilibrium if the inertia force is included in the free body diagram. FI + FD + FS = P(t) Where:

FI = mass x acceleration xmdt

xdm &&== 2

2

FD = Dashpot constant x velocity = xc& FS = Spring stiffness x displacement = kx ∴ Equation of motion: )(tPkxxcxm =++ &&&

P(t)

C

K

M

x(t)

P(t)

FI

FD

FS


Elastic Stiffness for Elements: k = P/∆ → Force Required for Unit displacement 1) Springs in Series: P1 = P2 = P P1 = K1∆1 ; P2 = K2∆2 ; ∆ = ∆1 + ∆2

21

21

2

2

1

121 KKKK

KP

KP

PPPK+

=+

=∆+∆

=∆

=

21

21

KKKKK

+=∴

2) Springs in Parallel:

∆1 = ∆2 = ∆

∆∆+∆

=∆+

=∆

= 221121 KKPPPK

21 KKK +=∴

K1

P1 P2

K2 P, ∆

P, ∆ , K

P1, ∆1 , K1

P2, ∆2 , K2


3) Axial Stiffness of a Bar:

4) Torsion: 5) Beams in Flexure:

a) Cantilever:

b) Simple Beam:

P

L

EAPL

=∆

LEAK =∴

T

L

GJTL

=θ

LGJK =∴

P

L EI

PL3

3

=∆ 33

LEIK =∴

∆

P

L EI

PL48

3

=∆ 348

LEIK =∴

∆ ∆ ∆


c)

d)

e)

31

11

12LEIK = 3

2

22

12LEIK =

K1 and K2 are two springs in parallel (have the same displacement)

( )2131

2112 IIL

EKKK +=+=∴

P

L EI

PL12

3

=∆ 312

LEIK =∴

∆

EIPL3

3

=∆ 33LEIK =∴

∆

L

∆

P

∆

K2 K1


0 200 400 600 800 1000 1200 1400 1600 1800

0 2 4 6 8

0 2 4 6

Types of Dynamic Loading: a) Free Vibration:

b) Forced Harmonic Vibration:

c) Impulse Loading: (Driving of Piles) d) Random Vibration:

0X&

tPtP Ω= sin)( 0

(Machine Vibration)

P0

X0

P0

P(t)

P(t)

X(t)

P(t)

Time

Time

Time

Time


0 2 4 6 8

2-2 Free Vibration Response of a SDOF Equation of motion: 0=+kxxm&&

020 =+∴ xx ω&& ; m

k=2

0ω

Solution of the above D.E.: X(t) = Asin(ω0t) + Bcos(ω0t) Substituting with the initial conditions: X(0) = X0 and 0)0( XX && =

we obtain: X(t) = )+) 00 tosXtX ωωω

(csin( 00

0&

The above equation can be written in a compact form as:

X(t) = 20

2

0

0 XX+⎟⎟

⎠

⎞⎜⎜⎝

⎛ω

&

[ ])+0 φω tsin(

X(t)

0X&

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

00

01

/tan

ωφ

XX&

T = 2π/ω0

X0


2-3 Undamped Free Vibration of a Damped System Equation of motion: 0=++ kxxcxm &&&

0=++ xmkx

mcx &&&

020 =++ xx

mcx ω&&&

Define: 02 ω

ξmc

= (damping ratio)

Equation of motion: 02 200 =++ xxx ωξω &&&

For 0 < ζ < 1 The solution of the above equation becomes: X(t) = te 0ξω− [ ])+) tosAtA DD ωω (csin( 21

Where: 021 ωξω −=D

(Frequency of vibration of a damped system)

D

XXAωξω 000

1+

=&

02 XA =

The above equation can be written in a compact form as:

X(t) = tXe 0ξω− [ ])+φω tDsin(


-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 5 10 15 20

Time

X(t)

Response to Harmonic Motion

Equation of motion: Ω=++ tm

xxx cosP2 0200 ωξω &&& _____________(*)

The solution consists of a homogeneous solution and a particular one:

Homogeneous: XC(t) = te 0ξω− [ ])+) tosAtA DD ωω (csin( 21

Try a particular solution in the form: XP(t) = Csin(Ωt) + Dcos(Ωt)

TD = 2π/ωD

P(t) = P0cosΩt or P0sinΩt C

K

M

x(t)


Substituting for the particular solution into (*) and solving for the constants C and D we obtain: XP(t) = XP cos(Ωt – φ) Where:

2

0

22

0

20

0

21 ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω−

=

ωξ

ω

ωmP

XP

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω−

⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω

= −2

0

01

1

2tan

ω

ωξ

φ

Adding the complementary and particular solutions, the total response is:

[ ]2

0

22

0

20

0

21

21

cos((csin()( 0

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω−

)−Ω+)+)= −

ωξ

ω

φωωωξω

tmP

tosAtAetX DDt

= transient term + steady state term

Amplitude of forced response

The angle at which the response lags the applied load

Decays exponentially and occurs at the damped frequency of the structure

Takes place at the forced frequency. Lags behind the force by phase angle φ.


Practically in design the steady state response is only considered. Define DMF Dynamic Magnification Factor:

2

0

22

0

20

0

21

1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω−

=

ωξ

ω

ωmP

X P

i.e. 20

00

ωmP

KP

X St == DMFXX StP .=∴

2

0

22

0

21

1

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛ Ω−

=

ωξ

ω

DMF

Dynamic Magnification Factor

0

1

2

3

4

5

6

7

8

9

10

11

12

0 0.5 1 1.5 2 2.5 3Ω/ω0

Dyn

amic

Mag

nific

atio

n Fa

ctor

ζ = 0

ζ = 0.05

ζ = 0.1

ζ = 0.2


Example 2-1 A simply supported beam of length “L” carries an electric motor of

weight W = 12 KN at its mid – span.

The flexural rigidity “EI” of the beam is such that the static

deflection under the weight of the machine is 3 mm. The

equivalent viscous damping is such that after 10 cycles, the free

vibration amplitude due to initial displacement X0 is reduced to 60

% of X0. The motor operates at 600 RPM and causes a harmonic

force due to unbalance F0= 2.5 KN. Neglecting the mass of the

beam, determine:

a) The value of the undamped frequency ω0.

b) The value of the damped frequency ωD.

c) The magnitude of the maximum dynamic displacement “X”

for the given values of the parameters due to the operation of

the machine.

d) The maximum total deflection including gravity and dynamic

loading.


2-4 Equation of Motion of a SDOF under Earthquake Motion

The figure above shows a simulation for a SDOF system

subjected to an earthquake ground motion described by a time

history ground acceleration Ϋg(t). X(t) represents the displacement

response – relative to ground of this SDOF system.

The equation of motion of the SDOF is given as:

0))()(( =+++ kxxctxtym g &&&&&

)()( tymkxxctxm g&&&&& −=++ As such, due to earthquake, the SDOF is subjected to a dynamic force )()( tymtP g&&−= Dividing the previous equation by the mass “m”, we get:

X(t)

K

M

Ϋg(t)

C


)()( tyxmkx

mctx g&&&&& −=++

Defining the natural frequency of the system mk

=0ω and its

damping ratio02 ω

ξmc

= , the equation of motion can be written as:

xxx 2002 ωξω ++ &&& )(tyg&&−=

The solution of the above 2nd order D.E. can be obtained by

Time History Analysis

Response Spectrum Analysis

Provides the entire time response.

Can be applied to linear and non-linear

systems.

Involves significant computations.

Provides only maximum responses.

Applicable only to linear systems.

Simple


2-5 Time History Analysis Closed form mathematical solutions exist for the free

vibration response of a SDOF system as well as the responses to

harmonic and impulse loading. Dynamic problems which involve

forces that can not be expressed in simple mathematical forms

(example earthquake forces) can not be solved in a closed form.

For such problems, the response is obtained by applying numerical

integration techniques to the equation of motion. In general, the

numerical integration techniques can be categorised to:

1) Explicit Algorithms.

2) Implicit Algorithms.

One of the most widely applied implicit algorithms will be

discussed in details in this section.

Implicit Algorithms

Newmark method U

Wilson – θ method

Newmark method

We will assume:

( )[ ] txxxx nnnn ∆+−+= ++ 11 1 &&&&&& γγ _______________ (1) ( )[ ] 211

21 txxtxxx nnnnn ∆+−+∆+= ++ &&&&& ββ _______________ (2)


γ and β are implicit parameters that can be determined to obtain

integration accuracy and stability.

γ = 1/2 and β = 1/6 ÿ correspond to linear acceleration.

γ = 1/2 and β = 1/4 ÿ correspond to constant average acceleration.

Equation of motion at t = (n+1)∆t:

)(1111 tFkxxcxm nnnn ++++ =++ &&& _______________ (3) For constant – average acceleration scheme (γ = 1/2 and β = 1/4):

[ ]( )2/11 txxxx nnnn ∆++= ++ &&&&&& _______________ (4)

[ ]( )4/211 txxtxxx nnnnn ∆++∆+= ++ &&&&& _______________ (5) From (5):

[ ]( ) nnnnn xttxxxx &&&&& −∆∆−−= ++ 211 /4 _______________ (6) Substituting (6) into (4), we get:

( ) [ ]( )[ ]( )2//42/ 211 txttxxxtxxx nnnnnnn ∆−∆∆−−+∆+= ++ &&&&&&& ___ (7) Equation (7) can be simplified to give:

( )( ) nnnn xtxxx && −∆−= ++ /211 _______________ (8) Substituting (6) and (8) into (3), we get:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

∆+⎟⎟

⎠

⎞⎜⎜⎝

⎛+

∆+

∆+=⎟

⎠⎞

⎜⎝⎛ +

∆+

∆++ n

nn

nnnn x

txcx

tx

txmtFxk

tc

tm

&&&& 244)(24

211

2 ___ (9)


Step–By–Step Solution Using Newmark method (γ = ½, β = ¼)

A- Initial Calculations:

1. Evaluate k, m and c.

2. Initialize 00 ,, xx & and 0x&&

3. Select a time step ∆t.

4. Calculate: a0 = 4/∆t2, a1 = 2/∆t, a2 = 4/∆t, a3 = ∆t/2 .

5. Evaluate camakk 10ˆ ++=

B- For Each time step:

1. Calculate effective load at time t = (n+1)∆t; 1ˆ +nF

( ) ( )nnnnnnn xxacxxaxamFF &&&& +++++= ++120

11ˆ

2. Solve 11 ˆˆ ++ = nn Fxk

3. Calculate acceleration and velocity at t = (n+1)∆t

[ ] nnnnn xxaxxax &&&&& −−−= ++2

10

1

[ ]nnnn xxaxx &&&&&& ++= ++ 13

1


Example 2-2

The frame structure shown below is subjected to a strong

ground motion pulse with a peak acceleration A0 = 1.2g. The latter

is modeled as a triangular pulse load of duration t1 equal to half the

period of vibration of the structure.

Assuming zero damping, determine by numerical integration

the maximum displacement of the structure, Xmax, and also the

time, tmax, at which it occurs. Use a time step ∆t=T/10 sec.

W 250x33

X W = 200 KN

4.5

m

3.0

m

I = ∞

Ϋg(t) Ϋg(t)

A0

t1/2 t1 t

t1= 0.5T A0= 1.2g


2-6 Concept of Elastic Response Spectrum

As a central concept of earthquake engineering, the response

spectrum provides a convenient means to summarize the peak

response of all possible linear SDOF systems to a particular

component of ground motion.

Construction of an Elastic Response Spectrum

The following steps are conducted in order to construct the

response spectrum of a given ground motion.

1. Consider a SDOF subjected to a ground motion given by the

time history analysis acceleration Ϋg(t). The SDOF system

has a certain period kmT π

ωπ 22

0

== and a damping ratio ζ.

2. Equation of Motion:

0))()(( =+++ kxxctxtym g &&&&&

)()( tymkxxctxm g&&&&& −=++ Or xxx 2

002 ωξω ++ &&& )(tyg&&−=

x

K/2

C

K/2

M

Ϋg(t)

x is the relative displacement between the mass and the ground


Knowing the time history record Ϋg(t), the response of the structure x(t) can be evaluated using any of the numerical methods.

The maximum value Sd for the response displacement is recorded.

This maximum displacement Sd(T,ζ) depends on the period and the

damping ratio of the SDOF.

Sd(T,ζ) is called the spectral displacement.

3. The maximum velocity response Sv(T,ζ) can be obtained by

the relation: Sv(T,ζ) = ω0Sd(T,ζ)

4. The maximum acceleration response Sa(T,ζ) can be obtained

by the relation: Sa(T,ζ) = ω0Sv(T,ζ) = ω02Sd(T,ζ)

5. Steps (1) to (4) are repeated for different values for the

natural period of the system T as well as for different

damping ratios ζ.


The following observations could be noticed:

1. Systems with very short period, say T < Ta = 0.025 s have

Sa= üg0 and Sd is very small. Why??

2. Systems with very long period, say T > 15 s have Sd

approaches üg0 and Sa is very small. Why??

3. For short–period systems, Sa can be idealised as constant.

4. For medium–period systems, Sv can be idealised as constant.

5. For long–period systems, Sd can be idealised as constant.

6. Sa, Sv and Sd are plotted together on a log-log scale as

function of T and ζ.

Figure 1 Response Spectrum (ζ=0, 2, 5 and 10%) for El Centro earthquake

N.B. Each earthquake has its own response spectrum


Based on the above observations one can conclude the following: • The long period region is called the displacement –

sensitive region as the structural response is most directly related to ground displacement.

• The medium period region is called the velocity –sensitive region as the structural response is most directly related to ground velocity.

• The short period region is called the acceleration –sensitive region as the structural response is most directly related to ground acceleration.

Elastic Design Spectrum

Based on the analysis of a large ensemble of ground motions recorded on firm ground an elastic design spectrum was developed by researchers as shown.

Figure 2 Elastic Design Spectrum


The amplification factors for two different non exceedance probabilities, 50% (median) and 84.1% (median plus one standard deviation) are given in Table 1 for several values of damping.

Median

(50 Percentile) One sigma

(84.1 Percentile) Damping, ζ (%) αA αV αD αA αV αD

1 3.21 2.31 1.82 4.38 3.38 2.73 2 2.74 2.03 1.63 3.66 2.92 2.42 5 2.12 1.65 1.59 2.71 2.30 2.01

10 1.64 1.37 1.2 1.99 1.84 1.69 20 1.17 1.08 1.01 1.26 1.37 1.38

Table 1 Amplification Factors: Elastic Design Spectra


K

M

H

2-7 Seismic Response of Structures Using Response Spectrum for a SDOF System

The elevated tank shown in the opposite figure

can be treated as a SDOF system having a period T

given by:

kmT π

ωπ 22

0

==

The maximum response of the tank to an earthquake (e.g. El

Centro earthquake) can be obtained by applying the following

steps:

1. A damping ratio ζ is assumed for the structure.

2. Using T and ζ, the response spectrum of El Centro can be

used to evaluate the maximum relative displacement Sd

experienced by the mass “M” during the earthquake.

3. The maximum velocity Sv(T,ζ) = ω0Sd(T,ζ)

4. The maximum acceleration Sa(T,ζ) = ω0Sv(T,ζ)

5. The maximum base shear Vmax = MSa = Mω02Sd

6. The maximum overturning moment Mmax= MHSa= MHω02Sd


Example 2-3

Consider a steel spherical elevated water tank supported by a single circular pipe. The diameter of the tank is 5 m and it is made of steel 0.01 m thick. The circular steel pipe is 1 m in diameter with a wall thickness of 0.02 m wall thickness. The height of the water tank (center of tank to ground level) is 22.5 m. Assuming the structure can be treated as a SDOF system with the total mass concentrated at the water tank level and the foundation condition is taken as rigid,

a) Calculate the maximum displacement, base shear force and the overturning moment caused by the 1940 El Centro earthquake. (Assume 2% damping ratio)

b) How does the maximum displacement and base shear change if the thickness of the supporting pipe is doubled?

CHAPTER 3

SEISMIC ANALYSIS OF MULTI DEGREES OF FREEDOM STRUCTURES

Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-1

3-1 Multi – Degree of Freedom Systems a) Single Storey Building

• Since all columns have the same displacement, they can be

treated as springs in parallel:

333

123312LEI

LEI

LEIKK i +⎟

⎠⎞

⎜⎝⎛+== ∑

• By neglecting the mass of the columns compared to the mass

of the floor, the frame can be treated as a SDOF

C

K = ΣKi

M

312

LEIK = 3

3LEIK = 3

3LEIK =3

3LEIK = 3

12LEIK =


b) 2-Storey Shear Building

K1 is the stiffness of the 1st storey.

K2 is the stiffness of the 2nd storey.

2 DOF System

M1 P1(t)

U2

U1

K1

K2

P2(t) M2

K2

K1

C1 M1

C2 M2

U1 U2

K2(U2 – U1) K1U1

M1

P1(t) C2(Ů2 – Ů1) C1Ů1

M1Ü1 K2(U2 – U1)

M2 P2(t)

C2(Ů2 – Ů1)

M2Ü2


Equations of motion:

M1Ü1 + K1U1 + C1Ů1 - K2(U2 – U1) - C2(Ů2 – Ů1) = P1(t)

M2Ü2 + K2(U2 – U1) + C2(Ů2 – Ů1) = P2(t)

The two above equations can be written in matrix form as:

⎭⎬⎫

⎩⎨⎧

=⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−++

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡)()(

00

2

1

2

1

22

221

2

1

22

221

2

1

2

1

tPtP

UU

CCCCC

UU

KKKKK

UU

MM

&

&

&&

&&

Or:

[ ] [ ] [ ] )(tFUCUKUM =++ &&&

For earthquake excitations:

• P1(t)= -M1Ϋg(t)

• P2(t)= -M2Ϋg(t)


3-2 Undamped Free Vibration of MDOF Systems (Determination of Natural Frequencies and Mode Shapes)

[ ] [ ] 0=+ UKUM && _____________________________(1)

Assuming that: )sin( θωφ += tU

Or: =U )( θω +tie φ

UU 2ω−=∴ &&

Now equation (1) becomes:

[ ] [ ][ ] 02 =− φω MK (Eigen Value Problem) ________________(2)

The non-trivial solution of the above equation is obtained when:

det [ ] [ ] 02 =− MK ω (characteristic equation)

For n degrees of freedom, the above will lead to n values of ω →

ω1, ω2, ……… ωn.

Each value of ωr can be back-substituted into equation (2) to obtain

the equivalent mode shape φr.


M1

4m

4m

M2

10 m

Example 3-3

Find the natural frequencies and

mode shapes for the following two

storey reinforced concrete structure.

Where:

• M1 = M2 = 30,000 Kg

• Ic= 0.005 m4

• ES= 200,000 MPa


3-3 Mathematical Properties of the Mode Shapes 1. If [ ]K and [ ]M are positive definite matrices, therefore for n

degrees of freedom, ω12, ω2

2, ……… ωn2 are real roots.

2. If iφ and jφ are two different mode shapes

[ ] [ ] 0==∴ jT

ijT

i KM φφφφ

3. jφ form a complete set of vectors; i.e. any function can be

expressed by these shape functions.

4. Normalization of φ :

Evaluate [ ] jTjj Mm φφ=ˆ j

jj m

φφˆ1ˆ =∴ [ ] 1ˆˆ =∴ j

T

j M φφ

5. [ ] *ii

Ti mM =φφ [ ] *2

iiiT

i mK ωφφ =∴

6. If [ ]

nxn

n

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

φφφ

φ

ˆ........ˆˆ

ˆ21

→ Ortho-normalized modes

[ ] [ ][ ] [ ]IMT

=∴ φφ ˆˆ and [ ] [ ][ ]

nxnn

TK

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

2

22

21

ˆˆ

ω

ωω

φφ


3-4 Modal Analysis Procedure to Evaluate Seismic Response of MDOF Systems

[ ] [ ] [ ] 1111 )( nxnxnxnnxnxnnxnxn tFUCUKUM =++ &&&

Assume that:

nnnx tXtXtXU φφφ )(......)()( 22111 ++=

i.e. [ ] 11

1 )( nxnxn

n

iiinx XtXU φφ ==∑

=

By pre-multiplying the equation of motion by Tφ , the following

equation is obtained:

[ ] [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] )(tFXCXKXM TTTT φφφφφφφ =++ &&&

The product [ ] [ ][ ]φφ MT will lead to the following diagonal matrix:

[ ] [ ][ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

n

T

mm

mm

M000000

3

2

1

φφ

Where [ ] iTii Mm φφ=

Meanwhile, the product [ ] [ ][ ]φφ KT will also lead to the following

diagonal matrix:


[ ] [ ][ ]

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

2

233

222

211

000000

nn

T

mm

mm

K

ωω

ωω

φφ

Assuming that the product [ ] [ ][ ]φφ CT also lead to a diagonal matrix,

a total number of n decoupled equations of motion are obtained.

Each equation of motion corresponds to a certain mode shape “i”

and is given as:

i

iiiiiii m

tFxxx )(ˆ2 2 −

=++ ωξω &&& where [ ] 11 )()(ˆnx

Txnii tFtF φ=

For earthquake excitations:

)(........)()()( 21 tYMtYMtYMtF gnggT &&&&&& −−−=

As such, the equation of motion of the ith mode can be written as:

)(2 2 tYmLxxx g

i

iiiiiii

&&&&& −=++ ωξω

Where: 2

1ji

n

jji Mm φ∑

=

=

ji

n

jji ML φ∑

=

=1

φji is the component “j” of the mode “i”

iii mL /=Γ is called the modal participation factor for

mode “i”


Using the response spectrum technique:

( ) dii Sx Γ=max

( ) vii Sx Γ=max&

( ) aii Sx Γ=max&&

jidiji SU φΓ=max

jiviji SU φΓ=max

&

jiaiji SU φΓ=max

&&

jjiaiji MSF φΓ=max

ai

in

jjjiaii S

mLMSV

2

1max =Γ= ∑

=

φ

Since the maximum responses of the ith modes do not occur at the

same time, a certain rule for modal combination has to be adopted.

Modal Combination:

Let the modal maximum responses be: Q1, Q2,….Qj…

1. Square root of sum of squares (SRSS): ..... 222

21 +++= jQQQQ

2. Absolute Sum: .......21 +++= jQQQQ

3. Complete Quadratic Combination (CQC):

∑∑∑≠= +

+=ji ij

jin

ii

QQQQ 2

1

22

1 ε where ji

jiij ωω

ωωξ

ε+−

=1

e.g. for 2 modes: 212

2122

21

2

12

ε+++=

QQQQQ


Sequence of Modal Combination:

Mode 1 Mode 2 Mode 3 Combination

Forces

Shear

Moment


Example 3-4

Use the modal analysis procedure to evaluate the response of

the frame shown in example 3-3 to El Centro earthquake.


3-5 Earthquake Response of MDOF Using Time History

Analysis

The equation of motion for each vibration mode is given as:

)(2 2 tYmLxxx g

i

iiiiiii

&&&&& −=++ ωξω

The above equation can be solved using the Newmark’s method to

obtain xi(t). The nodal displacements associated with mode “i” can

be then obtained as follows:

jiiji tXU φ)(=

The time history variation of the nodal displacements due to all

mode shapes is then given as:

∑=

=n

ijiinxj tXU

11

)( φ

Also: ∑=

=n

ijiinxj tXU

11

)( φ&&&&

Knowing the time history variation of the displacements all other

quantities such as the internal forces can be evaluated.


Typical time history responses at the top level of a three storey

building due to each one of the three modes of vibration together

with the total time history response at the same level are shown

below.


The time history variations of the base shear force associated with

each one of the three modes of vibration as well as the time history

variation of the total base shear force are given below.

CHAPTER 4

CODE PROCEDURES FOR EARTHQUAKE RESISTANT STRUCTURES

Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-1

Inelastic Behaviour and ductility Most of the modern seismic design codes allow inelastic behaviour to occur in

structures when they are subjected to strong earthquake motions. A load reduction

factor “R” is usually applied to the earthquake load intensity.

• Structures are designed under earthquake loads which are less than what are

expected if the structures remained elastic.

• Inelastic behaviour is expected to occur when the structure is subjected to a

strong ground motion.

• If the structure possesses enough ductility, the energy of the earthquake will

dissipate through inelastic hysterisis loops that yielded members undergo.

Accordingly, ductility is a very important property that governs the earthquake

response of the structure.


Local ductility µl = ∆m/∆y


The global ductility µg of a building is usually defined by the following relation:

µg = y

g

∆

∆

where : ∆g is the maximum displacement that can be sustained by the

structure before collapse.

∆y is the displacement at which the first plastic hinge forms in the

structure.

∆g Global ductility


The relation between the load factor R and the global ductility µg can be

established using one of the following concepts:

1) Equal Displacement

2) Equal Energy

d

c

b a

F

∆

R = d/c µg = a/b

(b-a) c =1/2 d (ad/c)

(d/c)2 = 2(b-a)/a

R2 = 2(µg -1)

R = √2(µg-1)

d

c

b a

F

∆

R = d/c

µg = a/b

From similarity: µg = R


V = 0

)(RR

WIMTS

d

Eva

V 0

)0.2(RR

WIMS

d

Ev

Ta : Fundamental period of the structure

S (Ta) : Design spectral response acceleration

Mv : Factor to account for higher mode effect on base shear

IE : Importance factor

Rd : Load reduction factor

Ro : Calibration factor

W : Dead Load + 25% Snow Load + 60% Storage Load

Load Reduction Factor (Rd)

Global Ductility µg

Local Ductility, µl


• According to the NBCC 2005 Code, the seismic hazard at various locations

in Canada is defined by five ground motion properties:

Sa(0.2), Sa(0.5), Sa(1.0), Sa(2.0), and PGA

Sa(T) represents the 5% damped spectral acceleration for period “T” in seconds,

expressed as a ratio of the gravitational acceleration.

Seismic Data for Toronto and Montreal is given below:

Seismic Data Location

Sa (0.2) Sa (0.5) Sa (1.0) Sa (2.0) PGA

Toronto 0.26 0.13 0.055 0.015 0.17

Montreal 0.69 0.34 0.14 0.048 0.43

• Depending on the soil conditions at the site, the design spectral acceleration

S(T) is determined using the set of equations provided on the following

page.


Design Spectral Acceleration


Definition of Soil Type


• Mv depends on the period of the structure “Ta”. Mv increases with an

increase in Ta (for flexible buildings). It reaches a value of 2.5 for

Ta ≥ 2.0 seconds

• The values of Rd, Ro are given in Table 4.1.8.9 for different structure

systems.

Notice, that there is a limit on the height of the building for some of the lateral

resisting systems depending on the seismic zone, which is defined by the

parameters IEFaSa(0.2) and IEFvSa(1.0)


Estimate of Ta

Moment Resisting Frames => Ta = 0.085(hn)3/4 steel ------ (1)

= 0.075(hn)3/4 concrete

= 0.1N others

Braced Frame => Ta = 0.025hn ----------- (2)

Shear Walls => Ta = 0.05(hn)3/4 ---------- (3)

The period can be calculated using computer model leading to a value “Ta”.

However, “Ta” should not be taken greater than

(1) For Moment resisting Frame, 1.5Ta as calculated in (1)

(2) For Braced Frame, 2.0 Ta as calculated in (2)

(3) For Shear Walls, 2.0 Ta as calculated in (3)


Force distribution

Fx = ∑=

−n

iii

xxt

hW

hWFV

1)(

)(

Ft = 0.07Ta V

0.25V

For Ta ≤ 0.7 sec. => Ft = 0

Reduction in Moment

Mx = Jx ∑Fi (hi – hx)

Jx = 1.0 for hx ≥ 0.6hn

Jx = J + n

xh

hJ6.0

)1( − for hx < 0.6hn

J depends on Ta and the type of lateral resisting system. (Table 4.1.8.11)

Ft

hx

hn Fx


Definition of Irregularities


A structure is considered to be irregular if it has any of the type of irregularities

described in the above table.

Equivalent Static Forces Analysis may be used if any of the following criteria is

met:

a) IEFaSa(0.2) <0.35

b) Regular structure, with hn≤ 60.0 m and Ta ≤ 2.0 sec. (in both two

orthogonal directions)

c) Structure with structural irregularities of Type 1, 2, 3, 4, 5, 6, or 8 with

hn≤ 20 m and Ta ≤ 0.5 sec. (in both two orthogonal directions)

*Otherwise dynamic analysis has to be carried.


Accidental Torsion

Bx = δmax/δave

δmax = maximum storey displacement at the extreme points induced by equivalent

static force applied at distances ± 0.1Dnx

δave = average displacement at the extreme points induced by equivalent static

force applied at distances ± 0.1Dnx

When only equivalent static analysis is required (i.e. no dynamic analysis is

required)

Tx = Fx (ex + 0.1 Dnx)

Tx = Fx (ex - 0.1 Dnx)


Direction of Loading

• If components of the lateral system are oriented along a set of orthogonal

axes,

1) Independent analyses about each of the principal axis shall be performed.

2) Case of no orthogonal axis and IEFaSa (0.2) < 0.35,

Independent analyses about any two orthogonal axes are permitted.

x

y

100% 100%

Independent

permitted

100% 100%

Independent

permitted

x

y 100%

100%

Independent


3) Case of no orthogonal axis and IEFaSa (0.2) ≥ 0.35,

x

y

30% 100%

together

100% 30%

together

30%

together

100% 100%

together

30%


Dynamic Analysis Procedure

Non-linear Analysis Linear Analysis

Modal Response Spectra

Time History

Ground motion history compatible with S(T)

Time History

Spectral acceleration values S(T) defined by the Code


Scaling used in Dynamic Analysis

1) Evaluate the inelastic base shear force “V” from the Code equation

(taking RdR0 into account)

2) Evaluate Ve' from Linear Dynamic Analysis (based on S(T))

3) Ve (elastic) = Ve'.IE (Dynamic analysis)

4) Vd' (inelastic) =

0RRV

d

e (Dynamic analysis)

5) Compare Vd' to V

6) Redo the dynamic analysis after scaling dynamic excitation by a ratio

α = Vd/Ve' to obtain the members internal forces.

7) To estimate deflection and interstorey drift, multiply the dynamic

analysis results obtained from (6) by the ratio RdR0

Vd' ≥ V Vd

' < V

Vd = Vd'

Bx < 1.7 Bx ≥ 1.7

Vd = 0.8V Vd = V


*The dynamic analysis should include the effect of accidental torsion

Bx < 1.7

center of mass shifted by

(± 0.05Dnx)Fx

Bx ≥ 1.7 (± 0.1Dnx)Fx


Time History Analysis

Selection of a Ground Input Acceleration ag(t)

Clause 4.1.8.12(3)

• ag(t) is such that its response spectra is compatible with S(T)

Its response spectrum should be equal or exceed the target spectra

throughout of the period range of interest.

• This can be done by scaling or modifying actual records of similar

magnitude and similar distances that contribute to seismic hazard at the site.


• Halchuk and Adams (2004) have conducted a deaggregation of seismic hazard for selected Canadian cities.

Halchuk K, S. and Adams, J. ‘Deaggregation of Seismic Hazard for Selected Canadian Cities” 13th World Conference on Earthquake Engineering, Vancouver, B. C., Canada, August 1-16,2004, paper No. 2470. Deaggregation: Dividing the total hazard into contributions based on distance and magnitude.

Halchuk and Adams (2004)


Halchuk and Adams (2004)


• Atkinson and Bersnev (1998) have developed various ground motion time

histories which are compatible with the uniform hazard spectra used in

NBCC 2005.

Atkinson, G., and Bersnev, I., (1998), “Compatible ground-motion time

histories for new national seismic hazard maps”, Canadian Journal for Civil

Engineering 25: 305-318.

Records for both eastern and Western Canada were developed in this study.

o For long period structures

Selected earthquake records should

(a) Have the same deaggregation parameters (D,M) as those specified for

S(1.0) of the location.

(b) Be scaled to match S (T1), where T1 is the fundamental period of the

structure.

o For short period structures

Selected earthquake records should

(a) Have the same deaggregation parameters (D,M) as those specified for

S(0.2) of the location.

(b) Be scaled to match S(T1), where T1 is the fundamental period of the

structure.

CHAPTER 5

SEISMIC ANALYSIS USING COMPUTER MODELING

Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-1

Two- Dimensional Seismic Analysis Using SAP2000 Given: 2-D moment resistant steel frame with

dimensions shown in Figure 1.

E = 2x108 kN/m

Required:

1- Free vibration: period, & mode shapes.

2- Time history Analysis.

3- Response Spectrum analysis.

4 m

4 m

4 m

m1 = 28645.2 kg

m2 = 42967.8 kg

m3 = 57290.4kg

10 m

Figure 1

W 690 x 419

W 690 x 419

W 690 x 419

W 6

10 x

551

W

530

x 4

47

W 4

60 x

286

W 6

10 x

551

W

530

x 4

47

W 4

60 x

286


Solution:

1- Select the units from the drop box

2- From File menu select New Model From Template, this will display

different modal templates, choose portal frame type and fill the properties

of the portal frame:

o Number of stories: 3

o Number of bays: 1

o Story height: 4

o Bay width: 10

3- The screen will be titled vertically showing the frame in 2-dimension (x-z

plane) & 3- dimension, close the 3 dimension window.

4- Click on Set Elements on the main toolbar & check the labels box

in the joints & frames area to see the joint & frame numbering.

5- Click on joints 1 and 6. From the Assign menu select Joint, from joints

select Restraints then choose fixed support from the fast restraints.

6- From the Define menu select Materials…, the define material dialog box

will show up, highlight STEEL then click Modify / Show Material to

check the properties of the material.

7- Verify that modulus of elasticity is 2.0 E+11 then click OK twice.

8- From the Define menu select Frame Sections…to display the define

frame sections dialog box.

9- In the Click to: choose Import I / Wide Flange, another box will

display for Section Properties File, from Sap2000n folder in your hard

drive choose Cisc.pro (see Fig. 2). Double click on the Cisc.pro and

choose from the given sections: W410x100, W610x551, W530x447, &


W460x286. (you can press the Ctrl key on the keyboard to choose all of

them at one step).

10- Select the column of the first story (element 1, 4), then from Assign

menu select Frame, Section…, choose W610x551, make the same for the

column of the second, third stories (element 2, 5, & element 3,6). After

each selection for beam or column section you can click on the

undeformed shape icon to see the element numbering again.

11- From Set Elements click on the frame labels to remove it.

12-The final Figure will be as shown in Fig. 3.

Fig. 2

Figure 2


13-Select joints 2 and 6. From Assign menu select Joint then select

Masses…, a dialog will display. In the dialog box of direction 1 write

14322.6.

14-Repeat step 16 for joints 3, 7 and write 21483.9 in the dialog box of

direction 1, also joints 4, 8 and write 28645.2 in the dialog box of

direction 1.

15-Select Nodes 2 & 6 then from the Assign menu choose Joint,

Constraints… this will display Constraints box.

Fig. 3


Fig. 4

16-In the Click to: choose Add Rod, accept the default Rod 1, Click Ok

twice to finish this part.

17-Repeat step 17, 18 for Nodes 3 , 7 & Nodes 4 , 8 to make Rod 2 & Rod

3.

18-Form the Define menu

select

Functions…Response

Spectrum to display

define response spectrum

functions box.

19-In the Click to: choose

User Spectrum, a dialog

box for response spectrum

function definition will

display.

20-In the Function Name

write montreal-est-C,

input the values of the

periods and spectral

accelerations

corresponding to the

response spectrum of Montréal-

east with soil type C and damping

of 0.05, see Fig. 4.

21-Form the Define menu select Functions… Time History to display

define time history functions box.


Fig. 5

22- In the Click to: choose Add Function From File, a dialog box for time

history function definition will display.

23-In the Function Name write

ELCENTRO, click Open File

to choose the file having the

data for your time history from

its location in your computer.

24-Write in the box of number of

points per line the number of

reading per line, in our

example we have 1 per line.

Then choose either Function

at Equal Period Step of or

Time and Function Values

depending on how your data

file is organized. In our case we

choose Time and Function values,

see Fig. 5.

25-Form the Define menu select Analysis Cases…Response Spectrum to

display define response spectra box.

26-Choose Add New Case, a dialog box for response spectrum case data

will display.

27-In the Response Spectrum Case Data box:

o Accept the default Spectrum Case Name: ACASE1

o Analysis Case Type: Response Spectrum

o Accept the default Excitation Angle: 0

o Accept the default Modal Combination Option, CQC.


o In the Damping box: 0.05

o Accept the default Directional Combination option, SRSS.

o In the Function area select montreal-est-C in U1 direction

o For U1 direction write 1 at the scale factor.

o Click ADD button.

o Click Ok.

28- Choose Add New Case, a dialog box for response spectrum case data

will display.

29-In the Time History Case Data box:

o Accept the default History Case Name: ACASE2

o For modal damping: click Modify / Show and write 0.05 in

Damping for all modes.

o For Load type: choose acceleration.

o In Number of output time steps box: 1560.

o In Output Time Step Size box: 0.02

o In the Load Assignments area select ELCENTRO from Functions

for U1 direction

o For U1 direction write 1 at the scale factor and leave the arrival

time, angle 0.

o Click ADD button.

30-Click Ok twice to finish this part.

31-From the Analyze menu select Set Options… to display the analysis

option dialog box and choose the x-z plane.


Output:

We will divide the output file into:

1- Output can be obtained from the screen for the time history of a certain

quantity such as:

o Time history for a joint displacement, rotation, acceleration or

velocity.

o Time history for bending moment, shear force or axial force at a

specific location of an element.

o Time history for base shear force and overturning moment.

2- Output obtained from the 2Dframe.out output file.

Time History 1. Time history for displacement ux of Joint 4:

o Click on Joint 4, and then from Display menu click on Show Time

History Traces… this will display Time History Display

Definition. From the choose function part double click on joint 4 at

List of Functions, joint 4 will now appear at Plot Function.

o The default is drawing displacement ux for this joint, if you want to

change click on Define Function icon, another box will display for

Time History Functions.

o Highlight joint 4, click on Modify/Show TH Function, another box

for Time History joint function will appear. Choose the function

you want then click Ok twice.

o Now Highlight Joint 4 then click the Display icon, another display

window will appear as in Fig. 4. You can either plot this figure or


save it by clicking in file the menu and choose Print Graphic or

Print Tables To File…

o The same procedure for joint 4 can be applied for frame element by

choosing the required frame element.

2. Time history for Base Shear & Base Moment:

o Click on the Define Function icon, the box for Time History

Functions will display.

o In the Click to: choose Add Base Functions, another window will

display for Base functions, Click on Base Shear X and Base

Moment Y. Click Ok Twice.

o Now Base Shear X and Base Moment Y will be at List of

Functions, use the same procedure described for joint 4 to plot the

Base Shear X and Base Moment Y.

Fig.6


Fig. 8

2Dframe.out output file:

In this file one can find information about:

1. Natural frequencies and mode shapes

2. Modal Participation Factors.

3. Mass Participation Factors.

4. Joint Displacements due to SPEC1 Spectra.

5. Frame Element Internal forces due to SPEC1 Spectra.

N.B. as shown in Fig. 8, some specific can be selected for printing.


Three-Dimensional Seismic Analysis

The building model can be developed using many graphical interface structural programs like ETABS. ETABS is a special purpose analysis and design program developed specifically for building structures. An example for a 30-storey reinforced concrete building will be illustrated in this section. The building is assumed to be located in Toronto. The total height of the building is 92 m. Modeling Main Steps: 1- Develop the building geometry and axes using CAD or ETABS

2- Define materials properties 3- Define the building elements properties Shell elements are used to model the slabs and the shear walls, and frame

elements are used to model the columns and the rigid frames. The stiffness of

the pile foundations or shallow foundations are usually accounted for through

the use of equivalent vertical, horizontal and rotational springs.


To account for the cracked section properties, the section stiffness is reduced

using stiffness modifiers. The reducing factors can be evaluated from the CSA-

A23.3-04 Clause 21.2.5.2.1 Table 21.1 and Clause 21.2.5.2.2.

4- Construct the Model


5- Preliminary Dynamic Analysis The building is assumed to be located in Toronto. The soil profile is type D.

NBCC 2005 spectral values for Toronto

Sa (0.2) = 0.26

Sa (0.5) = 0.13

Sa(1.0) = 0.055

Sa(2.0) = 0.015

Using Soil Type D ------------ Fa = 1.30 and Fv = 1.40

The Design spectral acceleration values can be determined using NBCC 2005

Clause 4.1.8.4.6

S (0.2) = 3.315 m/sec2

S (0.5) = 1.785 m/sec2

S(1.0) = 0.7553 m/sec2

S(2.0) = 0.206 m/sec2

S(4.0) = 0.103 m/sec2

Notice that for dynamic analysis S has units of acceleration. It is obtained by

multiply “Sa” by “Fa” or “Fv” and the ground acceleration “g”

Conduct Analysis to obtain the following results

1- The building modal participating mass ratios, mode shapes and periods are obtained from the analysis:

Mode Period UX UY UZ SumUX SumUY SumUZ RX RY RZ SumRX SumRY SumRZ1 4.235024 28.7694 40.5759 0 28.7694 40.5759 0 58.0654 40.2014 0.5623 58.0654 40.2014 0.56232 3.50052 23.0926 9.3154 0 51.862 49.8913 0 13.4977 30.6581 40.7607 71.5631 70.8595 41.32313 1.108758 10.3055 3.987 0 62.1674 53.8783 0 0.0702 0.2505 0.1777 71.6334 71.11 41.50074 1.067125 17.4395 15.8869 0 79.6069 69.7652 0 26.0066 27.8469 34.5346 97.6399 98.9568 76.03545 0.953012 1.7659 12.1331 0 81.3728 81.8983 0 1.8824 0.5209 4.1669 99.5223 99.4778 80.20236 0.529784 3.843 0.7729 0 85.2159 82.6712 0 0.0372 0.179 0.95 99.5595 99.6568 81.1524


T = 4.235 sec (Combined mode X-Y - dominant Y )

T= 3.50 sec (Combined mode X-Y-Torsion - dominant X and Torsion)

2- Estimate the periods using NBCC 2005 (hn)

Ta (MRF) = = 2.23 sec

Shear Walls Ta = = 1.50 sec

Maximum allowable periods according to NBCC 2005

Ta (1.50)(2.23) = 3.345 sec for (MRF)

Ta (2.0)(1.50) = 3.0 sec for (Shear walls)

Use Ta = 3.0 sec to evaluate the base shear force based on the NBCC 2005

6- Code calculations

Mv = 2.50 (Table 4.1.8.9)

Assume moderately ductile shear wall building

Rd = 2.0 Ro = 1.50

IE = 1.0

W = 143883 kN =The dead load of the building including the self weight

of all elements, the super imposed load, and 0.25 of snow loads.

= 0.206

The base shear force is calculated using the NBCC 2005 equation:

Clause (4.1.8.11)


and should not be less than

V = 24,700 KN

Conduct static analysis using the equivalent static earthquake load to obtain Bx.

The static analysis involves the cases of eccentricity.

Bx = δmax/δave

δmax =0.25 m δave = 0.09 m

Bx = 2.70

7- Evaluation of spectrum scaling factor and re-analysis

Along X-Direction

From initial analysis

V'xex = 19,073 kN

V'xey = 12,619 kN

V'xe = = 22,869 KN

Vxe = V'xe * IE = 22,869 *1.0 = 22,869 KN

V'xd = = 7,623 KN


Compare V'xd to V

Bx ≥ 1.7

Vd = V = 24,700 KN

Scaling Factor

α = Vd/V'xe = 1.08

Multiply all values of S(T) by 1.08 and redo the analysis to obtain the

members forces.

Repeat these steps along the Y-Direction

From initial analysis

V'yex = 12,619 kN

V'yey = 20,158 kN

V'ye = = 23,782 KN

Vye = V'ye * IE = 23,782 *1.0 = 23,782 KN

V'yd = = 7,927 KN

V'xd ≥ V V'xd < V

Vd = V'xd Regular Irregular

Vd = 0.8V Vd = V


Compare V'yd to V

Bx≥ 1.7

Vd = V = 24,700 KN

Scaling Factor

α = Vd/V'ye = 1.04

Multiply all the S(T) by 1.04 and redo the analysis to obtain the members

forces.

Earthquake analysis cases

-Specx : response spectrum in X- Direction

-Specy : response spectrum in Y- Direction

-Specxeccp : response spectrum in X- Direction with eccentricity +0.10

-Specxeccn : response spectrum in X- Direction with eccentricity -0.10

-Specyeccp : response spectrum in Y- Direction with eccentricity +0.10

-Specyeccn : response spectrum in Y- Direction with eccentricity -0.10

-Specxudp : The storey shear force obtained from the case Spec X and

applied at a distance of +0.10 Dnx from the shear center.

V'yd ≥ V V'yd < V

Vd = V'yd Bx < 1.7 Bx ≥ 1.7

Vd = 0.8V Vd = V


Dnx = Plan dimension of the building at level x perbendicular to the

direction of seismic loading being considered

-Specxudn : The storey shear force obtained from the case Spec X and

applied at a distance of -0.10 Dnx from the shear center.

-Specyudp : The storey shear force obtained from the case Spec Y and

applied at a distance of +0.10 Dnx from the shear center.

-Specyudn : The storey shear force obtained from the case Spec Y and

applied at a distance of -0.10 Dnx from the shear center.

Earthquake X-Direction is an envelope to all the X-direction cases.

Earthquake Y-Direction is an envelope to all the Y-direction cases

8- Load Combinations Based on NBCC 2005. Clause 4.1.3.2.

Case 1 (E1): 1.40 Dead loads

Case 2 (E2): (1.25 Dead Loads or 0.90 Dead loads) + 1.50 Live Load +

(0.25 Snow or 0.40 Wind Load)

Case 3 (E3): (1.25 Dead loads or 0.90 Dead loads) + 1.50 Snow + (0.50 live

loads or 0.40 Wind)

Case 4 (E4): (1.25 Dead loads or 0.90 Dead loads) + 1.40 Wind X-Direction

+ (0.50 live loads or 0.40 Snow)

Case 5 (E5): (1.25 Dead loads or 0.90 Dead loads) + 1.40 Wind Y-Direction

+ (0.50 live loads or 0.40 Snow)


Earth Quake Combinations

Case 6 (E6): 1.0 Dead loads +( 1.0 Earthquake X-Direction + 0.30

Earthquake Y-Direction) + 0.50 live loads + 0.25 Snow

Case 7 (E7): 1.0 Dead loads + (1.0 Earthquake Y-Direction + 0.30

Earthquake X-Direction) + 0.50 live loads + 0.25 Snow

9- Results

a) Shear wall Design

-Magnify the moments obtained from the final analysis with the P-∆ factor U2

)(1

12

∑∑∆

−

=

Vfhpff

U

∆f = inter storey drift due to factored lateral load.

Σ pf = Σ factored axial loads in the story.

Σ Vf = factored total story shear.

h = story height.

- Magnify the shear forces obtained from the final analysis by a factor =

(Rd). (Ro)

b) Columns Design

- Magnify the moments obtained from the final analysis with the P-∆ factor U2

- Magnify the shear forces obtained from the final analysis by a factor =

(Rd). (Ro)


c) Frame Girders Design (Ductile MRF)

- Magnify the moments obtained from final analysis by a factor

= ( )(

)('

0'

0

DuctileMRFRRlsductilewalModeratelyRR

d

d ). (U2)

) ------- Reduction factors for Ductile MRF

d) Deflection and interstorey drift

- Magnify the deflection and interstorey forces obtained from the final

analysis by a factor = (Rd). (Ro)

CHAPTER 6

SEISMIC DESIGN OF STEEL STRUCTURES

Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-1

6-1 Concept of Capacity Design Capacity design was developed in the late 1960’s in New Zealand as an

approach to resist the effect of severe earthquakes.

The concept of capacity design is based on the following:

Inelastic action is unavoidable during severe earthquakes.

The designer dictates where inelastic response should occur.

Zones of possible inelastic actions (called plastic zones) are selected to be

regions where large plastic deformation can develop without significant

loss of strength (regions which possess enough ductility)

These regions are detailed to suppress premature failure modes, such as

local buckling or member instability in the case of steel structures, or

shear failure in the case of concrete structures.

The surrounding members are designed in such a way that their capacities

are greater than that the values corresponding to the maximum capacity

of the plastic zone. These members are intended to remain elastic during

an earthquake.


An illustration for the concept of plastic design is shown in Fig. 6.1. This

example involves a cantilever beam of total length “L” consisting of a brittle

segment (such as a fiber composite material) of length “a” and a ductile steel

segment of length “b”.

Fig. 6.1

Capacity design approach would aim at making the brittle material stronger

than needed to ensure that plastic hinging occurs first in the steel segment of the

cantilever. Therefore, the moment capacity of the brittle material should satisfy:

Mbrittle ≥ α L/b Mp (steel)

α is a number greater than 1.0 to account for the possible reserve strength of the

cantilever beyond its nominal yield strength.

brittle material P

steel

a b

L

(L/b) Mp (steel) Mp (steel)

θ


Steel is in general a ductile material

However, brittle failure of a steel member might result due to any of the

following effects:

(a) Overall buckling

(b) Local buckling

(c) Failure of connections

Two examples emphasizing the capacity design of moment resisting and concentric bracing ductile frame structures are presented in these notes.

σ

σy

ε

Stress-strain curve for a typical steel coupon


6-2 Strength Design Before performing a seismic (capacity) design of a lateral resisting system, the

structure is analyzed under the effects of various load combinations (DL, LL,

W, S, E) defined by the NBCC and the maximum factored forces and moments

resulting from various load combinations are determined. Members are

designed to satisfy the strength and serviceability criteria defined in the design

Code.

The members and connections will be then re-designed according to the

capacity design provisions (clauses 27 ..)

6-3 Types of Resistance

Ultimate (Mr) Nominal (Mp) Probable (Mp’)

Φr = 0.9 Φp = 1.0 Φp’ = 1.1 Ry


Moment Resisting Frames (Energy dissipates through yielding of beams)


6-4 Seismic Design Provisions for Ductile Moment Resisting Frames,

Rd = 5.0, R0 = 1.5 (S16-01)

• No Limit on the height of the building

• Plastic hinges can form only in beams (flexural members). Plastic hinges

in columns are permitted only at the base, except in single storey

buildings.

Beams (27.2.2)

o Class 1 sections

o Laterally braced according to the requirements of clause 13.7 (b)

y

rrL =

yFk1550017250 +

k is the ratio between the smaller factored moment to the larger factored

moment at opposite ends of the beam; k is positive for double curvature, and

shall be based on the both gravity and seismic loads. Bending moment due to

seismic loads may be taken as linear variable from maximum value to zero at

both ends of the beam.


Columns (27.2.3.2)

Class 1 or Class 2

∑Mrc' ≥ ∑ [1.1 Ry Mpb + Vh (x+ dc/2)]

where:

∑Mrc' = Sum of the column factored flexural resistances at the intersection

of the beam and column

Mpb = Nominal plastic moment resistance of the beam

Vh = Shear acting at plastic hinge when 1.18 Ry Mpb is reached

x = Distance from centre of the beam plastic hinge to the column face

Mrc' = 1.18 φ Mpc [1-

y

f

CCφ

] ≤ φ Mpc

Mpc = nominal plastic moment resistance of the column

Cf = results from summation of Vh at the considered level.

More requirements are specified in Clause 27.2.3.1 if the plastic hinge develops

in the column.

Panel Zone ( 27.2.4)

When plastic hinges form in adjacent beams, the panel zone shall resist forces

arising from beam moments of:

∑ [1.1 Ry Mpb + Vh (x+ dc/2)]


Resistance of Panel Zone

(a) Vr = 0.55φ dc w' Fyc [1+ '

23wdd

tb

bc

cc] ≤ 0.66φ dc w' Fyc

The above equation can be used if the following conditions are satisfied

1) When IEFaSa(0.2) ≥ 0.55, the sum of the panel zone depth and width

divided by the panel zone thickness does not exceed 90.

2) Satisfy the width-to-thickness limit of Clause 13.4.1.1(a)

3) Doubler plates are groove or fillet welded. Doubler plate can be included

in calculating the thickness to width ratio if it is connected to the column

web near the centre of the panel.


Moderately Ductile Moment-Resisting Frames (Rd = 3.5, R0 = 1.5)

Same provisions as ductile Moment Resisting Frames except that:

(i) The beams shall be Class 1 or 2 sections

(ii) The bracing beam shall meet the requirements of Clause 1.7 (a) instead of

13.7 (b)

y

crrL =

yFk1500025000 +


Evaluation of Forces in Panel Zone

Equilibrium Equations

0)1()1()1()1( 4

4

3

3

2

2

1

1 =−

−−

−−

+− ββββ

MMMM (1)

Vh = [ M1 + M2 - )1()1( 4

44

3

33β

ββ

β−

−−

MM ]/0.95db (2)

11

95.0l

dc=β , 2

295.0l

dc=β , 3

395.0h

db=β , 4

495.0h

db=β

V3 C3

M3

V1 M1

C1

C4

V4 M4

C2

M2 V2

Vh

V1

V3

V4

V2

l1/2 l2/2

0.95 dc

0.95 db

h3/2

h4/2


Concentric Braced Frames

(Energy dissipates through yielding of bracing members)


Type Moderately Ductile Limited Ductility

If IEFaSa(0.2) < 0.35

No limit

If IEFaSa(0.2) < 0.35

No limit

Tension-Compression

(a)

Otherwise H 40 m

When H > 32 m, seismic

force shall be increased

by 3% per meter above

32 m

Otherwise H 60 m

When H > 48 m, seismic


by 2% per meter above

48 m

Chevron

(b)

Same as (a)

Same as (a)

Tension

only

H 20 m

If H > 16 m, seismic


by 3% per meter of

height above 16 m

H 40 m

If H > 32 m, seismic


by 3% per meter of

height above 32 m

K-truss

N.A.

N.A.


Moderately ductile concentrically braced frames (Rd = 3.0, R0 =1.3)

Diagonal Bracing

Limit on the width-to-thickness ratio

o For IEFaSa(0.2) ≥ 0.35

Sections kL/r ≤ 100 kL/r = 200

Rectangular section

and

HSS

300/√Fy Class 1

Legs of angle

and

flanges of Channel

145/√Fy 170/√Fy

Other elements Class 1 Class 2

o For IEFaSa(0.2) < 0.35

HSS => Class 1

Other sections => Class 2

Legs of angle = 170/√Fy


Bracing Connection

Factored resistance of the bracing connections shall be equal or exceed both

AgRyFy and 1.2 the times the nominal compressive resistance of the bracing

members (1.2Cp)

The tensile force should not exceed the combined effect of gravity in the

bracing and the effect of seismic loads corresponding to RdR0 = 1.3

(Redistribution due to buckling should be taken into account when calculating

the seismic loads corresponding to RdR0 = 1.3)

Beams

The capacity design of the beam should consider two conditions:

1- Compression bracing = 1.2Cp

Tension bracing = AgRyFy

2- Compression bracing buckled and retain a force = 0.2 AgRyFy

Tension bracing = AgRyFy

Columns

• Factored resistance of columns shall equal or exceed the effects of

gravity and the brace forces described above for connections design

• Class 1 or 2

• Bending resistance 0.2 ZFy


Seismic Design of a Moderately Ductile MRF and Moderately Ductile braced frame

1. Eight storeys 2. Located Montreal, Quebec. 3. Loading according to NBCC 05

Loads: Dead Load: (NBCC 4.1.5) 1. self weight = 0.10 (thickness of slab) * 25 (Concrete unit weight)= 2.5 kN/m2 2. partition = 1.0 kN/m2 3. mechanical service = 0.5 kN/m2 4. flooring = 0.5 kN/m2 5. structural elements (beams & columns) = 0.25 kN/m2 D.L. = Σ = 4.75 kN/m2 Live load: (NBCC 4.1.6) 1. L.L. on typical office floor = 2.4 kN/m2

6000 6000 8000 6000 6000

Braced Frame

Braced Frame

N

6x6000mms

Plan of Example Building

Moment

Resisting Frame

Moment

Resisting Frame


Roof load 1. self weight = 0.10 (thickness of slab) x 25 (Concrete unit weight) = 2.5kN/m2 2. mechanical loading = 1.6 kN/m2 3. roofing = 0.5 kN/m2 4. structural elements (beams & columns) = 0.25 kN/m2

D.L. = Σ = 4.85 kN/m2 5. snow load = 2.2 kN/m2

Analysis under Factored Loads The following load cases were considered:

i. 1.25 DL + 1.5 LL + 0.5 Snow + P-∆ effect ii. 1.25 DL + 0.5 LL + 1.5 Snow + P-∆ effect iii. DL + E (No eccen.) + 0.5LL + 0.25SL + P-∆ effect (both directions) iv. DL + E (Eccen. ±e) + 0.5LL + 0.25SL + P-∆ effect (both directions)

Earthquake load (N-S) (+e Eccen.)

Earthquake load (N-S) (-e Eccen.)

Earthquake load (E-W) (No Eccen.)

Earthquake load (E-W)

(+e Eccen.)

Earthquake load (E-W) (-e Eccen.)

Earthquake load (N-S) (No Eccen.)

Mass center


Seismic Loads Calculations Using the Equivalent Static Approach (NBCC 05)

The total weight of the floors are given in the table below, Floor Height (m) Weight ( kN)

8 29.7 6220.8

7 26.1 5472

6 22.5 5472

5 18.9 5472

4 15.3 5472

3 11.7 5472

2 8.1 5472

1 4.5 5472

Σ W = 44524.8 kN

Due to symmetry, the centre of mass in each storey coincides with the centre of geometry.

Lateral Earthquake force – V

V = S(Ta) . Mv . IE . W / (Rd . Ro) ≥ S(2.0) . Mv . IE . W / (Rd . Ro)

• Time period Ta

North-south direction: Steel moment fames, T = 0.085 (hn)3/4 = 1.08 sec. NBCC 4.1.8.11 East-west direction: Braced frames, T = 0.025 (hn) = 0.74 sec NBCC 4.1.8.11

• For Montréal, the 5% damped spectral response acceleration ratios, Sa(T), are provided in Table-C2 of NBCC.

Sa(0.2) Sa(0.5) Sa(1.0) Sa(2.0) PGA

0.69 0.34 0.14 0.048 0.14

• Site class: C (very dense soil and soft rock) is assumed. • From table 4.1.8.4.b, the value of acceleration-based site coefficient Fa = 1


• From table 4.1.8.4.c, the value of the velocity based coefficient Fv = 1.0 • The design spectral acceleration value S(T)

T =< 0.2 sec.; S(T) = Fa . Sa(0.2) = 1 * 0.69 = 0.69 T = 0.5 sec.; S(T) = Fv . Sa(0.5) = 1 * 0.34 = 0.34 take the smallest

= Fa . Sa(0.2) = 1 * 0.69 = 0.69 T = 1.0 sec.; S(T) = Fv . Sa(1.0) = 1 * 0.14 = 0.14 T = 2.0 sec.; S(T) = Fv . Sa(2.0) / 2 = 1 * 0.048 = 0.024

Using linear interpolation North-south direction S(1.08) = 0.13 East-west direction S(0.74) = 0.24

• From table 4.1.8.11., Sa(0.2)/Sa(2.0) = 14.375 > 8.0 , so higher mode factor Mv =

1, o Base overturning reduction factor Jy = 0.884 (N-S direction) o Base overturning reduction factor Jx = 0.92 (E-W direction)

• Normal importance category for this office building is assumed, IE = 1.0. • From table 4.1.8.9.

North-south direction, moderately ductile moment-resisting frames, Rd = 3.5, Ro = 1.5, East-west direction, moderately ductile concentrically braced fames, Rd = 3.0, Ro = 1.3

• Lateral Earthquake force V North-south direction V = S(Ta).Mv.IE.W/(Rd.Ro) = 1102.6 kN ≥ S(2.0). Mv. IE.W / (Rd.Ro) East-west direction V = S(Ta).Mv.IE.W/(Rd.Ro) = 2740.2 kN ≥ S(2.0). Mv. IE.W / (Rd.Ro)

• Concentrated force at the top of the building Ft = 0.07.Ta .V North-south direction = 83.4 kN East-west direction = 142.4 kN

• Σ Wi . hi = 770808 kN • Fi = (V - Ft). Wi . hi / (Σ Wi . hi)


Floor Level

North-south direction Force (kN)

East-west direction Force (kN)

8 244.2 + 83.4 622.6 + 142.4

7 188.8 481.2

6 162.7 414.9

5 136.7 348.5

4 110.6 282.1

3 84.6 215.7

2 58.5 149.3

1 32.5 82.9

The structure shall be designed to resist overturning effects caused by the earthquake forces determined in Sentence (6) 4.1.8.11 and the overturning moment at level x, M, shall be determined using the following equation:

Mx = Jx Σ Fi ( hi – hx)

where Jx = 1.0 for hx ≥ 0.6 hn, and Jx = J + (1-J) (hx / 0.6 hn) for hx < 0.6 hn, and (N – S) direction For floors 5 to 8, Jy = 1.0 Floor 4 : Jy = 0.884 + (1-0. 884) (15.3 / 17.82) = 0.983 Floor 3 : Jy = 0. 884 + (1-0. 884) (11.7 / 17.82) = 0.960 Floor 2 : Jy = 0. 884 + (1-0. 884) (8.1 / 17.82) = 0.937 Floor 1 : Jy = 0. 884 + (1-0. 884) (4.5 / 17.82) = 0.913 Ground : Jy = 0. 884 + (1-0. 884) (0 / 17.82) = 0.884 (E – W) direction For floors 5 to 8, Jx = 1.0 Floor 4 : Jx = 0.92 + (1-0.92) (15.3 / 17.82) = 0.988 Floor 3 : Jx = 0.92 + (1-0.92) (11.7 / 17.82) = 0.973 Floor 2 : Jx = 0.92 + (1-0.92) (8.1 / 17.82) = 0.956 Floor 1 : Jx = 0.92 + (1-0.92) (4.5 / 17.82) = 0.940 Ground : Jx = 0.92 + (1-0.92) (0 / 17.82) = 0.920


Floor

Overturning

moment

(kN.m)

(N – S)

direction

Overturning

moment

(kN.m)

(E – W)

direction

Reduced

overturning

moment

(kN.m)

(N – S)

direction

Reduced

overturning

moment

(kN.m)

(E – W)

direction

8 0 0 0 0

7 1179.7 2754.1 1179.7 2754.1

6 3039.3 7240.8 3039.3 7240.8

5 5484.8 13221.3 5484.8 13221.3

4 8422.5 20456.4 8284.4 20224.9

3 11758.7 28707.2 11290.3 27918.4

2 15399.7 37734.7 14425.3 36088.1

1 19251.5 47299.9 17582.3 44471.5

Ground 24212.9 59629.8 21404.2 54859.5

Evaluation of Seismic Loads Using Response Spectrum Analysis The natural frequencies and mode shapes of the building were evaluated using the program SAP 2000 and the results are shown below: Ve' (N-S) = 3868kN Ve' (E-W) = 5150 kN Ve (N-S) = Ve

'.IE = 3868 kN Ve (E-W) = Ve

'.IE = 5150 kN

Vd' (N-S) =

0RRV

d

e = 737 kN < 1103kN

Vd' (E-W) =

0RRV

d

e = 1321 kN < 2740 kN


First mode shape (Period =1.46075 sec)

Second mode shape (Period = 1.32939 sec)


Modal Participating Mass Ratios

OutputCase StepType StepNum Period UX UY UZ

Text Text Unitless Sec Unitless Unitless Unitless

modal Mode 1 1.460747 0.68881 9.718E-14 2.785E-09

modal Mode 2 1.329386 1.543E-08 0.70367 5.495E-12

modal Mode 3 1.269189 0.00006013 0.00024 2.891E-09

modal Mode 4 0.534707 0.14181 8.891E-13 3.087E-11

modal Mode 5 0.463849 0.00001184 0.00014 2.35E-10

modal Mode 6 0.450689 6.655E-09 0.2019 1.728E-09

modal Mode 7 0.300466 0.0645 2.248E-13 5.162E-08

modal Mode 8 0.266385 0.000004504 0.000001035 5.093E-08

modal Mode 9 0.24875 7.493E-10 4.123E-10 0.09393

modal Mode 10 0.248726 1.234E-09 0.04821 8.926E-08

modal Mode 11 0.241159 3.006E-07 2.695E-11 0.00004203

modal Mode 12 0.233314 6.89E-08 2.802E-10 0.0009


3-D view for the structure

Torsional Sensitivity Level 8 (East-west direction) Torsional sensitivity can be determined by calculating the ratio, Bx = δmax / δave δmax = 0.01456 m δave = (0.01456 + 0.01290) / 2 = 0.01373 Bx = 0.01456/ 0.01373 = 1.06045 Level 8 (North-south direction) Torsional sensitivity can be determined by calculating the ratio, Bx = δmax / δave δmax = 0.02080 m δave = (0.02080+ 0.01807) / 2 = 0.019435 Bx = 0.02080 / 0.019435 = 1.0702


Interstorey Drift East-west direction The maximum interstorey drift = 0.00228 * Rd Ro = 0.008268 m < 0.025hs (floor height) < 0.09 m

North-south direction The maximum interstorey drift = 0.00724 * Rd Ro = 0.003801m < 0.025hs (floor height) < 0.09 m

3-D view for the structure


Strength design according to CISC 95 • North-south direction:

Two moment resisting frames along the outer edges

Elevation of Moment Resisting Frame


• East-west direction: Two braced frames

Elevation of Braced Frame

S8

S7

S6

S5

S4

S3

S2

S1

F8

F7

F6

F5

F4

F3

F2

F1


Capacity design of the Rigid Frames An interior joint of the moment resisting frame is considered to illustrate the concept of capacity design. (Floor F4)

Beam

W690x140 Column

W760x147 d b t w

684 254 18.9 12.4

753 265 17.0 13.2

Beam length = L1 = L2 = 6000 mms Column height = h1 = h2 = 3600 mms β1 = β2 = 0.95dc/L1 = 0.1192 β3 = β4 = 0.95db/ h1 = 0.1805 Beams W 690x140

(y

crrL

= 55.65) > ( 350)5.0)(15000(25000 −

= 50) Not good

W 690 x 140

Level 4

W 760 x 147

W 690 x 140

Mp Mp/2


Assume the cross beam will be repeated every 2.0 m

(y

crrL

= 37.105) < ( 350)667.0)(15000(25000−

= 42.8) OK

Mr1 = Mr2 = φ MP = 1410 kN.m; Mp1 = 1410/0.9 = 1566.6 kN

M′p1 = M′

p2 = (1.1). (1.1). 1410 / 0.9 = 1895 kN.m

Assume beam reach their ultimate strengths,

∑ Mrc′ ≥ ∑ 1.1. Ry . Mpb + Vh (x + dc/2)

Example for F4

Vh = VhE + VG

VhE = 2Mpb' /Lh

Lh = L-2S

S = x + dc/2

VG = wL/2 => w = (DL +0.5 LL)

x = db / 2 for welded connection

w = 35.7 kN/m

Lh x

L

dc/2

S

Mp 2Mp/3 Mp/3


Example for F4 (see joint location on Page 6-25)

S = (x + dc/2) = 684/2 + 753/2 = 718.5 mm

Lh = 6000 – (2) (718.5) = 4563 mm

VhE = (2) (1895) / 4.563 = 830 kN

w = [4.75 + (0.5)(2.4)](6) = 35.7 kN/m

VG = (35.7). (6) / 2.0 = 107 kN

Unbalanced moment = ∑ (1.1 Ry . Mp + Vh (x + dc/2) )

= (1895) . (2) + (830) . (0.7185) . (2) = 4982 kN.m

Mrc′ should be greater than 4982 / 2 = 2491 kN.m

Mrc′ = (1.18). (φ ) . Mpc [ 1 -

y

f

CC ]

Take Cf = (107)(10) = 1070 kN

φ . Mpc = 1580 kN.m

Cy = A . fy = (18700 ) x (350) / 1000 = 6545 kN

Mrc′ = (1.18). (1580). [1 -

)6545)(9.0(1070 ] = 1830 kN.m < 2491 kN.m Not good

107

107

107

107

107

107

107

107

376

1083

1895

918

755 kN.m

529

1228466

830

107 107 kN 305 kN


Try W920 x 381 φ . Mpc = 5280 kN.m

Cy = 17010 kN

Mrc′ = (1.18). (5280). [1 -

)17010)(9.0(1070 ] = 5794 kN.m > 2491 kN.m OK

Panel Zone

0.0]1111

[4

44

3

3

2

2

1

1 =−

−−

−−

+− β

ββββ

MMMM

Vh = bdMMMM 95.0/]1

.1

.[4

44

3

3321 β

βββ

−−

−−+

β1 = β2 = 0.95dc/L1 = 20.06000

)951)(95.0(=

β3 = β4 = 0.95db/ h1 = 18.03600

)684)(95.0(=

==⇒−

=− 43

3

)18.01(2

)20.01()2)(2491( MM

M 2552 kN.m

Vh = kN5943)]684.0).(95.0/[(]18.01

)2552(18.018.01

)2552(18.024912491[ =−

−−

−+

Vr = 59434468])4.24)(684()951(

)6.43)(310).(3(1)[35.0).(4.24).(951).(9.0).(55.0(2

<=+x Not good

Try: W 920 x 585

W920x585 d b t w

960 427 55.9 31

Vr = okx

59436170])50)(684()960(

)9.55)(427).(3(1)[35.0).(31).(960).(9.0).(55.0(2

>=+

Vr should not exceed (0.66φdcw'Fy), i.e. should not exceed


Vr (0.66) (960) (0.9) (31) (0.35) 6188 kN

Check clause 13.4.1.1 a = 960 – (2) .(55.9) = 848.2 h = [ 684 – (2) (18.9)] = 647

h / w = 647 / 31 = 20.87

a / h = 848.2 / 647 = 1.31 > 1.0 OK

kv = 5.34 + (4) / (a/h)2 = 7.57

y

v

Fk

439 = 64.9

∴ y

v

Fk

wh 439≤

∴Fs = 0.66 Fy OK

a

h


Revision of braced frame design for ductility Consider S8 HSS 178 x 178 x 6

L = 5380 mm

b = 177.8 mm

t = 6.35 mm

r = 69.6 mm

A = 4250 mm2

σy = 350 MPa

Cr = 789 kN

Tr = 1339 kN

Cf = 248 kN

Tf = 99 kN

Slenderness ratio (Clause 27.5.3.1) kL/r = 5380/69.6 = 77.3 < 200 ok Wall slenderness (Clause 27.5.3.2)

(b – 4t) / t = 0.2435.6

)35.6).(4(8.177=

−

Limit = 63.17330=

yσ not good

The brace does not satisfy local buckling limit on wall slenderness Revised section HSS 152x152x10 A = 5210 mm2

b = 152.4 mm2

t = 9.53 mm2

r = 57.6 mm2

Tr = 1641 kN


Cr = 916 kN

kL/r = 93.4 < 200 ok

(b – 4t) / t = 11.99 <17.63 ok

Clause (27.5.4.2), connection between the brace and the column should be designed to resist a force “Fc” which is the largest of:

a) Tu = Ag. Ry. yσ = (1.1). (5210) .(350) = 2005 kN

b) Cu = kNC p 1222)916.()9.0(

2.12.1 ==

Fc should not exceed the effect of gravity load + effect of seismic loads with Rd. Ro = 1.3

Fc should be less than (65 + )3.1(

)3.1).(3).(362( )

Fc 1151 kN Connection should be designed to resist a force = 1151 kN Beam design Verification The capacity design for the beam should consider two conditions: 1. Compression bracing = 1.2 Cp

Tension bracing = Ag. Ry. Fy

2. Compression bracing buckled and retain a force = 0.2 Ag. Ry .Fy

Tension bracing = Ag. Ry .Fy

The beam need not to resist load effects exceeding to Rd. Ro = 1.3


Capacity design of strut beam

T7b = Ag . Ry . Fy = (5210). (1.1). (0.35) = 2005 kN

C7b = 0.2 Ag . Ry . Fy = 401 kN

T6b = (9260)(1.1)(0.35) = 3565 kN

C6b = 713 kN

- P = (T6b – C7b) cosӨ

P = - (3565 - 401) 0.743 = -2350 kN

Mf = (wDL + 0.5 wLL). L2 / (8) = 272 kN.m

Try W 460x 89

Cr = 3591 kN, Mr = 624 kN.m (laterally supported)

=+r

f

r MM

CP )85.0(

ok0.1624

272)85.0(

35912350

=+

HSS 152x152x10 After capacity design

C7b

T6b C6b

T7b

p

HSS 203x203x13 After capacity design

Ө

Example beam of level 7

F6


Intersection beam Example beam at F7

T8 = T7 = Ag . Ry . Fy = 2005 kN C8 = C7 = 1.2 Cp = 1222 kN (effective length 5.38 m) ± P = 0.5 (Ti + Ci) cosӨi - 0.5α ( Ti+1 + Ci+1) cosӨi+1 , α = 0.75 ± P = 0.5 (2005 + 1222) (0.743) - (0.75).(0.5).(2005+1222).(0.743) = 300 kN Mf = MDL + 0.5 MLL

HSS 152x152x10T8b

C7b T7b

C8b

p p → → F7

2.0 m

HSS 152x152x10


Try W 310x 24, Mf = 45.7 kN.m ( From SAP2000 analysis) Cr = Tr = 958 kN, Mr = 78.3 kN.m ( 2.0 m unbraced length)

okMM

CP

r

f

r99.0

3.787.45

958396

=+=+

Capacity design of column Example storey (6)

(1) Forces Based on Brace capacities T = Ag Fy Ry C = 1.2 Cp

2005

1222

3565

3040

3565

HSS 152x152x10

HSS 152x152x10

HSS 203x203x13

HSS 203x203x13

HSS 203x203x13

F8

F6

F4


1 2 3 4 5 6 7

Cr or Tu

SinӨ

SRSS Earthquake(RdRo) =

1.3

EQ (Smallest

of 3 and 4)

D+0.5L+P∆ EQ + D +0.5L+P∆

1341 1341 358 358 406 764

817 4543 2080 2080 596 2676

2385

2033 7273 4660 4660 1003 5663

2385

SRSS combines the maximum brace contribution at any level above the considered

level with the square root of the sum of the squares of all other brace contributions

about that level.

Example force at storey S4:

SRSS = 2033 + 2385 + 222 )1341()817()2385( ++ = 7273 kN

The column at this level should be designed to resist an axial load = 5663 kN resulting

from the combination of the EQ and the (DL+0.5LL+P∆).

Column should be Class 1 or 2 and should resist Mf = 0.2 Mp (S16-01-27.5.5.2)

2005

1222

3565

3040

3565

CHAPTER 7

SEISMIC DESIGN OF REINFORCED CONCRETE STRUCTURES

Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-1

MOMENT RESISTING FRAMES

(MRF)

Ductile MRF => Rd = 4.0

R0 = 1.7

Moderately Ductile MRF => Rd = 2.0

R0 = 1.4

Conventional MRF => Rd = 1.5

R0 = 1.3


Factored, Nominal and Probable Resistances:

Mr = Factored Flexural Resistance

Φc = 0.65 / Φs = 0.85

Mn = Nominal Flexural Resistance

Φc = 1.00 / Φs = 1.00

Mp = Probable Flexural Resistance

Φc = 1.00 / Φs = 1.25

20.1≈rMnM

For Beams, 47.1≈rMpM

For Columns, 57.1≈rMpM


Seismic Provisions for Ductile MRF


Beams (21.3.1.1) -Ductile Flexural Members

- Axial compressive Force 10

'cg fA

- Lclear ≥ 4d

- db

0.3

-b 250 mm -b width of supporting member + 2 x 3/4d

¾ d max.

b

¾ d max.

Plan View


Beams (cont..)

vejoM +

int ≥ 2int

vejoM −

vetionanyM +

sec ≥ 4intjoM

vetionanyM −

sec ≥ 4intjoM

(21.3.2.2)

M-ve joint

M+ve joint

ρmin = 1.4bwd/fy ρmax = 0.025 min. 2 continuous bars (21.3.2.1)


Beams (cont..)

*Lap splices shall not be used: -within the joints

-within a distance 2d from the face of the joint

-within a distance d from any plastic hinge caused by

inelastic lateral displacements

(21.3.2.3)

S ≤ d/4 8 db 24dst. 300 mm

2d

≤ 50 mm

S ≤ d/2 S ≤ d/4 100 mm


Beams (cont..)

Hoops Detailing

Cross -tie

All extensions > 6db 60 mm


Beams (cont..) Shear Strength Requirement: Vr ≥ Vf Vr ≥ The smaller of Vp

V determined from analysis with RdR0 = 1

Vp => shear corresponding to the probable moment resistance at the

forces of the joints while the member is loaded with the tributary

transverse load

Vr can be calculated using Clause 11 with β = 0 and θ = 450

(Vr = Vc + Vs)

Vc = φcλβ√f'c bwdv

Vs = φsAvfydvcotθ / S

Av = area of shear reinforcement within a distance s

dv = effective shear depth = maximum of 0.9d or 0.72h

λ = factor to account for density of concrete

θ = angle of inclination of compressive stress with longitudinal bar

β = factor to account for the shear resistance of cracked concrete


Columns (21.4.1) -Ductile Frame members subjected to Flexural and Axial Loads - width 300 mm -b/h ≥ 0.4

-P > 10

'cg fA

trcM

LnbM (include slab RFT)

brcM

RnbM (include slab RFT)

Strong Column Weak Beam ∑ Mrc ≥ 1.1∑Mnb


Columns (cont..) *Longitudinal RFT:

0.01 ≤ ρg ≤ 0.06

ld

1.3ld

Tension Lap splice within the centre half of the member length

Tension Lap splice at any section


Columns (cont..) Transverse RFT:

Ash ≥ 0.2kn kp (Ag/Ach) (f'c/fyh) s hc , kn = 2−nl

nl

Ash ≥ 0.09 ( f'c/fyh) s hc kp =

0PPf

* Stirrups:

Ach = Core area hc = dimension of concrete core perpendicular to the direction of hoop bars

Spacing > 6 db 150 mm

l0

l0

< 1.5 h clear height /6

Spacing > ¼ b 100+(350-hx)/3 6db

IF Pf ≤ 0.5φc f'c Ag

IF Pf > 0.5φc f'c Ag

< 2h clear height /6

l0


Column Cross sections:

hx hx

hx (21.4.4.4)

hx = Distance between laterally supported hx > max. of 200 mm

core dimension/3


Shear Strength Requirements

Vcol = u

bcoltcol

lMM +

≥colrV Vf

≥colrV The smaller of Vcol

V determined from analysis using RdR0 = 1

Vr can be calculated using Clause 11 with β ≤ 0.1 and θ ≥ 450

K2

K1

lu

veprM +

veprM −

tcolM =

)(21

1 vepr

vepr MM

kkk −+ ++

bcolM


Beam-Column Joints: • Column stirrups should continue through the joint except if the joint is

confined by Structural members as shown;

h

< 3/4h

b

< 3/4b


Applied shear on Joints:

Vfb = 1.25 Asfy - Vcolumn

Vcolumn

1.25 As fy


Vfb = 1.25 As1fy + 1.25As2fy - Vcolumn

Vcolumn

C2 =1.25As2fy

1.25As2fy

C1 =1.25As1fy


Allowable shear in beam-column joints (21.5.4.1) (a) Continued joints:

2.2 λ φc√f'c AJ

(b) Joints confined on 3 faces or two

opposite faces:

1.6 λ φc√f'c AJ

or

(c) All others:

1.3 φc√f'c AJ

AJ is lesser of Ag of column or bw hcol


* ldh = max of 8db 150 mm

0.2 'c

y

f

f db

* without the standard 900 hook

=> use 2.5ldh if depth of concrete below bar ≤ 300 mm

=> use 3.5 ldh otherwise (21.5.5.4)

ldh

12db

(21.5.5.2)


Seismic Design of a Ductile Moment Resisting Frame

7.40 m

2.20 3.00 2.20

7.40 m 7.40 m

• Analysis of the Moment Resisting Frame (MRF) was carried out using

plane frame finite element analysis program.

• Three load cases were considered:

o Dead Load (D)

o Live Load (L)

o Earthquake Load (E)

• To account for reduction of member stiffness due to concrete cracking

(Clause 21.2.5.2.1)

o IBeam = 0.4 Ig

o IColumn= g g0.5 0.6* * I 1.0 (0.6 0.7 I )' *

s

c g

ff A

⎛ ⎞+ ≤ →⎜ ⎟

⎝ ⎠


Design of the First Story Beam

CLA B

a b c

B.M. values are recorded in Table 1

Table 1: Bending moment along the beam [kN.m] (no redistribution) AB a b BA BC c D -117.45 113.43 88.26 -174.40 -159.10 85.10 L -55.77 54.28 42.10 -83.45 -76.02 40.58

E (±) 120.60 48.28 48.13 119.00 119.50 48.49 Load Cases 1.25D+1.5L -230.47 223.21 173.48 -343.18 -312.91 167.251.0D+1.0E 3.15 161.71 136.39 -55.40 -39.60 133.591.0D-1.0E -238.05 65.15 40.13 -293.40 -278.60 36.61

1.0D+0.5L+1.0E -24.74 188.85 157.44 -97.13 -77.61 153.881.0D+0.5L-1.0E -265.94 92.29 61.18 -335.13 -316.61 56.90Design moments

Max. Moment 3.15 223.21 173.48 -55.40 -39.60 167.25 Min. Moment -265.94 65.15 40.13 -343.18 -316.61 36.61

Negative moments are calculated at the face of the columns - For better, more economic design; moment distribution will be applied to reduce –ve B.M. - For a ductile MRF, CSA allows for a maximum redistribution of 20% (clause 9.4.2) - For simplicity only D & L B.M. will be redistributed B.M. values after redistribution are recorded in Table 2


Table 2: Bending moment along the beam [kN.m] (after redistribution)

AB a b BA BC c D -93.96 140.35 117.62 -139.52 -127.28 116.92 L -44.62 67.11 59.16 -66.76 -60.82 55.78

E (±) 120.60 48.28 48.13 119.00 119.50 48.49 Load Cases 1.25D+1.5L -184.37 276.10 235.76 -274.54 -250.32 229.83 1.0D+1.0E 26.64 188.63 165.75 -20.52 -7.78 165.41 1.0D-1.0E -214.56 92.07 69.49 -258.52 -246.78 68.43

1.0D+0.5L+1.0E 4.33 222.18 195.33 -53.90 -38.19 193.30 1.0D+0.5L-1.0E -236.87 125.62 99.07 -291.90 -277.19 96.32 Design moments

Max. Moment 26.64 276.10 235.76 -20.52 -7.78 229.83 Min. Moment -236.87 92.07 69.49 -291.90 -277.19 68.43

Design for flexural reinforcement max

600 2524 24

jb

ld = = = (Clause 21.5.5.6)

Maximum bar size that can be used is 25M (25.2 mm in diameter) Section at joint B From Table 2 Mf = -291.9 kN.m. Assuming flexural lever arm of 0.75 h = 0.75*600 = 525 mm

6

2291.9*10 1635.3 0.85*400*525

sA mm= =

Slab reinforcement within (3*hslab) from the side of the beam is effective 4-10 M bars are effective (As = 400 mm2) As required = 1635.3 – 400 = 1235.3 mm2

Try 4- 20 M (As = 1200 mm2) Note: it would be unwise to be too conservative when designing top reinforcement. Since beam shear, joint shear, column moments and column shears will be increased.


Clause 21.3.2.2: 0.5*ve veR RM M+ −≥ at column face

min max0.25*R RM M≥ at any section Assume As

+ve = 4-15M As = 1200 +400 = 1600 mm2

d = 700 – 30 – 11.3 – 19.2/2 = 649 mm

0.72 0.72*700 504

0.9 0.9*649 584.1 v

hd

d⎧ ⎧ ⎧

= = =⎨ ⎨ ⎨ ⇐⎩ ⎩ ⎩

A’s = 4 * 200 = 800 mm2

d ’ = 30 – 11.3 – 16/2 = 49.3 mm f’c = 30 MPa α1 = 0.85 – 0.0015 * f’c = 0.805 β 1 = 0.97 – 0.0025 * f’c = 0.895

4 - 20 M 10 M @ 300 mm top&bottom

4- 15 M

110

700

400

C

0.0035

Ts = As* Φs*fy = 1600 * 0.85 * 400 = 544*103 N C’s = As* ( Φs*fs – α1 * Φc*f’c ) = 800 * (0.85 * fs – 0.805 * 0.65 * 30) N C s = α1 * β1* Φc*f’c * b * C = 0.805 * 0.895 * 0.65 *30 * 400 * C 0 xF = ⇒∑ -Ts + C’s + C s = 0 ⇒ 544*103 – 680 * (fs – 18.47) – 5619.7 * C = 0 …. (1)

Compatibility of strains'

' 49.3 1 0.0035 1s ccdc c

ε ε⎛ ⎞ ⎛ ⎞⇒ = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

's s

49.3f = E * 700 1s cε ⎛ ⎞= −⎜ ⎟

⎝ ⎠ .… (2)


Solving (1) and (2) ⇒ C = 69.74 mm ,and fs = 205.1 MPa a = 0.895*69.74 = 62.42 mm

( )' 'f 333.3 kN.m. > M = 291.9 kN.m

2ve

cR saM C d C d d− ⎛ ⎞= − + − =⎜ ⎟

⎝ ⎠

Clause 21.3.2.1

min

max

1.4* * 1.4*400*649 908.6 o.k400

0.025* * 0.025*400*649 6400 > o.k

ws s

y

s s

b dA Af

A bw d A

= = = <

= = =

Bottom Bars for +ve reinforcement Beffective of the T-Section = 2*7.4/10+0.4 = 1.88 m For Section a Try As = 8-15 M d ≈ 635 mm

s

'c 1

* * 1600*0.85*400 18.43 mm* * * 0.65*0.805*30*1880

s y

c

A faf b

φφ α

= = = < tflange=110 mm

f

18.34*( ) 1600*0.85*400*(635 ) 340.42 kN.m. > M = 276.1 kN.m.2 2

vesR

aM T d− = − = − =

O.K.

MR

+ve at column face As= 4-15 M =800 mm2

d ≈ 650 mm a = 8.72 mm ⇒ MR

+ve = 175.6 > 0.5 * 333.3 = 166.6 kN.m


MR for section with As-ve = 2- 20 M (needed later for bar cut-off)

for simplicity ignore bottom reinforcement

a = 33.77 mm ⇒ MR = 129.5 kN.m

2- 20 M

400

Design of transverse reinforcement

In order to avoid brittle shear failure of beams; shear deformation should

always be in the elastic range. Shear design is based on the probable moment

of resistance of the beam, so we assure that when plastic hinges are formed

at the ends of the beam, the beam is still in the elastic range regarding shear

deformation.

Mp-ve = 1.47 * 333.33 = 490.00 kN.m

Mp+ve = 1.47 * 175.60 = 258.00 kN.m


137.5 kN 137.5 kN

490.0 kN.m 258.0 kN.m

2.2 3 2.2

238.6 kN 36.4 kN

P = 1.0 D + 0.5 L = 111 + 0.5* 53 = 137.5 kN

238.6 kN

101.1 kN

36.4 kN

490.0 kN.m

34.9 kN.m

338.2 kN.m

258.0 kN.m

Region of Plastic Hinging L = 3.0 m

m m m

A B

Note: when reversing the direction of the lateral load ⇒ B.M.D. & S.F.D. will be mirrored i.e. MB = 490 kN.m and VB = 238.6 kN


Section at column face Vf = 238.6 kN Clause 21.3.4.2 β =0 (excessive cracking is expected at high lateral drift due to E.Q.) θ = 45o

s * * * *cot( )v y v

sA f dV

sφ θ

=

Try 4-legged 10 M, Av=400 mm2

3

0.85*400*400*584.1*1 332.9 mm238.6*10

S = =

Check shear design requirements

(i) Maximum shear (clause 11.3.3) '

, max c

f

0.25* * * * = 0.25*0.65*30*400*584.1=1139.0 kN < V = 238.6 O.K.

r w vcV f b dφ=

(ii) Minimum amount of stirrups (11.2.8.2)

'

* 400*400 1217 mm O.K.0.06* 30 *4000.06* *

v y

wc

A fSf b

≤ ≤ ≤

(iii) Spacing limit (11.3.8.3)

( )

'c f

max

max

0.125* * * * 0.125*0.65*30*400*584.1 569.5 kN > VNo reduction in S is needed

S =

w vcf b dφ = =

⇒600

S O.K.0.7*584.1 = 408.87 mm ⎧

>⎨ ⇐⎩

Check “Anti-buckling” requirements (Clause 21.3.3.2)

Hoops are required for a distance of (2d) from the face of the column. The maximum hoop spacing (S) is

(i) d/4 = 649/4 = 162.25 (ii) 8*d bar, longitudinal = 8*19.5 = 156 ⇐ governs (iii) 24* d bar, hoops = 24*11.3 = 271 (iv) 300


Use 4-legged 10 M hoops @150 mm for a distance of (2d) = 1.25 m Section at mid span Vf = 101.1 kN Use 2-legged U-shape stirrups 3

0.85*200*400*584.1*1 392.86 mm101.1*10

S = =

Check plastic hinging region (Clause 21.3.3.1) Hoops have to be provided for the regions where plastic hinging may

occur and for a distance (d) beyond it

from previoiusly drawn B.M.D. LPlastic Hinging ≈ 3.0 m

hoops has to be provided for a distance of

3.0 + 0.649 = 3.65 m (from both column faces as the B.M.D.

is mirrored case of reverse lateral loading)

7.4 – 2*3.65 = 0.1 m (insignificant length. Provide hoops for the whole span) Reinforcement cut-off

Table 3.1 (Page 3-15 Part II of the CSA A23.3)

For M 20

Ld= 530 mm (for bottom bars k1=1.0)

= 1.3 * 530 = 689 mm (for top bars k1=1.3)

For M 15

Ld= 390 mm (for bottom bars k1=1.0)

= 1.3 * 390 = 507 mm (for top bars k1=1.3)


The theoretical cut-off location of the top 2-M 20 is @ 1.51 m from the column face.

Clause 12.10.4 states that an embedment length of at least (d) or (12 dbar) is provided

beyond the theoretical cut-off point.

cot( ) 1.51 0.584*cot(35) 2.34 m

max 0.69 0.584 1.28 m12*

r v

dbar

L dL d

ld

θ+ = + = ⇐⎧⎪= ⎧⎨ + = + =⎨⎪

⎩⎩

Bar Splice (Clause 12.15.1) For the 2-20 M bars LSplice = 1.3 * Ld = 1.3* 689 ≈ 900 mm


2-20M L =3.1 m

4-15M L=8.15 m

2-20M L=5.5 m

490.0

258.0338.2

338.2258.0

490.0

276.1

2-10M L=3.0 m

4-15M L=7.9 m

2-20M L =3.65 m

2-20M L =8.3 m

340.7175.6

4-15M L =5.7 m

1.25 D + 1.50 L

Formation of Plastic H

inges at bends of the beam

(Cut-off of bottom

renforcement)(C

ut-off of top renforcement)

0.92-20M

L=8.3 m2-20M

L=5.5 m2-10M

L=3.0 m

2- Legged10 M

hoops@

150 mm

1.71

22

11

theoretical cut-off Point ( M = 175.6 )

1.51

M = 129.5f

1.51

M = 129.5r

f

CL

235.8

1

L = 2.34+0.50+0.24 ˜ 3.1 m1

L = 300 mm

(taken as 500 - column dim

ension)dh

12 d = 240 mm

b

L = 3.45+0.50+0.24+0.45 ˜ 4.65 m2

2

L = 2*( 3.45+0.45)+0.50 ˜ 8.30 m3

3 175.6

M r

M f

4

L = 2*( 2.00+0.584*cot(35)) ˜ 5.7 m4

4-15M L =5.7 m

4


4 - 20 M 10 M @ 300 mm top&bottom

4- 15 M

110

400

Section 1-1

2- 20 M + 2- 10 M 10 M @ 300 mm top&bottom

8- 15 M

110

700

Section 2-2

400


Design of the first story interior column From F.E. analysis, the straining actions at top & bottom of the column in

consideration are depicted in Table 3.

Factored axial load and B.M. are shown in Table 4.

Table 3: Straining actions at the column ends

Location PD PL PE (±) MD ML ME (±) VD VL

VE (±)

Units kN kN kN kN.m kN.m kN.m kN kN kN.m Bottom -1329.68 -540.89 2.68 2.62 1.37 213.81 2.56 1.33 93.33

Top -1329.68 -540.89 2.68 6.35 3.27 112.84 2.56 1.33 93.33

Table 4: Factored Axial loads and Moments 1 2 3 4 5 1.25D+1.5L 1.0D+1.0E 1.0D-1.0E 1.0D+0.5L+1.0E 1.0D+0.5L-1.0E

Pf,Bottom [kN] -2473.44 -1327.00 -1332.36 -1597.45 -1602.81

Pf,top [kN] -2473.44 -1327.00 -1332.36 -1597.45 -1602.81

Mf,bottom [kN.m] 5.33 216.43 -211.19 217.11 -210.51

Mf,top [kN.m] 12.85 119.20 -106.49 120.83 -104.85


Preliminary selection of column Try 500 X 500 mm

As = 8 – 25 M bars = 4000 mm2

As,min = 0.01*500*500 = 2500 < As

O.K.

As,max = 0.06*500*500 = 15000 > As

O.K.

(Clause 21.4.3.1)

500

500

8 - 25 M


Check column capacity Draw the section interaction diagram and check if all load combinations is within the limits

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

0 50 100 150 200 250 300 350 400 450 500

B.M [kN.m]

P [k

N] 1

5 4

3 2

Strong column-weak beam requirements (clause 21.4.2.2) To ensure the formation of plastic hinges in the beams CSA A23.3 requires that: Σ Mnc ≥ Σ MPb

For case 2 Mr = 411.0 kN.m Mn= 1.2* Mr = 493.2 kN.m

For case 3 Mr = 415.0 kN.m Mn= 1.2* Mr = 498.0 kN.m For case 4 Mr = 425.5kN.m Mn= 1.2* Mr = 510.6 kN.m For case 5 Mr = 429.0kN.m Mn= 1.2* Mr = 514.8 kN.m Case 1 does not involve lateral load Σ MPb = 490 + 258 = 748.0 kN.m Σ Mnc = 2*493.2 = 986.4 kN.m Σ Mnc > Σ MPb O.K.


Design of transverse reinforcement of the column (Clause 21.4.5.1) Vmax = 2.56 + 0.5*1.33 + 93.33 = 96.55 kN Shear in columns when plastic hinges are formed in beams Pr Pr

1( )* (490 258)* 374 2top

ccol

c

kM M M kNk

+ −= + = + =∑

The first story column is connected to a strong foundation, hence it is expected that the column will hinge at the bottom

,max

1.57*432 678.24 .ColumnPM kN m= =

374 678 277 3.8ColumnV kN+

= =

Vdesign = 277 kN Shear Design (clause 21.4.5.2) 0.65*0.1* 30*500*(0.72*500) = 64.08 cV kN= 277 - 64.08 = 212.9 sV kN= Using transverse reinforcement as depicted in figure Av = (2+2*cos(45)) *100 = 341 mm2

3

0.85*341*400*0.72*500*cot(45) 196.0 212.9*10

S mm= =

'c f

max

0.125* * * * 0.125*0.65*30*500*0.72*500 438.75 kN > V600 mm

S = S O.K.0.7*0.72*500 = 252 mm

w vcf b dφ = =

⎧⇒ >⎨ ⇐⎩

'

* 341*400 830 mm O.K.0.06* 30 *5000.06* *

v y

wc

A fSf b

≤ ≤ ≤

Confinement requirements (Clause 21.4.4.2)

8 1.33

2 6l

nl

nkn

= = =−


'1* ( ) *

= 0.805*30*(500*500-4000)+400*4000=7541 kN1602 0.2127541

o g st stc y

fp

o

P f A A f A

PkP

α= − +

= = =

Hence the total area of reinforcement is :

'

2

2

0.2* * * * * *

500 30 = 0.2*1.33*0.212* * *420*420 400

= 2.5175 * S

g csh n p c

ch yh

fAA k k S hA f

S

=

'

, min

2 2

300.09* * * 0.09*420* * 2.835*400

341 341 mm S= 120.3 mm2.835

csh c

yh

sh

fA S h S Sf

for A

= = =

= ⇒ =

Anti-buckling requirements

Maximum spacing is :

(i) b/4 = 500/4 = 125 mm (ii) 6*d bar * 25 = 150 mm (iii) 350 350 190100 100 153 mm

3 3x

xhS − −

= + = + =

h x

Use 10M hoops @ 120 mm As this is the first story column, hoops shall be provided for the whole clear height (clause 21.4.4.6).


Splice details

Columns bars should be spliced at mid-height with a tension lap splice

Ld=820 mm (Table 3.1 A23.3 page 3-15) Lap length = 1.3*820 = 1066 mm

500

500

8 - 25 M

1066

Column Reinforcement Details


Check Joint Shear

VCol.

F1F2

f 1f 2

Vjoint = F1+F2-Vcol.

1

2

1.25* * 1.25*800*400 400 1.25*1200*400 600

374*2 197 3.8

col

F As fy kNF kN

V kN

= = == =

= =

Vjoint = 600 + 400 – 197 = 803 kN

', Joint

Joint

2.2 * * *

= 2.2*1.0*0.65* 30 *500*400 = 1566.5 kN > OK

r c jcV f A

V

λ φ=

258

374

374

490

374

258

374

374

490

374


Shear Walls


Ductile Shear Walls

Rd = 3.5, R0 = 1.6

Moderately Ductile Shear Walls

Rd = 2.0, R0 = 1.4

Conventional Constructional Shear Walls

Rd = 1.5, R0 = 1.3


Check of Ductility of Shear Walls (21.6.7)

o Walls effectively continuous in cross section

o Plastic hinge at the base

θid = 004.0)2/(

0 ≥−

∆−∆

ww

wfdf

lhRR γ

θic = 025.0)002.02

( ≤−clwcuε

θid = inelastic rotational demand on a wall

θic = inelastic rotational capacity of a wall

εcu = 0.0035

lw = length of wall

hw = vertical height of wall

γw = wall overstrength factor

= loadfactored

resistancenominaltoscorrespondloadtheofRatio 1.3

C

-ve

+ve 0.0035


Ductile Flexural walls (21.6.1.1)

hw

lw

Plastic hinge Region = 1.5 lw

For hw/lw > 2.0 Rd = 3.5

For hw/lw ≤ 2.0 Rd = 2.0

( 21.6.3.1)

Effective flange width

> ½ distance to adjacent wall 25% of the wall height above the Section under consideration


For all elevations above the plastic hinge region the design SF and BM shall

be increased by a ratio of the Factored Moment resistance to the factored

moment (21.6.2.2.C)

Factored resistance

Factored moment

Original design BM

BM* factored resistance/factored moment


Rectangular wall (Plastic hinge Region)

If C> min. of 4 bw => bw > lu/10 (21.6.3.4) 0.3lw lu : Clear distance between floors 0.0025 ≤ ρdistributed ≤ 0.0600 (21.6.4.3) 0.0015 bwlw ≤ Asconcentrated (min. 4 bars) ≤ 0.06 x Area of concentrated RFT

(21.6.4.3) C 0.55lw If C> 0.14 γwlw => confine the compression region as a column γw = 1/φs = 1.18 (over strength factor)

Concentrated RFT Spacing of distributed RFT ≤ 300 mm (21.6.5.2)

lw

Strain diagram 0.0035

C

bw


db 1/10 bw (21.6.4.4) 25 cm wall => db 25 mm 20 cm wall => db 20 mm *Tie concentrated RFT as columns: Spacing of Ties least of 6db (21.6.6.9) 24 dst ½ bw *Horizontal RFT ratio 0.0025 (21.6.5.1) *Splices:- Concentrated RFT: * max. 50% at the same location *1.5 ld * half the height of the each floor must be clear of splices Distributed RFT: * 1.5ld


Rectangular wall (Outside Plastic hinge Region)

0.0025 ≤ ρdistributed ≤ 0.0600

(21.6.4.3) 0.0010 bwlw ≤ Asconcentrated (min. 4 bars) ≤ 0.06 x Area of (21.6.4.4) concentrated RFT (21.6.4.3) *db 1/10 bw (21.6.4.4) *Ties for concentrated RFT are the same as those for columns. *Horizontal RFT ratio 0.0025 * Splices: 1.5ld (21.6.4.1)

Spacing of distributed RFT ≤ 450 mm (21.6.5.1)

lw

bw


Shear Strength Requirements for Ductile Walls Vr > Vf Vr ≥ the smallest of Vp -> shear corresponding to the development of the probable nominal capacity of the wall V determined from the analysis with RdR0 = 1 For θid ≥ 0.015 => shear demand 0.10φcf'

cbwdv β = 0 For θid = 0.015 => shear demand 0.15φcf'

cbwdv β = 0.18 For Ps = 0.1f'

cAs θ = 450 For Ps = 0.2f'

cAs θ = 350

CHAPTER 8

LESSONS LEARNED FROM THE PERFORMANCE OF BUILDING DURING PAST EARTHQUAKES


Post earthquake reconnaissance reports have provided overviews of

damage patterns and detailed descriptions of specific failure modes and

structural weaknesses. In addition, many reports have attempted to explain

the behaviour of individual buildings by correlating analytical and

experimental predictions with observed damage. The major lessons learned

from past earthquakes have repeated themselves over and over and have

confirmed the principles applied in modern seismic codes.

Lessons learned

1) Designing to Code does not always provide safeguard against

excessive damage in severe earthquakes.

2) Well-designed, well detailed and well constructed buildings usually

resist earthquake loads without excessive damage.

3) Poor construction practice can lead to severe damage and collapse

4) Ground failure can cause severe damage and collapse.

5) Buildings subjected to successive earthquakes may suffer progressive

weakening.

6) Ductility and redundancy provide safety against collapse.

7) Stiff elements, not considered in the design, affect the seismic

response of a building.

Primary examples are moment resisting frames that are filled with stiff

masonry bricks as shown in next page.


Typical damage Inflicted to Infill Walls and Concrete frame Elements

(From EERI, Pub. No. 86-02)

Framed bays with infill walls become stiffer once the frame is in contact

with infill wall. This increases the stiffness of the frame and consequently

the dynamic properties of the structural system.

If the infill walls are distributed unsymmetrically in plan, torsional modes of

vibration can be excited. The infill walls will crack if they are not designed

for the interaction forces occurring between frame and wall. The interaction

forces may also damage the frame.


8) Problems with soft stories:

A soft story can be generated

when a shear system is replaced at

the first story level with a number of

columns. Earthquakes have shown

that the stiffness discontinuity with

height often directs forces to places

where strength is minimal.

Structures perform better in

earthquakes when they have uniform,

or gradually changing, stiffness and

strength over the height. Abrupt

changes in stiffness or strength are

responsible for some of the dramatic

earthquake failure.

Soft Story Effect (From EERI, Pub. No. 86-02)

Soft story


9) Problems with vertical geometric irregularities

10) Importance of Horizontal Diaphragms:

Horizontal diaphragms (floor system) are responsible of carrying

seismic forces to the vertical-load resisting building elements. Failure of a

horizontal diaphragm leads to concentration of forces on certain elements of

the lateral resisting system.

11) Problems with horizontal (plan) irregularities:

a) Torsion irregularity

b) Reentrant corners

c) Diaphragms discontinuity

d) Non-perpendicular systems


12) Problems with corner columns

13) Problems with exterior panels and parapets

14) Unreinforced masonry buildings usually perform very poorly

15) Precast concrete elements must be well tied together.

16) Steel buildings generally perform well(except the extensive number of

failures in welded beam-column connections observed during the 1994

Northridge earthquake)

17) Damage caused by fires and damage caused to non-structural

components and building contents often exceed the consequences of

inadequate structural performance.

Torsion irregularity

opening

Reentrant corner

Non-perpendicular systems

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