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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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-2
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-3
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-4
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:
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-5
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-6
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-7
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-8
(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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-9
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-10
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 +=
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-11
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-12
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-13
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 1-14
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-1
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-2
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-3
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 =∴
∆ ∆ ∆
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-4
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-5
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-6
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-7
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(
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-8
-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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-9
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-10
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-11
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-12
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-13
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-14
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-15
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-16
)()( 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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-17
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-18
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-19
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-20
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-21
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-22
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-23
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-24
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-25
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-26
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-27
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-28
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-29
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-30
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-31
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-32
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?
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-33
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-34
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty . Page: 2-35
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 =
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-2
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-3
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-4
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-5
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-6
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-7
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-8
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-9
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-10
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:
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-11
[ ] [ ][ ]
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
=
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”
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-12
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-13
Sequence of Modal Combination:
Mode 1 Mode 2 Mode 3 Combination
Forces
Shear
Moment
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-14
Example 3-4
Use the modal analysis procedure to evaluate the response of
the frame shown in example 3-3 to El Centro earthquake.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-15
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-16
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-17
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-18
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-19
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-20
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty Page: 3-21
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-2
Local ductility µl = ∆m/∆y
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-3
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-4
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-5
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-6
• 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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-7
Design Spectral Acceleration
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-8
Definition of Soil Type
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-9
• 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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-10
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-11
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-12
Definition of Irregularities
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-13
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-14
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-15
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-16
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%
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-17
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-18
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-19
*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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-20
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-21
• 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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-22
Halchuk and Adams (2004)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty_____ Page: 4-23
• 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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-2
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, &
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-3
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-4
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-5
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-6
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-7
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-8
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-9
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-10
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-11
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-12
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-13
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-14
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)
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-15
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-16
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-17
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-18
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)
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-19
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)
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 5-20
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.
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-2
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-3
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-4
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-5
Moment Resisting Frames (Energy dissipates through yielding of beams)
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-6
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.
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-7
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)]
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-8
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.
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-9
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 +
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-10
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-11
Concentric Braced Frames
(Energy dissipates through yielding of bracing members)
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-12
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
force shall be increased
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
force shall be increased
by 3% per meter of
height above 16 m
H 40 m
If H > 32 m, seismic
force shall be increased
by 3% per meter of
height above 32 m
K-truss
N.A.
N.A.
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-13
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-14
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 Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-15
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-16
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 Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-17
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-18
• 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)
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-19
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-20
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-21
First mode shape (Period =1.46075 sec)
Second mode shape (Period = 1.32939 sec)
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-22
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-23
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-24
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-25
Strength design according to CISC 95 • North-south direction:
Two moment resisting frames along the outer edges
Elevation of Moment Resisting Frame
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-26
• 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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-27
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-28
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-29
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-30
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-31
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-32
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-33
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-34
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-35
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-36
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
Seismic Analysis and Design of Buildings By: Prof. A A. El Damatty . Page: 6-37
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-2
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 Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-3
Seismic Provisions for Ductile MRF
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-4
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-5
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-6
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-7
Beams (cont..)
Hoops Detailing
Cross -tie
All extensions > 6db 60 mm
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-8
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-9
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-10
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-11
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-12
Column Cross sections:
hx hx
hx (21.4.4.4)
hx = Distance between laterally supported hx > max. of 200 mm
core dimension/3
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-13
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-14
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-15
Applied shear on Joints:
Vfb = 1.25 Asfy - Vcolumn
Vcolumn
1.25 As fy
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-16
Vfb = 1.25 As1fy + 1.25As2fy - Vcolumn
Vcolumn
C2 =1.25As2fy
1.25As2fy
C1 =1.25As1fy
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-17
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-18
* 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 Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-19
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
⎛ ⎞+ ≤ →⎜ ⎟
⎝ ⎠
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-20
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-21
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-22
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-23
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-24
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-25
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-26
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-27
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)
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-28
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-29
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-30
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-31
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-32
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-33
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.
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-34
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
= = =−
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-35
'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).
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-36
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-37
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-38
Shear Walls
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-39
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-40
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-41
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-42
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-43
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-44
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-45
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
Seismic Analysis and Design of Buildings By: Prof. A. A. El Damatty____ Page: 7-46
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 8-1
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 8-2
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.
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 8-3
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 8-4
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
Seismic Analysis and Design of Buildings By: Prof. A. A El Damatty . Page: 8-5
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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