Lecture Notes OF Dynamic Analysis

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 Title Publisher

Chopra, Anil K. Dynamics of structures: theory and applications to earthquake engineering / Anil K. Chopra. Englewood Cliffs, N.J. : Prentice Hall, 1995.

3) Steel Structures Author Title Publisher

Bruneau, Michel, Ph.D. Ductile design of steel structures / Michel Bruneau, Chia-Ming Uang, Andrew Whittaker. New York : McGraw-Hill, 1998.

Author Title Publisher

Canadian Institute of Steel Construction Handbook of Steel Construction th Universal Offset Limited, 9 Edition, 2007.

4) Concrete and Masonry Structures Authors Title Publisher

Thomas Paulay, and M. J. N. Priestley Seismic Design of Reinforced Concrete and Masonry Buildings/ Thomas Paulay, and M. J. N. Priestley Willey- Interscience, 1992

Author Title Publisher

Cement Association of Canada Concrete Design Handbook rd Ottawa, 3 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.



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 V p and Vs. The velocities usually satisfy the following relation: V p V s

≈

√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 X f , hence: Xf = V p ∆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

X1 ∆T 1V p

3 −1

∆T 3V p

3 −1 X3

∆T 2V p

3 −1

X2

X1, X2, and X3 are recording stations

Fig. 1 Determination of Epicenter



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.



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:



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)



Page: 1-6 .



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



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)



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. Extensive damage Buildings abandoned

No damage 50m 200 m

1000 m Fig. 4



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 -1/2

of terms proportional to R e

-1

-2

, Re , R e and ln R e ; where R e 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:

a0 =C 1e

C 2 M C ( Re +C 3 ) 4



Page: 1-11 .

where: a0 is the peak ground acceleration M is the Richter magnitude of the earthquake R e 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



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:

-1.95 + 0.32I

(PGA) = 10

m/sec



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.



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



Page: 2-1 .

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

1. 2. 3. 4. 5.

Mass (m = W/g) Applied load P(t) Inertia force = F I Damping force = F D Elastic force = F S

Single Degree of Freedom Model: x(t)

C

P(t)

M K

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. FD FS

P(t)

FI

FI + FD + FS = P(t)

Where: FI = mass x acceleration = m

d 2 x 2

dt

= m x&&

FD = Dashpot constant x velocity = c x& FS = Spring stiffness x displacement = kx

∴ Equation of motion:

&& + c x& + kx = P (t ) m x



Elastic Stiffness for Elements:

k = P/∆

→ Force Required for Unit displacement

1) Springs in Series: K 1

K 2 P2

P1

P1 = P2 = P P1 = K 1∆1 ; K =

P

=

P

∆ ∆1 + ∆2

P2 = K 2∆2 ;

=

P P P 2 1

=

+

∆ = ∆1 + ∆2

K 1 K 2 K 1 + K 2

K 1 K 2

∴ K =

K 1 K 2

K 1 + K 2

2) Springs in Parallel: P1,

1

, K 1 P,

P2,

2

, K 2

∆1 = ∆2 = ∆ P P + P K ∆ + K 2∆2 K = = 1 2 = 1 1

∆

∆

∴ K = K 1 + K 2

∆

,K

P,

Page: 2-2 .



Page: 2-3 .

3) Axial Stiffness of a Bar:

P

∆=

PL EA

∴ K =

L

EA L

4) Torsion: T

θ =

TL GJ

∴ K =

L

GJ L

5) Beams in Flexure: a) Cantilever:

∆=

PL3 3 EI

∴ K =

P

3 EI

L

L 3

b) Simple Beam: P

∆=

PL3 48 EI

∴ K =

48 EI L 3

L



c)

Page: 2-4 .

P

∆=

PL3 12 EI

∴ K =

L

12 EI L 3

P

d)

∆=

PL3 3 EI

∴ K =

3 EI L

L

3

e)

K 1

K 1

=

K 2

12 EI 1

K 2

L 13

=

12 EI 2 L 32

K 1 and K 2 are two springs in parallel (have the same displacement)

∴ K = K 1 + K 2 =

12 E 3 1

L

( I 1 + I 2 )



Page: 2-5 .

Types of Dynamic Loading: & X 0

a) Free Vibration: X(t)

X0

0

2

4

6

Time

8

b) Forced Harmonic Vibration: P(t)

P0

P (t ) = P 0 sin Ω t

2

(Machine Vibration)

c) Impulse Loading:

P(t)

Time

4

6

P0

(Driving of Piles)

Time

d) Random Vibration: P(t) 0

200

400

600

800

1000

1200

1400

1600

1800

Time



Page: 2-6 .

2-2 Free Vibration Response of a SDOF

&& + kx = 0 Equation of motion: m x

∴ x&& + ω 02 x = 0

;

ω 02

=

k m

Solution of the above D.E.: X(t) = Asin(ω0t) + Bcos(ω0t)

& (0) = X & Substituting with the initial conditions: X(0) = X 0 and X 0 & X 0

X(t) = ω sin(ω 0t ) + X 0cos(ω 0t ) 0

we obtain:

The above equation can be written in a compact form as: 2 & ⎛ X ⎞

X(t) =

− ⎛ X 0 ⎞ ⎟⎟ φ = tan 1⎜⎜ & ⎝ X 0 / ω 0 ⎠

⎜⎜ 0 ⎟⎟ + X 02 [sin(ω t + φ )] 0 ⎝ ω 0 ⎠ & X 0 T=2 /

X(t)

0

X0

2

4

6

8



Page: 2-7 .

2-3 Undamped Free Vibration of a Damped System Equation of motion:

c = ξ Define: 2mω 0

&& + c x& + kx = 0 m x c k && + x& + x = 0 x m m c && + x& + ω 02 x = 0 x m (damping ratio)

2 Equation of motion: x&& + 2ω 0ξ x& + ω 0 x

=0

For 0 < < 1

The solution of the above equation becomes: X(t) = e

−ξω 0t

Where: ω D

[ A1sin(ω Dt ) + A2cos(ω Dt )]

=

1 − ξ 2 ω 0

(Frequency of vibration of a damped system)

A1 =

& + ξω X X 0 0 0 ω D

A2 = X 0

The above equation can be written in a compact form as: −ξω 0t

X(t) = Xe

[sin(ω D t + φ )]



Page: 2-8 .

1 0.8 0.6 0.4 0.2 ) t ( X

0

-0.2 0

5

10

15

20

-0.4

TD = 2 /

-0.6

D

-0.8 -1 Time

Response to Harmonic Motion x(t)

C

P(t) = P0cos t or P0sin t M K 2 Equation of motion: x&& + 2ω 0ξ x& + ω 0 x

=

P0

cos Ω t _____________ (*)

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

Homogeneous: XC(t) =

e

−ξω 0t

[ A1sin(ω Dt ) + A2 cos(ω Dt )]

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



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:

P 0 X P =

mω 02

⎡ ⎛ Ω ⎞ ⎢1− ⎜⎜ ⎟⎟ ⎢⎣ ⎝ ω 0 ⎠

2

Amplitude of forced response

2

⎤ ⎡ ⎛ Ω ⎞⎤ 2 ⎥ + ⎢2ξ ⎜⎜ ⎟⎟⎥ ⎥⎦ ⎣ ⎝ ω 0 ⎠⎦

⎡ ⎛ Ω ⎞ ⎤ ⎢ 2ξ ⎜⎜ ⎟⎟ ⎥ ⎢ ⎝ ω 0 ⎠ ⎥ φ = tan−1 ⎢ 2⎥ ⎞ ⎢1− ⎛ ⎜⎜ Ω ⎟⎟ ⎥ ⎢⎣ ⎝ ω 0 ⎠ ⎥⎦

The angle at which the response lags the applied load

Adding the complementary and particular solutions, the total response is: P 0 cos(Ωt − φ ) 2 mω 0 −ξω t X (t ) = e 0 [ A1sin(ω Dt ) + A2cos(ω Dt )] +

⎡ ⎛ Ω ⎞ ⎢1− ⎜⎜ ⎟⎟ ⎢⎣ ⎝ ω 0 ⎠

=

transient term Decays exponentially and occurs at the damped frequency of the structure

+

2

2

⎤ ⎡ ⎛ Ω ⎞⎤ 2 ⎥ + ⎢2ξ ⎜⎜ ⎟⎟⎥ ⎥⎦ ⎣ ⎝ ω 0 ⎠⎦

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



Page: 2-10 .

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

X P =

P 0

1

mω 02

⎡ ⎛ Ω Ω ⎞ ⎢1 − ⎜⎜ ⎟⎟ ⎢⎣ ⎝ ω 0 ⎠

2

2

2 ⎤ ⎡ ⎛ Ω Ω ⎞⎤ ⎥ + ⎢2ξ ⎜⎜ ⎟⎟⎥ ⎥⎦ ⎢⎣ ⎝ ω 0 ⎠⎦⎥

P 0 P 0 = = X St i.e. K mω 02

. ∴ X P = X St DMF 1

DMF =

2

2 2 ⎡ ⎛ Ω ⎤ ⎡ ⎤ ⎛ Ω Ω ⎞ Ω ⎞ ⎢1 − ⎜⎜ ⎟⎟ ⎥ + ⎢2ξ ⎜⎜ ⎟⎟⎥ ⎢⎣ ⎝ ω 0 ⎠ ⎥⎦ ⎣ ⎝ ω 0 ⎠⎦

Dynamic Dynamic Magnific Magnific ation Factor 12 11

r 10 o t c a 9 F n 8 o i t a 7 c i f i n 6 g a 5 M c 4 i m a 3 n y D 2

ζ=0 ζ = 0.05 ζ = 0.1 ζ = 0.2

1 0 0

0.5

1

1.5 0

2

2.5

3



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 X 0 is reduced to 60 % of X0. The motor operates at 600 RPM and causes a harmonic force due to unbalance F 0= 2.5 KN. Neglecting the mass of the beam, determine: a) The value of the undamped frequency b) The value of the damped frequency

ω0.

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.



Page: 2-12 .



Page: 2-13 .



Page: 2-14 .



Page: 2-15 .

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

Xt

K Ϋg(t)

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:

&& g (t ) + x&&(t )) + c x& + kx = 0 m( y &&(t ) + c x& + kx = −m y&& g (t ) m x As such, due to earthquake, the SDOF is subjected to a dynamic && g (t ) force P (t ) = −m y Dividing the previous equation by the mass “m”, we get:



Page: 2-16 .

c k &&(t ) + x& + x = − y&& g (t ) x m m Defining the natural frequency of the system ω 0 = damping ratio ξ =

&& + 2ω 0ξ x& x

k m

and its

c 2mω 0 , the equation of motion can be written as:

+ ω 02 x = − y&& g (t ) The solution of the nd above 2 order D.E. can be obtained by

Time History Analysis

Response Spectrum Analysis

Provides the entire time response.

Provides only maximum responses.

Can be applied to linear and non-linear systems.

Applicable only to linear systems.

Involves significant computations.

Simple



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:

&&n + γ x&&n+1 ∆t x& n+1 = x& n + (1 − γ ) x x n

+1

= x n + x& n ∆t +

(12 − β ) x&&n + β x&&n+1 ∆t 2

_______________ (1) _______________ (2)



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:

&&n+1 + c x& n+1 + kxn+1 = F n+1(t ) m x

_______________ (3)

For constant – average acceleration scheme ( γ = 1/2 and β = 1/4):

x& n+1 = x& n x n +1

+ x&&n + x&&n+1 (∆t / 2)

= x n + x& n ∆t + [ x&&n + x&&n +1 ](∆t 2 / 4)

_______________ (4) _______________ (5)

From (5):

&&n +1 x

= [ x n +1 − x n − x& n ∆t ](4 / ∆t 2 ) − x&&n

_______________ (6)

Substituting (6) into (4), we get: x& n +1

= x& n + x&&n (∆t / 2) + [[ x n +1 − x n − x& n ∆t ](4 / ∆t 2 ) − x&&n ](∆t / 2) ___ (7)

Equation (7) can be simplified to give: x& n +1

= ( x n +1 − x n )(2 / ∆t ) − x& n

_______________ (8)

Substituting (6) and (8) into (3), we get: n ⎞ ⎛ 4 xn 4 x& n 2 x ⎛ 4m + 2c + k ⎞ xn+1 = F n+1(t ) + m⎛ n n ⎞ ⎜⎜ 2 + + x&& ⎟⎟ + c⎜⎜ + x& ⎟⎟ ___ (9) ⎜ 2 ⎟ ⎝ ∆t ∆t ⎠ ⎝ ∆t ∆t ⎠ ⎝ ∆t ⎠



Page: 2-19 .

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

A- Initial Calculations: 1. Evaluate k, m and c. 0 0 0 2. Initialize x , x& and x&&

3. Select a time step

∆t.

4. Calculate: a 0 = 4/∆t , a1 = 2/∆t, a2 = 4/∆t, a3 = ∆t/2 . 2

ˆ = k + a m + a c 5. Evaluate k 0 1

B- For Each time step: ˆ n +1 1. Calculate effective load at time t = (n+1) ∆t; F ˆ n+1 = F n+1 + m(a x n F 0

+ a2 x& n + x&&n ) + c(a1 x n + x& n )

ˆ x n +1 = F ˆ n +1 2. Solve k

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

&&n+1 x

= a0 [ x n+1 − x n ] − a2 x& n − x&&n

x& n +1

= x& n + a3 [ x&&n+1 + x&&n ]



Page: 2-20 .

Example 2-2

The frame structure shown below is subjected to a strong ground motion pulse with a peak acceleration A 0 = 1.2g. The latter is modeled as a triangular pulse load of duration t 1 equal to half the period of vibration of the structure. Assuming zero damping, determine by numerical integration the maximum displacement of the structure, X max, and also the time, tmax, at which it occurs. Use a time step

∆t=T/10 sec. X

W = 200 KN I=∞ W 250x33

m 0 . 3

m 5 . 4

Ϋg(t)

Ϋg(t)

A0

t1= 0.5T A0= 1.2g t1/2

t1

t



Page: 2-21 .



Page: 2-22 .



Page: 2-23 .



Page: 2-24 .



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

2π m = = T 2 π has a certain period ω 0 k and a damping ratio ζ. M x C K/2

K/2

x is the relative displacement between the mass and the ground

Ϋg(t)

2. Equation of Motion:

&& g (t ) + x&&(t )) + c x& + kx = 0 m( y &&(t ) + c x& + kx = −m y&& g (t ) m x && + 2ω 0ξ x& + ω 02 x Or x

= − y&& g (t )



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 S d for the response displacement is recorded. This maximum displacement S d(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 S v(T,ζ) can be obtained by the relation: Sv(T,ζ) = ω0Sd(T,ζ) 4. The maximum acceleration response S a(T,ζ) can be obtained 2

by the relation: Sa(T,ζ) = ω0Sv(T,ζ) = ω0 Sd(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.



Page: 2-27 .



Page: 2-28 .

The following observations could be noticed:

1. Systems with very short period, say T < T a = 0.025 s have Sa= üg0 and Sd is very small. Why?? 2. Systems with very long period, say T > 15 s have S d approaches üg0 and Sa is very small. Why?? 3. For short–period systems, S a can be idealised as constant. 4. For medium–period systems, S v can be idealised as constant. 5. For long–period systems, S d 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 ( =0, 2, 5 and 10%) for El Centro earthquake

N.B. Each earthquake has its own response spectrum



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



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. Table 1 Amplification Factors: Elastic Design Spectra

Damping, ζ (%) 1 2 5 10 20

Median (50 Percentile)

One sigma (84.1 Percentile)

αA

αV

αD

αA

αV

αD

3.21 2.74 2.12 1.64 1.17

2.31 2.03 1.65 1.37 1.08

1.82 1.63 1.59 1.2 1.01

4.38 3.66 2.71 1.99 1.26

3.38 2.92 2.30 1.84 1.37

2.73 2.42 2.01 1.69 1.38



Page: 2-31 .

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

The elevated tank shown in the opposite figure can be treated as a SDOF system having a period T

H

given by:

T =

2π

= 2π

K

m

ω 0 k 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 S d experienced by the mass “M” during the earthquake. 3. The maximum velocity S v(T,ζ) = ω0Sd(T,ζ) 4. The maximum acceleration S a(T,ζ) = ω0Sv(T,ζ) 2

5. The maximum base shear V max = MSa = Mω0 Sd 2

6. The maximum overturning moment M max= MHSa= MHω0 Sd



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?



Page: 2-33 .



Page: 2-34 .



Page: 2-35 .

CHAPTER 3 SEISMIC ANALYSIS OF MULTI DEGREES OF FREEDOM STRUCTURES



Page: 3-1

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

K =

12 EI

K =

3

L

3 EI L3

K =

3 EI 3

L

K =

3 EI 3

L

K =

12 EI L3

• Since all columns have the same displacement, they can be treated as springs in parallel: K =

∑

K i

=

12 EI L3

3 EI ⎞ 12 EI + 3⎛ ⎜ 3 ⎟+ 3 ⎝ L ⎠ L

• By neglecting the mass of the columns compared to the mass of the floor, the frame can be treated as a SDOF

C M K = K i



b) 2-Storey Shear Building M2 U2

Page: 3-2

P2(t)

K 2 M1 U1

P1(t)

K 1

K 1 is the stiffness of the 1 st storey. K 2 is the stiffness of the 2nd storey. 2 DOF System

U1 C1

C2

M1

M2

K 2

K 1 K 1U1 C1Ů1

U2

M1 M1Ü1

K 2(U2 – U1) C2(Ů2 – Ů1) P1(t) K 2(U2 – U1) C2(Ů2 – Ů1)

M2 M2Ü2

P2(t)



Page: 3-3

Equations of motion: M1Ü1 + K 1U1 + C1Ů1 - K 2(U2 – U1) - C2(Ů2 – Ů1) = P1(t) M2Ü2 + K 2(U2 – U1) + C2(Ů2 – Ů1) = P2(t) The two above equations can be written in matrix form as: && ⎫ ⎡ K 1 + K 2 − K 2 ⎤⎧U 1 ⎫ ⎡C 1 + C 2 − C 2 ⎤⎧U & ⎫ ⎧ P 1 (t )⎫ ⎡ M 1 0 ⎤⎧U 1 1 + + ⎨ ⎬ ⎨ ⎬ ⎨ ⎬=⎨ ⎬ ⎢ 0 M ⎥ U ⎢ − K ⎥ ⎢ ⎥ & & & K 2 ⎦⎩U 2 ⎭ ⎣ − C 2 C 2 ⎦⎩U 2 ⎭ ⎩ P 2 (t )⎭ 2 ⎦⎩ 2 ⎭ ⎣ 2 ⎣

Or: &&}+ [ K ]{U } + [C ]{U & } = { F (t )} [ M ]{U

For earthquake excitations:

• P1(t)= -M1Ϋg(t) • P2(t)= -M2Ϋg(t)



Page: 3-4

3-2 Undamped Free Vibration of MDOF Systems (Determination of Natural Frequencies and Mode Shapes) && + [ K ]{U } = {0} [ M ] U

_____________________________(1)

Assuming that: {U } = {φ }sin(ω t + θ )

{U } = ei(ω t +θ ) {φ }

Or:

&& = −ω 2 {U } ∴ U

Now equation (1) becomes:

[ K ] − ω 2 [M ] {φ } = {0} (Eigen Value Problem) ________________(2) The non-trivial solution of the above equation is obtained when: det [ K ] −ω [M ] = 0 (characteristic equation) 2

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 .



Example 3-3

Page: 3-5

M2

Find the natural frequencies and mode shapes for the following two

m 4

storey reinforced concrete structure. Where:

M1

• M1 = M2 = 30,000 Kg • Ic= 0.005 m4

m 4

• ES= 200,000 MPa 10 m



Page: 3-6



Page: 3-7



Page: 3-8



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, ω22, ……… ωn2 are real roots.

2.

If

{φ i } and

φ j are

two

different

mode

shapes

∴{φ i }T [ M ]{φ j } = {φ i }T [K ]{φ j } = 0 3.

φ j form a complete set of vectors; i.e. any function can be

expressed by these shape functions. 4.

Normalization of {φ }: Evaluate

5.

6.

ˆ j = {φ j } [M ]{φ j }∴{φ ˆ j }= m

{φ i }T [ M ]{φ i } = mi*

T

1 {φ j } ∴{φ ˆ j }T [M ]{φ ˆj }=1 ˆ j m

∴{φ i }T [ K ]{φ i } = ω i2mi*

⎡{φ ˆ1} {φ ˆ2 } ........ {φ ˆn }⎤ ⎢ ⎥ ⎥ ˆ] = ⎢ [ φ ⎢ ⎥ → Ortho-normalized modes If ⎢ ⎥ ⎢⎣ ⎥⎦nxn ⎡ω 12 ⎢ 2 T ω 2 ˆ] [ K ][φ ˆ] = ⎢ T [ φ ˆ ˆ ∴[φ ] [ M ][φ ] = [ I ] and ⎢ ⎢ ⎢⎣

⎤ ⎥ ⎥ ⎥ 2⎥ ω n ⎥⎦ nxn


By: Prof. A. A. El Damatty Page: 3-10

3-4 Modal Analysis Procedure to Evaluate Seismic Response of MDOF Systems &&} + [ K ]nxn{U } + [C ]nxn{U & } = { F (t )} [ M ]nxn{U nx1 nx1 nx1 nx1

Assume that:

{U }nx1 = X 1(t ){φ 1} + X 2 (t ){φ 2} +...... X n (t ){φ n} i.e. {U }nx1 =

n

∑ X (t ){φ } = [φ ] i

i

{X }nx1

nxn

i=1

By pre-multiplying the equation of motion by {φ } , the following T

equation is obtained: &&} + [φ ]T [ K ][φ ]{ X } + [φ ] T [C ][φ ]{ X & } = [φ ]T { F (t )} [φ ]T [ M ][φ ]{ X

The product [φ ] [M ][φ ]will lead to the following diagonal matrix: T

⎡m1 ⎢ [φ ]T [ M ][φ ] = ⎢ ⎢ ⎢ ⎣

0

0 0

m2

m3

0⎤ 0 ⎥⎥ 0⎥

⎥

mn ⎦

Where mi = {φ i } [M ]{φ i } T

T Meanwhile, the product [φ ] [K ][φ ] will also lead to the following

diagonal matrix:


⎡m1ω 12 ⎢ [φ ]T [ K ][φ ] = ⎢ ⎢ ⎢ ⎣⎢

0

0 0

m2ω 22

m3ω 32


⎤ ⎥ ⎥ ⎥ ⎥ mnω n2 ⎦⎥ 0 0 0

T Assuming that the product [φ ] [C ][φ ] 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 x

− F ˆi ( t )

2

+ 2ω i ξ i x& i + ω i x i =

mi

ˆi (t ) = [φ i ]T { F (t )} where F nx1 1 xn

For earthquake excitations:

{ F (t )}T = {− M 1Y && g (t ) − M 2Y && g (t ) ........ − M nY && g (t )} As such, the equation of motion of the i th mode can be written as: &&i x

+ 2ω iξ i x& i + ω i2 xi = −

Where:

mi

=

Li && Y g ( t ) mi

n

∑ M φ j

2

ji

j =1

Li

=

n

∑ M φ j

ji

j =1

φ ji is the component “j” of the mode “i” Γi = Li / m i is called the modal participation factor for mode “i”



Using the response spectrum technique:

( x i )max = Γi S d ( x& i )max = Γi S v ( x&&i )max = Γi S a U ji

max

= Γi S d φ ji

{U & }max = Γ S {φ } ji

i

v

ji

{U && }max = Γ S {φ } ji

F ji

i

max

a

ji

= Γi S a φ ji M j n

L2i

j =1

mi

{V i }max = ∑ Γi S aφ ji M j =

S a

Since the maximum responses of the i th 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: Q 1, Q2,….Q j… 2 2 2 1. Square root of sum of squares (SRSS): Q = Q1 + Q2 + ...Q j + ..

2. Absolute Sum: Q = Q1 + Q2 + .... Q j + ... 3. Complete Quadratic Combination (CQC): Q

2

n

= ∑ Qi2 + ∑ ∑ i =1

e.g. for 2 modes:

i ≠ j

Q2

Qi Q j

1 + ε ij2 where

= Q12 + Q22 + 2

ε ij

Q1Q2

1 + ε 122

=

1 ω i − ω j ξ ω i

+ ω j



Sequence of Modal Combination: Mode 1

Mode 2

Mode 3

Forces

Shear

Moment

Combination



Example 3-4 Use the modal analysis procedure to evaluate the response of the frame shown in example 3-3 to El Centro earthquake.











3-5 Earthquake Response of MDOF Using Time History Analysis The equation of motion for each vibration mode is given as: &&i x

+ 2ω iξ i x& i + ω i2 xi = −

Li && Y g ( t ) mi

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: U ji

= X i (t ) φ ji

The time history variation of the nodal displacements due to all mode shapes is then given as: n

{U } 1 = ∑ X (t ){φ } j nx

i

ji

i =1

&& Also: {U j }nx1 =

n

∑ X && (t ){φ } i

ji

i=1

Knowing the time history variation of the displacements all other quantities such as the internal forces can be evaluated.



Typical time history responses at the top level of a three storey building due to each one of the three modes of vibration together with the total time history response at the same level are shown below.



The time history variations of the base shear force associated with each one of the three modes of vibration as well as the time history variation of the total base shear force are given below.

CHAPTER 4 CODE PROCEDURES FOR EARTHQUAKE RESISTANT STRUCTURES


By: Prof. A. A. El Damatty_____Page: 4-1

Inelastic Behaviour and ductility

Most of the modern seismic design codes allow inelastic behaviour to occur in structures when they are subjected to strong earthquake motions. A load reduction factor “R” is usually applied to the earthquake load intensity.

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

expected if the structures remained elastic.

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

strong ground motion.

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

dissipate through inelastic hysterisis loops that yielded members undergo.

Accordingly, ductility is a very important property that governs the earthquake response of the structure.



Local ductility µl = ∆m/∆y


∆g


Global ductility

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

µg =

∆ g ∆ y

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.



The relation between the load factor R and the global ductility µ g can be established using one of the following concepts:

1) Equal Displacement

F R = d/c µg = a/b d

From similarity: µg = R

c

∆

a b

2) Equal Energy R = d/c µg = a/b

F

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

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

2

R = 2(µg -1) c

R = √2(µg-1) a b

∆



Load Reduction Factor (R d)

Global Ductility µg

V=

Local Ductility, µ l

S (T a ) M v I E W

V

Rd R0

S ( 2.0) M v I E W Rd R0

Ta

: Fundamental period of the structure

S (Ta)

: Design spectral response acceleration

Mv

: Factor to account for higher mode effect on base shear

IE

:

R d

: Load reduction factor

R o

: Calibration factor

W

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

Importance factor



• 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: Location

Seismic Data Sa (0.2)

Sa (0.5)

Sa (1.0)

Sa (2.0)

PGA

Toronto

0.26

0.13

0.055

0.015

0.17

Montreal

0.69

0.34

0.14

0.048

0.43

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

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



Design Spectral Acceleration



Definition of Soil Type



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

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

Ta

≥ 2.0

seconds

• The values of R d, R o are given in Table 4.1.8.9 for different structure

systems.

Notice, that there is a limit on the height of the building for some of the lateral resisting systems depending on the seismic zone, which is defined by the parameters IEFaSa(0.2) and IEFvSa(1.0)



Estimate of Ta



Moment Resisting Frames => T a = 0.085(hn)

3/4

steel

------ (1)

3/4

= 0.075(hn) concrete = 0.1N



Braced Frame => Ta = 0.025hn



Shear Walls

=> Ta = 0.05(hn)

others

----------- (2)

3/4

---------- (3)

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

However, “Ta” should not be taken greater than

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

(2) For Braced Frame,

(3) For Shear Walls,

2.0 Ta as calculated in (2)

2.0 Ta as calculated in (3)



Force distribution Ft Fx =

(V − F t )W x h x n

∑ (W i hi ) i =1

Ft = 0.07Ta V

hn

Fx 0.25V hx

For Ta

≤ 0.7

sec. => Ft = 0

Reduction in Moment

Mx = Jx ∑Fi (hi – hx)

Jx = 1.0 for hx

Jx = J +

≥ 0.6hn

(1 − J ) h x 0.6hn

for hx < 0.6hn

J depends on Ta and the type of lateral resisting system.

(Table 4.1.8.11)



Definition of Irregularities



A structure is considered to be irregular if it has any of the type of irregularities described in the above table.

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

a) IEFaSa(0.2) <0.35

b) Regular structure, with hn ≤ 60.0 m and Ta

≤ 2.0

sec. (in both two orthogonal directions)

c) Structure with structural irregularities of Type 1, 2, 3, 4, 5, 6, or 8 with hn ≤ 20 m and T a

≤ 0.5

sec. (in both two orthogonal directions)

*Otherwise dynamic analysis has to be carried.



Accidental Torsion

Bx = δmax/δave

δmax = maximum storey displacement at the extreme points induced by equivalent static force applied at distances

± 0.1Dnx

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

± 0.1Dnx

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

Tx = Fx (ex + 0.1 Dnx)

Tx = Fx (ex - 0.1 Dnx)



Direction of Loading

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

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

y

100% 100%

x

Independent

2) Case of no orthogonal axis and I EFaSa (0.2) < 0.35, Independent analyses about any two orthogonal axes are permitted.

permitted 100%

permitted 100%

Independent

100%

y 100%

Independent

x



3) Case of no orthogonal axis and I EFaSa (0.2)

≥ 0.35,

y 100%

30%

together

30%

30%

100%

together

100%

together

100% 30% together

x



Dynamic Analysis Procedure

Linear Analysis

Modal Response Spectra

Time History

Ground motion history Spectral acceleration values compatible with S(T) S(T) defined by the Code

Non-linear Analysis

Time History



Scaling used in Dynamic Analysis

1) Evaluate the inelastic base shear force “V” from the Code equation (taking R dR 0 into account) '

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

3) Ve (elastic) = V e .IE

V e

'

4) Vd (inelastic) =

Rd R0

(Dynamic analysis)

(Dynamic analysis)

'

5) Compare Vd to V

Vd

'

'

≥ V

Vd = Vd

Vd < V

'

Bx < 1.7 Vd = 0.8V

Bx

≥

1.7

Vd = 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 R dR 0



*The dynamic analysis should include the effect of accidental torsion

Bx ≥ 1.7 ( ± 0.1Dnx)Fx

Bx < 1.7 center of mass shifted by ( ± 0.05Dnx)Fx



Time History Analysis

Selection of a Ground Input Acceleration a g(t) Clause 4.1.8.12(3)

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

Its response spectrum should be equal or exceed the target spectra throughout of the period range of interest.

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

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



• Halchuk and Adams (2004) have conducted a deaggregation of seismic

hazard for selected Canadian cities. Halchuk K, S. and Adams, J. ‘Deaggregation of Seismic Hazard for Selected Canadian th Cities” 13 World Conference on Earthquake Engineering, Vancouver, B. C., Canada, August 1-16,2004, paper No. 2470.

Deaggregation: Dividing the total hazard into contributions based on distance and magnitude.

Halchuk and Adams (2004)



Halchuk and Adams (2004)



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

histories which are compatible with the uniform hazard spectra used in NBCC 2005.

Atkinson, G., and Bersnev, I., (1998), “Compatible ground-motion time histories for new national seismic hazard maps”, Canadian Journal for Civil Engineering 25: 305-318.

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

o

For long period structures Selected earthquake records should

(a) Have the same deaggregation parameters (D,M) as those specified for S(1.0) of the location.

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

o

For short period structures Selected earthquake records should

(a) Have the same deaggregation parameters (D,M) as those specified for S(0.2) of the location.

(b) Be scaled to match S(T 1), where T1 is the fundamental period of the structure.

CHAPTER 5 SEISMIC ANALYSIS USING COMPUTER MODELING


By: Prof. A. A El Damatty Page: . 5-1

Two- Dimensional Seismic Analysis Using SAP2000

m3 = 57290.4kg

Given:

2-D moment resistant steel frame with

6 8 2 x 0 6 4 W

dimensions shown in Figure 1. 8

E = 2x10 kN/m

7 4 4 x 0 3 5 W

1 5 5 x 0 1 6 W

W 690 x 419

m2 = 42967.8 kg

W 690 x 419

m1 = 28645.2 kg

W 690 x 419

10 m

Figure 1

Required:

1- Free vibration: period, & mode shapes. 2- Time history Analysis. 3- Response Spectrum analysis.

6 8 2 x 0 6 4 W

7 4 4 x 0 3 5 W

1 5 5 x 0 1 6 W

4m

4m

4m



Solution:

1- Select the units from the drop

box

2- From File menu select New Model From Template, this will display different modal templates, choose portal frame type and fill the properties of the portal frame: o

Number of stories: 3

o

Number of bays: 1

o

Story height: 4

o

Bay width: 10

3- The screen will be titled vertically showing the frame in 2-dimension (x-z plane) & 3- dimension, close the 3 dimension window. 4- Click on Set Elements

on the main toolbar & check the labels box

in the joints & frames area to see the joint & frame numbering. 5- Click on joints 1 and 6. From the Assign menu select Joint, from joints select Restraints then choose fixed support

from the fast restraints.

6- From the Define menu select Materials…, the define material dialog box will show up, highlight STEEL then click Modify / Show Material to check the properties of the material. 7- Verify that modulus of elasticity is 2.0 E+11 then click OK twice. 8- From the Define menu select Frame Sections…to display the define frame sections dialog box. 9- In the Click to: choose Import I / Wide Flange, another box will display for Section Properties File, from Sap2000n folder in your hard drive choose Cisc.pro (see Fig. 2). Double click on the Cisc.pro and choose from the given sections: W410x100, W610x551, W530x447, &



W460x286. (you can press the Ctrl key on the keyboard to choose all of them at one step). 10- Select the column of the first story (element 1, 4), then from Assign menu select Frame, Section…, choose W610x551, make the same for the column of the second, third stories (element 2, 5, & element 3,6). After each selection for beam or column section you can click on the undeformed shape icon to see the element numbering again. 11- From Set Elements click on the frame labels to remove it. 12-The final Figure will be as shown in Fig. 3.

Figure 2

Fig. 2



Fig. 3

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.



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 Montréaleast with soil type C and damping

Fig. 4

of 0.05, see Fig. 4. 21-Form the Define menu select Functions… Time History to display define time history functions box.



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

Fig. 5

file is organized. In our case we choose Time and Function values, see Fig. 5. 25-Form the Define menu select Analysis Cases…Response Spectrum to display define response spectra box. 26-Choose Add New Case, a dialog box for response spectrum case data will display. 27-In the Response Spectrum Case Data box: o

Accept the default Spectrum Case Name: ACASE1

o

Analysis Case Type: Response Spectrum

o

Accept the default Excitation Angle: 0

o

Accept the default Modal Combination Option, CQC.



o

In the Damping box: 0.05

o

Accept the default Directional Combination option, SRSS.

o

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

o

For U1 direction write 1 at the scale factor.

o

Click ADD button.

o

Click Ok.

28- Choose Add New Case, a dialog box for response spectrum case data will display. 29-In the Time History Case Data box: o

Accept the default History Case Name: ACASE2

o

For modal damping: click Modify / Show and write 0.05 in Damping for all modes.

o

For Load type: choose acceleration.

o

In Number of output time steps box: 1560.

o

In Output Time Step Size box: 0.02

o

In the Load Assignments area select ELCENTRO from Functions for U1 direction

o

For U1 direction write 1 at the scale factor and leave the arrival time, angle 0.

o

Click ADD button.

30-Click Ok twice to finish this part. 31-From the Analyze menu select Set Options… to display the analysis option dialog box and choose the x-z plane.



Output:

We will divide the output file into: 1- Output can be obtained from the screen for the time history of a certain quantity such as: o

Time history for a joint displacement, rotation, acceleration or velocity.

o

Time history for bending moment, shear force or axial force at a specific location of an element.

o

Time history for base shear force and overturning moment.

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

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

Click on Joint 4, and then from Display menu click on Show Time History Traces… this will display Time History Display Definition. From the choose function part double click on joint 4 at List of Functions, joint 4 will now appear at Plot Function.

o

The default is drawing displacement ux for this joint, if you want to change click on Define Function icon, another box will display for Time History Functions.

o

Highlight joint 4, click on Modify/Show TH Function, another box for Time History joint function will appear. Choose the function you want then click Ok twice.

o

Now Highlight Joint 4 then click the Display icon, another display window will appear as in Fig. 4. You can either plot this figure or



save it by clicking in file the menu and choose Print Graphic or Print Tables To File… o

The same procedure for joint 4 can be applied for frame element by choosing the required frame element.

2. Time history for Base Shear & Base Moment: o

Click on the Define Function icon, the box for Time History Functions will display.

o In

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

display for Base functions, Click on Base Shear X and Base Moment Y. Click Ok Twice. o

Now Base Shear X and Base Moment Y will be at List of Functions, use the same procedure described for joint 4 to plot the

Base Shear X and Base Moment Y.

Fig.6



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.

Fig. 8



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 30storey reinforced concrete building will be illustrated in this section. The building is assumed to be located in Toronto. The total height of the building is 92 m. Modeling Main Steps:

1- Develop the building geometry and axes using CAD or ETABS

2- Define materials properties 3- Define the building elements properties Shell elements are used to model the slabs and the shear walls, and frame elements are used to model the columns and the rigid frames. The stiffness of the pile foundations or shallow foundations are usually accounted for through the use of equivalent vertical, horizontal and rotational springs.



To account for the cracked section properties, the section stiffness is reduced using stiffness modifiers. The reducing factors can be evaluated from the CSAA23.3-04 Clause 21.2.5.2.1 Table 21.1 and Clause 21.2.5.2.2.

4- Construct the Model



5- Preliminary Dynamic Analysis The building is assumed to be located in Toronto. The soil profile is type D. NBCC 2005 spectral values for Toronto Sa (0.2) = 0.26 Sa (0.5) = 0.13 Sa(1.0) = 0.055 Sa(2.0) = 0.015 Using Soil Type D

------------ Fa = 1.30 and

Fv = 1.40

The Design spectral acceleration values can be determined using NBCC 2005 Clause 4.1.8.4.6

S (0.2) = 3.315 m/sec

2

S (0.5) = 1.785 m/sec

2

S(1.0) = 0.7553 m/sec

2

2

S(2.0) = 0.206 m/sec

2

S(4.0) = 0.103 m/sec

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

1

4.235024

28.7694

40.5759

0

28.7694

40.5759

SumUZ 0

58.0654

RX

40.2014

RY

0.5623

RZ

SumRX 58.0654

SumRY 40.2014

SumRZ 0.5623

2

3.50052

23.0926

9.3154

0

51.862

49.8913

0

13.4977

30.6581

40.7607

71.5631

70.8595

41.3231

3

1.108758

10.3055

3.987

0

62.1674

53.8783

0

0.0702

0.2505

0.1777

71.6334

71.11

41.5007

4

1.067125

17.4395

15.8869

0

79.6069

69.7652

0

26.0066

27.8469

34.5346

97.6399

98.9568

76.0354

5

0.953012

1.7659

12.1331

0

81.3728

81.8983

0

1.8824

0.5209

4.1669

99.5223

99.4778

80.2023

6

0.529784

3.843

0.7729

0

85.2159

82.6712

0

0.0372

0.179

0.95

99.5595

99.6568

81.1524



T = 4.235 sec (Combined mode X-Y - dominant Y ) T= 3.50 sec (Combined mode X-Y-Torsion - dominant X and Torsion)

2- Estimate the periods using NBCC 2005 (h n) 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 R d = 2.0

R o = 1.50

IE = 1.0 W = 143883 kN =The dead load of the building including the self weight of all elements, the super imposed load, and 0.25 of snow loads.

= 0.206

The base shear force is calculated using the NBCC 2005 equation: Clause (4.1.8.11)



and should not be less than

V = 24,700 KN

Conduct static analysis using the equivalent static earthquake load to obtain B x. The static analysis involves the cases of eccentricity.

Bx = δmax/δave

δmax =0.25

δave =

m

0.09 m

Bx = 2.70

7- Evaluation of spectrum scaling factor and re-analysis

Along X-Direction

From initial analysis

V'xex = 19,073 kN V'xey = 12,619 kN V'xe =

= 22,869 KN

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

= 7,623 KN



Compare V'xd to V

V'xd

≥

V

V'xd < V

Vd = V'xd Irregular

Regular Vd = 0.8V

Bx

Vd = V

≥ 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



Compare V'yd to V

V'yd

≥

V

V'yd < V

Vd = V'yd Bx < 1.7 Vd = 0.8V

Bx

≥ 1.7

Vd = 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 D nx from the shear center.



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 D nx from the shear center. -Specyudp : The storey shear force obtained from the case Spec Y and applied at a distance of +0.10 D nx from the shear center. -Specyudn : The storey shear force obtained from the case Spec Y and applied at a distance of -0.10 D nx from the shear center. Earthquake X-Direction is an envelope to all the X-direction cases. Earthquake Y-Direction is an envelope to all the Y-direction cases 8- Load Combinations Based on NBCC 2005. Clause 4.1.3.2. Case 1 (E1): 1.40 Dead loads Case 2 (E2): (1.25 Dead Loads or 0.90 Dead loads) + 1.50 Live Load + (0.25 Snow or 0.40 Wind Load) Case 3 (E3): (1.25 Dead loads or 0.90 Dead loads) + 1.50 Snow + (0.50 live loads or 0.40 Wind) Case 4 (E4): (1.25 Dead loads or 0.90 Dead loads) + 1.40 Wind X-Direction + (0.50 live loads or 0.40 Snow) Case 5 (E5): (1.25 Dead loads or 0.90 Dead loads) + 1.40 Wind Y-Direction + (0.50 live loads or 0.40 Snow)



Earth Quake Combinations Case 6 (E6): 1.0 Dead loads +( 1.0 Earthquake X-Direction + 0.30 Earthquake Y-Direction) + 0.50 live loads + 0.25 Snow Case 7 (E7): 1.0 Dead loads + (1.0 Earthquake Y-Direction + 0.30 Earthquake X-Direction) + 0.50 live loads + 0.25 Snow

9- Results

a) Shear wall Design -Magnify the moments obtained from the final analysis with the P- ∆ factor U2 1

U 2 = 1− ( ∆f

∆ f ∑ pf

h∑Vf

)

= inter storey drift due to factored lateral load.

Σ pf = Σ factored Σ Vf =

axial loads in the story.

factored total story shear.

h = story height.

- Magnify the shear forces obtained from the final analysis by a factor = (Rd). (Ro)

b) Columns Design - Magnify the moments obtained from the final analysis with the P- ∆ factor U2 - Magnify the shear forces obtained from the final analysis by a factor = (Rd). (Ro)



c) Frame Girders Design (Ductile MRF) - Magnify the moments obtained from final analysis by a factor = (

Rd R0 ( Moderatelyductilewal ls ) ' ' Rd R0 ( DuctileMRF )

). (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


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.



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”.

steel

brittle material

a

P

b L

(L/b) M p (steel) M p (steel)

θ

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: M brittle

≥

α L/b M p (steel)

α is a number greater than 1.0 to account for the possible reserve strength of the cantilever beyond its nominal yield strength.



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

σ σy

ε Stress-strain curve for a typical steel coupon

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



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 (M p)

Probable (M p’)

Φr = 0.9

Φ p = 1.0

Φ p’ = 1.1 R y



Moment Resisting Frames (Energy dissipates through yielding of beams)

Page: 6-5 .



Page: 6-6 .

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

R d = 5.0, R 0 = 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)

Lr r y

=

17250 + 15500k F y

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.



Page: 6-7 .

Columns (27.2.3.2)

Class 1 or Class 2

∑Mrc'

≥

∑ [1.1 R y M pb + Vh (x+ dc/2)]

where:

∑Mrc' = Sum of the column factored flexural resistances at the intersection of the beam and column M pb = Nominal plastic moment resistance of the beam Vh = Shear acting at plastic hinge when 1.18 R y M pb is reached x = Distance from centre of the beam plastic hinge to the column face

Mrc = 1.18 φ M pc [1'

C f φ C y

]

≤

φ M pc

M pc = nominal plastic moment resistance of the column Cf = results from summation of V h 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 R y M pb + Vh (x+ dc/2)]



Page: 6-8 .

Resistance of Panel Zone

(a) Vr = 0.55φ dc w Fyc [1+ '

3bc t c2 d c d b w

'

]

≤

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.



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)

Lcr r y

=

25000 + 15000k F y



Page: 6-10 .

Evaluation of Forces in Panel Zone

M3

C3

V3 M1 V1 Vh

C1

0.95 d b

C2 V2 M2

V4 M4 C4

0.95 dc

V3 V1

h3/2 h4/2

V2 V4

l1/2

l2/2

Equilibrium Equations

M 1

+

(1 − β 1 )

M 2

−

(1 − β 2 )

Vh = [ M1 + M2 -

β 1

=

0.95d c l 1

, β 2

=

M 3 (1 − β 3 )

β 3 M 3 (1 − β 3 )

0.95d c l 2

M 4

−

−

=

(1 − β 4 )

β 4 M 4 (1 − β 4 )

, β 3

=

(1)

]/0.95d b

0.95d b h3

0

, β 4

=

(2)

0.95d b h4



Page: 6-11 .

Concentric Braced Frames (Energy dissipates through yielding of bracing members)


Type Tension-Compression

Moderately Ductile

Page: 6-12 .

Limited Ductility

If IEFaSa(0.2) < 0.35

If IEFaSa(0.2) < 0.35

No limit

No limit

Otherwise H (a)


40 m

Otherwise H

60 m

When H > 32 m, seismic When H > 48 m, seismic force shall be increased force shall be increased by 3% per meter above by 2% per meter above 32 m

48 m

(b)

Same as (a)

Same as (a)

Tension

H

H

only

If H > 16 m, seismic

If H > 32 m, seismic

force shall be increased

force shall be increased

by 3% per meter of

by 3% per meter of

height above 16 m

height above 32 m

N.A.

N.A.

Chevron

20 m

K-truss

40 m



Page: 6-13 .

Moderately ductile concentrically braced frames (R d = 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

300/√Fy

Class 1

145/√Fy

170/√Fy

Class 1

Class 2

HSS Legs of angle and flanges of Channel Other elements

o

For IEFaSa(0.2) < 0.35

HSS => Class 1 Other sections => Class 2 Legs of angle = 170/ √Fy



Page: 6-14 .

Bracing Connection

Factored resistance of the bracing connections shall be equal or exceed both AgR yFy 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 R dR 0 = 1.3 (Redistribution due to buckling should be taken into account when calculating the seismic loads corresponding to R dR 0 = 1.3)

Beams

The capacity design of the beam should consider two conditions:

1- Compression bracing = 1.2C p Tension bracing

= AgR yFy

2- Compression bracing buckled and retain a force = 0.2 A gR yFy Tension bracing

= AgR yFy

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



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

Moment Resisting Frame

Moment Resisting Frame Braced Frame

6x6000mms Braced Frame

6000

6000

8000

6000

6000

N

Plan of Example Building 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 2 5. structural elements (beams & columns) = 0.25 kN/m 2 D.L. = Σ = 4.75 kN/m Live load: (NBCC 4.1.6) 2 1. L.L. on typical office floor = 2.4 kN/m



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/m 2 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/m

2

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)

Mass center

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

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

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

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

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

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



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(T a) . Mv . IE . W / (R d . R o) ≥ S(2.0) . Mv . IE . W / (R d . R o) •

•

Time period Ta North-south direction: 3/4 Steel moment fames, T = 0.085 (h n) = 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, S a(T), are provided in Table-C2 of NBCC. Sa(0.2) 0.69

Sa(0.5) 0.34

Sa(1.0) 0.14

Sa(2.0) 0.048

PGA 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 F a = 1



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) = F a . 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, R d = 3.5, R o = 1.5, East-west direction, moderately ductile concentrically braced fames, R d = 3.0, R o = 1.3

•

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

•

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)



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 h x ≥ 0.6 hn, and Jx = J + (1-J) (h x / 0.6 h n) 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) Floor 3 : Jy = 0. 884 + (1-0. 884) (11.7 / 17.82) Floor 2 : Jy = 0. 884 + (1-0. 884) (8.1 / 17.82) Floor 1 : Jy = 0. 884 + (1-0. 884) (4.5 / 17.82) Ground : Jy = 0. 884 + (1-0. 884) (0 / 17.82)

(E – W) direction For floors 5 to 8, Jx = 1.0 Floor 4 : Jx = 0.92 + (1-0.92) (15.3 / 17.82) Floor 3 : Jx = 0.92 + (1-0.92) (11.7 / 17.82) Floor 2 : Jx = 0.92 + (1-0.92) (8.1 / 17.82) Floor 1 : Jx = 0.92 + (1-0.92) (4.5 / 17.82) Ground : Jx = 0.92 + (1-0.92) (0 / 17.82)

= 0.983 = 0.960 = 0.937 = 0.913 = 0.884

= 0.988 = 0.973 = 0.956 = 0.940 = 0.920


Floor


Page: 6-20 .

Overturning

Overturning

Reduced

Reduced

moment

moment

overturning

overturning

(kN.m)

(kN.m)

moment

moment

(N – S)

(E – W)

(kN.m)

(kN.m)

direction

direction

(N – S)

(E – W)

direction

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) = '

Vd (E-W) =

V e Rd R0 V e Rd R0

= 737 kN

< 1103kN

= 1321 kN < 2740 kN



First mode shape (Period =1.46075 sec)

Second mode shape (Period = 1.32939 sec)

Page: 6-21 .



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

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

5.093E-08



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

Page: 6-23 .



Page: 6-24 .

Interstorey Drift East-west direction The maximum interstorey drift = 0.00228 * R d R o = 0.008268 m < 0.025hs (floor height) < 0.09 m North-south direction The maximum interstorey drift = 0.00724 * R d R o = 0.003801m < 0.025hs (floor height) < 0.09 m

3-D view for the structure



Strength design according to CISC 95 •

North-south direction: Two moment resisting frames along the outer edges

Elevation of Moment Resisting Frame

Page: 6-25 .


East-west direction: Two braced frames •

F8

S8 F7

S7 F6

S6 F5

S5 F4

S4 F3

S3 F2

S2 F1

S1

Elevation of Braced Frame


Page: 6-26 .



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)

W 690 x 140

W 690 x 140

Level 4

W 760 x 147

Beam W690x140 684 254 18.9 12.4

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

Mp

(

Lcr r y

= 55.65) > (

Mp/2

25000 − (15000)(0.5) 350

= 50) Not good

Column W760x147 753 265 17.0 13.2



Assume the cross beam will be repeated every 2.0 m

Mp

(

Lcr r y

= 37.105) < (

2Mp/3

Mp/3

25000 − (15000)(0.667) 350

= 42.8) OK

Mr1 = Mr2 = φ MP = 1410 kN.m; M p1 = 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. R y . M pb + Vh (x + dc/2)

S

w = 35.7 kN/m

x

dc/2

Lh

L

Example for F4 Vh = VhE + VG '

VhE = 2M pb /Lh Lh = L-2S S = x + dc/2 VG = wL/2 => w = (DL +0.5 LL) x = d b / 2 for welded connection

Page: 6-28 .



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 305 kN 107

107 kN

107

376

107

755 kN.m 918

107

529

107

1228 107

466

107

1083 830 107

107 1895

Unbalanced moment =

∑ (1.1 R y . M p + 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 ) . M pc [ 1 -

C f C y

]

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

φ . M pc = 1580 kN.m C y = A . f y = (18700 ) x (350) / 1000 = 6545 kN ′

Mrc = (1.18). (1580). [1 -

1070 (0.9)(6545)

] = 1830 kN.m < 2491 kN.m Not good



Page: 6-30 .

Try W920 x 381

φ . M pc = 5280 kN.m C y = 17010 kN 1070

′

Mrc = (1.18). (5280). [1 -

] = 5794 kN.m > 2491 kN.m OK

(0.9)(17010)

Panel Zone M 1 [ 1 − β 1

+

M 2

M 3

−

1 − β 2

Vh = [ M 1 + M 2

−

1 − β 3 −

β 4 M 4 1 − β 4

M 3 . β 3 1 − β 3

−

] = 0.0

M 4 . β 4 1 − β 4

] / 0.95d b

β1 = β2 = 0.95dc/L1 =

(0.95)(951)

β3 = β4 = 0.95d b/ h1 =

(0.95)(684)

( 2491)(2) (1 − 0.20)

=

2 M 3 (1 − 0.18)

Vh = [2491 + 2491 −

=

6000

0.20

=

3600

⇒ M 3 = M 4 =

0.18(2552) 1 − 0.18

−

0.18

2552 kN.m

0.18(2552) 1 − 0.18

Vr = (0.55).(0.9).(951).(24.4).(0.35)[1 +

] /[( 0.95).(0.684)] = 5943 kN

(3).(310)(43.6)

2

(951) x (684)(24.4)

] = 4468 < 5943 Not good

Try: W 920 x 585

d b t w

Vr = (0.55).(0.9).(960).(31).(0.35)[1 +

W920x585 960 427 55.9 31 (3).(427)(55.9) 2 (960) x(684)(50)

] = 6170 > 5943 ok

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



Vr

(0.66) (960) (0.9) (31) (0.35)

6188 kN

Check clause 13.4.1.1

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

h

a

h / w = 647 / 31 = 20.87

a / h = 848.2 / 647 = 1.31 > 1.0

k v = 5.34 + (4) / (a/h)2 = 7.57

439

∴

k v F y

h w

∴Fs =

≤

= 64.9

439

k v F y

0.66 Fy

OK

OK

Page: 6-31 .



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) 177.8 − ( 4).(6.35)

(b – 4t) / t = Limit =

=

6.35

330 σ y

= 17.63 not

24.0

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 mm

2

r = 57.6 mm2 Tr = 1641 kN

Page: 6-32 .



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. R y. σ y = (1.1). (5210) .(350) = 2005 kN b) Cu = 1.2C p

=

1.2 (0.9)

.(916) = 1222 kN

Fc should not exceed the effect of gravity load + effect of seismic loads with R d. R o = 1.3 Fc should be less than (65 + Fc

(362).(3).(1.3) (1.3)

)

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 C p Tension bracing = Ag. R y. Fy

2. Compression bracing buckled and retain a force = 0.2 Ag. R y .Fy Tension bracing = Ag. R y .Fy The beam need not to resist load effects exceeding to R d. R o = 1.3



Capacity design of strut beam

HSS 152x152x10 After capacity design

C7b

T7b p

F6

Ө

C6b

T6b

HSS 203x203x13 After capacity design

Example beam of level 7 T7b = Ag . R y . Fy = (5210). (1.1). (0.35) = 2005 kN C7b = 0.2 Ag . R y . 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) P C r

+

(0.85) M f M r

=

2350 3591

(0.85) 272 +

624

= 1.0

ok

Page: 6-34 .



Page: 6-35 .

Intersection beam

Example beam at F7 HSS 152x152x10 C8b

F7

p

T8b

→

→

p

T7b

C7b

HSS 152x152x10

2.0 m

T8 = T7 = Ag . R y . Fy = 2005 kN C8 = C7 = 1.2 C p = 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



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) P C r

+

M f M r

=

396

+

958

45.7

=

78.3

0.99 ok

Capacity design of column Example storey (6)

HSS 152x152x10

2005

HSS 152x152x10

1222

F8

F6 HSS 203x203x13

HSS 203x203x13

3565 3040

F4 HSS 203x203x13

3565

(1) Forces Based on Brace capacities

T = Ag Fy R y C = 1.2 C p

Page: 6-36 .



1

2005

Page: 6-37 .

2 Cr or Tu SinӨ

3 SRSS

4 Earthquake (RdRo) = 1.3

5 EQ (Smallest of 3 and 4)

6 D+0.5L+P∆

7 EQ + D +0.5L+P∆

1341

1341

358

358

406

764

4543

2080

2080

596

2676

7273

4660

4660

1003

5663

1222

817

3565

2385

3040

2033 2385

3565

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 +

( 2385 ) 2

+

(817 ) 2

+

(1341 ) 2 = 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 M p (S16-01-27.5.5.2)

CHAPTER 7 SEISMIC DESIGN OF REINFORCED CONCRETE STRUCTURES


Page: 7-1 By: Prof. A. A. El Damatty____

MOMENT RESISTING FRAMES (MRF)

Ductile MRF

=> R d = 4.0 R 0 = 1.7

Moderately Ductile MRF

=> R d = 2.0 R 0 = 1.4

Conventional MRF

=> R d = 1.5 R 0 = 1.3



Factored, Nominal and Probable Resistances:

Mr = Factored Flexural Resistance

Φc = 0.65 /

Φs = 0.85

Mn = Nominal Flexural Resistance

Φc = 1.00 /

Φs = 1.00

M p = Probable Flexural Resistance

Φc = 1.00 /

Φs = 1.25

M n ≈ 1.20 M r

For Beams,

For Columns,

M p M r

≈ 1.47

M p ≈ 1.57 M r



Seismic Provisions for Ductile MRF



(21.3.1.1)

Beams

-Ductile Flexural Members

- Axial compressive Force

≥

- Lclear

-

-b

-b

b d

A g f 'c 10

4d

0.3

250 mm

width of supporting member + 2 x 3/4d

¾ d max.

b ¾ d max. Plan View



Beams (cont..)

M

-ve joint

+ve joint

M

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

+ ve M jo int ≥

− ve M jo int

2

ve M + any sec tion ≥

− ve

M any sec tion ≥

M jo int 4

M jo int 4

(21.3.2.2)



Beams (cont..)

≤ 50 mm

S ≤ d/4 8 d b 24dst. 300 mm

2d S ≤ d/2

S ≤ d/4 100 mm

*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)



Beams (cont..)

Hoops Detailing Cross -tie

All extensions >

6d b 60 mm



Beams (cont..)

Shear Strength Requirement: Vr ≥ Vf

Vr ≥ The smaller of

{ V p { V determined from analysis with R dR 0 = 1

V p => 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 = 45

0

(Vr = Vc + Vs) Vc = φcλβ√f c bwdv '

Vs = φsAvf ydvcotθ / 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



(21.4.1)

Columns

-Ductile Frame members subjected to Flexural and Axial Loads - width

300 mm

-b/h ≥ 0.4

-P >

A g f 'c 10

M rct

M nb L M nb R

(include slab RFT)

(include slab RFT)

M rcb

∑ Mrc ≥ 1.1∑Mnb

Strong Column Weak Beam



Columns (cont..) *Longitudinal RFT: 0.01

≤ ρg ≤ 0.06

ld

1.3ld

Tension Lap splice within the centre half of the member length

Tension Lap splice at any section



Columns (cont..) Transverse RFT: ' Ash ≥ 0.2k n k p (Ag/Ach) (f c/f yh) '

Ash ≥ 0.09 ( f c/f yh) s hc

s hc ,

k n =

nl nl − 2

k p =

P f P 0

* Stirrups:

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

l0 Spacing >

6 d b 150 mm

IF Pf ≤ 0.5φc f c Ag '

Spacing > ¼ b 100+(350-hx)/3 6d b

l0 < 1.5 h

clear height /6

IF Pf > 0.5φc f c Ag '

l0 < 2h

clear height /6



Column Cross sections:

hx

hx

hx

hx > max. of 200 mm core dimension/3

(21.4.4.4)

hx = Distance between laterally supported



Shear Strength Requirements

K 2

− ve pr

ve pr

M col t = k 1 k 1 + k 2 lu

K 1

M col b

Vcol =

M col t + M col b l u

V r col ≥ Vf V r col ≥ The smaller of { Vcol { V determined from analysis using R dR 0 = 1

Vr can be calculated using Clause 11 with

β ≤ 0.1 and θ ≥ 450

+

−

( M pr ve + M pr ve )



Beam-Column Joints:

• Column stirrups should continue through the joint except if the joint is confined by Structural members as shown;

b < 3/4h

< 3/4b

h



Applied shear on Joints:

Vcolumn

1.25 As f y

Vfb = 1.25 Asf y - Vcolumn



Vcolumn

C2 =1.25As2f y 1.25As2f y C1 =1.25As1f y

Vfb = 1.25 As1f y + 1.25As2f y - Vcolumn



Allowable shear in beam-column joints

(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 A g of column or b w hcol

(21.5.4.1)



ldh

12d b

(21.5.5.2)

* ldh = max of { 8d b {150 mm

f y { 0.2

d b

f c'

0

* without the standard 90 hook => use 2.5ldh if depth of concrete below bar ≤ 300 mm => use 3.5 ldh otherwise

(21.5.5.4)



Seismic Design of a Ductile Moment Resisting Frame

2.20

3.00

2.20

7.40 m

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= ⎜ 0.5 + 0.6* ⎟ * Ig ≤ f ' c * Ag

⎛ ⎝

f s

⎞ ⎠

1.0

(0.6 → 0.7 Ig)



Design of the First Story Beam A

B a

CL

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 D -117.45 113.43 88.26 -174.40 -159.10 L -55.77 54.28 42.10 -83.45 -76.02 E (±) 120.60 48.28 48.13 119.00 119.50 Load Cases 1.25D+1.5L -230.47 223.21 173.48 -343.18 -312.91

1.0D+1.0E 1.0D-1.0E 1.0D+0.5L+1.0E 1.0D+0.5L-1.0E Design moments Max. Moment Min. Moment

3.15 -238.05 -24.74 -265.94 3.15 -265.94

161.71 65.15 188.85 92.29 223.21 65.15

136.39 40.13 157.44 61.18 173.48 40.13

-55.40 -293.40 -97.13 -335.13 -55.40 -343.18

c 85.10 40.58 48.49

-39.60 -278.60 -77.61 -316.61

167.25 133.59 36.61 153.88 56.90

-39.60 -316.61

167.25 36.61

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



Table 2: Bending moment along the beam [kN.m] (after redistribution) AB a b BA BC D -93.96 140.35 117.62 -139.52 -127.28 L -44.62 67.11 59.16 -66.76 -60.82 E (±) 120.60 48.28 48.13 119.00 119.50 Load Cases 1.25D+1.5L -184.37 276.10 235.76 -274.54 -250.32 1.0D+1.0E 26.64 188.63 165.75 -20.52 -7.78 1.0D-1.0E -214.56 92.07 69.49 -258.52 -246.78 1.0D+0.5L+1.0E 4.33 222.18 195.33 -53.90 -38.19 1.0D+0.5L-1.0E -236.87 125.62 99.07 -291.90 -277.19 Design moments Max. Moment 26.64 276.10 235.76 -20.52 -7.78 Min. Moment -236.87 92.07 69.49 -291.90 -277.19

c 116.92 55.78 48.49 229.83 165.41 68.43 193.30 96.32 229.83 68.43

Design for flexural reinforcement d b max =

l j 24

=

600 24

= 25

(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 A s =

291.9*10 6 0.85*400*525

= 1635.3 mm2

Slab reinforcement within (3*hslab) from the side of the beam is effective 2 4-10 M bars are effective (As = 400 mm ) As required = 1635.3 – 400 = 1235.3 mm

2

2

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


Clause 21.3.2.2: M R+ ve ≥ 0.5* M R− ve M R min ≥ 0.25* M R max Assume As


at column face at any section

+ve

= 4-15M 2

As = 1200 +400 = 1600 mm d = 700 – 30 – 11.3 – 19.2/2 = 649 mm ⎧0.72 h ⎧0.72 * 700 ⎧ 504 =⎨ =⎨ d v = ⎨ 0.9 d 0.9 *649 ⎩ ⎩ ⎩ 584.1 ⇐ 2 A’s = 4 * 200 = 800 mm 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

0 1 1

0 0 7

C

4- 15 M 0.0035

400

Ts = As* Φs*f y = 1600 * 0.85 * 400 = 544*10 N C’s = As* ( Φs*f s – α1 * Φc*f’c ) = 800 * (0.85 * f s – 0.805 * 0.65 * 30) N C s = α1 * β1* Φc*f’c * b * C = 0.805 * 0.895 * 0.65 *30 * 400 * C 3

∑ F = 0 x

⇒

-Ts + C’s + C s = 0

3

⇒ 544*10 – 680 * (f s – 18.47) – 5619.7 * C = 0

⎛ d ' ⎞ Compatibility of strains ⇒ ε = ε cc ⎜1 − ⎟ = 0.0035 c ⎠ ⎝ ⎛ 49.3 ⎞ fs= E s *ε s' = 700 ⎜1 − ⎟ c ⎠ ⎝ ' s

…. (1)

⎛ 49.3 ⎞ ⎜1− ⎟ c ⎠ ⎝ .… (2)



Solving (1) and (2) ⇒ C = 69.74 mm ,and f s = 205.1 MPa a = 0.895*69.74 = 62.42 mm a⎞ ⎛ M R− ve = Cc ⎜ d − ⎟ + Cs' ( d − d ' ) = 333.3 kN.m. > M f = 291.9 kN.m 2⎠ ⎝ Clause 21.3.2.1 A s min =

1.4* bw * d y f

=

1.4* 400 *649 400

= 908.6 < As o.k

A s max = 0.025* bw * d = 0.025* 400 * 649 = 6400 > As o.k

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 a=

A s * φ s * f y φ c * α 1 * f c' * b

=

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

= 18.43 mm < tflange=110 mm

a 18.34 M R− ve = T s *( d − ) = 1600 *0.85* 400 *(635 − ) = 340.42 kN.m. > Mf = 276.1 kN.m. 2 2

O.K. +ve

MR

at column face

d ≈ 650 mm a = 8.72 mm

As= 4-15 M =800 mm +ve

2

⇒ MR = 175.6 > 0.5 * 333.3 = 166.6 kN.m

Seismic Analysis and Design of Buildings -ve

MR for section with As


= 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.

-ve

M p = 1.47 * 333.33 = 490.00 kN.m +ve

M p

= 1.47 * 175.60 = 258.00 kN.m


A


137.5 kN

B

137.5 kN

490.0 kN.m

258.0 kN.m

2.2 m

3m

2.2 m

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 258.0 kN.m 338.2 kN.m

Region of Plastic Hinging L = 3.0 m

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 V B = 238.6 kN



Section at column face

Vf = 238.6 kN Clause 21.3.4.2

β =0 (excessive cracking is expected at high lateral θ = 45

V s =

drift due to E.Q.) o

φ s * Av * fy * d v * cot(θ )

s 2 Try 4-legged 10 M, Av=400 mm S =

0.85*400*400*584.1*1 238.6*103

= 332.9 mm

Check shear design requirements (i)

Maximum shear (clause 11.3.3) Vr , max = 0.25*φ c * f c' * bw * dv = 0.25*0.65*30*400*584.1=1139.0 kN < Vf = 238.6 O.K.

(ii)

Minimum amount of stirrups (11.2.8.2) S ≤

(iii)

Av * f y 0.06* f c' * bw

≤

400* 400 0.06 * 30 * 4 00

≤ 1217 mm

O.K.

Spacing limit (11.3.8.3) 0.125* φ c * f c' * bw * d v = 0.125*0.65*30* 400 *584.1 = 569.5 kN > Vf

( No reduction in Smax is needed ) ⎧600 ⇒ Smax = ⎨ ⎩0.7*584.1 = 408.87 mm Check “Anti-buckling” requirements

⇐

> S O.K.

(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) (ii) (iii) (iv)

d/4 = 649/4 = 162.25 8*d bar, longitudinal = 8*19.5 = 156 24* d bar, hoops = 24*11.3 = 271 300

⇐ governs



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 S =

0.85*200*400*584.1*1 101.1*103

= 392.86 mm

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 k 1=1.0) = 1.3 * 530 = 689 mm (for top bars k 1=1.3) For M 15 Ld= 390 mm (for bottom bars k 1=1.0) = 1.3 * 390 = 507 mm (for top bars k 1=1.3)



The theoretical cut-off location of the top 2-M 20 is @ 1.51 m from the column face. Clause 12.10.4 states that an embedment length of at least (d) or (12 d bar ) is provided

beyond the theoretical cut-off point. ⎧ Lr + d v cot(θ ) = 1.51 + 0.584 * cot(35) = 2.34 m ⎪ L = ⎨ ⎧d d + max ⎨ l = 0.69 + 0.584 = 1.28 m ⎪ bar 12 * d ⎩ ⎩

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

⇐


L L L L L 1 2 = 3 = 2 = 1 = d h = d 2 2 3 2 3 b * * .4 . 3 0 = ( ( 5 0 2 2 3 + 4 + 4 . 0 .4 0 0 m 0 m . 0 5 . + + 5 5 0 ( m 0 0 0 m + + t . .4 0 0 a 5 k 8 5 .2 .2 e 4 ) 4 n + 4 * 0 + ˜ a c 0 . o 5 . 3 s t 5 4 ( 5 .1 0 3 0 5 ˜ ˜ m 0 ) ) 8 4 c 3 . ˜ . o 6 5 0 l 5 u . 7 m m m m n d i m e n s i o n )


2 5 8 . 0

4

2 -2 0 M L = 3 .1 m 1

4 4 -1 -1 5 5 M M L L = 4 = 8 5 .1 . 5 7 m m

2 -2 0 M

L 2 = 3 . 6 2 -1 5 0 m M L = 3 . 0 m

1

@1 2 0 1 L 5 M e 0 g h g m o e d m o p s 2

1 . 5 1

3 3 8 .2

L = 5 . 5 m

M f

= 1 2 9 . 5

2

0 . 9 M 3 3 8 .2

2 -2 0 M

4 -1 5 4 M -1 5 L M = 7 L . 9 4 = m 5 . 7 m

1

4 9 0 . 0

r

= 1 2 9 . 5 1 . 5 1

2 -2 0 M 2 5 8 . 0

L = 8 . 3 m

3

2 -1 0 M

C L

L = 3 . 0 m 2 -2 0 M

2 -2 0 M

L = 5 . 5 m

L = 8 . 3 m

4 9 0 . 0

F o r m a t i o n o f P l a s t i c H i n g e s a t b e n d s o f t h e b e a m ( C u t o f f o f t o p r e n f o r c e m e n t )

1 7 5 . 6

1 .2 5 D + 1 . 5 0 L

1 . 7 1 t h e o r e t i c a l c u t o f f P o i n t ( M

276.1

235.8

f

= 1 7 5 . 6 )

3 4 0 . 7

1 7 5 . 6 MM f r

( C u t o f f o f b o t t o m r e n f o r c e m e n t )



4 - 20 M


0 1 1

4- 15 M 400

Section 1-1

2- 20 M + 2- 10 M


0 1 1

0 0 7

8- 15 M 400

Section 2-2



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.

Location Units Bottom Top

Table 3: Straining actions at the column ends PE PD PL (±) MD ML

kN -1329.68 -1329.68

kN -540.89 -540.89

kN kN.m 2.68 2.62 2.68 6.35

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

Pf,Bottom [kN] Pf,top [kN] Mf,bottom [kN.m] Mf,top [kN.m]

kN.m 1.37 3.27

ME (±)

VD

VL

VE (±)

kN.m 213.81 112.84

kN 2.56 2.56

kN 1.33 1.33

kN.m 93.33 93.33

4 1.0D+0.5L+1.0E

5 1.0D+0.5L-1.0E

-2473.44

-1327.00

-1332.36

-1597.45

-1602.81

-2473.44

-1327.00

-1332.36

-1597.45

-1602.81

5.33

216.43

-211.19

217.11

-210.51

12.85

119.20

-106.49

120.83

-104.85



Preliminary selection of column 8 - 25 M

Try 500 X 500 mm

As = 8 – 25 M bars = 4000 mm

2

0 0 5

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



Check column capacity Draw the section interaction diagram and check if all load combinations is within the limits 5000

4500

4000

3500

3000

] N k 2500 [ P

1

2000

5

4

1500

3

1000

2

500

0 0

50

100

150

200

250

300

350

400

450

500

B.M [kN.m]

Strong column-weak beam requirements (clause 21.4.2.2) To ensure the formation of plastic hinges in the beams CSA A23.3 requires that: Σ Mnc ≥ Σ MPb For case 2 Mr = 411.0 kN.m Mn= 1.2* Mr = 493.2 kN.m For case 3 Mr = 415.0 kN.m Mn= 1.2* Mr = 498.0 kN.m For case 4 Mr = 425.5kN.m Mn= 1.2* Mr = 510.6 kN.m For case 5 Mr = 429.0kN.m Mn= 1.2* Mr = 514.8 kN.m Case 1 does not involve lateral load

Σ MPb = 490 + 258 = 748.0 kN.m Σ Mnc = 2*493.2 = 986.4 kN.m Σ Mnc > Σ MPb

O.K.



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

col top

= ( M Pr+ + M Pr− ) *

k c

∑ k

1

= (490 + 258)* = 374 kN 2

c

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

VColumn =

= 1.57*432 = 678.24 kN .m 374 + 678 3.8

= 277 kN

Vdesign = 277 kN Shear Design

(clause 21.4.5.2)

Vc = 0.65*0.1* 30*500*(0.72*500) = 64.08 kN V s = 277 - 64.08 = 212.9 kN

Using transverse reinforcement as depicted in figure 2 Av = (2+2*cos(45)) *100 = 341 mm

S=

0.85*341*400*0.72*500*cot(45) 212.9*103

= 196.0 mm

0.125*φ c * f c' * bw * d v = 0.125*0.65*30*500*0.72*500 = 438.75 kN > Vf

⎧600 mm ⇒ Smax = ⎨ ⎩0.7*0.72*500 = 252 mm S ≤

Av * f y ' c

0.06 * f * bw

≤

> S O.K.

⇐

341* 400 0.06 * 30 * 500

≤ 830 mm

Confinement requirements (Clause 21.4.4.2) nl 8 = = 1.33 k n = nl − 2 6

O.K.



Po = α 1 * f c' ( Ag − Ast ) + f y * Ast

= 0.805*30*(500*500-400 0.805*30*(500*500-4000)+400*4000= 0)+400*4000=7541 7541 kN k p =

P f P o

=

1602 7541

= 0.212

Hence the total area of reinforcement is : A sh = 0.2 * kn * kp *

A g Ach

*

= 0.2*1.33*0.212*

f c' f yh

* S * hc

5002 420

2

*

30 400

* 420 * S

= 2.5175 * S min = 0.09 * S * hc * A sh , min

for

f c' yh f

= 0.09 * 420 *

A sh = 341 mm 2 ⇒

S=

341 2.835

30 400

* S = 2.835 * S

= 120.3 mm2

Anti-buckling requirements Maximum spacing is : (i) (ii)

b/4 = 500/4 = 125 mm 6*d bar * 25 = 150 mm 350 − h x 350 − 19 1 90 = 100 + = 153 mm (iii) S x = 100 + 3

3

x

h

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



Splice details Columns bars should be spliced at mid-height with a tension lap splice Ld=820 mm (Table 3.1 A23.3 page 3-15) Lap length = 1.3*820 = 1066 mm

8 - 25 M

0 0 5

500

6 6 0 1

Column Reinforcement Details



Check Joint Shear

VCol.

F2

F1

f 2

f 1

V joint = F1+F2-Vcol.

374

F 1 = 1.25 * As * fy = 1.25 * 800 * 400 = 400 kN

490

258

F 2 = 1.25*1 5*1200*400 0*400 = 600 kN 374 374

Vcol =

374*2 3.8

= 197 kN

V joint = 600 + 400 – 197 = 803 kN Jointt = 2.2 λ *φ c * Vr , Join

f c' * Aj

= 2.2* 2.2*1. 1.0* 0*0. 0.65 65* * 30 *500* 400 400 = 1566.5 kN > V Joint OK 374 374 258

490 374



Shear Walls



Ductile Shear Walls R d = 3.5, R 0 = 1.6

Moderately Ductile Shear Walls R d = 2.0, R 0 = 1.4

Conventional Constructional Shear Walls R d = 1.5, R 0 = 1.3



Check of Ductility of Shear Walls

θid =

(21.6.7)

o

Walls effectively continuous in cross section

o

Plastic hinge at the base

∆ f Rd R0 − ∆ f γ w (hw − l w / 2)

θic = (

ε cu l w 2c

≥ 0.004

− 0.002) ≤ 0.025

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 =

Ratio of the load corresponds to nominal resistance factored load

1.3

-ve +ve

C

0.0035



Ductile Flexural walls

hw

(21.6.1.1)

For hw/lw > 2.0

R d = 3.5

For hw/lw ≤ 2.0

R d = 2.0

Plastic hinge Region = 1.5 l w

lw

( 21.6.3.1)

Effective flange width

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



For all elevations above the plastic hinge region the design SF and BM shall be increased by a ratio of the Factored Moment resistance to the factored moment

(21.6.2.2.C)

BM* factored resistance/factored moment

Original design BM

Factored resistance

Factored moment



Rectangular wall (Plastic hinge Region)

Spacing of distributed RFT (21.6.5.2) ≤ 300 mm

Concentrated RFT

bw

lw

Strain diagram

0.0035 C

If C> min. of {4 b w => bw > lu/10 {0.3lw

(21.6.3.4)

lu : Clear distance between floors

0.0025

≤ ρdistributed ≤ 0.0600

0.0015 bwlw

(21.6.4.3)

≤ 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)


d b

1/10 bw

25 cm wall => d b 20 cm wall => d b


(21.6.4.4) 25 mm 20 mm

*Tie concentrated RFT as columns: Spacing of Ties

least of { 6d b {24 dst { ½ bw

*Horizontal RFT ratio

0.0025

(21.6.6.9)

(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: * 1.5l d



Rectangular wall (Outside Plastic hinge Region)

Spacing of distributed RFT (21.6.5.1) ≤ 450 mm

bw lw

0.0025

≤ ρdistributed ≤ 0.0600 (21.6.4.3)

0.0010 bwlw (21.6.4.4)

*d b

≤ Asconcentrated (min. 4 bars) ≤ 0.06 x Area of concentrated RFT

1/10 bw

(21.6.4.4)

*Ties for concentrated RFT are the same as those for columns.

*Horizontal RFT ratio

* Splices: 1.5ld

(21.6.4.3)

0.0025

(21.6.4.1)



Shear Strength Requirements for Ductile Walls

Vr > Vf

Vr ≥ the smallest of { V p -> shear corresponding to the development of the probable nominal capacity of the wall { V determined from the analysis with R dR 0 = 1

For θid ≥ 0.015 => shear demand β = 0

0.10φcf c bwdv

For θid = 0.015 => shear demand β = 0.18

0.15φcf c bwdv

'

'

θ = 450

'

θ = 350

For Ps = 0.1f cAs

For Ps = 0.2f cAs

'

CHAPTER 8 LESSONS LEARNED FROM THE PERFORMANCE OF BUILDING DURING PAST EARTHQUAKES



Post earthquake reconnaissance reports have provided overviews of damage patterns and detailed descriptions of specific failure modes and structural weaknesses. In addition, many reports have attempted to explain the behaviour of individual buildings by correlating analytical and experimental predictions with observed damage. The major lessons learned from past earthquakes have repeated themselves over and over and have confirmed the principles applied in modern seismic codes.

Lessons learned 1) Designing to Code does not always provide safeguard against excessive damage in severe earthquakes. 2) Well-designed, well detailed and well constructed buildings usually resist earthquake loads without excessive damage. 3) Poor construction practice can lead to severe damage and collapse 4) Ground failure can cause severe damage and collapse. 5) Buildings subjected to successive earthquakes may suffer progressive weakening. 6) Ductility and redundancy provide safety against collapse. 7) Stiff elements, not considered in the design, affect the seismic response of a building.

Primary examples are moment resisting frames that are filled with stiff masonry bricks as shown in next page.



Typical damage Inflicted to Infill Walls and Concrete frame Elements (From EERI, Pub. No. 86-02)

Framed bays with infill walls become stiffer once the frame is in contact with infill wall. This increases the stiffness of the frame and consequently the dynamic properties of the structural system.

If the infill walls are distributed unsymmetrically in plan, torsional modes of vibration can be excited. The infill walls will crack if they are not designed for the interaction forces occurring between frame and wall. The interaction forces may also damage the frame.



8) Problems with soft stories: A soft story can be generated when a shear system is replaced at the first story level with a number of columns. Earthquakes have shown that the stiffness discontinuity with height often directs forces to places where Structures

strength perform

is

minimal. better

in

Soft story

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)



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

Lecture Notes OF Dynamic Analysis

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