Fundamental Concepts of Earthquake Engineering (Solutions, Instructor Solution Manual) [1 ed.] 9781439800881, 9781420064957, 1420064959

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SOLUTIONS MANUAL FOR Fundamental Concepts of Earthquake Engineering

by Roberto Villaverde

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2009 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-13: 978-1-4398-0088-1 (Softcover) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copy-right.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

SOLUTIONS MANUAL FOR Fundamental Concepts of Earthquake Engineering

by Roberto Villaverde

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

SOLUTION MANUAL

FUNDAMENTAL CONCEPTS OF EARTHQUAKE ENGINEERING

Roberto Villaverde University of California at Irvine

1

TABLE OF CONTENTS CHAPTER 4 ................................................................................................................................... 3 CHAPTER 6 ................................................................................................................................. 27 CHAPTER 7 ................................................................................................................................. 33 CHAPTER 8 ................................................................................................................................. 51 CHAPTER 9 ................................................................................................................................. 69 CHAPTER 10 ............................................................................................................................... 81 CHAPTER 12 ............................................................................................................................. 108 CHAPTER 17 ............................................................................................................................. 118 Problem 17.3 ............................................................................................................................... 122 Problem 17.4 ............................................................................................................................... 124 Problem 17.5 ............................................................................................................................... 126 Problem 17.6................................................................................................................................ 127 Problem 17.8................................................................................................................................ 131 Problem 17.12.............................................................................................................................. 146 Problem 17.15.............................................................................................................................. 158

2

CHAPTER 4 Problem 4.1 Determine the velocity of propagation of longitudinal waves traveling along a laterally constrained rod when the rod is made of (a) steel; (b) cast iron; and (c) concrete with f 'c = 4,000 psi. Solution: Young’s moduli, Poisson ratios, and unit weights for steel, cast iron, and concrete with f 'c =4,000 psi are as shown in Table P4.1 Table P4.1. Properties of steel, cast iron, and concrete Material Modulus of elasticity Poisson ratio Unit weight (psi) (pcf) 6 Steel 0.27 490 30×10 6 Cast iron 0.25 485 26×10 0.15 150 Concrete 57,000 f ′ c

Therefore, for the steel rod, the constrained modulus of elasticity and the propagation velocity of longitudinal waves are respectively equal to (see Equations 4.6 and 4.7) E (1 − μ) 30 × 106 (1 − 0.27) M= = = 37.5 × 106 psi (1 − 2μ)(1 + μ) [1 − 2(0.27)](1 + 0.27) 37.5 × 106 (144) M = = 18,838 ft/s = 5.74 km/s ρ 490 / 32.2 and similarly for the cast iron and reinforced concrete rods, E (1 − μ) 26 × 106 (1 − 0.25) M= = = 31.2 × 106 psi (1 − 2μ)(1 + μ) [1 − 2(0.25)](1 + 0.25) vc =

vc =

M=

31.2 × 106 (144) M = = 17,271 ft/s = 5.26 km/s ρ 485 / 32.2

57,000 4,000 (1 − 0.15) E (1 − μ) = 3.8 × 106 psi = (1 − 2μ)(1 + μ) [1 − 2(0.15)](1 + 0.15)

vc =

3.8 × 106 (144) M = = 10,838 ft/s = 3.30 km/s ρ 150 / 32.2

Problem 4.2 A rod of infinite length is subjected to an initial longitudinal displacement given by u0 = 2(1 − x) 0 ≤ x ≤1 u0 = 2 + x -2 ≤ x ≤ 0 Draw plots of the rod’s longitudinal displacement u against the position variable x at times t = 1, 2, 3, and 4 seconds. Consider that the velocity of propagation of longitudinal waves in the rod is equal to 0.5 m/s.

3

Solution: Noticing that

u0 = 0 at x = −2 and x = +1 u0 = 2 at x = 0 the form of the initial pulse is as shown below. Note also that the initial displacement generates two identical waves traveling in opposite directions. Furthermore, since the velocity of propagation is 0.5 m/s, the distance traveled by these waves are as indicated in the Table P4.2. Table P4.2. Distance traveled by waves at different times Time (s) Distance (m) 1.0 0.5 2.0 1.0 3.0 1.5 4.0 2.0

Therefore, the position of the initial displacement pulse at times of 1.0, 2.0, 3.0, and 4.0 seconds is as indicated in Figure P4.2. 2

u

x

2

x

2

x

2

x

2

t =0s

t =1s

t =2s

t =3s

t =4s -5

-4

-3

-2

-1

0

1

2

3

4

5

x

Figure P4.2. Position of displacement pulse at various times

Problem 4.3 Repeat Problem 4.2 considering an initial longitudinal velocity instead of an initial displacement and that this initial velocity is given by v0 = A - 2 ≤ x ≤ 2

v0 = 0

elsewhere

where A is a constant. Solution: According to Equation 4.19 and a zero initial displacement, the displacement in the rod is given by

1 u ( x, t ) = 2vc

x + vc t

∫ v (ζ)dζ 0

x − vc t

which may be considered as the superposition of the two displacement waves 1 u ( x, t ) = 2vc

x + vc t

1 ∫0 v0 (ζ)dζ − 2vc

4

x − vc t

∫ v (ζ)dζ 0

0

where ζ is a dummy variable. Considering first the first integral and the fact that vc = 0.5 m/s, one has that x + 0.5t

∫ v ( ζ ) dζ

u ( x, t ) =

0

0

Then, for –2.0 ≤ x +0.5t ≤ 2.0, x + 0.5t

u ( x, t ) =

∫ Adζ = A[ζ]

x + 0.5 t 0

= A( x + 0.5t )

0

In this case, therefore, the displacement u varies linearly with x for a constant t, and the upper and lower limits for which this relationship is valid are xu + 0.5t = 2.0 or xu = 2.0 − 0.5t xl + 0.5t = −2.0 or xl = −2.0 − 0.5t Hence, for the particular times of 0, 1, 2, 3, and 4 seconds, the displacement u and the lower and upper limits are as indicated in Table P4.3. Table P4.3a. Displacement function and integration limits for first integral at different times t u xl xu 0 -2.0 2.0 Ax 1 A(x + 0.5) -2.5 1.5 2 A(x + 1.0) -3.0 1.0 3 A(x + 1.5) -3.5 0.5 4 A(x + 2.0) -4.0 0

The shape and position of this wave at 0, 1, 2, 3 and 4 seconds (drawn with dashed lines) are as shown in Figure P4.3. Table P4.3b. Displacement function and integration limits for second integral at different times t u xl xu 0 -Ax -2.0 2.0 1 -A(x - 0.5) -1.5 2.5 2 -A(x - 1.0) -1.0 3.0 3 -A(x - 1.5) -0.5 3.5 4 -A(x - 2.0) 0 4.0

Considering now the second integral, one has that x − 0.5 t

u ( x, t ) = −

∫ v ( ζ ) dζ 0

0

Then, for –2.0 ≤ x - 0.5t ≤ 2.0, x − 0.5 t

u ( x, t ) = −

∫ Adζ = − A[ζ]

x − 0.5t 0

= − A( x − 0.5t )

0

In this case, the displacement u also varies linearly with x for a constant t, but the upper and lower limits for which the relationship is valid are xu − 0.5t = 2.0 or xu = 2.0 + 0.5t xl − 0.5t = −2.0 or xl = −2.0 + 0.5t

5

Hence, for the particular times of 0, 1, 2, 3, and 4 seconds, the displacement u and the lower and upper limits are as shown in Table P4.3b. The shape and position of this additional wave at 0, 1, 2, 3, and 4 seconds (drawn with solid lines) are also shown in Figure P4.3. u 2A

2A

-2A

-2A

x 2A

2A

x -2A

x -2A

2A

t =2s

-2A

2A

x -2A

t =1s

-2A

2A

2A

t =0s

t =3s

-2A

2A

t =4s -5

-4 -2A

-3

-2

-1

0

1

2

3

4 -2A

5

x

Figure P4.3. Shape and position of waves generated by intial velocity at different times

Problem 4.4 A long bar with a density of 7850 kg/m3 and a modulus of elasticity of 7.85 kPa is subjected to an initial axial disturbance defined by u0 ( x ) = 0

⎧cos x if x ≤ π / 2 v0 ( x) = ⎨ elsewhere ⎩0 where u0 and v0 respectively denote initial displacement and initial velocity. Determine the displacement induced by the disturbance at a distance x = -π/3 and time t = 2π/3. Solution: According to Equation 4.5, the velocity of propagation of longitudinal waves in the given bar is equal to E 7.85 × 103 = = 1.0 m/s ρ 7850 Similarly, according to Equation 4.19, the displacement induced by an initial velocity v0 and a zero initial displacement is given by vc =

1 u ( x, t ) = 2vc

x + vc t

∫ v (ζ)dζ 0

x − vc t

where ζ is a dummy variable. Hence, for the case under consideration x +t 1 1 u ( x, t ) = ∫ cos ζdζ = − [sin ζ ]xx +− tt 2 x −t 2 provided π π − ≤ x+t ≤ 2 2 and

6

− Otherwise,

π π ≤ x−t ≤ 2 2

u ( x, t ) = 0

Since for x = -π/3 and t = 2π/3 one has that x+t =− and x−t = − then

π 2π π + = 3 3 3

π 2π π − = −π (less than - ) 3 3 2

π π π 2π 1 1 1 u (− , ) = − [sin ζ ]xx +−tt = − [sin ζ ]π− π/ 3/ 2 = [sin( ) − sin(− )] = 0.933 m 3 3 2 2 2 3 2

Problem 4.5 During the process of driving a 40-m long reinforced concrete pile into a foundation soil, a piledriving hammer imparts a force impulse that can be assumed to vary as a half sine wave with an amplitude of 2500 kN and a duration of 0.012 seconds. To monitor the stresses in the pile during the driving process, strain gages are attached at the middle of the pile and at 5 m from its tip. Determine the maximum tensile and compressive stresses recorded by the gauges (a) when the pile is being driven through a soft soil that offers no resistance to the penetration of the pile; and (b) when the pile tip encounters rigid bedrock. Consider a modulus of elasticity of 20,000 MPa, a diameter of 500 mm, and a density of 2300 kg/m3. Solution: The compression force imparted to the pile top is

P(t ) = −2500 sin(

π t) 0.012

which, since the wave velocity in the pile is equal to vc =

E 20,000 × 106 = = 2,949 m/s ρ 2,300

may also be written as P(−vct ) = −2500 sin[−

π (−vct )] 0.012(2,949)

or as

π ( x − 2949t )] 35.39 Hence, the equation for the stress wave is of the form π P P σ( x − vct ) = = = −12,732 sin[− ( x − 2949t )] 2 35.39 A π(0.5) / 4 Note, also, that the length of the pulse is Pulse length = vc (0.012) = 2949(0.012) = 35.39 m P( x − vct ) = −2500 sin[−

7

(a) Soft soil i) Pile mid-height The maximum compressive stress is attained when the incident wave is positioned as shown in the left-hand side of Figure P4.5. Similarly, since the reflected wave is a tension wave when the pile tip encounters no resistance, and since the incident wave has completely disappeared when the center of the reflected wave reaches the mid-height of the pile, the maximum tension stress is attained when the reflected wave is positioned as shown in the right-hand side of Figure P4.5. In consequence, Maximum compressive stress = 12,732 kN/m 2 Maximum tension stress = 12,732 kN/m 2 2.3 m

20 m

20 m 17.7 m

σmax

σmax

(-)

(+) 17.7 m

20 m

20 m

2.3 m Tension stresses

Compression stresses

Figure P4.5a. Position of incident and reflected waves at the time of maximum tension and compression stresses at pile’s midheight

35 m

(+)

σmax

(-) (+)

5m

Tension stresses

Compression stresses

Figure P4.5b. Position of incident and reflected waves at the time of maximum tension and compression stresses at strain gage 5 m above pile tip ii) 5m from pile tip

At any time before the tail of the incident wave reaches the strain gauge 5 m above the pile tip, the incident and reflected waves are positioned as shown in Figure P4.5b. Therefore, the maxi8

mum compression stress at 5 m from the pile tip is given by the superposition of the incident and reflected waves. To compute this maximum stress, consider that, in accordance with the equation developed above, the stress at x = 35 m induced by the incident wave is given by π σi ( x − vct ) = −12,732 sin[ − (35 − 2949t )] = −12,732 sin(261.8t − 3.11) 35.39 Similarly, if one defines x′ and t′ as x′ = x − 40 t ′ = t − 0.014 the equation for the reflected stress wave (tension) is π σ( x′ − vct ′) = 12,732 sin[− ( x′ − 2949t ′)] 35.39 and thus at x′ = 5 m, the stress induced by the reflected wave is π σr (5, t ′) = 12,732 sin[ − (5 − 2949t ′)] = 12,732 sin(261.8t ′ − 0.444) 35.39 which in terms of t may be written as σ r (5, t ) = 12,732 sin( 261.8t − 4.11) It should be noted that the time at which the reflected wave appears for the first time is 40 t= = 0.014 s 2949 and the equation for the reflected wave is thus only valid for t ≥ 0.014 s. As a result, the stress induced by the incident and reflected waves is given by σ(35, t ) = −12,732[sin(261.8t − 3.11) − sin(261.8t − 4.11)] which may also be expressed as σ(35, t ) = −12,732[sin(261.8t ) cos(3.11) − cos(261.8t ) sin(3.11) − sin( 261.8t ) cos 4.11) + cos(261.8t ) sin 4.11)] = −12,732[−0.433 sin(261.8t ) − 0.856 cos(261.8t ) = 12,732 0.4332 + 0.8562 cos(261.8t − θ) = 12,214 cos(261.8t − θ), if t ≥ 0.014 s where 0.856 = 1.102 rad 0.433 It may be seen, thus, that the total stress is maximum when 261.8t − 1.102 = 0 or t = 0.004 s < 0.014 s (not possible) or when 261.8t − 1.102 = π or t = 0.016 s > 0.014 s In consequence, max σ(35, t ) = -12,214 kN/m 2 (compression) θ = tan −1

The maximum tensile stress occurs when the incident wave completely disappears; that is, when the tail of the incident wave reaches the pile tip, which, in turn, occurs when t = 40/2949 +0.012 = 0.026 s (see right-hand side of Figure P4.5b). Therefore, max σr (5, t ) = 12,732 sin[ 261.8(0.026) − 4.11] = 5,478 kN/m 2 (tension)

9

(b) Rigid bedrock i) Pile mid-height When the pile rests on rigid bedrock, the reflected wave is a compression wave. Therefore, the maximum compression stress at the mid-height of the pile is produced by either the incident wave alone or the reflected wave alone. Similarly, no tension stresses are produced. Hence, Maximum compression stress = 12, 732 kN/m 2 Maximum tension stress = 0 ii) 5m from pile tip

The maximum compression stress at 5 m from the pile tip is given by the maximum value of the superposition of the incident and reflected waves. But according to the equations derived above, this superposition is given by σ(35, t ) = −12,732[sin(261.8t − 3.11) + sin( 261.8t − 4.11)] = −12,732[sin(261.8t ) cos(3.11) − cos(261.8t ) sin(3.11) + sin(261.8t ) cos 4.11) − cos(261.8t ) sin 4.11)] = −12,732[−1.566 sin( 261.8t ) − 0.792 cos(261.8t ) = 12,732 1.5662 + 0.7922 cos(261.8t − θ) = 22,343 cos(261.8t − θ), if t ≥ 0.014 s where θ = tan −1

0.792 = 0.468 rad 1.566

Hence, Maximum compression stress = 22,343 kN/m 2 and, since, once again, the reflected wave is a compressive one, Maximum tension stress = 0 Problem 4.6 The 25-story building shown in Figure E4.3 and considered in Example 4.3 is subjected to the lateral ground displacement pulse shown in Figure P4.6. Modeling the building as a shear beam, obtain the lateral displacements at the top of the building and at the level of its setback two seconds after the application of the pulse. ug

A0

0.5

Time (s)

Figure P4.6. Loading pulse in Problem 4.6

Solution: From the results in Example 4.3, vs1 = 100 ft/s α = 0.354 and according to Equation 4.42,

vs 2 = 70.71 ft/s vr = 0.477vi

10

2 2 vi = vi = 0.477vi 1+ α 1 + 0.354 where vs1 and vs2 respectively represent the wave propagation velocities of the lower and upper parts of the building, α is given by Equation 4.36, and vi, vt and vr respectively denote the amplitudes of the incident, transmitted, and reflected waves. Also, the length of the pulse in the lower and upper parts of the building is λ1 = vs1 (0.5) = 100(0.5) = 50 ft λ 2 = vs 2 (0.5) = 70.71(0.5) = 35.36 ft Displacement at building top The time at which the pulse front reaches the discontinuity is x 180 td = = = 1.8s vs1 100 and the distance traveled by the pulse front at t = 2.0 seconds is x = 180 + 70.71(2.0 − 1.8) = 194.1 ft < 300 ft which means that at t = 2.0 seconds, the pulse front has not reached yet the top of the building. Hence, vx =300' = 0 Displacement at setback level At t = 2.0 seconds, the tail of the pulse has traveled a distance x = vs1 (2.0 − 0.5) = 100(1.5) = 150 ft Similarly, the front of the pulse has traveled a distance x = 180 + vs 2 (2.0 − 1.8) = 180 + 70.71(0.2) = 180 + 14.14 = 194.14 ft which means that the incident wave has crossed the discontinuity and a reflected wave is generated. Note also that the reflected wave has traveled a distance x = vs1 (2.0 − 1.8) = 100(0.2) = 20 ft On the basis of this information, the shape, amplitude, and position of the incident and reflected waves at t = 2.0 s is as indicated in Figure P4.6b. From this figure, one obtains 30 Displacement just above setback level = 1.477( )A 0 = 0.866A 0 50 30 30 Displacement just below setback level = A 0 + (0.477)A0 = 0.866A 0 50 50 vt =

120'

1.477A0

30 0.477A 0 50

14.14' 20'

30' 30 A 50 0

180' 150' x

0.477A0

1.477 30 A0 50

Incident wave

Reflected wave

Figure P4.6b. Shape, amplitude, and position of incident and reflected waves at t = 2.0 s

11

Problem 4.7 Assume that during an earthquake the horizontal motion at the ground surface of a deep deposit of soil is generated by a shear wave traveling parallel to the ground surface. Assume further that records of the associated ground velocity have been obtained at a point on the surface along two perpendicular directions. Derive an equation to determine the rotational motion experienced by the ground about an axis perpendicular to the ground surface passing through the point were the records were obtained. Express the desired equation in terms of the two horizontal components of the ground velocity and the shear wave velocity of the soil deposit. Solution: With reference to Figure P4.7 below and Equation 4.49, the displacements induced by the wave along the positive Y and X directions may respectively be expressed as u = f ( y − vs t ) v = g ( x − vst ) where f and g denote arbitrary functions, vs is the shear wave velocity of the soil deposit, and t represents time. Also, according to Equation 4.91, the rotation of the ground about a vertical axis is given by 1 ∂v ∂u ωz = ( − ) 2 ∂x ∂y which may also be written as 1 ∂g ∂ζ ∂f ∂ξ ωz = ( − ) 2 ∂ζ dx ∂ξ dy 1 = [ g ′( x − vst ) − f ′( y − vst )] 2 where ξ = y - vst and ζ = x - vst, and ∂g g ′( x − vst ) = ∂ζ ∂f f ′( y − vst ) = ∂ξ It may be noted, however, that ∂g ∂g ∂ζ = = −vs g ′ ∂t ∂ζ ∂t and ∂f ∂f ∂ξ = = −vs f ′ ∂t ∂ξ ∂t Hence, g′ and f′ are equal to 1 ∂g 1 = − v& g′ = − vs ∂t vs and 1 ∂f 1 = − u& f′=− vs ∂t vs

12

where u& and v& respectively represent the ground velocities along the X and Y directions. In consequence, the ground rotation may also be expressed as 1 ωz = − (v& − u& ) 2vs or as ωz =

1 (u& − v&) 2vs

Y

Shear wave

Y component of shear wave Particle displacement

u v

X component of shear wave X

Particle displacement

Figure P4.7. Components of ground velocity at a point induced by shear wave

Problem 4.8 Consider two harmonic displacement waves, both with a unit amplitude. The first wave has a wavelength of 10 meters and a period of 1.0 second. The corresponding parameters of the second wave are 11 meters and 1.1 seconds, respectively. Plot the variation with distance for the first 300 meters of the displacement induced by the superposition of the two waves 2.0 seconds after the initiation of the waves and determine from this plot the wavelength of the apparent wave. Solution: According to Equations 4.76 and 4.77, the amplitudes of the two given waves are given by 2π 2π 2π 2π v1 = sin( x − t ) = sin( x − t) λ1 10 1.0 T1 2π 2π 2π 2π v2 = sin( x − t ) = sin( x − t) λ2 11 1.1 T2 Thus, the superposition of the two waves at t = 2.0 s is given by v = sin[(0.2 x − 4)π] + sin[(0.182 x − 3.636)π] which leads to the plot shown in Figure P4.8. From this plot, it may be seen that the length of the apparent wave is approximately equal to λ app = 220 m which coincide with the value calculated according to 2π 4π 2 λ app = = = = 220 m Δk / 2 2π − 2π 1 − 1 λ1 λ 2 10 11

13

220 m

2.5 2 1.5 1 0.5 0 0

50

100

150

200

250

300

-0.5 -1 -1.5 -2 -2.5

Figure P4.8. Superposition of two harmomic waves and apparent wave length

Problem 4.9 The ground motion records shown in Figure P4.9 are obtained from sensors located at the same depth in two vertical boreholes spaced 5 m apart. The recorded motions resulted from downward (solid line) and upward (dashed line) mechanical blows imparted at the level of the sensors in a third borehole that was colinear with the other boreholes. The trigger that started the recording of the motions was installed in the borehole that was the nearest to the borehole where the disturbance was applied. Determine the average shear wave velocity of the soil between the sensors.

Figure P4.9. Ground motion records in Problem 4.9

Solution: 1

Disturbance

2

3

Geophone

Geophone

5m

Figure P4.9b. Identification of boreholes and location of geophones

14

As illustrated in Figure P4.9b, the disturbance is applied at Borehole 1 and the arrival of the P wave at Borehole 2 starts the recording equipment. Then, the geophone at Borehole 3 records the passing of the P and S wave. From Figure P4.9, it may be seen that the P wave arrives at Borehole 2 12 ms after the application of the disturbance and at Borehole 3 14 ms after the application of the disturbance. The velocity of propagation of the P wave is thus equal to 5 vp = = 2500 m/s (14 − 12)103 It may also be seen that the time gap between the arrival of the P and S waves is 35-14 = 21 ms. Hence, 5 5 − = 21 × 10− 3 s vs v p from which the velocity of propagation of the S wave results as 1 vs = = 217 m/s 1 21 × 10− 3 + 2500 5 Problem 4.10 Find the amplitude of the vertical and horizontal displacements at depths of 10 and 150 m below the ground surface generated by the passing of a Rayleigh wave with a wavelength of 100 m. Consider a medium with a Poisson ratio of 0.30. Solution: Calculation of α2 For a Poisson ratio of 0.30 and according to Equation 4.189, v 1 − 2μ 1 − 2(0.3) α2 = s = = = 0.286 v p 2(1 − μ) 2(1 − 0.3) Calculation of V 2 According to Equation 4.207, one has thus that V 6 − 8V 4 − (16α 2 − 24)V 2 − 16(1 − α 2 ) = 0 V 6 − 8V 4 − (16 × 0.286 − 24)V 2 − 16(1 − 0.286) = 0

V 6 − 8V 4 + 19.424V 2 − 11.424 = 0 which by trial and error leads to 2

V 2 = 0.8601

2

Calculation of q and s From Equations 4.187 and 4.188 and the calculated values of α2 and V 2, one obtains ω2 ω2 ω2 q 2 = 2 (1 − α 2V 2 ) = 2 (1 − 0.286 × 0.8601) = 0.754 2 vR vR vR s2 =

ω2 ω2 ω2 2 − V = − = ( 1 ) ( 1 0 . 8601 ) 0 . 140 vR2 vR2 vR2

However, since

15

vR =

λ 2π / k ω = = T 2π / ω k

then ω2 = k2 vR2 and q 2 = 0.754k 2 s 2 = 0.140k 2 Also, since for a wavelength of 100 m one has that 2π 2 π k= = = 0.0628 λ 100 and k 2 = 0.0039 Hence, q 2 = 0.754(0.0039) = 0.0030 s 2 = 0.140(0.0039) = 0.0006 and q = 0.0546 s = 0.0235 Displacement amplitudes (horizontal) According to Equation 4.213, the amplitude of the horizontal displacement at depth z is given by 2qs − sz u = A1k[e − qz − 2 e ] s + k2 Hence, for the case under consideration, one has that 2(0.0546)(0.0235) − 0.0235 z u = A1 (0.0628)[e − 0.0546 z − e ] 0.0006 + 0.0039 = 0.0628 A1[e −0.0546 z − 0.5703e −0.0235 z ] Thus, at z = 10 m, u = 0.00806 A1 and at z = 150 m, u = −0.00104 A1 Displacement amplitudes (vertical) According to Equation 4.214, the amplitude of the vertical displacement at depth z is equal to 2k 2 − sz − qz w = A1q[−e + 2 e ] s + k2 Therefore, for the case under consideration, one has that 2(0.0039) w = 0.0546 A1[−e− 0.0546 z + e − 0.0235 z ] 0.0006 + 0.0039 −0.0546 z = 0.0546 A1[−e + 1.7333e −0.0235 z ] Thus, at z = 10 m, w = 0.04319 A1 and at z = 150 m, 16

w = 0.00277 A1

Problem 4.11 A 40-m layer of soil has a unit weight of 20 kN/m3, a modulus of elasticity of 2 × 106 kPa, and a Poisson’s ratio of 0.20. Below this soil layer there is a homogeneous semi-infinite rock formation with a unit weight of 25 kN/m3, a modulus of elasticity of 25 × 106 kPa, and a Poisson’s ratio of 0.25. Determine the wavelength of the Love wave that would travel through this twolayer medium with a velocity equal to the average of the shear wave velocities of the two layers. Solution: From Equations 4.48 and 4.96, the shear wave moduli and the shear wave velocities in the soil layer and rock formation are E 2 × 106 G1 = = = 0.833 × 106 kPa 2(1 + μ1 ) 2(1 + 0.20)

vs1 =

G1 0.833 × 106 (9.81) = = 639 m/s ρ1 20

G2 =

E2 25 × 106 = = 10 × 106 kPa 2(1 + μ 2 ) 2(1 + 0.25)

G2 10 × 106 (9.81) vs 2 = = = 1981 m/s ρ2 25 In consequence, the propagation velocity of the Love wave is 1 vs = (639 + 1981) = 1310 m/s 2 Also, the wavelength of Love waves is given by (see Equation 4.249) tan(kH vL2 / vs21 − 1 =

G2 1 − vL2 / vs22 G1 vL2 / vs21 − 1

which for the case under consideration becomes tan[40k (1310 / 639) 2 − 1] = or

10 × 106 1 − (1310 / 1981) 2 0.833 × 106 (1310 / 639) 2 − 1

tan(71.586k ) = 5.032

Hence, the wave number is k=

1 tan −1 (5.032) = 0.019 71.586

and the wavelength is λ=

2π 2π = = 327 m k 0.019

17

Problem 4.12 A harmonic one-dimensional shear wave with a frequency of 2.0 Hz travels through a sandstone with a shear modulus of elasticity of 2.0 x 106 psf, a unit weight of 118 pcf, and a quality factor of 30. Determine the percentage by which the amplitude of the wave is reduced after it travels a distance of 10,000 feet. Solution: According to Equation 4.151, the shear wave velocity in the given sandstone is G 2 × 106 × 32.2 = = 738.8 ft/s ρ 118 Then, according to Equation 4.283, for a frequency f of 2.0 Hz, a quality factor Q of 30, a wave velocity v of 738.8 ft/s, and r = 10,000 feet, vs =

A(10000, t ) = A0e

−

πf r Qv

= A0e

−

π ( 2.0 ) 10000 30 ( 738.8 )

= 0.0587 A0

Similarly, for r = 0, A(0, t ) = A0 Thus,

and

A(10000, t ) ) = 0.0587 A(0, t ) Reduction percentage = (1 − 0.0587) × 100% = 94.1 %

18

CHAPTER 5 Problem 5.1 The P and S waves from an earthquake in central California arrived at three different stations at the times indicated in Table P5.1. Using the map of central California given in Figure P5.1, determine and plot in the map the location of the earthquake epicenter.

Figure P5.1. Map of central California showing location of seismographic stations A, B, and C Table P5.1. Arrival times of P and S waves at three different stations in central California Station P wave S wave Hr Min Sec Hr Min Sec A B C

15 15 15

45 46 46

54.2 07.6 04.5

15 15 15

46 46 46

07.1 28.0 25.5

Solution: From the data in Table P5.1 Time lag between arrival of P and S waves at Station B: 28.0 - 7.6 = 20.4 s Time lag between arrival of P and S waves at Station A: 67.1 - 54.2 = 12.9 s Time lag between arrival of P and S waves at Station C: 25.5 - 4.5 = 21.0 s Therefore, according to Equation 5.9, Distance between epicenter and Station A= 8 (12.9) = 103.2 km Distance between epicenter and Station C= 8(21.0) = 168.0 km Distance between epicenter and Station B= 8 (20.4) = 163.2 km Drawing circular arcs with centers at the Stations C, B, and A with radii respectively equal to 103.2, 163.2, and 168.0 km in Figure P5.1, the epicenter is located at the center of the triangle formed by the intersection of the three circular arcs shown in Figure P5.1b.

19

r = 103.2 km

Epicenter

r = 168.0 km r = 163.2 km

Figure P5.1b. Circular arcs and location of epicenter

Problem 5.2 Figure P5.2 shows the initial portion of a seismogram recorded in Australia from an earthquake in Oaxaca, Mexico. Identify in the figure the arrival times of the P, S, and surface waves and explain the reasons for the selection.

Figure P5.2. Seismogram considered in Problem 5.2

Solution: The times at which the P, S, and surface waves arrive are identified in Figure P5.2b below. The reasons for the selection are: (a) As the P wave travels faster than the S wave, the P wave is always the first to arrive. (b) The arrival of the S wave is identified by an abrupt change in amplitude a few seconds after the arrival of the P wave. (c) The surface wave is the third wave to arrive since surface waves travel with a slower velocity than S waves. (d) Since surface waves attenuate at a slower rate than body waves, the arrival of the surface wave is identified by a sudden increase in the motion amplitude after the arrival of the S wave. P wave

S wave

Surface wave

Figure P5.2b. Arrival of P, S, and surface waves 20

Problem 5.3 The seismogram shown in Figure P5.3 was obtained from a Wood-Anderson seismograph located 100 km from the hypocenter of the recorded earthquake. Determine (a) the local magnitude of the earthquake, and (b) the average shear wave velocity for the region surrounding the recording station considering that the earthquake started at 07hr 19min 32.0sec and the P wave arrived at the recording site at 07hr 19min 45.2sec.

Figure P5.3. Seismogram considered in Problem 5.3

Solution: (a) The seismogram is obtained at 100 km from the earthquake hypocenter. Then, the amplitude at 100 km from the earthquake hypocenter is A = 21.5 × 103 μm Therefore, according to Equation 5.11 A 21.5 × 103 μm M = log = log = log(215,000) = 4.33 A0 1 μm (b) From Figure P5.3, the time lag between the arrival of the P and S waves is tsp = 24 s

Similarly, the time it takes for the P wave to arrive to the recording station is t p = 45.2 − 32.0 = 13.2 s Therefore, since tsp =

one has that Vs =

Δ Δ − vs v p

Δ 100 Δ = = = 2.69 km/s tsp + Δ / V p tsp + t p 24.0 + 13.2

Problem 5.4 A seismogram from an earthquake exhibits a peak trace amplitude of 75 mm. The seismogram is obtained from a Wood-Anderson seismograph at a seismological station 500 km away from the epicenter of the earthquake. If the distance correction equation for this station is given by r − log A0 = 1.110 log( ) + 0.00189(r − 100) + 3 100 where A0 is amplitude in mm and r hypocentral distance in km, what was the local magnitude of the recorded earthquake? Solution: Using the distance correction equation for an epicentral distance of 500 km, one obtains

21

500 ) + 0.00189(500 − 100) + 3 = 4.53 mm 100 Therefore, according to the definition of local magnitude (see Equation 5.11), A M = log = log A − log A0 = log(75) + 4.53 = 6.4 A0 − log A0 = 1.110 log(

Problem 5.5 The seismogram shown in Figure P5.5 was obtained in southern California from a Wood- Anderson seismograph located 75 km from the hypocenter of the recorded earthquake. Determine (a) the local magnitude of the earthquake using Richter's empirical attenuation curves for Southern California; and (b) the velocity of propagation of the P waves assuming that the Poisson ratio for the medium along which the earthquake waves traveled is equal to 0.30.

Figure P5.5. Seismogram considered in Problem 5.5

Solution: (a) According to Equation 5.12, for southern California, log A0 = 5.12 − 2.56 log Δ where Δ represents hypocentral distance. Therefore, for the problem under consideration, A0 = 105.12 − 2.56 log(75) = 100.32 = 2.09 μm and according to Equation 5.11, 9 × 103 A M L = log = log = 3.6 2.09 A0 (b) According to Equation 4.164, for μ = 3, vp 2(1 − μ) 2(1 − 0.3) = = = 1.871 1 − 2μ 1 − 0.6 vs Hence, for the problem under consideration one has that 1 1 1 1 1 − )Δ = (1.871 − 1)(75) = 36 tsp = ( − )Δ = ( vs v p v p / 1.871 v p vp from which one obtains 0.875(75) vp = = 1.82 km/s 36 Problem 5.6 A seismogram from an earthquake shows a peak trace amplitude of 125 mm. The seismogram was obtained from a seismograph with a magnification factor of 1000 at a seismological station 600 km away from the earthquake epicenter. Determine the local magnitude of the recorded earthquake using a distance correction defined by

22

r − log A0 = 1.110 log( ) + 0.00189(r − 100) + 3 100 where A0 is amplitude in mm and r denotes hypocentral distance in km. Solution: From the given seismogram, the peak amplitude of the ground motion at the recording site (with no amplification) is 125 mm Peak amplitude = = 125 × 10− 3 mm 1000 Hence, the peak amplitude recorded by a Wood-Anderson seismograph would have been A = 2800(125 × 10−3 ) = 350 mm = 3.5 × 105 μm Similarly, for an epicentral distance of 600 km and according to the given distance correction equation, one has that 600 − log A0 = 1.110 log( ) + 0.00189(600 − 100) + 3 = 4.81 100 from which one finds that the zero-magnitude amplitude is A0 = 10−4.81 = 1.55 × 10−5 mm = 1.55 × 10-2 μm Thus, according to Equation 5.11, the local magnitude is 3.5 × 105 A = log = 7.3 M L = log 1.55 × 10− 2 A0 Problem 5.7 The seismogram shown in Figure P5.7 was obtained in southern California by a Wood-Anderson seismograph at a seismographic station located 165 km from the earthquake epicenter. (a) Determine the local magnitude of the earthquake using Richter’s attenuation equation for southern California. (b) If the earthquake started at 07hr 19 min 32.0sec, and if the first P wave arrived at the recording site at 07hr 19min 45.2sec, what is the average Poisson ratio for the region surrounding the recording station?

Figure P5.7. Seismogram considered in Problem 5.7

Solution: (a) For an epicentral distance of 165 km, Richter’s attenuation equation (Equation 5.12) yields, log A0 = 5.12 − 2.56 log Δ = 5.12 − 2.56 log(165) = −0.56 μm Hence, according to Equation 5.11, M L = log A − log A0 = log(23.0 × 103 ) + 0.56 = 4.9 (b) From Figure P5.7, the time gap between the arrivals of the S and P waves is tsp = 24 s

and the time it takes the P wave to arrive to the recording station is 23

t p = 45.2 − 32.0 = 13.2 s Hence, the velocity of propagation of the P and S waves are 165 vp = = 12.5 km/s 13.2 165 vs = = 4.44 km/s 13.2 + 24 Also, from Equation 4.164, one has that vp 2(1 − μ) 12.5 = = = 2.82 1 − 2μ 4.44 vs In consequence, 2(1 − μ) = 7.95 1 − 2μ from which one obtains μ = 0.43 Problem 5.8 The peak ground displacement recorded by a long-period seismograph located at a distance of 3,000 km from an earthquake’s epicenter was 7.6 millimeters. Determine the earthquake’s surface wave magnitude. Solution: The epicentral distance in degrees is (1° = 111 km), 3000 Δ= = 27° 111 Hence, according to Equation 5.14, the surface wave magnitude is M s = log a + 1.66 log Δ + 2.0 = log(7.6 × 103 ) + 1.66 log(27°) + 2 = 8.3 Problem 5.9 Estimate the moment magnitude of an earthquake generated at a fault that slips, on average, 9.2 m and ruptures an area 44 km wide and 75 km long. Solution: According to Hanks-Kanamori’s equation (Equation 5.22), the seismic moment is given by 2 M w = log M 0 − 10.7 3 where M0 is the seismic moment expressed in dyne-cm and equal to (see Equation 5.16) M 0 = GAD

where G ≈ 3×1011 dyne/cm2, A is the ruptured area, and D is the average amount of slip. Hence, for the case under consideration, M 0 = 3 × 1011 (44 × 105 × 75 × 105 )(9.2 × 102 ) = 9.11× 1027 dyne - cm and

24

2 M w = log(9.11 × 1027 ) − 10.7 = 7.9 3 Problem 5.10 Estimate the moment magnitude of earthquakes from faults whose ruptured area is more than 30 km long and 25 km deep and slip more than 1.0 m. Solution: From Equation 5.16,

N (30 × 103 m)(25 × 103 m)(1 m) = 2.25 × 1019 N - m 2 m and thus, according to Equation 5.23, 2 2 M w = log M 0 − 6.0 > log(2.25 × 1019 ) − 6.0 > 6.9 3 3 M 0 = GAD > 3 × 1010

Problem 5.11 A building is to be constructed in Southern California at a site whose location with respect to nearby earthquake faults is indicated in Table P5.11. Also given in this table is the local magnitude of the largest earthquake that is likely to occur at these faults. Find the maximum ground displacement for which the building should be designed. Table P5.11. Magnitudes and distances to sites considered in Problem 5.11 Fault Closest distance to site ML (k78 ) San Andreas 8.0 San Jacinto Santa Monica Whittier Newport-Inglewood

7.5 7.25 7.0 7.0

81 26 32 7

Solution: From the definition of local magnitude (see Equation 5.11), one has that A M L = log A0 and A = A010M L where for southern California earthquakes (see Equation 5.12) A0 = 105.12 − 2.56 log Δ Table P5.11b. Values of A0 and peak ground motion amplitudes ML Δ (km) A0 (μm) A/2800 (mm) 8.0 78 1.89 67.5 7.5 81 1.72 19.4 7.25 26 31.45 199.8 7.0 32 18.48 66.0 7.0 7 904.79 3231.4

25

Hence, for each of the given magnitudes and epicentral distances, the values of A0 and the peak ground motion amplitudes (i.e., A/2800) are as shown in Table P5.11b. The peak ground displacement for which the building should be designed is thus 3.23 m. Problem 5.12 The undamped natural frequency and damping ratio of a commercial accelerograph are 25 Hz and 60 per cent, respectively. What would be the error introduced by the accelerograph in the recording of a sinusoidal ground acceleration with a frequency of 12.5 Hz due to the nonuniform frequency response of the instrument? Solution: According to Equation 5.28, the relationship between the response of a single-degree-of-freedom system under a sinusoidal excitation and the ground acceleration is given by ω2 &u&g (t − φ) = − n u (t ) D where üg(t) signifies ground acceleration, u(t) denotes the displacement response of the system, ω represents the system’s natural frequency, φ is a phase angle, and 1 D= ω ω [1 − ( ) 2 ]2 + [2ξ ]2 ωn ωn which for the accelerograph under consideration results as 1 D= = 1.041 12.5 2 12.5 2 2 ] ) ] + [2(0.6) [1 − ( 25 25 Therefore, since D deviates from unity by 4.1%, the error introduced by the accelerograph is Error = 4.1%

26

CHAPTER 6 Problem 6.1 Suppose the predominant horizontal ground shaking during an earthquake can be assumed sinusoidal with a frequency of 5 Hz. A rigid body with a mass of 1 kilogram rests freely on the ground and just starts slipping during an earthquake. If the coefficient of friction between the ground and the body is around 0.5, estimate the peak displacement of the ground during the earthquake. Solution: Since the body is rigid, its acceleration is equal to the ground acceleration. That is, u&&b = u&&g . Al-

so, since the force in the body when it just starts moving is approximate equal to the friction force, one has that F = μN = 0.5mg and correspondingly, the peak acceleration with which the body will move is F u&&b = = 0.5 g m Thus, since it can be assumed that the ground shaking is sinusoidal with a frequency of 5 Hz, the ground acceleration is given by u&&g (t ) = u&&b (t ) = 0.5 g sin ωt which may also be expressed as u&&g (t ) = −ω2u0 sin ωt

where u0 denotes ground displacement amplitude. Hence, one has that ω2u0 = 0.5 g and thus the peak displacement of the ground is approximately equal to 0.5 g 0.5(9.81) u0 = 2 = = 0.00497 m = 4.97 mm ω (2π) 2 (5) 2 Problem 6.2 Figure P6.2 shows the elastic response spectrum for the S38W accelerogram recorded at the Union Bank Building in Los Angeles, California, during the 1971 San Fernando earthquake. Estimate from this response spectrum the maximum ground displacement at the recording site during that earthquake.

Figure P6.2. Elastic response spectrum for the accelerogram recorded in the S38W direction at the Union Bank Building in Los Angeles, California, during the 1971 San Fernando earthquake 27

Solution: Since for very long periods, the spectral displacement and the peak ground displacement are approximately equal, from the right hand side of the response spectrum curve, one has that Peak ground displacement ≈ 3 inches Problem 6.3 The total-deformation response spectrum for elastoplastic systems with 5 per cent damping of the N21E accelerogram recorded at the Castaic station during the 1971 San Fernando earthquake is shown in Figure P6.3. The recording station was located at a distance of 48 km from the earthquake epicenter. On the basis of this spectrum and Richter’s attenuation equation for the magnitude of earthquakes in Southern California, estimate the local magnitude of the earthquake.

Figure P6.3. Total deformation nonlinear response spectrum for elastoplastic systems with 5 percent damping corresponding to the N21E accelerogram recorded at the Castaic station during the 1971 San Fernando earthquake

Solution: For very small frequencies, the spectral displacements are approximately equal to the peak ground displacement. Therefore, from the response spectrum, one has that Peak ground displacement ≈ 2 in. = 5.1 cm = 5.1 × 104 μm Now, according to Equation 5.12, one has that for Δ = 48 km log A0 = 5.12 − 2.56 log Δ = 5.12 − 2.56 log(48) = 0.82 from which it is found that A0 = 100.82 = 6.55 μm Thus, since the magnified displacement recorded by a Wood-Anderson seismograph would be A = 5.1× 104 × 2800 = 142.8 × 106 μm the local magnitude of the earthquake is approximately equal to A 142.8 × 106 = log = 7.3 M L = log A0 6.55 Problem 6.4 Using the response spectrum in Figure P6.4a, construct a design spectrum by drawing an envelope to the 5 per cent damping response spectrum; that is, by drawing lines perpendicular to the displacement, velocity, and acceleration axes through the points of maximum spectral displacement, maximum spectral velocity, and maximum spectral acceleration for 5 per cent damp-

28

ing. On the basis of this design spectrum, determine the design lateral earthquake force for the power transformer shown in Figure P6.4b. The transformer is supported by a steel pedestal and can be idealized as a single-degree-of-freedom system.

Envelope

20 ft

W = 1140 Lb c.g.

E= I=

6

30 X 10 psi 72.5 in4

Figure P6.4. (a) Response spectrum for N11W component of Eureka, California, earthquake of December 21, 1954 and (b) simplified model of power transformer Solution: The stiffness and mass of the transformer are 3EI 3 × 106 (72.5) = 472 Lb/in k= 3 = (20 × 12)3 L W 1140 = 2.95 Lb - s 2 / in m= = g 32.2 × 12 and thus its natural period is 2π 2π 2π = = Tn = = 0.5 s ωn 472 / 2.95 k/m Similarly, the envelope to the 5 percent damping response spectrum is as shown in Figure P6.4b, and the spectral acceleration corresponding to a natural period of 0.5 s in the design spectrum defined by this envelope is SA = 0.55 g Consequently, the transformer should be designed for a lateral force equal to FE = mSA =

W 1140 SA = (0.55 g ) = 627 lb g g

Problem 6.5 Two identical 30-story towers are to be erected adjacent to each other. Their fundamental periods are estimated to be about 3 seconds, and their lateral force resisting structures are designed to withstand inelastic deformations of up to 6 times their yield values. Using the design spectrum in Figure P6.5, which represents an envelope to yield deformation spectra in a tetralogarithmic representation for a ductility factor of six, estimate the minimum clear distance that must be left between the two towers to avoid collision.

29

SV (in/s)

SD (in)

SA (g)

4.0

0.2

2.0

f (Hz)

Figure P6.5. Design spectrum considered in Problem 6.5

Solution: From the given yield deformation spectrum for a system with a natural frequency of 1/3.0=0.22 Hz, one has that SV = 4.0 in/s and thus SV 4.0 uy = = = 1.91 in ωn 2π / 3 Consequently, for a structure that is capable of sustaining inelastic deformations 6 times its yield deformation (i.e., μ = 6.0), the maximum deformation is umax = μu y = 6.0(1.91) = 11.46 in

Considering that each tower may vibrate in opposite directions, a collision may be avoided if Clearance between towers = 2(11.46) = 22.9 in Problem 6.6 An elastoplastic single-degree of freedom system with a mass of 2.95 kip-s2/in., a stiffness constant of 472 kip/in., a damping ratio of 5 per cent, and a yield strength of 330 kips is subjected to an earthquake ground motion whose yield deformation inelastic response spectrum is shown in Figure P6.6. If this spectrum is for elastoplastic systems with 5 per cent damping, what is the maximum relative displacement the system's mass will experience under this ground motion?

Figure P6.6. Yield deformation response spectrum for 5-percent damping elastoplastic systems of the EW accelerogram recorded during the 1972 Managua earthquake

Solution: The initial natural frequency of the system is equal to

30

f =

1 k 1 472 = = 2.0 Hz 2π m 2π 2.95

and its yield deformation equal to Fy

330 = 0.70 in k 472 From the given spectrum, the ductility factor for a system with a natural frequency of 2.0 Hz and a yield deformation of 0.70 is μ = 3.0 Therefore, the maximum deformation or maximum relative displacement is umax = μu y = 3.0(0.7) = 2.10 in uy =

=

Problem 6.7 The yield deformation inelastic response spectrum shown in Figure P6.7 is representative of the earthquake ground motions expected in a given area. On the basis of this response spectrum, find the yield strength that will limit the displacements of an elastoplastic single-degree-of-freedom system with a mass of 2.95 kip-s2/in, a stiffness constant of 472 kip/in, and a damping ratio of 5 per cent to values no grater than 5 times the displacement at which it yields.

Figure P6.7. Yield deformation response spectrum for 5-percent damping elastoplastic systems of the EW accelerogram recorded during the 1940 El Centro earthquake

Solution: To be able to limit the displacements of the system to 5 times its yield deformation, it is necessary to have a ductility factor μ of 5.0. Hence, from the given yield deformation spectrum, for a natural frequency of 1 k 1 472 = = 2.0 Hz fn = 2π m 2π 2.95 and a ductility factor of 5.0, the yield deformation is u y = 0.35 in

In consequence, the necessary yield strength is Fy = ku y = 472(0.35) = 165 kips

31

Problem 6.8 Determine the ordinate corresponding to a frequency of 1.0 Hz in the amplitude Fourier spectrum of a ground motion whose accelerations in m/s2 are defined by ⎧⎪e −t for t > 0 u&&g (t ) = ⎨ ⎪⎩ 0 otherwise

Solution: According to Equation 6.81, the Fourier spectrum of the given ground acceleration is ∞

∞

∞

0

0

0

F (ω) = ∫ u&&g (t )e − iωt dt = ∫ e − t e − iωt dt = ∫ e − (1+ iω) t dt

− 1 − (1+ iω) t ∞ 1 [e ]0 = 1 + iω 1 + iω Therefore, the amplitude response spectrum is given by 1 1 1 | F (ω) |= = = 2 | 1 + iω | 1+ ω 1 + (2π) 2 f 2 and for f = 1.0 Hz, 1 = 0.157 m/s | F (ω = 2π) |= 1 + ( 2 π) 2 =

Problem 6.9 During the calculation of the response spectrum ordinates for a given accelerogram, it is found that at the end of the accelerogram the displacement and velocity responses of an undamped single-degree-of-freedom system with a natural frequency of 2 Hz are 30 cm and 18 cm/s, respectively. What is the ordinate corresponding to a period of 0.5 seconds in the amplitude Fourier spectrum of the same accelerogram? Solution: According to Equation 6.105, the amplitude Fourier spectrum is given by F (ω) = u& 2 (td ) + ω2u 2 (td )

where u& (td ) and u (td ) respectively represent the velocity and displacement responses of an undamped single-degree-of-freedom system at the end of the ground motion. Since in this case, u& (td ) = 18 cm/s

and u (td ) = 30 cm

then, for a period of 0.5 seconds, which corresponds to ω = 2π/0.5 = 4π rad/s, the Fourier amplitude is F (ω = 4π) = (18) 2 + (4π) 2 (30) 2 = 377.4 cm/s

32

CHAPTER 7 Problem 7.1 It is determined from paleoseismicity studies that a nearby fault has slipped up to 5 m in the past. On the basis of this information, estimate the maximum moment magnitude of the earthquakes that can occur at this fault. Solution: According to the empirical equations suggested by Wells and Coppersmith (see Equation 7.3), the maximum magnitude of the earthquakes that can occur at a fault that has slipped 5 m in the past is M w = 6.69 + 0.74 log D = 6.69 + 0.74 log(5) = 7.2

Problem 7.2 A site in southern California is overlain by soil deposits with an average shear wave velocity of 225 m/s over the top 30 m. The site is located 25 km away from a known earthquake fault. Find the mean and the mean plus one standard deviation of the peak horizontal ground acceleration and peak horizontal ground velocity that would be observed at the site in the event of a shallow earthquake with a moment magnitude of 7.0 at that fault. Solution: Using Joyner-Boore-Fumal attenuation equation in Table 7.1for M w = 7.0 r = R 2 + h 2 = 252 + 5.57 2 = 25.61 km

and VS = 225 m/s

the logarithm of the mean peak ground acceleration and the standard deviation of this logarithm are ln AH = −0.242 + 0.527( M w − 6) − 0.778 ln r − 0.371ln(VS / 1396) = −0.242 + 0.527(7.0 − 6) − 0.778 ln(25.61) − 0.371ln(225 / 1396) = −1.561 σln AH = 0.520

Thus, the mean peak ground acceleration and the mean plus one standard deviation of the peak ground acceleration are AH 0.50 = e −1.56 = 0.210g AH 0.84 = e −1.561+ 0.520 = 0.353 g

Similarly, from the Joyner-Boore attenuation equation in Table 7.2 for r = R 2 + 42 = 252 + 42 = 25.32 km

and

S =1

one obtains log VH = 2.09 + 0.49( M w − 6) − log r − 0.26r + 0.17 S = 2.09 + 0.49(7.0 − 6) − log(25.32) − 0.0026( 25.32) + 0.17(1) = 1.281cm/s

33

σlog VH = 0.33 VH 0.50 = 101.281 = 19.1cm/s VH 0.84 = 101.281+ 0.33 = 40.83 cm/s

Problem 7.3 Generate the expected 5% damping pseudoacceleration response spectrum for the site described in Problem 7.2 and earthquakes with a moment magnitude of 7.0. Solution: From the attenuation equation and coefficients in Table 7.1, the mean 5% damping pseudoacceleration response spectrum for M w = 7.0 r = 25.61 km VS = 225 m/s

results as shown in Figure P7.3. 0.60

Spectral acceleration / g

0.50

0.40

0.30

0.20

0.10

0.00 0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Period (s)

Figure P7.3. Acceleration response spectrum for site described in Problem 7.2

Problem 7.4 An earthquake generated at a strike-slip fault in Manjil, Iran, had a moment magnitude of 7.4. Find the mean and the mean plus one standard deviation of the peak horizontal ground acceleration and peak horizontal ground velocity produced by this earthquake at a site overlain by soft rock and located 4 km from the fault. The basement rock at the area near the fault is estimated to be at a depth of 1400 m. Solution: Campbell’s attenuation equation in Table 7.3 for M w = 7.4 D = 1.4 km r = 42 + 1.42 = 4.24 km F =0 S SR = 1 S HR = 0 fV ( D) = 0

34

yields ln( AH ) = −3.512 + 0.904 M w − 1.328 ln r 2 + [0.149 exp(0.647 M w )]2 + (0.440 − 0.171ln r ) S SR = −3.512 + 0.904(7.4) − 1.328 ln 4.242 + [0.149 exp(0.647 × 7.4)]2 + [0.440 − 0.171ln(4.24)]1.0

= −0.495

from which one gets ( AH ) 0.5 = e −0.495 = 0.609 g

Also, since for AH > 0.21 g,

σln AH = 0.39

then ( AH )0.84 = e −0.495 + 0.39 = 0.900 g

Similarly, ln(VH ) = ln( AH ) + 0.26 + 0.29 M w − 1.44 ln[r + 0.0203 exp(0.958M w )] + 1.89 ln[r + 0.361exp(0.576 M w )] + (0.0001 − 0.000565M w )r − 0.15S SR + 0.75 tanh(0.51D)(1 − S HR ) = −0.495 + 0.26 + 0.29(7.4) − 1.44 ln[4.24 + 0.0203 exp(0.958 × 7.4)] + 1.89 ln[4.24 + 0.361exp(0.576 × 7.4)] + (0.0001 − 0.000565 × 7.4) 4.24 − 0.15(1.0) + 0.75 tanh(0.51 × 1.4)(1.0) = 3.795 σln(VH ) = σln2 AH + 0.06 2 = 0.39 2 + 0.06 2 = 0.395

and, thus, (VH )0.5 = e3.795 = 44.5 cm/s (V H ) 0.84 = e 3.795 + 0.395 = 66 .0 cm/s

Problem 7.5 Obtain the expected 5% damping pseudoacceleration response spectrum for the site described in Problem 7.4 and earthquakes with a moment magnitude of 7.4. Solution: From the attenuation equation and coefficients in Table 7.3 and for M w = 7.4 D = 1.4 km

r = 42 + 1.42 = 4.24 km S SR = 1 S HR = 0 f SA ( D) = 0

the mean 5-per-cent damping pseudoacceleration response spectrum results as shown in Figure P7.5.

35

1.4

Spectral acceleration / g

1.2

1

0.8

0.6

0.4

0.2

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Period (s)

Figure P7.5. Acceleration response spectrum for site described in Problem 7.4

Problem 7.6 Find the mean and the mean plus one standard deviation of the peak horizontal ground acceleration at a site overlying deposits of firm soil and located at a distance of 100 km from a subduction zone in central Chile when an earthquake with a moment magnitude of 7.9 and a focal depth of 31 km occurs at this subduction zone. Solution: Using Youngs et al. attenuation equation for ground motions on soil in Table 7.4 with M w = 7.9 r = 100 km H = 31 km ZT = 0

one gets ln AH = −0.6687 + 1.438M w − 2.329 ln[r + 1.097 exp(0.617 M w )] + 0.00648H = −0.6687 + 1.438(7.9) − 2.329 ln[100 + 1.097 exp(0.617 × 7.9)] + 0.00648(31) = −1.906

and σln AH = 1.45 − 0.1M w = 1.45 − 0.1(7.9)

=0.66 Thus, the mean and mean plus one standard deviation peak horizontal ground acceleration in question are ( AH )0.5 = e −1.906 = 0.149 g ( AH )0.84 = e −1.906 + 0.66 = 0.288 g

Problem 7.7 Determine the expected 5% damping acceleration response spectrum for the site described in Problem 7.6 and earthquakes with a moment magnitude of 7.9. Solution: For M w = 7.9

36

r = 100 km H = 31 km ZT = 0

the attenuation equation for ground motions on soil in Table 7.4 yields the mean 5% damping acceleration response spectrum shown in Figure P7.7. 0.35

Spectral acceleration / g

0.30

0.25

0.20

0.15

0.10

0.05

0.00 0

0.5

1

1.5

2

2.5

3

3.5

4

Period (s)

Figure P7.7. Acceleration response spectrum for site described in Problem 7.6

Problem 7.8 The number of earthquakes recorded in a seismically active region over a period of 120 years is shown in Table P7.8. Basing the calculations on the central values of the given magnitude ranges, (a) estimate the parameters of the Gutenberg-Richter law for the region; and (b) calculate neglecting earthquakes with a magnitude of less than 4.0 the probability that an earthquake in the region will have a magnitude between 6.5 and 7.5. Table P7.8. Earthquake data for Problem 7.8 Magnitude Number of earthquakes in range 120 years 3.5-4.5 420 4.5-5.5 27 5.5-6.5 4 6.5-7.5 1

Solution: (a) On the basis of the given data, the logarithm of the number of earthquakes per year that exceed a magnitude of 4.0, 5.0, 6.0, or 7.0 is as shown in Table P7.8b. Performing then a correlation analysis of magnitude versus log λm, one obtains log λ m = 3.960 − 0.877m

Table P7.8b. Number of earthquakes per year that exceed given magnitudes log λm Magnitude No. of earthquakes No. of earthquakes per year (m) with magnitude > m with magnitude > m (λm) 4 452 3.767 0.576 5 32 0.267 -0.574 6 5 0.042 -1.380 7 1 0.008 -2.079

37

Therefore, the parameters of the Gutenberg-Richter law for the region in question are a = 3.960 b = 0.877

(b) Using the Gutenberg-Richter law determined above, the number of earthquakes per year with a magnitude greater that 4.0, 6.5 or 7.5 is λ 4.0 = 103.960 − 0.877 ( 4.0 ) = 2.831 λ 6.5 = 103.960 − 0.877 ( 6.5) = 0.018 λ 7.5 = 103.960 − 0.877 ( 7.5) = 0.002

Therefore, the probability of that an earthquake in the region in question will have a magnitude between 6.5 and 7.5 is P[6.5 < M < 7.5] =

0.018 − 0.002 = 0.006 2.831

Problem 7.9 The average number of earthquakes exceeding a magnitude Mw in a year in a seismic region is given by log N = 4.97 − 0.87 M w

What is the probability that an earthquake of magnitude 5.0 or greater occur at least once in a 50year interval? Solution: For Mw > 0 and Mw > 5.0 one has that

ln N 0 = 4.97 − 0.87(0) = 4.97 N 0 = e 4.97 = 144.0269 ln N 5 = 4.97 − 0.87(5) = 0.62 N 5 = e0.62 = 1.8589

Thus, the probability of occurrence of an earthquake of magnitude greater than 5.0 in one year is approximately equal to λ=

N5 1.8589 = = 0.0129 N 0 144.0269

and, if the Poisson probability distribution is used, the probability of an earthquake of magnitude 5.0 or greater occur at least once in 50 years is P ( X t ≥ 1) = 1 − e − λt = 1 − e −0.0129(50) = 0.475

Problem 7.10 It is found from the analysis of the occurrences of earthquakes near a construction site over a period of 200 years and the use of an attenuation equation to estimate peak ground accelerations that eight times over the considered period the peak ground acceleration at the site is greater than 0.25g. What is the probability that the site be subjected to a peak ground acceleration greater than 0.25g in a time interval of 100 years? Solution: According to the given data,

38

mean occurrence rate = λ =

8 = 0.04 200

Therefore, if the Poisson probability distribution is used, the probability that the site be subjected to a peak ground acceleration greater than 0.25g in a time interval of 100 years is P[ X 100 ≥ 1] = 1 − e − λt = 1 − e −0.04(100 ) = 0.982

Problem 7.11 A regression analysis between the peak ground acceleration (expressed as a fraction of the acceleration of gravity) at a site and the frequency with which this acceleration has been exceeded in the past yields the empirical equation a = 0.33 log TR − 0.323

where a represents peak ground acceleration and TR denotes return period. Using this equation and assuming that the number of occurrences of an earthquake with given intensity is given by a Poisson process, determine (a) the probability that the ground acceleration at the given site exceed 1.0g in a time interval of 100 years; and (b) the peak ground acceleration corresponding to a probability of exceedance of 1 per cent in a time interval of 50 years. Solution: (a) Substitution of a = 1.0g into the given empirical equation yields log TR =

1.0 + 0.323 = 4.0091 0.33

from which one finds that TR = 10 4.0091 = 10,211.5 years

Thus, the mean annual rate of exceedance is λ=

1 1 = = 9.793 × 10 − 5 TR 10,211.5

and, if a Poisson process is considered, the probability of exceeding 1.0g in 100 years is equal to P ( X 100 ≥ 1) = 1 − e − λt = 1 − e − (9.793×10

−5

)(100 )

= 0.0097

(b) Similarly, the probability of exceedance of 1 per cent in 50 years is given by P ( X 50 ≥ 1) = 1 − e − λt = 1 − e −50λ

Thus, one has that 1 − e −50λ = 0.01

from which one obtains λ=

1 1 ln( ) = 0.0002 50 0.99

and TR =

1 1 = = 4,975 years λ 0.0002

Hence, the peak ground acceleration corresponding to a probability of exceedance of 1 per cent in 50 years is equal to a = 0.33 log(4,975) − 0.323 = 0.897 g

39

Problem 7.12 The frequency with which earthquakes occur at a fault 40 km from a construction site is given in Table P7.12 in terms of their moment magnitude. Using this information and the approach described in Section 7.6, construct a frequency-peak ground acceleration curve and estimate on the basis of this curve the peak ground acceleration at the construction site that has a 30% probability of being exceeded in a time interval of 50 years. Assume an average shear wave velocity for the site soil of 30 m/s. Base your calculations on the central values of the given magnitude ranges. Table P7.12. Earthquake data for Problem 7.12 Magnitude Number of earthquakes in range 100 years 4.75 - 5.75 250 5.75 - 6.75 78 6.75 - 8.25 2

Solution: Using Boore-Joyner-Fumal attenuation equation to estimate the peak ground acceleration at the site, one obtains the logarithm of the mean recurrence rates corresponding to the given magnitude ranges shown in Table P17.12b. Table P7.12b. Mean recurrence rates for given magnitudes Mw Mean PGA/g No. of exceedances per year (λ) log λ 5.25 0.053 (250+78+2)/100 = 3.3 0.519 6.25 0.089 (78+2)/100 = 0.8 -0.097 7.5 0.172 2/100 = 0.02 -1.699

Similarly, by performing a regression analysis with the PGA and log λ values in this table, one gets the following equation to relate PGA and mean recurrence rate, which when plotted represents the desired frequency-peak ground acceleration curve: log λ = −18.751 PGA + 1.537

But from the fact that a probability of exceedance in 50 years is 30%, one has that P[ X 50 ≥ 1] = 1 − e − λ (50) = 0.3

and, in consequence, λ=−

ln(0.70) = 0.007 50

Hence, upon substitution of this value into the equation developed above, the peak ground acceleration corresponding to the aforementioned probability of exceedance is given by PGA = −

log(0.007) − 1.537 = 0.196 g 18.751

Problem 7.13 The data shown in Table P7.13 represent the estimated values of the peak ground accelerations induced by the earthquakes that have occurred in the vicinity of a construction site over a period of 20 years. Using these data and the approach described in Section 7.7, find: (a) an expression to determine the mean annual frequency with which a given peak ground acceleration would be exceeded at the site; (b) the peak ground acceleration at the site corresponding to a return period of 500 years; and 40

(c) the peak ground acceleration at the site corresponding to an exceedance probability of 1% in 50 years. Table P7.13. Peak ground acceleration data for Problem 7.13

Year

Peak ground acceleration/g

1978 1979 1980 1981 1982 1982 1983 1983 1984

0.392 0.343 0.523 0.623 0.387 0.098 0.413 0.312 0.367

Year

Peak ground acceleration/g

Year

Peak ground acceleration/g

1984 1985 1986 1986 1987 1988 1989 1989 1990

0.444 0.000 0.423 0.333 0.354 0.000 0.345 0.498 0.600

1991 1992 1993 1993 1994 1995 1995 1996 1997

0.432 0.000 0.586 0.000 0.427 0.419 0.118 0.405 0.478

Solution: (a) Following the procedure described in Box 7.2, the given peak ground acceleration data is first rearranged in ascending order. Then, the cumulative frequencies of occurrence and the values of the standard variable s for each of the peak ground accelerations in the list are calculated as indicated in Box 7.1. The rearranged peak accelerations and the resulting cumulative frequencies and values of the standard variable s are shown in the Table 7.13b.

Table P17.13b. Rearranged peak accelerations and cumulative frequencies of standard variable s ln aj j/21 j aj s j = − ln[− ln( j / 21)] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0 0 0 0.343 0.354 0.387 0.392 0.405 0.413 0.419 0.423 0.427 0.432 0.444 0.478 0.498 0.523 0.586 0.600 0.623

-1.07002 -1.03846 -0.94933 -0.93649 -0.90387 -0.88431 -0.86988 -0.86038 -0.85097 -0.83933 -0.81193 -0.73814 -0.69716 -0.64817 -0.53444 -0.51083 -0.47321

0.047619 0.095238 0.142857 0.190476 0.238095 0.285714 0.333333 0.380952 0.428571 0.476190 0.523810 0.571429 0.619048 0.666667 0.714286 0.761905 0.809524 0.857143 0.904762 0.952381

41

-1.11334 -0.85500 -0.66573 -0.50575 -0.36122 -0.22535 -0.09405 0.03554 0.16570 0.29849 0.43598 0.58050 0.73486 0.90272 1.0892 1.30220 1.55443 1.86982 2.30175 3.02023

Performing now a regression analysis between the values of ln aj and sj, one obtains a straight line with Slope = 0.1758 Intercept = −0.9365

Hence, the parameters of the Type II extreme-value probability distribution are 1 = 5.688 0.1758 σ = e −0.9365 = 0.392

γ=

and thus the relationship between frequency of exceedance and peak ground acceleration is ⎡ ⎛ a ⎞ −5.688 ⎤ λ a = 1 − exp ⎢− ⎜ ⎥ ⎟ ⎥⎦ ⎢⎣ ⎝ 0.392 ⎠

(b) Since for a return period of 500 years, λa =

1 = 0.002 500

from the frequency-peak ground acceleration expression just developed the peak ground acceleration corresponding to a return period of 500 years is −

a = 0.392[− ln(1 − 0.002)]

1 5.688

= 1.17 g

(c) Similarly, for a probability of exceedance of 1 percent in 50 years one has that P ( X 50 ≥ 1) = 1 − eλ a (50 ) = 0.01

and thus that λa = −

ln(0.99) = 2.01 × 10 − 4 50

Therefore, the peak ground acceleration corresponding to a probability of exceedance of 1 percent in 50 years is equal to −

a = 0.392[− ln(1 − 2.01 × 10 − 4 )]

1 5.688

= 1.75 g

Problem 7.14 A multi-story building is to be built in the city of Anaheim, California, which lies over deposits of alluvium with shear wave velocities of 790, 1000, and 1670 ft/s at depths between 0 and 38 ft, 38 and 80 ft, and 80 and 95 ft, respectively. The historical seismic activity within a 100-mile radius from the building site is summarized in Table P7.14, where the magnitude of all the earthquakes in record and the corresponding epicentral distances from the site to the causative seismic sources are listed. Using these data, the attenuation laws proposed by Boore, Joyner, and Fumal, and the semi-probabilistic approach described in Section 7.7, construct frequency-intensity curves for the expected peak ground accelerations at the site. Then, on the basis of these curves and considering that the life expectancy of the building is 45 years, (a) draw exceedance probability curves for time intervals of 50, 100, and 500 years; (b) obtain the peak ground acceleration at the building site for an exceedance probability of 10%; and (c) determine the peak ground accelerations at the building site and the associated probabilities of exceedance corresponding to return periods of 100, 500, and 1000 years.

42

Table P7.14. Seismicity data for Problem 7.14 Latitude North 33.000 33.700 34.000 32.700 34.100 34.100 33.400 34.300 34.100 34.300 32.800 34.200 34.300 33.800 34.200 33.700 33.700 33.700 33.500 34.900 34.700 33.750 33.750 33.200 34.000 34.000 34.000 34.180 34.180 33.950 33.617 33.750 33.750 33.750 33.700 33.575 33.683 33.700 33.750 33.850 33.750 33.617 33.783 34.100 33.408 33.699

Longitude West 117.300 117.900 119.000 117.200 116.700 117.900 116.300 118.600 119.400 117.600 116.800 117.400 117.500 117.000 117.100 117.400 117.400 117.400 116.500 118.900 119.000 117.000 117.000 116.700 117.250 119.500 118.500 116.920 116.920 118.632 117.967 118.083 118.083 118.083 118.067 117.983 118.050 118.067 118.083 118.267 118.083 118.017 118.133 116.800 116.261 117.511

Date 11/22/1800 12/08/1812 09/24/1827 05/27/1862 02/07/1889 08/28/1889 02/09/1890 04/04/1893 05/19/1893 07/30/1894 10/23/1894 7/22/1899 7/22/1899 12/25/1899 9/2/1907 4/11/1910 5/13/1910 5/15/1910 9/30/1916 10/23/1916 10/23/1916 4/21/1918 6/6/1918 1/1/1920 7/23/1923 2/18/1926 8/4/1927 1/16/1930 1/16/1930 8/31/1930 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/11/1933 3/13/1933 3/14/1933 10/2/1933 10/24/1935 3/25/1937 5/31/1938

Depth (km) 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 10.000 10.000

43

Site Earthquake intensity magnitude MM 6.500 IV 6.900 IX 5.500 III 5.900 III 5.300 III 5.200 VI 6.300 III 5.400 IV 5.500 II 5.900 V 5.700 II 5.500 V 6.500 VI 6.600 V 5.300 IV 5.000 V 5.000 V 6.000 VI 5.000 II 5.200 II 5.200 II 6.600 V 5.000 III 5.000 II 6.000 V 5.000 II 5.000 IV 5.200 III 5.100 III 5.200 IV 6.300 VII 5.000 VI 5.100 VII 5.000 VI 5.100 VI 5.200 VI 5.500 VII 5.100 VI 5.100 VII 5.000 V 5.300 VII 5.100 VI 5.400 VII 5.100 III 6.000 II 5.500 VI

Approximate distance mi (km) 69 (111) 11 (017) 64 (102) 90 (144) 71 (144) 17 (027) 98 (157) 50 (081) 87 (140) 35 (057) 97 (156) 37 (060) 38 (062) 52 (084) 52 (083) 31 (050) 31 (050) 31 (050) 84 (136) 92 (148) 85 (137) 52 (084) 52 (084) 83 (133) 39 (062) 92 (148) 35 (057) 61 (098) 61 (098) 42 (068) 17 (027) 13 (020) 13 (020) 13 (020) 14 (023) 20 (032) 15 (024) 14 (023) 13 (020) 21 (033) 13 (020) 18 (029) 14 (023) 65 (105) 99 (160) 25 (040)

Table P7.14 (continued). Seismicity data for Problem P7.14 Site Latitude Longitude Depth Earthquake intensity North West Date (km) Magnitude MM 34.083 116.300 5/18/1940 0.000 5.400 II 34.067 116.333 5/18/1940 0.000 5.200 II 34.067 116.333 5/18/1940 0.000 5.000 II 34.867 118.933 9/21/1941 0.000 5.200 II 33.783 118.250 11/14/1941 0.000 5.400 VI 34.267 116.967 8/29/1943 0.000 5.500 III 33.976 116.721 6/12/1944 10.000 5.100 III 33.994 116.712 6/12/1944 10.000 5.300 III 33.950 116.850 9/28/1946 0.000 5.000 III 34.017 116.500 7/24/1947 0.000 5.500 III 34.017 116.500 7/25/1947 0.000 5.000 II 34.017 116.500 7/25/1947 0.000 5.200 II 34.017 116.500 7/26/1947 0.000 5.100 II 33.267 119.450 11/18/1947 0.000 5.000 I 33.933 116.383 12/4/1948 0.000 6.500 III 32.817 118.350 12/26/1951 0.000 5.900 III 34.950 118.867 7/21/1952 0.000 5.300 II 35.000 118.833 7/23/1952 0.000 5.400 II 35.000 119.833 7/23/1952 0.000 5.200 II 34.900 118.950 8/1/1952 0.000 5.100 II 34.519 118.198 8/23/1952 13.100 5.000 III 35.150 118.633 1/27/1954 0.000 5.000 I 34.983 118.983 5/23/1954 0.000 5.100 I 34.941 118.987 11/15/1961 10.700 5.000 I 34.932 118.976 3/1/1963 13.900 5.000 I 33.710 116.925 9/23/1963 16.500 5.000 III 34.712 116.503 9/25/1965 10.600 5.200 II 33.343 116.346 4/28/1969 20.000 5.800 II 33.291 119.193 10/24/1969 10.000 5.100 II 34.270 117.540 9/12/1970 8.000 5.400 V 34.411 118.401 2/9/1971 8.400 6.400 V 34.411 118.401 2/9/1971 8.000 5.800 IV 34.411 118.401 2/9/1971 8.000 5.800 IV 34.411 118.401 2/9/1971 8.000 5.300 IV 34.308 118.454 2/9/1971 6.200 5.200 IV 34.065 119.035 2/21/1973 8.000 5.900 IV 33.986 119.475 8/6/1973 16.900 5.000 II 34.516 116.495 6/1/1975 4.500 5.200 II 33.944 118.681 1/1/1979 11.300 5.000 IV 34.327 116.445 3/15/1979 2.500 5.200 II 33.501 116.513 2/25/1980 13.600 5.500 II 33.671 119.111 9/4/1981 0.000 5.400 III No. of records: 88 Maximum magnitude during time interval: 6.9 Maximum intensity during time interval: IX Distance from site to nearest historical earthquake: 11 mi (17 km) Number of years in considered time interval: 186 44

Approximate distance mi (km) 93 (150) 91 (147) 91 (147) 91 (147) 20 (33) 61 (98) 68 (110) 69 (111) 61 (98) 81 (131) 81 (131) 81 (131) 81 (131) 98 (157) 87 (141) 76 (123) 93 (150) 95 (153) 95 (153) 93 (150) 49 (078) 98 (158) 99 (159) 97 (156) 96 (155) 57 (092) 99 (160) 96 (155) 84 (135) 35 (057) 48 (077) 48 (077) 48 (077) 48 (077) 44 (071) 66 (107) 90 (145) 92 (149) 45 (072) 90 (144) 84 (135) 70 (113)

Solution: Considering the magnitudes and epicentral distances given in Table P7.14, Joyner-Boore-Fumal attenuation equations, and an average shear wave velocity for the upper 30 m of the site soil equal to vs =

790(38) + 1000( 42) + 1670(15) = 1021.8 ft/s = 311 m/s 95

one obtains the peak ground accelerations listed in ascending order in Table P7.14b below. Then, following the procedure described in Box 7.2 one arrives to the ln aj and sj values given in this same table. Performing thereafter a regression analysis with these ln aj and sj values, one finds that the parameters of the straight line that best fits these values are Slope = 0.6149 Intercept = −4.8849

In consequence, the parameters of the Type II probability distribution that best fits the given seismicity data are 1 = 1.6263 0.6149 σ = e −4.8849 = 0.00756 γ=

and the corresponding equation for this probability distribution is −1.6263 ⎤ ⎡ ⎛ a ⎞ λ a = 1 − exp ⎢− ⎜ ⎥ ⎟ ⎥⎦ ⎢⎣ ⎝ 0.00756 ⎠

where λa denotes the annual probability of exceedance. (a) Using this equation in combination with the equation for a Poisson distribution, the probabilities of exceeding a given peak ground acceleration in 50, 100, and 500 years are given by P ( X 50 ≥ 1) = 1 − e −50λ a P ( X 100 ≥ 1) = 1 − e −100λ a P ( X 500 ≥ 1) = 1 − e −500λ a

which lead to the exceedance probability curves shown in Figure P7.14. 1.0 0.9 t = 500 yr

Probability of exceedance

0.8 0.7 t = 100 yr 0.6 0.5 t = 50 yr 0.4 0.3 0.2 0.1 0.0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Peak ground acceleration / g

Figure P7.14. Exceedance probability curves for 50, 100, and 500 years

45

2

j

Year

144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186

1961 1954 1926 1965 1920 1975 1979 1916 1889 1952 1940 1893 1935 1946 1980 1947 1963 1944 1969 1981 1937 1862 1827 1943 1907 1951 1890 1930 1973 1927 1948 1970 1800 1923 1894 1971 1938 1899 1918 1910 1941 1933 1812

Table 7.14b. Sorted seismicity data ln aj j/187 Mw Dist. (km) aj s j = − ln[− ln( j / 187)] 5.0 156 0.0159 -4.1414 0.7701 1.3421 5.1 159 0.0165 -4.1044 0.7754 1.3689 5.0 148 0.0166 -4.0984 0.7807 1.3963 5.2 160 0.0173 -4.0570 0.7861 1.4243 5.0 133 0.0180 -4.0174 0.7914 1.4529 5.2 149 0.0183 -4.0009 0.7968 1.4821 5.2 144 0.0188 -3.9739 0.8021 1.5120 5.2 137 0.0196 -3.9322 0.8075 1.5426 5.3 144 0.0198 -3.9221 0.8128 1.5739 5.4 153 0.0200 -3.9120 0.8182 1.6061 5.4 150 0.0203 -3.8971 0.8235 1.6391 5.5 140 0.0225 -3.7942 0.8289 1.6730 5.1 105 0.0228 -3.7810 0.8342 1.7079 5.0 98 0.0228 -3.7810 0.8396 1.7438 5.5 135 0.0232 -3.7636 0.8449 1.7807 5.5 131 0.0237 -3.7423 0.8503 1.8189 5.0 92 0.0240 -3.7297 0.8556 1.8583 1.8991 5.3 111 0.0243 -3.7173 0.8610 5.8 155 0.0244 -3.7132 0.8663 1.9413 5.4 113 0.0253 -3.6770 0.8717 1.9852 6.0 160 0.0264 -3.6344 0.8770 2.0307 5.9 144 0.0272 -3.6045 0.8824 2.0781 5.5 102 0.0288 -3.5474 0.8877 2.1276 5.5 98 0.0297 -3.5166 0.8930 2.1794 5.3 83 0.0304 -3.4933 0.8984 2.2336 5.9 123 0.0308 -3.4802 0.9037 2.2906 6.3 157 0.0314 -3.4609 0.9091 2.3506 5.2 68 0.0377 -3.2781 0.9144 2.4141 5.9 107 0.0343 -3.3726 0.9198 2.4815 5.0 57 0.0347 -3.3610 0.9251 2.5534 6.5 141 0.0380 -3.2702 0.9305 2.6303 5.4 57 0.0429 -3.1489 0.9358 2.7132 6.5 111 0.0457 -3.0857 0.9412 2.8031 6.0 62 0.0551 -2.8986 0.9465 2.9012 0.0558 -2.8860 0.9519 3.0093 5.9 57 6.4 77 0.0576 -2.8542 0.9572 3.1299 5.5 40 0.0593 -2.8251 0.9626 3.2662 6.6 84 0.0598 -2.8167 0.9679 3.4231 6.6 84 0.0598 -2.8167 0.9733 3.6082 6.0 50 0.0651 -2.7318 0.9786 3.8340 5.4 33 0.0651 -2.7318 0.9840 4.1244 6.3 27 0.1217 -2.1062 0.9893 4.5326 6.9 17 0.2338 -1.4533 0.9947 5.2284

(b) For an exceedance probability of 10% in 45 years, one has that P ( X 45 ≥ 1) = 1 − e −45λ a = 0.1

from which one obtains

46

λa = −

ln(0.9) = 0.00234 45

Thus, upon substitution of this value in the equation for the annual probability of exceedance derived above and after solving for the peak ground acceleration, one arrives to −

a = 0.00756[− ln(1 − 0.00234)]

1 1.6263

= 0.31g

(c) In terms of return periods, the peak ground acceleration is given by 1

a = 0.00756[ − ln(1 −

1 − 1.6263 )] TR

and the associated probabilities of exceedance by P ( X 45 ≥ 1) = 1 − e −45 / TR

where TR signifies return period. Therefore, for return periods of 100, 500, and 1000 years, the peak ground accelerations expected at the site and the associated probabilities of exceedance are as shown in Table P7.14. Table P7.14c. Probabilities of exceedance for return periods of 100, 500, and 1000 years P(Xt ≥ 1) a TR 100 0.128g 0.362 500 0.345g 0.086 1000 0.529g 0.044

Problem 7.15 The site shown in Figure P7.15 is overlain by deposits of firm soil and located near two seismic sources: Fault A and Fault B. Fault A is a strike-slip fault while Fault B is a reverse one. The location of these two faults with respect to the site is as shown in the figure, where the given coordinates are in kilometers. The moment magnitudes of the largest earthquakes that can be produced at these two faults are 8.0 for Fault A and 9.0 for Fault B. Their seismicity is described, respectively, by the following Gutenberg-Richter laws: log λ m = 4.0 − 0.7 M w log λ m = 3.0 − 0.75M w

Using the probabilistic approach described in Section 7.8, Campbell’s attenuation equations, and a lower threshold magnitude of 5.0 for both faults, (a) determine the mean annual probabilities that the peak ground acceleration at the site exceed 0.10g and 0.40g; (b) construct an approximate frequency-peak ground acceleration curve for the site using these two probabilities; and (c) using this frequency-intensity curve, estimate the probability that the peak ground acceleration at the site exceed 0.25g in 50 years. (20,30) B ult Fa

Site

(60,15)

(0,20) (10,10)

Fau lt A

(40,0)

Figure P7.15 Site and nearby faults in Problem 7.15 (coordinates in km) 47

Solution: (a) From the given location of the two faults with respect to the site, the closest distance from the site to the source is 25 km for Fault A and 47.7 km for Fault B. The respective maximum distances are 50.2 km and 60.2 km. The fault lengths are 31.6 km for Fault A and 22.4 km for Fault B. From the given Gutenberg-Richter laws and the lower threshold magnitude of 5.0, it is similarly found that the annual frequency of the lower threshold magnitude is (see Section 7.5.3) λ m0 = exp[2.303(4.0) − 2.303(0.7)(5.0)] = 3.163 events/yr

for Fault A

λ m0 = exp[2.303(3.0) − 2.303(0.75)(5.0)] = 0.178 events/yr

for Fault B

Now, for the given site, SSR and SHR are both equal to zero, and F = 0 for Fault A and F = 1 for Fault B. Therefore, according to Campbell’s attenuation equation, the mean peak ground acceleration is given by ln( AH ) = −3.512 + 0.904m − 1.328ln r 2 + [0.149 exp(0.647 m)]2

for Fault A, and by ln( AH ) = −3.512 + 0.904m − 1.328ln r 2 + [0.149 exp(0.647 m)]2 + (1.125 − 0.112 ln r − 0.0957 m) = −2.387 + 0.808m − 0.112 ln r − 1.328ln r 2 + [0.149 exp(0.647 m)]2

for Fault B. Note that for consistency with the form of Equation 7.38, Mw has been substituted by m in the preceding equations. In like manner, the standard deviation of ln(AH) for both faults is given by σln AH = 0.173 − 0.140 ln( AH ) = 0.173 − 0.140 ln(0.10) = 0.495

when AH = 0.10g, and

σln AH = 0.39

when AH = 0.40g. Probability that peak ground acceleration exceed 0.10 g. For AH = 0.10g and in terms of magnitude and source-to-site distance, the standard variable z given by Equation 7.39 may be expressed as z=

ln AH − E[ln AH ] ln(0.10) + 3.512 − 0.904m + 1.328ln r 2 + [0.149 exp(0.647m)]2 = 0.495 σln AH

= 2.443 − 1.826m + 2.683 ln r 2 + [0.149 exp(0.647 m)]2

for Fault A, and ln AH − E[ln AH ] ln(0.10) + 2.387 − 0.808m + 0.112 ln r + 1.328ln r 2 + [0.149 exp(0.647 m)]2 z= = 0.495 σln AH = 0.171 − 1.632m + 0.226 ln r + 2.683ln r 2 + [0.149 exp(0.647m)]2

for Fault B. According to Equation 7.38, the probability that the peak ground acceleration exceed 0.10g is therefore equal to P[Y > 0.10 g ] =

2(2.303)(0.7) 2π (31.6)

50.2 8.0

−z

∫ ∫ ∫

r = 25 m = 5 s = −∞

for Fault A, and

48

re

−[

s2 +1.612( m − 5 )] 2

r 2 − 252

dsdmdr = 0.041(8.553) = 0.348

P[Y > 0.10 g ] =

60.2

2(2.303)(0.75) 2π ( 22.4)

−z

9.0

∫ ∫ ∫

re

r = 47.7 m = 5 s = −∞

−[

s2 +1.727( m − 5 )] 2

r 2 − 47.7 2

dsdmdr = 0.062(2.126) = 0.131

for Fault B. From Equation 7.33 and the values of λm0 obtained earlier, one finds thus that λ1 y = λ m0 P[Y > 0.10 g ] = (3.163)(0.348) = 1.101 events/yr

for Fault A, and λ 2 y = λ m0 P[Y > 0.10 g ] = (0.178)(0.131) = 0.023 events/yr

for Fault B. Upon substitution of these two values into Equation 7.31, the mean annual occurrence rate for the site is consequently equal to λ y = 1.101 + 0.023 = 1.124 events/yr

Probability that peak ground acceleration exceed 0.40 g. Proceeding similarly for AH = 0.40g, one arrives to z=

ln AH − E[ln AH ] ln(0.40) + 3.512 − 0.904m + 1.328ln r 2 + [0.149 exp(0.647m)]2 = 0.39 σln AH

= 6.656 − 2.308m + 3.405 ln r 2 + [0.149 exp(0.647m)]2 P[Y > 0.40 g ] =

2(2.303)(0.7) 2π (31.6)

50.2 8.0

−z

∫ ∫ ∫

re

r = 25 m = 5 s = −∞

−[

s2 +1.612( m − 5 )] 2

r 2 − 252

dsdmdr = 0.041(0.112) = 0.005

λ1 y = λ m0 P[Y > 0.10 g ] = (3.163)(0.005) = 0.016 events/yr

for Fault A, ln AH − E[ln AH ] ln(0.40) + 2.387 − 0.808m + 0.112 ln r + 1.328ln r 2 + [0.149 exp(0.647 m)]2 z= = σln AH 0.39 = 3.771 − 2.072m + 0.287 ln r + 3.405ln r 2 + [0.149 exp(0.647 m)]2 P[Y > 0.40 g ] =

2(2.303)(0.75) 2π ( 22.4)

60.2

9.0

−z

∫ ∫ ∫

r = 47.7 m = 5 s = −∞

re

−[

s2 +1.727( m − 5 )] 2 2

r − 47.7

2

dsdmdr = 0.062(0.019) = 1.178 × 10− 3

λ 2 y = λ m0 P[Y > 0.40 g ] = (0.178)(1.178 × 10−3 ) = 2.097 × 10 −4 events/yr

for Fault B, and λ y = 0.016 + 2.097 × 10−4 = 0.016 events/yr

for the site. (b) With the determined mean recurrence rates for the peak ground accelerations of 0.10g and 0.40g, it is now possible to developed a frequency-intensity curve by passing a straight line between the two known points (ln 1.124, 0.10) and (ln 0.016, 0.40). The slope of this straight line is m=

and thus its equation is given by

ln 1.124 − ln 0.016 = −14.174 0.10 − 0.40

ln λ − ln 0.016 = −14.174(a − 0.40)

or by ln λ a = −14.174a + 1.535

49

where λa denotes the mean annual rate of exceedance corresponding to a peak ground acceleration a. (c) For a peak ground acceleration of 0.25g, the developed equation for the mean annual rate of exceedance yields ln λ 0.25 = −14.174(0.25) + 1.535 = −2.009

from which one obtains λ 0.25 = e −2.009 = 0.134

According to Equation 7.17, the probability that the peak ground acceleration at the site exceed 0.25g in 50 years is thus equal to P ( X 50 ≥ 1) = 1 − e −0.134(50 ) = 0.999

50

CHAPTER 8 Problem 8.1 A soil deposit may be considered to be homogeneous, elastic, and with a horizontal free surface and rock interface. It has a depth of 100 feet, a unit weight of 120 pcf, a shear modulus of elasticity of 4800 psi, and a damping ratio of 15 per cent in all modes. During an earthquake, the base of the deposit is subjected to an acceleration that may be approximated by a harmonic function with an amplitude of 0.5g and a frequency of 36 rad/s. Determine (a) the acceleration induced by this ground motion at the level of the deposit’s free surface, and (b) the factor by which the peak acceleration at the base is amplified by the soil deposit. Consider that the deposit’s response is significant in its first three modes only. Solution: According to Equations 8.52, 8.53, and 8.50, the natural frequencies, mode shapes, and participation factors in the first three modes of the soil deposit are ω1 =

(2 − 1)π G π (4800 × 12 2 )(32.2) = = 6.76 rad/s 2H ρ 2(100) 120

ω2 =

(4 − 1)π G = 3(6.76) = 20.29 rad/s 2H ρ

ω3 =

(6 − 1) π G = 5(6.76) = 33.82 rad/s 2H ρ

π 2 −1 x π ) = sin( x) H 2 200 4 −1 x 3π π ) = sin( x) U 2 ( x) = sin( 2 200 H U1 ( x) = sin(

6 −1 x 5π π ) = sin( x) 2 200 H 4 4 α2 = = (4 − 1)π 3π

U 3 ( x) = sin( α1 =

4 4 = ( 2 − 1)π π

α3 =

4 4 = (6 − 1)π 5π

Also, since for the case under consideration the acceleration at the base of the soil deposit is given by u&&g (t ) = 0.5 g sin(36t )

and since for a harmonic displacement the normal coordinate ηr is given by (see Equation 8.47) ηr (t ) = −α r

u&&g (t ) sin(Ωt − θr ) 1 ωr2 [1 − (Ω / ωr ) 2 ]2 + [2ξ r Ω / ωr ]2

where Ω is the excitation frequency and θr is a phase angle equal to θr = tan −1[

2ξ r Ω / ωr ] 1 − (Ω / ωr ) 2

the modal coordinates for the first three modes of the system are η1 (t ) = −

4 1 0.5 g sin(36t − θ1 ) = −5.083 × 10− 4 g sin(36t + 0.06) 2 π 6.76 [1 − (36 / 6.76) 2 ]2 + [2(0.15)36 / 6.76]2

51

η2 (t ) = −

4 1 0.5 g sin(36t − θ2 ) = −2.329 × 10− 4 g sin(36t + 0.24) 2 3π 20.29 [1 − (36 / 20.29) 2 ]2 + [2(0.15)36 / 20.29]2

η3 (t ) = −

4 1 0.5 g sin(36t − θ3 ) = −3.218 × 10 − 4 g sin(36t + 1.18) 2 3π 33.82 [1 − (36 / 33.82) 2 ]2 + [ 2(0.15)36 / 33.82]2

Thus, according to Equation 8.42 and considering only the first three modes, the displacement response of the soil deposit at the level of the free surface (H = 100 ft) is given by u (100, t ) = −5.083 × 10−4 g sin(36t + 0.06) + 2.329 × 10−4 g sin(36t + 0.24) − 3.218 × 10−4 g sin(36t + 1.18)

where it has been considered that U1(100) = 1.0, U2(100) = -1.0, and U3(100) = 1.0. Correspondingly, the acceleration response at the same level is u&&(100, t ) = (36) 2 5.083 × 10−4 g sin(36t + 0.06) − (36) 2 2.329 × 10 −4 g sin(36t + 0.24) + (36) 2 3.218 × 10 −4 g sin(36t + 1.18)

= 0.66 g sin(36t + 0.06) − 0.30 sin(36t + 0.24) + 0.42 g sin(36t + 1.18) = 0.66 g[1.0 sin 36t + 0.06 cos 36t ] − 0.30 g[0.97 sin 36t + 0.24 cos 36t ] + 0.42 g[0.38 sin 36t + 0.92 cos 36t ] = 0.53g sin 36t + 0.36 g cos 36t = 0.532 + 0.362 g cos(36t − 0.60) = 0.64 g cos(36t − 0.60)

Thus, that the peak absolute acceleration at the free surface of the soil deposit is grater than the peak acceleration at the base by the factor A.F . =

max[0.53g sin 36t + 0.36 g cos 36t + 0.5 sin 36t ] 1.032 + 0.362 g = = 2.18 max[0.5 g sin 36t ] 0.5 g

Problem 8.2 The response spectrum for the ground motion recorded at a rock site is shown in Figure P8.2a. Estimate the maximum acceleration that would be observed at the surface of the soil deposit shown in Figure P8.2b and the factor by which the soil deposit would amplify the ground motion recorded on rock. Assume a linear behavior and a damping ratio of 5%. Error! No topic specified.

Figure P8.2a. Response spectrum considered in Problem 8.2 Ground surface Soil G = 18.375 MPa

ρ = 1.23 Mg/m

3

11.0 m

Rigid bedrock

Figure P8.2b. Soil deposit considered in Problem 8.2

Solution: According to Equations 8.52, 8.53, and 8.50, the natural frequency, mode shape, and participation factor in the first mode of the soil deposit are ω1 =

(2 − 1)π G π 18,375 = = 17.4 rad/s 2H ρ 2(11) 1.23

52

π 2 −1 x π ) = sin( x) 2 22 H 4 4 α1 = = (2 − 1)π π

U1 ( x) = sin(

It may also be seen from the response spectrum in Figure P8.2a that for a natural frequency of 17.4/2π = 2.8 Hz and a damping ratio of 5 percent, there corresponds a spectral acceleration of SA1 = 2.2 g

Therefore, upon the application of Equation 8.42 with only one mode considered, an estimate of the peak absolute acceleration at the surface of the soil deposit is (see Equation 10.56) max[u&&(11, t ) + u&&g (t )] = α1U1 (11) max[u&&1 (11, t ) + u&&g (t )] = α1U1 (11) SA1 =

4 (1.0)2.2 g = 2.8 g π

Furthermore, since the peak ground acceleration is given by the spectral accelerations for large natural frequencies, the amplification factor is approximately equal to A.F . =

2.8 g = 2.3 1.2 g

Problem 8.3 Determine the natural frequencies and mode shapes of the layered soil profile shown in Figure P8.3. Use a one-dimensional lumped-mass model and assume each layer exhibits a linear forcedeformation behavior. Depth (m) 0.0 11.0

34.5

Material

Density 3 (Mg/m )

Shear wave velocity (m/s)

Upper clay formation

1.23

153

Clay with sand lenses

1.25

270

Gravel and sand

1.37

316

Clay

1.75

402

Bedrock

1.76

1050

46.0

70.0

Figure P8.3. Soil profile considered in Problem 8.3

Solution: According to Equations 8.54 and 8.56, the masses and stiffnesses of the equivalent shear lumped-mass system are as indicated in Table P8.3a. Table P8.3a. Mass and stiffnesses of equivalent lumped-mass system Node Mass (Mg) Stiffness (kN/m) 4 6.765 2618 3 21.453 3878 2 22.565 11896 1 28.878 11784

Therefore, the mass and stiffness matrices of the lumped system are

53

0 0 0 ⎤ ⎡28.878 ⎢ 0 22.565 0 0 ⎥⎥ [M ] = ⎢ Mg ⎢ 0 0 21.453 0 ⎥ ⎢ ⎥ 0 0 6.765⎦ ⎣ 0

0 0 ⎤ ⎡ 23680 - 11896 ⎢- 11896 15774 - 3878 0 ⎥⎥ [K ] = ⎢ kN/m ⎢ 0 - 3878 6496 - 2618⎥ ⎢ ⎥ 0 - 2618 2618 ⎦ ⎣ 0

which, in turn, after solving the corresponding eigenvalue problem leads to the natural frequencies, mode shapes and participation factors shown in Table P8.3b. Table P8.3b. Dynamic properties of equivalent lumped-mass system Parameter Mode 1 2 3 4 Frequency (rad/s) 8.112 17.754 24.142 35.285 Period (s) 0.77 0.35 0.26 0.18 Participation factor 7.796 -4.018 1.038 1.289 0.044 -0.107 0.051 0.136 0.080 -0.132 0.030 -0.141 Mode shape 0.160 0.036 -0.137 0.029 0.193 0.191 0.272 -0.013

Problem 8.4 Determine the 5% damping acceleration response spectrum for the ground motion at the bottom and surface of the layered soil deposit shown in Figure P8.3, when the bedrock is excited by the N-S component of the ground acceleration recorded in Foster City during the Loma Prieta, California, earthquake of October 17, 1989 (record may be downloaded from one of the World Wide Web sites suggested in Chapter 5). Use a one-dimensional lumped-mass model, assuming that each layer behaves linearly, and that the damping ratio of the system is 10% in all modes. Solution: The 5% damping acceleration response spectrum for the N-S component of the ground acceleration recorded at Foster City during the 1989 Loma Prieta earthquake is shown in Figure P8.4a. The corresponding acceleration response spectrum for the ground motion generated at the surface of the soil deposit shown in Figure P8.3 when the base of the deposit is subjected to the aforementioned ground motion is shown in Figure P8.4b. This response spectrum is obtained from the time-history acceleration response of the top mass of an equivalent lumped-mass model for the soil deposit in question. The masses and stiffnesses considered in this model are listed in Table P8.4 (obtained from the solution to Problem 8.3). The damping matrix considered is obtained by assuming it proportional to the mass and stiffness matrices of the system, with constants of proportionality determined from the consideration of a damping ratio of 10% in the system’s first and fourth modes. In other words, it was considered that the damping matrix of the equivalent lumped-mass model was given by [C ] = 1.3191[ M ] + 0.00461[ K ]

where [C], [M], and [K] respectively represent the damping, mass, and stiffness matrices oaf the system. Table P8.4. Mass and stiffnesses of equivalent lumped-mass system Node Mass (Mg) Stiffness (kN/m) 4 6.765 2618 3 21.453 3878

54

2 1

22.565 28.878

11896 11784

1.4

Spectral acceleration / g

1.2

1.0

0.8

0.6

0.4

0.2

0.0 0

1

10

100

Frequency (Hz)

Figure P8.4a. Acceleration response spectrum for 5%-damping of the N-S component of the ground acceleration recorded at Foster City during the 1989 Loma Prieta earthquake 6.0

Spectral acceleration / g

5.0

4.0

3.0

2.0

1.0

0.0 0

1

10

100

Frequency (Hz)

Figure P8.4b. Acceleration response spectrum for 5%-damping of acceleration at the surface of the soil deposit in Figure P8.3

Problem 8.5 Determine the peak acceleration at the surface of the layered soil deposit shown in Figure P8.5 when the bedrock is excited by a ground motion represented by the elastic response spectrum shown in Figure P8.2a. Use a lumped-mass, one-dimensional model, assuming the soil deposit behaves linearly with 10% damping in all modes.

55

Depth (ft) 0 10

Unit weight 3 (lb/ft )

Shear wave velocity (ft/s)

100

714

120

897

125

1200

125

1300

135

1500

150

8000

160 230

730

1130 Bedrock

Figure P8.5. Soil deposit considered in Problem 8.5

Solution: Considering that mi =1/2(ρiHi + ρiHi+1) and ki = vsi2ρi/Hi, where ρi, Hi, and vsi respectively denote density, depth, and shear wave velocity, the masses and stiffnesses of an equivalent lumped-mass model are as indicated in Table P8.5a. The natural frequencies, mode shapes, and participation factors of such an equivalent lumped-mass system result as indicated in Table P8.5b. Table P8.5a. Mass and stiffness of equivalent lumped-mass system Node Mass (Slugs) Stiffness (lb/ft) 5 15.5 158322 4 295.0 19990 3 415.4 79858 2 1106.4 13121 1 1809.0 23583 Table P8.5b. Dynamic properties of equivalent lumped-mass system Parameter Mode 1 2 3 4 5 Frequency (rad/s) 1.97 4.77 8.31 17.70 103.70 Participation factor 56.059 -22.317 0.775 0.025 0.000 0.009 -0.021 0.002 0.000 0.000 0.021 0.007 -0.015 -0.013 0.000 0.022 0.010 -0.004 0.043 0.000 Mode shape 0.023 0.015 0.048 -0.011 -0.013 0.023 0.015 0.049 -0.011 0.248

Thus, the spectral ordinates corresponding to the first three natural frequencies of the system and a damping ratio of 10 percent in the response spectrum in Figure P18.2 are SA(1.97,0.10) ≈ 1.2 g

SA(4.77,0.10) ≈ 1.8 g

SA(8.31,0.10) ≈ 1.7 g

In consequence, the peak modal accelerations in the first three modes of the system are

56

⎧0.009⎫ ⎧0.622⎫ ⎪0.021⎪ ⎪1.407 ⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ {&y&}1 = 56.059⎨0.022⎬1.2 g = ⎨1.460 ⎬ g ⎪0.023⎪ ⎪1.554 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.023⎪⎭ ⎪⎩1.554 ⎪⎭ ⎧- 0.021⎫ ⎧ 0.863 ⎫ ⎪ 0.007 ⎪ ⎪− 0.288⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ {&y&}2 = −22.317 ⎨ 0.010 ⎬1.8 g = ⎨− 0.387 ⎬ g ⎪ 0.015 ⎪ ⎪− 0.598⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.015 ⎪⎭ ⎪⎩− 0.599⎪⎭ ⎧ 0.002 ⎫ ⎧ 0.000 ⎫ ⎪- 0.015⎪ ⎪− 0.018⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ {&y&}3 = 0.775⎨- 0.004⎬1.7 g = ⎨ 0.056 ⎬ g ⎪ 0.048 ⎪ ⎪− 0.015⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.049 ⎪⎭ ⎪⎩− 0.015⎪⎭

and thus, combining the modal accelerations with the square-root-of-the-sum-of-the-squares rule, the peak acceleration at the surface of the soil deposit is approximately equal to &y&s = 1.5542 + 0.5992 + 0.0152 g = 1.67 g

Problem 8.6 Find and plot the amplification function of the soil deposit considered in Problem 8.2 when the bedrock beneath is (a) perfectly rigid, and (b) elastic with a shear wave velocity of 1500 m/s and density of 1.75 Mg/m3. Assume that both the soil and the bedrock behave as a viscoelastic solid with a damping ratio of 20%. Solution: (a) Rigid bedrock The shear wave velocity of the soil deposit is equal to vs =

G 18.375 × 106 = = 122.2 m/s ρ 1230

Hence, according to Equation 8.72 and the assumption of a viscoelastic solid with a damping ratio of 20%, the amplification function of the soil deposit is given by 1

| H (ω) |=

2

cos (ωH / vs ) + (ξωH / vs ) =

2

=

1 2

cos [(11 / 122.2)ω] + [0.2(11) / 122.2]2 ω2

1 cos (0.09ω) + 3.241 × 10− 4 ω2 2

(b) Elastic bedrock According to Equation 8.65, the complex shear wave velocities of the soil deposit and the elastic rock are equal to

57

vss* =

Gs (1 + 2ξ s i ) = vss 1 + 2ξ s i = 122.2 1 + 0.4i ρs

vsr* =

Gr (1 + 2ξ r i ) = vsr 1 + 2ξ r i = 1500 1 + 0.4i ρr

Therefore, the complex impedance ratio for the soil deposit is (see Equation 8.80) α*z =

ρ s vss* 1230(122.2) 1 + 0.4i = = 0.057 ρr vsr* 1750(1500) 1 + 0.4i

and the amplification function is given by (see Equation 8.86) | H (ω) |=

1 cos(ωH

=

/ vss* )

+

iα*z

sin(ωH

/ vss* )

1

= cos(

11ω 11ω ) ) + 0.057i sin( 122.2 1 + 0.4i 122.2 1 + 0.4i

1 cos(0.084ω 1 − 0.4i ) + 0.057i sin(0.084ω 1 − 0.4i )

The plots of the two amplification functions are shown in Figure P8.6 below. 3.5 Rigid bedrock 3

Elastic bedrock

|H(ω)|

2.5

2

1.5

1

0.5

0 0

10

20

30

40

50

60

70

80

90

100

ω (rad/s)

Figure P8.6. Amplification functions for soil deposit in Figure P8.2b considering rigid and elastic bedrock

Problem 8.7 Find and plot the amplification function of the layered soil deposit shown in Figure P8.7. Assume that each layer and the rock all behave as a viscoelastic solid with the properties indicated in the same figure. Ground surface Layer 1 15 ft

28 ft

Layer 2

γ = 120 pcf vs = 900 ft/s ξ = 18% γ = 125 pcf vs = 1400 ft/s ξ = 20% γ = 115 pcf

52 ft

Layer 3

vs = 400 ft/s

ξ = 10% Elastic bedrock

γ = 150 pcf vs = 10,000 ft/s ξ = 15%

Figure P8.7. Soil profile considered in Problem 8.7

58

Solution: According to Equation 8.65, the complex shear wave velocities for the soil deposit’s three layers and the underlying bedrock are equal to vs*1 = vs1 (1 + ξ1i ) = 900(1 + 0.18i ) vs*2 = vs 2 (1 + ξ 2i ) = 1400(1 + 0.20i ) vs*3 = vs 3 (1 + ξ3i ) = 400(1 + 0.10i ) vs*4 = vs 4 (1 + ξ 4i ) = 10,000(1 + 0.15i )

Therefore, the complex impedance ratios between the layers are (see Equation 8.99) α1* =

ρ1vs*1 120(900)(1 + 0.18i ) = = 0.617(0.996 − 1.923 × 10 − 2 i ) = 0.614 − 1.186 × 10 − 2 i * 125 ( 1400 )( 1 + 0 . 20 i ) ρ 2 vs 2

α*2 =

ρ2vs*2 125(1400)(1 + 0.20i ) = = 3.804(1.010 + 9.900 × 10 − 2 i ) = 3.841 + 0.377 × 10− 2 i ρ3vs*3 115(400)(1 + 0.10i )

α*3 =

ρ3vs*3 115(400)(1 + 0.10i ) = = 0.031(0.993 − 4.890 × 10 − 2 i ) = 0.031 − 0.152 × 10− 2 i * 150 ( 10 , 000 )( 1 + 0 . 15 i ) ρ 4 vs 4

which together with Equations 8.100 and 8.101 lead to the following amplitudes of the incident and reflected waves at the surface of the second layer, third layer, and the bedrock in terms of the amplitude of the incident wave at the surface of the first layer: A2 =

* * A1 B (1 + α1* )eik1 H1 + 1 (1 − α1* )e − ik1 H 1 2 2

15

15

i (1− 0.18i ) ω −i (1− 0.18i ) ω 2 2 A B = 1 (1 + 0.614 − 0.01186i )e 900(1+ 0.18 ) + 1 (1 − 0.614 + 0.01186i )e 900(1+ 0.18 ) 2 2 ( 0.0029 + 0.0161i ) ω = A1[(0.807 − 0.0059i )e + (0.193 + 0.0059i )e − ( 0.0029 + 0.0161i ) ω ] * * A B B2 = 1 (1 − α1* )eik1 H 1 + 1 (1 + α1* )e − ik1 H 1 2 2 = A1[(0.193 + 0.0059i )e( 0.0029 + 0.0161i ) ω + (0.807 − 0.0059i )e − ( 0.0029 + 0.0161) ω ] * * A B A3 = 2 (1 + α*2 )eik 2 H 2 + 2 (1 − α*2 )e − ik 2 H 2 2 2 28

28

i (1− 0.2 i ) ω −i (1− 0.2 i ) ω 2 2 A B = 2 (1 + 3.841 + 0.00377i )e 1400(1+ 0.2 ) + 2 (1 − 3.841 − 0.00377i )e 1400 (1+ 0.2 ) 2 2 ( 0.0038 + 0.0192 i ) ω = A2 (2.42 + 0.0019i )e − B2 (1.42 + 0.0019i )e − ( 0.0038 + 0.0192i ) ω * * A B B3 = 2 (1 − α*2 )eik 2 H 2 + 2 (1 + α*2 )e − ik 2 H 2 2 2 = − A2 (1.42 + 0.0019i )e( 0.0038 + 0.0192i ) ω + B2 (2.42 + 0.0019i )e − ( 0.0038 + 0.0192i ) ω * * A B A4 = 3 (1 + α*3 )eik 3 H 3 + 3 (1 − α*3 )e − ik 3 H 3 2 2 52

52

i (1− 0.1i ) ω −i (1− 0.1i ) ω 2 2 A B = 3 (1 + 0.031 − 0.00152i )e 400(1+ 0.1 ) + 3 (1 − 0.031 + 0.00152i )e 400 (1+ 0.1 ) 2 2 ( 0.0129 + 0.1287 i ) ω = A3 (0.5155 − 0.0008i )e + B3 (0.4845 + 0.0008i )e − ( 0.0129 + 0.1287i ) ω

where it has been considered that B1 = A1 and ki* = ω/vsi*. In view of Equation 8.102 and the foregoing expressions, one has that

59

a4 (ω) =

B A4 A3 = (0.515 − 0.0008i )e( 0.0129 + 0.1287i ) ω + 3 (0.484 + 0.0008i )e − ( 0.0129 + 0.1287 i ) ω A1 A1 A1

and, thus, according to Equation 8.109, the transfer function between the displacement at the surface of the soil deposit and the bedrock surface when no soil is present is given by H14' (ω) =

1 A3 B (0.515 − 0.0008i )e ( 0.0129 + 0.1287i ) ω + 3 (0.484 + 0.0008i )e − ( 0.0129 + 0.1287i ) ω A1 A1

A plot of the modulus of this transfer function (amplification function) is shown in Figure P8.7b. 4 3.5 3

|H(ω)|

2.5 2 1.5 1 0.5 0 0

5

10

15

20

25

30

35

40

45

50

ω (rad/s)

Figure P8.7b. Amplification function for soil deposit in Figure P8.7

Problem 8.8 A site is underlain by 550 ft of soil, which in turn is underlain by rigid rock. The average shear wave velocity of the soil is 1500 ft/s2 and its average unit weight is 125 lb/ft3. Determine using the frequency-domain approach described in Section 8.4.4 the acceleration time history at the surface of this site when the rock underneath is excited horizontally by the N-S component of the ground acceleration recorded in the Corralitos station during the Loma Prieta, California, earthquake of October 17, 1989. Assume the soil behaves as a viscoelastic solid with a damping ratio of 10%. Solution: The acceleration time history at the base of the soil deposit and the amplitude Fourier spectrum A(ω) of this time history are shown in Figure P8.8a and P8.8b. According to Equation 8.70 and the assumption of a viscoelastic solid with a damping ratio of 10 per cent, the transfer function of the given soil deposit is equal to H (ω) =

1 1 1 = = cos[ω(1 − ξ i ) H / vs ] cos[ω(1 − 0.1i )(550 / 1500)] cos[(0.3667 − 0.0367i )ω]

where it has been considered that the transfer function for accelerations is the same as for displacements. The modulus of this transfer function is shown in Figure P8.8c. The frequency domain representation of the acceleration at the surface of the soil deposit, Y(ω), is thus given by Y (ω) = A(ω) H (ω) =

A(ω) cos[(0.3667 − 0.0367i )ω]

60

The modulus of this function is depicted in Figure P8.8d. By taking the inverse Fourier transform of Y(ω) one obtains thus the acceleration time history at the surface of the soil deposit. This time history is displayed in Figure P8.8e. It should be noted that in the calculation of the acceleration time history shown in Figure P8.8e it is considered that for negative frequencies Y(ω) is given by Y (−ω) = A(−ω) H * (ω)

where H*(ω) denotes the complex conjugate of the transfer function H(ω). There are two reasons for this consideration. The first is that in accordance with Equation 6.72 Y(-ω) should be the complex conjugate of Y(ω) to obtain a real-valued acceleration time history at the soil deposit surface. The second is that in this case H(-ω) is not equal to the complex conjugate of H(ω) and it is thus necessary to consider the complex conjugate of H(ω) explicitly for negative frequencies. 0.8

0.6

Acceleration /g

0.4

0.2

0 0

2

4

6

8

10

12

14

16

18

20

-0.2

-0.4

-0.6

Time (s)

Figure P8.8a. N-S component of ground motion recorded at Corralitos station during the 1989 Loma Prieta, California, earthquake 90 80 70

|A (ω)|

60 50 40 30 20 10 0 0

10

20

30

40

50

ω (rad/s)

Figure P8.8b. Amplitude Fourier spectrum of ground motion in Figure P8.8a

61

7

6

|H(ω)|

5

4

3

2

1

0 0

10

20

30

50

40

ω (rad/s)

Figure P8.8c. Modulus of transfer function of soil deposit in Problem 8.8 100 90 80

|A (ω)H (ω)|

70 60 50 40 30 20 10 0 0

5

10

15

20

25

30

35

40

45

50

ω (rad/s)

Figure P8.8d. Modulus of Fourier transform of ground motion at surface of soil deposit in Problem 8.8 0.8

0.6

Acceleration / g

0.4

0.2

0.0 0

2

4

6

8

10

12

14

16

18

20

-0.2

-0.4

-0.6

-0.8

Time (s)

Figure P8.8e. Time history of accelerations generated at the surface of the soil deposit in Problem 8.8

62

Problem 8.9 Estimate and plot the variation with depth of the maximum value of the shear modulus for the soil profile shown in Figure P8.3. Solution: As the shear wave velocities and densities are given and the measured wave velocities correspond to low strains (i.e., maximum shear moduli), a shear modulus can be determined directly from G = vs2ρ

where G denotes shear modulus, vs shear wave velocity, and ρ density. Accordingly, the desired shear moduli and the variation of shear modulus with depth are as indicated in Figure P8.9. Depth (m) 0.0 11.0

34.5

Material

Upper clay formation

Density 3 (Mg/m )

Shear wave velocity (m/s)

Shear modulus 2 (KN/m )

1.23

153

28,793

Clay with sand lenses

1.25

270

91,125

Gravel and sand

1.37

316

136,803

Clay

1.75

402

282,807

1.76

1050

G

46.0

70.0 Bedrock

H

Figure P8.9. Variation of shear modulus with depth

Problem 8.10 The ground water in a 50-ft-deep deposit of sand is located 10 ft below the ground surface. The unit weight of the sand above the groundwater table is 100 lb/ft3. Below the groundwater table, the saturated unit weight is 120 lb /ft3. Assuming the sand’s void ratio and effective angle of internal friction are 0.6 and 36°, respectively, estimate and plot the variation with depth of the maximum shear modulus for this sand deposit. Solution: Above the ground table, the effective vertical stress is σv = 100h (lb/ft 2 )

where h is the soil depth in feet measured from the surface. Thus, according to Equation 8.116, the effective confining pressure above the water table is equal to σ0 =

σv 100h (3 − 2 sin φ) = (3 − 2 sin 360 ) = 60.81h lb/ft 2 = 0.42h lb/in 2 3 3

Likewise, according to Equation 8.112, the maximum shear modulus is Gmax =

1230(2.973 − e) 2 1 / 2 1230(2.973 − 0.6) 2 σ0 = (0.42h)1 / 2 = 2805h1 / 2 lb/in 2 1+ e 1 + 0.6

Below the water table, the submerged unit weight is γ b = γ sat − γ w = 120 − 62.4 = 57.6 lb/ft 3

and the effective vertical stress is σv = 100(10) + 57.6(h − 10) (lb/ft 2 ) = 424 + 57.6h (lb/ft 2 )

63

Hence, the mean effective confining pressure is equal to σ0 =

424 + 57.6h σv (3 − 2 sin φ) = (3 − 2 sin 360 ) = 257.8 + 35.0h (lb/ft 2 ) = 1.79 + 0.24h (lb/in 2 ) 3 3

and the maximum shear modulus equal to Gmax =

1230(2.973 − e) 2 1 / 2 1230(2.973 − 0.6) 2 σ0 = (1.79 + 0.24h)1 / 2 = 4329(1.79 + 0.24h)1 / 2 (lb/in 2 ) 1+ e 1 + 0 .6

The variation of the shear modulus with depth is shown in Figure P8.10. 2

G max (lb/in ) 0

2000

4000

6000

8000

10000

12000

14000

16000

18000

0 5 10

Depth (ft)

15 20 25 30 35 40 45 50

Figure P8.10. Variation of shear modulus with depth

Problem 8.11 A 24-m-thick layer of sand is underlain by bedrock. The groundwater table is located at a depth of 4 m below the ground surface. Estimate the maximum shear modulus and damping ratio of the sand at a depth of 12 m below the ground surface considering a shear strain level of 0.05 %. The sand has a void ratio of 0.6, specific gravity of solids of 2.68, and angle of internal friction of 36o. Solution: Above the water table, the unit weight of the sand is equal to γ=

2.68 Gs γw = 9,810 = 16,432 N/m3 1+ e 1 + 0.6

Below the water table, the submerged unit weight is γ b = γ sat − γ w =

Gs − 1 2.68 − 1 γw = 9,810 = 10,300 N/m 3 1+ e 1 + 0.6

where γsat and γw respectively denote saturated unit weight and unit weight of water. The effective vertical stress at depth of 12 meters is therefore equal to σv = 16,432(4) + 10,300(8) = 148,128 N/m3

and, according to Equation 8.116, the mean effective confining pressure equal to σ0 =

σv 148,128 (3 − 2 sin φ) = (3 − 2 sin 360 ) = 90,083 N/m 2 3 3

Furthermore, from Equation 8.112, the maximum shear modulus is

64

Gmax

1230(2.973 − e) 2 1 / 2 1230(2.973 − 0.6) 2 = σ0 = (90,083)1 / 2 = 1,299 kN/m 2 1+ e 1 + 0 .6

while from the average curve in Figure 8.32, the G/Gmax ratio corresponding to a strain level of 0.05% is G / Gmax = 0.52

In consequence, the shear modulus at depth of 12 m is equal to G = 0.52(1,299) = 675 kN/m 2

Reading directly form the average curve in Figure 8.33, one similarly finds that the damping ratio for a strain of 0.05 % is approximately equal to ξ = 9.8 %

Problem 8.12 A clay deposit extends to a depth of 60 ft below the ground surface. The clay has a void ratio of 1.0, specific gravity of solids of 2.78, plasticity index of 50 %, and overconsolidation ratio of 2.0. The groundwater table coincides with the ground surface. Evaluate the deposit’s shear modulus and damping ratio at a depth of 30 ft at a strain level of 0.1 %. Solution: The clay’s saturated unit weight is equal to γ b = γ sat − γ w =

Gs − 1 (2.78 − 1)(62.4) γw = = 55.54 lb/ft 3 1+ e 1 + 1 .0

Hence, according to Equations 8.124 and 8.123 for a plasticity index of 50% and an overconsolidation ratio of 2.0, the earth pressure coefficent at rest is given by K 0( overconsolidated ) = K 0 OCR = [0.4 + 0.007(PI)] OCR = [0.4 + 0.007(50)] 2.0 = 1.06

Also, from Table 8.2 for a plasticity index of 50 %,

K = 0.36

Consequently, the effective vertical stress at a depth of 30 feet is equal to σv = 55.54(30) = 1,666 lb/ft 2

and according to Equation 8.122 the mean effective principal stress is equal to 1 1 σ0 = (1 + 2 K 0 ) σv = [1 + 2(1.06)](1,666) = 1,732 lb/ft 2 = 12.03 lb/in 2 3 3

Furthermore, the modified version of Hardin and Black’s formula for clays yields, Gmax = 1230

1.97 1.97 (OCR ) K σ01 / 2 = 1230 (2.0)0.36 (12.03)1 / 2 = 10,786 lb/in 2 2 2 0.3 + 0.7e 0.3 + 0.7(1.0)

Thus, since from Figure 8.36 for a strain of 0.1 % one has that G / Gmax = 0.19

the shear modulus at a depth of 12 feet is equal to G = 0.19(10,786) = 2049 lb/in 2

Similarly, the damping ratio for a strain of 0.1 %, obtained directly from the average curve in Figure 8.37, is equal to ξ = 10 %

65

Problem 8.13 Evaluate the moduli and damping ratios for the layers of the soil profile shown in Figure P8.13 for shear strains of 10-5 and 10-2. For each layer, consider that the confining pressure at midheight is representative of the confining pressure over the entire layer. 2m

Void ratio = 0.6 Specific gravity of solids = 2.68 Angle of internal friction = 35°

Sand

G.W.T.

Sand Void ratio = 0.7 Specific gravity of solids = 2.65

8m

Angle of internal friction = 30°

Normally 3 m consolidated clay

Specific gravity of solids = 2.73 Water content = 50 % Plasticity index = 32 %

Bedrock

Figure P8.13. Soil profile considered in Problem 8.13

Solution: Sand layer above water table The unit weight for the sand layer above the water table is equal to γ=

2.68 Gs γw = 9,810 = 16,432 N/m3 1+ e 1 + 0.6

and the effective vertical stress at a depth of 1 meter is σv = 16,432(1) = 116,432 N/m 3

Hence, the mean effective principal stress is equal to (see Equation 8.116), σ0 =

σv 116,432 (3 − 2 sin φ) = (3 − 2 sin 350 ) = 71,910 N/m 2 3 3

and the maximum shear modulus equal to (see Equation 8.112), Gmax =

1230(2.973 − e) 2 1 / 2 1230(2.973 − 0.6) 2 σ0 = (71,910)1 / 2 = 1,161 kN/m 2 1+ e 1 + 0.6

Also, since from the average curve in Figure 8.32, the G/Gmax ratios are G / Gmax = 1.0 for 10-5 strain G / Gmax = 0.75 for 10-2 strain

the shear modulus for the sand layer above the water table is equal to G = 1,161 kN/m 2 for 10-5 strain

and G = 0.75(1,161) = 871 kN/m 2 for 10-2 strain

Similarly, from the average curve in Figure 8.33, the damping ratio is equal to ξ = 0.6 % for 10-5 strain

and ξ = 5.8 % for 10-2 strain

66

Sand layer below water table The submerged unit weight of the sand layer below the water table is given by γ b = γ sat − γ w =

Gs − 1 2.65 − 1 γw = 9,810 = 9,521 N/m3 1+ e 1 + 0.7

and the effective vertical stress at depth of 6 meters is equal to σv = 16,432(2) + 9,521(4) = 70,948 N/m3

Hence, the mean effective principal stress is (see Equation 8.116), σ0 =

σv 70,948 (3 − 2 sin 300 ) = 47,299 N/m 2 (3 − 2 sin φ) = 3 3

and the maximum shear modulus is equal to (see Equation 8.112), Gmax =

1230(2.973 − e) 2 1 / 2 1230(2.973 − 0.7) 2 ( 47,299)1 / 2 = 813 kN/m 2 σ0 = 1+ e 1 + 0.7

As a result, since from the average curve in Figure 8.32 the G/Gmax ratios are equal to G / Gmax = 1.0 for 10-5 strain G / Gmax = 0.75 for 10-2 strain

the shear modulus for the sand below the water table is equal to G = 813 kN/m 2 for 10-5 strain

and G = 0.75(813) = 610 kN/m 2 for 10-2 strain

In like manner, from the average curve in Figure 8.33, the damping ratio is equal to ξ = 0.6 % for 10-5 strain

and ξ = 5.8 % for 10-2 strain

Clay layer The void ratio for the clay layer is e = wGs = 0.5(2.73) = 1.37

and its saturated unit weight is equal to γ b = γ sat − γ w =

Gs − 1 2.73 − 1 γw = 9,810 = 7161 N/m3 1+ e 1 + 1.37

Also, the earth pressure coefficient at rest for a plasticity index of 32 % is equal to (see Equation 8.123) K 0 = 0.4 + 0.007(PI) = 0.4 + 0.007(32) = 0.62

and the effective vertical stress at a depth of 11.5 feet is equal to σv = 16,432(2) + 9,521(8) + 7,161(1.5) = 119,773 N/m 2

Hence, the mean effective principal stress is (see Equation 8.122), 1 1 σ0 = (1 + 2 K 0 ) σv = [1 + 2(0.62)]119,773 = 89,430 N/m 2 3 3

and, from the modified version of Equation 8.121,the maximum shear modulus is equal to Gmax = 1230

1.97 1.97 (1.0) K (89,430)1 / 2 = 449 kN/m 2 (OCR ) K σ10 / 2 = 1230 2 2 0.3 + 0.7(1.37) 0.3 + 0.7e

In consequence, since from the average curve in Figure 8.36 the G/Gmax ratios are equal to G / Gmax = 1.0 for 10-5 strain

67

G / Gmax = 0.45 for 10-2 strain

the shear modulus is equal to G = 449 kN/m 2 for 10-5 strain

and G = 0.45( 449) = 202 kN/m 2 for 10-2 strain

Similarly, from the average curve in Figure 8.37, the damping ratio is equal to ξ = 2.5 % for 10-5 strain

and ξ = 5.5 % for 10-2 strain

68

CHAPTER 9 Problem 9.1 Construct using the peak ground acceleration and response spectrum shape approach an inelastic design spectrum for a site underlain by soft clays considering a damping ratio of 5 per cent and a ductility factor of 6. The maximum ground acceleration that future earthquakes may generate at the site is estimated to be 0.20g for a probability of exceedance of 10 per cent in 50 years. Solution: Multiplying by 0.20g and dividing by a ductility factor of 6 the ordinates of the response spectrum shape for soil Type 3 given in Figure 9.4, one obtains the ordinates listed in Table P9.1 and the elastic and inelastic design spectra shown in Figure P9.1. Table P9.1. Ordinates in elastic and inelastic response spectra for given periods in Problem 9.1 Period (s) SA/PGA SA/g SA/gμ 0 1 0.20 0.200 0.2 2.5 0.50 0.083 0.9 2.5 0.50 0.083 1 2.25 0.45 0.075 1.25 1.80 0.36 0.060 1.5 1.50 0.30 0.050 1.75 1.29 0.26 0.043 2 1.13 0.23 0.038 2.25 1.00 0.20 0.033 2.5 0.90 0.18 0.030 2.75 0.82 0.16 0.027 3 0.75 0.15 0.025 0.60

0.50

Spectral acceleration / g

Elastic spectrum 0.40

0.30

0.20

Inelastic spectrum

0.10

0.00 0

0.5

1

1.5

2

2.5

3

Period (s)

Figure P9.1. Constructed elastic and inelastic response spectra in Problem 9.1

Problem 9.2 A site is located 50 km from an active normal fault over a deposit of alluvial soils. It is estimated that the fault is capable of generating an earthquake with a magnitude of 8.0. On the basis of this information and the ground acceleration-response spectrum shape approach, construct an inelastic design spectrum for the site considering a damping ratio of 5 per cent and a ductility factor of 2. 69

Solution: Using Campbell’s attenuation equation with MW = 8.0, r =50 km, F = 0 (for normal faults), and SSR = SHR = 0 (for alluvial soils), one has that ln AH = −3.512 + 0.904 M w − 1.328 ln r 2 + [0.149 exp(0.674M w )]2 = −3.512 + 0.904(8.0) − 1.328 ln 502 + [0.149 exp(0.674 × 8.0)]2

= −1.638

and thus the expected peak ground acceleration at the site is AH = e −1.638 = 0.194 g

Multiplying by 0.194g and dividing by a ductility factor of 2 the ordinates of the response spectrum shape for soil Type 3 in Figure 9.4, one obtains the ordinates listed in Table P9.2 and the elastic and inelastic design spectra shown in Figure P9.2. Table P9.2. Ordinates in elastic and inelastic response spectra for given periods in Problem 9.2 Period (s) SA/PGA SA/g SA/gμ 0 1 0.194 0.194 0.2 2.5 0.485 0.243 0.9 2.5 0.485 0.243 1 2.25 0.437 0.218 1.25 1.80 0.349 0.175 1.5 1.50 0.291 0.146 1.75 1.29 0.249 0.125 2 1.13 0.218 0.109 2.25 1.00 0.194 0.097 2.5 0.90 0.175 0.087 2.75 0.82 0.159 0.079 3 0.75 0.146 0.073 0.600

Spectral acceleration / g

0.500 Elastic spectrum 0.400

0.300

0.200

Inelastic spectrum

0.100

0.000 0

0.5

1

1.5

2

2.5

3

Period (s)

Figure P9.2. Constructed elastic and inelastic response spectra in Problem 9.2

Problem 9.3 Construct employing the peak ground acceleration-response spectrum shape approach a design spectrum for elastoplastic structures with 5 per cent damping, a ductility factor of 4, and located at a site that is 100 km from an active fault. It is estimated that the fault is capable of generating an earthquake with a moment magnitude of 7.5. In addition, the site is underlain by soft clays 70

for which the average shear wave velocity over the upper 30 meters is 600 m/s. Use JoynerBoore-Fumal attenuation equation introduced in Chapter 7 to estimate the peak ground acceleration at the site. List the spectral accelerations corresponding to the natural periods of 0.1, 0.5, and 3.0 seconds. Solution: Using Joyner-Boore-Fumal attenuation equation with Mw = 7.5, r = 100 km, and Vs = 600 m/s, one has that V ln PGA = −0.242 + 0.527( M w − 6.0) − 0.778 ln R 2 + 5.57 2 − 0.371ln( S ) 1396 600 = −0.242 + 0.527(7.5 − 6.0) − 0.778 ln 100 2 + 5.57 2 − 0.371ln( ) 1396 = −2.722

and thus the expected peak ground acceleration at the site is PGA = e −2.722 = 0.066 g

Multiplying by 0.0.066 and dividing by a ductility factor of 4 the ordinates of the response spectrum shape for soil Type 3 in Figure 9.4, one obtains the ordinates listed in Table P9.3 and the elastic and inelastic design spectra shown in Figure P9.3. Table P9.3. Ordinates in elastic and inelastic response spectra for given periods in Problem 9.3 Period (s) SA/PGA SA/g SA/gμ 0 1 0.066 0.066 0.1 1.75 0.116 0.054 0.2 2.5 0.165 0.041 0.9 2.5 0.165 0.041 1 2.25 0.149 0.037 1.25 1.80 0.119 0.030 1.5 1.50 0.099 0.025 1.75 1.29 0.085 0.021 2 1.13 0.074 0.019 2.25 1.00 0.066 0.017 2.5 0.90 0.059 0.015 2.75 0.82 0.054 0.014 3 0.75 0.050 0.012 0.180 0.160

Spectral acceleration / g

0.140 Elastic spectrum 0.120 0.100 0.080 0.060 Inelastic spectrum

0.040 0.020 0.000 0

0.5

1

1.5

2

2.5

3

Period (s)

Figure P9.3. Constructed elastic and inelastic response spectra in Problem 9.3

71

Problem 9.4 The graph shown in Figure P9.4 represents a site-specific average design spectrum for elastoplastic structures with 2% damping and a ductility factor of 5. On the basis of this design spectrum, estimate the lateral strength for which a single-degree-of-freedom structure with a weight of 10 kips, natural period of 2.5 seconds, and damping ratio of 2% should be designed if (a) the structure cannot resist inelastic deformations, and (b) the structure is capable of withstanding inelastic deformations as large as 5 times its yield deformation. SD (in)

PSV (in/s)

SA/g

4.0

0.5

2

8

33

f (Hz) Figure P9.3. Inelastic design spectrum considered in Problem 9.3

Solution: The natural frequency and stiffness of the structure are respectively equal to 1/2.5 = 0.4 Hz and k =(

10 2π 2 2π = 0.163 kip/in ) m = ( )2 T 2.5 386.4

Hence, from the given inelastic design spectrum the ordinate corresponding to a natural frequency of 0.4 Hz is ( SD)in =

PSV 4.0 = = 1.59 in ω 2π(0.4)

and thus the ordinate in the corresponding elastic design spectrum would be ( SD) el = μ( SD)in = 5(1.59) = 7.95 in

In consequence, the lateral strength of the structure should be equal to F = k ( SD)el = 0.163(7.96) = 1.30 kips

if the structure is not capable of resisting inelastic deformations, and F = k ( SD)in = 0.163(1.592) = 0.26 kips

if the structure can resist inelastic deformations up to 5 times its yield deformation. Problem 9.5 The peak ground acceleration at a given rock site is estimated to be 0.6g for a probability of exceedance of 10% in 100 years. Construct an inelastic design spectrum for 2% damping and a ductility factor of 8 using Newmark-Hall approach. Consider the average v/a and ad/v2 ratios proposed by Mohraz to estimate the peak ground velocity and peak ground displacement at the site. Tetralogarithmic graph paper is provided in Figure P9.5. Solution: 72

From Table 9.2 for a rock site, ad / v 2 = 5.3

v / a = 24 in/s/g

Therefore, for a = 0.6g one has that

v = 24(0.6) = 14.4 in/s d=

5.3(14.4) 2 = 4.74 in 0.6(386.4)

Similarly, from Table 9.1 the mean amplification factors for a 2% damping elastic spectrum are A = 2.74

V = 2.03

D = 1.63

Consequently, the peak spectral ordinates using Newmark-Hall approach are ( SA)el = 2.74(0.6 g ) = 1.64 g ( SV ) el = 2.03(14.4) = 29.2 in/s ( SD)el = 1.63(4.74) = 7.7 in

and the peak ordinates in the corresponding inelastic spectrum for a ductility factor of 8 are ( SA)in =

1.64 g 2(8) − 1

= 0.42 g

29.2 = 3.6 in/s = 9.3 cm/s 8 7.7 ( SD)in = = 0.96 in = 2.4 cm 8

( SV )in =

Drawing: (1) lines perpendicular to the acceleration, velocity, and displacement axis with ordinates equal to 0.42g, 9.3 cm/s and 2.4 cm; (2) a line perpendicular to the acceleration axis with an ordinate equal to 0.6g, and (3) a transition line between the frequencies of 8 and 33 Hz, the design spectrum shown in Figure P9.5 is obtained.

Figure P9.5. Constructed inelastic design spectrum in Problem 9.5

73

Problem 9.6 A site in California is underlain by stiff soils and is located about 60 km from the San Andreas Fault. Considering that there are no other nearby faults and that the largest earthquake that can occur at the San Andreas Fault is one with a moment magnitude of 8.0 and a focal depth of 20 km, construct a 5% damping design spectrum for this site and the design of buildings capable of withstanding inelastic deformations as large as 5 times their yield deformations. The average shear wave velocity over the upper 30 meters of the soils beneath the site is 900 m/s. Use JoynerBoore-Fumal attenuation equation and Newmark-Hall approach. Tetralogarithmic graph paper is provided in Figure P9.5. Solution: Using Joyner-Boore-Fumal equation for horizontal peak ground acceleration with Mw = 8.0, R = 60 km, b1 = bss, and Vs = 900 m/s, one has that V ln PHGA = −0.313 + 0.527( M w − 6) − 0.778 ln R 2 + 5.57 2 − 0.371ln( s ) 1396 900 = −0.313 + 0.527(8 − 6) − 0.778 ln 60 2 + 5.57 2 − 0.371ln( ) 1396 = −2.285

from which one finds that the mean peak ground acceleration for the site is PHGA = e −2.285 = 0.102 g

Similarly, using Joyner-Boore attenuation equation for horizontal peak ground velocity with Mw = 8.0, R = 60 km, and S =1, one obtains log PHGV = 2.09 + 0.49( M w − 6) − log R 2 + 4.02 − 0.0026 R 2 + 4.02 + 0.17 S = 2.09 + 0.49(8 − 6) − log 602 + 4.02 − 0.0026 602 + 4.02 + 0.17(1.0) = 1.305

and thus PHGV = 101.305 = 20.2 cm/s

The peak ground displacement may be found using Newmark-Hall relationship for this purpose. The results is d=

6.0v 2 6.0(20.2) 2 = = 24.5 cm a 0.102(981)

Thus, since from Table 9.1 the mean values of the amplification factors proposed by Newmark and Hall for 5% damping are A = 2.12

V = 1.65

D = 1.39

the peak ordinates of the elastic design spectrum are ( SA)el = 2.12(0.102 g ) = 0.216 g ( SV )el = 1.65(20.2) = 33.3 cm/s ( SD)el = 1.39(24.5) = 34.1 cm

and the peak ordinates in the corresponding inelastic spectrum for a ductility factor of 5 are ( SA)in =

0.102 g

( SV )in =

2(5) − 1

= 0.034 g

33.3 = 6.7 cm/s 5

74

( SD)in =

34.1 = 6.8 cm 5

Plotting these ordinates in a tetralogarithmic graph together with the transition lines described in Box 9.3, the design spectrum shown in the Figure P9.6 is obtained.

Figure P9.6. Constructed inelastic design spectrum in Problem 9.6

Problem 9.7 Construct an elastic design spectrum for the site described in Problem 9.6 using alternatively (a) Newmark-Hall approach and the average v/a and ad/v2 ratios proposed by Mohraz, and (b) Joyner-Boore-Fumal attenuation equation presented in Chapter 7. Plot the two spectra in a single spectral acceleration versus natural period graph with arithmetic scales. Solution: From the solution to Problem 9.6, one has that

PHGA = 0.102 g

2

and thus, according to the v/a and ad/v ratios proposed by Mohraz for alluvium, the peak ground velocity and peak ground displacement are equal to (see Table 9.2) v = 48(0.102) = 4.90 in/s = 12.4 cm/s d=

3.9(4.90) 2 = 2.38 in = 6.0 cm 0.102(386.4)

Similarly, from Table 9.1, the mean values of the amplification factors proposed by Newmark and Hall for 5% damping are A = 2.12

V = 1.65

D = 1.39

Therefore, the peak ordinates for Newmark-Hall elastic 5% damping design spectra are 75

( SA)el = 2.12(0.102 g ) = 0.216 g ( SV )el = 1.65(12.4) = 20.5 cm/s ( SD)el = 1.39(6.0) = 8.3 cm

The spectral accelerations in the Newmark-Hall spectrum are thus given by SA = 0.216 g

in the acceleration sensitive range, by SA = ωSV =

2π 128.8 0.131g (20.5) = cm/s2 = T T T

in the velocity sensitive range, and by SA = ω2 SD =

4π 2 327.7 0.334 g (8.3) = cm/s2 = 2 2 T T T2

in the displacement sensitive range. The period interval is 0.125 =1/8 ≤ T ≤ 0.131/0.216 = 0.60 s for the acceleration sensitive range, 0.60 ≤ T ≤ 0.334/0.131 = 2.5 s for the velocity sensitive range, and T ≥ 2.5 s for the displacement sensitive range. The spectral accelerations in the transition range between T = 1/0.33 = 0.03 s (where the spectral acceleration is equal to the peak ground acceleration) and T = 0.125 are given by SA = 0.102 + 1.2(T − 0.03)

Now, according to Joyner-Boore-Fumal attenuation equation, the spectral ordinates for a design spectrum with 5 % damping are given by ln SA = b1SS + b2 (M w - 6 ) + b3(M w - 6 )2 + b5 ln R 2 + h 2 + bV ln (VS /VA ) = b1SS + b2 (8 - 6) + b3 (8 - 6) 2 + b5 ln 60 2 + h 2 + bV ln(900 /VA )

where the coefficients b1, b2, etc. are given in Table 7.1. Substitution of the different coefficients for the different periods leads thus to the spectral accelerations shown Table P9.7. The Newmark-Hall and Joyner-Boore-Fumal design spectra are plotted in Figure P9.7. 0.250

Newmark-Hall

Spectral acceleration/g

0.200 Joyner-Boore-Fumal 0.150

0.100

0.050

0.000 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Period (s)

Figure P9.7. Constructed elastic response spectra in Problem 9.7

76

Table P9.7. Spectral ordinates for given periods obtained in Problem 9.7 Period b1SS b2 b3 b5 bV VA h ln SA SA/g 0 -0.313 0.527 0 -0.778 -0.371 1396 5.57 -2.285 0.102 0.1 1.006 0.753 -0.226 -0.934 -0.212 1112 6.27 -2.176 0.113 0.11 1.072 0.732 -0.23 -0.937 -0.211 1291 6.65 -2.150 0.116 0.12 1.109 0.721 -0.233 -0.939 -0.215 1452 6.91 -2.129 0.119 0.13 1.128 0.711 -0.233 -0.939 -0.221 1596 7.08 -2.106 0.122 0.14 1.135 0.707 -0.23 -0.938 -0.228 1718 7.18 -2.071 0.126 0.15 1.128 0.702 -0.228 -0.937 -0.238 1820 7.23 -2.056 0.128 0.16 1.112 0.702 -0.226 -0.935 -0.248 1910 7.24 -2.036 0.131 0.17 1.09 0.702 -0.221 -0.933 -0.258 1977 7.21 -2.014 0.133 0.18 1.063 0.705 -0.216 -0.93 -0.27 2037 7.16 -1.985 0.137 0.19 1.032 0.709 -0.212 -0.927 -0.281 2080 7.1 -1.965 0.140 0.2 0.999 0.711 -0.207 -0.924 -0.292 2118 7.02 -1.947 0.143 0.22 0.925 0.721 -0.198 -0.918 -0.315 2158 6.83 -1.914 0.147 0.24 0.847 0.732 -0.189 -0.912 -0.338 2178 6.62 -1.886 0.152 0.26 0.764 0.744 -0.18 -0.906 -0.36 2173 6.39 -1.865 0.155 0.28 0.681 0.758 -0.168 -0.899 -0.381 2158 6.17 -1.827 0.161 0.3 0.598 0.769 -0.161 -0.893 -0.401 2133 5.94 -1.823 0.162 0.32 0.518 0.783 -0.152 -0.888 -0.42 2104 5.72 -1.807 0.164 0.34 0.439 0.794 -0.143 -0.882 -0.438 2070 5.5 -1.795 0.166 0.36 0.361 0.806 -0.136 -0.877 -0.456 2032 5.3 -1.794 0.166 0.38 0.286 0.82 -0.127 -0.872 -0.472 1995 5.1 -1.780 0.169 0.4 0.212 0.831 -0.12 -0.867 -0.487 1954 4.91 -1.781 0.168 0.42 0.14 0.84 -0.113 -0.862 -0.502 1919 4.74 -1.784 0.168 0.44 0.073 0.852 -0.108 -0.858 -0.516 1884 4.57 -1.789 0.167 0.46 0.005 0.863 -0.101 -0.854 -0.529 1849 4.41 -1.791 0.167 0.48 -0.058 0.873 -0.097 -0.85 -0.541 1816 4.26 -1.803 0.165 0.5 -0.122 0.884 -0.09 -0.846 -0.553 1782 4.13 -1.802 0.165 0.55 -0.268 0.907 -0.078 -0.837 -0.579 1710 3.82 -1.823 0.162 0.6 -0.401 0.928 -0.069 -0.83 -0.602 1644 3.57 -1.858 0.156 0.65 -0.523 0.946 -0.06 -0.823 -0.622 1592 3.36 -1.887 0.151 0.7 -0.634 0.962 -0.053 -0.818 -0.639 1545 3.2 -1.927 0.146 0.75 -0.737 0.979 -0.046 -0.813 -0.653 1507 3.07 -1.956 0.141 0.8 -0.829 0.992 -0.041 -0.809 -0.666 1476 2.98 -1.993 0.136 0.85 -0.915 1.006 -0.037 -0.805 -0.676 1452 2.92 -2.025 0.132 0.9 -0.993 1.018 -0.035 -0.802 -0.685 1432 2.89 -2.063 0.127 0.95 -1.066 1.027 -0.032 -0.8 -0.692 1416 2.88 -2.103 0.122 1 -1.133 1.036 -0.032 -0.798 -0.698 1406 2.9 -2.146 0.117 1.1 -1.249 1.052 -0.03 -0.795 -0.706 1396 2.99 -2.211 0.110 1.2 -1.345 1.064 -0.032 -0.794 -0.71 1400 3.14 -2.283 0.102 -1.428 1.073 -0.035 -0.793 -0.711 1416 3.36 -2.348 0.096 1.3 1.4 -1.495 1.08 -0.039 -0.794 -0.709 1442 3.62 -2.409 0.090 1.5 -1.552 1.085 -0.044 -0.796 -0.704 1479 3.92 -2.469 0.085 1.6 -1.598 1.087 -0.051 -0.798 -0.697 1524 4.26 -2.530 0.080 1.7 -1.634 1.089 -0.058 -0.801 -0.689 1581 4.62 -2.582 0.076 1.8 -1.663 1.087 -0.067 -0.804 -0.679 1644 5.01 -2.643 0.071 1.9 -1.685 1.087 -0.074 -0.808 -0.667 1714 5.42 -2.689 0.068 2 -1.699 1.085 -0.085 -0.812 -0.655 1795 5.85 -2.745 0.064

77

Problem 9.8 The peak ground acceleration, peak ground velocity, and peak ground displacement expected at a given alluvial site are estimated to be 0.843g, 50.74 in/s, and 12.81 in, respectively. Construct an inelastic design spectrum for this site using Newmark-Hall average amplification factors to construct the elastic spectrum and Miranda’s reduction factors to obtain the inelastic spectrum from the elastic spectrum. Consider a damping ratio of 3% and a ductility factor of 4. Tetralogarithmic graph paper is provided in Figure P9.5. Solution: From Table 9.1, the mean amplification factors for 3% damping are A = 2.46

V = 1.86

D = 1.52

Therefore, the peak ordinates in the elastic Newmark-Hall design spectrum are ( SA)el = 2.46(0.843g ) = 2.07 g ( SV )el = 1.86(50.74) = 94.38 in/s ( SD)el = 1.52(12.81) = 19.47 in

and thus the spectral accelerations are given by

SA = 2.07 g

in the acceleration sensitive range, SA = ωSV =

2π 593.0 1.53g (94.38) = in/s 2 = T T T

in the velocity sensitive range, and SA = ω2 SD =

4π2 768.6 1.99 g (19.47) = in/s 2 = 2 2 T T T2

in the displacement sensitive range. The period intervals that define the acceleration, velocity, and displacement sensitive regions are, respectively, 0.125 = 1/8 ≤ T ≥ 1.53/2.07 = 0.74 s 0.60 ≤ T = 1.99/1.53 = 1.3 s T ≥ 1.3 s

In the transition period between T = 1/33=0.03 s and T = 0.125 s, the spectral accelerations are given by SA = 0.843 + 12.916(T − 0.03)

Lastly, Miranda’s reduction factors are given by R =1+

μ −1 Φ

where μ denotes ductility factor and for alluvial soils and the site under consideration 1 2 1 − exp[−2(ln T − ) 2 ] (12 − μ)T 5T 5 1 2 1 =1+ − exp[−2(ln T − ) 2 ] 8T 5T 5

Φ =1+

In terms of the equations established above for the spectral accelerations and reductions factors, the ordinates in the inelastic design spectrum for a series of natural periods are determined. The obtained inelastic acceleration design spectrum is shown Figure P9.10.

78

2.5

Spectral acceleration/g

2.0

1.5 Elastic 1.0

0.5

Inelastic

0.0 0

1

2

3

4

5

6

7

8

9

10

Period (s)

Figure P9.8. Constructed elastic and inelastic response spectra in Problem 9.8

Problem 9.9 A single-degree-of-freedom structure has a total weight of 10 kips, a natural period of 1.5 s and a damping ratio of 5%. The structure will be built on a site that is characterized by an expected peak ground acceleration of 0.4g and the elastic response spectrum shape shown in Figure 9.4 for rock and stiff soils. Determine the lateral strength for which this structure should be designed assuming that the structure will have the capability of resisting inelastic deformations of up to 5 times its yield strength. Use alternatively the criterion established in Section 9.2 and Miranda’s reduction factors to account for the structure’s ability to withstand inelastic deformations. Solution: From the graph in Figure 9.4 for rock and stiff soils and a period of 1.5 seconds, SA = 0.7 PGA

and thus the corresponding ordinate in the elastic spectrum is SA = 0.7(0.4 g ) = 0.28 g

(a) Using the criterion in Section 9.2 and considering a ductility factor of 5, the structure’s lateral strength result as Fy = ku y = k

( SD)el m( SA)el 10 0.28 g = = = 0.56 kips μ μ g 5

(b) Miranda’s reduction factor is given by Rμ = 1 +

μ −1 4 =1+ Φ Φ

where for rock sites and a natural period of 1.5 s, 3 1 1 3 − exp[− (ln T − ) 2 ] (10 − μ)T 2T 2 5 1 1 3 3 =1+ − exp[− (ln1.5 − ) 2 ] (10 − 5)1.5 2(1.5) 2 5 = 0.82

Φ =1+

79

If the elastic spectral ordinate is instead reduced using Miranda’s reduction factor, the structure’s lateral strength is equal to Fy =

m( SA)el 10 0.28 g = = 0.48 kips Rμ g 1 + 4 / 0.82

80

CHAPTER 10 Problem 10.1 A 5-story shear building has the properties shown in Table P10.1. Determine using the response spectrum method the maximum story shears induced in this building when the base of the building is subjected to a ground motion represented by an acceleration response spectrum constructed utilizing the spectral shape corresponding to Soil Type 3 in Figure 8.17 and a peak ground acceleration of 0.40 g. Consider a damping ratio of 5% in all modes. Recall that the spectral shapes in Figure 8.17 were generated for systems with a damping ratio of 5%. Table P10.1. Properties of building in Problem 10.1 Story Height Stiffness (MN/m) Mass (Mg) (m) 1 4 249.88 179 2 3 237.05 170 3 3 224.57 161 4 3 212.09 152 5 3 199.62 143

Solution: Mass and stiffness matrices 0 0 0 ⎤ ⎡179 0 ⎢ 0 170 0 0 0 ⎥⎥ ⎢ [M ] = ⎢ 0 0 161 0 0 ⎥ Mg ⎢ ⎥ 0 0 152 0 ⎥ ⎢ 0 ⎢⎣ 0 0 0 0 143⎥⎦ 0 0 0 ⎤ ⎡ 486.93 − 237.05 ⎢− 237.05 461.62 − 224.57 0 0 ⎥⎥ ⎢ [K ] = ⎢ 0 0 ⎥ MN/m − 224.57 436.66 − 212.09 ⎥ ⎢ 0 − 212.09 411.71 − 199.62⎥ ⎢ 0 ⎢⎣ 0 0 0 − 199.62 199.62 ⎥⎦

Dynamic properties In terms of these mass and stiffness matrices, the dynamic properties of the building are as indicated in Table P10.1. Table P10.1. Dynamic properties of building in Problem 10.1a Mode 1 2 3 4 Frequency2 (rad2/s2) 124.090 949.250 2336.300 3845.300 Period (s) 0.56 0.20 0.13 0.10 Participation factor 26.335 -8.958 -4.759 2.660 0.013 -0.034 -0.044 0.041 0.026 -0.046 -0.013 -0.035 Mode shape 0.036 -0.025 0.043 -0.013 0.045 0.015 0.026 0.048 0.049 0.046 -0.039 -0.028

81

5 4997.50 0.09 1.226 0.025 -0.042 0.047 -0.037 0.014

Spectral accelerations From the design spectrum for Soil Type 3 in Figure 8.15, the spectral accelerations corresponding to the natural periods of the system are shown in Table P10.1b. Table P10.1b. Spectral accelerations corresponding to building’s natural periods in Problem 10.1 Period (s) SA/PGA SA(g) SA(m/s2) 0.56 2.5 1.00g 9.81 0.20 2.5 1.00g 9.81 0.13 2.3 0.92g 9.02 0.10 2.0 0.80g 7.85 0.09 1.9 0.76g 7.46

Maximum modal displacements By the application of Equation 10.54, the maximum modal displacements result as ⎧0.013⎫ ⎧0.02723⎫ ⎪0.026⎪ ⎪0.05338⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ Γ 26.335 ⎪ ⎪ max{u}1 = 12 {φ}1 SA1 = ⎨0.036⎬9.81 = ⎨0.07597 ⎬ m 124.090 ⎪ ω1 ⎪0.09273⎪ 0.045⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪0.049⎭⎪ ⎩⎪0.10178⎭⎪ ⎧− 0.034⎫ ⎧ 0.00315 ⎫ ⎪− 0.046⎪ ⎪ 0.00421 ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ Γ − 8.958 ⎪ ⎪ max{u}2 = 22 {φ}2 SA2 = ⎨− 0.025⎬9.81 = ⎨ 0.00231 ⎬ m 949.250 ⎪ ω2 ⎪- 0.00137⎪ 0.015 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪ 0.046 ⎭⎪ ⎩⎪- 0.00429⎭⎪ ⎧- 0.044⎫ ⎧ 0.00082 ⎫ ⎪- 0.013⎪ ⎪ 0.00024 ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ Γ - 4.759 ⎪ ⎪ max{u}3 = 32 {φ}3 SA3 = ⎨ 0.043 ⎬9.02 = ⎨- 0.00080⎬ m 2336.300 ⎪ ω3 ⎪- 0.00048⎪ 0.026 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪- 0.039⎭⎪ ⎩⎪ 0.00071 ⎭⎪ ⎧ 0.041 ⎫ ⎧ 0.00022 ⎫ ⎪- 0.035⎪ ⎪− 0.00019⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ Γ 2.660 ⎪ ⎪ max{u}4 = 42 {φ}4 SA4 = ⎨- 0.013⎬7.85 = ⎨− 0.00007⎬ m 3845.300 ⎪ ω4 ⎪ 0.00026 ⎪ 0.048 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪- 0.028⎭⎪ ⎩⎪− 0.00015⎭⎪ ⎧ 0.025 ⎫ ⎧ 0.00004 ⎫ ⎪- 0.042⎪ ⎪- 0.00008⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ Γ 1.226 ⎪ ⎪ max{u}5 = 52 {φ}5 SA5 = ⎨ 0.047 ⎬7.46 = ⎨ 0.00009 ⎬ m 4997.50 ⎪ ω5 ⎪- 0.00007⎪ - 0.037⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪ 0.014 ⎭⎪ ⎩⎪ 0.00003 ⎭⎪

82

Maximum modal lateral forces Considering Equation 10.57 and these modal displacements, the maximum modal lateral forces are 0 0 0 ⎤ ⎧0.02723⎫ ⎧0.604⎫ ⎡ 486.93 − 237.05 ⎥ ⎪0.05338⎪ ⎪1.126 ⎪ ⎢− 237.05 461.62 − 224.57 0 0 ⎪⎪ ⎪⎪ ⎪⎪ ⎥ ⎪⎪ ⎢ max{Fs }1 = [ K ] max{u}1 = ⎢ 0 0 ⎥ ⎨0.07597 ⎬ = ⎨1.518 ⎬ MN − 224.57 436.66 − 212.09 ⎥ ⎢ 0 − 212.09 411.71 − 199.62⎥ ⎪0.09273⎪ ⎪1.749 ⎪ ⎢ 0 ⎪ ⎪ ⎪ ⎪ ⎢⎣ 0 0 0 − 199.62 199.62 ⎥⎦ ⎩⎪0.10178⎭⎪ ⎩⎪1.806 ⎭⎪ ⎡ 486.93 ⎢− 237.05 ⎢ max{Fs }2 = [ K ] max{u}2 = ⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0 ⎡ 486.93 ⎢− 237.05 ⎢ max{Fs }3 = [ K ] max{u}3 = ⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0 ⎡ 486.93 ⎢− 237.05 ⎢ max{Fs }5 = [ K ] max{u}5 = ⎢ 0 ⎢ ⎢ 0 ⎢⎣ 0

0 0 0 ⎤ ⎧ 0.00315 ⎫ ⎧ 0.535 ⎫ − 237.05 461.62 − 224.57 0 0 ⎥⎥ ⎪⎪ 0.00421 ⎪⎪ ⎪⎪ 0.680 ⎪⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎨ 0.00231 ⎬ = ⎨ 0.353 ⎬ MN − 224.57 436.66 − 212.09 ⎥ 0 − 212.09 411.71 − 199.62⎥ ⎪- 0.00137 ⎪ ⎪− 0.198⎪ ⎪ ⎪ ⎪ ⎪ 0 0 − 199.62 199.62 ⎥⎦ ⎪⎩- 0.00429⎪⎭ ⎪⎩− 0.582⎪⎭ 0 0 0 ⎤ ⎧ 0.00082 ⎫ ⎧ 0.342 ⎫ − 237.05 461.62 − 224.57 0 0 ⎥⎥ ⎪⎪ 0.00024 ⎪⎪ ⎪⎪ 0.094 ⎪⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎨- 0.00080⎬ = ⎨− 0.299⎬ MN − 224.57 436.66 − 212.09 ⎥ 0 − 212.09 411.71 − 199.62⎥ ⎪- 0.00048⎪ ⎪− 0.170⎪ ⎪ ⎪ ⎪ ⎪ 0 0 − 199.62 199.62 ⎥⎦ ⎪⎩ 0.00071 ⎪⎭ ⎪⎩ 0.237 ⎪⎭ 0 0 0 ⎤ ⎧ 0.00004 ⎫ ⎧ 0.040 ⎫ − 237.05 461.62 − 224.57 0 0 ⎥⎥ ⎪⎪- 0.00008⎪⎪ ⎪⎪− 0.066⎪⎪ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎨ 0.00009 ⎬ = ⎨ 0.069 ⎬ MN − 224.57 436.66 − 212.09 ⎥ 0 − 212.09 411.71 − 199.62⎥ ⎪- 0.00007 ⎪ ⎪ − 0.051⎪ ⎪ ⎪ ⎪ ⎪ 0 0 − 199.62 199.62 ⎥⎦ ⎪⎩ 0.00003 ⎪⎭ ⎪⎩ 0.019 ⎪⎭

Maximum modal story shears Adding the lateral forces to obtain story shears, one obtains ⎧0.604 + 1.126 + 1.518 + 1.749 + 1.806⎫ ⎧6.804⎫ ⎪ 1.126 + 1.518 + 1.749 + 1.806 ⎪ ⎪6.199⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{V }1 = ⎨ 1.518 + 1.749 + 1.806 ⎬ = ⎨5.073⎬MN ⎪ ⎪ ⎪3.555⎪ 1.749 + 1.806 ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭ ⎪⎩1.806 ⎪⎭ 1.806 ⎧0.535 + 0.680 + 0.353 − 0.198 − 0.582⎫ ⎧ 0.787 ⎫ ⎪ ⎪ ⎪ 0.252 ⎪ 0.680 + 0.353 − 0.198 − 0.582 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{V }2 = ⎨ 0.353 − 0.198 − 0.582 ⎬ = ⎨− 0.428⎬MN ⎪ ⎪ ⎪ − 0.781⎪ − 0.198 − 0.582 ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭ ⎪⎩− 0.582⎪⎭ − 0.582 ⎧0.342 + 0.094 − 0.299 − 0.170 + 0.237 ⎫ ⎧ 0.204 ⎫ ⎪ ⎪ ⎪− 0.138⎪ 0.094 − 0.299 − 0.170 + 0.237 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{V }3 = ⎨ − 0.299 − 0.170 + 0.237 ⎬ = ⎨− 0.232⎬MN ⎪ ⎪ ⎪ 0.067 ⎪ − 0.170 + 0.237 ⎪ ⎪ ⎪ ⎪ 0.237 ⎩⎪ ⎭⎪ ⎩⎪ 0.237 ⎭⎪

83

⎧0.153 − 0.123 − 0.045 + 0.154 − 0.082⎫ ⎧ 0.056 ⎫ ⎪ − 0.123 − 0.045 + 0.154 − 0.082 ⎪ ⎪− 0.097⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{V }4 = ⎨ − 0.045 + 0.154 − 0.082 ⎬ = ⎨ 0.026 ⎬MN ⎪ ⎪ ⎪ 0.071 ⎪ 0.154 − 0.082 ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭ ⎪⎩− 0.082⎪⎭ − 0.082 ⎧0.040 − 0.066 + 0.069 − 0.051 + 0.019⎫ ⎧ 0.011 ⎫ ⎪ − 0.066 + 0.069 − 0.051 + 0.019 ⎪ ⎪− 0.029⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{V }5 = ⎨ 0.069 − 0.051 + 0.019 ⎬ = ⎨ 0.037 ⎬MN ⎪ ⎪− 0.033⎪ ⎪ − 0.051 + 0.019 ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎪⎭ ⎪⎩ 0.019 ⎪⎭ 0.019

Maximum story shears by SRSS rule Combining then the modal story shear with the square-root-of-the-sum-of-the-squares rule (except for the insignificant fourth and fifth modes, there is no closely spaced natural frequencies), the maximum story shears are thus equal to

{V }max

⎧(6.804 2 + 0.787 2 + 0.2042 + 0.0562 + 0.0112 )1 / 2 ⎫ ⎧6.852⎫ ⎪ ⎪ ⎪ 2 2 2 2 2 1/ 2 ⎪ ⎪(6.199 + 0.252 + 0.138 + 0.097 + 0.029 ) ⎪ ⎪6.207 ⎪ ⎪ ⎪ ⎪ ⎪ = ⎨(5.0732 + 0.4282 + 0.232 2 + 0.0262 + 0.037 2 )1 / 2 ⎬ = ⎨5.096 ⎬ MN ⎪ (3.5552 + 0.7812 + 0.067 2 + 0.0712 + 0.0332 )1 / 2 ⎪ ⎪ 3.641⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ (1.8062 + 0.5822 + 0.237 2 + 0.0822 + 0.0192 )1 / 2 ⎪⎭ ⎪⎩1.914 ⎪⎭

Problem 10.2 Using the response spectrum method and the design spectra shown in Figure P10.2, calculate the maximum lateral forces exerted on a three-story shear building with the properties listed in Table P10.2. Assume a damping matrix proportional to the building's mass and stiffness matrices and a damping ratio of 5% in its first two modes.

Figure P10.2. Design spectrum specified in Problem 10.2

84

Table P10.2. Mass and stiffness properties of three-story shear building in Problem 10.2 Story Stiffness (kN/m) Mass (Mg) 1 355.3 3.0 2 236.9 1.5 3 118.4 1.0

Solution: Dynamic properties The mass and stiffness matrices of the system are 0⎤ ⎡3.0 0 ⎢ [ M ] = ⎢ 0 1.5 0 ⎥⎥ ⎢⎣ 0 0 1.0⎥⎦

0 ⎤ ⎡ 592.2 − 236.9 ⎢ [ K ] = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎢⎣ 0 − 118.4 118.4 ⎥⎦

Then, solving the eigenvalue problem, one obtains the dynamic properties given in Table P10.2b Table P10.2b. Dynamic properties of building in Problem 10.2 Mode 1 2 3 Period (s) 1.00 0.50 0.33 Participation factor 1.0 1.0 1.0 0.5 0.4 0.1 Mode shape 1.0 0.2 -0.2 1.5 -0.6 0.1

Damping ratios Assuming a damping matrix proportional to the mass and stiffness matrices and considering damping ratios of 5 % for the first two modes, Equation 10.93 leads to α=

2ω1ω2 ξ1 2(2π / 1.00)(2π / 0.5) = 0.05 = 0.419 ω1 + ω2 2π / 1.00 + 2π / 0.5 β=

2ξ1 2(0.05) = = 0.005 ω1 + ω2 2π / 1.00 + 2π / 0.5

and thus, according to Equation 10.92, one has that ξ3 =

1 α 1 0.419 ( + βω3 ) = [ + 0.005(2π / 0.33)] = 0.058 2 ω3 2 2π / 0.33

Spectral accelerations From the given design spectrum, the spectral accelerations are SA(1.0,0.05) ≈ 0.35 g = 3.43 m/s 2 SA(0.5,0.05) ≈ 0.65 g = 6.38 m/s 2 SA(0.33,0.058) ≈ 0.70 g = 6.87 m/s 2

Modal displacements Applying Equation 10.54, ⎧0.04⎫ ⎧0.5⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ 1.0 ⎬3.43 = ⎨0.09⎬ m max{u}1 = 2 {φ}1 SA(ω1 , ξ1 ) = 2 ⎨ ω1 (2π / 1.00) ⎪ ⎪ ⎪0.13⎪ ⎩1.5 ⎭ ⎩ ⎭

85

⎧ 0 .4 ⎫ ⎧ 0.02 ⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ 0.2 ⎬6.38 = ⎨ 0.01 ⎬ m max{u}2 = 2 {φ}2 SA(ω2 , ξ 2 ) = 2 ⎨ ω2 (2π / 0.5) ⎪ ⎪ ⎪− 0.02⎪ ⎩ − 0 .6 ⎭ ⎩ ⎭ ⎧ 0.002 ⎫ ⎧ 0.1 ⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ − 0.2⎬6.87 = ⎨− 0.004⎬ m max{u}3 = 2 {φ}3 SA(ω3 , ξ3 ) = 2 ⎨ ω1 ( 2π / 0.33) ⎪ ⎪ ⎪ 0.002 ⎪ ⎩ 0.1 ⎭ ⎩ ⎭

Applying Equation 10.57, ⎡ 592.2 max{Fs }1 = [ K ] max{u}1 = ⎢⎢− 236.9 ⎢⎣ 0 ⎡ 592.2 max{Fs }2 = [ K ] max{u}2 = ⎢⎢− 236.9 ⎢⎣ 0

− 236.9 355.3 − 118.4 − 236.9 355.3 − 118.4

⎤ ⎧0.04⎫ ⎧2.37 ⎫ ⎪ ⎪ ⎪ ⎪ − 118.4⎥⎥ ⎨0.09⎬ = ⎨ 7.11⎬ kN 118.4 ⎥⎦ ⎪⎩0.13⎪⎭ ⎪⎩4.74⎪⎭ 0 ⎤ ⎧ 0.02 ⎫ ⎧ 9.47 ⎫ ⎪ ⎪ ⎪ ⎪ − 118.4⎥⎥ ⎨ 0.01 ⎬ = ⎨ 1.18 ⎬ kN 118.4 ⎥⎦ ⎪⎩− 0.02⎪⎭ ⎪⎩− 3.55⎪⎭ 0

0 ⎤ ⎧ 0.002 ⎫ ⎧ 2.13 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ ⎪ ⎢ max{Fs }3 = [ K ] max{u}3 = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎨− 0.004⎬ = ⎨− 2.13⎬ kN ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩ 0.002 ⎪⎭ ⎪⎩ 0.71 ⎪⎭

Maximum lateral forces Combining the maximum lateral forces with the SRSS rule, the maximum lateral forces are {Fs }max

⎧(2.37 2 + 9.47 2 + 2.132 )1 / 2 ⎫ ⎧9.99⎫ ⎪ ⎪ ⎪ ⎪ = ⎨ (7.112 + 1.16 2 + 2.132 )1 / 2 ⎬ = ⎨7.51⎬ kN ⎪ (4.74 2 + 3.552 + 0.712 )1 / 2 ⎪ ⎪5.96⎪ ⎭ ⎩ ⎭ ⎩

Problem 10.3 Repeat Problem 10.2 considering a damping matrix proportional to the building's stiffness matrix and a damping ratio of 5% in its fundamental mode. Solution: Damping ratios From Equation 10.95, one has that β=

2ξ1 2(0.05) = = 0.016 ω1 2π / 1.00

and, thus, the damping ration in the second and third mode are 1 1 ξ 2 = βω2 = (0.016)(2π / 0.5) = 0.10 2 2 1 1 ξ3 = βω3 = (0.016)(2π / 0.33) = 0.15 2 2

Spectral accelerations From the given design spectrum, the spectral accelerations are SA(1.0,0.05) ≈ 0.35 g = 3.43 m/s 2 SA(0.5,0.10) ≈ 0.40 g = 3.92 m/s 2

86

SA(0.33,0.15) ≈ 0.28 g = 2.75 m/s 2

Modal displacements Applying Equation 10.54, one gets ⎧0.5⎫ ⎧0.04⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ max{u}1 = 2 {φ}1 SA(ω1 , ξ1 ) = 1.0 ⎬3.43 = ⎨0.09⎬ m 2 ⎨ ω1 (2π / 1.00) ⎪ ⎪ ⎪0.13⎪ ⎩1.5 ⎭ ⎩ ⎭ ⎧ 0 .4 ⎫ ⎧ 0.010 ⎫ 1 Γ1 ⎪ ⎪ ⎪ ⎪ max{u}2 = 2 {φ}2 SA(ω2 , ξ 2 ) = 0.2 ⎬3.92 = ⎨ 0.005 ⎬ m 2 ⎨ ( 2π / 0.5) ⎪ ω2 ⎪ ⎪− 0.015⎪ ⎩ − 0 .6 ⎭ ⎩ ⎭ ⎧ 0.1 ⎫ ⎧ 0.0008 ⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ − 0.2⎬2.75 = ⎨− 0.0015⎬ m max{u}3 = 2 {φ}3 SA(ω3 , ξ3 ) = 2 ⎨ ω1 ( 2π / 0.33) ⎪ ⎪ ⎪ 0.0008 ⎪ ⎩ 0.1 ⎭ ⎩ ⎭

Applying then Equation 10.57, one arrives to 0 ⎤ ⎧0.04⎫ ⎧2.37 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ ⎪ ⎢ max{Fs }1 = [ K ] max{u}1 = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎨0.09⎬ = ⎨ 7.11⎬ kN ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩0.13⎪⎭ ⎪⎩4.74⎪⎭ 0 ⎤ ⎧ 0.010 ⎫ ⎧ 4.74 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ ⎪ ⎢ max{Fs }2 = [ K ] max{u}2 = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎨ 0.005 ⎬ = ⎨ 1.18 ⎬ kN ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩− 0.015⎪⎭ ⎪⎩− 2.37 ⎪⎭ 0 ⎤ ⎧ 0.0008 ⎫ ⎧ 0.83 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ ⎪ ⎢ max{Fs }3 = [ K ] max{u}3 = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎨− 0.0015⎬ = ⎨− 0.82⎬ kN ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩ 0.0008 ⎪⎭ ⎪⎩ 0.27 ⎪⎭

Maximum lateral forces Combining then the maximum lateral forces with the SRSS rule, one obtains {Fs }max

⎧(2.37 2 + 4.74 2 + 0.832 )1 / 2 ⎫ ⎧5.36⎫ ⎪ ⎪ ⎪ ⎪ = ⎨ (7.112 + 1.182 + 0.832 )1 / 2 ⎬ = ⎨7.25⎬ kN ⎪(4.742 + 2.37 2 + 0.27 2 )1 / 2 ⎪ ⎪5.30⎪ ⎭ ⎩ ⎭ ⎩

Problem 10.4 Repeat Problem 10.2 considering a damping matrix proportional to the building's mass matrix and a damping ratio of 5% in its fundamental mode. Solution: Damping ratios From Equation 10.94, one has that α = 2ξ1ω1 = 2(0.05)(2π × 1.00) = 0.628

and, thus, the damping ratio in the second and third mode are ξ2 =

1 α 1 0.628 = = 0.025 2 ω 2 2 2 π / 0 .5

87

ξ3 =

1 α 1 0.628 = = 0.017 2 ω3 2 2π / 0.33

Spectral accelerations From the given design spectrum, the spectral accelerations are SA(1.0,0.05) ≈ 0.35 g = 3.43 m/s 2 SA(0.5,0.025) ≈ 1.0 g = 9.81 m/s 2 SA(0.33,0.017) ≈ 1.4 g = 13.73 m/s 2

Modal displacements Applying Equation 10.54, one obtains ⎧0.5⎫ ⎧0.04⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ max{u}1 = 2 {φ}1 SA(ω1 , ξ1 ) = 1.0 ⎬3.43 = ⎨0.09⎬ m 2 ⎨ ω1 (2π / 1.00) ⎪ ⎪ ⎪0.13⎪ ⎩1.5 ⎭ ⎩ ⎭ 0 . 4 0 . 025 ⎧ ⎫ ⎧ ⎫ 1 Γ1 ⎪ ⎪ ⎪ ⎪ max{u}2 = 2 {φ}2 SA(ω2 , ξ 2 ) = 0.2 ⎬9.81 = ⎨ 0.012 ⎬ m 2 ⎨ (2π / 0.5) ⎪ ω2 ⎪ ⎪− 0.037 ⎪ ⎩− 0.6⎭ ⎩ ⎭ ⎧ 0.004 ⎫ ⎧ 0.1 ⎫ Γ1 1 ⎪ ⎪ ⎪ ⎪ max{u}3 = 2 {φ}3 SA(ω3 , ξ3 ) = − 0.2⎬13.73 = ⎨− 0.008⎬ m 2 ⎨ ω1 (2π / 0.33) ⎪ ⎪ ⎪ 0.004 ⎪ ⎩ 0.1 ⎭ ⎩ ⎭

Then, applying Equation 10.57 one gets 0 ⎤ ⎧0.04⎫ ⎧2.37 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ ⎪ ⎢ max{Fs }1 = [ K ] max{u}1 = ⎢− 236.9 355.3 − 118.4⎥⎥ ⎨0.09⎬ = ⎨ 7.11⎬ kN ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩0.13⎪⎭ ⎪⎩4.74⎪⎭ ⎡ 592.2 max{Fs }2 = [ K ] max{u}2 = ⎢⎢− 236.9 ⎢⎣ 0 ⎡ 592.2 max{Fs }3 = [ K ] max{u}3 = ⎢⎢− 236.9 ⎢⎣ 0

0 ⎤ ⎧ 0.025 ⎫ ⎧ 11.96 ⎫ − 236.9 ⎪ ⎪ ⎪ ⎪ 355.3 − 118.4⎥⎥ ⎨ 0.012 ⎬ = ⎨ 2.72 ⎬ kN − 118.4 118.4 ⎥⎦ ⎪⎩− 0.037 ⎪⎭ ⎪⎩− 5.80⎪⎭ 0 ⎤ ⎧ 0.004 ⎫ ⎧ 4.26 ⎫ − 236.9 ⎪ ⎪ ⎪ ⎪ 355.3 − 118.4⎥⎥ ⎨− 0.008⎬ = ⎨− 4.26⎬ kN − 118.4 118.4 ⎥⎦ ⎪⎩ 0.004 ⎪⎭ ⎪⎩ 1.42 ⎪⎭

Maximum lateral forces Combining the maximum lateral forces with the SRSS rule, one arrives to {Fs }max

⎧(2.37 2 + 4.742 + 0.832 )1 / 2 ⎫ ⎧12.92⎫ ⎪ ⎪ ⎪ ⎪ = ⎨ (7.112 + 1.182 + 0.832 )1 / 2 ⎬ = ⎨ 8.72 ⎬ kN ⎪(4.742 + 2.37 2 + 0.27 2 )1 / 2 ⎪ ⎪ 7.62 ⎪ ⎭ ⎩ ⎭ ⎩

Problem 10.5 Estimate the maximum acceleration response of the two-story shear building shown in Figure P10.5 when the base of the building is subjected to the ground motion represented by the elastic response spectrum shown in Figure P8.2a. Assume elastic behavior and a damping ratio of 5% for all modes.

88

Mass = 1.0 Gg Stiffness = 20.0 MN/m

Mass = 6.76 Gg Stiffness = 296.4 MN/m

Figure P10.5. Two-story shear building in Problem 10.5

Solution: Mass and damping matrices ⎡316.4 − 20⎤ [K ] = ⎢ 20 ⎥⎦ ⎣ − 20

⎡6.76 0 ⎤ [M ] = ⎢ ⎥ ⎣ 0 1.0⎦

Eigenvalue problem The solution of the corresponding eigenvalue problem leads to 316.4 − 6.76ω2 − 20

[ K ] − ω2 [ M ] =

− 20 =0 20 − ω2

or (316.4 − 6.76ω2 )(20 − ω2 ) = 0

or ω4 − 66.8ω2 + 876.9 = 0

from which one gets 66.8 1 + (66.8) 2 − 4(876.9) = 33.4 ± 15.4 2 2 ω12 = 18.0 rad/s ω22 = 48.8 rad/s f1 = 0.68 Hz f 2 = 1.11 Hz

ω2 =

Writing the equation of motion explicitly in terms of mode shapes and natural frequencies, one obtains ⎡6.76 0 ⎤ ⎧ 1 ⎫ ⎡316.4 − 20.⎤ ⎧ 1 ⎫ ⎧0⎫ − ω2 ⎢ ⎥⎨ ⎬ + ⎢ ⎥⎨ ⎬ = ⎨ ⎬ ⎣ 0 1.0⎦ ⎩φ2 ⎭ ⎣ − 20. 20. ⎦ ⎩φ2 ⎭ ⎩0⎭

and thus from the lower part, one has that − ω2φ2 − 20 + 20φ2 = 0

which leads to φ2 =

20 20 − ω2

Hence, φ21 =

20 = 10 20 − 18.0

φ22 =

and thus the mode shapes are

89

20 = −0.69 20 − 48.8

⎧ 1.0 ⎫ {φ}2 = ⎨ ⎬ ⎩− 0.69⎭

⎧ 1.0 ⎫ {φ}1 = ⎨ ⎬ ⎩10.0⎭

Participation factors According to Equation 10.32, the participation factors are equal to 6.76 × 1.0 + 1.0 × 10.0 = 0.16 6.76 × 1.02 + 1.0 × 10.02 6.76 × 1.0 − 1.0 × 069 Γ2 = = 0.84 6.76 × 1.02 + 1.0 × 0.692 Γ1 =

Response spectrum ordinates The corresponding response spectrum ordinates are thus equal to SA(0.68,0.05) ≈ 0.65 g SA(1.11,0.05) ≈ 1.1g

Maximum acceleration response Applying, then, Equation 10.56, one gets ⎧ 1.0 ⎫ ⎧0.10⎫ max{&y&}1 = Γ1{φ}1 SA( f1 , ξ1 ) = 0.16⎨ ⎬0.65 g = ⎨ ⎬g ⎩10.0⎭ ⎩1.04 ⎭ ⎧ 0.92 ⎫ ⎧ 1.0 ⎫ max{&y&}2 = Γ2{φ}2 SA( f 2 , ξ 2 ) = 0.84⎨ ⎬g ⎬1.1g = ⎨ ⎩− 0.64⎭ ⎩− 0.69⎭

and after combining the modal responses with the square-root-of-the sum-of the-squares rule (no closely spaced natural frequencies), one arrives to ⎧0.102 + 0.922 ⎫ ⎧0.92⎫ {&y&}max = ⎨ =⎨ ⎬g 2 2⎬ ⎩1.04 + 0.64 ⎭ ⎩1.22 ⎭

Problem 10.6 A system is modeled as an elastic two-degree-of-freedom system with a damping ratio of 10% in each mode and masses for the first and second degrees of freedom of 6.8 Gg and 1.0 Gg, respectively. The natural frequencies and mode shapes of this two-degree-of-freedom model are given in Table P10.6. Estimate using the response spectrum method the shear force at the base of the system when its base is subjected to a ground motion represented by the response spectrum shown in Figure E10.4b. Table P10.6. Dynamic properties of building in Problem 10.6 Mode 1 2 Frequency (Hz) 0.674 1.112 Participation factor 0.162 0.838 Mode shape 1.000 1.000 9.747 -0.694

Solution: From the response spectrum in Figure E10.4b, one obtains

SA(0.674,0.10) ≈ 0.15 g SA(1.112,0.10) ≈ 0.35 g

Thus, by virtue of Equation 10.56, the maximum modal accelerations are

90

⎧1.000 ⎫ ⎧0.238⎫ 2 max{&y&(t )}1 = Γ1{φ}1 SA( f1 , ξ1 ) = 0.162⎨ ⎬0.15 g = ⎨ ⎬ m/s ⎩9.747 ⎭ ⎩2.323⎭ ⎧ 1.000 ⎫ ⎧ 2.876 ⎫ 2 max{&y&(t )}2 = Γ2{φ}2 SA( f 2 , ξ 2 ) = 0.838⎨ ⎬0.35 g = ⎨ ⎬ m/s − 0 . 694 − 1 . 996 ⎩ ⎭ ⎩ ⎭

and the maximum modal lateral forces are ⎡6.8 0 ⎤ ⎧0.238⎫ ⎧1.62 ⎫ max{Fs }1 = [ M ] max{&y&}1 = ⎢ ⎬=⎨ ⎬ MN ⎥⎨ ⎣ 0 1.0⎦ ⎩2.323⎭ ⎩2.32⎭ ⎡6.8 0 ⎤ ⎧ 2.876 ⎫ ⎧ 19.56 ⎫ max{Fs }2 = [ M ] max{&y&}2 = ⎢ ⎬ MN ⎬=⎨ ⎥⎨ ⎣ 0 1.0⎦ ⎩− 1.996⎭ ⎩− 2.00⎭

In consequence, the maximum modal base shears are max(V0 )1 = 1.62 + 2.32 = 3.94 MN max(V0 ) 2 = 19.56 − 2.00 = 17.56 MN

Combining the maximum modal base shears with the SRSS rule (no closely spaced natural frequencies), the maximum base shear is thus equal to (V0 ) max = 3.942 + 17.562 = 18.0 MN

Problem 10.7 The yield deformation response spectra shown in Figures P10.7a and P10.7b are representative of the earthquake ground motions expected at a given area. Using the response spectrum method, determine the maximum story shears in the building described in Problem 10.2 when the building is subjected to a ground motion defined by these response spectra. Consider that the building stories are capable of withstanding deformations that are five times as large as their respective yield deformations. Consider, in addition, that the damping matrix of the building is of the Rayleigh type (that is, proportional to mass and stiffness matrices) and that the damping ratio in its fundamental mode is 2% of critical. Use linear interpolation to find the spectral ordinates for damping ratios other than 2 and 5%.

Figure P10.7. Yield deformation response spectra for elastoplastic systems with (a) 2% damping and (b) 5% damping of N10W component of ground accelerations recorded during the 1971 Santiago, Chile, earthquake Solution: Using Equation 10.96, one gets α = ξ1ω1 = 0.02(2π)(1.0) = 0.126

91

β=

ξ1 0.02 = = 0.00318 ω1 2π(1.0)

Hence, the damping ratios in the second and third modes of the system are equal to 1 α 1 0.126 ( + βω2 ) = [ + 0.00318(2π)(2.0)] = 0.025 2 ω2 2 2π( 2.0) 1 0.126 1 α + 0.00318(2π)(3.0)] = 0.033 ξ3 = ( + βω3 ) = [ 2 2π(3.0) 2 ω3

ξ2 =

and from the given inelastic spectra for μ =5, the spectral displacements are SD1 = 0.5 in. = 0.0127 m SD2 = 0.18 in. = 0.0046 m SD3 = 0.10 in. = 0.0025 m

The maximum modal displacements are thus equal to ⎧0.006⎫ ⎧0.5⎫ ⎪ ⎪ ⎪ ⎪ max{u}1 = ⎨1.0 ⎬0.0127 = ⎨0.013⎬ m ⎪0.019⎪ ⎪1.5 ⎪ ⎩ ⎭ ⎩ ⎭ ⎧ 0.4 ⎫ ⎧ 0.002 ⎫ ⎪ ⎪ ⎪ ⎪ max{u}2 = ⎨ 0.2 ⎬0.0046 = ⎨ 0.001 ⎬ m ⎪ − 0.6 ⎪ ⎪− 0.003⎪ ⎩ ⎭ ⎩ ⎭ 0 . 1 2 . 5 ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ max{u}3 = ⎨− 0.2⎬0.0025 = ⎨− 5.0⎬ × 10- 4 m ⎪ 2.5 ⎪ ⎪ 0.1 ⎪ ⎩ ⎭ ⎭ ⎩

In consequence, the maximum lateral forces are equal to ⎡ 592.2 max{Fs }1 = [ K ] max{u}1 = ⎢⎢− 236.9 ⎢⎣ 0 ⎡ 592.2 max{Fs }2 = [ K ] max{u}2 = ⎢⎢− 236.9 ⎢⎣ 0

0 ⎤ ⎧0.006⎫ ⎧0.473⎫ − 236.9 ⎪ ⎪ ⎪ ⎪ 355.3 − 118.4⎥⎥ ⎨0.013⎬ = ⎨0.948⎬ kN − 118.4 118.4 ⎥⎦ ⎪⎩0.019⎪⎭ ⎪⎩0.710⎪⎭ − 236.9 0 ⎤ ⎧ 0.002 ⎫ ⎧ 0.947 ⎫ ⎪ ⎪ ⎪ ⎪ 355.3 − 118.4⎥⎥ ⎨ 0.001 ⎬ = ⎨ 0.237 ⎬ kN − 118.4 118.4 ⎥⎦ ⎪⎩− 0.003⎪⎭ ⎪⎩− 0.474⎪⎭

0 ⎤ ⎧ 2.5 ⎫ ⎧ 0.266 ⎫ ⎡ 592.2 − 236.9 ⎪ ⎪ ⎪ −4 ⎪ ⎢ ⎥ max{Fs }3 = [ K ] max{u}3 = ⎢− 236.9 355.3 − 118.4⎥ ⎨− 5.0⎬10 = ⎨− 0.266⎬ kN ⎪ 0.089 ⎪ ⎢⎣ 0 − 118.4 118.4 ⎥⎦ ⎪⎩ 2.5 ⎪⎭ ⎩ ⎭

and the corresponding modal story shears equal to ⎧0.473 + 0.948 + 0.710⎫ ⎧ 2.131⎫ ⎪ ⎪ ⎪ ⎪ {V }1 = ⎨ 0.948 + 0.710 ⎬ = ⎨1.658 ⎬ kN ⎪ ⎪ ⎪0.710⎪ 0.710 ⎭ ⎩ ⎭ ⎩ ⎧0.947 + 0.237 − 0.474⎫ ⎧ 0.710 ⎫ ⎪ ⎪ ⎪ ⎪ {V }2 = ⎨ 0.237 − 0.474 ⎬ = ⎨− 0.237 ⎬ kN ⎪ ⎪ ⎪− 0.474⎪ − 0.474 ⎩ ⎭ ⎩ ⎭

92

⎧0.266 − 0.266 + 0.089⎫ ⎧ 0.089 ⎫ ⎪ ⎪ ⎪ ⎪ {V }3 = ⎨ − 0.266 + 0.089 ⎬ = ⎨− 0.177 ⎬ kN ⎪ ⎪ ⎪ 0.089 ⎪ 0.089 ⎭ ⎩ ⎭ ⎩

Combining, then, the modal story shears with the SRSS rule (no closely spaced natural frequencies), one arrives to {V }max

⎧( 2.1312 + 0.7102 + 0.089 2 )1 / 2 ⎫ ⎧2.248⎫ ⎪ ⎪ ⎪ ⎪ = ⎨(1.6582 + 0.237 2 + 0.177 2 )1 / 2 ⎬ = ⎨1.684 ⎬kN ⎪(0.7102 + 0.474 2 + 0.0892 )1 / 2 ⎪ ⎪0.858⎪ ⎭ ⎩ ⎭ ⎩

Problem 10.8 Determine using the response spectrum method, the maximum acceleration at the top floor of the building-penthouse system analyzed in Example 10.1 when the building is subjected to the same ground motion considered in Example 10.1 but the masses and stiffnesses of the penthouse are reduced by a factor of ten. Use alternatively the following rules to combine the system’s modal responses: (a) square root of the sum of the squares; (b) complete quadratic combination; and (c) double sum in conjunction with S-M formula to compute the modal correlation coefficients. Assume a damping ratio of 1% for all modes and consider only the first three modes of vibration. Solution:

The dynamic properties of the system considering m4 = 0.0045 Mg, m5 = 0.0015 Mg, K4 = 0.36 kN/m, and K5 = 0.09 kN/m results as shown in Table P10.8. Table P10.8. Dynamic properties of building in Problem 10.8 Mode 1 2 3 4 Frequency (rad/s) 6.174 6.473 10.950 12.664 Period (s) 1.018 0.971 0.574 0.496 Participation factor 1.525 -1.476 -0.116 -0.943 0.161 -0.172 -0.038 -0.420 0.327 -0.340 -0.039 -0.208 Mode shape 0.502 -0.497 0.019 0.633 5.687 4.838 12.860 -1.046 15.590 16.043 -12.879 0.625

5 18.975 0.331 -0.316 -0.316 0.632 -0.317 0.099 -0.020

From the response spectrum in Figure E10.1(b) and considering only the first three modes, one then gets SA1 = SA(6.174,0.01) = 0.39 g

SA2 = SA(6.473,0.01) = 0.38 g

SA3 = SA(10.950,0.01) = 1.20 g

SD1 = SD(6.174,0.01) = 0.100 m SD2 = SD(6.473,0.01) = 0.090 m SD3 = SD(10.950,0.01) = 0.100 m PSV1 = PSV (6.174,0.01) = 0.62 m/s PSV2 = PSV (6.473,0.01) = 0.58 m/s PSV3 = PSV (10.950,0.01) = 1.09 m/s

Also, according to Equations 10.79 through 10.81, av = 1.095 + 0.647ξ − 0.382ξ 2 = 1.095 + 0.647(0.01) − 0.382(0.01) 2 = 1.102 bv = 0.193 + 0.838ξ − 0.621ξ 2 = 0.193 + 0.838(0.01) − 0.621(0.01) 2 = 0.201 2π 0.201 SV1 = avT1bv PSV1 = 1.102( ) (0.62) = 0.69 m/s 6.174

93

2π 0.201 ) (0.58) = 0.63 m/s 6.473 2π 0.201 SV3 = avT3bv PSV3 = 1.102( ) (1.09) == 1.07 m/s 10.950 SV2 = avT2bv PSV2 = 1.102(

By virtue of Equation 10.56, one has thus that max( &y&5 )1 = Γ1φ15 SA(ω1 , ξ1 ) = 1.525(15.590)0.39g = 9.27 g max( &y&5 ) 2 = Γ2φ25 SA(ω2 , ξ 2 ) = −1.476(16.043)0.38g = −9.00 g max( &y&5 )3 = Γ3φ35 SA(ω3 , ξ3 ) = −0.116(−12.879)1.20g = 1.79 g

Square root of the sum of the squares (SRSS) Combining the modal accelerations with the SRSS rule, ( &y&5 ) max = g 9.27 2 + 9.002 + 1.792 = 13.0 g

Complete quadratic combination According to Equation 10.73, the modal correlation coefficients are given by ρmn =

3/ 2 3/ 2 8ξ 2 (1 + rmn )rmn 0.0008(1 + rmn )rmn = 2 2 2 2 (1 − rmn ) + 4ξ 2 rmn (1 + rmn ) 2 (1 − rmn ) + 0.0004rmn (1 + rmn ) 2

where rmn = ωm/ωn. Therefore, for the first three modes, the modal correlation coefficients are ⎡1.000 0.151 0.001⎤ [ρmn ] = ⎢⎢0.151 1.000 0.001⎥⎥ ⎢⎣0.001 0.001 1.000⎥⎦

Applying Equation 10.67 with the first three modes considered, one then gets ( &y&5 ) max = g 9.27 2 + 9.002 + 1.792 + 2(0.151)(9.27)(−9.00) + 2(0.001)(9.27)(1.79) + 2(0.001)(−9.00)(1.79) = 12.04 g

Singh-Maldonado rule The constants defined by Equations 10.75 through 10.78, arranged in matrix form, are ⎡0.002 0.010 0.466⎤ [ Dmn ] = ⎢⎢0.012 0.002 0.424⎥⎥ ⎢⎣ 4.611 3.470 0.002⎥⎦

8.606 0.466⎤ ⎡ 0.500 ⎢ [ Amn ] = ⎢− 9.303 0.500 0.537 ⎥⎥ ⎢⎣ − 1.463 − 1.534 0.500⎥⎦ ⎡ 0.5 − 9.303 − 1.463⎤ [Cmn ] = ⎢⎢8.606 0.5 − 1.534⎥⎥ ⎢⎣0.466 0.537 0.5 ⎥⎦

− 0.224 − 0.012⎤ ⎡ 0 ⎢ [ Bmn ] = ⎢0.224 0 − 0.013⎥⎥ ⎢⎣0.012 0.013 0 ⎥⎦

Then, by virtue of Equation 10.74, the modal correlation coefficients, also arranged in matrix form, are ⎡ 1.0 0.782 0.181⎤ [ρmn ] = ⎢⎢0.782 1.0 0.157⎥⎥ ⎢⎣ 0.181 0.157 1.0 ⎥⎦

Thus, ( &y&5 ) max = g 9.27 2 + 9.00 2 + 1.792 + 2(0.782)(9.27)(−9.00) + 2(0.181)(9.27)(1.79) + 2(0.157)(−9.00)(1.79) = 6.37 g

94

Problem 10.9 Determine using the modal acceleration method the displacement response of the five-story building described in Problem 10.1 when the building is subjected to the same ground motion defined therein. Consider alternatively (a) only the first mode of vibration; and (b) only the first two modes of vibration. Solution: With only the first mode considered According to Equation 10.109, the maximum displacements are given by max{u (t )} = {[

1

1

∑∑ ρ

mn

m =1 n =1

max[u (t )]m max[u (t )]n + U s2 [max u&&g (t )]2 − 2U sVP ]1 / 2 }

where Us and VP are the elements of the vectors (see Equation 10.110) {U s } = [ K ]−1[ M ]{J } −

1

Γr

∑ω r =1

2 r

{φ}r

{VP } =

1

∑ Γ {φ} (SV r

r

2 r

− ωr2 SDr2 )

r =1

But from the solution to Problem 10.1, one has that ⎧0.027232 ⎫ ⎧0.00074⎫ ⎪ ⎪ ⎪ 2⎪ ⎪0.05338 ⎪ ⎪0.00285⎪ 1 1 ⎪ ⎪ ⎪ ⎪ { ρ mn max[u (t )]m max[u (t )]n } = {(max[u (t )]1 ) 2 } = ⎨0.07597 2 ⎬ = ⎨0.00577⎬ m m=1 n =1 ⎪0.092732 ⎪ ⎪0.00860⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.101782 ⎪⎭ ⎪⎩0.01036⎪⎭

∑∑

0 0 0 ⎤ ⎡179 0 ⎢ 0 170 0 0 0 ⎥⎥ ⎢ [M ] = ⎢ 0 0 161 0 0 ⎥ Mg ⎢ ⎥ 0 0 152 0 ⎥ ⎢ 0 ⎢⎣ 0 0 0 0 143⎥⎦

⎧1⎫ ⎪1⎪ ⎪⎪ ⎪⎪ {J } = ⎨1⎬ ⎪1⎪ ⎪⎪ ⎪⎩1⎪⎭

0 0 0 ⎤ ⎡ 486.93 − 237.05 ⎢− 237.05 461.62 − 224.57 0 0 ⎥⎥ ⎢ [K ] = ⎢ 0 − 224.57 436.66 − 212.09 0 ⎥ × 103 kN/m ⎢ ⎥ − 212.09 411.71 − 199.62⎥ 0 ⎢ 0 ⎢⎣ 0 − 199.62 199.62 ⎥⎦ 0 0

⎡0.00400 ⎢0.00400 ⎢ [ K ]−1 = ⎢0.00400 ⎢ ⎢0.00400 ⎢⎣0.00400

0.00400 0.00822 0.00822 0.00822 0.00822

0.00400 0.00822 0.01267 0.01267 0.01267

95

0.00400 0.00822 0.01267 0.01739 0.01739

0.00400⎤ 0.00822⎥⎥ 0.01267⎥ × 10− 3 ⎥ 0.01739⎥ 0.02240⎥⎦

⎧ 0.00321⎫ ⎪0.00585⎪ ⎪⎪ ⎪⎪ [ K ]−1[ M ]{J } = ⎨0.00788⎬ ⎪0.00927⎪ ⎪ ⎪ ⎪⎩0.00999⎪⎭ ⎧0.013⎫ ⎧0.00276⎫ ⎪0.026⎪ ⎪0.00552⎪ 1 ⎪⎪ ⎪⎪ ⎪⎪ 26.335 ⎪⎪ Γr { } φ = = 0 . 036 0 . 00764 ⎨ ⎬ ⎨ ⎬ r 2 124.090 ⎪ r =1 ωr ⎪ ⎪ 0.045 0.00955⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.049⎪⎭ ⎪⎩0.01040⎪⎭

∑

⎧ 0.00321⎫ ⎧0.00276⎫ ⎧ 0.00045 ⎫ ⎪0.00585⎪ ⎪0.00552⎪ ⎪ 0.00033 ⎪ 1 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ Γr {U s } = [ K ]−1[ M ]{J } − { } φ = − = 0 . 00788 0 . 00764 0 . 00024 ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ r 2 r =1 ωr ⎪0.00927 ⎪ ⎪0.00955⎪ ⎪− 0.00028⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.00999⎪⎭ ⎪⎩0.01040⎪⎭ ⎪⎩ − 0.00041⎪⎭

∑

From the data given in Problem 10.1, the relationship between pseudovelocity and spectral acceleration, and Equations 10.79 through 10.81, one also has that max u&&g (t ) = 0.40 g = 0.40(9.81) = 3.92 m/s 2 PSV1 = SA1 / ω1 = 9.81 / 11.140 = 0.881 m/s av = 1.095 + 0.647ξ − 0.382ξ 2 = 1.095 + 0.647(0.05) − 0.382(0.05) 2 = 1.126 bv = 0.193 + 0.838ξ − 0.621ξ 2 = 0.193 + 0.838(0.05) − 0.621(0.05) 2 = 0.233 2π 0.233 SV1 = avT1bv PSV1 = 1.126( ) (0.881) = 0.868 m/s 11.140

and thus ⎧0.013⎫ ⎧0.008⎫ ⎪0.026⎪ ⎪0.016⎪ 1 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ {VP } = Γr {φ}r ( SVr2 − ω2r SDr2 ) = 26.335⎨0.036⎬(0.8682 − 0.8812 ) = −⎨0.022⎬ r =1 ⎪0.045⎪ ⎪0.027⎪ ⎪ ⎪ ⎪ ⎪ ⎩⎪0.049⎭⎪ ⎩⎪0.029⎭⎪

∑

Therefore, ⎧0.00074⎫ ⎧ 0.000452 ⎫ ⎧ 0.00045 ⎫⎧0.008⎫ ⎧0.00075⎫ ⎪0.00285⎪ ⎪ ⎪ 0.00033 ⎪⎪0.016⎪ ⎪0.00286⎪ 2 ⎪ ⎪⎪ ⎪⎪ ⎪⎪ 0.00033 ⎪⎪ ⎪⎪ ⎪⎪⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ max{u 2 (t )} = ⎨0.00577⎬ + ⎨ 0.000242 ⎬3.922 + 2⎨ 0.00024 ⎬⎨0.022⎬ = ⎨0.00578⎬ ⎪0.00860⎪ ⎪− 0.000282 ⎪ ⎪− 0.00028⎪⎪0.027⎪ ⎪0.00859⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.01036⎪⎭ ⎪⎩ − 0.000412 ⎪⎭ ⎪⎩ − 0.00041⎪⎭⎪⎩0.029⎪⎭ ⎪⎩0.01034⎪⎭

and thus

96

⎧0.027⎫ ⎪0.053⎪ ⎪⎪ ⎪⎪ max{u (t )} = ⎨0.076⎬ m ⎪0.093⎪ ⎪ ⎪ ⎪⎩0.102⎪⎭

With only the first two modes considered According to Equation 10.109, the maximum displacements are given by max{u (t )} = {[

2

2

∑∑ ρ m =1 n =1

mn

max[u (t )]m max[u (t )]n + U s2 [max u&&g (t )]2 − 2U sVP ]1 / 2 }

But 2

2

∑∑ ρ

{

mn

max[u (t )]m max[u (t )]n } = {(max[u (t )]1 ) 2 } + {(max[u (t )]2 ) 2 } + 2ρ12{max[u (t )]1 max[u (t )]2 }

m =1 n =1

and in view of Equations 10.79 through 10.81 and 10.74, r12 =

ω1 11.140 = = 0.362 ω2 30.810

D12 = 1 + 4ξ1ξ 2 r12 − 2(1 − 2ξ12 − 2ξ 22 ) r122 + 4ξ1ξ 2 r123 + r124 = 1 + 4(0.05)(0.05)(0.362) − 2[1 − 2(0.05) 2 − 2(0.05) 2 ]0.3622 + 4(0.05)(0.05)(0.362)3 + 0.3624 = 0.762 A12 = [ r122 + 4ξ1ξ 2 r123 − (1 − 4ξ12 ) r124 ] / D12 = [0.362 2 + 4(0.05)(0.05)0.3623 − (1 − 4 × 0.052 )0.3624 ] / 0.762 = 0.150 B12 = (r122 − 1) / ω22 D12 = (0.3622 − 1) /(30.8102 × 0.762) = −0.0012 Cmn = [ −1 + 4ξ 22 + 4ξ1ξ 2 r12 + r122 ] / D12 = [ −1 + 4(0.05) 2 + 4(0.05)(0.05)(0.362) + 0.3622 ] / 0.762 = −1.122 PSV1 = SA1 / ω1 = 9.81 / 11.140 = 0.881 m/s PSV2 = SA2 / ω2 = 9.81 / 30.810 = 0.318 m/s 2π 0.233 SV1 = avT1bv PSV1 = 1.126( ) (0.881) = 0.868 m/s 11.140 2π 0.233 SV2 = avT2bv PSV2 = 1.126( ) (0.318) = 0.247 m/s 30.810 SD1 = PSV1 / ω1 = 0.881 / 11.140 = 0.079 m SD2 = PSV2 / ω2 = 0.318 / 30.810 = 0.010 m ρ12 =

A12 SD12 + B12 ( SV12 − SV22 ) + C12 SD22 SD1SD2

| 0.150(0.079) − 0.0012(0.8682 − 0.247 2 ) − 1.122(0.010) 2 | 0.079(0.010) = 0.009

=

2

and thus 2

2

∑∑ ρ

{

mn

max[u (t )]m max[u (t )]n }

m =1 n =1

97

⎧0.027232 ⎫ ⎧ 0.003152 ⎫ ⎧0.02723⎫⎧ 0.00315 ⎫ ⎧0.00075⎫ ⎪ ⎪0.05338⎪⎪ 0.00421 ⎪ ⎪0.00287⎪ 2⎪ ⎪ 2 ⎪ ⎪0.05338 ⎪ ⎪ 0.00421 ⎪ ⎪⎪ ⎪⎪⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ = ⎨0.07597 2 ⎬ + ⎨ 0.002312 ⎬ + 2(0.009)⎨0.07597⎬⎨ 0.00231 ⎬ = ⎨0.00578⎬ m ⎪0.09273⎪⎪- 0.00137⎪ ⎪0.00860⎪ ⎪0.092732 ⎪ ⎪- 0.00137 2 ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.10178⎪⎭⎪⎩- 0.00429⎪⎭ ⎪⎩0.01038⎪⎭ ⎪⎩0.101782 ⎪⎭ ⎪⎩- 0.004292 ⎪⎭

Similarly, ⎧ 0.00321⎫ ⎪0.00585⎪ ⎪⎪ ⎪⎪ [ K ]−1[ M ]{J } = ⎨0.00788⎬ ⎪0.00927⎪ ⎪ ⎪ ⎪⎩0.00999⎪⎭ ⎧0.013⎫ ⎧− 0.034⎫ ⎧0.00308⎫ ⎪0.026⎪ ⎪− 0.046⎪ ⎪0.00595⎪ 2 ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪⎪ 26.335 ⎪ Γr ⎪ 8.958 ⎪⎪ { } φ = − = 0 . 036 0 . 025 0 . 00788 − ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ r 2 124.090 ⎪ 949.25 ⎪ r =1 ωr ⎪ ⎪ ⎪ 0.045 0.015 0.00940⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.049⎪⎭ ⎪⎩ 0.046 ⎪⎭ ⎪⎩0.00996⎪⎭

∑

⎧ 0.00321⎫ ⎧0.00308⎫ ⎧ 0.00013 ⎫ ⎪0.00585⎪ ⎪0.00595⎪ ⎪− 0.00010⎪ 2 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ Γr {U s } = [ K ]−1[ M ]{J } − { } φ = − = 0 . 00788 0 . 00788 0 . 00000 ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ r 2 r =1 ωr ⎪0.00927⎪ ⎪0.00940⎪ ⎪− 0.00014⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.00999⎪⎭ ⎪⎩0.00996⎪⎭ ⎪⎩ 0.00002 ⎪⎭

∑

⎧0.013⎫ ⎧− 0.034⎫ ⎪0.026⎪ ⎪− 0.046⎪ 2 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ {VP } = Γr {φ}r ( SVr2 − ω2r SDr2 ) = 26.335⎨0.036⎬(0.8682 − 0.8812 ) − 8.958⎨− 0.025⎬(0.247 2 − 0.3182 ) r =1 ⎪0.045⎪ ⎪ 0.015 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩0.049⎪⎭ ⎪⎩ 0.046 ⎪⎭ ⎧− 0.00418⎫ ⎪− 0.00048⎪ ⎪⎪ ⎪⎪ = ⎨ 0.01304 ⎬ ⎪ 0.03237 ⎪ ⎪ ⎪ ⎪⎩ 0.04548 ⎪⎭

∑

Consequently, 2

2

∑∑ ρ

max{u 2 (t )} = {

m =1 n =1

mn

max[u (t )]m max[u (t )]n + U s2 [max u&&g (t )]2 − 2U sVP }

98

⎧0.00075⎫ ⎧ 0.000132 ⎫ ⎧ 0.00013(−0.00418) ⎫ ⎧0.003⎫ ⎪0.00287⎪ ⎪ ⎪− 0.00010(−0.00048)⎪ ⎪ 0.001⎪ 2⎪ ⎪⎪ ⎪⎪ ⎪⎪− 0.00010 ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ = ⎨0.00578⎬ + ⎨ 0.00000 ⎬3.92 2 − 2⎨ 0.00000(0.01304) ⎬ = ⎨0.006⎬ ⎪ − 0.00014(0.03237) ⎪ ⎪0.007 ⎪ ⎪0.00860⎪ ⎪− 0.000142 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ 0.00002(0.04548) ⎪⎭ ⎪⎩ 0.011⎪⎭ ⎪⎩0.01038⎪⎭ ⎪⎩ 0.000022 ⎪⎭ and thus ⎧0.027 ⎫ ⎪0.054⎪ ⎪⎪ ⎪⎪ max{u (t )} = ⎨0.076⎬ m ⎪0.093⎪ ⎪ ⎪ ⎩⎪0.102⎭⎪

Problem 10.10 Figure P10.10a shows the plan view of a one-story building formed with a perfectly rigid roof slab and four columns. The roof height is 12 ft and the columns are rigidly attached to the roof slab and fixed at their other end. The columns have a circular cross section with the flexural rigidities indicated in Figure P10.10a, where EI = 2592 kip-ft2. The roof supports a uniformly distributed vertical load, self weight included, of 144 lb/ft2. Determine using the response spectrum method and considering the degrees of freedom indicated in Figure P10.10a the maximum acceleration response of the building when the base of the building is subjected to the ground motion recorded during the 1952 Taft earthquake. Figure P10.10b shows the response spectra for the three components of this ground motion. Assume a damping ratio of 2% for all modes.

y 4EI

2EI uy(t) uθ(t)

25 ft

c.m.

2EI

ux(t)

x

EI 25 ft

Figure P10.10a. (a) Plan view of one-story structure in Problem 10.10 and (b) elastic response spectra for 2% damping of the three components of ground acceleration recorded at Taft Lincoln School Tunnel during the 1952 Kern County, California, earthquake Solution:

Stiffness matrix Using the definition of stiffness influence coefficients and referring to the three degrees of freedom indicated in Figure P10.10a, one has that k xx = k yy =

9(12 EI ) 9(12 × 2592) = = 162 kip/ft H3 123

99

kθθ

12(18EI ) L2 216(2592) 252 = = = 50,625 kip/ft 4 4 123 H3 k xy = k yx = 0

k xθ = kθx = k yθ = kθy = −

3(12 EI ) L 36(2592) 25 =− = −675 kip/ft 3 2 2 123 H

and thus the stiffness matrix of the system is − 675 ⎤ u x 0 ⎡ 162 ⎢ [K ] = ⎢ 0 162 − 675 ⎥⎥ u y ⎢⎣− 675 − 675 50625⎥⎦ uθ

Mass matrix The total mass of the system is equal to 0.144(25)(25)/32.2 = 2.8 kip-s2/ft and the corresponding mass moment of inertia is equal to 2.8 (252 +252)/12 = 291.67 kip-ft-s2. Hence, the mass matrix of the system is given by 0 ⎤ ux ⎡2.8 0 ⎢ [ M ] = ⎢ 0 2.8 0 ⎥⎥ u y ⎢⎣ 0 0 291.67⎥⎦ uθ

Natural frequencies and mode shapes From the solution of the eigenvalue problem, the natural frequencies and mode shapes of the system are ω1 = 6.993 rad/s ⎧0.408⎫ ⎪ ⎪ {φ}1 = ⎨0.408⎬ ⎪0.015⎪ ⎩ ⎭

ω2 = 7.606 rad/s ⎧− 0.423⎫ ⎪ ⎪ {φ}2 = ⎨0.423 ⎬ ⎪0 ⎪ ⎭ ⎩

ω3 = 13.510 rad/s ⎧− 0.109⎫ ⎪ ⎪ {φ}3 = ⎨− 0.109⎬ ⎪0.057 ⎪ ⎩ ⎭

Influence vectors According to the definition introduced in Section 10.8.2, the influence vectors are ⎧1 ⎫ ⎪ ⎪ {J } x = ⎨0⎬ ⎪0⎪ ⎩ ⎭

⎧0⎫ ⎪ ⎪ {J } y = ⎨1 ⎬ ⎪0⎪ ⎩ ⎭

Participation factors Since T

0 ⎤ ⎧0.408⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨0.408⎬ = 1.0 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0.015⎪⎭

T

0 ⎤ ⎧− 0.423⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨0.423 ⎬ = 1.0 ⎢ ⎪ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0 ⎭

T

0 ⎤ ⎧− 0.109⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨− 0.109⎬ = 1.0 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0.057 ⎪⎭

⎧0.408⎫ ⎪ ⎪ T * M 1 = {φ}1 [ M ]{φ}1 = ⎨0.408⎬ ⎪0.015⎪ ⎩ ⎭

⎧− 0.423⎫ ⎪ ⎪ T * M 2 = {φ}2 [ M ]{φ}2 = ⎨0.423 ⎬ ⎪0 ⎪ ⎩ ⎭ ⎧− 0.109⎫ ⎪ ⎪ T * M 3 = {φ}3 [ M ]{φ}3 = ⎨− 0.109⎬ ⎪0.057 ⎪ ⎩ ⎭

then according to Equation 10.121 the modal participation factors of the system are

100

Γ1x =

{φ}1T [ M ]{J }x M 1*

T

0 ⎤ ⎧1⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨0⎬ = 1.143 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

T

0 ⎤ ⎧1⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨0⎬ = −1.183 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

T

0 ⎤ ⎧1⎫ ⎡ 2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨0⎬ = −0.306 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

T

0 ⎤ ⎧0⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨1⎬ = 1.143 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

⎧0.408⎫ 1 ⎪ ⎪ = ⎨0.408⎬ 1.0 ⎪ ⎪ ⎩0.015⎭

Γ2 x =

{φ}T2 [ M ]{J }x M 2*

⎧− 0.423⎫ 1 ⎪ ⎪ = ⎨0.423 ⎬ 1.0 ⎪ ⎪ ⎩0 ⎭

Γ3 x =

{φ}T3 [ M ]{J }x M 3*

⎧− 0.109⎫ 1 ⎪ ⎪ = ⎨− 0.109⎬ 1.0 ⎪ ⎪ ⎩0.057 ⎭

{φ}1T [ M ]{J }y M 1*

⎧0.408⎫ 1 ⎪ ⎪ = ⎨0.408⎬ 1.0 ⎪ ⎪ ⎩0.015⎭

{φ}T2 [ M ]{J }y M 2*

⎧− 0.423⎫ 1 ⎪ ⎪ = ⎨0.423 ⎬ 1.0 ⎪ ⎪ ⎩0 ⎭

Γ1 y =

T

Γ2 y =

T

Γ3 y =

{φ}T3 [ M ]{J } y M 3*

⎧− 0.109⎫ 1 ⎪ ⎪ = ⎨− 0.109⎬ 1.0 ⎪ ⎪ ⎩0.057 ⎭

0 ⎤ ⎧0⎫ ⎡ 2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨1⎬ = 1.183 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

0 ⎤ ⎧0⎫ ⎡2.8 0 ⎪ ⎪ ⎢ 0 2.8 0 ⎥⎥ ⎨1⎬ = −0.306 ⎢ ⎢⎣ 0 0 291.67⎥⎦ ⎪⎩0⎪⎭

Response spectrum ordinates From the response spectra in Figure P10.10b, the spectral ordinates for the N21E component that correspond to the natural frequencies and damping ratios of the system are SAN 21E (6.993,0.02) = 0.31g

SAN 21E (7.606,0.02) = 0.49 g

SAN 21E (13.510,0.02) = 0.56 g

and similarly for the S69E component, these ordinates are SAS 69 E (6.993,0.02) = 0.31g

SAS 69 E (7.606,0.02) = 0.34 g

SAS 69 E (13.510,0.02) = 0.45 g

To follow the notation introduced in Section 10.8.7, it will be considered that the N21E component is applied along the principal direction 1 and the S69E component along the principal direction 2. Maximum modal responses under single component ground motion If the N21E component is applied in the direction of the x axis, then by virtue of Equation 10.137 the modal accelerations under this component of motion are equal to ⎧0.145⎫ ⎧0.408⎫ ⎪ ⎪ ⎪ ⎪ {R1x1} = max{&y&(t )}1x1 = Γ1x {φ}1 SAN 21E (ω1 , ξ1 ) = 1.143⎨0.408⎬0.31g = ⎨0.145⎬ g ⎪0.005⎪ ⎪0.015⎪ ⎩ ⎭ ⎩ ⎭ ⎧ 0.245 ⎫ ⎧− 0.423⎫ ⎪ ⎪ ⎪ ⎪ {R1x 2 } = max{&y&(t )}1x 2 = Γ2 x {φ}2 SAN 21E (ω2 , ξ 2 ) = −1.183⎨0.423 ⎬0.49 g = ⎨− 0.245⎬ g ⎪ 0 ⎪ ⎪0 ⎪ ⎩ ⎭ ⎩ ⎭ {R1x 3} = max{&y&(t )}1x 3

⎧− 0.109⎫ ⎧ 0.019 ⎫ ⎪ ⎪ ⎪ ⎪ = Γ3 x {φ}3 SAN 21E (ω3 , ξ3 ) = −0.306⎨− 0.109⎬0.56 g = ⎨ 0.019 ⎬ g ⎪0.057 ⎪ ⎪− 0.010⎪ ⎩ ⎭ ⎩ ⎭

101

Similarly, if the same component is applied along the y axis, the corresponding modal accelerations are ⎧0.145⎫ ⎧0.408⎫ ⎪ ⎪ ⎪ ⎪ {R1 y1} = max{&y&(t )}1 y1 = Γ1 y {φ}1 SAN 21E (ω1 , ξ1 ) = 1.143⎨0.408⎬0.31g = ⎨0.145⎬ g ⎪0.005⎪ ⎪0.015⎪ ⎩ ⎭ ⎩ ⎭ ⎧− 0.245⎫ ⎧− 0.423⎫ ⎪ ⎪ ⎪ ⎪ {R1 y 2 } = max{&y&(t )}1 y 2 = Γ2 y {φ}2 SAN 21E (ω2 , ξ 2 ) = 1.183⎨0.423 ⎬0.49 g = ⎨ 0.245 ⎬ g ⎪ 0 ⎪ ⎪0 ⎪ ⎩ ⎭ ⎩ ⎭ {R1 y 3} = max{&y&(t )}1 y 3

⎧− 0.109⎫ ⎧ 0.019 ⎫ ⎪ ⎪ ⎪ ⎪ = Γ3 y {φ}3 SAN 21E (ω3 , ξ3 ) = −0.306⎨− 0.109⎬0.56 g = ⎨ 0.019 ⎬ g ⎪0.057 ⎪ ⎪− 0.010⎪ ⎩ ⎭ ⎩ ⎭

Moreover, if the S69E component is applied along the direction of the x axis, the modal accelerations result as ⎧0.145⎫ ⎧0.408⎫ ⎪ ⎪ ⎪ ⎪ {R2 x1} = max{&y&(t )}2 x1 = Γ1x {φ}1 SAS 69 E (ω1 , ξ1 ) = 1.143⎨0.408⎬0.31g = ⎨0.145⎬ g ⎪0.005⎪ ⎪0.015⎪ ⎩ ⎭ ⎭ ⎩ ⎧ 0.170 ⎫ ⎧− 0.423⎫ ⎪ ⎪ ⎪ ⎪ {R2 x 2 } = max{&y&(t )}2 x 2 = Γ2 x {φ}2 SAS 69 E (ω2 , ξ 2 ) = −1.183⎨0.423 ⎬0.34 g = ⎨− 0.170⎬ g ⎪ 0 ⎪ ⎪0 ⎪ ⎩ ⎭ ⎭ ⎩ ⎧− 0.109⎫ ⎧ 0.015 ⎫ ⎪ ⎪ ⎪ ⎪ {R2 x 3} = max{&y&(t )}3 x = Γ3 x {φ}3 SAS 69 E (ω3 , ξ3 ) = −0.306⎨− 0.109⎬0.45 g = ⎨ 0.015 ⎬ g ⎪0.057 ⎪ ⎪− 0.008⎪ ⎩ ⎭ ⎩ ⎭

and if applied along the direction of the y axis, as ⎧0.145⎫ ⎧0.408⎫ ⎪ ⎪ ⎪ ⎪ {R2 y1} = max{&y&(t )}2 y1 = Γ1 y {φ}1 SAS 69 E (ω1 , ξ1 ) = 1.143⎨0.408⎬0.31g = ⎨0.145⎬ g ⎪0.005⎪ ⎪0.015⎪ ⎩ ⎭ ⎭ ⎩ ⎧− 0.245⎫ ⎧− 0.423⎫ ⎪ ⎪ ⎪ ⎪ {R2 y 2 } = max{&y&(t )}2 y 2 = Γ2 y {φ}2 SAS 69 E (ω2 , ξ 2 ) = 1.183⎨0.423 ⎬0.34 g = ⎨ 0.245 ⎬ g ⎪ 0 ⎪ ⎪0 ⎪ ⎩ ⎭ ⎩ ⎭ ⎧− 0.109⎫ ⎧ 0.019 ⎫ ⎪ ⎪ ⎪ ⎪ {R2 y 3} = max{&y&(t )}2 y 3 = Γ3 y {φ}3 SAS 69 E (ω3 , ξ3 ) = −0.306⎨− 0.109⎬0.45 g = ⎨ 0.019 ⎬ g ⎪0.057 ⎪ ⎪− 0.010⎪ ⎭ ⎩ ⎭ ⎩ Modal correlation coefficients The system possesses closely spaced natural frequencies. Hence, it is necessary to use the double-sum rule to combine the modal responses and one of the formulas introduced in Section 10.3.2 to determine the modal correlations coefficients in it. For simplicity, the expression developed by Der Kiureghian for equal damping ratios (Equation 10.73) will be employed. Accordingly, since all the damping ratios are equal to 0.02, and since r12 = ω1/ω2 = 6.993/7.606 = 0.919; r13 = ω1/ω3 = 6.993/13.510 = 0.518; and r23 = ω2/ω3 = 7.606/13.510 = 0.563; such modal correlation coefficients are equal to

102

8ξ 2 (1 + r12 )r123 / 2 8(0.02) 2 (1 + 0.919)(0.919)3 / 2 = 0.18 = ρ12 = ρ21 = (1 − r122 ) 2 + 4ξ 2 r12 (1 + r12 ) 2 [1 − (0.919) 2 ]2 + 4(0.02) 2 (0.919)(1 + 0.919) 2 ρ13 = ρ31 =

8ξ2 (1 + r13 )r133 / 2 8(0.02) 2 (1 + 0.518)(0.518)3 / 2 = 0.00 = (1 − r132 ) 2 + 4ξ 2 r13 (1 + r13 ) 2 [1 − (0.518) 2 ]2 + 4(0.02) 2 (0.518)(1 + 0.518) 2

8ξ 2 (1 + r23 ) r233 / 2 8(0.02) 2 (1 + 0.563)(0.563)3 / 2 = 0.00 = ρ23 = ρ32 = (1 − r232 ) 2 + 4ξ 2 r23 (1 + r23 ) 2 [1 − (0.563) 2 ]2 + 4(0.02) 2 (0.563)(1 + 0.563) 2

Square of maximum responses under single component ground motion According to Equations 10.148 through 10.153 and the above modal correlation coefficients, the square of the maximum acceleration responses when each of the ground motion components is applied by itself along the direction of the x or y axis result equal to {R12x } = {R12x1 + R12x 2 + R12x 3 + 2ρ12 R1x1R1x 2 } ⎧0.1452 ⎫ ⎧ 0.2452 ⎫ ⎧ 0.019 2 ⎫ ⎧ (0.145)(0.245) ⎫ ⎧ 0.0942 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪ 2⎪ 2 ⎪ = ⎨0.145 ⎬ + ⎨− 0.245 ⎬ + ⎨ 0.019 ⎬ + 2(0.18)⎨(0.145)(−0.245)⎬ = ⎨ 0.0942 ⎬ ⎪0.0052 ⎪ ⎪ 0 ⎪ ⎪− 0.0102 ⎪ ⎪ (0.005)(0) ⎪ ⎪0.000125⎪ ⎭ ⎭ ⎩ ⎩ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ {R12y } = {R12y1 + R12y 2 + R12y 3 + 2ρ12 R1 y1R1 y 2 }

⎧0.1452 ⎫ ⎧− 0.2452 ⎫ ⎧ 0.019 2 ⎫ ⎧(0.145)(−0.245)⎫ ⎧ 0.0686 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪ 2 ⎪ 2 ⎪ = ⎨0.145 ⎬ + ⎨ 0.245 ⎬ + ⎨ 0.019 ⎬ + 2(0.18)⎨ (0.145)(0.245) ⎬ = ⎨ 0.0942 ⎬ ⎪0.0052 ⎪ ⎪ 0 ⎪ ⎪− 0.0102 ⎪ ⎪ (0.005)(0) ⎪ ⎪0.000125⎪ ⎭ ⎭ ⎩ ⎩ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ {R22x } = {R22x1 + R22x 2 + R22x 3 + 2ρ12 R2 x1R2 x 2 } = ⎧0.1452 ⎫ ⎧ 0.170 2 ⎫ ⎧ 0.0152 ⎫ ⎧ (0.145)(0.170) ⎫ ⎧ 0.0590 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪ 2⎪ 2 ⎪ = ⎨0.145 ⎬ + ⎨− 0.170 ⎬ + ⎨ 0.015 ⎬ + 2(0.18)⎨(0.145)(−0.170)⎬ = ⎨ 0.0590 ⎬ ⎪0.0052 ⎪ ⎪ 0 ⎪ ⎪− 0.0082 ⎪ ⎪ (0.005)(0) ⎪ ⎪0.000089⎪ ⎭ ⎭ ⎩ ⎩ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ {R22y } = {R22y1 + R22y 2 + R22y 3 + 2ρ12 R2 y1 R2 y 2 }

⎧0.1452 ⎫ ⎧− 0.2452 ⎫ ⎧ 0.019 2 ⎫ ⎧(0.145)(−0.245)⎫ ⎧ 0.0686 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪ 2 ⎪ 2 ⎪ = ⎨0.145 ⎬ + ⎨ 0.245 ⎬ + ⎨ 0.019 ⎬ + 2(0.18)⎨ (0.145)(0.245) ⎬ = ⎨ 0.0942 ⎬ ⎪0.0052 ⎪ ⎪ 0 ⎪ ⎪− 0.0102 ⎪ ⎪ (0.005)(0) ⎪ ⎪0.000125⎪ ⎭ ⎭ ⎩ ⎩ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ {R12xy } = {R1x1R1 y1 + R1x 2 R1 y 2 + R1x 3 R1 y 3 + ρ12 R1x1R1 y 2 + ρ21R1x 2 R1 y1}

⎧(0.145)(−0.245)⎫ ⎧ (0.145)(145) ⎫ ⎧(0.245)(−0.245)⎫ ⎧ (0.019)(0.019) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = ⎨(0.145)(0.145)⎬ + ⎨( −0.245)(0.245)⎬ + ⎨ (0.019)(0.019) ⎬ + 0.18⎨ (0.145)(0.245) ⎬ ⎪ (0.005)(0) ⎪ ⎪ ⎪(−0.010)(−0.010) ⎪ ⎪(0.005)(0.005)⎪ ⎪ (0)(0) ⎩ ⎭ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩ ⎧ (0.245)(0.145) ⎫ ⎧ − 0.0386 ⎫ ⎪ ⎪ ⎪ ⎪ + 0.18⎨( −0.245)(0.145)⎬ = ⎨ − 0.0386 ⎬ ⎪ (0)(0.005) ⎪ ⎪0.000125⎪ ⎭ ⎩ ⎭ ⎩ {R22xy } = {R2 x1R2 y1 + R2 x 2 R2 y 2 + R2 x 3 R2 y 3 + ρ12 R2 x1R2 y 2 + ρ 21R2 x 2 R2 y1}

⎧(0.145)(−0.245)⎫ ⎧(0.145)(0.145)⎫ ⎧(0.170)(−0.245)⎫ ⎧ (0.015)(0.019) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = ⎨(0.145)(0.145)⎬ + ⎨(−0.170)(0.245)⎬ + ⎨ (0.015)(0.019) ⎬ + 0.18⎨ (0.145)(0.245) ⎬ ⎪ (0.005)(0) ⎪ ⎪ ⎪(−0.008)(−0.010) ⎪ ⎪(0.005)(0.005)⎪ ⎪ (0)(0) ⎩ ⎭ ⎭ ⎩ ⎭ ⎩ ⎭ ⎩

103

⎧ (0.170)(0.145) ⎫ ⎧ − 0.0223 ⎫ ⎪ ⎪ ⎪ ⎪ + 0.18⎨( −0.170)(0.145) ⎬ = ⎨ − 0.0184 ⎬ ⎪ (0)(0.005) ⎪ ⎪0.000105⎪ ⎭ ⎩ ⎭ ⎩

Maximum acceleration response According to Equation 10.159, the maximum acceleration response of the system results thus equal to ⎧u&&x (t ) + u&&gx (t ) ⎫ ⎪ ⎪ max{&y&(t )} = max ⎨u&&y (t ) + u&&gy (t )⎬ ⎪ ⎪ u&&θ (t ) ⎩ ⎭ 1 1 = {[ ( R12x + R22y + R12y + R22x ) + [ ( R12x + R22y − R12y − R22x )]2 + ( R12xy − R22xy ) 2 ]1 / 2 } 2 2 ⎧[ 1 (0.0942 + 0.0686 + 0.0686 + 0.0590) + [ [ 1 (0.0942 + 0.0686 − 0.0686 − 0.0590)]2 + (−0.0386 + 0.0223) 2 ]1 / 2 ⎫ 2 ⎪ 2 ⎪ ⎪1 2 2 1/ 2 ⎪ 1 = ⎨[ 2 (0.0942 + 0.0942 + 0.0942 + 0.0590) + [ [ 2 (0.0942 + 0.0942 − 0.0942 − 0.0590)] + (−0.0386 + 0.0184) ] ⎬ ⎪ ⎪ −3 10 −3 [ 12 (0.125 + 0.125 − 0.125 − 0.089)]2 + (0.125 − 0.105) 2 ]1 / 2 ⎪ [ 2 (0.125 + 0.125 + 0.125 + 0.089) + [10 ⎪ ⎩ ⎭

⎧ ⎫ 0.1452 + 0.0240 ⎧ 0.411⎫ ⎪ ⎪ ⎪ ⎪ 0.1708 + 0.0268 =⎨ ⎬ g = ⎨0.444⎬ g ⎪0.016⎪ ⎪ (0.2320 + 0.0269) × 10− 3 ⎪ ⎩ ⎭ ⎩ ⎭

Problem 10.11 The industrial pipe shown in Figure P10.11 is clamped at its two ends and has a 90° bend at midlength. It supports two heavy valves of mass m each as shown in the figure. Neglecting axial deformations and the pipe mass, determine the pipe’s acceleration response when its left and right supports are respectively subjected to horizontal ground displacements ug1(t) and ug2(t). The ground displacement ug2(t) is equal to the ground displacement ug1(t) except that it begins τ seconds later. Express the answers in terms of the mass m, flexural rigidity EI, length L, phase lag τ, and &z&r1 (t ), the acceleration response of a single-degree of freedom system to the ground acceleration üg1(t). Assume the pipe has no damping. y2 (t)

EI EI

L/2

ug2(t)

y1 (t)

m L/2

EI m

EI ug1(t) L/2

L/2

Figure P10.11. Two-dimensional pipe in Problem 10.11

Solution: With reference to the global system of coordinates shown in Figure P10.11b, the stiffness matrices of the two vertical and two horizontal structural elements are given by

104

y

4 2

7 B

6 D

C 5

9

1

A 8

3 x Figure P10.11. Global system of coordinated and degrees of freedom in pipe in Problem 10.11 6L ⎡ 12 ⎢ 6 L 4 L2 EI [ K ] A = [ K ]B = 3 ⎢ L ⎢ − 12 6 L ⎢ 2 ⎣− 6 L 2 L

− 12

− 6L⎤ − 6 L 2 L2 ⎥⎥ 12 6L ⎥ ⎥ 6L 4 L2 ⎦

6 L − 12 6 L ⎤ ⎡ 12 ⎢ 6 L 4 L2 − 6 L 2 L2 ⎥ EI ⎢ ⎥ [ K ]C = [ K ] D = L3 ⎢− 12 − 6 L 12 − 6 L ⎥ ⎢ ⎥ 2 − 6 L 4 L2 ⎦ ⎣ 6L 2L

Structure stiffness matrix With the degrees of freedom numbered as indicated in the same figure, the stiffness matrix of the entire structure (support degrees of freedom included) results thus as 0 0 0 0 0 ⎤1 − 12 − 6L 6L ⎡ 24 ⎢ 0 24 0 0 0 0 6 L ⎥⎥ 2 − 12 − 6L ⎢ ⎢ − 12 0 12 0 0 0 0 ⎥3 − 6L − 6L ⎥ ⎢ 0 12 0 0 0 − 12 − 6L − 6L⎥ 4 ⎢ 0 EI [K ] = 3 ⎢ 0 0 0 8L2 0 2 L2 2 L2 0 ⎥5 − 6L ⎥ L ⎢ 2 2 0 0 0 8L 2L 0 2 L2 ⎥ 6 − 6L ⎢ 0 ⎢− 6 L − 6 L 0 0 2 L2 2 L2 8L2 0 0 ⎥7 ⎥ ⎢ 0 0 2 L2 0 0 4 L2 0 ⎥8 − 6L ⎢ 6L ⎥ ⎢ 6L 0 0 2 L2 0 0 4 L2 ⎦ 9 − 6L ⎣ 0

which, after eliminating the massless unconstrained rotational degrees of freedom (degrees of freedom 5, 6 and 7) through a static condensation, may be reduced to 36 156 L − 12 L ⎤ 1 ⎡ 528 − 144 − 300 ⎢ − 144 528 36 − 300 − 12 L 156 L ⎥⎥ 2 ⎢ 36 201 123L 3L ⎥ 3 −9 EI ⎢ − 300 [ Kˆ ] = ⎥ ⎢ 201 3L − 300 −9 − 123L ⎥ 4 28 L3 ⎢ 36 ⎢ 156 L − 12 L − 123L 3L 97 L2 − L2 ⎥ 8 ⎥ ⎢ 3L 97 L2 ⎥⎦ 9 − 123L − L2 ⎢⎣− 12 L 156 L 105

Hence, if it is considered that the supports are fixed against rotations and thus that the rotations at degrees of freedom 8 and 9 are equal to zero, the stiffness matrix that relates the forces and displacements at the first four degrees of freedom results as 36 ⎤ 1 ⎡ 528 − 144 − 300 ⎢ − 144 528 36 − 300⎥⎥ 2 EI ⎢ ~ [K ] = 36 201 −9 ⎥3 28 L3 ⎢− 300 ⎥ ⎢ 201 ⎦ 4 − 300 − 9 ⎣ 36

Hence, the submatrices [K], [Kg] and [Kgg] defined by Equation 10.165 are equal to ⎡ 11 − 3⎤ ⎢− 3 11 ⎥ ⎣ ⎦ 36 ⎤ 3EI ⎡− 25 3 ⎤ EI ⎡− 300 [K g ] = = 3⎢ ⎥ 3 ⎢ − 300⎦ 7 L ⎣ 3 − 25⎥⎦ 28L ⎣ 36 EI ⎡201 − 9 ⎤ 3EI ⎡ 67 − 3⎤ [ K gg ] = ⎢ ⎥= ⎢ ⎥ 28L3 ⎣ − 9 201⎦ 28L3 ⎣− 3 67 ⎦ [K ] =

EI ⎡ 528 − 144⎤ 12 EI ⎢ ⎥= 28 L3 ⎣− 144 528 ⎦ 7 L3

Influence matrix Since [ K ]−1 =

L3 192 EI

⎡11 3 ⎤ ⎢ 3 11⎥ ⎣ ⎦

then, according to Equation 10.174, the influence matrix of the system is equal to [ L] = −[ K ]−1[ K g ] = −

L3 3EI 192 EI 7 L3

3 ⎤ 1 ⎡19 3 ⎤ ⎡11 3 ⎤ ⎡− 25 = ⎢ 3 11⎥ ⎢ 3 − 25⎥⎦ 32 ⎢⎣ 3 19⎥⎦ ⎣ ⎦⎣

and the influence vectors are {J }1 =

1 ⎧19⎫ ⎨ ⎬ 32 ⎩3 ⎭

{J }2 =

1 ⎧3 ⎫ ⎨ ⎬ 32 ⎩19⎭

Mass matrix The mass matrix corresponding to degrees of freedom 1 and 2 is equal to ⎡1 0⎤ [M ] = m⎢ ⎥ ⎣0 1 ⎦

Equation of motion If damping is neglected, then the equation of motion is of the form [ M ]{u&&} + [ K ]{u} = −[ M ][ L]{u&&g (t )}

Hence, after substituting the mass, stiffness, and influence matrices determined above, it becomes ⎡1 0⎤ ⎧u&&1 (t ) ⎫ 12 EI m⎢ ⎬+ ⎥⎨ 3 ⎣0 1⎦ ⎩u&&2 (t ) ⎭ 7 L

⎡ 11 − 3⎤ ⎧u1 (t ) ⎫ m ⎡1 0⎤ ⎡19 3 ⎤ ⎧⎪u&&g1 (t ) ⎫⎪ m ⎡19 3 ⎤ ⎧⎪u&&g1 (t ) ⎫⎪ = − = − ⎬ ⎬ ⎢− 3 11 ⎥ ⎨u (t )⎬ ⎢ ⎥⎢ ⎥⎨ ⎢ ⎥⎨ 32 ⎣0 1⎦ ⎣ 3 19⎦ ⎪⎩u&&g 2 (t )⎪⎭ 32 ⎣ 3 19⎦ ⎪⎩u&&g 2 (t )⎪⎭ ⎣ ⎦⎩ 2 ⎭

which may also be expressed as ⎡1 0⎤ ⎧u&&1 (t ) ⎫ 12 EI ⎢0 1⎥ ⎨u&& (t ) ⎬ + 3 ⎣ ⎦ ⎩ 2 ⎭ 7 mL

⎡ 11 − 3⎤ ⎧u1 (t ) ⎫ 1 ⎧19⎫ 1 ⎧3⎫ ⎢− 3 11 ⎥ ⎨u (t )⎬ = − 32 ⎨ 3 ⎬u&&g1 (t ) − 32 ⎨19⎬u&&g 2 (t ) ⎣ ⎦⎩ 2 ⎭ ⎩ ⎭ ⎩ ⎭

Natural frequencies and mode shapes From the solution of the eigenvalue problem, the natural frequencies and mode shapes of the system are 106

ω1 =

96 EI rad/s 7mL3 ⎧1⎫ {φ}1 = ⎨ ⎬ ⎩1⎭

168 EI rad/s 7 mL3 ⎧1 ⎫ {φ}2 = ⎨ ⎬ ⎩− 1⎭ ω2 =

Participation factors According to Equation 10.182, the participation factors are then equal to T

⎧1⎫ ⎡1 ⎨⎬ ⎢ T {φ}1 [ M ]{J }1 1 ⎩1⎭ ⎣0 Γ11 = = {φ}1T [ M ]{φ}1 32 ⎧1⎫T ⎡1 ⎨⎬ ⎢ ⎩1⎭ ⎣0 T

⎧1 ⎫ ⎡1 ⎨ ⎬ ⎢ T {φ}2 [ M ]{J }1 1 ⎩− 1⎭ ⎣0 Γ21 = = {φ}T2 [ M ]{φ}2 32 ⎧1 ⎫T ⎡1 ⎨ ⎬ ⎢ ⎩− 1⎭ ⎣0 T

⎧1⎫ ⎡1 ⎨⎬ ⎢ T {φ}1 [ M ]{J }2 1 ⎩1⎭ ⎣0 Γ12 = = {φ}1T [ M ]{φ}1 32 ⎧1⎫T ⎡1 ⎨⎬ ⎢ ⎩1⎭ ⎣0 T

⎧1 ⎫ ⎡1 ⎨ ⎬ ⎢ T {φ}2 [ M ]{J }2 1 ⎩− 1⎭ ⎣0 Γ22 = = {φ}T2 [ M ]{φ}2 32 ⎧1 ⎫T ⎡1 ⎨ ⎬ ⎢ ⎩− 1⎭ ⎣0

0⎤ ⎧19⎫ ⎨ ⎬ 1⎥⎦ ⎩ 3 ⎭ 0⎤ ⎧1⎫ ⎨⎬ 1⎥⎦ ⎩1⎭

=

0⎤ ⎧19⎫ ⎨ ⎬ 1⎥⎦ ⎩ 3 ⎭ 0⎤ ⎧ 1 ⎫ ⎨ ⎬ 1⎥⎦ ⎩− 1⎭ 0⎤ ⎧ 3 ⎫ ⎨ ⎬ 1 ⎥⎦ ⎩19⎭ 0⎤ ⎧1⎫ ⎨⎬ 1⎥⎦ ⎩1⎭ 0⎤ ⎧ 3 ⎫ ⎨ ⎬ 1⎥⎦ ⎩19⎭ 0⎤ ⎧ 1 ⎫ ⎨ ⎬ 1⎥⎦ ⎩− 1⎭

11 32

=

=

8 32

11 32

=−

8 32

Acceleration response According to Equation 10.190, the acceleration response of the system is thus given by ⎧ &y&1 (t ) ⎫ ⎨ ⎬= ⎩ &y&2 (t )⎭

2

2

∑∑ Γ {φ} [u&& rl

r

gl (t )

l =1 r =1

+ &z&rl (t )]

= Γ11{φ}1[u&&g1 (t ) + &z&11 (t )] + Γ21{φ}2 [u&&g1 (t ) + &z&21 (t )] + Γ12{φ}1[u&&g 2 (t ) + &z&12 (t )] + Γ22{φ}2 [u&&g 2 (t ) + &z&22 (t )] 8 ⎧1 ⎫ 11 ⎧1⎫ 8 ⎧1 ⎫ 11 ⎧1⎫ ⎨ ⎬[u&&g1 (t ) + &z&11 (t )] + ⎨ ⎬[u&&g1 (t ) + &z&21 (t )] + ⎨ ⎬[u&&g 2 (t ) + &z&12 (t )] − ⎨ ⎬[u&&g 2 (t ) + &z&22 (t )] 32 ⎩1⎭ 32 ⎩− 1⎭ 32 ⎩1⎭ 32 ⎩− 1⎭ However, since u&&g 2 (t ) = u&&g1 (t − τ) , one has that =

&z&12 (t ) = &z&11 (t − τ)

&z&22 (t ) = &z&21 (t − τ)

and thus the acceleration response of the system is given by ⎧ &y&1 (t ) ⎫ 11 ⎧1⎫ 8 ⎧1 ⎫ ⎨ ⎬= ⎨ ⎬[u&&g1 (t ) + &z&11 (t )] + ⎨ ⎬[u&&g1 (t ) + &z&21 (t )] 32 ⎩− 1⎭ ⎩ &y&2 (t )⎭ 32 ⎩1⎭ +

11 ⎧1⎫ 8 ⎧1 ⎫ ⎨ ⎬[u&&g1 (t − τ) + &z&11 (t − τ)] − ⎨ ⎬[u&&g1 (t − τ) + &z&21 (t − τ)] 32 ⎩1⎭ 32 ⎩− 1⎭

107

CHAPTER 12 Problem 12.1 Solve Problem 10.1 using the equivalent lateral force procedure. Solution: Considering as a first approximation the largest ordinate in the specified response spectrum, the base shear of the building under consideration is equal to V=

SA W = 2.5(0.40)(7897.1) = 7897.1 kN g

Similarly, according to Equation 12.19, this base shear, and the floor weights and heights given in Table P10.1, the lateral forces Fsj and the story shears Vj result as listed in Table P12.1. Table P12.1a. Lateral forces and story shears in building considered in Problem 12.1 Floor Wj Wjhj Fsj Vj hj Δhj (kN) (m) (kN-m) (kN) (kN) (m) 5 1402.8 3 16 22445.3 2322.4 2322.4 4 1491.1 3 13 19384.6 2005.7 4328.2 3 1579.4 3 10 15794.1 1634.2 5962.4 2 1667.7 3 7 11673.9 1207.9 7170.3 1 1756.0 4 4 7024.0 726.8 7897.1 Sum 7897.1 76321.8 7897.1

Now, with the given story stiffness and the calculated base shears, the lateral displacements are as indicated in Table P12.1b. Table P12.1.b. Story drifts and lateral displacements in building considered in Problem 12.1 Stiffness Shear Drift Displ. Story (kN/m) (kN) (m) (m) 5 199620 2322.4 0.01 0.12 4 212090 4328.2 0.02 0.11 3 224570 5962.4 0.03 0.09 2 237050 7170.3 0.03 0.06 1 249880 7897.1 0.03 0.03

According to the calculated lateral forces and displacements, one has then that 5

∑w u j

2 j

= 58.48 kN - m 2

j

= 740.12 kN - m

j =1

and 5

∑F u sj

j =1

Thus, according to Equation 12.23, an estimate of the structure’s fundamental natural period is

108

5

∑w u j

T1 = 2π

2 j

j =1 5

g

∑F u sj

= 2π

58.48 = 0.56 s 9.81(740.12)

j

j =1

and from Figure 8.17 the spectral acceleration corresponding to this natural period and Soil Type 3 is 2.5g, which is the value assumed above. Therefore, the desired story shears are as indicated in Table P12.1a. Problem 12.2 Solve Problem 10.2 using the equivalent lateral force procedure. Consider equal story heights of 3.0 m. Solution: Assuming as a first approximation a fundamental natural period equal to the number of floors over 10 (see Equation 17.29), that is, 0.3 seconds, the spectral acceleration corresponding to this natural period in the design spectrum shown in Figure P10.2 is SA = 0.65 g

Thus, the base shear for the building under consideration is V=

SA W = 0.65(54.0) = 35.1 kN g

According to Equation 12.19, this base shear and the given weights and story heights, the lateral forces result then as shown in Table P12.2. Table P12.2a. Lateral forces and story shears in building considered in Problem 12.2 Floor Wjhj Fsj Vj hj Wj Δhj (kN) (m) (kN-m) (kN) (kN) (m) 3 9.8 3 9 88.3 11.7 11.7 2 14.7 3 6 88.3 11.7 23.4 1 29.4 3 3 88.3 11.7 35.1 Sum 54.0 264.9 35.1

Similarly, with the given story stiffness and the calculated base shears, the lateral displacements are as indicated in Table P12.2b. Table 12.2 b. Story drifts and lateral displacements in building considered in Problem 12.2 Stiffness Shear Drift Displ. Story (kN/m) (kN) (m) (m) 3 118.4 11.7 0.10 0.30 236.9 2 23.4 0.10 0.20 355.3 1 35.1 0.10 0.10

Thus, one has that 3

∑w u j

2 j

= 1720.44 kN - m 2

j =1

109

3

∑F u sj

j

= 6923.47 kN - m

j =1

and according to Equation 12.23, an estimate of the fundamental natural period is 3

∑w u j

T1 = 2π

2 j

j =1 3

g

∑F u sj

= 2π

1720.44 = 1.00 s 9.81(6923.47)

j

j =1

which is substantially different from the assumed one. Performing, then, another iteration with this value of the natural period, one obtains SA = 4 g

SA W = 4(54.0) = 216.0 kN g

V= Floor 3 2 1 Sum

Δhj (m) 3 3 3

Wj (kN) 9.8 14.7 29.4 54.0

Story 3 2 1

Wjhj (kN-m) 88.3 88.3 88.3 264.9

hj (m) 9 6 3

Shear (kN) 72.0 144.0 216.0

Stiffness (kN/m) 118.4 236.9 355.3 3

∑w u j

Drift (m) 0.61 0.61 0.61

2 j

= 65.262 kN - m 2

j

= 262.629 kN - m

Fsj (kN) 72.0 72.0 72.0 216.0

Vj (kN) 72.0 144.0 216.0

Displ. (m) 1.82 1.22 0.61

j =1 3

∑F u sj

j =1

3

∑w u j

T1 = 2π

2 j

j =1 3

g

∑F u sj

= 2π

65.262 = 1.00 s 9.81(262.629)

j

j =1

The new value of the natural period is equal to the previous one and consequently no more iterations are needed. Hence, the desired lateral forces are ⎧ Fs1 ⎫ ⎧72.0⎫ ⎪ ⎪ ⎪ ⎪ ⎨ Fs 2 ⎬ = ⎨72.0⎬ kN ⎪ F ⎪ ⎪72.0⎪ ⎩ s3 ⎭ ⎩ ⎭

110

Problem 12.3 Determine using the equivalent lateral force procedure the axial forces, shearing forces, and bending moments of the columns of the first story of the frame building shown in Figure P12.3. The floor weights of the building are as indicated in the figure. The building will be designed to resist ground motions represented by the 2% damping design spectrum shown in Figure P10.2. Assume that the shearing forces at the interior columns are twice those carried by the exterior columns and that the four columns have the same cross-sectional area. 2000 kips 15 ft

2500 kips

15 ft

2500 kips

15 ft

2500 kips

15 ft

20 ft

20 ft

20 ft

Figure P12.3. Frame building considered in Problem 12.3

Solution: Assuming as a first approximation a fundamental natural period equal to the number of floors over 10 (see Equation 17.29), that is, 0.4 seconds, the spectral acceleration corresponding to this natural period in the design spectrum shown in Figure P10.2 is SA = 1.7 g

Thus the base shear for the building under consideration is V=

SA W = 1.7(9500) = 16,150 kips g

Considering Equation 12.19, this base shear and the given weights and story heights, the lateral forces and story shears result then as shown in Table P12.3. Table P12.3. Lateral forces and story shears in building considered in Problem 12.3 Floor Wjhj Fsj Vj hj Wj Δhj (kips) (ft) (kip-ft) (kips) (kips) (ft) 4 2,000 15 60 120,000.0 5,617.4 5,617.4 3 2,500 15 45 112,500.0 5,266.3 10,883.7 2 2,500 15 30 75,000.0 3,510.9 14,394.6 1 2,500 15 15 37,500.0 1,755.4 16,150.0 Sum 9,500 345,000.0 16,150.0

The shearing force in the exterior columns of the first story is thus equal to (Vc1 )int =

V1 16,150 = = 2,692 kips 6 6

and the shearing force in the corresponding interior columns is (Vc1 )ext =

16,150 = 5,383 kips 3

Similarly, the bending moment in the interior columns is ( M c1 )ext = (Vc1 )ext h1 / 2 = 2,692(15) / 2 = 20,190 kip - ft

111

while that in the interior columns is ( M c1 )int = (Vc1 )int h1 / 2 = 5,383(15) / 2 = 40,375 kip - ft

The axial forces may be determined using the well-known flexural formula Pc1 =

M o1 y A I

where Mo1 is the overturning moment at the ground level, I is the centroidal moment of inertia, and y is the distance from the centroid to the column under consideration. Accordingly, since in this case the moment of inertia is given by 2 2 2 2 I = 2 A( yint ) + 2 A( yext ) = 2 A( yint + yext ) = 2 A(10 2 + 30 2 ) = 2,000 A

and the overturning moment by M o1 = 3,400(60) + 4,250(45 + 30 + 15) = 586,500 kip - ft

the axial force in the interior columns is equal to ( Pc1 )int =

M o1 yint 586,500(10) A= = 2,932.5 kips I 2000

and the axial force in the exterior columns equal to ( Pc1 ) ext =

M o1 yext 586,500(30) A= = 8,797.5 kips I 2000

Problem 12.4 A two-story shear-wall structure has the dimensions and configuration shown in Figure P12.4. Each story is 12 ft high and the thickness of all walls is 8 in. The floor weights are 350 and 250 kips, respectively for the first and second levels. Determine using the equivalent lateral force procedure the shearing forces exerted on the longitudinal shear walls of the first story when the structure is subjected to a ground motion along its longitudinal direction represented by the 5% damping design spectrum shown in Figure P10.2. Assume the lateral stiffness of shear walls is equal to GA/h, where G is shear modulus of elasticity, A is cross section area, and h is height. Y

6 ft

6 ft

12 ft

8 ft

40 ft 8 ft

12 ft

24 ft

X

24 ft

60 ft

Figure P12.4. Floor plan of two-story shear wall structure considered in Problem 12.4

Solution: Base shear Using Equation 17.28 with Ct = 0.020 and x = 0.75, an estimate of the structure’s fundamental natural period is Ta = Ct hnx = 0.020( 24)0.75 = 0.22 s

Therefore, from the design spectrum in Figure P10.2, the spectral acceleration is SA(0.22,0.05) = 0.3 g

and the corresponding base shear is

112

V=

SA W = 0.3(350 + 250) = 180 kips g

Lateral forces and story shears For this base shear and the given floor weights and story heights, the lateral forces and story shear results then as indicated in Table P12.4. Table P12.4. Lateral forces and story shears in building considered in Problem 12.4 Floor Wjhj Fsj Vj hj Wj Δh j (kip) (ft) (kip-ft) (kip) (kip) (m) 2 250 12 24 6000.0 105.9 105.9 1 250 12 12 4200.0 74.1 180.0 Sum 600 10200.0 180.0

Horizontal distribution of story shears It may be noted that because the lateral forces are applied at the floor’s centers of mass, the two story shears will pass through these centers. However, the story shears will not be distributed uniformly to the shear walls since the structure is asymmetric and will be thus subjected to translational and rotational motions. To determine the shear forces induced in the walls, it is therefore necessary to determine the location of the center of twist, the magnitude of the torsional moments, and the shearing forces induced by the translational and torsional motions considered separately. Location of center of twist If it is assumed that the lateral stiffness of the shear walls is equal to GA/h, where G is shear modulus of elasticity, A is cross section area, and h is height, then for both stories the lateral stiffness of the long longitudinal, short longitudinal, long transverse, and short transverse walls are respectively equal to 48(8 / 12) G = 2.67G 12 12(8 / 12) = G = 0.67G 12 24(8 / 12) = G = 1.33G 12 16(8 / 12) = G = 0.89G 12

K LL = K SL

K LT K ST

If the structure is subjected now to a longitudinal shear Vi applied at the center of twist of any of the two floors, a uniform displacement Δi is induced and reaction forces appear in the longitudinal walls as shown in Figure P12.4b. Then the equilibrium of moments about the center of mass yields, Vi e y = K LL Δ i (20) − K SL Δ i ( 20)

Similarly, the equilibrium of forces in the longitudinal direction leads to Vi = ( K LL + K SL )Δ i

Therefore, from these two equations, it is found that the distance from the center of mass to the center of twist in the transverse direction (eccentricity along the y axis) is equal to ey =

20( K LL − K SL ) 20(2.67 − 0.67) = = 12.0 ft K LL + K SL 2.67 + 0.67

113

Y

Δi

KSL Δi

C.M. X ey

C.T. Vi

KLL Δi

Figure P12.4b. Lateral displacement and reactive forces in walls

Torsional moments The torsional moments induced by this eccentricity in the second and first stories are thus respectively equal to M T 2 = V2e y = 105.9(12) = 1,270.8 kip - ft M T 1 = V1e y = 180.0(12) = 2,160.0 kip - ft

Shearing forces induced by translational motion alone If the story shear in the ith story is now considered applied at the center of twist, this story shear will induce in the ith story a uniform lateral displacement Δi and reacting shearing forces in the longitudinal walls (see Figure P12.4b). Considering the equilibrium of forces in the longitudinal direction, the magnitude of this displacement is Δi =

Vi K LL + K SL

Hence, the shearing forces induced by the translational motion in the long and short longitudinal walls of the second story are equal to K LL 2.67 V2 = 105.9 = 84.7 kips K LL + K SL 2.67 + 0.67 K SL 0.67 ′ 2 = K SL Δ 2 = VSL V2 = 105.9 = 21.2 kips K LL + K SL 2.67 + 0.67

′ 2 = K LL Δ 2 = VLL

while those in the first story are equal to 2.67 K LL V1 = 180.0 = 144.0 kips K LL + K SL 2.67 + 0.67 K SL 0.67 ′ 1 = K SL Δ1 = VSL V1 = 180.0 = 36.0 kips K LL + K SL 2.67 + 0.67

′ 1 = K LL Δ1 = VLL

Shearing forces induced by torsional motion alone If the torsional moment in the ith story is now applied to the center of mass, this moment will induce a rotational motion and reacting shearing forces in all four walls (see Figure P12.4c). By equilibrium of moments, then, one has that M Ti = K LL Δ LLi y LL + K SL Δ SLi ySL + K LT Δ LTi xLT + K ST Δ STi xST

where ΔLLi, ΔLLi, ΔLLi, and ΔLLi represent the displacements induced in the walls by the torsional moment, and yLL, ySL, xLT, and xST signify the distances from the center of mass to the walls. But from the figure and under the assumption of small angels, the wall displacements are given by Δ LLi = θi y LLi

Δ SLi = θi ySLi

114

Δ LTi = θi xLTi

Δ STi = θi xSTi

where θi denotes the angle of rotation. In consequence, MTi may be written as 2 2 2 2 M Ti = K LL y LL θi + K SL ySL θi + K LT xLT θi + K ST xST θi

from which one obtains M Ti 2 2 + + K LT xLT + K ST xST 1 M Ti = 2 2 G 2.67(20) + 0.67(20) + 1.33(30) 2 + 0.89(30) 2 M Ti = 3334G

θi =

2 K LL y LL

2 K SL ySL

Y

ΔSLi

KSLΔSLi KSTΔSTi

ΔLTi

MTi C.M.

KLTΔLTi

θi

X

θi

ΔSTi KLL ΔLLi

ΔLLi

Figure P12.4c. Rotation motion and reactive forces in walls

The shearing forces induced in the longitudinal walls of the second story by the torsional motion alone are thus equal to K LL yLL 2.67(20) MT 2 = 1270.8 = 20.4 kips 3334G 3334 K y 0.67(20) = K SL ySL θ2 = SL SL M T 2 = 1270.8 = 5.1kips 3334G 3334

′′ 2 = K LL Δ LL 2 = K LL y LL θ2 = VLL ′′ 2 = K SL Δ SL 2 VSL

while those induced in the first story are K LL y LL 2.67(20) M T1 = 2160.0 = 34.6 kips 3334G 3334 K y 0.67(20) ′′ 1 = K SL Δ LL1 = K SL ySL θ1 = SL SL M T 1 = VSL 2160.0 = 8.7 kips 3334G 3334

′′ 1 = K LL Δ LL1 = K LL xLL θ1 = VLL

Total wall shears Adding the shears determined above for the cases of translational motion only and torsional motion only, the shearing forces exerted on the shear walls of the second and first stories are thus equal to ′ 2 + VLL ′′ 2 = 84.7 + 20.4 = 105.1kips VLL 2 = VLL ′ 2 + VSL ′′ 2 = 21.2 + 5.1 = 26.3 kips VSL 2 = VSL ′ 1 + VLL ′′ 1 = 144.0 + 34.6 =178.6 kips VLL1 = VLL ′ 1 + VSL ′′ 1 = 36.0 + 8.7 = 44.7 kips VSL1 = VSL

115

Problem 12.5 As shown in Figure P12.5, a building has four stories, four bays 4-m wide along its X direction, and three bays also 4-m wide along its Y direction. The structure of the building is constructed with steel moment-resisting frames along both of these directions. The four frames along the X direction are all identical to the one shown in Figure P12.5a. These frames are in turn connected at each floor level by beams parallel to the Y axis. The moments of inertia of the columns about their two principal axes are as indicated in Figure P12.5a, where I = 0.750 × 10-4 m4. The moments of inertia of all beams, longitudinal and transverse, are equal to 2I. The modulus of elasticity of the steel used in the construction is 200,000 MN/m2. The floor weights are also as indicated in the figure. Determine using the equivalent lateral force procedure: (a) the equivalent lateral forces acting on the floors of the building along the direction of the Y axis; (b) the lateral displacements induced by these lateral forces in columns D and H along the Y direction at the level of the second floor; (c) the lateral displacements generated by the same set of forces in columns A and D along the X direction, also at the level of the second floor; and (d) the shear forces along the X and Y directions on columns D and H at the level of the first story. The building will be designed to resist ground motions represented by an acceleration response spectrum constructed utilizing the spectral shape corresponding to soil Type 2 in Figure 8.17 and a peak ground acceleration of 0.40 g. Y

Z w4 = 96 kN 2I

2I 2I w3 = 1440 kN

4.0 m

3I

4I 2I w2 = 1920 kN

4.0 m

E

F

H

G

D 4.0 m

2I

4I

3I

5I

C

4I

3I

4.5 m

4.0 m

w1 = 1920 kN B

6I

8I

9I

6I

5I

6.0 m

4.0 m A

X 4.0 m

4.0 m

4.0 m

4.0 m

X 4.0 m

(a) Elevation view

4.0 m

4.0 m

4.0 m

(b) Plan view

Figure P12.5. (a) Elevation and (b) plan views of four-story building considered in Problem 12.5

Solution: Using Equation 17.28 with Ct = 0.072 and x = 0.8, an estimate of the structure’s fundamental natural period is Ta = Ct hnx = 0.072(18.5) 0.8 = 0.74 s

Therefore, for a peak ground acceleration of 0.40g and the ordinate corresponding to a period of 0.74 s and soil Type 2 in Figure 8.17, the design spectral acceleration for the building is SA = 1.96(0.4 g ) = 0.78 g

and the corresponding base shear is V0 =

SA W = 0.78(5376) = 4,215 kN g

Hence, upon the application of Equation 12.19 and organizing the calculations as shown in Table P12.5a, one arrives to the equivalent lateral forces listed in the last column of this table. Using then a computer program for the static analysis of three-dimensional frame structures for the analysis of the structure under the determined lateral forces applied at the floor’s geometric centers, the desired displacements and shearing forces result as shown in the Tables P12.5b and P2.5c.

116

Table P12.5a. Lateral forced in building considered in Problelm 12.5 Floor Wj Wjhj Fsj hj Δhj (kN) (m) (m) (kN-m) (kN) 4 96 4.0 18.5 1776.0 137.8 3 1440.0 4.0 14.5 20880.0 1619.7 2 1920.0 4.5 10.5 20160.0 1563.9 1 1920.0 6.0 6 11520.0 893.6 Sum 5376.0 54336.0 4215.0 Table P12.5b. Lateral displacements at level of second floor induced by equivalent lateral forces Column Displacement X direction Column Displacement Y direction (m) (m) A -0.00574 D 0.131 D 0.00574 H 0.094 Table P12.5c. Shearing forces at columns of first story induced by equivalent lateral forces. Column Shearing force (kN) X direction Y direction D 8.81 146.5 H 7.49 110.7

117

CHAPTER 17 Problem 17.1 Calculate the base shear, story shears, overturning moments, and story drifts for the shear building shown in Figure P17.1 using the equivalent lateral force procedure prescribed by the International Building Code. The building will be used for offices, will be located in Los Angeles, California, over a deposit of stiff soil, and will be constructed with moment-resistant steel frames. Consider m1 = 2 kip-s2/in and k1 = 4000 kip/in. Height (ft)

Level

60

6

1

6 / 21

50

5

1

11 / 21

40

4

1

15 / 21

30

3

1

18 / 21

20

2

1

20 / 21

10

1

1

21 / 21

mi / m1

ki / k1

45 ft

Figure P17.1. Six-story building considered in Problem 17.1

Solution: Building weight

W = 6( 2)386.4 = 4,636.8 kips

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting steel frames, Ta = Ct hnx = 0.028(60)0.8 = 0.74 s

Base shear The latitude and longitude of Los Angeles, California, are: 33° 56' N and 118° 24' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.50 S1 = 0.60

Also, as the building site may be classified as Site D (see Table 17.3), from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.5(0.60) = 0.90

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are S DS =

2 2 S MS = (1.50) = 1.00 3 3

118

S D1 =

2 2 S M 1 = (0.90) = 0.60 3 3

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = II Importance factor = I =1.0 Seismic design category = D Response modification coefficient (for ordinary frames) = R = 3.5 Hence, according to Equation 17.23 Cs =

S DS 1.00 = = 0.29 R / I 3.5 / 1.0

which is greater than (Cs ) min =

0.5S1 0.5(0.60) = = 0.09 R/I 3.5 / 1.0

However, it is greater than (Cs ) max =

S D1 0.60 = = 0.23 T ( R / I ) 0.74(3.5/1.0)

and thus it will be considered that Cs = 0.23

In consequence, the base shear is equal to V = CsW = 0.23( 4636.8) = 1,066.5 kips

Lateral forces and story shears From Table 17.14, one obtains 1 1 k = 1.0 + (T − 0.5) = 1.0 + (0.60 − 0.5) = 1.05 2 2 and thus from Equations 17.32 and 17.33, the lateral forces (Fx), story shears (Vx), and overturning

moments (M0) are as indicate in Table P17.1a. It should be noted that to facilitate the calculations, the overturning moments may be considered given by M 0i = M 0(i +1) + Vi hsx

where M0x and M0(x+1) are respectively the overturning moments on the xth and (x+1)th stories. Table P17.1a. Lateral forces, story shears, and overturning moments in building in Problem 17.1 Level

wx (kip)

hsx (ft)

hx (ft)

wxhxk (kip-ftk)

C vx =

w x h xk N

∑w h

k i i

Fx = C vxV (kip)

Vx (kip)

M0 (kip-ft)

319.60 260.57 202.95 147.05 93.38 42.96 1,066.50

319.60 580.17 783.12 930.16 1023.54 1066.50

3196.0 8997.7 16828.9 26130.5 36365.9 47030.9

i =1

6 5 4 3 2 1 Sum

772.8 772.8 772.8 772.8 772.8 772.8 4,636.8

10 10 10 10 10 10

60 50 40 30 20 10

75787.31 61789.33 48125.39 34869.28 22142.20 10187.49 252,901.00

0.300 0.244 0.190 0.138 0.088 0.040 1.000

Story drifts and lateral displacements Since in this case the story stiffnesses are given, the story drifts may be calculated according to

119

Δx =

Vx kx

The elastic components of the lateral displacements may be calculated by simply adding the corresponding story drifts. According to Equation 17.36, the total displacements are given by δx =

C d δ xe I

where from Table 17.1 for the building under consideration Cd = 3.0

and, as determined earlier, I = 1.0. The results are presented in Table P17.1b, where Δx denotes story drift, δxe elastic displacement, and δx total displacement. Table P17.1b. Lateral displacements and story drifts in building in Problem 17.1 Level Vx kx δ xe δx Δx (kip) (kip/in) (in) (in) (in) 6 319.60 1142.86 0.280 1.637 4.912 5 580.17 2095.24 0.277 1.358 4.073 4 783.12 2857.14 0.274 1.081 3.242 3 930.16 3428.57 0.271 0.807 2.420 2 1023.54 3809.52 0.269 0.535 1.606 1 1066.50 4000.00 0.267 0.267 0.800

Problem 17.2 Compute the lateral displacements and the overturning moment at the base of the two-story shear building shown in Figure P17.2 using the equivalent lateral force procedure prescribed by the International Building Code. The building will be located in San Francisco, California, over a deep deposit of soft clay, will be used as a hospital, and will be structured with moment-resisting reinforced concrete frames. Determine the fundamental natural period of the building using an eigenvalue analysis. m2 = 2.5 kip-s2/in k2 = 500 kip/in 10 ft 2

m1 = 3.75 kip-s /in

k1 = 1000 kip/in

12 ft

10 ft

Figure P17.2. Two-story building considered in Problem 17.2

Solution: Building weight

W = (3.75 + 2.5)386.4 = 2415 kips

Natural period From an eigenvalue analysis, the building’s fundamental natural period is T = 0.60 s

and from Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting concrete frames, the approximate value of the fundamental natural period is

120

Ta = Ct hnx = 0.016(22) 0.9 = 0.26 s

Base shear The latitude and longitude of San Francisco, California, are 37° 46' N and 122° 26' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.50 S1 = 0.70

Also, as the building site may be classified as Site E (see Table 17.3), from Tables 17.4 and 17.5, one obtains, Fa = 0.9

Fv = 2.4

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 0.9(1.50) = 1.35 S M 1 = Fv S1 = 2.4(0.70) = 1.68

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 S MS = 3 2 S D1 = S M 1 = 3

S DS =

2 (1.35) = 0.90 3 2 (1.68) = 1.12 3

Furthermore, from Table 17.6, and 17.1, one has that Occupancy category = IV Importance factor = I =1.5 Response modification coefficient (for ordinary frames) = R = 3.0 and from Table 17.7, Seismic design category = D It should be noted that the seismic design category is selected from Table 17.7 instead of Table 17.8 because the requirements imposed by the code to be able to do so are in this case satisfied. It should also be noted that the code requires that the value of the natural period determined from an eigenvalue analysis should not exceed cu times the approximate value, where, from Table 17.13 for SD1 ≥ 0.4, cu = 1.4. In consequence, the maximum value of the natural period that can be considered in the calculation of the base shear is Tmax = 1.4(0.26) = 0.36 s

Thus, according to Equation 17.23 Cs =

S DS 0.90 = = 0.45 R / I 3.0 / 1.5

which is greater than (Cs ) min =

0.5S1 0.5(0.70) = = 0.18 R/I 3.0 / 1.5

and less than (Cs ) max =

S D1 1.12 = = 1.56 T ( R / I ) 0.36(3.0/1.5)

In consequence, the base shear is equal to V = CsW = 0.45(2415) = 1087 kips

Lateral forces and overturning moment From Table 17.14, one obtains

121

k = 1.0 +

1 1 (T − 0.5) = 1.0 + (0.74 − 0.5) = 1.12 2 2

and thus from Equations 17.32 and 17.33, the lateral forces (Fx) are as indicate in Table P17.2a and the overturning moment at the base of the building results as M 0 = 1505(22) + 741(10) = 40,520 kip - ft Table P17.2a. Lateral forces, story shears, and overturning moments in building in Problem 17.2 Fx = C vxV Level wx hsx hx wxhxk w x h xk k C = vx (kip) (ft) (ft) (kip-ft ) N (kip)

∑w h

k i i

i =1

2 1 Sum

966 1149 2415

12 10

22 10

30,796 15,147 45,943

0.67 0.33 1.00

728 359 1,087

Lateral displacements As the story stiffnesses are given, the story drifts may be calculated according to Δx =

Vx kx

The elastic components of the lateral displacements may then be calculated by simply adding the corresponding story drifts. According to Equation 17.36, the total displacements are given by δx =

C d δ xe I

where from Table 17.1 for the building under consideration C d = 2 .5

and, as determined before, I = 1.5. The results are presented in Table P17.2b, where Δx denotes story drift, δxe elastic displacement, and δx total displacement. Table P17.2b. Lateral displacements and story drifts in building in Problem 17.2 Level kx Vx Δx δ xe δx (kip) (kip/in) (in) (in) (in) 2 728 500 1.46 2.55 4.25 1 1,087 1,000 1.09 1.09 1.82

Problem 17.3 Calculate the design shearing forces for the columns of the first story of the shear building shown in Figure P17.3 using the equivalent lateral force procedure prescribed by the International Building Code. The building will be located in Reno, Nevada, over stiff soil and will be used as a hotel. Its lateral load resisting system will be constructed with special reinforced concrete momentresisting frames. The given floor masses include dead and live load.

122

Mass 1.0 Mg

Stiffness 118.4 kN/m

3.0 m

1.5 Mg 236.9 kN/m

3.0 m

3.0 Mg 353.3 kN/m

4.5 m

Figure P17.3. Three-story building considered in Problem 17.3

Solution: Building weight

W = (3.0 + 1.5 + 1.0)9.81 = 53.96 kN

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting concrete frames, the approximate value of the fundamental natural period is Ta = Ct hnx = 0.047(10.5)0.9 = 0.39 s

Base shear The latitude and longitude of Reno, Nevada, are 39° 30' N 119° 47' W, respectively. Therefore, from the maps in Figures 17.1a and 17.2a, one has that S S = 1.92 S1 = 0.77

Also, as the building site may be classified as Site D (see Table 17.3), from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.92) = 1.92 S M 1 = Fv S1 = 1.5(0.77) = 1.16

and in view of Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.92) = 1.28 3 3 2 2 S D1 = S M 1 = (1.16) = 0.77 3 3

S DS =

Furthermore, from Table 17.6, and 17.1, one has that Occupancy category = II Importance factor = I =1.0 Response modification coefficient (for special moment concrete frames) = R = 8.0 and from Table 17.7, Seismic design category = E It should be noted that the seismic design category is selected from Table 17.8 because the code requires the selection of the most severe seismic design category. Thus, according to Equation 17.23

123

Cs =

S DS 1.28 = = 0.16 R / I 8.0 / 1.0

which is greater than (Cs ) min =

0.5S1 0.5(0.77) = = 0.05 R/I 8.0 / 1.0

and less than (Cs ) max =

S D1 0.77 = = 0.25 T ( R / I ) 0.39(8.0/1.0)

In consequence, the base shear is equal to V = CsW = 0.16(53.96) = 8.63 kN

and the design shearing force in each of the columns of the first story is Vc =

V 8.63 = = 4.32 kN 2 2

Problem 17.4 Determine the design axial forces in the columns of the first story of the building shown in Figure P17.4 using the equivalent lateral force procedure prescribed by the International Building Code. The building will be located in Las Vegas, Nevada, over a deep deposit of soft clays, will be used as a hospital, and will be structured with special moment-resistant reinforced concrete frames. w4= 1000 kips 4 8 ft w3 = 1200 kips 3 10 ft w2 = 1200 kips 2 10 ft w1 = 1500 kips 1 12 ft

45 ft

Figure P17.4. Four-story building considered in Problem 17.4

Solution: Building weight

W = 1,500 + 1,200 + 1,200 + 1,000 = 4,900 kips

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting concrete frames, Ta = Ct hnx = 0.016(40) 0.9 = 0.44 s

Base shear The latitude and longitude of Las Vegas, Nevada, are 36° 5' N and 115° 10' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 0.55 S1 = 0.16

Also, as the building site may be classified as Site E (see Table 17.3), from Tables 17.4 and 17.5, one obtains, 124

Fa = 1.6

Fv = 3.32

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.6(0.55) = 0.88 S M 1 = Fv S1 = 3.32(0.16) = 0.53

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (0.88) = 0.59 3 3 2 2 S D1 = S M 1 = (0.53) = 0.35 3 3

S DS =

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = IV Importance factor = I =1.5 Seismic design category = D Response modification coefficient (for special moment concrete frames) = R = 8.0 Hence, according to Equation 17.23 Cs =

S DS 0.59 = = 0.11 R / I 8.0 / 1.5

which is greater than (Cs ) min =

0.5S1 0.5(0.16) = = 0.02 R/I 8.0 / 1.5

and less than (Cs ) max =

S D1 0.35 = = 0.15 T ( R / I ) 0.44(8.0/1.5)

In consequence, the base shear is equal to V = CsW = 0.11(4,900) = 539 kips

Lateral forces and overturning moment From Table 17.14, one finds that for the case under consideration k = 1.0. Hence, according to Equations 17.32 and 17.33, the lateral forces (Fx) are as indicate in Table P17.4. As a result, the overturning moment at the level of the first story is equal to M 0 = 175.57(40) + 168.55(32) + 115.88(22) + 79.01(12) = 15,914 kip - ft

and the design axial force in each of the columns of the first story is Pc1 =

15,914 = 354 kips 45

Table P17.4. Lateral forces in building in Problem 17.4 Fx = C vxV Level hsx hx wxhxk wx w x h xk k C = vx (kip-ft ) (kip) (ft) (ft) N (kip)

∑w h

k i i

i =1

4 3 2 1 Sum

1000 1200 1200 1500 4900

8 10 10 12

40 32 22 12

40,000 38,400 26,400 18,000 122800

125

0.326 0.313 0.215 0.147 1.000

175.57 168.55 115.88 79.01 539.00

Problem 17.5 Determine the design story drifts of the building described in Problem P17.4 using the equivalent lateral force procedure prescribed by the International Building Code. Indicate, in addition, whether the calculated story drifts are within the limits established by the code. The inverse of the structure’s stiffness matrix is equal to ⎡0.250 ⎢ 1 ⎢0.250 −1 [K ] = 250 ⎢0.250 ⎢ ⎣0.250

0.250 0.583 0.583 0.583

0.250 0.583 0.917 0.917

0.250⎤ 0.583⎥⎥ in/kip 0.917 ⎥ ⎥ 1.417 ⎦

Solution: From the solution to Problem 17,4, the equivalent lateral forces for the structure in Figure P17.4 are, ⎧ 79.01 ⎫ ⎪115.88⎪ ⎪ ⎪ {Fs } = ⎨ ⎬ kips 168 . 55 ⎪ ⎪ ⎪⎩175.57⎪⎭

Therefore, the elastic components of the corresponding lateral displacements are ⎡0.250 ⎢ 1 ⎢0.250 δ = { xe } 250 ⎢0.250 ⎢ ⎣0.250

0.250 0.583 0.583 0.583

0.250 0.583 0.917 0.917

0.250⎤ ⎧ 79.01 ⎫ ⎧0.54⎫ 0.583⎥⎥ ⎪⎪115.88⎪⎪ ⎪⎪1.15 ⎪⎪ ⎬ in ⎨ ⎬=⎨ 0.917 ⎥ ⎪168.55⎪ ⎪1.61⎪ ⎥ 1.417 ⎦ ⎪⎩175.57 ⎪⎭ ⎪⎩1.96 ⎪⎭

and since for the structure under consideration one has that I =1.5 and Cd =5.5, the total displacements are equal to (see Equation 17.36) ⎧0.54⎫ ⎧1.98 ⎫ ⎪ ⎪ ⎪ ⎪ C 5.5 ⎪1.15 ⎪ ⎪4.22⎪ {δ x } = d {δ xe } = = ⎬ in ⎨ ⎬ ⎨ I 1.5 ⎪1.61⎪ ⎪ 5.91⎪ ⎪⎩1.96 ⎪⎭ ⎪⎩7.20⎪⎭

In consequence, the design story drifts are ⎧1.98 ⎫ ⎪2.24⎪ ⎪ ⎪ {Δ x } == ⎨ ⎬ in 1 . 69 ⎪ ⎪ ⎪⎩1.29 ⎪⎭

But from Table 17.2 for a structure with Occupancy Category IV the allowable story drifts are 0.015 times the story heights; that is, ⎧144⎫ ⎧2.16⎫ ⎪ ⎪ ⎪ ⎪ ⎪120⎪ ⎪1.80 ⎪ {Δ a } = 0.015⎨ ⎬ = ⎨ ⎬ in ⎪120⎪ ⎪1.80 ⎪ ⎪⎩ 96 ⎪⎭ ⎪⎩1.44 ⎪⎭

It may be seen, thus, that the story drift of the second story exceeds the limit imposed by the code.

126

Problem 17.6 The building shown in Figure P17.6 will be built in Anchorage, Alaska, on a parcel of land classified as Class B using special moment-resisting steel frames. The building height is 135 ft and the plan dimensions are 100 ft × 170 ft. The dead load at each floor is 120 lb/ft2. Determine using the equivalent lateral force procedure prescribed by the International Building Code the design base shear, lateral forces, and story shears along the transverse direction of the building. 9 8 9 @ 15' = 135'

7 6 5 4 3 2 1 4 @ 25' = 100'

Figure P17.6. Nine-story building considered in Problem 17.6

Solution: Building weight Assuming the building will be used for offices and a live load of 50 psf, the total floor load is equal to (see Section 17.6.2) w = 120 + 0.25(50) = 132.5 psf

and thus the total weight of the building is equal to

W = 9(132.5 × 100 × 170) = 20,272 kips

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting steel frames, Ta = Ct hnx = 0.028(135)0.8 = 1.42 s

Base shear The latitude and longitude of Anchorage, Alaska, are: 61° 10' N and 150° 1' W, respectively. Therefore, from the maps in Figures 17.3 and 17.4, one has that S S = 1.50 S1 = 0.55

Also, as the building will be built at a Site B, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.0

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.0(0.55) = 0.55

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.55) = 0.37 3 3 S DS =

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that 127

Occupancy category = II Importance factor = I =1.0 Seismic design category = D Response modification coefficient (for special moment frames) = R = 8.0 Hence, according to Equation 17.23 1.00 S DS = = 0.125 R / I 8.0 / 1.0

Cs =

which is greater than (Cs ) min =

0.5S1 0.5(0.55) = = 0.034 R/I 8.0 / 1.0

and greater than (Cs ) max =

S D1 0.37 = = 0.033 T ( R / I ) 1.42(8.0/1.0)

and thus it will be considered that Cs = (Cs ) min = 0.034

In consequence, the base shear is equal to V = CsW = 0.034(20,272) = 689 kips

Lateral forces and story shears From Table 17.14, one obtains 1 1 k = 1.0 + (T − 0.5) = 1.0 + (1.42 − 0.5) = 1.46 2 2

and thus from Equations 17.32 and 17.33, the lateral forces (Fx) and story shears (Vx) are as indicated in Table P17.6. Table P17.6. Lateral forces and story shears in building in Problem 17.6 Fx = C vxV Level wxhxk wx hsx hx Vx w x h xk k C = vx (kip) (kip-ft ) (kip) (ft) (ft) N (kip)

∑w h

k i i

i =1

9 8 7 6 5 4 3 2 1 Sum

2252.5 2252.5 2252.5 2252.5 2252.5 2252.5 2252.5 2252.5 2252.5 20,272.5

15 15 15 15 15 15 15 15 15

135 120 105 90 75 60 45 30 15

2,903,702 2,444,947 2,011,876 1,606,419 1,230,989 888,721 583,922 323,044 117,424 12,111,045

0.240 0.202 0.166 0.133 0.102 0.073 0.048 0.027 0.010 1.000

165.19 139.09 114.46 91.39 70.03 50.56 33.22 18.38 6.68

165.19 304.29 418.74 510.13 580.16 630.72 663.94 682.32 689.00

Problem 17.7 The 12-story moment-resisting steel frame shown in Figure P17.7 is used to model a building for seismic analysis. The effective weight on all levels of the building is 90 lb/ft2, except on the roof where it is 80 lb/ft2. The tributary area for the frame has a width of 30 ft. The overall dimension of the building in the direction perpendicular to the direction of analysis is 180 ft. The building 128

is located in Valdez, Alaska, on a site classified as class C. The building will be used for office space. Determine using the equivalent lateral force procedure prescribed by the International Building Code the (a) equivalent lateral forces, (b) story shears, (c) torsional moments, (d) overturning moments, (e) lateral displacements, and (f) story drifts. Consider that the modulus of elasticity for the steel elements is 30,000 kips/in2.

8 7 6 5 4 3 2 1

16W40 21W44 11 @ 12' = 132'

9

16W40 16W40

21W44 21W50 21W50 24W55 24W55 24W55 24W55

15'

10

16W40 14W159 14W109 14W90 14W68 14W48 14W48

11

14W109 14W82 14W68 14W53 14W43 14W43

12

3 @ 20' = 60'

Figure P17.7. Twelve-story building considered in Problem 17.7

Solution: Building weight The tributary area for the frame under analysis is 30 × 60 ft. Therefore, the total weight supported by the frame is W = 11(0.09 × 30 × 60) + 0.08 × 30 × 60 = 1,926 kips

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting steel frames, the approximate value of the fundamental natural period is Ta = Ct hnx = 0.028(147)0.8 = 1.52 s

Base shear The latitude and longitude of Valdez, Alaska, are: 61° 8' N and 146° 21' W, respectively. Therefore, from the maps in Figures 17.3 and 17.4, one has that S S = 1.50 S1 = 0.50

Also, as the building will be built at a Site C, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.3

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.3(0.50) = 0.65

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.65) = 0.43 3 3 S DS =

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = II Importance factor = I =1.0 129

Seismic design category = D Response modification coefficient (for ordinary moment frames) = R = 3.5 Hence, according to Equation 17.23 Cs =

1.00 S DS = = 0.29 R / I 3.5 / 1.0

which is greater than (Cs ) min =

0.5S1 0.5(0.50) = = 0.07 R/I 3.5 / 1.0

and greater than (Cs ) max =

S D1 0.43 = = 0.08 T ( R / I ) 1.52(3.5/1.0)

and thus it will be considered that Cs = (Cs ) max = 0.08

In consequence, the base shear is equal to V = CsW = 0.08(1,926) = 154.1 kips Table P17.7b. Lateral forces and story shears in building in Problem 17.7 Level

wx (kip)

hsx (ft)

hx (ft)

wxhxk (kip-ft)

C vx =

w x h xk N

∑w h

k i i

Fx = C vxV (kip)

Vx (kip)

M0x (kip-ft)

Tx (kip-ft)

26.0 25.7 22.3 19.1 16.1 13.2 10.6 8.1 5.9 3.9 2.3 0.9 154.1

26.0 51.7 74.0 93.1 109.2 122.4 133.0 141.1 147.0 150.9 153.2 154.1

311.5 931.2 1818.7 2935.4 4245.1 5713.6 7308.8 9001.6 10765.3 12576.2 14414.2 16725.7

1401.8 2788.6 3993.5 5025.4 5893.6 6607.9 7178.8 7617.5 7936.4 8149.1 8271.2 8321.4

i =1

12 11 10 9 8 7 6 5 4 3 2 1 Sum

144 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 15 1926 147

147 135 123 111 99 87 75 63 51 39 27 15

269,781 266,882 231,885 198,588 167,080 137,464 109,865 84,434 61,369 40,928 23,490 9,670 1,601,434

0.168 0.167 0.145 0.124 0.104 0.086 0.069 0.053 0.038 0.026 0.015 0.006 1.000

Lateral forces, story shears, overturning moments, and torsional moments From Table 17.14, one obtains 1 1 k = 1.0 + (T − 0.5) = 1.0 + (1.52 − 0.5) = 1.51 2 2

and thus from Equations 17.32 and 17.33, the lateral forces (Fx) and story shears (Vx) are as indicated in Table P17.7a. The overturning moments (M0x) and torsional moments are also presented in Table 17.7a. The overturning moments are calculated according to M 0i = M 0(i +1) + Vi hsx

As the building is symmetric, the torsional moments (Tx) are simply computed by multiplying the total story shears by the accidental eccentricity stipulated by the code. The total story shears are obtained by multiplying the calculated story shears by 6 (there are 6 spans of 30 feet each). The accidental eccentricity is equal to 0.05×180 = 9.0 feet. 130

Lateral displacements and story drifts With the lateral forces listed in Table P17.7a, the given beam and column sections, and using a computer program for the static analysis of frames, the elastic floor displacements (δxe) result as shown in Table P17.1b. The corresponding inelastic displacements (δx) are also listed in this table. These displacements are computed according to Equation 17.36, considering an importance factor I of 1.0 and a Cd value of 3.0. The story drifts (Δx) are determined by subtracting the displacements of the floors above and below each story. For a comparison with the allowable drift, this table also presents the story drift ratios Δx/hsx. Note that the computed displacements do not include the displacements produced by the torsional moments as these displacements are not significant. Table P17.7b. Lateral displacements and story drifts in building in Problem 17.7 Level hsx Δx δ xe δx Δx/hsx (in)

12 11 10 9 8 7 6 5 4 3 2 1

(in)

(in)

(in)

144 10.93 32.80 1.47 144 10.44 31.33 2.43 144 9.63 28.89 3.28 144 8.54 25.62 3.56 144 7.35 22.05 3.21 144 6.28 18.85 3.35 144 5.17 15.50 3.06 144 4.14 12.43 2.93 144 3.17 9.50 2.56 144 2.31 6.94 2.55 144 1.46 4.39 2.22 180 0.72 2.17 2.17

0.010 0.017 0.023 0.025 0.022 0.023 0.021 0.020 0.018 0.018 0.015 0.012

20'

15'

15'

Problem 17.8 Figure P17.8 shows the elevation and plan views of a three-story reinforced concrete warehouse that will be built in Carson City, Nevada. As shown in the figure, the building is constructed with shear walls with a height of 20 feet for the first story and 15 feet for the second and third stories. The total gravity load for the first, second, and third floors is 300, 250, and 220 kips, respectively. Determine using the equivalent lateral force procedure prescribed by the International Building Code the design lateral forces, story shears, lateral displacements, and story drifts along the transverse direction of the building. Assume a modulus of elasticity of 3,600 kips/in2 for the concrete used in the construction of the shear walls. Neglect the bending deformation of the shear walls.

Elevation

30'

Shear wall 8"

4 @ 25' = 100' Floor plan

Figure P17.8. Three-story building considered in Problem 17.8

Solution: Building weight

131

W = 300 + 250 + 220 = 770 kips

Lateral stiffnesses If the bending deformations are neglected, the lateral stiffness of a shear wall is given by GAi EAi = hi 2(1 + μ)hi

Ki =

where E denotes modulus of elasticity, μ Poisson ratio, Ai cross-section area, and hi wall height. Therefore, if a Poisson ratio of 0.15 is assumed, the stiffness of the first story is K1 = 5

3,600(360 × 8) = 93,913 kip/in 2(1 + 0.15)240

Similarly, the stiffnesses of the second and third stories are K 2 = K3 = 5

3,600(360 × 8) = 125,217 kip/in 2(1 + 0.15)180

Natural period From Equation 17.31, one has that for a structure with 5 shear walls, CW =

100 AB

⎛ hn ⎞ ⎜ ⎟ ⎜ ⎟ i =1 ⎝ hi ⎠ 5

∑

2

Ai ⎛h ⎞ 1 + 0.83⎜⎜ i ⎟⎟ ⎝ Di ⎠

=

2

100 ⎛ 50 ⎞ (5)⎜ ⎟ 30 × 100 ⎝ 50 ⎠

2

(8 / 12) × 30 ⎛ 50 ⎞ 1 + 0.83⎜ ⎟ ⎝ 30 ⎠

2

= 1.008

and thus, from Equation 17.30, the natural period of the structure is approximately given by Ta =

0.0019 CW

hn =

0.0019 1.008

50 = 0.095 s

Base shear The latitude and longitude of Carson City, Nevada, are: 39° 10' N and 119° 46' W, respectively. Therefore, from the maps in Figures 17.1a and 17.2a, one has that S S = 1.92 S1 = 0.77

Also, the building site may be classified as Site D since the soil properties are unknown. Hence, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

and thus, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.92) = 1.92 S M 1 = Fv S1 = 1.5(0.77) = 1.16

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.92) = 1.28 3 3 2 2 S D1 = S M 1 = (1.16) = 0.77 3 3

S DS =

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = II Importance factor = I =1.0 Seismic design category = E Response modification coefficient (for ordinary reinforced concrete walls) = R = 4.0 Hence, according to Equation 17.23 132

Cs =

1.28 S DS = = 0.32 R / I 4.0 / 1.0

which is greater than (Cs ) min =

0.5S1 0.5(0.77) = = 0.10 R/I 4.0 / 1.0

and is less than (Cs ) max =

S D1 0.77 = = 2.03 T ( R / I ) 0.095(4.0/1.0)

and thus the base shear is equal to V = CsW = 0.32(770) = 246 kips

Lateral forces and story shears From Table 17.14, one obtains k = 1.0 and thus from Equations 17.32 and 17.33, the lateral forces (Fx) and story shears (Vx) are as indicate in Table P17.8a. Table P17.8a. Lateral forces, story shears, and overturning moments in building in Problem 17.1 Level

wx (kip)

hsx (ft)

hx (ft)

wxhxk (kip-ft)

C vx =

w x h xk N

∑w h

k i i

Fx = C vxV (kip)

Vx (kip)

92.0 91.5 62.7 246.2

92.0 183.5 246.2

i =1

3 2 1 Sum

220 250 300 770

15 15 20

40 35 20

8,800 8,750 6,000 23,550

0.374 0.372 0.255 1.000

Story drifts and lateral displacements Since in this case the story stiffnesses are known, the story drifts may be calculated according to Δx =

Vx Kx

The elastic components of the lateral displacements may then be calculated by simply adding the corresponding story drifts. According to Equation 17.36, the total displacements are given by δx =

C d δ xe I

where from Table 17.1 for the building under consideration Cd = 4.0

and, as determined earlier, I = 1.0. The results are presented in Table P17.8b, where Δx denotes story drift, δxe elastic displacement, and δx total displacement. Table P17.7b. Lateral displacements and story drifts in building in Problem 17.1 Level Vx kx Δx δ xe δx (kip)

3 2 1

(kip/in)

(in)

(in)

(in)

92.0 125,217 0.00073 0.00482 0.0193 183.5 125,217 0.00147 0.00409 0.0164 246.2 93,913 0.00262 0.00262 0.0105

Problem 17.9

133

The building shown in Figure P17.9 will be located in Fresno, California, will be used for hazardous and water treatment facilities, and will be constructed with ordinary moment-resisting frames. The building site has been classified as Class B. The dead loads for the first, second, and third floors are 609.4 kips, 590.9 kips, and 217.2 kips, respectively. Determine using the modal response spectrum analysis procedure prescribed by the International Building Code the base shear, lateral forces, story shears, lateral displacements, and story drifts of the building along its transverse direction. Assume that the moments of inertia of the columns are Ix = 1530 in4 and Iy = 548 in4, and those of the beams are Ix = 1170 in4 and Iy = 60.3 in4. Similarly, assume that the modulus of elasticity of the steel used in the construction of the building structure is 29,000 kips/in2. 3

3 @ 30' = 90'

15' 2

15' 1

20' 3 @ 30' = 90' 6 @ 20' = 120' Elevation Floor plan

Figure P17.9. Three-story building considered in Problem 17.9

Solution: Floor weights and total weight Modeling the building with one of the intermediate frames with a tributary area of 90 ft × 20 ft, and considering 25 percent of a live load of 50 psf on the floors and no live load on the roof, the total floor and roof weights are w1 = 609.4 / 6 + 0.25(0.05)(90 × 20) = 124.1 kips w2 = 590.9 / 6 + 0.25(0.05)(90 × 20) = 120.9 kips w3 = 217.2 / 6 = 36.2 kips

Correspondingly, the total weight supported by the frame is

W = 124.1 + 120.9 + 36.2 = 281.2 kips

Natural periods and mode shapes With the weights determined above and the given moments of inertia, the solution of the corresponding eigenvalue problem leads to the following eigenvalues, natural periods, participation factors, and modal matrix: ω12 = 68.667 rad 2 /s 2 T1 = 0.758 s Γ1 = −3.046

ω22 = 815.47 rad 2 /s 2 ω32 = 2546.5 rad 2 /s 2 T2 = 0.220 s T2 = 0.124 s Γ2 = 0.803 Γ3 = 0.249

⎡0.212 − 0.369 − 0.210⎤ 0.324 ⎥⎥ [Φ] = ⎢⎢0.358 0.012 ⎢⎣0.427 0.525 − 0.481⎥⎦

Effective modal weights The application of Equation 17.41 with the mode shape amplitudes and floor weights determined

134

above yields the modal effective weights listed in Table P17.9a. From the percentages shown in this table, it may be seen that the effective weight in the first mode of the system is more than 90 per cent of the total weight of the building. Hence, in accordance with the code recommendations (see Section 17.7.1), only the first mode will be considered in the analysis. Table P17.9a. Modal effective weights of building considered in Problem 17.9 Total weight Effective weight, percentage Mode Wm (kips) 1 261.2 92.9 2 18.0 6.4 3 2.0 0.6 Total 281.2 99.9

Design response spectrum The latitude and longitude of Fresno, California, are 36° 46' N and 119° 43' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 0.44 S1 = 0.21

and from Tables 17.4 and 17.5 for a Site B, one obtains, Fa = 1.0

Fv = 1.0

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(0.44) = 0.44 S M 1 = Fv S1 = 1.0(0.21) = 0.21

and according to Equations 17.11 and 17.12 the corresponding design spectral accelerations are 2 S MS = 3 2 S D1 = S M 1 = 3

S DS =

2 (0.44) = 0.29 3 2 (0.21) = 0.14 3

By virtue of Equations 17.17 and 17.18 and the map in Figure 17.7a, the periods that delimit the four regions of the design spectrum are therefore equal to S D1 0.14 = 0.2 = 0.10 s S DS 0.29 S 0.14 TS = D1 = = 0.48 s S DS 0.29 TL = 12.0 s

T0 = 0.2

Using Equations 17.13 through 17.16, the spectra accelerations in the design spectrum for the system under analysis may be thus defined as follows (see Figure 17.6): (a) For periods less than or equal to 0.10 seconds, S a = S DS (0.4 + 0.6

T T ) = 0.29(0.4 + 0.6 ) = 0.12 + 1.74T T0 0.10

(b) For periods greater than or equal to 0.10 seconds but less than or equal to 0.48 seconds, S a = S DS = 0.29

(c) For periods greater than or equal to 0.48 seconds but less than or equal to 12.0 seconds, 135

Sa =

S D1 0.14 = T T

(d) For periods greater than or equal to 12.0 seconds, Sa =

S D1TL 0.14(12.0) 1.68 = = 2 T2 T2 T

In consequence, the spectral acceleration corresponding to the first mode of the system is S a1 =

0.14 0.14 = = 0.18 T 0.758

Base shear From Tables 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = III Importance factor = I =1.25 Seismic design category = B Response modification coefficient (for ordinary steel frames) = R = 3.5 In the light of Equations 17.40 and 17.39 and the effective weights and spectral acceleration determined above, the seismic response coefficient for the first mode is therefore equal to 0.18 S a1 = = 0.06 R / I 3.5 / 1.25

Cs1 =

and, correspondingly, the modal base shear in the first mode is V1 = Cs1W1 = 0.06(261.2) = 15.7 kips

As only the first mode is being considered, the total base shear in this case is thus equal to Vt = 15.7 kips

It should be recalled, however, that the code imposes a limit to the value of the base shear that may be considered when the modal analysis procedure is used. As described in Section 17.7.4, this limit is given by 85 percent of the value of the base shear obtained with the equivalent lateral force procedure considering a natural period equal to T = Cu Ta, where Cu is determined from Table 17.13 and Ta is the approximate value of the fundamental natural period of the system computed with Equation 17.28, whenever the calculated period exceeds the product of Cu and Ta. To obtain such a limit, consider then that for the system under analysis Ta = Ct (hn ) x = 0.028(50)0.8 = 0.64 s T = CuTa = 1.68(0.64) = 1.08 s

and that, in consequence, the calculated period of 0.758 s does not exceed the upper limit imposed by the code for the fundamental natural period. In the light of Equation 17.23, the seismic response coefficient calculated using the equivalent lateral force procedure is therefore equal to Cs =

S DS 0.29 = = 0.10 R / I 3.5 / 1.25

which is greater than (see Equation 17.25) (Cs ) min =

0.5S1 0.5(0.21) = = 0.04 R/I 3.5 / 1.25

but exceeds (see Equation 17.26) (Cs ) max =

S D1 0.14 == = 0.07 T (R / I ) 0.758(3.5 / 1.25)

Considering, then, this upper limit, from Equation 17.22 one has that V = CsW = 0.07(281.2) = 19.7 kips

136

which leads to a lower limit for the base shear of 0.85×19.7 = 16.7 kips. This value, in turn, exceeds the value of Vt determined above. The lateral forces and story shears (but not displacements and story drifts) will be calculated considering a base shear of 16.7 kips. Lateral forces and story shears The modal lateral forces are given by Equations 17.42 and 17.43. The values of the vertical distribution factor Cvxm, the lateral forces Fxm, and the story shears Vxm for the first mode are thus as indicated in Table P17.9b. As only the first mode is being considered, the values shown in this table also correspond to the design values of the lateral forces and story shears. Level, x

Table P17.9b. Lateral forces and story shears in first mode Fx1 = C vx1V1 wx wφ wxφx1 φx1 Cvx1 = 3 x x1 (kips) (kips) (kips) wi φi1

Vx1 (kips)

∑ i =1

3 2 1 Sum:

36.2 120.9 124.1 281.2

0.427 0.358 0.212

15.46 43.28 26.31 85.05

0.182 0.509 0.309 1.000

3.04 8.50 5.17 16.71

3.04 11.53 16.71

Lateral displacements and story drifts From Equation 17.45, the elastic components of the modal lateral displacements are given by 2 2 ⎛ g ⎞⎛ T F ⎞ ⎛ g ⎞⎛ 0.758 Fx1 ⎞⎟ δ xe1 = ⎜ 2 ⎟⎜⎜ 1 x1 ⎟⎟ = ⎜ 2 ⎟⎜⎜ ⎟ wx ⎝ 4π ⎠⎝ wx ⎠ ⎝ 4π ⎠⎝ ⎠

and, according to Equation 17.44, the total modal lateral displacements are equal to δ x1 =

Cd δ xe1 I

where for the case under consideration Cd = 3.0 (see Table 17.1) and I = 1.25. After substituting the floor weights, the natural period, and the lateral forces multiplied by 15.7/16.7, the lateral displacements and story drifts in the first mode of the system, and hence the design lateral displacements and story drifts, are as indicated in Table P17.9c. For comparison with the allowable ones, the story drift ratios are also included in this table. Table P17.9c. Modal lateral displacements and story drifts in first mode Level 0 wx hsx (kips) (in) . 9 4 F x δ 1 x δ ( k i p s )

137

e

x

1

1

( i n . )

( i n . )

Δx1 (in.) 3 2 1

Δx1/hsx 36.2 120.9 124.1

180 180 240

2.85 7.99 4.86

0.44 0.37 0.22

1.06 0.89 0.53

0.17 0.36 0.53

0.001 0.002 0.002

Problem 17.10 The three-story building shown in Figure P17.10 will be located in San Diego, California, and constructed with intermediate moment-resisting steel frames. An evaluation of the building site shows that the underlying soil deposit has a depth of about 250 feet and an average shear wave velocity of 1500 feet per second. The masses, natural frequencies, and modal matrix of the structure are as shown below. Determine using the modal response spectrum analysis procedure prescribed by the International Building Code the design base shear, story shear, overturning moments, and story drifts. m1 = m2 = m3 = 1132.2141 lb - s 2 / ft ω1 = 7.30916 rad/s ω2 = 23.58844 rad/s ω3 = 40.04150 rad/s 1.0 ⎤ ⎡0.2555 0.9413 ⎢ [Φ ] = ⎢0.6752 1.0 − 0.7321⎥⎥ ⎢⎣ 1.0 − 0.9421 0.2680 ⎥⎦ m3 14' m2 14' m1 14'

22'

Figure P17.10. Three-story building considered in Problem 17.10

Solution: Floor weights and total weight The floor weights are equal to

w1 = w2 = w3 = 1,132.214(32.2) = 36.46 kips

and thus the total weight of the building is

W = 3(36.46) = 109.38 kips

Natural periods From the given natural frequencies, the natural periods of the building are T1 = 0.860 s

T2 = 0.266 s

T3 = 0.157 s

Effective modal weights The application of Equation 17.41 with the given mode shapes and floor weights determined above yields the modal effective weights listed in Table P17.9a. From the percentages shown in this table, it may be seen that the effective weights in the first two modes of the system add up to more than 90 per cent of the total weight of the building. Thus, in accordance with the code rec-

138

ommendations (see Section 17.7.1), only the first two modes will be considered in the analysis. Table P17.10a. Modal effective weights of building considered in Problem 17.9 Effective weight, Wm Total weight percentage Mode (kips) 1 89.69 82.0 2 13.13 12.0 3 6.56 6.0 Total 109.38 100.0

Design response spectrum The latitude and longitude of San Diego, California, are 32° 44' N and 117° 10' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.50 S1 = 0.60

Also, from Table 17.3 and the given soil conditions, the building site corresponds to Site C. Hence, from Tables 17.4 and 17.5 for a Site C, one obtains, Fa = 1.0

Fv = 1.3

and, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.3(0.60) = 0.78

Moreover, from Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.78) = 0.52 3 3

S DS =

By virtue of Equations 17.17 and 17.18 and the map in Figure 17.7a, the periods that delimit the four regions of the design spectrum are therefore equal to S D1 0.52 = 0.2 = 0.10 s S DS 1.00 S 0.52 TS = D1 = = 0.52 s S DS 1.00 TL = 8.0 s

T0 = 0.2

Using Equations 17.13 through 17.16, the spectra accelerations in the design spectrum for the system under analysis may be thus defined as follows (see Figure 17.6): (a) For periods less than or equal to 0.10 seconds, S a = S DS (0.4 + 0.6

T T ) = 1.00(0.4 + 0.6 ) = 0.4 + 6T T0 0.10

(b) For periods greater than or equal to 0.10 seconds but less than or equal to 0.52 seconds, S a = S DS = 1.00

(c) For periods greater than or equal to 0.52 seconds but less than or equal to 8.0 seconds, Sa =

S D1 0.52 = T T

(d) For periods greater than or equal to 8.0 seconds,

139

Sa =

S D1TL 0.52(8.0) 4.16 = = 2 T2 T2 T

In consequence, the spectral accelerations corresponding to the first two modes of the system are S a1 =

0.52 0.52 = = 0.60 T 0.860 S a 2 = 1.00

Modal and total base shears Assuming the building will be used for offices, from Tables 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = II Importance factor = I =1.0 Seismic design category = D Response modification coefficient (for intermediate steel frames) = R = 4.5 In the light of Equations 17.40 and 17.39 and the effective weights and spectral accelerations determined above, the seismic response coefficients for the first two modes are therefore equal to S a1 0.60 = = 0.13 R / I 4.5 / 1.0 1.00 S = 0.22 Cs 2 = a 2 = R / I 4.5 / 1.0 Cs1 =

and, correspondingly, the modal base shears in the first two modes are V1 = Cs1W1 = 0.13(89.69) = 11.66 kips V2 = Cs 2W2 = 0.22(13.13) = 2.89 kips

Combining, then, the modal base shears with the square-root-of-the-sum-of-the-squares rule, the total base shear is approximately equal to Vt = V12 + V22 = 11.662 + 2.892 = 12.01 kips

It should be recalled, however, that the code imposes a limit to the value of the base shear that may be considered when the modal analysis procedure is used. As described in Section 17.7.4, this limit is given by 85 percent of the value of the base shear obtained with the equivalent lateral force procedure considering a natural period equal to T = Cu Ta, where Cu is determined from Table 17.13 and Ta is the approximate value of the fundamental natural period of the system computed with Equation 17.28, whenever the calculated period exceeds the product of Cu and Ta. To obtain such a limit, consider then that for the system under analysis Ta = Ct (hn ) x = 0.028(42)0.8 = 0.56 s T = CuTa = 1.4(0.56) = 0.78 s

and that, in consequence, the calculated period of 0.860 s does exceed the upper limit imposed by the code for the fundamental natural period. In the light of Equation 17.23, the seismic response coefficient calculated using the equivalent lateral force procedure is therefore equal to Cs =

1.00 S DS = = 0.22 R / I 4.5 / 1.0

which is greater than (see Equation 17.25) (Cs ) min =

0.5S1 0.5(0.60) = = 0.07 R/I 4.5 / 1.0

but exceeds (see Equation 17.26)

140

(Cs ) max =

S D1 0.52 == = 0.15 T (R / I ) 0.78(4.5 / 1.0)

This upper limit applies and, hence, from Equation 17.22, one has that V = CsW = 0.15(109.38) = 16.41 kips

which leads to a lower limit for the base shear of 0.85×16.41 = 13.95 kips. This value, in turn, exceeds the value of Vt determined above. The lateral forces, story shears, and overturning moments obtained by the modal analysis procedure will be thus scaled up by the factor (see Equation 17.46) Cm = 0.85

V 16.41 = 0.85 = 1.16 Vt 12.01

Lateral forces, story shears, and overturning moments Applying Equations 17.42 and 17.43, the modal values of the vertical distribution factor Cvxm, the lateral forces Fxm, the story shears Vxm and the overturning moments M0xm for the two modes being considered result as indicated in Tables P17.10b and P17.10c. It should be noted that these result correspond to modal base shears augmented by the factor Cm of 1.16 determined above. The corresponding design values, obtained by combining the modal values with the square root of the sum of the squares rule, are shown in Table P17.10d. Level, x

Table P17.10b. Lateral forces, story shears, and overturning moments in first mode Fx1 = C vx1V1 M0x1 Vx1 hsx wx wφ wxφx1 φx1 Cvx1 = 3 x x1 (kips) (k-ft) (kips) (ft) (kips) (kips) wi φi1

∑ i =1

3 2 1 Sum:

Level, x

36.46 36.46 36.46 109.38

14 14 14 42

1.000 0.675 0.256

36.46 24.62 9.32 70.40

0.518 0.350 0.132 1.000

7.01 4.73 1.79 13.53

7.01 11.74 13.53

98.14 262.50 451.92

Table P17.10c. Lateral forces, story shears, and overturning moments in second mode Fx 2 = C vx 2V2 M0x2 Vx2 hsx wx wφ wxφx2 φx2 Cvx 2 = 3 x x 2 (kips) (kips) (ft) (k-ft) (kips) (kips) wi φi 2

∑ i =1

3 2 1 Sum:

36.46 36.46 36.46 109.38

14 14 14 42

-0.942 1.000 0.941

-34.35 36.46 34.32 36.43

-0.943 1.001 0.942 1.000

-3.16 3.36 3.16 3.36

-3.16 0.20 3.36

-44.24 -41.44 5.60

Table P17.10d. Design values of lateral forces, story shears, and overturning moments Lateral forces (kips) Story shears (kips) Overturning moments (k-ft) Level Mode Mode Design Mode Mode Design Mode Mode Design 1 2 value 1 2 value 1 2 value 3 7.01 -3.16 7.69 7.01 -3.16 7.69 98.14 -44.24 107.65 2 4.73 3.36 5.80 11.74 0.20 11.74 262.50 41.44 265.75 1 1.79 3.16 3.63 13.53 3.36 13.93 451.92 5.60 451.95

Lateral displacements and story drifts From Equation 17.45, the elastic components of the modal lateral displacements are given by 141

2 ⎛ g ⎞⎛ T F ⎞ δ xem = ⎜ 2 ⎟⎜⎜ m xm ⎟⎟ ⎝ 4π ⎠⎝ wx ⎠

and according to Equation 17.44, the total modal lateral displacements are equal to δ xm =

Cd δ xem I

where for the case under consideration Cd = 4.0 (see Table 17.1) and I = 1.0. After substituting the floor weights, the natural periods, and the lateral forces divided by the factor Cm = 1.16 (recall the lower limit for the base shear does not apply for displacements and drifts), the lateral displacements and story drifts in the first two modes of the system result as indicated in Table P17.10e. After combining these modal values with the square root of the sum of the squares rule, the design values shown in Table E17.2f are obtained. For comparison with the allowable ones, the story drift ratios are also included in this table. Table P17.10e. Modal lateral displacements and story drifts in first and second modes First mode Second mode Lev- wx hsx Fx1 /Cmδxe1 δx1 el (kips) (in) (kips)(in) (in) Δx1 (in) Δx2

δxe2 (in)

Fx2 /Cm (kips)

δx2 (in)

(in) 3 36.46 168 6.04 1.20 4.80 1.56 -2.72 -0.052 -0.208 -0.43 2 36.46 168 4.08 0.81 3.24 2.00 2.90 0.055 0.220 0.01 1 36.46 168 1.54 0.31 1.24 1.24 2.72 0.052 0.208 0.21 Table P17.10f. Design values of lateral displacements and story drifts Level 3 2 1

Lateral displacements (in) Mode 1 Mode 2 Design value

4.80 3.24 1.24

-0.208 0.220 0.208

Mode 1

4.81 3.25 1.26

1.56 2.00 1.24

Story drifts (in) Mode 2 Design value

-0.43 0.01 0.21

1.62 2.00 1.26

Drift ratios

0.010 0.012 0.008

Problem 17.11 Repeat Problem 17.1 using the modal response spectrum analysis procedure prescribed by the International Building Code. Solution: Natural periods Solving the eigenvalue problem, the following natural periods and mode shapes are obtained: T1 = 0.644 s T4 = 0.122 s

T2 = 0.263 s T5 = 0.096 s

142

T3 = 0.166 s T6 = 0.079 s

⎡0.074 ⎢0.148 ⎢ ⎢0.222 Φ=⎢ ⎢0.297 ⎢ 0.371 ⎢ ⎢⎣0.445

− 0.167 − 0.254 − 0.329 − 0.381 0.385 ⎤ − 0.293 − 0.331 − 0.214 0.076 − 0.481⎥⎥ − 0.335 − 0.140 0.247 0.394 0.321 ⎥ ⎥ − 0.251 0.229 0.339 − 0.407 − 0.128⎥ 0.000 0.420 − 0.398 0.165 0.029 ⎥ ⎥ 0.460 − 0.280 0.109 − 0.025 − 0.003⎥⎦

Effective modal weights The application of Equation 17.41 with the mode shapes and floor weights determined above yields the modal effective weights listed in Table P17.11a. From the percentages shown in this table, it may be seen that the effective weights in the first two modes of the system add up to more than 90 per cent of the total weight of the building. Thus, in accordance with the code recommendations (see Section 17.7.1), only the first two modes will be considered in the analysis. Table P17.11a. Modal effective weights of building considered in Problem 17.9 Effective weight, Wm Total weight percentage Mode (kips) 1 3,745.1 80.77 2 529.6 11.42 3 195.8 4.22 4 94.1 2.03 5 49.0 1.06 6 23.2 0.50 Total 4,636.8 100.00

Design response spectrum From the SDS and SD1 values determined in the solution of Problem 17.1 and by virtue of Equations 17.17 and 17.18 and the map in Figure 17.7a, the natural periods that delimit the four regions of the design spectrum for this problem are 0.60 S D1 = 0.2 = 0.12 s S DS 1.00 S 0.60 = 0.60 s TS = D1 = S DS 1.00 TL = 8.0 s

T0 = 0.2

Using Equations 17.13 through 17.16, the spectra accelerations in the design spectrum for the system under analysis may be thus defined as follows (see Figure 17.6): (a) For periods less than or equal to 0.10 seconds, S a = S DS (0.4 + 0.6

T T ) = 1.00(0.4 + 0.6 ) = 0.4 + 5T T0 0.12

(b) For periods greater than or equal to 0.12 seconds but less than or equal to 0.60 seconds, S a = S DS = 1.00

(c) For periods greater than or equal to 0.60 seconds but less than or equal to 8.0 seconds, Sa =

S D1 0.60 = T T

(d) For periods greater than or equal to 8.0 seconds,

143

Sa =

S D1TL 0.60(8.0) 4.8 = = 2 T2 T2 T

In consequence, the spectral accelerations corresponding to the first two modes of the system are S a1 =

0.60 0.60 = = 0.93 T 0.644 S a 2 = 1.00

Modal and total base shears Since for this building one has that Occupancy category = II Importance factor = I =1.0 Seismic design category = D Response modification coefficient (for intermediate steel frames) = R = 3.5 the seismic response coefficients for the first two modes are equal to S a1 0.93 = = 0.27 R / I 3.5 / 1.0 S 1.00 = 0.29 Cs 2 = a 2 = R / I 3.5 / 1.0 Cs1 =

and, correspondingly, the modal base shears in the first two modes are V1 = Cs1W1 = 0.27(3,745.1) = 1,011.2 kips V2 = Cs 2W2 = 0.29(529.6) = 153.6 kips

Combining, then, these modal base shears with the square-root-of-the-sum-of-the-squares rule, the total base shear is approximately equal to Vt = V12 + V22 = 1,011.22 + 153.62 = 1,022.8 kips

It should be recalled, however, that the code imposes a limit to the value of the base shear that may be considered when the modal analysis procedure is used. As described in Section 17.7.4, this limit is given by 85 percent of the value of the base shear obtained with the equivalent lateral force procedure considering a natural period equal to T = Cu Ta, where Cu is determined from Table 17.13 and Ta is the approximate value of the fundamental natural period of the system computed with Equation 17.28, whenever the calculated period exceeds the product of Cu and Ta. To obtain such a limit, consider that for the system under analysis Ta = Ct (hn ) x = 0.028(60)0.8 = 0.74 s T = CuTa = 1.4(0.74) = 1.04 s

and that, in consequence, the calculated period of 0.644 s does not exceed the upper limit imposed by the code for the fundamental natural period. In the light of Equation 17.23, the seismic response coefficient calculated using the equivalent lateral force procedure is therefore equal to Cs =

S DS 1.00 = = 0.29 R / I 3 .5 / 1 .0

which is greater than (see Equation 17.25) (Cs ) min =

0.5S1 0.5(0.60) = = 0.09 R/I 3.5 / 1.0

and exceeds (see Equation 17.26) (Cs ) max =

S D1 0.60 == = 0.27 T (R / I ) 0.644(3.5 / 1.0)

This upper limit applies and, hence, from Equation 17.22, one has that

144

V = CsW = 0.27(4,636.8) = 1,251.9 kips

which leads to a lower limit for the base shear of 0.85×1,251.9 = 1,064.1 kips. This value, in turn, exceeds the value of Vt determined above. The lateral forces, story shears, and overturning moments obtained by the modal analysis procedure will be thus scaled up by the factor (see Equation 17.46) Cm = 0.85

V 1,251.9 = 0.85 = 1.04 Vt 1,022.8

Table P17.11b. Lateral forces, story shears, and overturning moments in first mode Fx1 = C vx1V1 Vx1 wx hsx wx φ x1 wxφx1 φx1 C = vx 1 6 (kips) (ft) (kips) (kips) (kips) wi φi1

Level, x

∑

M0x1 (k-ft)

i =1

6 5 4 3 2 1 Sum:

772.8 772.8 772.8 772.8 772.8 772.8 4,636.8

10 10 10 10 10 10

0.445 0.371 0.297 0.222 0.148 0.074

343.70 286.42 229.14 171.85 114.57 57.28 1202.95

0.286 0.238 0.190 0.143 0.095 0.048 1.00

303.92 253.26 202.61 151.96 101.31 50.65 1,063.71

303.92 557.18 759.80 911.75 1013.06 1,063.71

3,039.2 8,611.0 16,209.0 25,326.5 35,457.1 46,094.2

Lateral forces, story shears, and overturning moments Applying Equations 17.42 and 17.43, the modal values of the vertical distribution factor Cvxm, the lateral forces Fxm, the story shears Vxm and the overturning moments M0xm for the two modes being considered result as indicated in Tables P17.11b and P17.11c. It should be noted that these results correspond to modal base shears augmented by the factor Cm of 1.04. The corresponding design values, obtained by combining the modal values with the square root of the sum of the squares rule, are shown in Table P17.11d. Table P17.11c. Lateral forces, story shears, and overturning moments in second mode Fx 2 = C vx 2V2 Level, x Vx2 M0x2 wx hsx wx φ x 2 wxφx2 φx2 = C vx 2 6 (kips) (k-ft) (kips) (ft) (kips) (kips) wi φi 2

∑ i =1

6 5 4 3 2 1 Sum:

772.8 772.8 772.8 772.8 772.8 772.8 4,636.8

10 10 10 10 10 10

-0.167 -0.293 -0.335 -0.251 0.000 0.460

-129.25 -226.19 -258.50 -193.87 0.00 355.43 -452.38

0.286 0.500 0.571 0.429 0.000 -0.786 1.000

45.64 79.87 91.28 68.46 0.00 -125.51 159.74

45.64 456.4 125.51 1,711.5 216.79 3,879.5 285.25 6,732.0 285.25 9,584.6 159.74 11,182.0

Table P17.11d. Design values of lateral forces, story shears, and overturning moments Lateral forces (kips) Story shears (kips) Overturning moments (k-ft) Level Mode Mode Design Mode Mode Design Mode Mode Design 1 2 value 1 2 value 1 2 value 6 303.9 45.6 307.3 303.9 45.6 307.3 3,039.2 456.4 3,073.3 5 253.3 79.9 265.6 557.2 125.5 571.1 8,611.0 1,711.5 8,779.5 4 202.6 91.3 222.2 759.8 216.8 790.1 16,209.0 3,879.5 16,666.8 145

3 2 1

152.0 101.3 50.6

68.5 0.0 -125.5

166.7 101.3 135.3

911.7 1013.1 1,063.7

285.2 285.2 159.7

955.3 1052.5 1075.6

25,326.5 35,457.1 46,094.2

6,732.0 9,584.6 11,182.0

26,206.0 36,729.7 47,431.1

Table P17.11e. Modal lateral displacements and story drifts in first and second modes First mode Second mode Level wx Fx1 /Cm δxe1 δx1 hsx (kips) (in) (kips) (in) (in) Δx1 (in) Δx2

(in) 6 5 4 3 2 1

Fx2 /Cm (kips)

772.8 772.8 772.8 772.8 772.8 772.8

120 120 120 120 120 120

292.23 243.52 194.82 146.11 97.41 48.71

δxe2 (in)

δx2 (in)

1.53 1.28 1.02 0.77 0.51 0.26

4.60 3.83 3.06 2.30 1.53 0.77

0.04 0.12 -0.09 0.77 43.89 0.77 76.80 0.07 0.20 -0.03 0.77 87.77 0.08 0.23 0.06 0.77 65.83 0.06 0.17 0.17 0.77 0.00 0.00 0.00 0.32 0.77 -120.68 -0.11 -0.32 -0.32

Table P17.11f. Design values of lateral displacements and story drifts Level 6 5 4 3 2 1

Lateral displacements (in) Mode 1 Mode 2 Design value

4.60 3.83 3.06 2.30 1.53 0.77

0.12 0.20 0.23 0.17 0.00 -0.32

Mode 1

4.60 3.84 3.07 2.30 1.53 0.83

Story drifts (in) Mode 2 Design value

0.77 0.77 0.77 0.77 0.77 0.77

-0.09 -0.03 0.06 0.17 0.32 -0.32

0.77 0.77 0.77 0.79 0.83 0.83

Drift ratios

0.006 0.006 0.006 0.007 0.007 0.007

Lateral displacements and story drifts From Equation 17.45, the elastic components of the modal lateral displacements are given by 2 ⎛ g ⎞⎛ T F ⎞ δ xem = ⎜ 2 ⎟⎜⎜ m xm ⎟⎟ ⎝ 4π ⎠⎝ wx ⎠

and according to Equation 17.44, the total modal lateral displacements are equal to δ xm =

Cd δ xem I

where for the case under consideration Cd = 3.0 and I = 1.0. After substituting the floor weights, the natural periods, and the lateral forces divided by the factor Cm = 1.04 (recall the lower limit for the base shear does not apply for displacements and drifts), the lateral displacements and story drifts in the first two modes of the system result as indicated in Table P17.10e. After combining these modal values with the square root of the sum of the squares rule, the design values shown in Table P17.11f are obtained. For comparison with the allowable ones, the story drift ratios are also included in this table. Problem 17.12 Repeat Problem 17.5 using the modal response spectrum analysis procedure prescribed by the International Building Code. 146

Solution: Natural periods The mass and stiffness matrices of the system are 0 0 0 ⎤ ⎡3.882 ⎢ 0 3.106 0 0 ⎥⎥ ⎢ [M ] = kip - s 2 /in ⎢ 0 3.106 0 ⎥ 0 ⎥ ⎢ 0 0 2.588⎦ ⎣ 0

0 0 ⎤ ⎡1750 - 750 ⎢- 750 1500 - 750 0 ⎥⎥ ⎢ [K] = kip/in ⎢ 0 - 750 1250 - 500⎥ ⎢ ⎥ 0 - 500 500 ⎦ ⎣ 0

Therefore, from the solution of the eigenvalue problem, the natural periods and mode shapes are: T1 = 1.077 s

T2 = 0.419 s T3 = 0.287 s ⎡ 0.111 − 0.267 − 0.348 − 0.230⎤ ⎢0.240 − 0.312 0.049 0.406 ⎥⎥ [Φ ] = ⎢ ⎢0.334 − 0.066 0.349 − 0.290⎥ ⎢ ⎥ ⎣0.406 0.396 − 0.237 0.094 ⎦

T4 = 0.223 s

Table P17.12a. Modal effective weights of building considered in Problem 17.4 Effective weight, Wm Total weight percentage Mode (kips) 1 4,118.0 84.0 2 544.7 11.1 3 204.8 4.2 4 32.5 0.7 Total 4,900.0 100.00

Effective modal weights The application of Equation 17.41 with the mode shapes determined above and the given floor weights yields the modal effective weights listed in Table P17.12a. From the percentages shown in this table, it may be seen that the effective weights in the first two modes of the system add up to more than 90 per cent of the total weight of the building. Thus, in accordance with the code recommendations (see Section 17.7.1), only the first two modes will be considered in the analysis. Design response spectrum From the SDS and SD1 values determined in the solution of Problem 17.4 and by virtue of Equations 17.17 and 17.18 and the map in Figure 17.7a, the natural periods that delimit the four regions of the design spectrum for this problem are S D1 0.35 = 0.2 = 0.12 s S DS 0.59 S 0.35 TS = D1 = = 0.59 s S DS 0.59 TL = 6.0 s

T0 = 0.2

Using Equations 17.13 through 17.16, the spectra accelerations in the design spectrum for the system under analysis may be thus defined as follows (see Figure 17.6): (a) For periods less than or equal to 0.12 seconds,

147

S a = S DS (0.4 + 0.6

T T ) = 0.59(0.4 + 0.6 ) = 0.24 + 2.95T T0 0.12

(b) For periods greater than or equal to 0.12 seconds but less than or equal to 0.59 seconds, S a = S DS = 0.59

(c) For periods greater than or equal to 0.59 seconds but less than or equal to 6.0 seconds, Sa =

S D1 0.35 = T T

(d) For periods greater than or equal to 8.0 seconds, Sa =

S D1TL 0.35(6.0) 2.1 = = 2 T2 T2 T

In consequence, the spectral accelerations corresponding to the first two modes of the system are S a1 =

0.60 0.35 = = 0.32 T 1.077 S a 2 = 0.59

Modal and total base shears Since for the building under analysis one has that Occupancy category = IV Importance factor = I =1.5 Response modification coefficient (for special moment concrete frames) = R = 8.0 the seismic response coefficients for the first two modes are equal to S a1 0.32 = = 0.06 R / I 8.0 / 1.5 S 0.59 Cs 2 = a 2 = = 0.11 R / I 8.0 / 1.5 Cs1 =

and, correspondingly, the modal base shears in the first two modes are V1 = Cs1W1 = 0.06(4,118.0) = 247.1 kips V2 = Cs 2W2 = 0.11(544.7) = 59.9 kips

Combining, then, these modal base shears with the square-root-of-the-sum-of-the-squares rule, the total base shear is approximately equal to Vt = V12 + V22 = 247.12 + 59.92 = 254.3 kips

It should be recalled, however, that the code imposes a limit to the value of the base shear that may be considered when the modal analysis procedure is used. As described in Section 17.7.4, this limit is given by 85 percent of the value of the base shear obtained with the equivalent lateral force procedure considering a natural period equal to T = Cu Ta, where Cu is determined from Table 17.13 and Ta is the approximate value of the fundamental natural period of the system computed with Equation 17.28, whenever the calculated period exceeds the product of Cu and Ta. To obtain such a limit, consider that for the system under analysis Ta = 0.44 s, Cu = 1.4 and, as a result, T = CuTa = 1.4(0.44) = 0.62 s

This means that the calculated fundamental period of 1.077 s exceeds the upper limit imposed by the code and that this limit should be used instead. In the light of Equation 17.23, the seismic response coefficient calculated using the equivalent lateral force procedure is therefore equal to Cs =

0.59 S DS = = 0.11 R / I 8.0 / 1.5

which is greater than (see Equation 17.25)

148

(Cs ) min =

0.5S1 0.5(0.16) = = 0.02 R/I 8.0 / 1.5

and equals (see Equation 17.26) (Cs ) max =

S D1 0.35 == = 0.11 T (R / I ) 0.62(8.0 / 1.5)

The calculated value applies and, hence, from Equation 17.22, one has that V = CsW = 0.11(4,900) = 539.0 kips

which leads to a lower limit for the base shear of 0.85×539.0 = 458.2 kips. This value, in turn, exceeds the value of Vt determined above. The lateral forces obtained by the modal analysis procedure will be thus scaled up by the factor (see Equation 17.46) Cm = 0.85

V 539.0 = 0.85 = 1.80 Vt 254.3

Lateral forces Applying Equations 17.42 and 17.43, the modal values of the vertical distribution factor Cvxm and the lateral forces Fxm for the two modes being considered result as indicated in Tables P17.12b and P17.12c. It should be noted that these results correspond to modal base shears augmented by the factor Cm of 1.80. The corresponding design values, obtained by combining the modal values with the square root of the sum of the squares rule, are shown in Table P17.12d. Level, x

Table P17.12b. Lateral forces in first mode wx hsx wφ φx1 wxφx1 Cvx1 = 4 x x1 (kips) (ft) (kips) wi φi1

∑

Fx1 = C vx1V1

(kips)

i =1

4 3 2 1 Sum:

1,000 8 0.406 406.0 0.322 143.17 1,200 10 0.334 400.8 0.318 141.34 1,200 10 0.240 288.0 0.228 101.56 1,500 12 0.111 166.5 0.132 58.71 4,900 1,261.3 1.000 444.78 Table P17.12c. Lateral forces in second mode Fx 2 = C vx 2V2 Level, x wx hsx wx φ x 2 φx2 wxφx2 = C vx 2 4 (kips) (ft) (kips) (kips) wi φi 2

∑ i =1

4 3 2 1 Sum:

1,000 1,200 1,200 1,500 4,900

8 10 10 12

0.396 396 -0.066 -79.2 -0.312 -374.4 -0.267 -400.5 -458.1

-0.864 0.173 0.817 0.874 1.000

-93.20 18.64 88.12 94.26 107.82

Lateral displacements and story drifts From Equation 17.45, the elastic components of the modal lateral displacements are given by 2 ⎛ g ⎞⎛ T F ⎞ δ xem = ⎜ 2 ⎟⎜⎜ m xm ⎟⎟ ⎝ 4π ⎠⎝ wx ⎠

and according to Equation 17.44, the total modal lateral displacements are equal to δ xm =

Cd δ xem I

149

where for the case under consideration Cd = 5.5 and I = 1.5. After substituting the floor weights, the natural periods, and the lateral forces divided by the factor Cm = 1.8 (recall the lower limit for the base shear does not apply for displacements and drifts), the story drifts in the first two modes of the system result as indicated in Table P17.12d. After combining these modal values with the square root of the sum of the squares rule, the design values shown in Table P17.2e are obtained. The allowable drifts, calculated as 0.015 times the story heights, are also included in this table. It may be seen from Table 17.12e that all the story drifts satisfy the limit imposed by the code. Table P17.12d. Modal lateral displacements and story drifts in first and second modes Level Fx1 /Cm δxe1 Fx2 /Cm wx δx1 Δx1 δxe2 δx2 Δx2 (kips) (kips) (kips) (in) (in) (in) (in) (in) (in) 4 1,000 79.5 0.904 3.313 0.588 -51.8 -0.089 -0.326 -0.381 3 1,200 78.5 0.743 2.726 0.767 10.4 0.015 0.054 -0.203 2 1,200 56.4 0.534 1.959 1.053 49.0 0.070 0.257 0.037 1 1,500 32.6 0.247 0.906 0.906 52.4 0.060 0.220 0.220 Table P17.12e. Design and allowable story drifts Level hsx Δx Δa (in) (in) (in) 4 96 0.70 1.44 3 120 0.79 1.80 2 120 1.05 1.80 1 144 0.93 2.16

Problem 17.13 Repeat Problem 17.7 using the modal response spectrum analysis procedure prescribed by the International Building Code. Solution: Natural periods From the solution of the eigenvalue problem, the natural periods and mode shapes in the first three modes of the system are: T1 = 2.809 s

T2 = 1.044 s

150

T3 = 0.605 s

⎧ 0.051⎫ ⎪0.103⎪ ⎪ ⎪ ⎪0.163⎪ ⎪ ⎪ ⎪0.222⎪ ⎪0.288⎪ ⎪ ⎪ ⎪0.356⎪ {φ}1 = ⎨ ⎬ ⎪0.429⎪ ⎪0.497⎪ ⎪ ⎪ ⎪ 0.571⎪ ⎪0.637⎪ ⎪ ⎪ ⎪0.684⎪ ⎪0.712⎪ ⎩ ⎭

⎧ 0.152 ⎫ ⎪ 0.293 ⎪ ⎪ ⎪ ⎪ 0.430 ⎪ ⎪ ⎪ ⎪ 0.528 ⎪ ⎪ 0.583 ⎪ ⎪ ⎪ ⎪ 0.574 ⎪ {φ}2 = ⎨ ⎬ ⎪ 0.484 ⎪ ⎪ 0.317 ⎪ ⎪ ⎪ ⎪ 0.039 ⎪ ⎪− 0.277 ⎪ ⎪ ⎪ ⎪− 0.539⎪ ⎪− 0.696⎪ ⎭ ⎩

⎧ 0.255 ⎫ ⎪ 0.456 ⎪ ⎪ ⎪ ⎪ 0.572 ⎪ ⎪ ⎪ ⎪ 0.538 ⎪ ⎪ 0.325 ⎪ ⎪ ⎪ ⎪− 0.007⎪ {φ}3 = ⎨ ⎬ ⎪− 0.387⎪ ⎪− 0.625⎪ ⎪ ⎪ ⎪− 0.584⎪ ⎪− 0.220⎪ ⎪ ⎪ ⎪ 0.274 ⎪ ⎪ 0.634 ⎪ ⎭ ⎩

Effective modal weights The application of Equation 17.41 with these mode shapes and the floor weights determined in the solution of Problem 17.7 yields the modal effective weights listed in Table P17.13a. It may be seen from this table that the effective weights in the first three modes of the system add up to more than 90 per cent of the total weight of the building. Thus, in accordance with the code recommendations (see Section 17.7.1), only the first three modes will be considered in the analysis. Table P17.13a. Modal effective weights of building considered in Problem 17.4 Effective weight, Wm Total weight percentage Mode (kips) 1 1,461.8 75.9 2 263.8 13.7 3 92.4 4.8 Total 1,818.0 94.4

Design response spectrum From the SDS and SD1 values determined in the solution of Problem 17.7 and by virtue of Equations 17.17 and 17.18 and the map in Figure 17.7a, the natural periods that delimit the four regions of the design spectrum for this problem are S D1 0.43 = 0 .2 = 0.09 s S DS 1.00 S 0.43 TS = D1 = = 0.43 s S DS 1.00 TL = 16.0 s

T0 = 0.2

Using Equations 17.13 through 17.16, the spectra accelerations in the design spectrum for the system under analysis may be thus defined as follows (see Figure 17.6): (a) For periods less than or equal to 0.09 seconds, S a = S DS (0.4 + 0.6

T T ) = 1.00(0.4 + 0.6 ) = 0.4 + 6.7T 0.09 T0

(b) For periods greater than or equal to 0.09 seconds but less than or equal to 0.43 seconds, S a = S DS = 1.00

151

(c) For periods greater than or equal to 0.43 seconds but less than or equal to 16.0 seconds, Sa =

S D1 0.43 = T T

(d) For periods greater than or equal to 16.0 seconds, Sa =

S D1TL 0.43(16.0) 6.9 = = 2 T2 T2 T

In consequence, the spectral accelerations corresponding to the first three modes of the system are 0.43 0.43 = = 0.15 T1 2.809 0.43 0.43 Sa 2 = = = 0.41 T2 1.044 0.43 0.43 Sa3 = = = 0.71 T3 0.605

S a1 =

Modal and total base shears Since for the building in Problem 17.7 one has that Occupancy category = II Importance factor = I =1.0 Response modification coefficient (for special moment concrete frames) = R = 3.5 the seismic response coefficients for the first three modes are equal to S a1 0.15 = = 0.04 R / I 3.5 / 1.00 S 0.41 = 0.12 Cs 2 = a 2 = R / I 3.5 / 1.0 S 0.71 = 0.20 Cs 3 = a 2 = R / I 3 .5 / 1 .0

Cs1 =

and, correspondingly, the modal base shears in the first three modes are V1 = Cs1W1 = 0.04(1,461.8) = 58.5 kips V2 = Cs 2W2 = 0.12(263.8) = 31.7 kips V3 = Cs 3W3 = 0.20(92.4) = 18.5 kips

Combining, then, these modal base shears with the square-root-of-the-sum-of-the-squares rule, the total base shear is approximately equal to Vt = V12 + V22 + V32 = 58.52 + 31.7 2 + 18.52 = 69.1 kips

It should be recalled, however, that the code imposes a limit to the value of the base shear that may be considered when the modal analysis procedure is used. As described in Section 17.7.4, this limit is given by 85 percent of the value of the base shear obtained with the equivalent lateral force procedure considering a natural period equal to T = Cu Ta, where Cu is determined from Table 17.13 and Ta is the approximate value of the fundamental natural period of the system computed with Equation 17.28, whenever the calculated period exceeds the product of Cu and Ta. To obtain such a limit, consider that for the system under analysis Ta = 1.52 s, Cu = 1.4 and, as a result, T = CuTa = 1.4(1.52) = 2.13 s

152

This means that the calculated fundamental period of 2.809 s exceeds the upper limit imposed by the code and that this limit should be used instead. In the light of Equation 17.23, the seismic response coefficient calculated using the equivalent lateral force procedure is therefore equal to Cs =

S DS 1.00 = = 0.29 R / I 3 .5 / 1 .0

which is greater than (see Equation 17.25) (Cs ) min =

0.5S1 0.5(0.50) = = 0.07 R/I 3.5 / 1.0

and exceed (see Equation 17.26) (Cs ) max =

S D1 0.43 == = 0.06 T (R / I ) 2.13(3.5 / 1.0)

This maximum value applies and, hence, from Equation 17.22, one has that V = CsW = 0.06(1,926) = 115.6 kips

which leads to a lower limit for the base shear of 0.85×115.6 = 98.2 kips. This value, in turn, exceeds the value of Vt determined above. The lateral forces obtained by the modal analysis procedure will be thus scaled up by the factor (see Equation 17.46) Cm = 0.85

115.6 V = 0.85 = 1.42 Vt 69.1

Lateral forces, story shears, overturning moments and torsional moments Applying Equations 17.42 and 17.43, the modal values of the vertical distribution factor Cvxm, lateral forces Fxm, story shears Vxm, overturning moments, M0xm, and torsional moments Txm for the three modes being considered result as shown in Tables P17.13b through P17.13d. It should be noted that these results correspond to modal base shears augmented by the factor Cm of 1.42. The corresponding design values, obtained by combining the modal values with the square root of the sum of the squares rule, are listed in Table P17.13e. Table P17.13b. Lateral forces, story shears, and overturning and torsional moments in first mode M0x1 Level wx hsx Tx1 wφ φx1 wxφx1 Fx1 = Cvx1V1 Vx1 Cvx1 = 12 x x1 ip) (kip) (ft) (k (kip-ft) (kip-ft) (kips) (kip) wi φi1

∑ i =1

12 11 10 9 8 7 6 5 4 3 2 1 Sum

144 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 12 162 15 1926 147

0.712 0.684 0.637 0.571 0.497 0.429 0.356 0.288 0.222 0.163 0.103 0.051

103 111 103 92 81 70 58 47 36 26 17 8 751

0.137 0.148 0.137 0.123 0.107 0.093 0.077 0.062 0.048 0.035 0.022 0.011 1.000

153

11.3 12.3 11.4 10.2 8.9 7.7 6.4 5.2 4.0 2.9 1.8 0.9 83.070

11.3 23.6 35.0 45.3 54.2 61.9 68.3 73.4 77.4 80.3 82.2 83.1

136.2 419.6 839.9 1383.0 2033.1 2775.5 3594.6 4475.7 5404.4 6368.1 7354.0 8600.0

612.9 1275.2 1891.5 2444.1 2925.5 3340.8 3685.8 3964.8 4179.3 4336.6 4436.4 4485.8

Table P17.13c. Lateral forces, story shears, and overturning and torsional moments in second mode M0x2 Level wx hsx Vx2 Tx2 wφ φx2 wxφx2 Fx 2 = Cvx 2V2 Cvx 2 = 12 x x 2 (kip) (kip) (ft) (kip-ft) (kip-ft) (kips) (kip) wi φi 2

∑ i =1

12 11 10 9 8 7 6 5 4 3 2 1 Sum

144 12 -0.696 162 12 -0.539 162 12 -0.277 162 12 0.039 162 12 0.317 162 12 0.484 162 12 0.574 162 12 0.583 162 12 0.528 162 12 0.430 162 12 0.293 162 15 0.152 1,926 147

-100 -87 -45 6 51 78 93 94 86 70 47 25 318

-0.315 -0.274 -0.141 0.020 0.161 0.246 0.292 0.297 0.269 0.219 0.149 0.077 1.000

-14.2 -12.4 -6.3 0.9 7.3 11.1 13.2 13.4 12.1 9.9 6.7 3.5 45.014

-14.2 -26.5 -32.9 -32.0 -24.7 -13.7 -0.5 12.9 25.0 34.8 41.5 45.0

-170.3 -488.8 -883.5 -1267.4 -1564.3 -1728.1 -1733.9 -1579.6 -1280.0 -862.1 -363.7 311.6

-766.3 -1433.4 -1776.1 -1727.6 -1335.8 -737.2 -26.4 694.7 1348.1 1880.4 2243.1 2430.8

Table P17.13d. Lateral forces, story shears, and overturning and torsional moments in third mode Level wx hsx Vx3 M0x3 Tx3 wφ φx3 wxφx3 Fx 3 = Cvx3V3 Cvx3 = 12 x x 3 (kip) (ft) (kip) (kip-ft) (kip-ft) (kips) (kip) wi φi 3

∑ i =1

12 11 10 9 8 7 6 5 4 3 2 1 Sum

144 12 162 12 162 12 162 12 162 12 162 12 162 12 1 62 12 162 12 162 12 162 12 162 15 1,926 147

0.634 0.274 -0.220 -0.584 -0.625 -0.387 -0.007 0.325 0.538 0.572 0.456 0.255

91 44 -36 -95 -101 -63 -1 53 87 93 74 41 188

0.486 0.236 -0.189 -0.503 -0.539 -0.334 -0.006 0.280 0.463 0.493 0.393 0.220 1.000

12.8 6.2 -5.0 -13.2 -14.2 -8.8 -0.2 7.4 12.2 13.0 10.3 5.8 26.270

12.8 153.2 689.2 19.0 380.8 1024.5 14.0 548.8 755.8 0.8 558.0 41.6 -13.4 397.4 -722.8 -22.2 131.6 -1196.1 -22.3 -136.1 -1204.6 -15.0 -315.6 -807.5 -2.8 -349.0 -150.5 10.2 -227.0 549.2 20.5 19.0 1107.0 26.3 413.1 1418.6

Lateral displacements and story drifts From Equation 17.45, the elastic components of the modal lateral displacements are given by 2 ⎛ g ⎞⎛ T F ⎞ δ xem = ⎜ 2 ⎟⎜⎜ m xm ⎟⎟ ⎝ 4π ⎠⎝ wx ⎠

and, according to Equation 17.44, the total modal lateral displacements are equal to δ xm =

Cd δ xem I

where for the case under consideration Cd = 3.0 and I = 1.0. After substituting the floor weights, the natural periods, and the lateral forces divided by the factor Cm = 1.42, the lateral displacements and story drifts in the first three modes of the system result as indicated in Table P17.13f. After combining these modal values with the square root of the sum of the squares rule, the design values shown 154

in Table P17.2g are obtained. The allowable drifts, calculated as 0.02 times the story heights, are also included in this table. Table P17.13e. Design values of lateral forces, story shears, and overturning and torsional moments Level Lateral forces Story shears Overturning Torsional (kips) moments (k-ft) moments (k-ft) (kips) 12 22.2 22.2 266.5 1199.1 11 18.5 40.3 748.3 2174.9 50.0 1336.9 10 14.0 2702.5 55.4 1957.2 9 16.7 2993.3 61.0 2595.9 8 18.2 3296.2 67.1 3272.2 7 16.1 3624.3 71.8 3993.3 6 14.6 3877.8 76.0 4756.7 5 16.1 4105.4 81.4 5564.9 4 17.6 4394.0 88.1 6430.2 3 16.5 4758.6 94.3 7363.0 2 12.5 5093.0 6.8 98.1 8615.6 5295.6 1 Table P17.13f. Modal lateral displacements and story drifts in first and second modes Level δxe1 δx1 Δx1 δxe2 δx2 Δx2 δxe3 δx3 Δx3 (in) (in) (in) (in) (in) (in) (in) (in) (in) 12 6.09 18.26 0.72 -1.05 -3.15 -0.71 0.32 0.95 0.54 11 5.85 17.54 1.22 -0.81 -2.44 -1.19 0.14 0.41 0.74 10 5.44 16.32 1.69 -0.42 -1.25 -1.43 -0.11 -0.33 0.55 9 4.88 14.63 1.89 0.06 0.18 -1.26 -0.29 -0.88 0.06 8 4.25 12.75 1.75 0.48 1.43 -0.76 -0.31 -0.94 -0.36 7 3.67 11.00 1.86 0.73 2.19 -0.41 -0.19 -0.58 -0.57 6 3.05 9.14 1.75 0.87 2.60 -0.04 0.00 -0.01 -0.50 5 2.46 7.39 1.71 0.88 2.64 0.25 0.16 0.49 -0.32 4 1.89 5.68 1.52 0.80 2.39 0.44 0.27 0.81 -0.05 3 1.39 4.17 1.53 0.65 1.95 0.62 0.29 0.86 0.17 2 0.88 2.64 1.33 0.44 1.33 0.64 0.23 0.69 0.69 1 0.44 1.31 1.31 0.23 0.69 0.69 0.13 0.38 0.38 Table P17.13g. Design lateral displacements and design and allowable story drifts Level hsx δx Δx Δa = 0.02hsx (in) (in) (in) (in) 12 144 18.56 1.15 2.88 11 144 17.72 1.86 2.88 10 144 16.37 2.28 2.88 2.88 9 144 14.66 2.27 2.88 8 144 12.86 1.94 2.88 7 144 11.23 1.99 2.88 6 144 9.50 1.82 5 144 7.86 1.76 2.88 2.88 4 144 6.22 1.58 2.88 3 144 4.68 1.66 2.88 2 144 3.03 1.63 3.60 1 180 1.53 1.53

155

Problem 17.14 A twenty five-story building is located in the city of Santa Monica, California, has a height of 300 ft, is supported by a 100 ft × 100 ft mat foundation that rests near the surface of a homogeneous soil deposit, and supports a total gravity load of 15,000 kips. The soil deposit exhibits a shear modulus of elasticity of 2.4 × 106 lb/ft2, a unit weight of 120 lb/ft3, and a Poisson ratio of 0.45. From a fixed-base analysis using the equivalent lateral force procedure in the International Building Code, it is found that the building has a fundamental natural period of 3.0 s, and is subjected to a base shear of 1,200 kips, an overturning moment at is base of 240,000 kip-ft, and a lateral displacement at its roof level of 3.0 in. Determine following the provisions of the International Building Code the reduction in the base shear of the building and the increase or decrease in its roof displacement when the flexibility of the foundation soil is taken into account. Solution: Effective weight and effective height For the building under consideration, the effective gravity load and effective height are equal to W = 0.7W = 0.7(15,000) = 10,500 kips h = 0.7 hn = 0.7(300) = 210 ft

Effective stiffness and characteristic foundation lenghts According to Equations 17.58, 17.61, and 17.62, ⎛ W ⎞ ⎛ 10,500 ⎞ k = 4π2 ⎜⎜ 2 ⎟⎟ = 4π 2 ⎜ = 1,430.4 kip/ft 2 ⎟ ⎝ 32.2 × 3.0 ⎠ ⎝ gT ⎠ A0 100 × 100 ra = = = 56.4 ft π π

rm = 4

4 I 0 4 4 × 1004 / 12 = 57.1 ft = π π

Design spectral accelerations The latitude and longitude of Santa Monica, California, are: 34° 1' N and 118° 29' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.50 S1 = 0.60

Also, the building site may be classified as Site D since the soil properties are unknown. Hence, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

and thus, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.5(0.60) = 0.90

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.90) = 0.60 3 3 S DS =

Lateral and rotational stiffness of foundation soil

156

From Table 17.15 for SD1 ≥ 0.30, one has that G/G0 = 0.42. Hence, the shear modulus for the foundation soil adjusted for strain level is G = 0.42G0 = 0.42( 2,400) = 1,008.0 kip/ft 2

Consequently, Equations 17.63 and 17.64 yield the following lateral and rotational static soil stiffnesses: Ky = Kθ =

8Gra 8(1,008.0)(56.4) = = 293,425 kip/ft 2−μ 2 − 0.45

8Grm3 8(1,008.0)(57.1)3 = = 909,860,685 kip - ft/rad 3(1 − μ) 3(1 − 0.45)

The corresponding dynamic stiffnesses (see Equations 13.193 and 13.195) may be determined by multiplying these static stiffnesses by the coefficients αx and αφ given in Figure 13.25 as a function of the dimensionless frequency a0 defined by Equation 13.64 and the material damping tan δ, which according to Equation 13.192 is equal to twice the value of the damping ratio ξ. In this case, tan δ = 2(0.05) = 0.10, and the dimensionless frequencies corresponding to the lateral and rotational characteristic lengths are respectively equal to a0 =

2πra T

γ 2π(56.4) 0.120 = = 0.23 Gg 3.0 (1,008.0)(32.2)

a0 =

2πrm T

γ 2π(57.1) 0.120 = = 0.23 Gg 3 .0 (1,008.0)(32.2)

For these dimensionless frequencies, the coefficients αx and αφ are both very close to unity and, thus, it will be assumed that the dynamic soil stiffnesses are equal to the static stiffnesses. Effective period

From Equation 17.57, the effective period of the building is thus equal to 2 K yh 2 ⎞ ⎛ ⎞ k ⎛⎜ ~ ⎟ = T 1 + 1,430.4 ⎜1 + 293,425 × 210 ⎟ T = T 1+ 1+ ⎜ ⎟ ⎜ ⎟ 293,425 ⎝ 909,860,685 ⎠ Ky ⎝ Kθ ⎠ = 1.036T = 1.036(3.0) = 3.11 s

Effective damping ratio In this case S DS / 2.5 = 1.00 / 2.5 = 0.40 ≥ 0.20 h / L0 = 210 / 100 = 2.10 > 1.0 r = rm = 57.1 ft h / r = 210 / 57.1 = 3.7 ~ T 3.11 = = 1.036 3 .0 T

Thus, from Figure 17.10 one obtains β0 ≈ 0.005

and the effective damping ratio of the building results as 0.05 0.05 ~ β = β0 + ~ = 0.005 + = 0.05 3 1.0363 (T / T )

which is equal to the specified minimum of 0.05.

157

Base shear reduction In accordance with Section 17.6.2, the seismic response coefficient of the fixed-base structure is given by Cs =

S D1 T (R / I )

and thus the seismic response coefficient for the structure on flexible soil may be expressed as S S D1 ~ Cs = ~ D1 = = 0.965Cs 1 . 036 T (R / I ) T (R / I )

In the light of Equation 17.49, the reduction in base shear for the building being considered due to soil-structure interaction effects is therefore equal to 0.4 0.4 ⎡ ⎡ ~ ⎛ 0.05 ⎞ ⎤ ⎛ 0.05 ⎞ ⎤ ⎟⎟ ⎥W = Cs ⎢1 − 0.965⎜ ΔV = ⎢Cs − Cs ⎜⎜ ⎟ ⎥ 0.7W = 0.025CsW ⎢⎣ ⎝ 0.05 ⎠ ⎥⎦ ⎢⎣ ⎝ β ⎠ ⎥⎦

which represents a reduction of 2.5 percent over the base shear on the fixed-base building. The corresponding reduced base shear is ~ V = V − ΔV = CsW − 0.025CsW = (1 − 0.025)CsW = 0.975V = 0.975(1,200) = 1,170 kips

Roof displacement increase or decrease From Equation 17.69, the roof displacement of the building on flexible soil is equal to ~ ⎞ 1,170 ⎛ 240,000 × 300 ⎞ ~ V ⎛ M 0 hn + 0.25 ⎟⎟ = 0.321 ft = 3.85 in δn = ⎜⎜ + δ x ⎟⎟ = ⎜⎜ V ⎝ Kθ ⎠ ⎠ 1,200 ⎝ 909,860,685

which represents an increase of 28 percent over the value obtained when the building is considered with a fixed base. Problem 17.15 The tower shown in Figure P17.15 is located in the city of Irvine, California, over a site classified as Class E. It supports a water tank that weighs 2,000 kN. The tower foundation is a reinforced concrete circular mat with a diameter of 12 m, a depth of 1.0 m, and a unit weight of 24 kN/m3. The foundation soil has a shear modulus of elasticity of 72 MN/m2, a unit weight of 17.7 kN/m3, and a Poisson ratio of 0.45. When modeled as a single-degree-of-freedom system with a fixed base, the tower exhibits a natural period of 0.25 s. Modeling it as a single degree-offreedom and using the equivalent lateral force procedure prescribed by the International Building Code, determine the base shear and maximum lateral displacement in the tower considering the flexibility of the foundation soil. Assume a response modification coefficient of 2.5, a deflection amplification factor of 2.5, and an importance factor of 1.0. Neglect the weight of the tower.

15 m

1.0 m 6m

6m

Figure P17.15. Water tower considered in Problem 17.15

158

Solution: Base shear The latitude and longitude of Irvine, California, are: 33° 40' N and 117° 49' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.50 S1 = 0.60

Also, the tank site is classified as Class E. Hence, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

and thus, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.5(0.60) = 0.90

and, according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.90) = 0.60 3 3

S DS =

For this structure, one has that Importance factor = I =1.0 Response modification coefficient = R = 2.5 Then, according to Equation 17.23, the seismic response coefficient is equal to Cs =

S DS 1.00 = = 0.40 R / I 2 .5 / 1 .0

which is greater than (Cs ) min =

0.5S1 0.5(0.60) = = 0.12 R/I 2.5 / 1.0

and is less than (Cs ) max =

S D1 0.60 = = 0.96 T ( R / I ) 0.25(2.5/1.0)

In consequence, the fixed-base base shear is V = CsW = 0.40(2,000) = 800 kN

Lateral displacement For the structure under consideration, T = 0.25 s, Fx = V = 800 kN, and Cd = 2.5. Hence, from Equation 17.45, the elastic component of the lateral displacement is equal to 2 2 ⎛ g ⎞⎛ T Fx ⎞⎟ ⎛ 9.81 ⎞⎛⎜ 0.25 × 800 ⎞⎟ = 0.0062 m = δ xe = ⎜ 2 ⎟⎜⎜ ⎜ ⎟ 2 ⎝ 4π ⎠⎝ wx ⎟⎠ ⎝ 4π ⎠⎜⎝ 2,000 ⎟⎠

and, according to Equation 17.44, the fixed-base total lateral displacement is equal to δx =

Cd δ xe 2.5(0.0062) = = 0.0155 m 1 .0 I

Effective stiffness According to Equation 17.58, the tower’s effective stiffness is equal to ⎛ W ⎞ ⎛ 2,000 ⎞ k = 4π2 ⎜⎜ 2 ⎟⎟ = 4π 2 ⎜ = 128,778 kN/m 2 ⎟ ⎝ 9.81 × 0.25 ⎠ ⎝ gT ⎠

159

Lateral and rotational stiffness of foundation soil From Table 17.15 for SD1 ≥ 0.30, one has that G/G0 = 0.42. Hence, the shear modulus for the foundation soil adjusted for strain level is G = 0.42G0 = 0.42(72,000) = 30,240 kN/m 2

Hence, according to Equations 17.63 and 17.64, the lateral and rotational static soil stiffnesses are: Ky = Kθ =

8Gr 8(30,240)(6) = = 936,465 kN/m 2−μ 2 − 0.45

8Gr 3 8(30,240)(6)3 = = 31,669,527 kN - m/rad 3(1 − μ) 3(1 − 0.45)

The corresponding dynamic stiffnesses (see Equations 13.193 and 13.195) may be determined by multiplying these static stiffnesses by the coefficients αx and αφ given in Figure 13.25 as a function of the dimensionless frequency a0 defined by Equation 13.64 and the material damping tan δ. In this case, tan δ = 2(0.05) = 0.10 and the dimensionless frequency is equal to a0 =

2πr T

2π(6) γ 17.7 = = 1.16 0.25 (30,240)(9.81) Gg

For this material damping and dimensionless frequency, the coefficients ax and aφ are equal to α x = 0.95

α φ = 0.80

and, thus, the dynamic soil stiffnesses are equal to k y = α x K y = 0.95(936,465) = 889,642 kN/m kθ = α φ K θ = 0.80(31,669,527) = 25,335,622 kN - m/rad Effective period

From Equation 17.57 and the stiffnesses determined above, the effective period of the system is equal to 2 128,778 ⎛ 889,642 ×152 ⎞ k ⎛ k y h ⎞⎟ ~ ⎜1 + ⎟ = T 1+ T = T 1 + ⎜1 + 889,642 ⎜⎝ k y ⎜⎝ kθ ⎟⎠ 25,335,622 ⎟⎠ = 1.513T = 1.513(0.25) = 0.378 s

Effective damping ratio In this case, S DS / 2.5 = 1.00 / 2.5 = 0.40 ≥ 0.20 h / r = 15 / 6 = 2.5 ~ T 0.378 = = 1.513 T 0.25

Therefore, from Figure 17.10 one obtains β0 ≈ 0.074

and the effective damping ratio results as 0.05 0.05 ~ β = β0 + ~ = 0.074 + = 0.088 3 1.5133 (T / T )

which is greater than the specified minimum of 0.05 and less than the maximum of 0.20. Base shear on flexible soil In accordance with Section 17.6.2, the seismic response coefficient of the structure on flexible soil is given by 160

1.00 S ~ = 0.40 Cs = DS = R / I 2.5 / 1.0

which is greater than 0.5S1 0.5(0.60) ~ (Cs ) min = = = 0.12 R/I 2.5 / 1.0

and is less than ~ (Cs ) max =

S D1 0.60 = = 0.63 T ( R / I ) 0.378(2.5/1.0)

In the light of Equation 17.49, the reduction in the base shear for the system being considered due to soil-structure interaction effects is therefore equal to 0.4 ⎡ ⎡ ⎛ 0.05 ⎞0.4 ⎤ ~ ⎛ 0.05 ⎞ ⎤ ⎢ ⎥ ⎜ ⎟ ΔV = Cs − Cs ⎜ ⎟ W = 0.40 ⎢1 − ⎜ 0.088 ⎟ ⎥W = 0.08(2,000) = 160 kN ⎢⎣ ⎠ ⎥⎦ ⎝ β ⎠ ⎥⎦ ⎢⎣ ⎝

which is less than 0.3V = 0.3(800) = 240 kN. The corresponding reduced base shear is consequently equal to ~ V = V − ΔV = 800 − 160 = 640 kN

which represents a reduction of 20 percent over the base shear on the fixed-base building. Maximum lateral displacement on flexible soil The overturning moment at the base of the tower using the unmodified lateral forces is M 0 = 800(15) = 12,000 kN - m

Therefore, from Equation 17.69, the tower’s top displacement when on flexible soil is equal to ~ ⎞ 640 ⎛ 12,000(15) ~ V ⎛ M 0 hn ⎞ δn = ⎜⎜ + δ x ⎟⎟ = + 0.0155 ⎟⎟ = 0.0181 m ⎜⎜ V ⎝ kθ ⎠ ⎠ 800 ⎝ 25,335,622

which represents an increase of 16.7 percent over the value obtained when the tower is considered fixed at its base. Problem 17.16 The steam boiler shown in Figure P17.16 will be installed on the roof of a four-story building that is 40 ft high. The boiler weighs 20 kips and will be anchored to the roof slab using four bolts, one-inch in diameter each. The bolts have an embedment length of 6 in. The fundamental period of the boiler and its anchors is 0.04 seconds. The building is located in Fairbanks, Alaska, over a site classified as Class C and will be used as a hotel. Determine using the provisions of the International Building Code the shear and tension demands on the anchor bolts. Fp 20 k 1" φ anchor bolts 6" embed. (4 total)

4.0'

2.5'

Figure P17.16. Steam boiler considered in Problem 17.16 161

Solution: Design spectral accelerations The latitude and longitude of Fairbanks, Alaska, are 64° 49' N and 147° 52' W, respectively. Therefore, from the maps in Figures 17.3 and 17.4, one has that S S = 1.12 S1 = 0.31

Also, as the building site is classified as Class C, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.49

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.12) = 1.12 S M 1 = Fv S1 = 1.49(0.31) = 0.46

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.12) = 0.75 3 3 2 2 S D1 = S M 1 = (0.46) = 0.31 3 3

S DS =

Equivalent static force From Table 17.19, the component amplification ap and the component response modification factor Rp for boilers are equal to a p = 1 .0

R p = 2 .5

and from Table 17.17 the component importance factor is I p = 1 .0

Accordingly, since for the building and equipment location under consideration, z = h = 40 ft

from Equation 17.74 one obtains Fp =

0.4a p S DS W p Rp / I p

z (1 + 2 ) h

40 0.4(1.0)(0.75) (1 + 2 )W p = 0.36W p 40 2.5 / 1.0 = 0.36(20) = 7.2 kips =

which is greater than ( Fp ) min = 0.3S DS I pW p = 0.3(0.75)(1.0)W p = 0.225W p = 0.225(20) = 4.5 kips

and less than ( Fp ) max = 1.6 S DS I pW p = 1.6(0.75)(1.0)W p = 1.20W p = 1.20(20) = 24.0 kips

Shear force and tension demands Thus, the shear force demand on each anchor is equal to V=

7.2 = 1.8 kips 4

In like fashion, the overturning moment due to the lateral force is equal to M 0 = 7.2( 4) = 28.8 kip - ft

162

and thus, in the light of Equation 17.77, earthquake effects alone produce an uplift force on each anchor equal to ( Ft ) E =

1 28.8 0.2(0.75)(20) + = 6.5 kips 2 2 .5 4

Consequently, when combined with the gravity load in accordance with Equation 17.2, the tension demand on the anchors is ( Ft ) D + E = 6.5 − 0.9

20 = 2.0 kips 4

Problem 17.17 The electric generator shown in Figure P17.17 will be installed in the third floor of a 5-story emergency command center. The command center is located in the downtown area of Crescent City, California, over a site classified as Class E. The fundamental natural period of the building is 0.4 seconds and the height of each of its stories is 12 feet. The generator weighs 15 kips and will be mounted on four vibration isolators, each with a lateral stiffness of 3 kips per inch. Determine using the provisions of the International Building Code the shear and tension demands on the vibration isolators. 15 k Fp Vibration isolators (4 total)

4.5 ft

4.0 ft

Figure P17.17. Electric generator considered in Problem 17.17

Solution: Design spectral acclerations The latitude and longitude of Crescent City, California, are 41° 46' N and 124° 12' W, respectively. Therefore, from the maps in Figures 17.1a and 17.2a, one has that S S = 1.50 S1 = 0.75

Also, as the building site is classified as Class E, from Tables 17.4 and 17.5, one obtains, Fa = 0.9

Fv = 2.4

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 0.9(1.50) = 1.35 S M 1 = Fv S1 = 2.4(0.75) = 1.80

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 S MS = 3 2 S D1 = S M 1 = 3

S DS =

2 (1.35) = 0.90 3 2 (1.80) = 1.20 3

163

Equivalent static force From Table 17.19, the component amplification ap and the component response modification factor Rp for spring-isolated equipment are a p = 2.5

R p = 2.0

and from Table 17.17 the component importance factor (for an emergency command center) is I p = 1.5

Accordingly, since for the building and equipment location under consideration, z = 3(12) = 36 ft

from Equation 17.74 one obtains Fp =

0.4a p S DS W p Rp / I p

z (1 + 2 ) h

36 0.4(2.5)(0.90) (1 + 2 )W p = 1.485W p 60 2.0 / 1.5 = 1.485(15) = 22.3 kips =

which is greater than ( Fp ) min = 0.3S DS I pW p = 0.3(0.90)(1.5)W p = 0.405W p = 0.405(15) = 6.1kips

and less than ( Fp ) max = 1.6 S DS I pW p = 1.6(0.90)(1.5)W p = 2.16W p = 2.16(15) = 32.4 kips

It should be noted, however, that the code requires that the force Fp be doubled for components mounted on vibration isolators if the nominal clearance between the equipment support frame and restraint is greater than 0.25 in. (see note under Table 17.19). Therefore, since no restrain is specified, the lateral force on the component is Fp = 2(22.3) = 44.6 kips

Shear force and tension demands Thus, the shear force demand on each isolator is equal to V=

44.6 = 11.2 kips 4

In like fashion, the overturning moment due to the lateral force is equal to M 0 = 44.6(4.5) = 200.7 kip - ft

and thus, in the light of Equation 17.77, earthquake effects alone produce an uplift force on each isolator equal to ( Ft ) E =

1 200.7 0.2(0.90)(15) + = 25.8 kips 2 4.0 4

Consequently, when combined with the gravity load in accordance with Equation 17.2, the tension demand on the anchors is ( Ft ) D + E = 25.8 − 0.9

15 = 22.4 kips 4

Problem 17.18 A vertical steel pipeline is connected to a five-story building as shown in Figure P17.18. The building is located in Long Beach, California, over a deposit of stiff sandy soils and is used as a hospital. It is constructed with ordinary moment-resisting steel frames and carries a total gravity load of 1,000 kips per floor. The total weight of the pipeline is 6 kips. Determine using the pro164

visions of the International Building Code the design lateral force and the relative displacement demand for the pipeline. Assume that the moments of inertia of the beams and columns of the building are respectively equal to 1170 in4 and 1530 in4. Similarly, consider that the modulus of elasticity of the structural steel is 29,000 kips/in2. 5

5 @ 12 ft = 60 ft

4 3 2 1

Pipeline

3 @ 24 ft = 78 ft

Figure P17.18. Building and pipeline considered in Problem 17.18 Solution:

Building weight The total weight supported by the building is

W = 5(1,000) = 5,000 kips

Natural period From Equation 17.28 and the period coefficient and exponent in Table 17.12 for momentresisting steel frames, the approximate value of the fundamental natural period is Ta = Ct hnx = 0.028(60)0.8 = 0.74 s

Base shear The latitude and longitude of Long Beach, California, are 33° 49' N and 118° 9' W, respectively. Therefore, from the maps in Figures 17.1b and 17.2b, one has that S S = 1.75 S1 = 0.60

Also, as the building site may be classified as Class C (see Table 17.3), from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.3

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.75) = 1.75 S M 1 = Fv S1 = 1.3(0.60) = 0.78

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.75) = 1.17 3 3 2 2 S D1 = S M 1 = (0.78) = 0.52 3 3 S DS =

Furthermore, from Table 17.6, 17.7, 17.8, and 17.1, one has that Occupancy category = IV Importance factor = I =1.5 Seismic design category = D Response modification coefficient (for ordinary moment frames) = R = 3.5 165

Hence, according to Equation 17.23 Cs =

1.17 S DS = = 0.50 R / I 3.5 / 1.5

which is greater than (Cs ) min =

0.5S1 0.5(0.60) = = 0.13 R/I 3.5 / 1.5

and greater than (Cs ) max =

S D1 0.52 = = 0.30 T ( R / I ) 0.74(3.5/1.5)

and thus Cs = (Cs ) max = 0.30

In consequence, the base shear is equal to V = CsW = 0.30(5,000) = 1,500 kips

Lateral forces From Table 17.14, one obtains 1 1 k = 1.0 + (T − 0.5) = 1.0 + (0.74 − 0.5) = 1.12 2 2

and thus from Equations 17.32 and 17.33, the lateral forces Fx on the building are as indicated in Table P17.18a. Table P17.18a. Lateral forces and story shears in building in Problem 17.18 Level

wx (kip)

hsx (ft)

hx (ft)

wxhxk (kip-ft)

C vx =

w x h xk N

∑w h

k i i

Fx = C vxV (kip)

i =1

5 4 3 2 1 Sum

1,000 1,000 1,000 1,000 1,000 1926

12 12 12 12 12 60

60 48 36 24 12

98,068.47 76,381.85 55,342.50 35,142.82 16,168.99 281,104.64

0.349 0.272 0.197 0.125 0.058 1.000

523.30 407.58 295.31 187.53 86.28 1,500.0

Lateral displacements and story drifts With the lateral forces listed in Table P17.18a, the given beam and column moments of inertia, and using a computer program for the static analysis of frames, the elastic floor displacements (δxe) result as shown in Table P17.18b. The corresponding inelastic displacements (δx) are also listed in this table. These displacements are computed according to Equation 17.36, considering an importance factor I of 1.5 and a deflection amplification factor Cd of 3.0. Table P17.18b. Elastic and total ateral displacements in building of Problem 17.18 Level δ xe δx (in) (in) 5 29.6 59.2 4 26.1 52.3 3 20.4 40.8 2 13.0 25.9 1 5.0 10.0

166

Component equivalent static force From Table 17.19, the component amplification ap and the component response modification factor Rp for piping are a p = 2.5

R p = 4 .5

and from Table 17.17 the component importance factor (for a hospital) is I p = 1.5

Also, for the building and pipeline location under consideration, z = 60 ft

Accordingly, from Equation 17.74, the design lateral force is Fp =

0.4a p S DS W p Rp / I p

z (1 + 2 ) h

60 0.4(2.5)(1.17) (1 + 2 )W p = 1.17W p 60 4 .5 / 1 .5 = 1.17(6) = 7.02 kips

=

which is greater than ( Fp ) min = 0.3S DS I pW p = 0.3(1.17)(1.5)W p = 0.53W p = 0.53(6) = 3.2 kips

and less than ( Fp ) max = 1.6S DS I pW p = 1.6(1.17)(1.5)W p = 2.81W p = 2.81(6) = 16.9 kips

Relative displacement demand From Equation 17.80, the relative displacement demand is calculated according to D p = δ xA − δ yA

where δxA and δyA are the displacements at levels x and y of the structure determined as indicated in Section 17.6.7. For the case under consideration these displacements are equal to δ xA = 59.2 in δ yA = 10.0 in

and thus Equation 17.80 leads to D p = 59.2 − 10.0 = 49.2 in

However, since for the building under consideration the allowable drift Δa is equal to 0.010hsx, the calculated value of Dp is greater than the upper limit of (see Equation 17.81) ( D p ) max = (hx − hy )

Δ aA 0.010hsx = (720 − 144) = 5.8 in hsx hsx

In consequence, the relative displacement demand is D p = ( D p ) max = 5.8 in

Problem 17.19 A three-story office building will be implemented with a base isolation system. The building will be located on Anchorage, Alaska, on a site soil classified as Class D. The building has a regular configuration with no vertical or horizontal irregularities. Its plan dimensions are 150 ft × 100 ft and its weight for seismic design is 6,200 kips. The lateral load resisting system consists of ordinary reinforced concrete moment frames. The center of mass and the center of rigidity coincide at each floor. The fundamental natural period of the building when its base is considered fixed is 0.35 s. It is estimated that for an effective performance the isolation system should 167

have effective isolated periods at the design and maximum displacements equal to 2.5 and 3.0 times the fixed-base natural period, respectively, and an effective damping ratio of 15%. A variation of ±15% from the mean stiffness values of the isolation elements is considered acceptable. For the purpose of a preliminary design, determine using the provisions of the International Building Code: (a) the minimum design displacement for the isolation system; (b) the base shear for designing the isolation system and the structural elements below the isolation system; and (c) the base shear for the design of the superstructure. Solution: Design spectral accelerations The latitude and longitude of Anchorage, Alaska, are 61° 10' N and 150° 1' W, respectively. Therefore, from the maps in Figures 17.3 and 17.4, one has that S S = 1.50 S1 = 0.55

Also, as the building will be built on a Site D, from Tables 17.4 and 17.5, one obtains, Fa = 1.0

Fv = 1.5

Hence, according to Equations 17.9 and 17.10, the spectral accelerations adjusted for soil effects are S MS = Fa S s = 1.0(1.50) = 1.50 S M 1 = Fv S1 = 1.5(0.55) = 0.83

and according to Equations 17.11 and 17.12, the corresponding design spectral accelerations are 2 2 S MS = (1.50) = 1.00 3 3 2 2 S D1 = S M 1 = (0.83) = 0.55 3 3 S DS =

Minimum design displacement for isolation system From Table 17.20 for a damping ratio of 15 percent, one gets BD = BM = 1.35

Thus, considering Equations 17.86 and 17.87, the calculated values of SD1 and SM1, and the assumed values for the design and maximum natural periods of the isolation system (TD and TM), the design displacement and maximum displacement at the center of rigidity of the isolation system are gS D1TD 386.4(0.55)(2.5 × 0.35) = 3.49 in = 4π2 (1.35) 4π2 BD 386.4(0.83)(3.0 × 0.35) gS T DM = M2 1 M = = 6.32 in 4π BM 4π 2 (1.35) DD =

To compute the maximum displacements of the individual isolation elements, it is necessary to account for the code-mandated mass eccentricity of 5 percent of the building dimension perpendicular to the direction of analysis. That is, the total displacements of the isolation elements need be calculated for an eccentricity of e = 0.05(150) = 7.5 ft

Thus, assuming that the isolation system will have a uniform spatial distribution of lateral stiffness and using Equations 17.88 and 17.89, estimates of the total displacements for the isolation elements at the building edges are

168

12e ⎤ ⎡ ⎡ 150 12(7.5) ⎤ DTD = DD ⎢1 + y 2 = 3.49 ⎢1 + = 3.49(1.21) = 4.22 in 2⎥ 2 100 2 + 150 2 ⎥⎦ b +d ⎦ ⎣ ⎣ 12e ⎤ ⎡ DTM = DM ⎢1 + y 2 = 6.32(1.21) = 7.65 in. b + d 2 ⎥⎦ ⎣

Base shear for design of isolation system and structural elements below From Equations 17.84 and 17.85, the assumed values of TD and TM, and the seismic weight W, one has that k D min =

4π 2 W 4π2 6,200 = = 827.4 kip/in. TD2 g 0.8752 386.4

k M min =

4π 2 W 4π2 6,200 = = 574.6 kip/in. TM2 g 1.052 386.4

which, after consideration of the assumed ±15% variation about the mean stiffness values, lead to k D min 1.15 = (827.4) = 1,119.4 kip/in. 0.85 0.85 1.15 = (574.6) = 777.4 kip/in. 0.85

k D max = 1.15k Dmean = 1.15

k M max

Thus, from Equation 17.90, the design shear force for the isolation system and the structural elements below is Vb = k D max DD = 1,119.4(3.49) = 3,907 kips

which corresponds to a base shear coefficient of 3,907/6,200 = 0.63. Base shear for the design of superstructure The strength reduction factor for ordinary reinforced concrete frames is 3.0 (see Table 17.1) and, in consequence, the corresponding reduction factor for the base-isolated structure is (see Section 17.12.4.3) 3 RI = (3.0) = 1.13 ≤ 2.0 8

the base shear for the design of the superstructure is (see Equation 17.91) Vs =

k D max DD 1,119.4(3.49) = = 3,457 kips 1.13 RI

which is tantamount to a seismic coefficient of 3,457/6,200 = 0.56. The design base shear for a fixed-base structure with the same weight, a natural period equal to TD = 0.875 s, and an importance factor of 1.0 (for an office building) is equal to (see Equations 17.22 and 17.26) V = CsW =

0.55 S D1 W= 6,200 = 0.18(6,200) = 1,116 kips R/I 3.0 / 1.0

which is less than the upper limit of (V ) max =

S D1 0.55 W= 6,200 = 0.21(6,200) = 1,299 kips T (R / I ) 0.875(3.0 / 1.0)

Thus, the calculated base shear of 3,457 kips complies with the code requirement that the base shear for the isolated structure should not be less that the base shear for a fixed-base structure with the same weight and same natural period.

169

Problem 17.20 Repeat Example 17.6 considering instead linear viscous dampers. Solution: Base shear, lateral forces, and story drifts As a change in the force-velocity relationship of the dampers will not affect the considered increase in the damping ratio of the system (20 percent), the base shear, lateral forces, and story drifts of the frame when the dampers added are linear viscous dampers will be as calculated in Example 17.6.

Dampers’ damping constant From Equation 17.156, one has that

∑W

1j

j

4πW1

= 0.20

where ΣW1j and W1 are, respectively, the dissipated energy in one cycle of response and the maximum strain energy in the first mode of the structure under the first-mode displacements determined in Example 17.6. But according to Equation 16.25, ΣW1j may be calculated according to

∑W

1j

= λC N ω1α [(Δ11 cos θ1 ) α+1 + Δ12 cos θ 2 ) α+1 + Δ13 cos θ3 ) α+1 ]

j

where λ is given by Equation 17.165, θ1, θ2, and θ3 are the angles the dampers in the first, second, and third stories make with respect the horizontal, and Δ11, Δ12, and Δ13 are the story drifts of the first, second, and third stories when the structure is subjected to the displacements {δi1}. For the structure under consideration θ1 = 28.2˚, θ2 = θ3 = 27.6˚, and for viscous dampers α = 1.0 and thus λ = 4( 2 α )

Γ 2 (1 + α / 2) Γ 2 (1 + 1.0 / 2) 0.8862 = 4(2) = 8.0 = 3.140 Γ(2 + α) Γ(2 + 1.0) 2.0

In consequence,

∑W

1j

= C N (3.140)(4.4487){(0.023 cos 28.2o ) 2.0 + [(0.059 − 0.023) cos 27.6o ]2.0

j

+ [(0.088 − 0.059) cos 27.6o ]2.0 } = 0.029C N

Similarly, according to Equation 17.157 and the first-mode lateral forces computed in Example 17.6, the maximum strain energy is given by W1 =

1 2

∑F δ

i1 i1

i

1 = [33.1(0.023) + 84.2(0.059) + 67.4(0.088)] = 5.83 kN - m 2

Therefore, the dampers’ damping constant is equal to CN =

4π(0.20)(5.83) = 505.2 kN - s / m 0.029

Maximum damper forces According to Equation 16.22 and the horizontal story velocities calculated in Example 17.6, the maximum forces in the dampers in the fundamental and residual modes are given by ⎧0.192 cos 28.2o ⎫ ⎧ 85.5 ⎫ ⎧ ∇11M cos θ1 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ {F1D } = C N ⎨∇ 21M cos θ2 ⎬ = 505.2⎨0.300 cos 27.6o ⎬ = ⎨134.3⎬kN ⎪0.236 cos 27.6o ⎪ ⎪105.7 ⎪ ⎪∇ cos θ ⎪ 3⎭ ⎭ ⎩ 31M ⎩ ⎭ ⎩

170

⎧ 0.512 cos 28.2o ⎫ ⎧ 228.0 ⎫ ⎧ ∇1RM cos θ1 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ {FRD } = C N ⎨∇ 2 RM cos θ2 ⎬ = 505.2⎨− 0.456 cos 27.6o ⎬ = ⎨− 204.1⎬kN ⎪∇ ⎪ ⎪− 0.367 cos 27.6o ⎪ ⎪ − 164.3⎪ ⎩ 3 RM cos θ3 ⎭ ⎭ ⎩ ⎭ ⎩

which lead to the following maximum damper forces: ⎧ (85.52 + 228.02 )1 / 2 ⎫ ⎧243.5⎫ ⎪ ⎪ ⎪ ⎪ {FD } = ⎨(134.32 + 204.12 )1 / 2 ⎬ = ⎨244.3⎬kN 2 2 1/ 2 ⎪ ⎪ ⎪ ⎪ ⎩(105.7 + 164.3 ) ⎭ ⎩195.4 ⎭

Thus, three linear viscous dampers with an output force of at least 245 kN on each of the two seismic-force resisting frames will provide the additional damping of 20 percent. Problem 17.21 Repeat Example 17.6 using instead the response spectrum procedure prescribed in the provisions of the International Building Code. Solution: Effective weights According to the information given in Table E17.6, the effective weights in the first two modes of the structure add up to more than 90% of the structure’s total weight. Hence, only the first two modes will be considered in the subsequent analysis.

Base shear From the calculations in Example 17.6, the base shear in the fundamental mode is equal to V1 = 184.7 kN

For the calculation of the base shear in the second mode, it needs to be considered that the code stipulates that the inherent damping be assumed the same in all modes, that the effective ductility demands in the higher modes be all equal to 1.0, and that the hysteretic damping be taken as zero for the higher modes. Therefore, according to Equation 17.147, the effective damping in the second mode is equal to β2 D = β I + βV 2 + β HD = 0.05 + 0.20 + 0 = 0.25

and from Table 17.21 the numerical coefficient that corresponds to this effective damping is B2 D = 1.65

According to Equations 17.139 and 17.137, the seismic coefficient and base shear in the second mode of the structure are thus equal to ⎛ R ⎞ S DS ⎛ 8.0 ⎞ 1.0 ⎟ Cs 2 = ⎜⎜ =⎜ = 0.294 ⎟ ⎟ ⎝ Cd ⎠ Ω 0 B2 D ⎝ 5.5 ⎠ 3.0(1.65) V2 = Cs 2W2 = 0.294(0.147)(3,683.5) = 159.2 kN

and, in consequence, the total base shear is equal to V = V12 + V22 = (184.7) 2 + (159.2) 2 = 243.8 kN

This value, however, is less than the minimum of 262.5 kN calculated in Example 17.6. As a result, it will be considered that V = 262.5 kN

Design lateral forces Using Equations 17.41 and the dynamic properties listed in Table E17.6, the design lateral forces in 171

the first two modes of the system are {F1} = {wi φi1}

⎧1,450.0(0.265)⎫ ⎧33.1⎫ Γ1 ⎪ ⎪ ⎪ ⎪ V1 = 1.388⎨1,450.0(0.675)⎬0.062 = ⎨84.2 ⎬kN W1 ⎪ 783.5(1.000) ⎪ ⎪67.4⎪ ⎩ ⎭ ⎩ ⎭

⎧ 165.7 ⎫ ⎧1,450.0(−0.739)⎫ Γ2 ⎪ ⎪ ⎪ ⎪ {F2 } = {wi φi 2 } V2 = −0.526⎨1,450.0(−0.510)⎬0.294 = ⎨ 114.4 ⎬kN W2 ⎪− 121.2⎪ ⎪ 783.5(1.000) ⎪ ⎩ ⎭ ⎩ ⎭ and thus the total lateral forces are ⎧(33.12 + 165.7 2 )1 / 2 ⎫ ⎧169.0⎫ ⎪ ⎪ ⎪ ⎪ {Fx } = ⎨(84.2 2 + 114.42 )1 / 2 ⎬ = ⎨142.0⎬kN ⎪(67.42 + 121.22 )1 / 2 ⎪ ⎪138.7⎪ ⎭ ⎩ ⎭ ⎩

Floor deflections and story drifts From the calculation in Example 17.6, the floor deflections in the fundamental mode are equal to ⎧0.265⎫ ⎧0.047⎫ ⎪ ⎪ ⎪ ⎪ {δ1D } = D1D {φi1} = 0.177 ⎨0.675⎬ = ⎨0.119⎬m ⎪1.000 ⎪ ⎪0.177⎪ ⎩ ⎭ ⎩ ⎭

Also, according to Equation 17.144 and the value of B2D found above, the roof deflection in the second mode is equal to D2 D = (

g S T 9.81 0.6(0.435) )Γ2 D1 2 = − ( 2 )0.526 = −0.021 m 2 B2 D 1.65 4π 4π

which is greater than ( D2 D ) max = (

9.81 1.0(0.435) 2 g S DS T22 ) Γ = − ( ) 0 . 526 = −0.015 m 2 B2 D 1.65 4π 2 4π 2

and thus it will be considered that D2D = -0.015 m. In consequence, the second mode (see Equation 17.143) and total floor deflections are ⎧− 0.739⎫ ⎧ 0.011 ⎫ ⎪ ⎪ ⎪ ⎪ {δ 2 D } = D2 D {φi 2 } = −0.015⎨− 0.510⎬ = ⎨ 0.008 ⎬m ⎪ 1.000 ⎪ ⎪− 0.015⎪ ⎭ ⎩ ⎭ ⎩ ⎧(0.047 2 + 0.0112 )1 / 2 ⎫ ⎧0.048⎫ ⎪ ⎪ ⎪ ⎪ {δ D } = ⎨(0.1192 + 0.0082 )1 / 2 ⎬ = ⎨0.119⎬m ⎪(0.177 2 + 0.0152 )1 / 2 ⎪ ⎪0.178⎪ ⎭ ⎩ ⎭ ⎩

Similarly, the modal and total story drifts are 0.047 ⎫ ⎧0.047 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ {Δ1D } = ⎨0.119 − 0.047 ⎬ = ⎨0.072 ⎬m ⎪ ⎪ ⎪ ⎪ ⎩0.177 − 0.119⎭ ⎩0.058⎭ 0.011 ⎫ ⎧ 0.011 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ {Δ 2 D } = ⎨ 0.008 − 0.011 ⎬ = ⎨- 0.003⎬m ⎪− 0.015 − 0.008⎪ ⎪- 0.023⎪ ⎭ ⎭ ⎩ ⎩

172

⎧(0.047 2 + 0.0112 )1 / 2 ⎫ ⎧0.048⎫ ⎪ ⎪ ⎪ ⎪ {Δ D } = ⎨(0.072 2 + 0.0032 )1 / 2 ⎬ = ⎨0.072⎬m 2 2 1/ 2 ⎪ ⎪ ⎪ ⎪ ⎩(0.058 + 0.023 ) ⎭ ⎩0.062⎭

Drift limits According to the calculations in Example 17.6, the drift limit for the first story is 0.129 m and the drift limit for the second and third stories is 0.125 m. Thus, the building meets the story drift limits when it is considered with the added dampers. Dampers’ damping constant The dampers’ damping constant, CN, is determined in Example 17.6 in terms of the fundamental mode properties. As these properties are the same when the response spectrum method is used, the dampers’ damping constant remains the same. That is, the dampers’s damping constant remains equal to C N = 155.9 kN - s 0.5 / m 0.5

Maximum damper forces From Example 17.6, the roof displacement under the maximum considered earthquake in the first mode of the system is equal to D1M = (

g S T 9.81 0.9(2.45) )Γ1 M 1 1M = ( 2 )1.388 = 0.284 m 2 2.679 B1M 4π 4π

For the second mode, one has that β2M = β2D = 0.25, B2M = B2D = 1.65, and thus Equation 17.146 yields D2 M = (

9.81 0.9(0.435) g S T )Γ2 M 1 2 = −( 2 )0.526 = 0.031 2 B2 M 1.65 4π 4π

which is greater than ( D2 M ) max = (

9.81 1.5(0.435) 2 g S MS T22 ) Γ = − ( ) 0 . 526 = 0.022 2 1.65 B2 M 4π 2 4π 2

and thus it will be considered that D2M = 0.022 m. In consequence, the modal floor deflections under the maximum considered earthquake are equal to ⎧0.265⎫ ⎧0.075⎫ ⎪ ⎪ ⎪ ⎪ {δ1M } = D1M {φi1} = 0.284⎨0.675⎬ = ⎨0.192⎬m ⎪1.000 ⎪ ⎪0.284⎪ ⎩ ⎭ ⎩ ⎭

⎧− 0.739⎫ ⎧ 0.016 ⎫ ⎪ ⎪ ⎪ ⎪ {δ 2 M } = D2 M {φi 2 } = −0.022⎨− 0.510⎬ = ⎨ 0.011 ⎬m ⎪ ⎪ ⎪ ⎪ ⎩ 1.000 ⎭ ⎩− 0.022⎭

the corresponding modal story drifts equal to 0.075 ⎧ ⎫ ⎧0.075⎫ ⎪ ⎪ ⎪ ⎪ {Δ1M } = ⎨0.192 − 0.075⎬ = ⎨0.117 ⎬m ⎪0.284 − 0.192⎪ ⎪0.092⎪ ⎩ ⎭ ⎩ ⎭

0.016 ⎫ ⎧ 0.016 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ {Δ 2 M } = ⎨ 0.011 − 0.016 ⎬ = ⎨- 0.005⎬m ⎪− 0.022 − 0.011⎪ ⎪- 0.033⎪ ⎩ ⎭ ⎩ ⎭

and the corresponding modal story velocities equal to 173

⎧0.075⎫ ⎧0.192⎫ 2π 2π ⎪ ⎪ ⎪ ⎪ {∇1M } = {Δ1M } = ⎨0.117 ⎬ = ⎨0.300⎬m/s 2.45 ⎪ T1M ⎪ ⎪ ⎪ ⎩0.092⎭ ⎩0.236⎭ ⎧ 0.016 ⎫ ⎧ 0.231 ⎫ 2π 2π ⎪ ⎪ ⎪ ⎪ {Δ 2 M } = {∇ 2 M } = ⎨- 0.005⎬ = ⎨− 0.072⎬m/s 0.435 ⎪ T2 ⎪ ⎪ ⎪ ⎩- 0.033⎭ ⎩− 0.477 ⎭

According to Equation 16.22 and these horizontal story velocities, the maximum forces in the dampers in the fundamental and residual modes are then given by ⎧ (∇11M cos θ1 ) 0.5 ⎫ ⎧(0.192 cos 28.2o ) 0.5 ⎫ ⎧64.13⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ {F1D } = C N ⎨(∇ 21M cos θ 2 ) 0.5 ⎬ = 155.9⎨(0.300 cos 27.6o ) 0.5 ⎬ = ⎨80.38⎬kN ⎪(∇ ⎪(0.236 cos 27.6o ) 0.5 ⎪ ⎪71.30⎪ θ 0.5 ⎪ ⎭ ⎩ 31M cos 3 ) ⎭ ⎩ ⎭ ⎩

⎧ (∇12 M cos θ1 ) 0.5 ⎫ ⎧ (0.231cos 28.2o )0.5 ⎫ ⎧70.34⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ {FRD } = C N ⎨(∇ 22 M cos θ2 )0.5 ⎬ = 155.9⎨(0.072 cos 27.6o ) 0.5 ⎬ = ⎨39.38⎬kN 0.5 ⎪ ⎪ (∇ ⎪(0.477 cos 27.6o )0.5 ⎪ ⎪101.4 ⎪ ⎭ ⎩ 32 M cos θ3 ) ⎭ ⎩ ⎭ ⎩

which yield the following maximum damper forces: ⎧(64.132 + 70.342 )1 / 2 ⎫ ⎧ 95.2 ⎫ ⎪ ⎪ ⎪ ⎪ {FD } = ⎨(80.382 + 39.382 )1 / 2 ⎬ = ⎨ 89.5 ⎬kN ⎪(71.302 + 101.42 )1 / 2 ⎪ ⎪124.0⎪ ⎭ ⎩ ⎭ ⎩

Thus, three dampers with an output force of at least 124 kN on each of the two seismic-force resisting frames will provide the desired additional damping of 20 percent and reduce the story drifts to levels below the limits specified by the code. Nonetheless, these results need be confirmed with a response time-history analysis as the code requires a response time-history analysis for structures located at sites with a S1 equal to or greater than 0.6 (see Section 17.13.2).

174

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