Solutions Manual for Dynamics of Structures: Theory and Applications to Earthquake Engineering – 6th Edition (Anil K. Chopra) | Structural Dynamics Problem Solutions 2025/2026

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, Dynamics of Structures
Theory and Applications to Earthquake Engineering, 6th Edition
Anil K. Chopra
6TH EDITION


TABLE OF CONTENTS
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PART ONE - SINGLE-DEGREE-OF-FREEDOM SYSTEMS

Chapter 1 Equations of Motion, Problem Statement, and Solution Methods
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Chapter 2 Free Vibration
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Chapter 3 Response to Harmonic and Periodic Excitations

Chapter 4 Response to Arbitrary, Step, and Pulse Excitations

Chapter 5 Numerical Evaluation of Dynamic Response
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Chapter 6 Earthquake Response of Linear Systems

Chapter 7 Earthquake Response of Inelastic Systems
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Chapter 8 Generalized Single-Degree-of-Freedom Systems

PART TWO - MULTI-DEGREE-OF-FREEDOM SYSTEMS
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Chapter 9 Equations of Motion, Problem Statement, and Solution Methods

Chapter 10 Free Vibration
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Chapter 11 Damping in Structures

Chapter 12 Dynamic Analysis and Response of Linear Systems

Chapter 13 Earthquake Analysis of Linear Systems
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Chapter 14 Analysis of Nonclassically Damped Linear Systems

Chapter 15 Reduction of Degrees of Freedom
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Chapter 16 Numerical Evaluation of Dynamic Response

Chapter 17 Systems with Distributed Mass and Elasticity

Chapter 18 Introduction to the Finite Element Method
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PART THREE - EARTHQUAKE ENGINEERING

Chapter 19 Earthquake Response Spectrum Analysis
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Chapter 20 Seismic Analysis of Structures

Chapter 21 Structural Dynamics in Building Codes

Chapter 22 Soil-Structure-Foundation Interaction

, Problem 1.6 3. Write the equation of motion using Newton’s second law
of motion.
Repeat Problem 1.5 for the system shown in Fig. P1.6,
which differs in only one sense: its width varies from zero M 0  I 0
tu
at O to b at the free end. 2L 1
 mg sin   mL2
3 2
mL2  2mgL
 sin   0 (a)
2 3
d

4. Specialize for small .
For small , sin  , and Eq. (a) becomes
yg

Figure P1.6
mL2  2mgL
  0
Solution: 2 3
1. Find the moment of inertia about about O. or
en

4g
   0 (b)
3L
L

r
2
I0   dA 5. Determine natural frequency.
x
0
4g
iu
L n 
3L
L r
L r

  r 2 ( r  dr )
0
In each case the system is equivalent to the spring-
  4 mass system shown for which the equation of motion is
 L 
s2
4
1  w  
 mL 2   u  ku  0
2 g
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2. Draw a free body diagram of the body in an arbitrary k
displaced position.

ww
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2L/3
2L/3 u
xx

The spring stiffness is determined from the deflection u
 under a vertical force fS applied at the location of the
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lumped weight:

mg fS L3 48 EI
yy mg Simply-supported beam: u   k 
48 EI L3
fS L3 3 EI
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Cantilever beam: u   k 
3 EI L3
fS L3 192 EI
Clamped beam: u   k 
192 EI L3
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, Problem 1.7

Develop the equation governing the longitudinal motion of
the system of Fig. P1.7. The rod is made of an elastic
tu
material with elastic modulus E; its cross-sectional area is
A and its length is L. Ignore the mass of the rod and
measure u from the static equilibrium position.
d yg
en

Figure P1.7

Solution:
Draw a free body diagram of the mass:
iu

fS
s2
mü

u
p(t)
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Write equation of dynamic equilibrium:

mu  f S  p (t ) (a)

Write the force-displacement relation:
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fS 
 AE  u
  (b)
 L 
Substitute Eq. (b) into Eq. (a) to obtain the equation of
motion:
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mu 
 AE  u  p (t )
 
 L 
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