INTRODUCTION TO DYNAMICS OF 
                         STRUCTURES
                       A PROJECT DEVELOPED FOR THE 
               UNIVERSITY CONSORTIUM ON INSTRUCTIONAL SHAKE TABLES
                        http://ucist.cive.wustl.edu/
                              Required Equipment: 
                           • Instructional Shake Table
                           • Two Story Building 
                           • Three Accelerometers
                           • MultiQ Board
                           • Power Supply
                           • Computer
                           • Software: Wincon and Matlab 
                            Developed by: 
                   Mr. Juan Martin Caicedo (jc11@cive.wustl.edu)
                         Ms. Sinique Betancourt
                    Dr. Shirley J. Dyke (sdyke@seas.wustl.edu)
                     Washington University in Saint Louis
                     This project is supported in part by the 
                 National Science Foundation Grant No. DUE–9950340. 
                                                   Introduction to Dynamics of Structures
                                                    Structural Control & Earthquake Engineering Laboratory
                                                                  Washington University in Saint Louis
                      Objective: The objective of this experiment is to introduce you to principles in structural dynam-
                      ics through the use of an instructional shake table. Natural frequencies, mode shapes and damping
                      ratios for a scaled structure will be obtained experimentally. 
                      1.0  Introduction
                             The dynamic behavior of structures is an important topic in many fields. Aerospace engineers
                      must understand dynamics to simulate space vehicles and airplanes, while mechanical engineers
                      must understand dynamics to isolate or control the vibration of machinery. In civil engineering, an
                      understanding of structural dynamics is important in the design and retrofit of structures to with-
                      stand severe dynamic loading from earthquakes, hurricanes, and strong winds, or to identify the
                      occurrence and location of damage within an existing structure. 
                             In this experiment, you will test a small test building of two floors to observe typical dynamic
                      behavior and obtain its dynamic properties. To perform the experiment you will use a bench-scale
                      shake table to reproduce a random excitation similar to that of an earthquake. Time records of the
                      measured absolute acceleration responses of the building will be acquired. 
                      2.0  Theory: Dynamics of Structures
                             To understand the experiment it is necessary to understand concepts in dynamics of struc-
                      tures. This section will provide these concepts, including the development of the differential equa-
                      tion of motion and its solution for the damped and undamped case. First, the behavior of a single
                      degree of freedom (SDOF) structure will be discussed, and then this will be extended to a multi
                      degree of freedom (MDOF) structure. 
                             The number of degrees of freedom is defined as the minimum number of variables that are re-
                      quired for a full description of the movement of a structure. For example, for the single story
                      building shown in figure 1 we assume the floor is rigid compared to the two columns. Thus, the
                      displacement of the structure is going to be completely described by the displacement, x, of the
                      floor. Similarly, the building shown in figure 2 has two degrees of freedom because we need to
                      describe the movement of each floor separately in order to describe the movement of the whole
                      structure.
                      Introduction to Dynamics of Structures                                 1                              Washington University in St. Louis
                                                                            x
                                                                             2
                                                                                          p(t)
                                                                               m
                                  x                                         x
                                                p(t)                         1           k,c
                                    m                                          m
                                               k,c
                   y                                            y
                       x                                            x
              Figure 1. One degree of freedom structure.   Figure 2. Two degree of freedom structure.
              2.1  One degree of freedom
                  We can model the building shown in figure 1 as
              the simple dynamically equivalent model shown in                       x
              figure 3a. In this model, the lateral stiffness of the     k
              columns is modeled by the spring (k), the damping                     m        p(t)
              is modeled by the shock absorber (c) and the mass           c
              of the floor is modeled by the mass (m). Figure 3b
              shows the free body diagram of the structure. The       a. mass with spring and damper
              forces include the spring force fs()t , the damping
              force f()t, the external dynamic load on the struc-        f
                    d                                                     s
              ture, pt(), and the inertial force f()t. These forces                f       p(t)
              are defined as:                i                           fd        i
                                f =kx⋅                     (1)          b. free body diagram
                                 s
                                        ·                        Figure 3. Dynamically equivalent 
                                f=cx⋅                      (2)    model for a one floor building.
                                 d
                                        ··
                                f= mx⋅                     (3)
                                 i
                        ·                                                                       ··
              where the x  is the first derivative of the displacement with respect to time (velocity) and x  is the
              second derivative of the displacement with respect to time (acceleration).
                  Summing the forces shown in figure 3b we obtain 
                                                     ··          ·
                                            ΣF ==mx⋅      pt()–cx–kx                                (4)
                                                 ··   ·
                                               mx++cx kx =p()t                                      (5)
              where the mass m and the stiffness k are greater than zero for a physical system. 
              Introduction to Dynamics of Structures      2                  Washington University in St. Louis
                                2.1.1  Undamped system
                                         Consider the behavior of the undamped system (c=0). From differential equations we know
                                that the solution of a constant coefficient ordinary differential equation is of the form
                                                                                                                                         αt
                                                                                                                      xt()=e  (6)
                                and the acceleration is given by 
                                                                                                                   ··                  2αt
                                                                                                                  x()t       =αe.(7)
                                Using equations (6) and (7) in equation (5) and making pt() equal to zero we obtain
                                                                                                                    2αt                 αt
                                                                                                            mαe +ke =0                                                                                                             (8)
                                                                                                                αt           2
                                                                                                             e[]mα+k =0.(9)
                                Equation (9) is satisfied when 
                                                                                                                                    –k
                                                                                                                          2
                                                                                                                                    -----
                                                                                                                       α=                                                                                                        (10)
                                                                                                                                     m
                                                                                                                                        k
                                                                                                                    α = ±i ---- .(11)
                                                                                                                                       m
                                The solution of equation (5) for the undamped case is
                                                                                                                              ω it               –ω it
                                                                                                      xt()= Ae n +Be n                                                                                                           (12)
                                where A and B are constants based on the initial conditions, and the natural frequency ω  is de-
                                fined as                                                                                                                                                                                  n
                                                                                                                                       k
                                                                                                                    ω = ---- .                                                                                                   (13)
                                                                                                                        n             m
                                         Using Euler’s formula and rewriting equation (12) yields
                                                                                                        e i α t   = cosαt+isinαt                                                                                                 (14)
                                                             xt()=A()cos()ωt+isin()ω t                                              +B()cos()–ωt+isin()–ω t                                                                      (15)
                                                                                                 n                         n                                  nn
                                                             xt()=Acos()ωt+BAisin()ω t +cos()–ωt+Bisin()–ω t .                                                                                                                   (16)
                                                                                               n                            n                               nn
                                Using cos()–α= cos()α  and sin()–α= –sin()α  we have
                                                                xt()= Acos()ωt+BAisin()ω t +                                                   cos()ωt–Bisin()ω t                                                                (17)
                                                                                                  n                             n                            nn
                                or, 
                                Introduction to Dynamics of Structures                                                              3                                          Washington University in St. Louis