Types of Dynamic Loading • the loading can be described in terms of an analytical say that we know the loading time-history . known only in a discrete set of instants; in this case, we • the loading is measured experimentally, hence it is t p o function, e.g., p t Non Periodic Loadings here we see two sub-cases, Dealing with deterministic dynamic loadings we will study, in order of Periodic Loadings a periodic loading repeats itself with a fixed period T : Harmonic Loadings a force is modulated by a harmonic function of time, complexity, 6 characterized by a frequency ω and a phase φ : p ( t ) = p 0 sin( ω t − φ ) . p ( t ) = p 0 f ( t ) with f ( t ) ≡ f ( t + T ) .
Types of Dynamic Loading Dealing with deterministic dynamic loadings we will study, in order of complexity, Harmonic Loadings a force is modulated by a harmonic function of time, Periodic Loadings a periodic loading repeats itself with a fixed period T : Non Periodic Loadings here we see two sub-cases, • the loading can be described in terms of an analytical • the loading is measured experimentally, hence it is known only in a discrete set of instants; in this case, we say that we know the loading time-history . 6 characterized by a frequency ω and a phase φ : p ( t ) = p 0 sin( ω t − φ ) . p ( t ) = p 0 f ( t ) with f ( t ) ≡ f ( t + T ) . function, e.g., p ( t ) = p o exp( α t )
Types of Dynamic Loading Dealing with deterministic dynamic loadings we will study, in order of complexity, Harmonic Loadings a force is modulated by a harmonic function of time, Periodic Loadings a periodic loading repeats itself with a fixed period T : Non Periodic Loadings here we see two sub-cases, • the loading can be described in terms of an analytical • the loading is measured experimentally, hence it is known only in a discrete set of instants; in this case, we say that we know the loading time-history . 6 characterized by a frequency ω and a phase φ : p ( t ) = p 0 sin( ω t − φ ) . p ( t ) = p 0 f ( t ) with f ( t ) ≡ f ( t + T ) . function, e.g., p ( t ) = p o exp( α t )
Characteristics of a Dynamical Problem A dynamical problem is essentially characterized by the relevance of inertial forces , arising from the accelerated motion of structural or serviced masses. A dynamic analysis is required only when the inertial forces represent a significant portion of the total load. On the other hand, if the loads and the deflections are varying slowly , a static analysis will provide an acceptable approximation. We will define slowly 7
Characteristics of a Dynamical Problem A dynamical problem is essentially characterized by the relevance of inertial forces , arising from the accelerated motion of structural or serviced masses. A dynamic analysis is required only when the inertial forces represent a significant portion of the total load. On the other hand, if the loads and the deflections are varying slowly , a static analysis will provide an acceptable approximation. We will define slowly 7
Characteristics of a Dynamical Problem A dynamical problem is essentially characterized by the relevance of inertial forces , arising from the accelerated motion of structural or serviced masses. A dynamic analysis is required only when the inertial forces represent a significant portion of the total load. On the other hand, if the loads and the deflections are varying slowly , a static analysis will provide an acceptable approximation. We will define slowly 7
Characteristics of a Dynamical Problem A dynamical problem is essentially characterized by the relevance of inertial forces , arising from the accelerated motion of structural or serviced masses. A dynamic analysis is required only when the inertial forces represent a significant portion of the total load. On the other hand, if the loads and the deflections are varying slowly , a static analysis will provide an acceptable approximation. We will define slowly 7
Formulation of a Dynamical Problem
Formulation of a Dynamical Problem In a structural system the inertial forces depend on the time derivatives of displacements while the elastic forces, equilibrating the inertial ones, depend on the spatial derivatives of the displacements. ... the natural statement of the problem is hence in terms of partial differential equations . In many cases it is however possible to simplify the formulation of the structural dynamic problem to ordinary differential equations. 8
Formulation of a Dynamical Problem In a structural system the inertial forces depend on the time derivatives of displacements while the elastic forces, equilibrating the inertial ones, depend on the spatial derivatives of the displacements. ... the natural statement of the problem is hence in terms of partial differential equations . In many cases it is however possible to simplify the formulation of the structural dynamic problem to ordinary differential equations. 8
Lumped Masses In many structural problems we can say that the masses are concentrated in a discrete set of lumped masses (e.g., in a multi-storey building most of the masses is concentrated at the level of the storeys’floors) . Under this assumption, the analytical problem is greatly simplified: 1. the inertial forces are applied only at the lumped masses, 2. the only deflections that influence the inertial forces are the deflections of the lumped masses, 3. using methods of static analysis we can determine those deflections, thus consenting the formulation of the problem in terms of a set of ordinary differential equations, one for each relevant component of the inertial forces. 9
Lumped Masses In many structural problems we can say that the masses are concentrated in a discrete set of lumped masses (e.g., in a multi-storey building most of the masses is concentrated at the level of the storeys’floors) . Under this assumption, the analytical problem is greatly simplified: 1. the inertial forces are applied only at the lumped masses, 2. the only deflections that influence the inertial forces are the deflections of the lumped masses, 3. using methods of static analysis we can determine those deflections, thus consenting the formulation of the problem in terms of a set of ordinary differential equations, one for each relevant component of the inertial forces. 9
Dynamic Degrees of Freedom The dynamic degrees of freedom (DDOF) in a lumped mass, discretized system are the displacements components of the lumped masses associated with the relevant components of the inertial forces. 10
Dynamic Degrees of Freedom If a lumped mass can be regarded as a point mass then at most 3 translational DDOFs will suffice to represent the associated inertial force. On the contrary, if a lumped mass has a discrete volume its inertial force depends also on its rotations (inertial couples) and we need at most 6 DDOFs to represent the mass deflections and the inertial force. Of course, a continuous system has an infinite number of degrees of freedom. 11
Dynamic Degrees of Freedom If a lumped mass can be regarded as a point mass then at most 3 translational DDOFs will suffice to represent the associated inertial force. On the contrary, if a lumped mass has a discrete volume its inertial force depends also on its rotations (inertial couples) and we need at most 6 DDOFs to represent the mass deflections and the inertial force. Of course, a continuous system has an infinite number of degrees of freedom. 11
Generalized Displacements The lumped mass procedure that we have outlined is effective if a large proportion of the total mass is concentrated in a few points (e.g., in a multi-storey building one can consider a lumped mass for each storey). When the masses are distributed we can simplify our problem expressing the deflections in terms of a linear combination of assigned functions of position, the coefficients of the linear combination being the generalized coordinates (e..g., the deflections of a rectilinear beam can be expressed in terms of a trigonometric series). 12
Generalized Displacements The lumped mass procedure that we have outlined is effective if a large proportion of the total mass is concentrated in a few points (e.g., in a multi-storey building one can consider a lumped mass for each storey). When the masses are distributed we can simplify our problem expressing the deflections in terms of a linear combination of assigned functions of position, the coefficients of the linear combination being the generalized coordinates (e..g., the deflections of a rectilinear beam can be expressed in terms of a trigonometric series). 12
Generalized Displacements, cont. To fully describe a displacement field, we need to combine an infinity of linearly independent base functions , but in practice a good approximation can be achieved using just a small number of functions and degrees of freedom. Even if the method of generalized coordinates has its beauty, we must recognize that for each different problem we have to derive an ad hoc formulation, with an evident loss of generality. 13
Generalized Displacements, cont. To fully describe a displacement field, we need to combine an infinity of linearly independent base functions , but in practice a good approximation can be achieved using just a small number of functions and degrees of freedom. Even if the method of generalized coordinates has its beauty, we must recognize that for each different problem we have to derive an ad hoc formulation, with an evident loss of generality. 13
Finite Element Method The finite elements method (FEM) combines aspects of lumped mass and generalized coordinates methods, providing a simple and reliable method of analysis, that can be easily programmed on a digital computer. 14
Finite Element Method • In the FEM, the structure is subdivided in a number of non-overlapping pieces, called the finite elements , delimited by common nodes . • The FEM uses piecewise approximations (i.e., local to each element) to the field of displacements. • In each element the displacement field is derived from the displacements of the nodes that surround each particular element, using interpolating functions . • The displacement, deformation and stress fields in each element, as well as the inertial forces, can thus be expressed in terms of the unknown nodal displacements . • The nodal displacements are the dynamical DOFs of the FEM model. 15
Finite Element Method Some of the most prominent advantages of the FEM method are 1. The desired level of approximation can be achieved by further subdividing the structure. 2. The resulting equations are only loosely coupled, leading to an easier computer solution. 3. For a particular type of finite element (e.g., beam, solid, etc) the procedure to derive the displacement field and the element characteristics does not depend on the particular geometry of the elements, and can easily be implemented in a computer program. 16
Formulation of the equations of motion
Writing the equation of motion In a deterministic dynamic analysis, given a prescribed load, we want to evaluate the displacements in each instant of time. In many cases a limited number of DDOFs gives a sufficient accuracy; further, the dynamic problem can be reduced to the determination of the time-histories of some selected component of the response. The mathematical expressions, ordinary or partial differential equations, that we are going to write express the dynamic equilibrium of the structural system and are known as the Equations of Motion (EOM). The solution of the EOM gives the requested displacements. The formulation of the EOM is the most important, often the most difficult part of a dynamic analysis. 17
Writing the equation of motion In a deterministic dynamic analysis, given a prescribed load, we want to evaluate the displacements in each instant of time. In many cases a limited number of DDOFs gives a sufficient accuracy; further, the dynamic problem can be reduced to the determination of the time-histories of some selected component of the response. The mathematical expressions, ordinary or partial differential equations, that we are going to write express the dynamic equilibrium of the structural system and are known as the Equations of Motion (EOM). The solution of the EOM gives the requested displacements. The formulation of the EOM is the most important, often the most difficult part of a dynamic analysis. 17
Writing the equation of motion In a deterministic dynamic analysis, given a prescribed load, we want to evaluate the displacements in each instant of time. In many cases a limited number of DDOFs gives a sufficient accuracy; further, the dynamic problem can be reduced to the determination of the time-histories of some selected component of the response. The mathematical expressions, ordinary or partial differential equations, that we are going to write express the dynamic equilibrium of the structural system and are known as the Equations of Motion (EOM). The solution of the EOM gives the requested displacements. The formulation of the EOM is the most important, often the most difficult part of a dynamic analysis. 17
Writing the EOM, cont. We have a choice of techniques to help us in writing the EOM, namely: • the D’Alembert Principle, • the Principle of Virtual Displacements, • the Variational Approach. 18
D’Alembert Principle In structural dynamics, we may regard the mass as a constant, and thus u . acceleration of the particle, we can write an equation of equilibrium for the u as the inertial force that contrasts the with a dot. where each operation of differentiation with respect to time is denoted By Newton’s II law of motion, for any particle the rate of change of write 19 u momentum is equal to the external force, dt ( md ⃗ ⃗ p ( t ) = d dt ) , where ⃗ u ( t ) is the particle displacement. p ( t ) = m ¨ ⃗ ⃗ u , When we interpret the term − m ¨ ⃗ p ( t ) = m ¨ particle, ⃗ ⃗
D’Alembert principle, cont. The concept that a mass develops an inertial force opposing its acceleration is known as the D’Alembert principle, and using this principle we can write the EOM as a simple equation of equilibrium. particle, including the reactions of kinematic or elastic constraints, internal forces and external, autonomous forces. In many simple problems, D’Alembert principle is the most direct and convenient method for the formulation of the EOM . 20 The term ⃗ p ( t ) must comprise each different force acting on the
Principle of virtual displacements In a reasonably complex dynamic system, with e.g. articulated rigid bodies and external/internal constraints, the direct formulation of the EOM using D’Alembert principle may result difficult. In these cases, application of the Principle of Virtual Displacements is very convenient, because the reactive forces do not enter the equations of motion, that are directly written in terms of the motions compatible with the restraints/constraints of the system. 21
Principle of Virtual Displacements, cont. Considering, e.g., an assemblage of rigid bodies, the pvd states that necessary and sufficient condition for equilibrium is that, for every virtual displacement (i.e., any infinitesimal displacement compatible with the restraints) the total work done by all the external forces is zero. For an assemblage of rigid bodies, writing the EOM requires 1. to identify all the external forces, comprising the inertial forces, and to express their values in terms of the ddof ; 2. to compute the work done by these forces for different virtual displacements, one for each ddof ; 3. to equate to zero all these work expressions. The pvd is particularly convenient also because we have only scalar equations, even if the forces and displacements are of vectorial nature. 22
Variational approach q of indipendent coordinates convenience (and, to some extent, of personal taste). The method to be used in a particular problem is mainly a matter of q i Variational approaches do not consider directly the forces acting on d dt are, respectively, the kinetic and the potential energy of the system For example, the equation of motion of a generical system can be equations. and potential energy and lead, as well as the pvd , to a set of scalar the dynamic system, but are concerned with the variations of kinetic 23 derived in terms of the Lagrangian function , L = T − V where T and V expressed in terms of a vector ⃗ � ∂ L � = ∂ L , i = 1 , . . . , N . ∂ ˙ ∂ q i
Variational approach q of indipendent coordinates convenience (and, to some extent, of personal taste). The method to be used in a particular problem is mainly a matter of q i Variational approaches do not consider directly the forces acting on d dt are, respectively, the kinetic and the potential energy of the system For example, the equation of motion of a generical system can be equations. and potential energy and lead, as well as the pvd , to a set of scalar the dynamic system, but are concerned with the variations of kinetic 23 derived in terms of the Lagrangian function , L = T − V where T and V expressed in terms of a vector ⃗ � ∂ L � = ∂ L , i = 1 , . . . , N . ∂ ˙ ∂ q i
1 DOF System
1 DOF System y position. easy when the motion occurs in a small neighborhood of the equilibrium It is difficult to integrate the above equation in the general case, but it’s f y y mF y y 1 velocity of the particle, the equation of motion is Structural dynamics is all about the motion of a system in the F y y , where y is the displacement and y the non-linear, internal force F If our system has a constant mass m and it’s subjected to a generical, equilibrium. freedom system, without external forces, subjected to a perturbation of the We’ll start by studying the most simple of systems, a single degree of neighborhood of a point of equilibrium. 24
1 DOF System y the position. easy when the motion occurs in a small neighborhood of the equilibrium It is difficult to integrate the above equation in the general case, but it’s Structural dynamics is all about the motion of a system in the velocity of the particle, the equation of motion is If our system has a constant mass m and it’s subjected to a generical, equilibrium. freedom system, without external forces, subjected to a perturbation of the We’ll start by studying the most simple of systems, a single degree of neighborhood of a point of equilibrium. 24 non-linear, internal force F = F ( y , ˙ y ) , where y is the displacement and ˙ ¨ mF ( y , ˙ y ) = f ( y , ˙ y = 1 y ) .
1 DOF System y the position. easy when the motion occurs in a small neighborhood of the equilibrium It is difficult to integrate the above equation in the general case, but it’s Structural dynamics is all about the motion of a system in the velocity of the particle, the equation of motion is If our system has a constant mass m and it’s subjected to a generical, equilibrium. freedom system, without external forces, subjected to a perturbation of the We’ll start by studying the most simple of systems, a single degree of neighborhood of a point of equilibrium. 24 non-linear, internal force F = F ( y , ˙ y ) , where y is the displacement and ˙ ¨ mF ( y , ˙ y ) = f ( y , ˙ y = 1 y ) .
1 DOF System, cont. the equation of motion y In a position of equilibrium, y eq. , the velocity and the acceleration 25 are zero, and hence f ( y eq. , 0 ) = 0. The force can be linearized in a neighborhood of y eq. , 0: y ) = f ( y eq. , 0 ) + ∂ f ∂ y ( y − y eq. ) + ∂ f f ( y , ˙ y (˙ y − 0 ) + O ( y , ˙ y ) . ∂ ˙ Assuming that O ( y , ˙ y ) is small in a neighborhood of y eq. , we can write ¨ x + a ˙ x + bx = 0 � � where x = y − y eq. , a = − ∂ f y = 0 and b = − ∂ f � � ∂ ˙ ∂ y � � y = y eq , ˙ y = y eq , ˙ y = 0 .
of a linear differential equation of second order with constant static equilibrium), the equation of motion can be studied in terms coefficients. 26 In an infinitesimal neighborhood of y eq. ( y eq. being a position of
1 DOF System, cont. A linear constant coefficient differential equation has the of motion gives whose solutions are a 2 The general integral is hence 27 homogeneous integral x = A exp( st ) , that substituted in the equation s 2 + as + b = 0 � s 1 , 2 = − a 4 − b . 2 ∓ x ( t ) = A 1 exp( s 1 t ) + A 2 exp( s 2 t ) .
1 DOF System, cont. conditions, the nature of the solution depends on the sign of the real an infinitesimal perturbation of the equilibrium, so that in this case we have an unstable equilibrium . equilibrium . response is harmonic with constant amplitude. 28 Given that for a free vibration problem A 1 , A 2 are given by the initial part of s 1 , s 2 , because s i = r i + ı q i and exp( s i t ) = exp( ı q i t ) exp( r i t ) . If one of the r i > 0, the response grows infinitely over time, even for If both r i < 0, the response decrease over time, so we have a stable Finally, if both r i = 0 the roots s i are purely imaginary and the
1 DOF System, cont. The roots being a 2 with negative real part, the system is asymptotically stable, equilibrium is indifferent, the oscillations are harmonic, and the system is unstable. 29 � s 1 , 2 = − a 2 ∓ 4 − b , • if a > 0 and b > 0 both roots are negative or complex conjugate • if a = 0 and b > 0, the roots are purely imaginary, the • if a < 0 or b < 0 at least one of the roots has a positive real part,
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position completely described by the coordinate x t . • The energy-loss (the a x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the time-varying force p t . 30
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position • The energy-loss (the a x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the time-varying force p t . 30 completely described by the coordinate x ( t ) .
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the time-varying force p t . 30 completely described by the coordinate x ( t ) . • The energy-loss (the a ˙
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the time-varying force p t . 30 completely described by the coordinate x ( t ) . • The energy-loss (the a ˙
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the 30 completely described by the coordinate x ( t ) . • The energy-loss (the a ˙ time-varying force p ( t ) .
The famous box car In a single degree of freedom ( sdof ) system each property, m , a and b , can be conveniently represented in a single physical element • The entire mass, m , is concentrated in a rigid block, its position x term) is represented by a massless damper, its damping constant being c . • The elastic resistance to displacement ( b x ) is provided by a massless spring of stiffness k • For completeness we consider also an external loading, the 30 completely described by the coordinate x ( t ) . • The energy-loss (the a ˙ time-varying force p ( t ) .
The famous box car damping constant being c . In a single degree of freedom ( sdof ) system each property, m , a and • For completeness we consider also an external loading, the spring of stiffness k • The elastic resistance to displacement ( b x ) is provided by a massless 30 x term) is represented by a massless damper, its • The entire mass, m , is concentrated in a rigid block, its position b , can be conveniently represented in a single physical element completely described by the coordinate x ( t ) . • The energy-loss (the a ˙ time-varying force p ( t ) . x ( t ) x k f S ( t ) m p ( t ) f D ( t ) f I ( t ) p ( t ) c ( a ) ( b )
Equation of motion of the basic dynamic system The equation of motion, written using the D’Alembert Principle, requires the writing the resisting forces and the external force across the equal sign The equation of motion, merely expressing the equilibrium of these forces, acceleration, velocity and displacement. equilibrium of all the forces acting on the mass including the inertial one . 31 x ( t ) x k f S ( t ) m p ( t ) f D ( t ) f I ( t ) p ( t ) c ( a ) ( b ) The forces are the external force , p ( t ) , positive in the direction of motion and the resisting forces , i.e., the inertial force f I ( t ) , the damping force f D ( t ) and the elastic force, f S ( t ) , that are opposite to the direction of the f I ( t ) + f D ( t ) + f S ( t ) = p ( t )
EOM of the basic dynamic system, cont. According to D’Alembert principle, the inertial force is the product of the mass and acceleration Assuming a viscous damping mechanism, the damping force is the product of the damping constant c and the velocity, Finally, the elastic force is the product of the elastic stiffness k and the displacement, 32 f I ( t ) = m ¨ x ( t ) . f D ( t ) = c ˙ x ( t ) . f S ( t ) = k x ( t ) .
EOM of the basic dynamic system, cont. • The differential equation of dynamic equilibrium written in the left member of the EoM. direction of motion, i.e., resisting the motion and, by convention as well, are The resisting forces are, by convention, positive when opposite to the second order, with constant coefficients. • The equation of motion is a linear differential equation of the 33 • The resisting forces in the EoM f I ( t ) + f D ( t ) + f S ( t ) = m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = p ( t ) . f I ( t ) , f D ( t ) , f S ( t ) are proportional to the deflection x ( t ) or one of its time derivatives, ˙ x ( t ) , ¨ x ( t ) .
Influence of static forces Expressing the displacement as the sum of a constant, static W p t k x t s t k c x t m x t and substituting in the EOM we have x t s t x t displacement and a dynamic displacement, Considering the presence of a constant force W , the EOM is 34 x ( t ) ¯ ∆ st x x ( t ) k k ∆ st W m p ( t ) f S ( t ) p ( t ) f D ( t ) c f I ( t ) ( a ) ( b ) m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = p ( t ) + W .
Influence of static forces Considering the presence of a constant force W , the EOM is and substituting in the EOM we have displacement and a dynamic displacement, Expressing the displacement as the sum of a constant, static 34 ¯ x ( t ) ∆ st x x ( t ) k k ∆ st W m p ( t ) f S ( t ) p ( t ) f D ( t ) c f I ( t ) ( a ) ( b ) m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = p ( t ) + W . x ( t ) = ∆ s t + ¯ x ( t ) , m ¨ x ( t ) + c ˙ x ( t ) + k ∆ s t + k ¯ x ( t ) = p ( t ) + W .
the static forces, and must be computed using the superposition of Influence of static forces, cont. The equation of motion expressed with reference to the static effects. Note that the total displacements, stresses. etc. are influenced by with x t . referenced from the equilibrium position and denoted, for simplicity, For this reasons, all displacements in further discussions will be equilibrium position is not affected by static forces. 35 EOM can be written as x the Recognizing that k ∆ s t = W (so that the two terms, on opposite sides x ≡ ˙ x ≡ ¨ of the equal sign, cancel each other), that ˙ ¯ x and that ¨ ¯ m ¨ x ( t ) + c ˙ ¯ ¯ x ( t ) + k ¯ x ( t ) = p ( t ) .
the static forces, and must be computed using the superposition of Influence of static forces, cont. The equation of motion expressed with reference to the static effects. Note that the total displacements, stresses. etc. are influenced by referenced from the equilibrium position and denoted, for simplicity, For this reasons, all displacements in further discussions will be equilibrium position is not affected by static forces. 35 EOM can be written as x the Recognizing that k ∆ s t = W (so that the two terms, on opposite sides x ≡ ˙ x ≡ ¨ of the equal sign, cancel each other), that ˙ ¯ x and that ¨ ¯ m ¨ x ( t ) + c ˙ ¯ ¯ x ( t ) + k ¯ x ( t ) = p ( t ) . with x ( t ) .
Influence of static forces, cont. EOM can be written as effects. Note that the total displacements, stresses. etc. are influenced by referenced from the equilibrium position and denoted, for simplicity, For this reasons, all displacements in further discussions will be equilibrium position is not affected by static forces. The equation of motion expressed with reference to the static 35 x the Recognizing that k ∆ s t = W (so that the two terms, on opposite sides x ≡ ˙ x ≡ ¨ of the equal sign, cancel each other), that ˙ ¯ x and that ¨ ¯ m ¨ x ( t ) + c ˙ ¯ ¯ x ( t ) + k ¯ x ( t ) = p ( t ) . with x ( t ) . the static forces, and must be computed using the superposition of
Influence of support motion Displacements, deformations and stresses in a structure are induced also equipment due to vibrations of the building in which it is housed. foundation due to earthquake and the motion of the base of a piece of Important examples of support motion are the motion of a building by a motion of its support. 36 x tot ( t ) m Fixed reference axis k k 2 2 c x g ( t ) x ( t )
f I t mx tot t mx g t m x tot t m x g t p eff t Support motion is sufficient to excite a dynamic system: p eff t m x g t . mx t Writing the EOM for a null external load, p t 0, is hence Influence of support motion, cont. c x t k x t and velocities, the inertial force is proportional to the total acceleration, or, m x t c x t k x t 0 While the elastic and damping forces are still proportional to relative displacements 37 the total displacement is and the total acceleration is with respect to a inertial frame of reference, Considering a support motion x g ( t ) , defined x tot ( t ) m Fixed reference axis k k 2 2 x tot ( t ) = x g ( t ) + x ( t ) c ¨ x tot ( t ) = ¨ x g ( t ) + ¨ x ( t ) . x g ( t ) x ( t )
m x tot t m x g t p eff t Support motion is sufficient to excite a dynamic system: p eff t m x g t . Influence of support motion, cont. and velocities, the inertial force is proportional to the total acceleration, k x t c x t m x t or, 0 k x t c x t 0, is hence Writing the EOM for a null external load, p t While the elastic and damping forces are still proportional to relative displacements 37 and the total acceleration is the total displacement is with respect to a inertial frame of reference, Considering a support motion x g ( t ) , defined x tot ( t ) m Fixed reference axis k k 2 2 x tot ( t ) = x g ( t ) + x ( t ) c ¨ x tot ( t ) = ¨ x g ( t ) + ¨ x ( t ) . x g ( t ) x ( t ) f I ( t ) = − m ¨ x tot ( t ) = m ¨ x g ( t ) + m ¨ x ( t ) .
Support motion is sufficient to excite a dynamic system: p eff t m x g t . Influence of support motion, cont. with respect to a inertial frame of reference, or, and velocities, the inertial force is proportional to the total acceleration, While the elastic and damping forces are still proportional to relative displacements and the total acceleration is the total displacement is 37 Considering a support motion x g ( t ) , defined x tot ( t ) m Fixed reference axis k k 2 2 x tot ( t ) = x g ( t ) + x ( t ) c ¨ x tot ( t ) = ¨ x g ( t ) + ¨ x ( t ) . x g ( t ) x ( t ) f I ( t ) = − m ¨ x tot ( t ) = m ¨ x g ( t ) + m ¨ x ( t ) . Writing the EOM for a null external load, p ( t ) = 0, is hence m ¨ x tot ( t ) + c ˙ x ( t ) + k x ( t ) = 0 , m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = − m ¨ x g ( t ) ≡ p eff ( t ) .
Influence of support motion, cont. with respect to a inertial frame of reference, or, and velocities, the inertial force is proportional to the total acceleration, While the elastic and damping forces are still proportional to relative displacements and the total acceleration is the total displacement is 37 Considering a support motion x g ( t ) , defined x tot ( t ) m Fixed reference axis k k 2 2 x tot ( t ) = x g ( t ) + x ( t ) c ¨ x tot ( t ) = ¨ x g ( t ) + ¨ x ( t ) . x g ( t ) x ( t ) f I ( t ) = − m ¨ x tot ( t ) = m ¨ x g ( t ) + m ¨ x ( t ) . Writing the EOM for a null external load, p ( t ) = 0, is hence m ¨ x tot ( t ) + c ˙ x ( t ) + k x ( t ) = 0 , m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = − m ¨ x g ( t ) ≡ p eff ( t ) . Support motion is sufficient to excite a dynamic system: p eff ( t ) = − m ¨ x g ( t ) .
Free vibrations of a SDOF system
Free Vibrations The equation of motion, is a linear differential equation of the second order, with constant coefficients. Its solution can be expressed in terms of a superposition of a that is the solution of the so called homogeneous problem , where In the following, we will study the solution of the homogeneous problem, the so-called homogeneous or complementary solution, that is the free vibrations of the SDOF after a perturbation of the position of equilibrium. 38 m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = p ( t ) particular solution , depending on p ( t ) , and a free vibration solution, p ( t ) = 0.
Free vibrations of an undamped system m As m and k are positive quantities, s must be purely imaginary. above equation we have perpetuum . Nevertheless, it is an useful idealization. In this case, the homogeneous equation of motion is place, is just an ideal notion, as it would be a realization of motus 39 An undamped system, where c = 0 and no energy dissipation takes m ¨ x ( t ) + k x ( t ) = 0 which solution is of the form exp st ; substituting this solution in the ( k + s 2 m ) exp st = 0 noting that exp st ̸ = 0, we finally have � ( k + s 2 m ) = 0 ⇒ s = ± − k
Undamped Free Vibrations k The solution has an imaginary part? equation is 40 Introducing the natural circular frequency ω n ω 2 n = k m , the solution of the algebraic equation in s is � � √ � s = ± − k m = ± − 1 m = ± i ω 2 n = ± i ω n where i = √− 1 and the general integral of the homogeneous x ( t ) = G 1 exp( i ω n t ) + G 2 exp( − i ω n t ) .
Undamped Free Vibrations k The solution has an imaginary part? equation is 40 Introducing the natural circular frequency ω n ω 2 n = k m , the solution of the algebraic equation in s is � � √ � s = ± − k m = ± − 1 m = ± i ω 2 n = ± i ω n where i = √− 1 and the general integral of the homogeneous x ( t ) = G 1 exp( i ω n t ) + G 2 exp( − i ω n t ) .
Undamped Free Vibrations The solution is derived from the general integral imposing the (real) x 0 , we finally find ... substituting these values in the general solution and collecting x 0 2 i 2 i Solving the linear system we have x 0 41 (1) initial conditions x 0 x ( 0 ) = x 0 , ˙ x ( 0 ) = ˙ Evaluating x ( t ) for t = 0 and substituting in (1), we have � G 1 + G 2 = x 0 = ˙ i ω n G 1 − i ω n G 2 G 1 = ix 0 + ˙ G 2 = ix 0 − ˙ x 0 /ω n x 0 /ω n , , and ˙
Undamped Free Vibrations n t n t 2 i and the solution can be rewritten in terms of the elementary trigonometric functions x t x 0 x o n t n n t For every real initial conditions we can use indifferently one of the above representations. We usually choose to represent the homogeneous solu- tion as a linear combination of sine and cosine of circular frequency n i i 42 n t 2 2 i x 0 The exponentials with imaginary argument and the trigonometric functions are connected by the Euler formulas, n t i n t i n t 2 x ( t ) = exp( i ω n t )+exp( − i ω n t ) x 0 + exp( i ω n t ) − exp( − i ω n t ) ˙ ω n .
Undamped Free Vibrations 2 n tion as a linear combination of sine and cosine of circular frequency representations. We usually choose to represent the homogeneous solu- For every real initial conditions we can use indifferently one of the above functions and the solution can be rewritten in terms of the elementary trigonometric 2 i 42 are connected by the Euler formulas, The exponentials with imaginary argument and the trigonometric functions x 0 2 i 2 x ( t ) = exp( i ω n t )+exp( − i ω n t ) x 0 + exp( i ω n t ) − exp( − i ω n t ) ˙ ω n . cos ω n t = exp( i ω n t ) + exp( − i ω n t ) sin ω n t = exp( i ω n t ) − exp( − i ω n t ) , , x ( t ) = x 0 cos( ω n t ) + (˙ x o /ω n ) sin( ω n t ) .
Undamped Free Vibrations are connected by the Euler formulas, representations. We usually choose to represent the homogeneous solu- For every real initial conditions we can use indifferently one of the above functions and the solution can be rewritten in terms of the elementary trigonometric 2 i 2 42 The exponentials with imaginary argument and the trigonometric functions x 0 2 i 2 x ( t ) = exp( i ω n t )+exp( − i ω n t ) x 0 + exp( i ω n t ) − exp( − i ω n t ) ˙ ω n . cos ω n t = exp( i ω n t ) + exp( − i ω n t ) sin ω n t = exp( i ω n t ) − exp( − i ω n t ) , , x ( t ) = x 0 cos( ω n t ) + (˙ x o /ω n ) sin( ω n t ) . tion as a linear combination of sine and cosine of circular frequency ω n
Undamped Free Vibrations Our preferred representation of the general integral of undamped free vibrations is For the usual initial conditions, we have already seen that x 0 43 x ( t ) = A cos( ω n t ) + B sin( ω n t ) B = ˙ A = x 0 , . ω n
Undamped Free Vibrations with 44 evidence: Sometimes we prefer to write x ( t ) as a single harmonic, introducing a phase difference ϕ so that the amplitude of the motion , C , is put in x ( t ) = A cos ω n t + B sin ω n t = C cos( ω n t − φ ) = C (cos ω n t cos φ + sin ω n t sin φ ) From A = C cos φ and B = C sin φ we have tan φ = B / A , from √ A 2 + B 2 = C 2 (cos 2 φ + sin 2 φ ) we have C = A 2 + B 2 and eventually � A 2 + B 2 � C = x ( t ) = C cos( ω n t − φ ) , φ = arctan( B / A )
Undamped Free Vibrations It is worth noting that the coefficients A , B and C have the dimension adimensional. 45 x ( t ) arctan ˙ x 0 ρ x 0 t − θ ω T = 2 π ω of a length, the coefficient ω n has the dimension of the reciprocal of time and that the coefficient φ is an angle, or in other terms is
Free vibrations of a damped system
Behavior of Damped Systems The viscous damping modifies the response of a sdof system introducing a decay in the amplitude of the response. Depending on the amount of damping, the response can be oscillatory or not. The amount of damping that separates the two behaviors is denoted as critical damping . 46
The solution of the EOM or, after dividing both members by m , c c The equation of motion for a free vibrating damped system is whose solutions are 2 m 47 simplifying, we have that the parameter s must satisfy the equation m ¨ x ( t ) + c ˙ x ( t ) + k x ( t ) = 0 , substituting the solution exp st in the preceding equation and m s 2 + c s + k = 0 s 2 + c m s + ω 2 n = 0 �� �� c � 2 � 2 . s = − c 2 m ∓ − ω 2 n = ω n − ∓ − 1 2 m ω n 2 m ω n
Critical Damping The behavior of the solution of the free vibration problem depends critical damping , The value of c that make the radicand equal to zero is known as the 48 c � � 2 of course on the sign of the radicand ∆ = − 1: 2 m ω n ∆ < 0 the roots s are complex conjugate, ∆ = 0 the roots are identical, double root, ∆ > 0 the roots are real. √ c cr = 2 m ω n = 2 mk .
Critical Damping A single degree of freedom system is denoted as critically damped , under-critically damped or over-critically damped depending on the value of the damping coefficient with respect to the critical damping. Typical building structures are undercritically damped. 49
Critical Damping A single degree of freedom system is denoted as critically damped , under-critically damped or over-critically damped depending on the value of the damping coefficient with respect to the critical damping. Typical building structures are undercritically damped. 49
Damping Ratio c If we introduce the ratio of the damping to the critical damping, or the equation of free vibrations can be rewritten as c cr 50 critical damping ratio ζ , ζ = c = , c = ζ c cr = 2 ζω n m 2 m ω n ¨ x ( t ) + 2 ζω n ˙ x ( t ) + ω 2 n x ( t ) = 0 and the roots s 1 , 2 can be rewritten as � ζ 2 − 1 . s = − ζω n ∓ ω n
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