CIV 8/77 Chapter - /75 Introduction To discuss the dynamics of a single-degree-of freedom springmass system. To derive the finite element equations for the time-dependent stress analysis of the one-dimensional bar, including derivation of the lumped and consistent mass matrices. To introduce procedures for numerical integration in time, including the central difference method, Newmark's method, and dwil Wilson's method. To describe how to determine the natural frequencies of bars by the finite element method. To illustrate the finite element solution of a time-dependent bar problem. Introduction To develop the beam element lumped and consistent mass matrices. To illustrate the determination of natural frequencies for beams by the finite element method. To develop the mass matrices for truss, plane frame, plane stress, plane strain, axisymmetric, and solid elements. To report some results of structural dynamics problems solved using a computer program, including a fixed-fixed beam for natural frequencies, a bar, a fixed-fixed beam, a rigid frame, and a gantry crane-all subjected to time-dependent forcing functions.
CIV 8/77 Chapter - /75 Introduction This chapter provides an elementary introduction to timedependent problems. We will introduce the basic concepts using the single-degree-offreedom spring-mass system. We will include discussion of the stress analysis of the onedimensional bar, beam, truss, and plane frame. Introduction We will provide the basic equations necessary for structural dynamic analysis and develop both the lumped- and the consistent-mass matrices involved in the analyses of a bar, beam, truss, and plane frame. We will describe the assembly of the global mass matrix for truss and plane frame analysis and then present numerical integration methods for handling the time derivative. We will provide longhand solutions for the determination of the natural frequencies for bars and beams, and then illustrate the time-step integration process involved with the stress analysis of a bar subjected to a time dependent forcing function.
CIV 8/77 Chapter - 3/75 Dynamics of a Spring-Mass System In this section, we will discuss the motion of a single-degree-offreedom spring-mass system as an introduction to the dynamic behavior of bars, trusses, and frames. Consider the single-degree-of-freedom spring-mass system subjected to a time-dependent force F(t) as shown in the figure below. The term k is the stiffness of the spring and m is the mass of the system. Dynamics of a Spring-Mass System The free-body diagram of the mass is shown below. The spring force T = kx and the applied force F(t) act on the mass, and the mass-times-acceleration term is shown separately. Applying Newton s second law of motion, f = ma, to the mass, we obtain the equation of motion in the x direction: Ft () kxmx where a dot over a variable indicates differentiation with respect to time.
CIV 8/77 Chapter - 4/75 Dynamics of a Spring-Mass System The standard form of the equation is: mx kx F() t The above equation is a second-order linear differential equation whose solution for the displacement consists of a homogeneous solution and a particular solution. The homogeneous solution is the solution obtained when the right-hand-side is set equal to zero. A number of useful concepts regarding vibrations are available when considering the free vibration of a mass; that is when F(t) = 0. Dynamics of a Spring-Mass System et s define the following term: m k The equation of motion becomes: x x 0 where is called the natural circular frequency of the free vibration of the mass (radians per second). Note that the natural frequency depends on the spring stiffness k and the mass m of the body.
CIV 8/77 Chapter - 5/75 Dynamics of a Spring-Mass System The motion described by the homogeneous equation of motion is called simple harmonic motion. Atypical displacement/time curve is shown below. where x m denotes the maximum displacement (or amplitude of the vibration). Dynamics of a Spring-Mass System The time interval required for the mass to complete one full cycle of motion is called the period of the vibration (in seconds) and is defined as: The frequency in hertz (Hz = /s) is f = / = /().
CIV 8/77 Chapter - 6/75 Dynamics of a Spring-Mass System Dynamics of a Spring-Mass System
CIV 8/77 Chapter - 7/75 Dynamics of a Spring-Mass System Dynamics of a Spring-Mass System
CIV 8/77 Chapter - 8/75 Dynamics of a Spring-Mass System Dynamics of a Spring-Mass System
CIV 8/77 Chapter - 9/75 Direct Derivation of the Bar Element et s derive the finite element equations for a time-dependent (dynamic) stress analysis of a one-dimensional bar. Step - Select Element Type We will consider the linear bar element shown below. where the bar is of length, cross-sectional area A, and mass density (with typical units of lb-s /in 4 ), with nodes and e subjected to external time-dependent loads: f () t x Direct Derivation of the Bar Element Step - Select a Displacement Function A linear displacement function is assumed in the x direction. u a a x The number of coefficients in the displacement function, a i, is equal to the total number of degrees of freedom associated with the element. We can express the displacement function in terms of the shape functions: u u N N N x N x u
CIV 8/77 Chapter - 0/75 Direct Derivation of the Bar Element Step 3 - Define the Strain/Displacement and Stress/Strain Relationships The stress-displacement relationship is: du x [ B] d dx where: u [ B ] d u The stress-strain relationship is given as: [ D] [ D][ B] d x x Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations The bar element is typically not in equilibrium under a timedependent force; hence, f x f x. We must apply Newton s second law of motion, f = ma, to each node. Write the law of motion as the external force f xe minus the internal force equal to the nodal mass times acceleration.
CIV 8/77 Chapter - /75 Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations Therefore: e u e u f x f x m f x f x m t t where: m A A m u e f f x m 0 x t e f f x 0 m x u t Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations If we replace {f} with [k]{d} we get: f e () t kd md Where the elemental stiffness matrix is: k AE and the lumped-mass matrix is: 0 m A 0 d t d
CIV 8/77 Chapter - /75 Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations et s derive the consistent-mass matrix for a bar element. The typical method for deriving the consistent-mass matrix is the principle of virtual work; however, an even simpler approach is to use D Alembert s principle. e The effective body force is: X u The nodal forces associated with {X e } are found by using the following: T f N X dv [ ] { } b V Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations Substituting {X e } for {X} gives: f [ ] T b N udv The second derivative of the u with respect to time is: u [ N] d u [ N] d where u and u are the nodal velocities and accelerations, respectively. V
CIV 8/77 Chapter - 3/75 Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations Therefore: where: T f b N N d dv m d V T m N N dv V The mass matrix is called the consistent mass matrix because it is derived using the same shape functions use to obtain the stiffness matrix. Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations Substituting the shape functions in the above mass matrix equations give: x x x m dv V x x x x m A dx 0 x
CIV 8/77 Chapter - 4/75 Direct Derivation of the Bar Element Step 4 - Derive the Element Stiffness Matrix and Equations Substituting the shape functions in the above mass matrix equations give: x xx m A dx 0 xx x Evaluating the above integral gives: m A 6 Direct Derivation of the Bar Element Step 5 - Assemble the Element Equations and Introduce Boundary Conditions The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations. where Ft () K{ d} Md K N ( e k ) e N N e M m ( e F f ) ( ) e e
CIV 8/77 Chapter - 5/75 Numerical Integration in Time We now introduce procedures for the discretization of the equations of motion with respect to time. These procedures will allow the nodal displacements to be determined at different time increments for a given dynamic system. The general method used is called direct integration. There are two classifications of direct integration: explicit it and implicit. it We will formulate the equations for two direct integration methods. Numerical Integration in Time The first, and simplest, is an explicit method known as the central difference method. The second more complicated but more versatile than the central difference method, is an implicit method known as the Newmark-Beta (or Newmark s) method. The versatility of Newmark s method is evidenced by its adaptation in many commercially available computer programs.
CIV 8/77 Chapter - 6/75 Central Difference Method The central difference method is based on finite difference expressions for the derivatives in the equation of motion. For example, consider the velocity and the acceleration at time t: d i d i d d i i ( t) d i d i ( t) where the subscripts indicate the time step for a given time increment of t. Central Difference Method The acceleration can be expressed in terms of the displacements (using a Taylor series expansion) as: d i d d d i i i ( t) We generally want to evaluate the nodal displacements; therefore, we rewrite the above equation as: d i di di di ( t ) The acceleration can be expressed as: d i M Fi Kdi
CIV 8/77 Chapter - 7/75 Central Difference Method To develop an expression of d i+, first multiply the nodal displacement equation by M and substitute the above equation for d i into this equation. Md i M di M di Fi K di t Combining terms in the above equations gives: M d t F M t K d M d i i i i To start the computation to determine di, d i,and d i we need the displacement at time step i -. Central Difference Method Using the central difference equations for the velocity and acceleration and solving for {d i- }: didi( t) d id i Procedure for solution: d, d, and F( t). Given: 0 0 i i ( t). If the acceleration is not given, solve for d 0 d 0 M F0 Kd0
CIV 8/77 Chapter - 8/75 Central Difference Method Procedure for solution: 3. Solve for {d - }att t = -t ( t) dd0( t) d 0d 0 4. Solve for {d } at t = t using the value of {d - } from Step 3 M d t F M t K d M d i i i i d M t F M t K d M d 0 0 Central Difference Method Procedure for solution: 5. With {d 0 } given and {d } determined in Step 4 solve for {d } d M t F M t K d M d 0 6. Solve for d : d M F Kd 7. Solve for d : d d d 0 ( t) 8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps.
CIV 8/77 Chapter - 9/75 Central Difference Method Procedure for solution: Central Difference Method Example Problem Determine the displacement, acceleration, and velocity at 0.05 second time intervals for up to 0. seconds for the one- dimensional spring-mass system shown in the figure below. Consider the above spring-mass system as a single degree of freedom problem represented by the displacement d.
CIV 8/77 Chapter - 0/75 Central Difference Method Example Problem Procedure for solution: d 0 d 0. At time t =0: 0 0. If the acceleration is not given, solve for d 0 d 0 M F0 Kd0 000,000 00(0) d 0 6.83 in 3.83 s Central Difference Method Example Problem Procedure for solution: 3. Solve for {d - }att t = -t ( t) dd0( t) d 0d 0 (0.05) d 0 (0.05)0 (6.83) 0.0785 in 4. Solve for {d }att t = 0.05 s using the value of {d - }fromstep3 d M t F 0 t d0 d M K M d 0.05,000 3.830.05 00 0 3.830.0785 3.8 0.0785 in
CIV 8/77 Chapter - /75 Central Difference Method Example Problem Procedure for solution: 5. With {d 0 } given and {d } determined in Step 4 solve for {d } d M t F t d d0 M K M d 0.05,500 3.830.05 00 0.07853.830 3.8 0.74 in 6. Solve for d : d M F Kd d,500 000.0785 46.88 in 3.83 s Central Difference Method Example Problem Procedure for solution: 7. Solve for d : d d d d 0 ( t) 0.74 0.74 in 0.05 s 8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps.
CIV 8/77 Chapter - /75 Central Difference Method Example Problem Procedure for solution: 5. With {d } given and {d } determined in Step 4 solve for {d 3 } d3 M t F t d d M K M d 0.05,000 3.830.05 00 0.743.830.0785 3 3.8 0.546 in 6. Solve for d : d M F Kd d,000 000.74 30.56 in 3.83 s Central Difference Method Example Problem Procedure for solution: 7. Solve for d : d d d d 3 ( t) 0.546 0.0785 4.68 in 0.05 s 8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps.
CIV 8/77 Chapter - 3/75 Central Difference Method Example Problem The following table summarizes the results for the remaining time steps as compared with the exact solution. t (s) F(t) (lb) d (in/s ) d (in/s) d (in) d (exact) i i 0.00,000 6.83 0.00 0.000 0.0000 0.05,500 46.88.74 0.0785 0.078 0.0,000 30.56 4.68 0.74 0.603 0.5 500 3.99 5.79 0.546 0.55 0.0 0 -.68 6.07 0.854 0.850 0.5 0-3.63 5.9.54.3 i i Central Difference Method Example Problem Plotting the motion for about 4 s gives: 70.000 60.000 50.000 Velocity Acceleration Position 40.000 30.000 0.000000 0.000 0.000-0.000 0 0.5.5.5 3 3.5 4-0.000
CIV 8/77 Chapter - 4/75 Newmark s Method Newmark s equations are given as: d ( ) ( ) i di t di d i di di ( t) d i ( t) d i d i where and are parameters. The parameter is typically between 0 and ¼, and is often taken to be ½. For example, if = 0 and = ½ the above equation reduce to the central difference method. Newmark s Method To find {d i+ } first multiply the above equation by the mass matrix [M] and substitute the result into this the expression for acceleration. Recall the acceleration is: d M F K d i i i The expression [M]{d i+ } is: Md i M di ( t) M d i ( t) M d i d ( t) Fi K i Combining terms gives: M ( t) Kd i ( t) Fi Mdi M i M i ( t) d ( t) d
CIV 8/77 Chapter - 5/75 Newmark s Method Dividing the above equation by (Δt) gives: where: ( t) K' K M K' F' di i M F' F ( ) ( ) ( t) d t d t d i i i i i The advantages of using Newmark s method over the central difference method are that Newmark s method can be made unconditionally stable (if = ¼ and = ½) and that larger time steps can be used with better results. Newmark s Method Procedure for solution:
CIV 8/77 Chapter - 6/75 Newmark s Method Procedure for solution: d, d,and F ( t ). Given: 0 0. If the acceleration is not given, solve for d 0 d 0 M F0 Kd0 3. Solve for {d }att t = 0 K' d F' i Newmark s Method Procedure for solution: d d i d i d dd0( t) d 0 ( t) d 0 4. Solve for (original Newmark equation for rewritten for ): ( t) 5. Solve for d d d 0 ( t ) ( ) d 0 d 6. Repeat Steps 3, 4, and 5 to obtain the displacement, acceleration, and velocity for the next time step.
CIV 8/77 Chapter - 7/75 Newmark s Method Example Problem Determine the displacement, acceleration, and velocity at 0. second time intervals for up to 0.5 seconds for the one- dimensional spring-mass system shown in the figure below. Consider the above spring-mass system as a single degree of freedom problem represented by the displacement d. Use Newmark s method with = /6 and = ½. Newmark s Method Example Problem Procedure for solution: d, d,and F ( t ). Given: 0 0. If the acceleration is not given, solve for d 0 d 0 M F0 Kd0 i 00 70(0) 56 5 /.77 00 70(0) 56.5 in / s
CIV 8/77 Chapter - 8/75 Newmark s Method Example Problem Procedure for solution: 3. Solve for {d } at t = 0. s K' d F' K K M 6 ' 70 (.77),3 lb / in ( t) (0.) M F' F ( ) ( ) ( tt ) d t d t d 0 0 0 d.77 80 0 (0.)0 6(0.) 56.5 80 lb (0.) 6 F' 80lb K',3 lb in 0.48 in Newmark s Method Example Problem Procedure for solution: 4. Solve for d at t =0s 0. d d dd 0( t) d 0 ( t) d 0 ( t) 0.48 0 (0.)0 (0.) 56.5 35.4 6 (0.) 6 5. Solve for d d d 0 ( t ) ( ) d 0 d 0 (0.) ( )56.5 35.4 4.59 in s in s
CIV 8/77 Chapter - 9/75 Newmark s Method Example Problem Procedure for solution: 6. Repeat Steps 3, 4, and 5 to obtain the displacement, acceleration, and velocity for the next time step. Repeating Steps 3, Solve for {d } at t = 0. s M F' F d ( t) d ( t) d ( t) d.77 60 0.48 (0.)4.59 6(0.) 35.4 934 lb (0.) 6 F' 934 K',3 0.85 in Newmark s Method Example Problem Procedure for solution: Repeating Step 4: solve for d at t =0s 0. d dd ( t) d ( t) d ( t) d 0.85 0.48 (0.)4.59 (0.) 35.4 6 6 (0.).7 in s
CIV 8/77 Chapter - 30/75 Newmark s Method Example Problem Procedure for solution: 5. Solve for d d d ( t ) ( ) d d 4.59 (0.) ( )35.4.7 6.4 in s Newmark s Method Example Problem Procedure for solution: The following table summarizes the results for the time steps through t = 0.5 seconds. t (s) F(t) lb (in/s ) (in/s) d i (in) 0.0 00 56.6 0.0 0.0 0. 80 35.4 4.59 0.48 0. 60.7 6.4 0.85 0.3 48.6-6. 5.7.36 0.4 45.7-4..75.7 0.5 4.9-4. -.45.68
CIV 8/77 Chapter - 3/75 Newmark s Method Example Problem Procedure for solution: Plotting the motion for about 4 s gives: 80.000 60.000 Velocity Acceleration Position 40.000 0.000 0.000-0.000 0 0.5.5.5 3 3.5 4-40.000-60.000 Natural Frequencies of a One-Dimensional Bar Before solving the structural stress dynamic analysis problem, let s consider how to determine the natural frequencies of continuous elements. Natural frequencies are necessary in vibration analysis and important when choosing a proper time step for a structural dynamics analysis. Natural frequencies are obtained by solving the following equation: M d K d 0
CIV 8/77 Chapter - 3/75 Natural Frequencies of a One-Dimensional Bar The standard solution for {d} is given as: () ' i t dt d e where {d } is the part of the nodal displacement matrix called natural modes that is assumed to independent of time, i is the standard imaginary number, and is a natural frequency. Differentiating the above equation twice with respect to time gives: ' i t d d e Substituting the above expressions for {d} and equation of motion gives: M K i t it d' e d' e 0 d into the Natural Frequencies of a One-Dimensional Bar it Combining terms gives: e K Md Since e it is not zero, then: K M 0 ' 0 The above equations are a set of linear homogeneous equations in terms of displacement mode {d }. There exists a non-trivial solution if and only if the determinant of the coefficient matrix of is zero. K M 0
CIV 8/77 Chapter - 33/75 One-Dimensional Bar - Example Problem Determine the first two natural frequencies for the bar shown in the figure below. Assume the bar has a length, modulus of elasticity E, mass density, and cross-sectional area A. One-Dimensional Bar - Example Problem et s discretize the bar into two elements each of length as shown below. We need to develop the stiffness matrix and the mass matrix (either the lumped- mass of the consistent-mass matrix). In general, the consistent-mass matrix has resulted in solutions that compare more closely to available analytical and experimental results than those found using the lumped-mass matrix.
CIV 8/77 Chapter - 34/75 One-Dimensional Bar - Example Problem et s discretize the bar into two elements each of length as shown below. However, when performing a long hand solution, the consistentmass matrix is more difficult and tedious to compute; therefore, we will use the lumped-mass matrix in this example. One-Dimensional Bar - Example Problem et s discretize the bar into two elements each of length as shown below. The elemental stiffness matrices are: 3 AE AE k () k() The global stiffness matrix is: K 0 AE 0
CIV 8/77 Chapter - 35/75 One-Dimensional Bar - Example Problem et s discretize the bar into two elements each of length as shown below. The lumped-mass matrices are: 3 A 0 A 0 m () m() 0 0 The global lumped-mass matrix is: 0 0 A M 0 0 0 0 One-Dimensional Bar - Example Problem Substituting the above stiffness and lumped-mass matrices into the natural frequency equation: K M d ' 0 and applying the boundary condition {u } = 0 (or {d } = 0) gives: AE ' A 0d 0 0 d ' 3 0 Set the determinant of the coefficient matrix equal to zero as: AE A 0 0 0 where =
CIV 8/77 Chapter - 36/75 One-Dimensional Bar - Example Problem Dividing the above equation by A and letting E 0 gives: Evaluating the determinant of the above equations gives: 0.60 3.4 For comparison, the exact solution gives = 0.66 m, whereas the consistent-mass approach yields = 0.648 m. One-Dimensional Bar - Example Problem Therefore, for bar elements, the lumped-mass approach can yield results as good as, or even better than, the results from the consistent-mass approach. However, the consistent-mass approach can be mathematically proven to yield an upper bound on the frequencies, whereas the lumped-mass approach has no mathematical proof of boundedness. The first and second natural frequencies are given as: 0.77.85
CIV 8/77 Chapter - 37/75 One-Dimensional Bar - Example Problem The term may be computed as: 6 E 30 0 4.0 (0.00073)(00) s 6 Therefore, first and second natural frequencies are:.560 rad / s 3.76 0 rad / s 3 3 One-Dimensional Bar - Example Problem In general, an n-degree-of-freedom discrete system has n natural modes and frequencies. A continuous system actually has an infinite number of natural modes and frequencies. The lowest modes and frequencies are approximated most often; the higher frequencies are damped out more rapidly and are usually less important.
CIV 8/77 Chapter - 38/75 One-Dimensional Bar - Example Problem Substituting into the following equation Gives: AE ' A 0d 0 0 d ' 0 3.4 d' d' 0 d' 0.7 d' 0 () () () () 3 3 where the superscripts indicate the natural frequency. It is customary to specify the value of one of the natural modes {d } for a given i or i and solve for the remaining values. For example, if () () d ' 3 than the solution for d ' 0.7 One-Dimensional Bar - Example Problem () Similarly, if we substitute and let d ' 3 the solution of the above equations gives d ' 0.7 The modal responses for the first and second natural frequencies are shown in the figure below. () The first mode means that the bar is completely in tension or compression, depending on the excitation direction.
CIV 8/77 Chapter - 39/75 One-Dimensional Bar - Example Problem () Similarly, if we substitute and let d ' 3 the solution of the above equations gives d ' 0.7 The modal responses for the first and second natural frequencies are shown in the figure below. () The second mode means that bar is in compression and tension or in tension and compression. Time-Dependent One-Dimensional Bar - Example Consider the one-dimensional bar system shown in the figure below. Assume the boundary condition {d x } = 0 and the initial conditions {d = 0 and d 0 } 0 0 et = 0.00073 lb-s /in 4, A = in, E = 30 x 0 6 psi, and = 00 in.
CIV 8/77 Chapter - 40/75 Time-Dependent One-Dimensional Bar - Example The bar will be discretized into two elements as shown below. The elemental stiffness matrices are: 3 AE AE k () k() The global stiffness matrix is: K 0 AE 0 Time-Dependent One-Dimensional Bar - Example The bar will be discretized into two elements as shown below. The lumped-mass matrices are: 3 A 0 A 0 m () m() 0 0 The global lumped-mass matrix is: 0 0 A M 0 0 0 0
CIV 8/77 Chapter - 4/75 Time-Dependent One-Dimensional Bar - Example Substitute the global stiffness and mass matrices into the global dynamic equations gives: 0 u 0 0u R AE A u 0 0 u 0 0 u 3 0 0 u 3 F3( t) where R denotes the unknown reaction at node. Time-Dependent One-Dimensional Bar - Example For this example, we will use the central difference method, because it is easier to apply, for the numerical time integration. It has been mathematically shown that the time step t must be less than or equal to two divided by the highest natural frequency. t max For practical results, we should use a time step defined by: 3 t 4 max
CIV 8/77 Chapter - 4/75 Time-Dependent One-Dimensional Bar - Example An alternative guide (used only for a bar) for choosing the approximate time step is: t t c where is the element length, and is the longitudinal wave velocity. Evaluating the time step estimates gives: 3.5 t 0.40 0 3 4max 3.760 00 t 0.48 0 c x 6 30 0 0.00073 x c x E x 3 3 s s Time-Dependent One-Dimensional Bar - Example Guided by these estimates for time step, we will select t = 0.5 x 0-3 s. Procedure for solution:. At time t = 0: d0 d 0 0 0. If the acceleration is not given, solve for d 0 d 0 M F0 Kd0 d 0 u 0 0 AE 0 u3 A 0,000 0 t 0
CIV 8/77 Chapter - 43/75 Time-Dependent One-Dimensional Bar - Example Applying the boundary conditions u = 0 and simplifying gives: u 000 0 0 in u A 7,400 s d 0 3 t 0 u 0 and 3. Solve for d - at t = -t d d ( t) d d 0 0 0 3 ( t) 3 u 0 3 0 (0.5 0 ) 0 (0.50 ) u 0 0 7,400 0 0.8560 3 in Time-Dependent One-Dimensional Bar - Example 4. Solve for d at t = t using the value of d - from Step 3 d M t F M t K d M d 0 0 u 0 3 0 (0.073) 0 0.5 0 u 0.073 0,000 0 3 0 0.073 0 0 0.50 3 30 0 4 3 0 0 0.856 0 u 0 0 0 3 3 u3 0.073 0 0.065 0 0.030 u 0 in 3 u 0.858 0 3
CIV 8/77 Chapter - 44/75 Time-Dependent One-Dimensional Bar - Example 5. With d 0 given and d determined in Step 4 solve for d d M t F M t K d M d 0 u 0 3 0 (0.073) 0 0.5 0 u 0.073 0 000 0 3 0 0.073 00 0.50 3 30 0 4 3 0.858 0 0 0 3 u 0 0 0.060 3 3 u3 0.073 0 0.0650 0.04660 3 u 0.0 in 3 u.990 3 Time-Dependent One-Dimensional Bar - Example 6. Solve for d d M F Kd u 0 0 4 0 (30 0 ) 3 u3 0.073 0 000 0.858 0 u 3,56 in u 3 0,345 s
CIV 8/77 Chapter - 45/75 Time-Dependent One-Dimensional Bar - Example 7. Solve for d d d d 0 ( t) 3 0.0 0 3 u 0.990 0.44 3 u in 3 0.50 5.98 s Time-Dependent One-Dimensional Bar - Example 8. Repeat Steps 5, 6, and 7 to obtain the displacement, acceleration, and velocity for other time steps. Repeating Step 5: d M t F M t K d M d 3 u 0 3 0 (0.073) 0 0.5 0 073 u 3 0.073 0 000 0 3 3 3 0.0 0.073 0 0 0.5 0 3 300 4 3 3.99 0 0 0.8580 3 u.0960 in 3 u 5.3970
CIV 8/77 Chapter - 46/75 Time-Dependent One-Dimensional Bar - Example Repeating Step 6. Solve for d d M F Kd 3 u 0 0 4 0.0 (30 0 ) 3 u3 0.073 0 000.990 u 0,500 in u 3 4600 4,600 s Time-Dependent One-Dimensional Bar - Example Repeat Step 7: Solve ford d d d 3 ( t) 3.096 0 0 3 3 u 0.858 0 5.397 0.9 in 3 u 3 0.50 9.078 s
CIV 8/77 Chapter - 47/75 Beam Element Mass Matrices and Natural Frequencies We will develop the lumped- and consistent-mass matrices for time-dependent beam analysis. Consider the beam element shown in the figure below. The basic equations of motion are: Ft () K{ d} Md Beam Element Mass Matrices and Natural Frequencies The stiffness matrix is: k The lumped-mass matrix is: v v 6 6 EI 6 4 6 3 6 6 6 6 4 v v 0 0 0 0 0 0 0 m A 0 0 0 0 0 0 0
CIV 8/77 Chapter - 48/75 Beam Element Mass Matrices and Natural Frequencies The mass in lumped equally into each transitional degree of freedom; however, the inertial effects associated with any possible rotational degrees of freedom is assumed to be zero. A value for these rotational degrees of freedom could be assigned by calculating the mass moment of inertia about each end node using basic dynamics as: I 3 A A m 3 3 4 Beam Element Mass Matrices and Natural Frequencies The consistent-mass matrix can be obtained by applying T m N N dv V N N m N N N3 N4dAdx 0 A N3 N 4 where 3 3 N 3 x 3x N x x x 3 3 3 N x x 3 3 3 3 3 N4 x x 3
CIV 8/77 Chapter - 49/75 Beam Element Mass Matrices and Natural Frequencies The shape functions are shown below:..0 0.8 0.6 0.4 0. N N N3 N4 0.0 0.00.00-0. -0.4 Beam Element Mass Matrices and Natural Frequencies Substituting the shape functions into the above mass expression and integrating g gives: 56 54 3 A 4 3 3 [ m] 40 54 3 56 3 3 4
CIV 8/77 Chapter - 50/75 Beam Element - Example Determine the first natural frequency for the beam shown in the figure below. Assume the bar has a length, modulus of elasticity E, mass density, and cross-sectional area A. et s discretize the beam into two elements each of length. We will use the lumped-mass matrix. Beam Element - Example We can obtain the natural frequencies by using the following equation. K M 0 The boundary conditions are v = = 0 and v 3 = 3 = 0. Therefore, the global stiffness and lumped-mass matrices are: v EI 4 0 3 0 8 K M v A 0 0 0
CIV 8/77 Chapter - 5/75 Beam Element - Example Substituting the global stiffness and mass matrices into the global dynamic equations gives: K M 0 Dividing by A and simplify EI 4 0 0 A 3 0 8 0 0 0 4EI 4 A 4.90 EI A The exact solution for the first natural frequency is: 5.59 EI A Beam Element - Example Determine the first natural frequency for the beam shown in the figure below. Assume the bar has a length 3, modulus of elasticity E, mass density, and cross-sectional area A. et s discretize the beam into three elements each of length. We will use the lumped-mass matrix.
CIV 8/77 Chapter - 5/75 Beam Element Example We can obtain the natural frequencies by using the following equation. K M 0 The boundary conditions are v = = 0 and v 4 = 4 = 0. Therefore the elements of the stiffness matrix for element are: v v 6 6 () EI 6 4 6 k 3 6 6 6 6 4 Beam Element Example Element : v 3 3 6 6 () EI 6 4 6 k 3 6 6 6 6 4 Element 3: v v 3 3 4 4 6 6 (3) EI 6 4 6 k 3 6 6 6 6 4 v
CIV 8/77 Chapter - 53/75 Beam Element Example Assembling the global stiffness matrix as: K K v 66 6 EI 6 6 4 6 3 6 66 6 66 4 4 v v 3 3 3 3 0 6 EI 0 6 6 3 6 4 0 6 0 8 v Beam Element Example We can obtain the natural frequencies by using the following equation. K M 0 Therefore the elements of the mass matrix for element are: 0 0 0 () A 0 0 0 0 m 0 0 0 0 0 0 0 v v
CIV 8/77 Chapter - 54/75 Beam Element Example Element : v v Element 3: 3 3 3 3 4 4 0 0 0 0 0 0 () 0 0 0 0 A () A 0 0 0 0 m 0 0 0 m 0 0 0 0 0 0 0 0 0 0 0 The assembled mass matrix is: m v 3 3 0 0 0 A 0 0 0 0 0 0 0 0 0 0 0 v M v v 0 0 0 0 0 0 0 A 0 0 0 0 0 0 0 v v 3 3 Beam Element Example We can obtain the Frequency equation as: K M 0 0 6 0 0 0 EI 0 6 6 0 0 0 0 A 3 6 4 0 0 0 0 6 0 8 0 0 0 0 0 A 6 EI EI EI 3 0 6 6 EI EI EI EI EI EI 6 4 A 0 3 3 6 0 8 EI EI EI 0
CIV 8/77 Chapter - 55/75 Beam Element Example Simplifying and assuming: A 6 EI EI EI 3 0 6 6 EI EI EI EI EI EI 6 4 0 3 3 6 0 8 EI EI EI 0 Evaluating the 4x44 determinate t gives:,5 EI 48 EI 576EI, 96EI 5 8 8 3 3 4 4 4 4 4 3 3 4 4 4 96 EI 4 EI 6,9EI 0 5 8 Beam Element Example Simplifying: 4 3 3 4 4 44 EI,056 EI 7,63EI 0 5 8 4 64 EI,908E I 0 3 6 Dividing by 4E I 5.8754 EI 3 and solving for the two roots for 9.8754 EI 3
CIV 8/77 Chapter - 56/75 Beam Element Example Ignoring the negative root as it is not physically possible and solving explicitly for gives: 9.8754 EI 3 In summary: 4.90 EI Two elements: A Three elements: 9.8754 EI 5.46 EI A 3 5.46 EI A Exact solution: 5.59 EI A Beam Element Example 3 Determine the first natural frequency for the beam shown in the figure below. Assume the bar has a length = 30 in, modulus of elasticity 7 4 E = 3 x 0 7 psi, mass density = 0.00073 lb-s /in 4, and cross-sectional area A = in, moment of inertia I = 0.0833 in 4, and Poisson s ratio = 0.3. K M 0
CIV 8/77 Chapter - 57/75 Beam Element Example 3 Determine the first natural frequency for the beam shown in the figure below. et s discretize the beam into two elements each of length = 5 in. We will use the lumped-mass matrix. We can obtain the natural frequencies by using the following equation. K M 0 Beam Element Example 3 In this example, the elemental stiffness matrices are: v Element : 6 6 () EI 6 4 6 k 3 6 6 6 6 4 v Element : v 3 3 6 6 () EI 6 4 6 k 3 6 6 6 6 4 v
CIV 8/77 Chapter - 58/75 Beam Element Example 3 In this example, the elemental mass matrices are: v v Element : 0 0 0 () A 0 0 0 0 m 0 0 0 0 0 0 0 3 3 Element : 0 0 0 () A 0 0 0 0 m 0 0 0 0 0 0 0 v v Beam Element Example 3 The boundary conditions are u = = 0. Therefore the global stiffness and lumped mass matrices is: K v 3 3 v 3 3 4 0 6 0 0 0 0 8 6 EI A 0 0 0 0 M 3 6 6 0 0 0 6 6 4 0 0 0 0 v v
CIV 8/77 Chapter - 59/75 Beam Element Example 3 Substituting the global stiffness and mass matrices into the global dynamic equations gives: 4 0 6 0 0 0 EI 0 8 6 A 0 0 0 0 3 6 6 0 0 0 6 6 4 0 0 0 0 0 4 0 6 0 0 0 0 8 6 0 0 0 0 6 6 0 0 0 6 6 4 0 0 0 0 0 EI A 4 Beam Element Example 3 Evaluating the determinant of the above equations gives: 4 4 3 4 4 40 44 0 The solution for the first natural frequency is: 0.789 EI 4.0647 EI A A 0.789 4.0647
CIV 8/77 Chapter - 60/75 Beam Element Example 3 In this example, the length of each element is actual /, therefore: 0.789 EI 3.564 EI A A 4.0647 EI 6.588 EI A A The exact solution for the first natural frequency is: 3.56 EI A Beam Element Example 3 In the exact solution of the vibration of a clamped-free beam, the higher natural frequencies to the first natural frequency can be given as: 3 6.669 7.5475
CIV 8/77 Chapter - 6/75 Beam Element Example 3 The figure below shows the first, second, and third mode shapes corresponding to the first three natural frequencies for the cantilever beam. Beam Element Example 3 The table below shows various finite element solutions compared to the exact solution. (rad/s) (rad/s) Exact Solution 8,434 FE Solution Using elements 05,86 Using 6 elements 6,37 Using 0 elements 7.5,40 Using 30 elements 8.5,430 Using 60 elements 8.5,43
CIV 8/77 Chapter - 6/75 Truss Analysis The dynamics of trusses and plane frames are preformed by extending the concepts of bar and beam element. The truss element requires the same transformation of the mass matrix from local to global coordinates as that used for the stiffness matrix given as: T m T m T Truss Analysis Considering two-dimensional motion, the axial and the transverse displacement are given as: u u x 0 x 0 v v 0 x 0 xu v The shape functions for the matrix are: N x 0 x 0 0 x 0 x
CIV 8/77 Chapter - 63/75 Truss Analysis The consistent-mass matrix can be obtained by applying: 0 0 T [ m] [ N] [ N] dv A 0 0 m V 6 0 0 0 0 The lumped-mass matrix for two-dimensional motion is obtained by simply lumping mass at each node (mass is the same in both the x and y directions): 0 0 0 A 0 0 0 m 0 0 0 0 0 0 Plane Frame Analysis The plane frame element requires combining the bar and beam elements to obtain the local mass matrix. There are six degrees of freedom associated with a plane frame element.
CIV 8/77 Chapter - 64/75 Plane Frame Analysis The plane frame analysis requires first expanding and then combining the bar and beam mass matrices to obtain the local mass matrix. The bar and beam mass matrices are expanded to a 6 x 6 and superimposed Plane Frame Analysis Combining the local axis consistent-mass matrices for the bar and beam elements gives: 6 0 0 6 0 0 0 56 54 3 40 40 0 40 40 0 4 3 3 40 40 0 40 40 m A 6 0 0 6 0 0 0 54 3 56 40 40 0 40 40 3 3 4 0 40 40 0 40 40
CIV 8/77 Chapter - 65/75 Plane Frame Analysis The resulting lumped-mass matrix for a plane frame element is give as: 0 0 0 0 0 0 0 0 0 0 A 0 0 0 0 0 0 m 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Plane Frame Analysis The global mass matrix for the plane frame element arbitrarily oriented in x-y coordinates is transformed by: T m T mt where the transformation matrix is given as: T C S 0 0 0 0 S C 0 0 0 0 0 0 0 0 0 0 0 0 C S 0 0 0 0 S C 0 0 0 0 0 0
CIV 8/77 Chapter - 66/75 Plane Stress/Strain Elements The plane stress/strain constant-strain triangle consistent-mass matrix is obtained using the shape functions given below as: N N 0 N 0 N3 0 0 N 0 N 0 N3 The linear triangular shape functions are illustrated below: N N N 3 y y y x 3 x 3 x 3 Plane Stress/Strain Elements The consistent-mass matrix can be obtained by applying: T [ m ] [ N ] [ N ] dv V where dv = tda The CST global consistent-mass matrix is: 0 0 0 0 0 0 0 0 0 [ m] ta 0 0 0 0 0 0 0 0 0
CIV 8/77 Chapter - 67/75 Frame Example Problem Use SAP000 to determine the motion of the frame structure below. Frame Example Problem Assume the modulus of elasticity E = 3 x 0 7 psi. The nodal lumped mass values are obtained by dividing the total weight (dead loads included) of each floor or wall section by gravity.
CIV 8/77 Chapter - 68/75 Frame Example Problem For example, compute the total mass of the uniform vertical load on elements 4, 5, and 6: M M M 4 5 6 50 psf 30 ft 5 ft W4 58.8 lb s g 386.04 in s 04 psf 30 ft 5 ft W5.3 lb s g 386.04 in s 04 psf 30 ft 5 ft W6.3 lb s g 386.04 in s in in in Frame Example Problem Next, lump the mass equally to each node of the beam element. For this example calculation, a lumped mass of 9.4 lbs /in should be added to nodes 7 and 8 and a mass of 60.6 lbs /in should be added to nodes 3, 4, 5, and 6, all in the x direction.
CIV 8/77 Chapter - 69/75 Frame Example Problem In an identical manner, masses for the dead loads for additional wall sections should be added to their respective nodes. In this example, additional wall loads should be converted to mass added to the appropriate loads. Frame Example Problem For example, the additional mass due to the wall load on element 8 is: M 7 0 psf 0 ft 5 ft W7 7.77 lb s g 386.04 in s Therefore, an additional 3.88 lbs /in should be added to nodes 6 and 8. in
CIV 8/77 Chapter - 70/75 Frame Example Problem For example, the additional mass due to the wall load on element 8 is: M 8 0 psf 0 ft 5 ft W8 7.77 lb s g 386.04 in s Therefore, an additional 3.88 lbs /in should be added to nodes 4 and 6. in Frame Example Problem For example, the additional mass due to the wall load on element 8 is: M 9 0 psf 5 ft 5 ft W9.66 lb s g 386.04 in s Therefore, an additional 5.83 lbs /in should be added to node 4. in
CIV 8/77 Chapter - 7/75 Frame Example Problem The final values for the nodal masses on this frame are shown below. Frame Example Problem A trace of the displacements of nodes 8, 6, and 4 as a function of time can be generated using the SAP000 Display Menu. Nodes 8, 6, and 4 are listed in SAP 000 as 5, 6, and 7.
CIV 8/77 Chapter - 7/75 Frame Example Problem A plot of the displacements of nodes 8, 6, and 4 over the first 5 seconds of the analysis generate by SAP000 is shown below: Nodes 8, 6, and 4 are listed in SAP 000 as 5, 6, and 7. Frame Example Problem The maximum displacement of node 8 (node 5 in the SAP 000 plot) is 3.6 in. and the period of the vibration is approximately 35 3.5 seconds.
CIV 8/77 Chapter - 73/75 Frame Example Problem Use SAP000 to determine the motion of the frame structure below. Frame Example Problem A trace of the displacements as a function of time can be generated using the SAP000 Display Menu. A plot of the displacements of nodes 8, 6, and 4 over the first 4 seconds of the analysis generate by SAP000 is shown below:
CIV 8/77 Chapter - 74/75 Frame Example Problem The maximum displacement of node 8 is 0.84 in. and the period of the vibration is approximately.65 seconds. Development of the Plate Bending Element Problems 3. Do problems 6.5 and 6.c on pages 84-88 in your textbook A First Course in the Finite Element Method by D. ogan. 4. Do problems 6.4 and 6.6 on pages 84-88 in your textbook A First Course in the Finite Element Method by D. ogan using SAP000 or WinFElt.
CIV 8/77 Chapter - 75/75 End of Chapter 6