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UNIS Template Finite Element (FE) equilibrium equation: ku=f 1D spring element 1D bar element Beam element 1 Finite Element Method (FEM) formulation 2 Finite Element Method (FEM) formulation Nodal force: In FE formulation, loading has only been applied on nodes. Distributed load: In some cases, distributed loading is applied along elements, where the distributed loads can be converted to the nodes by using equivalent nodal forces. 3 Classroom activity #14 1 2 3 (2) (1) EA 2EA L L A step-bar system with uniform material density uniform Young’s modulus cross-section areas of for bar AB and for bar BC. The gravity () is considered as a distributed force. Determine (1) the displacement at the middle node B and (2) the reaction forces from the wall A (Exam 2014, 2017, Quiz 2015). A B C Solution Step 0: Analysis of the problem: Type of elements: Bar (loading follows axial direction of structural components) Number of elements (one thick bar and one thin bar elements) , : 2 Number of degree of freedom: Number of nodes (1, 2, 3) = 3 Step 1: uniformly distributed gravitational load: Element AB (1): Element BC (2): 4 Classroom Activity #14 – Cont’d Step 3: Expanded elemental equilibrium equations for global equilibrium equation: Step 2: Elemental equilibrium equations: Element A: Element B: 5 Classroom Activity #14 – Cont’d Step 3: Apply boundary and loading conditions: 6 Classroom Activity #15 1 2 3 (2) (1) EA 2EA L L A B C If there is a “stopper” is placed in the bottom, please solve the problem again. which has are two unknowns: , 7 Classroom Activity #15 – Cont’d 1 2 3 (2) (1) EA 2EA L L A B C From the second equation above: From the “dropped equation” in the global equilibrium 8 Chapter 7 Finite Element Method 7.5 Vibration Analysis MECH3361/9361 Mechanics of Solids II Qing Li School of Aerospace, Mechanical and Mechatronic Engineering 9 7.5 FE methods for vibrational analysis Vibrational analysis is often performed on mechanical systems to evaluate their dynamic behaviour and stability during vibrations Single degree-of-freedom systems: A system consisting of one mass, one spring, and one damper vibrating in one direction is a typical single DOF system, A single mass vibrating in a single direction is a single DOF system We are concerned with the motion of the mass m, i.e. x(t). The mass m is considered to be a rigid body. 10 7.5 FEM for vibrational analysis – Multiple degree of freedom A 3D single mass can have a maximum of 6 DOFs, i.e. 3 DOFs in translation (straight line motion) and 3 DOFs in rotation. Multiple degree of freedom: A system that must be described by two or more independent co-ordinates is called a multi-DOF system. A single mass vibrating in different directions constitutes a multi DOF system, as well as multiple masses vibrating in a single (or multiple) direction(s). General Multiple DOF: In an n-DOF system, the number of co-ordinates required to describe the motion of the masses is N. An N-DOF system has N number of natural frequencies (ωn). For each ωn, there is a corresponding vibration mode (i.e. mode shape). 11 7.5.3 FEM for vibrational analysis - Equations of motion Displacement of each mass (i.e. x(t)) is measured from its static equilibrium position. Thus we can ignore the gravity force of each mass and the initial deformation of the springs when we derive the vibration differential equations using Newton’s second law. Dynamic equilibrium positions Static equilibrium 12 7.5.3 FEM for vibrational analysis - Equations of motion Equations of motion: consider the free body diagrams of the 2 DOF system. Newton’s second law () can be applied to each mass. For mass 2: For mass 1: Express these two equations in matrix form 13 7.5.3 FEM for vibrational analysis - Equations of motion is the mass matrix is the stiffness matrix; is the unknown displacement vector Equations of motion: 14 7.5.4. Characteristic free vibration (synchronous) Synchronous solutions: In the above example, each mass undergoes harmonic motion at the same frequency (synchronous) and exhibits a certain vibration mode. Therefore, we assume the following synchronous solutions Substituting the synchronous solutions into the equation of motion: Characteristic equation: For a non-trivial solution (at least one of the A’s is non-zero): 15 7.5.5. Natural frequencies Natural frequencies can be found from the solutions to the characteristic equation. For simplicity, let’s look at a special system where and . The characteristic equation becomes: Divide by , and put , where is an assumed reference frequency. This is a quadratic equation, in terms of . Therefore, solving the quadratic equation 16 7.5.5. Natural frequencies Two natural frequencies (two DOF) are obtained, and . 1st natural circular frequency: 2nd natural circular frequency: 17 7.5.6. Mode Shape For each natural frequency , there is a corresponding mode shape. For a multi DOF spring-mass system, a mode shape is simply an amplitude ratio. Recall the synchronous solutions, we provided two synchronous solutions with different amplitudes and . Hence, the mode shape can be expressed mathematically as: . Mode shape expressions can be derived from either equations 18 7.5.6. Mode Shape – Cont’d Mode shape expressions can be derived from either equations Both ratios are equivalent for a given . For our special case in section 7.5.5, this can be confirmed by substituting the natural circular frequency : into both equations in F 19 7.5.6. Mode Shape – Cont’d Similarly, for For convenience, the amplitude is taken to be a unit (i.e. ). As a result, the mode shapes can also be expressed in terms of a vector form as 1st mode shape: 2nd mode shape: Remarks: these are actually the eigenvectors corresponding to the eigenvalues ( found from the characteristic equation 20 7.5.6. Mode Shape – Cont’d For and , the first order vibration mode (if A2=1, then A1=0.618) can be drawn For and , the second order vibration mode (if A2=1, then A1=-1.618) can be drawn. Notably, when vibrating at the 2nd order natural frequency, the two masses are out of phase, i.e. they are vibrating in different directions. 21 ijx i F j F i M j M EIL )( ii ,v)( jj ,v Bar Element Method 1 (Displacement Method) Method 2 (Energy Method) Displacement u ji uLxuLxxu )/()/1()( jijjii uuuNuNu )1()()()( LL uu ij BuuNNu dx d dx d dx du L E E BuEE Force or W=U k L EA A L E AF fuuBBu T 2 1 2 1 V TT dVEU Equilibrium eqn fku fku Stiffness matrix 11 11 L EA kk kk k 11 11 L EA k Bar Element Method 1 (Displacement Method) Method 2 (Energy Method) Displacement u Force or W=U Equilibrium eqn Stiffness matrix L L u u i j D = - = e ( ) Bu u N Nu = ÷ ø ö ç è æ = = = dx d dx d dx du e L E E D = = e s Bu E E = = e s D = D ÷ ø ö ç è æ = D = = k L EA A L E A F s ( ) f u u B B u T 2 1 2 1 = ú ú û ù ê ê ë é ò V T T dV E U f ku = ú û ù ê ë é - - = ú û ù ê ë é - - = 1 1 1 1 L EA k k k k k ú û ù ê ë é - - = 1 1 1 1 L EA k j i u L x u L x x u ) / ( ) / 1 ( ) ( + - = j i j j i i u u u N u N u x x x x x + - = + = ) 1 ( ) ( ) ( ) ( x θx y θ y θz z m x θ x y θ y θ z z m m x(t) y(t) m x(t) y(t) � � ����� ����� �� ������ ��� � � ��� ����� −�ω ��� ������������� UNIS Template Finite Element (FE) equilibrium equation: ku=f 1D spring element 1D bar element Beam element 1 Solution Step 0: Analysis of the problem: Number of elements (one bar and one spring elements) : 2 Classroom Activity #12 A spring – bar system with the spring constant and . It is assumed that an axial force is applied in node 2 and a stopper is placed next to node 2 with a gap of . Determine (1) the displacement at the middle node and (2) the reaction force from the stopper/walls. Number of degrees of freedom: Step 1: Elemental equilibrium equations: Number of nodes (1, 2, 3) = 3 Bar Element Spring Element 2 Classroom Activity #12 – Cont’d Step 2: Expanded elemental equilibrium equations for global equilibrium equation: 3 Classroom Activity #12 – Cont’d Step 3: Apply boundary and loading conditions: Note that includes all the forces applied at node 2, which comprises external force and reaction from the stopper . Equilibrium Thus 4 Assignment 3 – Q1 1D finite element method (10 marks) The two horizontal bars are connected by a linear spring. The system is fully clamped at ends A and D, as illustrated in Fig. 1. 1m 1m 1m EA=2 k s = 3 EA=1 x 2P 4P A B C D 0.5m 0.5m O Discretise the system into two bar elements and one spring element. Write each elemental equilibrium equation, and compile the global equilibrium equation. Calculate the displacements at B and C, and the reaction forces at A and D. Calculate the displacement at O (the midpoint of spring element) by using shape functions. Plot the distributions of displacement and strain in the system. If a stopper is placed to the left of node C with a distance of separation =0.1P as shown in