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ME 520 Fundamentals of Finite Element Analysis

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ME 520 Fundamentals of Finite Element Analysis
5-Bar Element Dr. Ahmet Zafer Şenalp Mechanical Engineering Department Gebze Technical University

Mechanical Engineering Department, GTU
5-Bar Element Bar Element Bar (truss) structures: Cross section examples for bar structures ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Bar Element Consider a uniform prismatic bar: L: length A: cross-sectional area E: elastic modulus u=u(x): displacement e=e(x): strain s=s(x): stress Strain-displacement relation: Stress-strain relation: FE Model i j ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Stiffness Matrix - Direct Method Assuming that the displacement u is varying linearly along the axis of the bar, i.e., The bar is acting like a spring in this case and we conclude that element stiffness matrix is: D: elongation, F: force in bar k: stiffness of the bar ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Stiffness Matrix - A Formal Approach We derive the same stiffness matrix for the bar using a formal approach which can be applied to many other more complicated situations: Define two linear shape functions as follows: B: Element strain-displacement matrix ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Stiffness Matrix - A Formal Approach Stress can be written as: Consider the strain energy stored in the bar: The work done by the two nodal forces is: For conservative system, we state that: which gives: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Stiffness Matrix - A Formal Approach k is the element stiffness matrix. The above expression is a general result which can be used for the construction of other types of elements. This expression can also be derived using other more rigorous approaches, such as the Principle of Minimum Potential Energy, or the Galerkin’s Method. Now, we evaluate for the bar element which is the same as we derived using the direct method. Strain energy in the element can be written as: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Degree of Freedom (dof) Number of components of the displacement vector at a node. For 1-D bar element: one dof at each node. Physical Meaning of the Coefficients in k: The jth column of k (here j = 1 or 2) represents the forces applied to the bar to maintain a deformed shape with unit displacement at node j and zero displacement at the other node. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 FE Model Find; the global stiffness matrix (b) displacements of nodes (c) the reaction forces (d) stresses at the elements Solution : Connectivity table: 1 2 1 2 3 E# N1 N2 1 2 3 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 Boundary conditions : Displacement boundary conditions : Force boundary conditions : a) Element Stiffness Matrices : ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 Construction of global stiffness matrix : Equilibrium (FE) equation for the whole system is; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 b) Applying boundary conditions; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 For the data given; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 1 c) From the 1st and 3rd row of the finite element equation, the reaction forces can be calculated as; d) Stresses in the elements: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Notes · In this case, the calculated stresses in elements 1 and 2 are exact within the linear theory for 1-D bar structures. It will not help if we further divide element 1 or 2 into smaller finite elements. · For tapered bars, averaged values of the cross-sectional areas should be used for the elements. · We need to find the displacements first in order to find the stresses, since we are using the displacement based FEM. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution procedure with matlab As linear bar element has 2 degrees of freedom (1 at each node) for a structure with n nodes, the global stiffness matrix K will be of size nxn. The global stiffness matrix K is obtained by making calls to the Matlab function LinearBarAssemble which is written for this purpose. Once the global stiffness matrix; K is obtained we have the following structure equation; At this step boundary conditions are applied manually to the vectors U and F. Then the matrix equation is solved by portioning and Gaussion elimination. Finally once the unknown displacements and and reactions are found, the force is obtained for each element as follows: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution procedure with matlab where f is the 2x1 element force vector and u is the 2x1 element displacement vector. The element stress is obtained by using local stiffness matrix and local displacement vector. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Matlab functions used LinearBarElementStiffness(E,A,L) This function returns the element stiffness matrix for a linear bar with modulus of elasticity E, cross-sectional area A, and length L. The size of the element stiffness matrix is 2 x 2. Function contents: function y = LinearBarElementStiffness(E,A,L) %LinearBarElementStiffness This function returns the element % stiffness matrix for a linear bar with % modulus of elasticity E, cross-sectional % area A, and length L. The size of the % element stiffness matrix is 2 x 2. y = [E*A/L -E*A/L ; -E*A/L E*A/L]; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Matlab functions used LinearBarAssemble(K,k,i,j) This function assembles the element stiffness matrix k of the linear bar with nodes i and j into the global stiffness matrix K. This function returns the global stiffness matrix K after the element stiffness matrix k is assembled. Function contents: function y = LinearBarAssemble(K,k,i,j) %LinearBarAssemble This function assembles the element stiffness % matrix k of the linear bar with nodes i and j % into the global stiffness matrix K. % This function returns the global stiffness % matrix K after the element stiffness matrix % k is assembled. K(i,i) = K(i,i) + k(1,1); K(i,j) = K(i,j) + k(1,2); K(j,i) = K(j,i) + k(2,1); K(j,j) = K(j,j) + k(2,2); y = K; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Matlab functions used LinearBarElementForces(k,u) This function returns the element nodal force vector given the element stiffness matrix k and the element nodal displacement vector u. Function contents: function y = LinearBarElementForces(k,u) %LinearBarElementForces This function returns the element nodal % force vector given the element stiffness % matrix k and the element nodal displacement % vector u. y = k * u; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Matlab functions used LinearBarElementStressess_azLinearBarElementStresses_az(k, u, L, E) This function returns the element nodal stress vector given the element stiffness matrix k, the element nodal displacement vector u, the length of the element and E Function contents: function y = LinearBarElementStresses_az(k, u, L, E) %LinearBarElementStresses This function returns the element nodal % stress vector given the element stiffness % matrix k, the element nodal displacement % vector u, the length of the element and E L_m=[-1/L 1/L]; y = E*L_m*u; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab Solution: Use the 7 steps to solve the problem using linear bar element. Step 1-Discretizing the domain: This problem is already discretized. The domain is subdivided into 2 elements and 3 nodes. The units used in Matlab calculations are N and milimeter. The element connectivity is: E# N1 N2 1 2 3 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab Step 2-Copying relevant files and starting Matlab Create a directory Copy LinearBarElementStiffness.m LinearBarAssemble.m LinearBarElementForces.m LinearBarElementStresses_az.m files under the created directory Open Matlab; Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab Start solving the problem in Command Window: >>clearvars >>clc Enter the data >>E=2e4 >>A(1)=400 >>A(2)=200 >>L=125 Step 3-Writing the element stiffness matrices: >>k1=LinearBarElementStiffness(E,A(1),L) k1 = >>k2=LinearBarElementStiffness(E,A(2),L) ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab >>k2=LinearBarElementStiffness(E,A(2),L) k2 = Step 4-Assembling the global stiffness matrix: Since the structure has 3 nodes, the size of the global stiffness matrix is 3x3. Therefore to obtain K we first set up a zero matrix of size 3x3 then make 2 calls to the Matlab function LinearBarAssemble since we have 2 bar elements in the system. Each call to the function will assemble one element. The following are the Matlab commands: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab >>K=zeros(3,3) K = >>K=LinearBarAssemble(K,k1,1,2) >>K=LinearBarAssemble(K,k2,2,3) ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab Step 5-Applying the boundary conditions: Finite element equation for the problem is; The boundary conditions for the problem are; Inserting the above conditions into finite element equation ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab Step 6-Solving the equations: Solving the above system of equations will be performed by partitioning (manually) and Gaussian elimination (with Matlab) First we partition the above equation; >>u2=50000/96000 Step 7-Post-processing: In this step we obtain the reactions at nodes 1 and 3 and the force in each spring using Matlab as follows. First we set up the global nodal displacement vector U, then we calculate the nodal force vector F. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab >>U=[0; u2; 0] U = 0.5208 >>F=K*U F = 1.0e+04 * 5.0000 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 1 with Matlab >>u_el_1=[U(1); U(2)] >>u_el_2=[U(2); U(3)] >>stress1=LinearBarElementStresses_az(k1, u_el_1, L, E) >>stress2=LinearBarElementStresses_az(k2, u_el_2, L, E) stress1 = stress2 = :element stresses ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab Create a directory Copy LinearBarElementStiffness.m LinearBarAssemble.m LinearBarElementForces.m LinearBarElementStresses_az.m files under the created directory Open Matlab; Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab Start solving the problem in Command Window: >>clearvars >>clc Enter the data >>E=210e6 >>A=0.003 >>L1=1.5 >>L2=1 Calculate stiffness matrices >>k1=LinearBarElementStiffness(E,A,L1) >>k2=LinearBarElementStiffness(E,A,L2) ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab Assemble the global stiffness matrix; >>K=zeros(3,3) >>K=LinearBarAssemble(K,k1,1,2) >>K=LinearBarAssemble(K,k2,2,3) K is calculated as: K = ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab FE Equation is: Apply BCs: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab Solving the above system of equations will be performed by partitioning (manually) and Gaussian elimination (with Matlab). First we partition the above equation by extracting the submatrix in row 2 and column 2 which turns to be a 1x1 matrix. Because of the applied displacement of m at node 3, we need to extract the submatrix in row 2 and column 3 which also turns out to be a 1x1 matrix. Therefore we obtain; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab The solution of the above system is obtained using Matlab as follows. Note that the backslash operator ‘\’ is used for Gaussian elimination. >>k=K(2,2) >>k0=K(2,3) >>u0=0.002 >>f=[-10] >>f0=f-k0*u0 >>u=k\f0 result is; u = 0.0012 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab In this step, we obtain the reactions at nodes 1 and 3, and the stress in each bar using Matlab as follows. First we set up the global displacement vector; U, then we calculate the golbal force vector F. >>U=[0; u; u0] >>F=K*U results; F = ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab Next step is to set up element nodal displacement vectors u1 and u2. Then we calculate the element force vectors f1 and f2 by making calls to the Matlab function LinearBarElementForces. >>u1=[0; U(2)] >>f1=LinearBarElementForces(k1,u1) result is; f1 = ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab >>u2=[U(2); U(3)] >>f2=LinearBarElementForces(k2,u2) result is; f2 = Finally, we call LinearBarElementStresses_az to calculate stresses. >>sigma1=LinearBarElementStresses_az(k1, u1, L1, E) results; sigma1 = 1.6667e+05 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 2 with Matlab >>sigma2=LinearBarElementStresses_az(k2, u2, L2, E) results; sigma2 = 1.7000e+05 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 3 Given; Find; The support reaction forces at the two ends of the bar Solution : Connectivity table: E# N1 N2 1 2 3 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 3 We first check to see if or not the contact of the bar with the wall on the right will occur. To do this, we imagine the wall on the right is removed and calculate the displacement at the right end, BCs; contact occurs The global FE equation is found to be, ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example 3 Applying boundary conditions; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Example: 3 Reaction forces: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 contact occurs BCs; ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 Create a directory Copy LinearBarElementStiffness.m LinearBarAssemble.m LinearBarElementForces.m LinearBarElementStresses_az.m files under the created directory Open Matlab; Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory. ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 Start solving the problem in Command Window: >>clearvars >>clc Enter the data >>P=6e4 >>E=2e4 >>A=250 >>L=150 >>DELTA=1.2 Calculate stiffness matrices >>k1=LinearBarElementStiffness(E,A,L) >>k2=LinearBarElementStiffness(E,A,L) ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 Assemble the global stiffness matrix; >>K=zeros(3,3) >>K=LinearBarAssemble(K,k1,1,2) >>K=LinearBarAssemble(K,k2,2,3) K is calculated as: K = 1.0e+04 * ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 FE Equation is: Apply BCs: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 >>k=K(2,2) >>k0=K(2,3) >>u0=1.2 >>f=[60000] >>f0=f-k0*u0 >>u=k\f0 result is; u = 1.5000 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 >>U=[0; u; u0] >>F=K*U results; F = -50000 60000 -10000 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 >>u1=[0; U(2)] >>f1=LinearBarElementForces(k1,u1) results; f1 = -50000 50000 >>u2=[U(2); U(3)] >>f2=LinearBarElementForces(k2,u2) f2 = 10000 -10000 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Solution of Example 3 with Matlab E# N1 N2 1 2 3 >>sigma1=LinearBarElementStresses_az(k1, u1, L, E) results; sigma1 = 200 >>sigma2=LinearBarElementStresses_az(k2, u2, L, E) sigma2 = -40 ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Distributed Load Uniformly distributed axial load q (N/mm, N/m, lb/in) can be converted to two equivalent nodal forces of magnitude qL/2. We verify this by considering the work done by the load q, ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Distributed Load Thus, from the U=W concept for the element, we have: The new nodal force vector is: ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

Mechanical Engineering Department, GTU
5-Bar Element Distributed Load ME Dr. Ahmet Zafer Şenalp Mechanical Engineering Department, GTU

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