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5-Bar Element Dr. Ahmet Zafer Şenalp Mechanical Engineering Department Gebze Technical University.

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Presentation on theme: "5-Bar Element Dr. Ahmet Zafer Şenalp Mechanical Engineering Department Gebze Technical University."— Presentation transcript:

1 5-Bar Element Dr. Ahmet Zafer Şenalp Mechanical Engineering Department Gebze Technical University ME 520 Fundamentals of Finite Element Analysis

2 Bar (truss) structures: Bar Element ME 520 Dr. Ahmet Zafer Şenalp 2Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Cross section examples for bar structures

3 Consider a uniform prismatic bar: L: length A: cross-sectional area E: elastic modulus u=u(x): displacement  =  (x): strain  =  (x): stress Strain-displacement relation: Stress-strain relation: Bar Element ME 520 Dr. Ahmet Zafer Şenalp 3Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element i j FE Model

4 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: Stiffness Matrix - Direct Method ME 520 Dr. Ahmet Zafer Şenalp 4Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element  : elongation, F: force in bar k: stiffness of the bar

5 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 Stiffness Matrix - A Formal Approach ME 520 Dr. Ahmet Zafer Şenalp 5Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

6 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: Stiffness Matrix - A Formal Approach ME 520 Dr. Ahmet Zafer Şenalp 6Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

7 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: Stiffness Matrix - A Formal Approach ME 520 Dr. Ahmet Zafer Şenalp 7Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

8 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. Degree of Freedom (dof) ME 520 Dr. Ahmet Zafer Şenalp 8Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

9 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 9Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Find; (a)the global stiffness matrix (b) displacements of nodes (c) the reaction forces (d) stresses at the elements Solution : Connectivity table: E#N1N FE Model 2 2 1

10 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 10Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Boundary conditions :  Displacement boundary conditions :  Force boundary conditions : a) Element Stiffness Matrices :

11 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 11Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Construction of global stiffness matrix : Equilibrium (FE) equation for the whole system is;

12 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 12Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element b) Applying boundary conditions;

13 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 13Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element For the data given;

14 Example 1 ME 520 Dr. Ahmet Zafer Şenalp 14Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element c) From the 1 st and 3 rd row of the finite element equation, the reaction forces can be calculated as; d) Stresses in the elements:

15 · 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. Notes ME 520 Dr. Ahmet Zafer Şenalp 15Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

16 ME 520 Dr. Ahmet Zafer Şenalp 16Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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: Solution procedure with matlab Solution procedure with matlab

17 ME 520 Dr. Ahmet Zafer Şenalp 17Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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. Solution procedure with matlab Solution procedure with matlab

18 Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 18Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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];

19 Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 19Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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;

20 Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 20Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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;

21 Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 21Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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;

22 ME 520 Dr. Ahmet Zafer Şenalp 22Mechanical Engineering Department, GTU 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: Solution of Example 1 with Matlab E#N1N Bar Element 5-Bar Element

23 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 23Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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.

24 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 24Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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)

25 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 25Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>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:

26 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 26Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>K=zeros(3,3) K = >>K=LinearBarAssemble(K,k1,1,2) K = >>K=LinearBarAssemble(K,k2,2,3) K =

27 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 27Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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

28 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 28Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>u2=50000/96000 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; 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.

29 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 29Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>U=[0; u2; 0] U = >>F=K*U F = 1.0e+04 *

30 Solution of Example 1 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 30Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>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

31 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 31Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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.

32 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 32Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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)

33 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 33Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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 =

34 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 34Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element FE Equation is: Apply BCs:

35 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 35Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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;

36 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 36Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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 =

37 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 37Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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 =

38 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 38Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Next step is to set up element nodal displacement vectors u 1 and u 2. Then we calculate the element force vectors f 1 and f 2 by making calls to the Matlab function LinearBarElementForces. >>u1=[0; U(2)] >>f1=LinearBarElementForces(k1,u1) result is; f1 =

39 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 39Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>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 = e+05

40 Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 40Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element >>sigma2=LinearBarElementStresses_az(k2, u2, L2, E) results; sigma2 = e+05

41 Example 3 ME 520 Dr. Ahmet Zafer Şenalp 41Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Given; Find; The support reaction forces at the two ends of the bar Solution : Connectivity table: E#N1N

42 Example 3 ME 520 Dr. Ahmet Zafer Şenalp 42Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element 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,

43 Example 3 ME 520 Dr. Ahmet Zafer Şenalp 43Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Applying boundary conditions;

44 Example: 3 ME 520 Dr. Ahmet Zafer Şenalp 44Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element Reaction forces:

45 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 45Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N contact occurs BCs;

46 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 46Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N 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.

47 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 47Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N 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)

48 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 48Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N 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 *

49 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 49Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N FE Equation is: Apply BCs:

50 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 50Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N >>k=K(2,2) >>k0=K(2,3) >>u0=1.2 >>f=[60000] >>f0=f-k0*u0 >>u=k\f0 result is; u =

51 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 51Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N >>U=[0; u; u0] >>F=K*U results; F =

52 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 52Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N >>u1=[0; U(2)] >>f1=LinearBarElementForces(k1,u1) results; f1 = >>u2=[U(2); U(3)] >>f2=LinearBarElementForces(k2,u2) results; f2 =

53 Solution of Example 3 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 53Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element E#N1N >>sigma1=LinearBarElementStresses_az(k1, u1, L, E) results; sigma1 = 200 >>sigma2=LinearBarElementStresses_az(k2, u2, L, E) results; sigma2 = -40

54 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, Distributed Load ME 520 Dr. Ahmet Zafer Şenalp 54Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

55 Thus, from the U=W concept for the element, we have: The new nodal force vector is: Distributed Load ME 520 Dr. Ahmet Zafer Şenalp 55Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element

56 Distributed Load ME 520 Dr. Ahmet Zafer Şenalp 56Mechanical Engineering Department, GTU 5-Bar Element 5-Bar Element


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