1. 6-Bar Elements in 2-D Space e-mail: Dr. Ahmet Zafer Şenalp e-mail: azsenalp@gmail.comazsenalp@gmail.com 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 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Cross section examples for bar structures

3. 2-D Case ME 520 Dr. Ahmet Zafer Şenalp 3Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

4. Transformation: In matrix form: 2-D Case ME 520 Dr. Ahmet Zafer Şenalp 4Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

5. Transformation matrix: For the two nodes of the bar element, we have: The nodal forces are transformed in the same way: 2-D Case ME 520 Dr. Ahmet Zafer Şenalp 5Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

6. In the local coordinate system, we have: Using transformations: Multiplying both sides by T T and noticing that T T T = I, we obtain: Thus, the element stiffness matrix k in the global coordinate system is: which is a 4´4 symmetric matrix. Stiffness Matrix in the 2-D Space ME 520 Dr. Ahmet Zafer Şenalp 6Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

7. Explicit form: Calculation of the directional cosines l and m: The structure stiffness matrix is assembled by using the element stiffness matrices in the usual way as in the 1-D case. Stiffness Matrix in the 2-D Space ME 520 Dr. Ahmet Zafer Şenalp 7Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

8. Element Stress: Stiffness Matrix in the 2-D Space ME 520 Dr. Ahmet Zafer Şenalp 8Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

9. Example 1 ME 520 Dr. Ahmet Zafer Şenalp 9Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Given; Element 1: E,A,L Element 2: E,A,L Find ; (a)displacement of node 2 (b)stress in each bar Solution: Connectivity table: E#N1N2Angle 11245 223135

10. Example 1 ME 520 Dr. Ahmet Zafer Şenalp 10Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Boundary conditions:  Displacement boundary conditions:  Force boundary conditions: a)Element Stiffness Matrices: In local coordinate systems, we have; These two matrices cannot be assembled together, because they are in different coordinate systems. We need to convert them to global coordinate system OXY.

11. Example 1 ME 520 Dr. Ahmet Zafer Şenalp 11Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Element 1: Element 2:

12. Example 1 ME 520 Dr. Ahmet Zafer Şenalp 12Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Assemble the structure FE equation: Applying the BC’s:

13. Example 1 ME 520 Dr. Ahmet Zafer Şenalp 13Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Stresses in the two bars: Check the results: Look for the equilibrium conditions, symmetry, antisymmetry, etc.

14. Solution procedure with matlab Solution procedure with matlab ME 520 Dr. Ahmet Zafer Şenalp 14Mechanical Engineering Department, GTU As plane truss element has 4 degrees of freedom (2 at each node) for a structure with n nodes, the global stiffness matrix K will be of size 2nx2n. The global stiffness matrix K is obtained by making calls to the Matlab function PlaneTrussAssemble 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 partioning and Gaussion elimination. Finally once the unkown displacements and and reactions are found, the force is obtained for each element as follows: 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

15. ME 520 Dr. Ahmet Zafer Şenalp 15Mechanical Engineering Department, GTU where f is the force in this element (a scalar) and u is the 4x1 element displacement vector. The element stress is obtained by dividing the element force by the cross- sectional area A. If there is an inclined support at one of the nodes of the truss then the global stiffness matrix needs to be modified using the following equation: where [T] is a 2nx2n transformation matrix that is obtained by making a call to the function PlaneTrussInclinedSupport. Solution procedure with matlab Solution procedure with matlab 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

16. ME 520 Dr. Ahmet Zafer Şenalp 16Mechanical Engineering Department, GTU The inclined support is assumed to be at node i with an angle of inclination alpha as shown below The new matrix K0 obtained is thus the global stiffness matrix for the structure. Solution procedure with matlab Solution procedure with matlab 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

17. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 17Mechanical Engineering Department, GTU PlaneTrussElementLength(x1,y1,x2,y2) This function returns the length of the plane truss element whose first node has coordinates (x1,y1) and second node has coordinates (x2,y2). Function contents: function y = PlaneTrussElementLength(x1,y1,x2,y2) %PlaneTrussElementLength This function returns the length of the % plane truss element whose first node has % coordinates (x1,y1) and second node has % coordinates (x2,y2). y = sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1)); 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

18. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 18Mechanical Engineering Department, GTU PlaneTrussElementStiffness(E,A,L,theta) This function returns the element stiffness matrix for a plane truss element with modulus of elasticity E, cross-sectional area A, length L, and angle theta (in degrees). The size of the element stiffness matrix is 4 x 4. Function contents: function y = PlaneTrussElementStiffness(E,A,L, theta) %PlaneTrussElementStiffness This function returns the element % stiffness matrix for a plane truss % element with modulus of elasticity E, % cross-sectional area A, length L, and % angle theta (in degrees). % The size of the element stiffness % matrix is 4 x 4. x = theta*pi/180; C = cos(x); S = sin(x); y = E*A/L*[C*C C*S -C*C -C*S ; C*S S*S -C*S -S*S ; -C*C -C*S C*C C*S ; -C*S -S*S C*S S*S]; 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

19. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 19Mechanical Engineering Department, GTU PlaneTrussAssemble(K,k,i,j) This function assembles the element stiffness matrix k of the plane truss element 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 = PlaneTrussAssemble(K,k,i,j) %PlaneTrussAssemble This function assembles the element stiffness % matrix k of the plane truss element 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(2*i-1,2*i-1) = K(2*i-1,2*i-1) + k(1,1); K(2*i-1,2*i) = K(2*i-1,2*i) + k(1,2); K(2*i-1,2*j-1) = K(2*i-1,2*j-1) + k(1,3); K(2*i-1,2*j) = K(2*i-1,2*j) + k(1,4); K(2*i,2*i-1) = K(2*i,2*i-1) + k(2,1); K(2*i,2*i) = K(2*i,2*i) + k(2,2); K(2*i,2*j-1) = K(2*i,2*j-1) + k(2,3); K(2*i,2*j) = K(2*i,2*j) + k(2,4); 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

20. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 20Mechanical Engineering Department, GTU K(2*j-1,2*i-1) = K(2*j-1,2*i-1) + k(3,1); K(2*j-1,2*i) = K(2*j-1,2*i) + k(3,2); K(2*j-1,2*j-1) = K(2*j-1,2*j-1) + k(3,3); K(2*j-1,2*j) = K(2*j-1,2*j) + k(3,4); K(2*j,2*i-1) = K(2*j,2*i-1) + k(4,1); K(2*j,2*i) = K(2*j,2*i) + k(4,2); K(2*j,2*j-1) = K(2*j,2*j-1) + k(4,3); K(2*j,2*j) = K(2*j,2*j) + k(4,4); y = K; 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

21. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 21Mechanical Engineering Department, GTU PlaneTrussElementForce(E,A,L,theta,u) This function returns the element force given the modulus of elasticity E, the cross-sectional area A, the length L, the angle theta (in degrees), and the element nodal displacement vector u. Function contents: function y = PlaneTrussElementForce(E,A,L,theta,u) %PlaneTrussElementForce This function returns the element force % given the modulus of elasticity E, the % cross-sectional area A, the length L, % the angle theta (in degrees), and the % element nodal displacement vector u. x = theta * pi/180; C = cos(x); S = sin(x); y = E*A/L*[-C -S C S]* u; 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

22. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 22Mechanical Engineering Department, GTU PlaneTrussElementStress(E,L,thetax,thetay,thetaz,u) This function returns the element stress given the modulus of elasticity E, the length L, the angles thetax, thetay, thetaz (in degrees), and the element nodal displacement vector u. It returns the element stress as a scalar. Function contents: function y = PlaneTrussElementStress(E,L,theta,u) %PlaneTrussElementStress This function returns the element stress % given the modulus of elasticity E, the % the length L, the angle theta (in % degrees), and the element nodal % displacement vector u. x = theta * pi/180; C = cos(x); S = sin(x); y = E/L*[-C -S C S]* u; 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

23. Matlab functions used ME 520 Dr. Ahmet Zafer Şenalp 23Mechanical Engineering Department, GTU PlaneTrussInclinedSupport(T,i,alpha) This function calculates the transformation matrix T of the inclined support at node i with angle of inclination alpha (in degrees). Function contents: function y = PlaneTrussInclinedSupport(T,i,alpha) %PlaneTrussInclinedSupport This function calculates the % tranformation matrix T of the inclined % support at node i with angle of % inclination alpha (in degrees). x = alpha*pi/180; T(2*i-1,2*i-1) = cos(x); T(2*i-1,2*i) = sin(x); T(2*i,2*i-1) = -sin(x) ; T(2*i,2*i) = cos(x); y = T; 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space

24. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 24Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Given; E=210 GPa A=1x10 -4 m 2 Find ; (a)global stiffness matrix of the structure (b) horizontal displacement at node 2 (c) horizontal and vertical displacements at node 3 (d) reactions at nodes 1 and 2 (a)stress in each element Solution: Use the 7 steps to solve the problem using space truss element.

25. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 25Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Step 1-Discretizing the domain: This problem is already discretized. The domain is subdivided into 3 elements and 3 nodes. The units used in Matlab calculations are kN and meter. The element connectivity is: E#N1N2 112 213 323

26. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 26Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Step 2-Copying relevant files and starting Matlab Create a directory Copy PlaneTrussAssemble.m PlaneTrussElementForce.m PlaneTrussElementLength.m PlaneTrussElementStiffness.m PlaneTrussElementStress.m PlaneTrussInclinedSupport.m files under the created directory Open Matlab; Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory. Start solving the problem in Command Window: >>clearvars >>clc E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

27. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 27Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Enter the data >>E=210e6 >>A=1e-4 >>L1=4 >>L2=PlaneTrussElementLength(0,0,2,3) >>L3= PlaneTrussElementLength(4,0,2,3) Step 3-Writing the element stiffness matrices: >>k1=PlaneTrussElementStiffness(E,A,L1, 0) >>theta2=atan(3/2)*180/pi >>theta3=180-theta2 >>k2=PlaneTrussElementStiffness(E,A,L2, theta2) >>k3=PlaneTrussElementStiffness(E,A,L3, theta3) E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

28. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 28Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space yields; k2 = 1.0e+03 * 1.7921 2.6882 -1.7921 -2.6882 2.6882 4.0322 -2.6882 -4.0322 -1.7921 -2.6882 1.7921 2.6882 -2.6882 -4.0322 2.6882 4.0322 k3 = 1.0e+03 * 1.7921 -2.6882 -1.7921 2.6882 -2.6882 4.0322 2.6882 -4.0322 -1.7921 2.6882 1.7921 -2.6882 2.6882 -4.0322 -2.6882 4.0322 E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

29. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 29Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Step 4-Assembling the global stiffness matrix: Since the structure has 3 nodes the size of the global stiffness matrix is 6x6. So to find global stiffness matrix; >>K=zeros(6,6) >>K=PlaneTrussAssemble(K,k1,1,2) >>K=PlaneTrussAssemble(K,k2,1,3) >>K=PlaneTrussAssemble(K,k3,2,3) yields; K = 1.0e+03 * 7.0421 2.6882 -5.2500 0 -1.7921 -2.6882 2.6882 4.0322 0 0 -2.6882 -4.0322 -5.2500 0 7.0421 -2.6882 -1.7921 2.6882 0 0 -2.6882 4.0322 2.6882 -4.0322 -1.7921 -2.6882 -1.7921 2.6882 3.5842 0.0000 -2.6882 -4.0322 2.6882 -4.0322 0.0000 8.0645 E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

30. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 30Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Step 5-Applying the boundary conditions: BCs are: E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2 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;

31. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 31Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2 >>k=[K(3,3) K(3,5) K(3,6); K(5,3) K(5,5) K(5,6); K(6,3) K(6,5) K(6,6)] >>f=[0; 5 ; -10] >>u=k\f u = 0.0011 0.0020 -0.0016

32. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 32Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Step 7-Post-processing: In this step we obtain the reactions and and the force in each element using Matlab as follows. First we set up the global nodal displacement vector U, then we calculate the nodal force vector F. >>U=[0 ; 0 ; u(1) ; 0; u(2:3)] yields; U = 0 0.0011 0 0.0020 -0.0016 E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

33. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 33Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space >>F=K*U yields; F = -5.0000 1.2500 -0.0000 8.7500 5.0000 -10.0000 E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

34. Solution of Example 2 with Matlab ME 520 Dr. Ahmet Zafer Şenalp 34Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Form element displacement vector >>u1=[U(1) ; U(2) ; U(3) ; U(4)] >>u2=[U(1) ; U(2) ; U(5) ; U(6)] >>u3=[U(3) ; U(4) ; U(5) ; U(6)] To find stresses recall the below function subroutine >>sigma1=PlaneTrussElementStress(E,L1,0,u1) >>sigma2=PlaneTrussElementStress(E,L2,theta2,u2) >>sigma3=PlaneTrussElementStress(E,L3,theta3,u3) yields; sigma1 = 5.8333e+04 sigma2 = -1.5023e+04 sigma3 = -1.0516e+05 E#N1N2 112 213 323 Given; E=210 GPa A=1x10 -4 m 2

35. ME 52035GYTE-Makine Mühendisliği Bölümü 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Given; Element 1 and 2: Element 3: Find; (a)displacements (b)reaction forces Solution: Connectivity table: E#N1N2Açı 11290 2230 31345 Example 3-Multipoint constraint

36. ME 520 Dr. Ahmet Zafer Şenalp 36Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Boundary conditions:  Displacement boundary conditions:  Force boundary conditions: a)Element Stiffness Matrices (In Global coordinate system): Element 1: E#N1N2 112 223 313

37. Example 3-Multipoint constraint ME 520 Dr. Ahmet Zafer Şenalp 37Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Boundary conditions:  Displacement boundary conditions:  Force boundary conditions: a)Element Stiffness Matrices (In Global coordinate system): Element 1: E#N1N2 112 223 313

38. Example 3-Multipoint constraint ME 520 Dr. Ahmet Zafer Şenalp 38Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Element 2: Element 3:

39. Example 3-Multipoint constraint ME 520 Dr. Ahmet Zafer Şenalp 39Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space The global FE equation is: BC’s:

40. Example 3-Multipoint constraint ME 520 Dr. Ahmet Zafer Şenalp 40Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Applying BC’s: Solving for unknowns:

41. Example 3-Multipoint constraint ME 520 Dr. Ahmet Zafer Şenalp 41Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space From the global FE equation, we can calculate the reaction forces: Check the results! A general multipoint constraint (MPC) can be described as, where A j ’s are constants and u j ’s are nodal displacement components. In the FE software, such as MSC/NASTRAN, users only need to specify this relation to the software. The software will take care of the solution.

42. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 42Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space Given: E=70 GPa A=0.004 m 2 Find: a)global stiffness matrix b)displacements at nodes 2, 3, and 4 c)reactions at nodes 1 and 4 d)stress in each element Solution: Connectivity table; E#N1N2 112 214 313 424 523 634

43. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 43Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 Create a directory Copy PlaneTrussAssemble.m PlaneTrussElementForce.m PlaneTrussElementLength.m PlaneTrussElementStiffness.m PlaneTrussElementStress.m PlaneTrussInclinedSupport.m files under the created directory Open Matlab; Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory. Start solving the problem in Command Window: >>clearvars >>clc

44. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 44Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 Enter the data >>E=70e6 >>A=0.004 >>L1=3.5 >>theta1=90 >>L2=4 >>theta2=0 >>L3=PlaneTrussElementLength(0,0,4,3.5) >>theta3=atan(3.5/4)*180/pi >>L4= L3 >>theta4=360-theta3 >>L5=4 >>theta5=0 >>L6=3.5 >>theta6=270

45. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 45Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 Calculate stiffness matrices; >>k1=PlaneTrussElementStiffness(E,A,L1, theta1) >>k2=PlaneTrussElementStiffness(E,A,L2, theta2) >>k3=PlaneTrussElementStiffness(E,A,L3, theta3) >>k4=PlaneTrussElementStiffness(E,A,L4, theta4) >>k5=PlaneTrussElementStiffness(E,A,L5, theta5) >>k6=PlaneTrussElementStiffness(E,A,L6, theta6) Assemble the global stiffness matrix; Since the structure has 4 nodes the size of the global stiffness matrix is 8x8. So to find global stiffness matrix; >>K=zeros(8,8) >>K=PlaneTrussAssemble(K,k1,1,2) >>K=PlaneTrussAssemble(K,k2,1,4) >>K=PlaneTrussAssemble(K,k3,1,3)

46. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 46Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 >>K=PlaneTrussAssemble(K,k4,2,4) >>K=PlaneTrussAssemble(K,k5,2,3) >>K=PlaneTrussAssemble(K,k6,3,4) yields; K = 1.0e+05 * 0.9984 0.2611 -0.0000 -0.0000 -0.2984 -0.2611 -0.7000 0 0.2611 1.0284 -0.0000 -0.8000 -0.2611 -0.2284 0 0 -0.0000 -0.0000 0.9984 -0.2611 -0.7000 0 -0.2984 0.2611 -0.0000 -0.8000 -0.2611 1.0284 0 0 0.2611 -0.2284 -0.2984 -0.2611 -0.7000 0 0.9984 0.2611 -0.0000 -0.0000 -0.2611 -0.2284 0 0 0.2611 1.0284 -0.0000 -0.8000 -0.7000 0 -0.2984 0.2611 -0.0000 -0.0000 0.9984 -0.2611 0 0 0.2611 -0.2284 -0.0000 -0.8000 -0.2611 1.0284

47. Next we need to modify the global stiffness matrix obtained above to take the effect of the inclined support at node 4. If there is an inclined support at one of the nodes of the truss (multipoint constraint) then the global stiffness matrix needs to be modified using the following equation: where [T] is a 2nx2n transformation matrix that is obtained by making a call to the function PlaneTrussInclinedSupport. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 47Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634

48. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 48Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 >>T=eye(8,8) results; T = 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 >>T=PlaneTrussInclinedSupport(T,4,45)

49. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 49Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 yields; T = 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 0.7071 0.7071 0 0 0 0 0 0 -0.7071 0.7071 >>K0=T*K*T’ yields;

50. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 50Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 K0 = 1.0e+05 * 0.9984 0.2611 -0.0000 -0.0000 -0.2984 -0.2611 -0.4950 0.4950 0.2611 1.0284 -0.0000 -0.8000 -0.2611 -0.2284 0 0 -0.0000 -0.0000 0.9984 -0.2611 -0.7000 0 -0.0264 0.3956 -0.0000 -0.8000 -0.2611 1.0284 0 0 0.0231 -0.3461 -0.2984 -0.2611 -0.7000 0 0.9984 0.2611 -0.0000 -0.0000 -0.2611 -0.2284 0 0 0.2611 1.0284 -0.5657 -0.5657 -0.4950 0 -0.0264 0.0231 -0.0000 -0.5657 0.7523 0.0150 0.4950 0 0.3956 -0.3461 -0.0000 -0.5657 0.0150 1.2745 Applying the BCs: FE equation is;

51. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 51Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 BCs are: Imposing BCs >>k=K0(3:7,3:7) yields; k = 1.0e+05 * 0.9984 -0.2611 -0.7000 0 -0.0264 -0.2611 1.0284 0 0 0.0231 -0.7000 0 0.9984 0.2611 -0.0000 0 0 0.2611 1.0284 -0.5657 -0.0264 0.0231 -0.0000 -0.5657 0.7523

52. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 52Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 >>f=[0 ;0 ;30 ;0 ;0 ] >>u=k\f yields, u = 1.0e-03 * 0.6053 0.1590 0.8129 -0.3366 -0.2367

53. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 53Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 Post-processing; >>U=[0 ; 0 ; u ; 0] >>F=K0*U yields; F = -3.7500 -26.2500 0.0000 30.0000 -0.0000 0 37.1231

54. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 54Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 Stresses; >>u1=[U(1) ; U(2) ; U(3) ; U(4) ] >>sigma1=PlaneTrussElementStress(E,L1,theta1,u1) results; sigma1 = 3.1791e+03 >>u2=[U(1) ; U(2) ; U(7) ; U(8) ] >>sigma2=PlaneTrussElementStress(E,L2,theta2,u2) results; sigma2 = -4.1425e+03 >>u3=[U(1) ; U(2) ; U(5) ; U(6) ] >>sigma3=PlaneTrussElementStress(E,L3,theta3,u3) results; sigma3 = 5.1380e+03

55. Solution of Example 4 (Multipoint constraint) with Matlab ME 520 Dr. Ahmet Zafer Şenalp 55Mechanical Engineering Department, GTU 6-Bar Elements in 2-D Space 6-Bar Elements in 2-D Space E#N1N2 112 214 313 424 523 634 >>u4=[U(3) ; U(4) ; U(7) ; U(8) ] >>sigma4=PlaneTrussElementStress(E,L4,theta4,u4) results; sigma4 = -6.9666e+03 >>u5=[U(3) ; U(4) ; U(5) ; U(6) ] >>sigma5=PlaneTrussElementStress(E,L5,theta5,u5) results; sigma5 = 3.6333e+03>>u6=[U(5) ; U(6) ; U(7) ; U(8) ] >>sigma6=PlaneTrussElementStress(E,L6,theta6,u6) results; sigma6 = -6.7311e+03