1. March 20, 2003 9:35 AM Little 109 CES 4141 Forrest Masters A Recap of Stiffness by Definition and the Direct Stiffness Method

2. Farther Down the Yellow Brick Road..

3. Our Emphasis This Week: Trusses.. n Composed of slender, lightweight members n All loading occurs on joints n No moments or rotations in the joints n Axial Force Members n Tension (+) n Compression (-)

4. Stiffness n K ij = the amount of force required at i to cause a unit displacement at j, with displacements at all other DOF = zero n A function of: –System geometry –Material properties (E, I) –Boundary conditions (Pinned, Roller or Free for a truss) n NOT a function of external loads K = AE/L

5. From Strength of Materials.. Combine two equations to get a stiffness element F = k *  Spring Axial Deformation Units of Force per Length

6. Go to the Board.. Let’s take a look at last week’s homework to shed some light on the Stiffness by Definition Procedure DOF

7. From Stiffness by Definition n We can create a stiffness matrix that accounts for the material and geometric properties of the structure n A square, symmetric matrix K ij = K ji n Diagonal terms always positive n The stiffness matrix is independent of the loads acting on the structure. Many loading cases can be tested without recalculating the stiffness matrix Stiffness by Definition only uses a small part of the information available to tackle the problem However..

8. Stiffness by Definition Only Considers.. n Stiffnesses from Imposed Displacements n Unknown Displacements n Known Loadings For each released DOF, we get one equation that adds to the stiffness, displacement and loading matrices But what about Reactions and Known Displacements? K * r = R Stiffness Matrix Unknown Displacements Known External Forces

9. A Better Method: Direct Stiffness Consider all DOFs Stiffness ByDirect Definition Stiffness..now we have more equations to work with PIN ROLLER 0 1 2 2

10. A Simple Comparison 6 5 4 3 2 1 Stiffness by Definition n 2 Degrees of Freedom Direct Stiffness n 6 Degrees of Freedom n DOFs 3,4,5,6 = 0 n Unknown Reactions (to be solved) included in Loading Matrix Remember.. More DOFs = More Equations

11. Node Naming Convention n Unknown or “Unfrozen” Degrees of Freedom are numbered first… r1, r2 n Unknown or “Unfrozen” Degrees of Freedom follow r3, r4, r5, r6 6 5 4 3 2 1 If Possible.. X-direction before Y-direction

12. Stiffness by Definition vs Direct Stiffness K 11 K 12 K 13 K 14 K 15 K 16 K 21 K 22 K 23 K 24 K 25 K 26 K 31 K 32 K 33 K 34 K 35 K 36 K 41 K 42 K 43 K 44 K 45 K 46 K 51 K 52 K 53 K 54 K 55 K 56 K 61 K 62 K 63 K 64 K 65 K 66 r1r1r1r1 r2r2r2r2 r3r3r3r3 r4r4r4r4 r5r5r5r5 r6r6r6r6 R1R1R1R1 R2R2R2R2 R3R3R3R3 R4R4R4R4 R5R5R5R5 R6R6R6R6 =65 4 3 2 1 Stiffness by Definition Solution in RED Direct Stiffness Solution in RED/YELLOW

13. The Fundamental Procedure n Calculate the Stiffness Matrix n Determine Local Stiffness Matrix, Ke n Transform it into Global Coordinates, KG n Assemble all matrices n Solve for the Unknown Displacements n Use unknown displacements to solve for the Unknown Reactions n Calculate the Internal Forces

14. To continue.. n You need your Direct Stiffness – Truss Application Handout to follow the remaining lecture. If you forgot it, look on your neighbor’s, please n I have your new homework (if you don’t have it already) Go to http://www.ce.ufl.edu/~kgurl for the handouthttp://www.ce.ufl.edu/~kgurl FOR MORE INFO..

15. Overview First, we will decompose the entire structure into a set of finite elements Next, we will build a stiffness matrix for each element (6 Here) Later, we will combine all of the local stiffness matrices into ONE global stiffness matrix Node 1 Node 2 1 42 35

16. Element Stiffness Matrix in Local Coordinates n Remember K ij = the amount of force required at i to cause a unit displacement at j, with displacements at all other DOF = zero n For a truss element (which has 2 DOF).. K11*v1 + K12*v2 = S1 K21*v1 + K22*v2 = S2 S1 S2 v2 v1 K11K12 K21K22v1v2 =S1S2 Gurley refers to the axial displacement as “v” and the internal force as “S” in the local coordinate system

17. Element Stiffness Matrix in Local Coordinates n Use Stiffness by Definition to finding Ks of Local System Node 1 Node 2 AE L K 22 K 12 AE L K 11 K 21 K 12 = - AE / L K 22 = AE / L K 11 = AE / L K 21 = - AE / L

18. Element Stiffness Matrix in Local Coordinates Cont.. Put the local stiffness elements in matrix form Simplified.. For a truss element

19. Displacement Transformation Matrix n Structures are composed of many members in many orientations n We must move the stiffness matrix from a local to a global coordinate system S1 S2 v2 v1 r1 r2 r4 r3 x y LOCAL GLOBAL

20. How do we do that? n Meaning if I give you a point (x,y) in Coordinate System Z, how do I find the coordinates (x’,y’) in Coordinate System Z’ xyx’ y’ Use a Displacement Transformation Matrix

21. To change the coordinates of a truss.. n Each node has one displacement in the local system concurrent to the element (v1 and v2) n In the global system, every node has two displacements in the x and y direction r1 r2 r4 r3 x y v1v2 v1 will be expressed by r1 and r2 v2 will be expressed by r3 and r4

22. Displacement Transformation Matrix Cont.. n The relationship between v and r is the vector sum: v1 = r1*cos  x + r2*cos  Y v2 = r3*cos  x + r4*cos  Y xx YY v1 r1 r2 We can simplify the cosine terms: Lx = cos  x Ly = cos  y v1 = r1*Lx + r2*Ly v2 = r3*Lx + r4*Ly Put in matrix form

23. Displacement Transformation Matrix Cont.. v1 = r1*Lx + r2*Ly v2 = r3*Lx + r4*Ly Transformation matrix, a gives us the relationship we sought So.. v = a*r

24. Force Transformation Matrix Similarly, we can perform a transformation on the internal forces S1 S2 R1 R2 R3 R4

25. Element Stiffness Matrix in Global Coordinates Let’s put it all together.. We know that the Internal force = stiffness * local displacement (S = k * v) Units: Force = (Force/Length) * Length local disp = transform matrix * global disp (v = a * r) Substitute local displacement Internal force = stiffness * transform matrix * global disp (S = k * a * r) Premultiply by the transpose of “a” a T * S= a T * k * a * r and substitute R = a T * S to get R = a T * k * a * r

26. Element Stiffness Matrix in Global Coordinates Cont.. is an important relationship between the loading, stiffness and displacements of the structure in terms of the global system R = a T * k * a * r Stiffnessterm n We have a stiffness term, Ke, for each element in the structure n We use them to build the global stiffness matrix, KG Ke = a T * k * a

27. Element Stiffness Matrix in Global Coordinates Cont.. Let’s expand all of terms to get a Ke that we can use. Ke = a T * k * a (14) From notes Great formula to plug into your calculator

28. Element Stiffness Matrix in Global Coordinates Cont.. n Let’s use a problem to illustrate the rest of the procedure n We will start by calculating KE’s for the two elements 6 5 4 3 2 1 3 ft 4 ft Node 1 Node 3 Node 2 Element 2 Element 1

29. Assembly of the Global Stiffness Matrix (KG) r1 r2 r3 r4 3 ft Element 1 L= 3 Lx =  x / L = (3-0) / 3 = 1 Ly =  y / L = (0-0) / 3 = 0 NearFar r1r2r3r4 r1r2r3r4 Pick a Near and a Far Plug Lx, Ly and L into equation 14 to get

30. Assembly of the Global Stiffness Matrix (KG) r1 r2 r5 r6 3 ft 4 ft 5 ft Element 2 L= 5 Lx =  x / L = (3-0) / 5 = 0.6 Ly =  y / L = (4-0) / 5 = 0.8 Near Farr1r2r5r6r1r2r5r6

31. The Entire Local Stiffness Matrix in Global Terms r1r2r5r6r1r2r5r6 r1r2r3r4r5r6 r1 r2 r3 r4 r5 r6 Notice that there aren’t any terms in the local matrix for r3 and r4 Shorthand Real Matrix

32. Assembly of the Global Stiffness Matrix (KG) Summing Ke1 and Ke2 r1r2r3r4r5r6 r1 r2 r3 r4 r5 r6 =KrR How does this relate to Stiffness by Definition?

33. Solution Procedure Now, we can examine the full system Reactions Known displacements @ reactions ( = 0 ) Unknown Deflections Loads acting on the nodes

34. Solution Procedure cont.. To find the unknowns, we must subtend the matrices K11 K22 K12 K21 = Two Important Equations Rk = AE ( K11*ru + K12*rk ) Ru = AE ( K21*ru + K22*rk ) (24) (25) Going to be ZERO. Why?

35. Solution Procedure cont.. 6 5 4 3 2 1 3 ft 4 ft 10 kips We will apply a load at DOF 2 Then use equation (24) Rk = AE ( K11*ru + K12*rk ) 0 = AE ( 0.405*r1 + 0.096*r2) -10 = AE ( 0.096*r1 + 0.128*r2) r1 = 22.52/AE r2 = -95.02/AE solved 0

36. Solution Procedure cont.. With the displacements, we can use equation (25) to find the reactions at the pinned ends Ru = AE ( K21*ru + K22*rk ) 0 R3 = -7.5 kipsR4 = 0 kips R5 = 7.5 kipsR6 = 10 kips

37. Internal Member Force Recovery n To find the internal force inside of an element, we must return to the local coordinate system n Remember the equation S = k * a * r ? But S1 always Equals –S2 so

38. Internal Member Force Recovery Cont.. n For Element 1 n For Element 2 r1 r2 r3 r4 = -7.5 kips r1 r2 r5 r6 = 12.5 kips

39. Conclusion We solved n Element Stiffnesses n Unknown Displacements n Reactions n Internal Forces I will cover another example in the laboratory

40. Matrices.. Start with a basic equation In order to solve x,y,z.. You must have three equations But you must put these equations in matrix form =

41. 41 A Sample Problem solved with Stiffness by Definition and Direct Stiffness

42. 42 For Stiffness by Definition, we are only concerned with the three DOF’s that are free to move: r1 r2 r3

43. 43 For Column 1, we setr1 = 1 and r2 = r3 = 0 A B C B’ Element Change in Length 16/10 Long 28/10Short 30 Unit Displacement

44. 44 For Column 2, we setr2 = 1 and r1 = r3 = 0 A B C B’ Element Change in Length 18/10 Short 26/10Short 30 Unit Displacement

45. 45 For Column 3, we setr3 = 1 and r1 = r2 = 0 A B C Element Change in Length 10 24/5Long 31Long C’ Unit Displacement

46. 46 r1 r2 r3 r1 r2 r3 The final stiffness matrix is as follows..

47. 47 For Direct Stiffness, we are concerned with all six DOF’s in the structural system: r1 r2 r3 r4 r5 r6

48. 48 In the Direct Stiffness Method, we will use this equation for each elements 1, 2 and 3: Near XNear YFar XFar Y Near X Near Y Far X Far Y DOF Location

49. 49 Element 1 r5 r6 r1 r2 r5 r6 r1 r2 L = 6 Lx = 0.6 Ly = -0.8

50. 50 Element 1 – Another View r1 r2 r3 r4 r5 r6 r1 r2 r3 r4 r5 r6

51. 51 Element 2 r1 r2 r3 r4 r1 r2 r3 r4 L = 8 Lx = 0.8 Ly = 0.6

52. 52 Element 3 r5 r6 r3 r4 r5 r6 r3 r4 L = 10 Lx = 1 Ly = 0

53. 53 Summing Elements 1 through 3 ++ Remember: We must take care to add the correct elements from the local stiffness matrix to the global stiffness matrix.

54. 54 Summing Elements 1 through 3 r1 r2 r3 r4 r5 r6 r1 r2 r3 r4 r5 r6

55. 55 Summing Elements 1 through 3 Look Familiar? We found the yellow portion in the Stiffness by Definition Method

56. X Stiffness by Definition vs Direct Stiffness K K completed r unknown Runknownrknown R known = = X Reactions Zero Unless Settlement Occurs