A Recap of Stiffness by Definition and the Direct Stiffness Method

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

2. Farther Down the Yellow Brick Road..

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

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

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

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 • We can create astiffness matrixthat accounts for the material and geometric properties of the structure • A square, symmetric matrix Kij = Kji • Diagonal terms always positive • The stiffness matrix is independent of the loads acting on the structure. Manyloading casescan 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.. K * r = R Stiffness Matrix Known External Forces Unknown Displacements • Stiffnesses from Imposed Displacements • Unknown Displacements • Known Loadings For each released DOF, we get one equation that adds to the stiffness, displacement and loading matrices But what aboutReactionsandKnown Displacements?

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

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

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

12. 6 5 2 1 4 3 Stiffness by Definition vs Direct Stiffness Stiffness by Definition Solution in RED Direct Stiffness Solution in RED/YELLOW =

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

14. To continue.. • You need your Direct Stiffness – Truss Application Handout to follow the remaining lecture. If you forgot it, look on your neighbor’s, please • I have your new homework (if you don’t have it already) Go to http://www.ce.ufl.edu/~kgurl for the handout 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 2 Node 1 2 4 1 3 5

16. Element Stiffness Matrix in Local Coordinates • Remember Kij = the amount of force required at i to cause a unit displacement at j, with displacements at all other DOF = zero • For a truss element (which has 2 DOF).. • K11*v1 + K12*v2 = S1 • K21*v1 + K22*v2 = S2 S2 v1 v2 = S1 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 K21 K11 AE L Node 2 AE L K22 Node 1 K12 • Use Stiffness by Definition to finding Ks of Local System K11 = AE / L K21 = - AE / L K12 = - AE / L K22 = 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 • Structures are composed of many members in many orientations • We must move the stiffness matrix from a local to a global coordinate system GLOBAL S2 r4 r3 v1 v2 r2 y S1 r1 x LOCAL

20. y x How do we do that? • 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’ y’ Use a Displacement Transformation Matrix x’

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

22. r2 v1 QY Qx r1 Displacement Transformation Matrix Cont.. • The relationship between v and r is the vector sum: v1 = r1*cos Qx + r2*cos QY v2 = r3*cos Qx + r4*cos QY Lx = cos Qx Ly = cos Qy We can simplify the cosine terms: 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 S2 R4 R3 R2 S1 R1

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” aT * S= aT * k * a * r and substitute R = aT * S to get R = aT * k * a * r

26. Element Stiffness Matrix in Global Coordinates Cont.. R = aT * k * a * r is an important relationship between the loading, stiffness and displacements of the structure in terms of the global system Stiffness term • We have a stiffness term, Ke, for each element in the structure • We use them to build the global stiffness matrix, KG Ke = aT * 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 = aT * k * a (14) From notes Great formula to plug into your calculator

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

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

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

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

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

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

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

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

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 kips R4 = 0 kips R5 = 7.5 kips R6 = 10 kips

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

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

39. Conclusion We solved • Element Stiffnesses • Unknown Displacements • Reactions • 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: r3 r2 r1

43. 43 For Column 1, we set r1 = 1 and r2 = r3 = 0 A C B B’ Element Change in Length 1 6/10 Long 2 8/10 Short 3 0 Unit Displacement

44. 44 For Column 2, we set r2 = 1 and r1 = r3 = 0 A C B’ Unit Displacement B Element Change in Length 1 8/10 Short 2 6/10 Short 3 0

45. 45 For Column 3, we set r3 = 1 and r1 = r2 = 0 C C’ A Unit Displacement B Element Change in Length 1 0 2 4/5 Long 3 1 Long

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

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

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

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

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