1. 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses

2. 2 Today One-dimensional bar element

3. 3 Example See Example 3.8 from D.L. Logan (2007). E=1.2 x 10 6 psi for all elements. See the Matlab code, Lect5_Truss2.m for the solution. Answer: σ 1 =-945 psi; σ 2 =1440 psi; σ 3 =-2850 psi

4. 4 Homework 4 Given a space truss structure as shown here. A=1.53e-3 m 2 ; E= 75 GPa Determine the nodal displacements, support reactions, and the element internal forces (use Matlab). Use SAP2000 and compare your results. Instead of use truss model in SAP2000, use rigid frame model and compare the results. Source: http://www.andrew.cmu.edu/course/ 24-ansys/htm/s4_truss.htm

5. 5 Truss Structures in SAP2000 In SAP2000, there is no special truss element To do analysis of a truss structure, use Frame element If the truss is an ideal truss (see the truss structure assumptions), you may either:  Set the geometric Section properties j, i33, and i22 all to zero, or  Release both bending rotations, R2 and R3, at both ends and release the torsional rotation, R1, at either end.

6. 6 One-dimensional Bar Element– Principal-of-Virtual Work Approach

7. 7 From direct to virtual work approach So far we have derived the spring element and the bar element by directly applied the physical laws (direct approach), i.e. Hooke’s law and force equilibrium law. More complex finite elements, however, may not be able to be formulated using the direct approach. So we turn to alternative approach for deriving the element stiffness equation. There are several methods for obtaining the stiffness equation, but in this course we are using the principle of virtual work.

8. 8 Basic concepts from the theory of elasticity Consider a 1D bar as shown here. Assume that the bar has:  Length L  Cross-sectional area A  Modulus of elasticity E  Coefficient of thermal expansion α x, u b=b(x)b=b(x) L X=0 X=LX=L P

9. 9 A, E and α may vary along the bar, i.e. A=A(x), E=E(x), and α= α(x) The bar is subjected to:  Distributed axial loal b=b(x), in the unit of [force]/[length], Note that this load is in the category of body force because it acts at every point in the body  Concentrated force P at the right end, and  Temperature change T 0 C (may also vary along the bar). The loads and temperature change causes the bar deforms, which can be described completely by an axial displacement field, x, u b=b(x)b=b(x) L X=0 X=LX=L P u=u(x)u=u(x)

10. 10 In the theory of elasticity, there are three basic set of equations, i.e.  Strain-displacement equations  Stress-strain equations  Equations of equilibrium To obtain strain-displacement equations, consider the free body diagram of the bar segment between sections at distances x and x+Δx: u(x): displacement of the section at distance x u(x+Δx): displacement of the section at distance x+Δx xx+Δxx+Δx u(x)u(x)u(x+Δx) ΔxΔx

11. 11 The average strain in the bar segment is If Δx approaches zero, then the average strain becomes the strain at any point in the section located at distance x, i.e. u(x)u(x)u(x+Δx) ΔxΔx

12. 12 Thus, the strain-displacement equation for the bar is The strain, ε, at each point of the bar consists of two parts, i.e.  The strain due to thermal expansion,  The strain due to external loads, If the bar is not restrained, then the thermal strain will not induce any thermal stress. The strain that induces stress is …(1)

13. 13 The stress-strain equation is given by Hooke’s law, viz. Now, consider the force equilibrium of the bar segment:  : average axial force along the bar segment Force equilibrium: σ(x)σ(x)σ(x+Δx) ΔxΔx …(2)

14. 14 N(x): internal normal (axial) force in the bar at distance x If Δx approaches zero, then σ(x)σ(x)σ(x+Δx) ΔxΔx

15. 15 Thus, the equation of equilibrium for the bar is Or shortly, We can obtain the governing equation for the 1D bar using Eqs. (1), (2) and (3).  Multiplying Eq. (2) by A results in …(3) …(4)

16. 16  Substituting Eq. (1) into Eq. (4), we obtain  Now substituting this equation into Eq. (3)  If E, A, α and T constant along the bar, then the equation simply becomes …(4) …(5)

17. 17 In conclusion, the governing equation for the bar is Given the external loads and displacement boundary condition, utilizing this 2 nd order differential equation we can obtain the displacement field, the strain field and the stress field. x, u b=b(x)b=b(x) L X=0 X=LX=L P

18. 18 10-Jan-16 Recall the example of bar structure Find the exact solutions for the displacement, strain, and stress along the bar and compare to the finite element solutions.

19. 19 10-Jan-16 For this problem, p(x)=0 and T=0. Thus, the governing equation can be written as The area of a cross section at a distance x: Boundary conditions: Displacement/geometrical/essential boundary condition x ; u F H Force/mechanical/natural boundary condition

20. 20 10-Jan-16 The solution of the governing equation is The strain field is The stress field is Displacement field

21. 21 Recall the finite element solutions

22. 22 10-Jan-16

23. 23 10-Jan-16

24. 24 Strain energy Energy  The capacity or ability to do work. Work  A product of a force and a distance in the direction where the force moves. Elastic strain energy (or elastic deformation energy)  Energy stored in a solid body in the deformed state.  The capacity of internal forces (or stresses) to do work through deformation (strains) in the solid body.

25. 25 Strain energy for a bar A bar has only one stress component along its axis (uniaxial stress condition) Consider an infinitesimal element subjected to normal stress σ x only

26. 26 Due to N x, the element extends Δ x Assume that the material is elastic-linear

27. 27 The work done by N x Thus, the strain energy of the infinitesimal element is

28. 28 The strain energy per unit volume is then The strain energy per unit volume is called strain energy density, U 0  U 0 is equal to the area under the stress-strain curve and above the strain axis. The strain energy for the whole solid body:

29. 29 The strain energy for the bar: x, u b=b(x)b=b(x) L X=0 X=LX=L P

30. 30 Problem example Consider the following two bars Suppose that the bars absorb the same amount of energy corresponding to the axial load Neglecting stress concentration, compare the axial loads acting on bar 1, P 1, and bar 2, P 2 1 2

31. 31 Solution Bar 1 Bar 2

32. 32 Solution (cont’d)

33. 33 Homework 5 (due date: FEM Mid- sem Exam Time) 1. Show how to obtain that the solutions for the displacement and strain fields of the tapered bar. x ; u F H

34. 34 2. Compute the displacement and stress at the right end of the bar, i.e. u(L T ) and σ(L T ) by modeling the bar using finite elements. The area of each element is taken to be equal to the section area at the midpoint of the element. a. Use one element, with L 1 =L T b. Use two elements, with L 1 =L 2 =L T /2 c. Use three elements, with L 1 =L 2 =L 3 =L T /3 d. Use four elements, with the length of each element L T /4

35. 35 3. Examine the convergence of the finite element solutions in Problem #2 by: a. Computing the relative error (%) for each solution, both for the displacement and the stress. b. Draw convergence graphs for the displacement and the stress, with the abscissa is the number of element and the ordinate is the solutions, including the exact solution. Give your comment regarding the finite element solutions. 4. Prove that the exact strain energy in the bar is

36. 36 The principle of virtual work A virtual (imaginary) displacement is an imaginary and very small change in the configuration of a system The principle of virtual work (or more precisely the principle of virtual displacement) is stated as follows: If a body in equilibrium is subjected to arbitrary virtual displacements associated with a compatible deformation of the body, the virtual work of external forces on the body is equal to the virtual strain energy of the internal stresses Compatible (admissible) displacements are those that satisfy the boundary conditions and ensure that no discontinuities, such as voids or overlaps

37. 37 The principle of the virtual work δU: the virtual strain energy of the internal stresses δW: the virtual work of external forces on the body X, u Z, w w δwδw δw is the virtual displacement

38. 38 The virtual strain energy for a bar  δε : virtual strain, which is the strain associated with the virtual displacement, i.e. The external virtual work: x, u b=b(x)b=b(x) L X=0 X=LX=L P

39. 39 Notice that Why is it not equal to the definition of actual strain energy? Discussion

40. 40 Formulation of bar element Consider the two-node bar element as follows b(x) : body force along the bar element in the unit of force per unit length (e.g. kN/m) u=u(x) : displacement field within the element, i.e. displacement as a function of point x within the element

41. 41 Displacements at nodes 1 and 2: Forces at nodes 1 and 2: The first and fundamental step in the finite element formulation is to assume the displacement field within the element in terms of its nodal displacements

42. 42 Here it is assumed that the displacement field is a linear function Let us now express the assumed function in terms of nodal displacements d 1 and d 2

43. 43 As a result,

44. 44 Written in matrix form Thus we can write

45. 45 Where N is also called the matrix of interpolation functions, because it interpolates the displacement field u=u(x) from the nodal displacements Matrix of shape functions Vector of nodal displacements