1 20-Oct-15 Last course Lecture plan and policies What is FEM? Brief history of the FEM Example of applications Discretization Example of FEM softwares Why study the theory of finite elements? Computational steps Introduction to Matlab Homework 1

2 20-Oct-15 Today Stiffness equations for the linear spring element and the bar element Application of the direct stiffness method for structural systems made up of springs and bars in 1D space Interpretation and properties of the stiffness matrix

Discretize the structure (problem domain)  Divide the structure or continuum into finite elements Once the structure has been discretized, the computational steps faithfully follow the steps in the direct stiffness method. The direct stiffness method (DSM):  The global stiffness matrix of the discrete structure are obtained by superimposing (assembling) the stiffness matrices of the element in a direct manner. Computational steps of the FEM- the direct stiffness method

Idealization Process Source: C. Felippa (2012)

DSM: Breakdown Steps Source: C. Felippa (2012)

DSM: Assembly & Solution Steps Source: C. Felippa (2012)

7 2015/10/20 Any question before we discuss the simplest element, i.e. spring element?

Linear spring Linear spring:  Obeys Hooke’s law: F : applied axial load δ : elongation or contraction of the spring (deformation) k s : spring constant (stiffness)  Resists forces only in the direction of the spring F F = k s δ

Linear spring as a finite element (1) Points 1 and 2 are the finite element nodes. x: element (local) coordinate system f 1, f 2 : element nodal forces, shown in the positive sense d 1, d 2 : element nodal displacement, shown in the positive sense Now let’s find the relationship between the nodal forces and the nodal displacement when the spring is in equilibrium state. f 1, d 1 f 2, d 2 x 12 ksks

Linear spring as a finite element (2) First, let d 1 >0 and d 2 =0 (compression)  Nodal forces consistent with static equilibrium and Hooke’s law are f1f1 f2f2 12 d1d1

11 20-Oct-15 Linear spring as a finite element (3) Next, let d 1 =0 and d 2 >0 (tension) f1f1 f2f2 12 d2d2

12 20-Oct-15 If both d 1 >0 and d 2 >0 In matrix format or

13 20-Oct-15 This is equation is called the (discrete) element equilibrium equation or the element stiffness equation, refer to the element (local) coordinate system Spring element stiffness matrix Linear spring as a finite element (4)

Example of a spring assemblage (1) 1, 2, 3: structural (global) node numbers (as opposed to element or local nodes) X: global coordinate system F 2 and F 3 are the given applied forces, while F 1 is the unknown reaction force. D 2 and D 3 are the unknown nodal displacement, while D 1 is the given nodal displacement (in this case D 1 =0) X F 1, D 1 k1k1 k2k2 F 2, D 2 F 3, D 3

Example of a spring assemblage (2) The number of degree of freedom (DOF)= 3.  The number of active DOF= 2. We will establish the stiffness equation for the spring assemblage by using the element stiffness equations 1 and 2. In this problem the local axes coincide with the global axis X F 1, D 1 k1k1 k2k2 F 2, D 2 F 3, D 3

Example of a spring assemblage (3) Equilibrium condition for element 1: Equilibrium condition for element 2: x 12 k1k1 x 12 k2k2

Example of a spring assemblage (4) Continuity or displacement compatibility condition: X F 1, D 1 k1k1 k2k2 F 2, D 2 F 3, D 3 x 12 k1k1 x 12 k2k2

The equilibrium equations expressed in terms of the global nodal displacements: Example of a spring assemblage (5)

Expanding the matrix equations to the size of the DOF: D1D1 D2D2 D3D D1D1 D2D2 D3D3

The addition of the two equations yields Example of a spring assemblage (7)

The equilibrium conditions for nodes 1, 2, and 3: x 12 k1k1 x 12 k2k2 Example of a spring assemblage (8) 1 2 3

Example of a spring assemblage (9) Substituting the last equations, we obtain: or Structural or global nodal force matrix Structural or global nodal displacement matrix Structural or global stiffness matrix

To facilitate the assembling of the global stiffness equation, let us define a matrix in which each column contains the numerical labels of the global nodal displacement associated with each element. Let call this matrix “code matrix”, denote M code The size of M code is Ndofe x Nelem, where  Ndofe: number of degrees of freedom per element  Nelem: total number of elements in the structure Assembling process (1)

Assembling process (2) X F 1, D 1 k1k1 k2k2 F 2, D 2 F 3, D 3 12 d1d1 d2d2

Assembling process (3) The rows and columns of each stiffness matrix are labeled according to the DOFs associated with them; the labels are copied from M code

Boundary Conditions (1) Given F 1, F 2 and F 3, can you now solve this equation to obtain D 1, D 2, and D 3 ? The support or boundary conditions must be specified, otherwise K will be singular. Types of boundary conditions:  Homogeneous b.c.’s: zero values of nodal displacement are specified.  Non-homogeneous b.c.’s: nonzero values are specified.

Example of homogeneous b.c.’s: D 1 =0 The global stiffness equation can be written as The unknown values: F 1, D 2, D 3 Boundary Conditions (2)

Boundary Conditions (3) Considering the 2 nd and the 3 rd equations, we can write the matrix equation as This equation is called reduced stiffness equation. For all homogeneous b.c.’s, the reduced stiffness equation can be obtained by simply deleting the rows and columns corresponding to the zero-displacement DOF’s from the original stiffness equation.

Boundary Conditions (4) Solving the reduced stiffness equation, we can obtain Substituting D 2 and D 3 into the original stiffness equation, we obtain

Example of nonhomogeneous b.c.’s: D 1 =δ The global stiffness equation can be written as The unknown values: F 1, D 2, D 3 Boundary Conditions (5)

Boundary Conditions (6) Considering the 2 nd and the 3 rd equations, or When dealing with non-homogeneous b.c.’s, we must transform the terms associated with the known displacements to the right-side force matrix.

Boundary Conditions (7) In general, specified boundary conditions are treated by partitioning the global stiffness equation as follows: Subscript a: associated with active (free, unconstrained) DOF’s Subscript p: associated with passive (locked, constrained) DOF’s Unknown vectors: D a, F p

Boundary Conditions (8) From this equation, we have

Example Given a system of three springs supporting three equal weights W. W=1 kN, k= 0.2 kN/mm Treating the springs as finite elements, determine the vertical displacement of each weight and the support reaction. Repeat the problem if the second weight is “forced” to move 5 mm upward. 3k 2k2k W W k W

Homework 2 (due date next class) The spring stiffness k 1 =1 kN/mm, k 2 =2 kN/mm, k 3 =3 kN/mm, k 4 =4 kN/mm, k 5 =5 kN/mm. Applying a force of 2 kN at node 2 in the positive x direction, determine the nodal displacements, the reactions, and the forces in each element. Use the direct stiffness method, implemented in Matlab. Instead of applying the force, now at node 2 given a fixed, known displacement – 2 mm. Repeat the calculations.

Any questions before we proceed to 1D bar element?

Linear-elastic Bar-- Assumptions The bar is geometrically straight. The material obeys Hooke’s law, i.e.  A: sectional area of the bar  E: modulus of elasticity of the material  σ: axial stress in the bar, σ x = F/A  ε x : axial strain of the bar FF A, E

Linear-elastic Bar-- Assumptions Forces are applied only at the ends of the bar. The bar supports axial loading only.  Bending, torsion, and shear are not transmitted to the element FF

Bar element (1) Now let’s find the relationship between the nodal forces and the nodal displacement when the spring is in equilibrium state. f 1, d 1 f 2, d 2 x12 EA L

40 20-Oct-15 Bar element (2) First, let d 1 >0 and d 2 =0 (compression)  Nodal forces consistent with static equilibrium and Hooke’s law are

41 20-Oct-15 Bar element (3) Next, let d 1 =0 and d 2 >0 (tension)

42 20-Oct-15 If both d 1 >0 and d 2 >0 In matrix format or

43 20-Oct-15 Example – A bar structure (1) The area of a cross section at a distance x: A=A(x)

44 20-Oct-15 Example – A bar structure (2) Discretization

45 20-Oct-15 Example – A bar structure (3) Generate element stiffness matrices  Apply the element stiffness matrix formula to the bar problem Element 1Element 2

46 20-Oct-15 Example – A bar structure (4) Element 1 D1D1 D2D2 D3D3 D1D1 D2D2 D3D3

47 20-Oct-15 Example – A bar structure (4) Element 2 D1D1 D2D2 D3D3 D1D1 D2D2 D3D3

48 20-Oct-15 Example – A bar structure (5) All elements are present

49 20-Oct-15 Example – A bar structure (6) Applying the loads  Nodal force vector Unknown reaction force D1D1 D2D2 D3D3

50 20-Oct-15 Example – A bar structure (7) Imposing the displacement boundary conditions  D 1 =0  In consequence, now there are only two unknown parameters D 2 and D 3  Thus the “active” DOF=2

51 20-Oct-15 Solving the global equation,  The unknowns are D 2, D 3, and H  Discarding row 1 and column 1 from K, we obtain

52 20-Oct-15 Solution Support reaction Example – A bar structure (9) Thus Agree with the global equilibrium FF

53 20-Oct-15 Example – A bar structure (10) “Disassembly” of the structural nodal displacement Element 2 Element 1

54 20-Oct-15 Example – A bar structure (11) Calculating stresses  Element 1  Element 2

Any questions before we discuss the interpretation and properties of the stiffness matrix?

Element stiffness matrix Recall the bar element The stiffness equation for the element The stiffness matrix Oct-15

From this eq., can you give a definition of stiffness matrix? A stiffness matrix is a matrix relating the nodal displacement vector to nodal force vector How did we derive this matrix? By directly applying the basic physical laws, i.e. the Hooke’s and equilibrium laws This direct approach is limited to simple elements Oct-15

Consider the stiffness equation If d 1 =1 and d 2 =0, then On the other hand, if d 1 =0 and d 2 = Oct-15 The 1 st column of k The 2 nd column of k

Thus, a column of k is the vector of loads that must be applied to an element at its nodes to maintain a deformation state in which the corresponding nodal d.o.f. has unit value while all other nodal d.o.f are zero This concept is general; it is applicable to all stiffness matrices, including the structural stiffness matrix Oct-15

Properties of stiffness matrices Nonnegative main diagonal values This is because it is physically unreasonable that a single load in a given direction would produce a displacement component in the opposite direction Symmetry k (or K) is symmetric if loads are linearly related to displacements Oct-15

Properties of stiffness matrices (cont’d) Singularity This is because the element or structure can under go “rigid body motion” Sparsity (for the global or structural stiffness matrix)  A matrix is called “sparse” if it contains many zeros  A practical FE structure may have thousands of d.o.f, and more than 99% of coefficients in its stiffness matrix may be zero Oct-15