# The elements used in commercial codes can be classified in two basic categories:

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1 CHAPTER 3 Truss Element 3.1 Introduction The single most important concept in understanding FEA, is the basic understanding of various finite elements that we employ in an analysis. Elements are used for representing a real engineering structure, and therefore, their selection must be a true representation of geometry and mechanical properties of the structure. Any deviation from either the geometry or the mechanical properties would yield erroneous results. The elements used in commercial codes can be classified in two basic categories: 1. Discrete elements: These elements have a well defined deflection equation that can be found in an engineering handbook, such as, Truss and Beam/Frame elements. The geometry of these elements is simple, and in general, mesh refinement does not affect the results. Discrete elements have a very limited application; bulk of the FEA application relies on the Continuous-structure elements. 2. Continuous-structure Elements: Continuous-structure elements do not have a well define deflection or interpolation function, it is developed and approximated by using the theory of elasticity. In general, a continuous-structure element can have any geometric shape, unlike a truss or beam element. The geometry is represented by 1-D, 2-D, or a 3-D solid element. Since elements in this category can have any shape, it is very effective in calculation of stresses at a sharp curve or geometry, i.e., evaluation of stress concentrations. Since discrete elements cannot be used for this purpose, continuous structural elements are extremely useful for finding stress concentration points in structures. Elements Figure 3.1 A Discrete element Structure Nodes As explained earlier, for analyzing an engineering structure, we divide the structure into small sections and represent them by appropriate elements. Nodes define geometry of the structure and elements are generated when the applicable nodes are connected. Results

2 are always obtained for node points and not for elements - which are then interpolated to provide values for the corresponding elements. For a static structure, all nodes must satisfy the equilibrium conditions and the continuity of displacement, translation and rotation. However, the equilibrium conditions may not be satisfied in the elements. In the following sections, we will get familiar with characteristics of the basic finite elements. 3.2 Structures & Elements Most 3-D structures can be analyzed using 2-D elements (idealization), which require relatively less computing time than the 3-D solid elements. Therefore, in FEA, 2-D elements are the most widely used elements. However, there are cases where we must use 3-D solid elements. In general, elements used in FEA can be classified as: - Trusses - Beams - Plates - Shells - Plane solids - Axisymmetric solids - 3-D solids Since Truss element is a very simple and discrete element, let us look at its properties and application first. 3.3 Truss Elements The characteristics of a truss element can be summarized as follows: Truss is a slender member (length is much larger than the cross-section). It is a two-force member i.e. it can only support an axial load and cannot support a bending load. Members are joined by pins (no translation at the constrained node, but free to rotate in any direction). The cross-sectional dimensions and elastic properties of each member are constant along its length. The element may interconnect in a 2-D or 3-D configuration in space. The element is mechanically equivalent to a spring, since it has no stiffness against applied loads except those acting along the axis of the member. FEA Lecture Notes by R. B. Agarwal 3-2

3 However, unlike a spring element, discussed in previous chapters, a truss element can be oriented in any direction in a plane, and the same element is capable of tension as well as compression. j i Figure 3.2 A Truss Element Stress Strain relation: As stated earlier, all deflections in FEA are evaluated at the nodes. The stress and strain in an element is calculated by interpolation of deflection values shared by nodes of the element. Since the deflection equation of the element is clearly defined, calculation of stress and strain is rather simple matter. When a load F is applied on a truss member, the strain at a point is found by the following relationship. du ε = dx x or, ε = δl/l L + δl L where, ε = strain at a point u = axial displacement of any point along the length L Figure 3.3 Truss member in Tension By hook s law, σ = Eε Where, E = young s modulus or modulus of elasticity. From the above relationship, and the relation, F = Aσ FEA Lecture Notes by R. B. Agarwal 3-3

4 the deflection, δl, can be found as δl = FL/AE (3.1) Where, F = Applied load A = Cross-section area L = Length of the element Treatment of Loads in FEA For a truss element, loads can be applied on a node only. If loads are distributed on a structure, they must be converted to the equivalent loads that can be applied at nodes. Loads can be applied in any direction at the node, however, the element can resist only the axial component, and the component perpendicular to the axis merely causes free rotation at the joint Finite Element Equation of a Truss Structure In this section, we will derive the finite element equation of a truss structure. The procedure presented here is the basis for all FEA analyses formulations, wherever h- element are used. Analogues to the previous chapter, we will use the direct or equilibrium method for generating the finite element equations. Assembly procedure for obtaining the global matrix will remain the same. In FEA, when we find deflections at nodes, the deflections are measured with respect to a global coordinate system, which is a fixed frame of reference. Displacements of individual nodes with respect to a fixed coordinate system are desirable in order to see the overall deformed structural shape. However, these deflection values are not convenient in the calculation of stress and strain in an element. Global coordinate system is good for predicting the overall deflections in the structure, but not for finding deflection, strain, and stress in an element. For this, it s much easier to use a local coordinate system. We will derive a general equation, which relates local and global coordinates. In Figure 3.4, the global coordinates x-y can give us the overall deflections measured with respect to the fixed coordinate system. These deflections are useful for finding the final shape or clearance with the surroundings of the structure. However, if we wish to find the strain in some element, say, member 2-7 in figure 3.4, it will be easier if we know the deflections of node 2 and 3, in the y direction. Thus, calculation of strain value is much easier when the local deflection values are known, and will be timeconsuming if we have to work with the x and y values of deflection at these nodes. FEA Lecture Notes by R. B. Agarwal 3-4

5 Therefore, we need to establish a trigonometric relationship between the local and global coordinate systems. In Figure 3.4, xy coordinates are global, where as, x y are local coordinates for element 4-7 y y 6 7 x 8 Nodes Elements x Fixed-Frame Origin Figure 3.4. Local and Global Coordinates Relationship Between Local and Global Deflections Let us consider the truss member, shown in Figure 3.5. The element is inclined at an angle θ, in a counter clockwise direction. The local deflections are δ 1 and δ 2. The global deflections are: u 1, u 2, u 3, and u 4. We wish to establish a relationship between these deflections in terms of the given trigonometric relations. u 2y R 1, δ 1 u 1y 2 1 θ u 1x u 2x δ 2, R 2 Figure 3.5 Local and Global Deflections By trigonometric relations, we have, δ 1 = u 1x cosθ + u 2 sinθ = c u 1x + s u 1y FEA Lecture Notes by R. B. Agarwal 3-5

6 where, cosθ = c, and sinθ = s δ 2 = u 2x cosθ + u 2y sinθ = c u 2x + s u 2y Writing the above equations in a matrix form, we get, δ 1 c s 0 0 u 1y = u 2x (3.2) δ c s u 2y u 1x Or, in short form, δ = T u Where T is called Transformation matrix. Along with equation (3.2), we also need an equation that relates the local and global forces Relationship Between Local and Global Forces By using trigonometric relations similar to the previous section, we can derive the desired relationship between local and global forces. However, it will be easier to use the workenergy concept for this purpose. The forces in local coordinates are: R 1 and R 2, and in global coordinates: f 1, f 2, f 3, and f 4, see Figure 3.6 for their directions. Since work done is independent of a coordinate system, it will be the same whether we use a local coordinate system or a global one. Thus, work done in the two systems is equal and given as, W = δ T R = u T f, or in an expanded form, R 1 f 1 W = δ 1 δ 2 = u 1x u 1y u 2x u 2y f 2 R 2 f 3 f 4 = {δ} T {R} = {u} T {f} Substituting δ = T u in the above equation, we get, FEA Lecture Notes by R. B. Agarwal 3-6

7 [[T] {u}] T {R} = {u} T {f}, or {u} T [T] T {R} = {u} T {f}, dividing by {u} T on both sides, we get, [T] T {R} = {f} (3.3) Equation (3.3) can be used to convert local forces into global forces and vice versa. F 4, u 2y f 2, u 1y 2 1 θ f 3 u 2x R 2, δ 2 R 1, δ 1 f 1, u 1x Figure 3.6 Local and Global Forces Finite Element Equation in Local Coordinate System Now we will derive the finite element equation in local coordinate system. This equation will be converted to global coordinate system, which can be used to generate a global structural equation for the given structure. Note that, we can not use the element equations in their local coordinate form, they must be converted to a common coordinate system, the global coordinate system. Consider the element shown below, with nodes 1 and 2, spring constant k, deflections δ 1, and δ 2, and forces R 1 and R 2. As established earlier, the finite element equation in local coordinates is given as, R 1 k -k δ 1 1 k 2 = δ 1, R 1 R 2 -k k δ 1 δ 2, R 2 Recall that, for a truss element, k = AE/L Figure 3.7 A Truss Element Let k e = stiffness matrix in local coordinates, then, FEA Lecture Notes by R. B. Agarwal 3-7

8 k e = AE/L -AE/L -AE/L AE/L Stiffness matrix in local coordinates Finite Element Equation in Global Coordinates Using the relationships between local and global deflections and forces, we can convert an element equation from a local coordinate system to a global system. Let k g = Stiffness matrix in global coordinates. In local system, the equation is: R = [k e ]{ δ} (A) We want a similar equation, but in global coordinates. We can replace the local force R with the global force f derived earlier and given by the relation: {f} = [T T ]{R} Replacing R by using equation (A), we get, {f} = [T T ] [[k e ]{ δ}], and δ can be replaced by u, using the relation δ = [T]{u}, therefore, {f} = [T T ] [k e ] [T]{u} {f} = [k g ] { u} Where, [k g ] = [T T ] [k e ] [T] Substituting the values of [T] T, [T], and [k e ], we get, c 0 [k g ] = s 0 AE/L -AE/L c s c -AE/L AE/L 0 0 c s 0 s FEA Lecture Notes by R. B. Agarwal 3-8

9 Simplifying the above equation, we get, c 2 cs -c 2 -cs cs s 2 -cs -s 2 [k g ] = -c 2 -cs c 2 cs (AE/L) -cs -s 2 cs s 2 This is the global stiffness matrix of a truss element. This matrix has several noteworthy characteristics: The matrix is symmetric Since there are 4 unknown deflections (DOF), the matrix size is a 4 x 4. The matrix represents the stiffness of a single element. The terms c and s represent the sine and cosine values of the orientation of element with the horizontal plane, rotated in a counter clockwise direction (positive direction). The following example will illustrate its application. FEA Lecture Notes by R. B. Agarwal 3-9

10 (1) AL Examples 3 For the truss structure shown: 150 AL (3) (2) ST Find displacements of joints 2 and kn Find stress, strain, & internal forces in each member. 2 A AL = 200 mm 2, A ST = 100 mm 2 All other dimensions are in mm. Solution Let the following node pairs form the elements: Element Node Pair (1) 1-3 (2) 2-1 (3) 2-3 Using Shigley s Machine Design book for yield strength values, we have, S y (AL) = kN/mm 2 S y (ST) = kN/mm 2 (375 Mpa) (586 Mpa) E (AL) = 69kN/mm 2, E (ST) = 207kN/mm 2 A (1) = A (2) = 200mm 2, A (3) =100mm 2 Find the stiffness matrix for each element Element (1) L (1 ) = 260 mm, u 1x u 3x E (1) = 69kN/mm 2, mm 3 A (1) = 200mm 2 θ = 0 c = cosθ = 1, c 2 = 1 s = sinθ = 0, s 2 = 0 cs = 0 u 1y (1) u 3y FEA Lecture Notes by R. B. Agarwal 3-10

11 EA/L = 69 kn/mm 2 x 200 mm 2 x 1/(260mm) = kn/mm c 2 cs - c 2 -cs [K g ] (1) = (AE/L) x cs s 2 -cs - s 2 -c 2 -cs c 2 cs -cs -s 2 cs s [K g ] (1) = (53.1) x Element 2 u 1y θ = 90 0 c = cos 90 0 = 0, c 2 = 0 1 u 1x s = sin 90 0 = cos 0 0 = 1, s 2 = 1 cs = 0 (2) EA/L = 69 x 200 x (1/150) = 92 kn/mm u 2x u 2y u 1x u 1y u 2x u 2y [k g ] (2) = (92) u 1x u 1y 2 u 2x u 2y FEA Lecture Notes by R. B. Agarwal 3-11

12 Element 3 θ = 30 0 c = cos 30 0 = 0.866, c 2 = 0.75 s = cos 60 0 =.5, s 2 = 0.25 u 3y 3 u 3x cs = u 2y (3) 300 mm EA/L = 207 x 100 x (1/300) = 69 kn/mm u 2x u 2y u 3x u 3y θ = u 2x u 2x u 2y (69) [k g ] (3) = u 3x u 3y Assembling the stiffness matrices Since there are 6 deflections (or DOF), u 1 through u 6, the matrix is 6 x 6. Now, we will place the individual matrix element from the element stiffness matrices into the global matrix according to their position of row and column members. Element [1] u 1x u 1y u 2x u 2y u 3x u 3y u 1x u 1y u 2x u 2y u 3x u 3y The blank spaces in the matrix have a zero value. FEA Lecture Notes by R. B. Agarwal 3-12

13 Element [2] u 1x u 1y u 2x u 2y u 3x u 3y u 1x u 1y u 2x u 2y u 3x u 3y Element [3] u 1x u 1y u 2x u 2y u 3x u 3y u 1x u 1y u 2x u 2y u 3x u 3y Assembling all the terms for elements [1], [2] and [3], we get the complete matrix equation of the structure. FEA Lecture Notes by R. B. Agarwal 3-13

14 u 1x u 1y u 2x u 2y u 3x u 3y u 1x F u 1y F u 2x = F u 2y F u 3x F u 3y F 1 Boundary conditions Node 1 is fixed in both x and y directions, where as, node 2 is fixed in x-direction only and free to move in the y-direction. Thus, u 1x = u 1y = u 2x = 0. Therefore, all the columns and rows containing these elements should be set to zero. The reduced matrix is: u u u3 y x y 0 = Writing the matrix equation into algebraic linear equations, we get, 29.9u 2y u 3x u 3y = u 2y u 3x u 3y = u 2y u 3x u 3y = -0.4 solving, we get u 2y = u 3x = u 3y = Sress, Strain and deflections Element (1) Note that u 1x, u 1y, u 2x, etc. are not coordinates, they are actual displacements. FEA Lecture Notes by R. B. Agarwal 3-14

15 L = u 3x = = L/L = /260 = 5.02 x 10-5 mm/mm = E = 69 x 5.02 x 10-5 = kn/mm 2 Reaction R = A = kn Element (2) L = u 2y = = L/L = /150 = 2.87 x 10-5 mm/mm = E = 69 x 2.87 x 10-5 = kn/mm 2 Reaction R = A = ( x 10-3 ) (200) = kn Element (3) Since element (3) is at an angle 30 0, the change in the length is found by adding the displacement components of nodes 2 and 3 along the element (at 30 0 ). Thus, L = u 3x cos u 3y sin 30 0 u 2y cos60 0 = cos sin cos60 0 = = L/L = /300 = x 10-5 = 3.87 x 10-5 mm/mm = E = 207 x x 10-5 = kn/mm 2 Reaction R = A = ( ) (100) = kn FEA Lecture Notes by R. B. Agarwal 3-15

16 Factor of Safety Factor of safety n is the ratio of yield strength to the actual stress found in the part. Element(1) S y n = = = 10.8 σ Element(2) S y n = = = 18.9 σ Element(3) S y n = = = σ The lowest factor of safety is found in element (3), and therefore, the steel bar is the most likely to fail before the aluminum bar does. Final Notes - The example presented gives an insight into how the element analysis works. The example problem is too simple to need a computer based solution; however, it gives the insight into the actual FEA procedure. In a commercial FEA package, solution of a typical problem generates a very large stiffness matrix, which will require a computer assisted solution. - In an FEA software, the node and element numbers will have variable subscripts so that they will be compatible with a computer-solution - Direct or equilibrium method is the earliest FEA method. FEA Lecture Notes by R. B. Agarwal 3-16

17 Example 2 Given: Elements 1 and 2: Aluminum Element 3: steel A (1) = 1.5in 2 A (2) = 1.0in 2 A (3) = 1.0in 2 Required: Find stresses and displacements using hand calculations 1 (1) y 4000 lb (3) (2) x. 30 Solution Calculate the stiffness constants: K K K AE lb = = = L 20 in 6 AE lb = = = L 40 in 6 AE lb = = = L 50 in Calculate the Element matrix equations. FEA Lecture Notes by R. B. Agarwal 3-17

18 Element (1) u 1y u 2y (1) u 1x 1 2 u 2x Denoting the Spring constant for element (1) by k 1, and the stiffness matrix by K (1), the stiffness matrix in global coordinates is given as, u 1x u 1y u 2x u 2y c 2 cs -c 2 -cs u 1x cs s 2 -cs -s 2 u 1y [K g ] (1) = K 1 c 2 -cs c 2 cs u 2x -cs -s 2 cs s 2 u 2y For element (1), θ = 0 0, therefore c =1, c 2 = 1 s = 0, s 2 =0, and cs = 0 u 1x u 1y u 2x u 2y u 1x u 1y [k (1) ] = k u 2x u 2y u 2y Element (2) 2 u 2x For this element, θ = 90 0, Therefore, (2) c= cosθ = 0, c 2 = 0 θ = 90 0 s = sinθ = 1, s 2 = 1 3 u 3x cs= 0 FEA Lecture Notes by R. B. Agarwal 3-18

19 The stiffness matrix is, u 3y u 3x u 3y u 2x u 2y c 2 cs -c 2 -cs u 3x cs s 2 -cs -s 2 u 3y [k g ] (2) = k 2 c 2 -cs c 2 cs u 2x -cs -s 2 cs s 2 u 2y u 3x u 3y u 2x u 2y u 3x u 3y [k g ] (2) = k u 2x u 2y Element 3 u 2y For element (3), θ = u 2x c= cos( ) = -0.6, c 2 =.36 (3) θ = s = sin( , s 2 =.64 4 u 4x cs = u 4x u 4y u 2x u 2y u 4y u 4x u 4y [k g ] (3) = k u 2x u 2y FEA Lecture Notes by R. B. Agarwal 3-19

20 Assembling the global Matrix Following the procedure for assembly described earlier, the assembled matrix is, [K g ] = 1 K1 0 K K1 0 K K3.48K K3.48K K3 K2 +.64K3 0 K2.48 K3.64K K2 0 K K3.48K K3.48K K3.64K K3.64K3 The boundary conditions are: u 1x = u 1y = u 3x = u 3y = u 4x = u 4y = 0 We will suppress the corresponding rows and columns. The reduced matrix is a 2 x2, given below, u 2x u 2y [K g ] = K K 3 u 2x -.48K 3 K K 3 u 2y The final equation is, K K K 3 u 2x = -.48K 3 K K 3 u 2x 8000 Substituting values for k 1, k 2, and k 3, we get FEA Lecture Notes by R. B. Agarwal 3-20

21 u 2x = u 2y 8000 u 2x = in. u 2y = in. 1 = P/A = K 1 u/a 1 = [(7.5 x 10 5 )( )] / (1.5) = 214 psi 2 = P/A = K 2 u/a 2 = [(2.5 x 10 5 )( )] / (1.0) = 3015 psi 3 = P/A = K 3 u/a 3 = [(6 x 10 5 )( cos sin )] / (1.0) = 6119 psi FEA Lecture Notes by R. B. Agarwal 3-21

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