VIEWS: 70 PAGES: 44 CATEGORY: College POSTED ON: 8/31/2010 Public Domain
One-Dimensional Problems We wish to use FEM for solving the following problems: x 10 kN x δ = 2 x 10-2 mm Objectives 1. To develop a system of linear equations for one-dimensional problem. 2. To apply FE method for solving general problems involving bar structures with different support conditions. General Loading Condition Consider a non-uniform bar subjected to a general loading condition, as shown. Note: The bar is constrained by a fix support at the top and is free at the other end. The positive x-direction is taken downward. Types of Loading a) Body force, f Distributed force per unit volume (N/m3) Example: self-weight due to gravity b) Traction force, T Force per unit area (N/m2) For a 1-D problem, force T = × (perimeter of area ) area Examples: Frictional forces, Viscous drag, and Surface shear. c) Point load, Pi Concentrated load (in Newton) acting at any point i. Finite element Modeling Element Discretization The first step is to subdivide the bar into several sections – a process called discretization. Note: The bar is discretized into 4 sections, each has a uniform cross-sectional area. The non-uniform bar is transformed into a stepped bar. We will use the stepped bar as a basis for developing a finite element model of the original non-uniform bar. Numbering Scheme To analyze the stepped bar systematically, a global numbering scheme is assigned as shown. The x-direction is considered as the global coordinate direction. Note: F1, …, F5 represent global forces acting on the points connecting all sections of the stepped bar. Q1, …, Q5 represent global displacements of the points resulting from the forces acting on these points. {Q} = [Q1 Q2 Q3 Q4 Q5 ] T The stepped bar is transformed into {F } = [F1 F2 F3 F4 F5 ] T a finite element model using 1-D (line) elements. Element Connectivity Consider a single line element. It lies in a local coordinate system, denoted by^ x. Note: Node number in local Element connectivity table coordinate is denoted by a number with a hat on top. ˆ 1 ˆ 2 ^ ^ ˆ x q1 and q2 are nodal displacements in the local coordinate direction. Connectivity between global and local nodes must be established for each element, as tabulated in the table shown. Natural Coordinate and Shape Functions Natural Coordinate Consider a single element. Local node 1 is at distance x1 from a datum, and node 2 is at x2, measured from the same datum point. We define a natural or intrinsic coordinate system, ξ, ξ= 2 (x − x1 ) − 1 (x2 − x1 ) Note: The ξ-coordinate will be used to define shape functions, required to establish interpolation function for the displacement field within the element. Shape Functions The displacement field, u(x), within the element is not known. For simplicity, it is assumed that the displacement varies linearly from node 1 to node 2 within the element. We establish a linear interpolation function to represent the linear displacement field within the element. To implement this, linear shape functions are defined, given by, 1− ? 1+ ? N1 (? ) = and N 2 (? ) = 2 2 The linear displacement field, u(x), within the element can now be expressed in terms of the linear shape functions and the local nodal displacement q1 and q2 as: ^ u ( x ) = N1q1 + N 2 q2 ^ 1− ξ 1+ξ u( x) = q1 + q2 2 2 In matrix form: u( x ) = [N ]{q} ^ where [N ] = [N1 N2 ] q1 and {q} = = [q1 q2 ] T q2 Isoparametric Formulation Coordinate x of any point on the element (measured from the same datum point as x1 and x2) can be expressed in terms of the same shape functions, N1 and N2 as x = N1 x1 + N 2 x2 1−ξ 1+ξ x= x1 + x2 2 2 When the same shape functions N1 and N2 are used to establish interpolation function for coordinate of a point within an element and the displacement of that point, the formulation is specifically referred to as an isoparametric formulation. Example 1 (a) Evaluate ξ, N1, and N2 at point P. (b) If q1 = 0.003 in and q2 = -0.005 in, determine the value of displacement u at point P. Solution (a) The ξ coordinate of point P is given by 2 ξP = ( x − x1 ) − 1 x2 − x1 2 = ( 24 − 20 ) − 1 36 − 20 ξ P = −0.5 The shape functions are: 1 − ξ 1 + 0.5 N1 = = = 0.75 2 2 1 + ξ 1 − 0.5 N2 = = = 0.25 2 2 (b) Displacement of point P uP = N1q1 + N 2 q2 = 0.75 ( 0.003) + 0.25 ( −0.005 ) uP = 0.001 in Strain-Displacement Relation Normal strain is related to displacement by du ε= dx Using the chain rule of differentiation du d ξ ε= d ξ dx The two terms of the above relation are obtained as follows 2 dξ 2 ξ= ( x − x1 ) − 1 ⇒ = ( x2 − x1 ) dx ( x2 − x1 ) 1− ξ 1+ ξ du ( − q1 + q2 ) u = q1 + q2 ⇒ = 2 2 dξ 2 Thus the normal strain relation can be written as 1 ε= [ −q1 + q2 ] ( x2 − x1 ) which can be written in matrix form as q1 ε = [B ] q2 where [B] is a row matrix called the strain-displacement matrix, given by 1 1 [ B] = [ −1 1] = [ −1 1] ( x2 − x1 ) le since x2 – x1 = element length = le. Stress-Strain Relation Normal stress is related to the normal strain by a Hooke’s law, σ = Eε where E is modulus of elasticity. Substitute for the normal strain ε, we get, q1 σ = E [ B] q2 Robert Hooke (1635-1703); (Experimental Philosopher) Element Stiffness Matrix We will use the potential energy approach to derive the element stiffness matrix [ k] for the 1-D element. Total potential energy of a body subjected to loads is given by, πp =U +Ω U = internal strain energy; Ω = potential energy of external forces. For the non-uniform bar, its total potential energy is given by 1 σ T ε A dx − ∫ u T fA dx − ∫ uT T dx − ∑ Qi Pi 2 ∫L πp = L L i Since the bar has been discretized into finite elements 1 πp = ∑ σ T ε A dx − ∑ ∫ uT fA dx − ∑ ∫ uT T dx − ∑ Qi Pi e 2 ∫e e e e e i We will derive the element stiffness matrix of the 1-D element using the internal strain energy term, U as follows, 1 2 ∫e Ue = σ T ε A dx Recall, the stress and strain are given by σ = E [ B ] {q} and ε = [ B ]{q} Substitute these into the expression for Ue, U e = ∫ E ([ B ] {q}) [ B ]{q} A dx 1 T 2 e 1 = ∫ {q} [ B ] E [ B ]{q} A dx T T 2 e 1 2 ( U e = {q} ∫ [ B ] E [ B ]A dx {q} T e T ) dξ 2 le Recall again, = ⇒ dx = dξ dx le 2 Substitute and simplifying the expression yields, 1 T le +1 U e = {q} [ B ] Ee [ B ] Ae ∫ dξ {q} T 2 2 −1 1 le = {q} [ B ] Ee [ B ] Ae ( 2 ){q} T T 2 2 1 1 −1 1 = {q} Aele Ee [ −1 1]{q} T 2 le 1 le 1 1 −1 = {q} Aele Ee 2 [−1 1]{q} T 2 le 1 1 T A E 1 −1 U e = {q} e e {q} 2 le −1 1 The internal strain energy for the 1-D element can now be written in the form, 1 U e = {q} [ k ] {q} T e 2 where [k]e represents the element stiffness matrix for the 1-D element, i.e. Ee Ae 1 −1 [k ] = e le −1 1 Note: Ee = elastic modulus; Ae = cross-sectional area; le = element length. Element Force Vector The forces acting on 1-D structures can be of body force, fb, traction force, T, and concentrated force, P. They may act individually in various combination. The total potential energy of the structure, 1 π p = ∑ ∫ σ T ε A dx − ∑ ∫ uT f b A dx − ∑ ∫ uTT dx − ∑ Qi Pi e 2 e e e e e i a) Due to body force, fb The potential energy due to body force fb in a single element is given by the second term, i.e. Ω f = ∫ uT f b A dx e = ∫ ( N1q1 + N 2 q2 ) f b A dx T e Ω f = Ae fb ∫ ( N1q1 + N 2 q2 ) dx T e Ae fb N1 dx Ω f = {q} T ∫e Rewrite, Ae fb ∫e N 2 dx Recall that, le dx = dξ 2 le +1 1 − ξ le Also, ∫e N1 dx = 2 ∫−1 2 dξ = 2 le +1 1 + ξ le ∫e N 2 dx = 2 ∫−1 2 dξ = 2 Substitute and simplifying, yields Ae f bl e 2 T Al f 1 Ω f = {q} = {q} ⋅ e e b T Ae f bl e 2 1 2 The potential energy due to the body force can now be expressed in the form, Ω f = {q} {f} T e where the force vector due to body force fb is, Ae le fb 1 {f} = e 2 1 Quiz: Can you give the physical interpretation of {f}e? b) Due to traction force, T The potential energy due to traction force T is given by, ΩT = ∫ uT T dx = ∫ ( N1q1 + N 2 q2 ) T dx T e e Recall, le +1 1 − ξ le le dx = dξ ∫e N1 dx = 2 ∫−1 2 dξ = 2 2 le +1 1 + ξ le ∫e N 2 dx = 2 ∫−1 2 dξ = 2 Rearranging and simplifying, le T N1 dx ∫e 2 ΩT = {q} = {q} T T T ∫e N 2 dx le 2 The last equation is in the form, ΩT = {q} {T } T e le 1 ΩT = {q} T i.e. ⋅T ⋅ 2 1 Thus, element traction force vector due to traction T, Tle 1 {T }e = 2 1 Quiz: Can you give the physical interpretation of this? Summary We have established, for 1-D problems, 1. Stress-strain relation 3. Element force vector due to body force, fb E q1 σ = [ −1 1] q2 Al f 1 le {f} e = ee b 2 1 2. Element stiffness matrix 4. Element force vector AE 1 −1 [k ] due to traction force, T e = e e −1 1 le Tl 1 {T } = e e 2 1 Example 2 A thin steel plate has a uniform thickness t = 1 in., as shown. Its elastic modulus, E = 30 x 106 psi, and weight density, ρ = 0.2836 lb/in3. The plate is subjected to a point load P = 100 lb at its midpoint and a traction force T = 36 lb/ft. Determine: a) Displacements at the mid-point and at the free end, b) Normal stresses in the plate, and c) Reaction force at the support. Solution 1. Transform the given plate into 2 sections, each having uniform cross-sectional area. Note: Area at midpoint is Amid = 4.5 in2. Average area of section 1 is A1 = (6 + 4.5)/2 = 5.25 in2. Average area of section 2 is A2 = (4.5 + 3)/2 = 3.73 in2. 2. Model each section using 1-D (line) element. 3. Write the element stiffness matrix for each element 5.25 × 30 ×106 1 −1 [k ] = (1) element 1: −1 1 12 3.75 × 30 ×106 1 −1 [k ] = (2) element 2: −1 1 12 4. Assemble global stiffness matrix, 5.25 −5.25 0 30 × 10 [ K ] = 12 −5.25 9.00 −3.75 6 0 −3.75 3.75 Note: The main diagonal must contain positive numbers only! 5. Write the element force vector for each element a) Due to body force, fb = 0.2836 lb/in3 5.25 × 12 × 0.2836 1 { fb } (1) element 1 = 2 1 3.75 × 12 × 0.2836 1 { fb } (2) element 2 = 2 1 Assemble global force vector due to body force, 5.25 8.9 12 × 0.2836 { Fb } = 9.00 = 15.3 2 3.75 6.4 b) Due to traction force, T = 36 lb/ft 36 ×12 1 1 12 {T } (1) element 1 = = 18 2 1 1 36 × 12 1 1 12 {T } (2) element 2 = = 18 2 1 1 Assemble global force vector due to traction force, 1 18 { FT } = 18 2 = 36 1 18 c) Due to concentrated load, P = 100 lb at node 2 0 { FP } = 100 0 6. Assemble all element force vectors to form the global force vector for the entire structure. 8.9 + 18 + 0 26.9 { F } = 15.3 + 36 + 100 = 151.3 lb 6.4 + 18 + 0 24.4 7. Write system of linear equations (SLEs) for entire model The SLEs can be written in condensed matrix form as [ K ]{Q} = {F } Expanding all terms and substituting values, we get 5.25 −5.25 0 Q1 26.9 30 ×10 −5.25 9.00 −3.75 Q2 = 151.3 6 12 0 −3.75 3.75 Q3 24.4 Note: 1. The global force term includes the unknown reaction force R1 at the support. But it is ignored for now. 2. The SLEs have no solutions since the determinant of [K] = 0; Physically, the structure moves around as a rigid body. 8. Impose boundary conditions (BCs) on the global SLEs There are 2 types of BCs: a) Homogeneous = specified zero displacement; b) Non-homogeneous = specified non-zero displacement. In this example, homogeneous BC exists at node 1. How to impose this BC on the global SLEs? DELETE ROW AND COLUMN #1 OF THE SLEs! 5.25 −5.25 0 Q1 26.9 30 × 10 6 −5.25 9.00 −3.75 Q2 = 151.3 12 0 −3.75 3.75 Q3 24.4 9. Solve the reduced SLEs for the unknown nodal displacements The reduced SLEs are, 30 ×106 9.00 −3.75 Q2 151.3 −3.75 3.75 Q = 24.4 12 3 Solve using Gaussian elimination method, yields Q2 1.339 ×10−5 = −5 in Q3 1.599 ×10 Quiz: Does the answers make sense? Explain… 10. Estimate stresses in each elements 1 q1 Recall, σ (e ) = E [ B ]{q} = E ⋅ [−1 1] le q2 element 1 1 0 σ (1) = 30 ×10 ⋅ [ −1 1] 6 −5 = 33.48 psi 12 1.339 ×10 element 2 1 1.339 × 10 −5 σ (2) = 30 × 10 ⋅ [ −1 1] 6 −5 = 6.5 psi 12 1.599 × 10 11. Compute the reaction force R1 at node 1 We now include the reaction force term in the global SLEs. From the 1st. equation we get, 5.25 −5.25 0 0 26.9 + R1 30 × 10 6 1.339 × 10−5 = 151.3 12 −5.25 9.00 −3.75 0 −3.75 3.75 1.599 × 10−5 24.4 We have, 0 30 × 10 6 R1 = [ 5.25 −5.25 0 ] 1.339 × 10−5 − 26.9334 12 1.599 × 10−5 R1 = −202.68 lb Example 3 A concentrated load P = 60 kN is applied at the midpoint of a uniform bar as shown. Initially, a gap of 1.2 mm exists between the right end of the bar and the support there. 1.2 mm If the elastic modulus E = 20 x 103 N/mm2, determine the: 250 mm2 a) displacements field, P b) stresses in the bar, and x c) reaction force at the support. 150 mm 150 mm Solution 1. Write the element stiffness matrices and assemble the global stiffness matrix. 1 −1 0 20 × 10 × 250 3 [ K] = −1 2 −1 150 0 −1 1 2. Write the element force vectors and assemble the global force vector. {F } = 0, T 60 ×10 , 0 3 3. Write the global system of linear equations. 500 −500 0 Q1 0 10 −500 1000 −500 Q2 = 103 60 3 15 0 −500 500 Q3 0 4. Impose the boundary conditions. We have; Q1 = 0; Q3 = 1.2 mm. Using Gaussian elimination method: a) Delete 1 st row and column. b) Delete 3 rd row and column and modify the force term. 500 −500 0 Q1 0 10 −500 1000 −500 Q2 = 103 60 3 15 0 −500 500 1.2 0 The reduced SLE becomes, Modification to force term 103 500 (1.2 ) [1000]{Q2 } = 10 60 + 3 15 15 7. Solve the reduced SLE, we get Q2 = 1.5 mm 8. Compute stresses in the bar, 1 0 σ 1 = 20 × 10 × 3 [ −1 1] 1.5 150 σ 1 = 200 MPa 1 1.5 σ 2 = 20 × 10 × 3 [ −1 1] 1.2 150 σ 2 = −40 MPa 9. Compute reaction forces at supports Using the 1 st and 3rd equations, we obtain, R1 = -50 x 103 N; R3 = -10 x 103 N. Exercise 1 A composite bar ABC is subjected to axial forces as shown. Given, the elastic moduli, E1 = 200 GPa and E2 = 70 GPa. Estimate: a) Displacement of end C; [Answer: δC = 6.62x10-2 mm] b) Stress in section 2, and c) Reaction force at support A. Verify your results with analytical solution. Exercise 2 Reconsider Exercise 2-1. Suppose a gap of δ = 2 x 10-2 mm exists between end C and a fixed support there. Estimate: a) Displacement of point B; b) Stress in section 1, and c) Reaction forces at both supports. 10 kN 2 x 10-2 mm Assignment 1 q Find a journal paper on the application of finite element method to model and simulate real world problems, from various journ- als on the internet. (e.g. : www.sciencedirect.com). q Download the paper (in PDF format), and print it. q Read the paper and make 5 pages summary on the content of the paper - typewritten. Summary must consist of an introduc- tion, the problem statement, the objectives, the methodology (boundary conditions, materials properties, etc), the results and conclusion of the overall paper. q Submit the summary and copy of the paper to me. Use cover page. q Due in: 4 weeks from date of announcement.