B
Course Content: A INTRODUCTION AND OVERVIEW Numerical method and Computer-Aided Engineering; Phsical problems; Mathematical models; Finite element method;. B Elements and nodes, natural coordinates, interpolation function, bar elements, constitutive equations, stiffness matri, boundar conditions, applied loads, theor of minimum potential energ; eamples. C PLANE ELASTICITY PROBLEM FORMULATIONS Constant-strain triangular (CST) elements; Plane stress, plane strain; Aismmetric elements; Stress calculations; Programming structure; Numerical eamples.
REVIEW OF MATRIX ALGEBRA
Matri Algebra In this course, we need to solve sstem of linear equations in the form a a a n a a a n......... a n a a n nn n n n... b b b n (-) where,,, n are the unknowns. Eqn. (-) can be written in a matri form as b A (-) where [A] is a (n n) square matri, {} and {b} are (n ) vectors.
(-) Note: Element located at i th row and j th column of matri [A] is denoted b a ij. For eample, element at the nd row and nd column is a. n n nn n n n n b b b a a a a a a a a a : b and, :,... : : : :...... A The square matri [A] and the {} and {b} vectors are is given b, 5
Matri Multiplication The product of matri [A] of size (m n) and matri [B] of size (n p) will results in matri [C], with size (m p). A B C ( m n) ( n p) ( m p) (-) Note: The (ij) th component of [C], i.e. c ij, is obtained b taking the DOT product, c ij ( ith row of [ A]) ( jth column of [ B]) (-5) Eample: 0 ( ) 5 0 ( ) 7-0 ( ) 5 7 6
Matri Transposition If matri [A] = [a ij ], then transpose of [A], denoted b [A] T, is given b [A] T = [a ji ]. Thus, the rows of [A] becomes the columns of [A] T. Eample: [A] 0 5 6 Then, [ A] T 5 0 6 Note: In general, if [A] is of dimension (m n), then [A] T has the dimension of (n m). 7
Transpose of a Product The transpose of a product of matrices is given b the product of the transposes of each matrices, in reverse order, i.e. T T T T ([ A ][ B][ C]) [ C] [ B] [ A] (-6) Determinant of a Matri Consider a square matri [], The determinant of this matri is give b, det (-7) 8
EXAMPLE Given that: } { 0 0 ] [ 0 0 ] [ 0 ] [ E D C A Find the product for the following cases: a) [A][C] b) [D]{E} c) [C] T [A] 9
Solution of Sstem of Linear Equations Sstem of linear algebraic equations can be solved for the unknown using the following methods: a) Cramer s Rule b) Inversion of Coefficient Matri c) Gaussian Elimination d) Gauss-Seidel Iteration Eample: Solve the following SLEs using Gauss elimination method. 8 6 (i) (ii) (iii) 0
Gauss Elimination Method Reducing a set of n equations in n unknowns to an equivalent triangular form (forward elimination). The solution is determined b back substitution process. Basic approach -An equation can be multiplied (or divided) b a nonzero scalar -An equation can be added to (or subtracted from) another equation -The position of an two equations in the set can be interchanged
Eample: Solve the following SLEs using Gaussian elimination. (iii) 6 (ii) 8 (i) Eliminate from eq.(ii) and eq.(iii). Multipl eq.(ii) b 0.5 we get, (iii) 6 (ii) *.5 (i)
Subtract eq.(ii)* from eq.(i), we obtain Add eq.(iii) with eq.(i), ields (iii) 6 (ii) ** 7.5 0 (i) * (iii) 5 0 (ii) ** 7.5 0 (i)
Eliminate from eq.(iii)*. Multipl eq.(ii)** b and eq.(iii)* b we get 0 0 0 (iii) ** Subtract eq.(iii)** from eq.(ii)***, we obtain 6 6.5 8 (i) (ii) *** 0 0 0.5 5.5 7 (i) (ii) ** (iii) *** From eq.(iii)*** we determine the value of, i.e. 5.5
Back substitute value of into eq.(ii)** and solve for, we get 7.5( ) Back substitute value of and into eq.(i) and solve for, we get 6 5
Eample Solve the following sstems of linear equations b using the Gaussian elimination method. 6 8 0 a) b) 6
We wish to use FEM for solving the following problems: (It is more efficient to solve these problems analticall) Calculate displacement of bar ABC, take E = 00GPa 0 kn d = 0 - mm 7
NATURAL COORDINATES IN -D To describe location of a point inside an element in terms of nodal coordinates, for efficient computational routine. A local coordinate sstem Coordinates ranges from zero to unit A function of the global Cartesian coordinate N N 8
NATURAL COORDINATES (-D) (, ) (, ) (, ) (, ) η (-,) (,) ζ (-,-) (,-) Global Cartesian coordinate (, ) Natural coordinate (ζ, η) Relies on the element geometr for its definition Coordinates ranges from zero to unit A function of the global Cartesian coordinate 9
INTERPOLATION FUNCTION Functions used to represent the behavior of a field variable within an element. Also called shape functions or approimating functions. Polnomials are widel used because the are relativel eas to manipulate. N i i N N i = interpolation functions Interpolation functions chosen must meet certain continuit requirements to ensure: - Continuit of field variables - Convergence to correct solution as element size decreases. Interpolation functions must have geometric isotrop. The epansion remains unchanged under linear transformation from one Cartesian coordinate sstem to another. 0
Continuit requirements Assume that the functions appearing under the integral in the element equations contain up to (r+) th order. To ensure convergence, N i must satisf Compatibilit requirement the functions must have C r continuit at element interface. Completeness requirement The functions must have C r+ continuit within an element. Definitions: C 0 - field variable is continuous at element interface C the first derivatives are continuous C the second derivatives are also continuous
Polnomials as interpolation functions independent variable, () No. of terms: T () n = n+ Eamples: n =, T () =, n () n i0 i i 0 independent variables, (, ) n () n i j,, i j n k No. of terms: k n n ( ) n n =, T () = 0
independent variables, (,, z) No. of terms: Eamples: N=, T () = N=, T () =0 n k j i z z n l k j i l n,,, 6 n n n n z z,, 0 9 8 7 6 5,, z z z z z
Geometric Isotrop for Interpolation Functions Geometric isotrop the polnomial epansion for the element must remains unchanged under a linear transformation from one Cartesian coordinate sstem to another. Global coordinate sstem (, ) Local coordinate sstem (r, s). ) P n with all terms have geometric isotrop ) P n that are incomplete, et contain appropriate terms to preserve smmetr have geometric isotrop Terms in smmetric pair can be omitted Eg. (, ) (, ).
Eample: In order to use the polnomial but with less terms, we can drop the pair (eg., ) 0 9 8 7 6 5, P Cubic 0 terms 9 8 6 5, P
Triangular elements for C 0 problems C 0 problems require onl continuit of at element interfaces. Number of nodes along element side (hence, number of nodal values of ) must be sufficient to uniquel determine variation of along that side. 6- node triangle quadratic variation of 6 5, - node triangle Linear variation of,
0- node triangle Cubic variation of 0 9 8 7 6 5, 0 6 5 8 7 9 Polnomials of order greater than are rarel used eg. 5-node (quartic) -node (quintic)
Deriving Interpolation Functions To epress the generalized coordinates (coefficients α i s) in terms of nodal dof. Concept: (, ) (, ) (, ) (, ) Field variable, Assign one value of Element has dof. Select a -term polnomial as interpolation function., p to each node
Evaluate the coefficients G G G Thus, Recall,
- is an interpolation function that applies to the whole element Epress field variable behavior in terms of generalized coordinates. N i - refer to individual node and individual dof. Collectivel represent field variable behavior at node i i. 0 at other nodes Issues: Sometimes G do not eist for all orientation of elements in global coordinate sstem. Cost of computing G is prohibitive. Tr to determine N i b inspection using natural coordinates.
INTERPOLATION FUNCTION A prescribed function to represent the variation of unknown field variables within the element in terms of known nodal variables. Linear shape functions are defined: N ξ ξ and N ξ ξ
INTERPOLATION FUNCTION A shape function is used to interpolate values of displacement, q() along the bar element between the two nodal displacements, q and q For linear interpolation function: N = N ξ ξ N ξ ξ N = ˆ ˆ u Nq Nq
LINEAR DISPLACEMENT FIELD where uˆ( ) q N N q q N q N N N q q q q When the same shape functions N and N are used to establish interpolation function for coordinate of a point within an element and the displacement of that point, the formulation is referred to as an isoparametric formulation. N N
EXAMPLE (a) Evaluate, N, and N at point P. (b) If q = 0.00 in and q = -0.005 in, determine the value of displacement u at point P.
Developing FE equations 5
Steps in solving a continuum problem b FEM Identif and understand the problem (This essential step is not FEM) Select the solution domain Select the solution region for analsis. Discretize the continuum Divide the solution region into finite number of elements, connected to each other at specified points / nodes. Select interpolation functions Choose the tpe of interpolation function to represent the variation of the field variables over the element. Derive element characteristic matrices and vectors Emplo direct, variational, weighted residual or energ balance approach. [k] (e) {} (e) = {f} (e) 6
Steps (Continued) Assemble the element characteristic matrices and vectors Combine the element matri equations and form the matri equations epressing the behavior of the entire solution region / sstem. Modif the sstem equations to account for the boundar conditions of the problem. Solve the sstem equations Solve the set of simultaneous equations to obtain the unknown nodal values of the field variables. Make additional computations, if desired Use the resulting nodal values to calculate other important parameters. 7
Define the problem Tasks To develop finite element equations for -D problem -Stiffness matri -Load vector Consider a non-uniform bar subjected to a general loading condition 8
Element Discretization Discretize the solution domain into finite number of elements -Label the global ais (ref.) -Label each element -Label each node Each element has a constant crosssection 9
Numbering Scheme Global nodal displacements Q T Q Q Q Q Q5 Global nodal forces F T F F F F F5 0
Element Connectivit To establish unique connection between local and global nodes for each element ˆ Global numbering ˆ ˆ ˆ Local numbering
Select Interpolation Function A shape function is used to interpolate values of displacement, q() along the bar element between the two nodal displacements, q and q For linear interpolation function: N = N ξ ξ N ξ ξ N = ˆ ˆ u Nq Nq
Strain-Displacement Relations du d du d d d q q du q q u q q d d d B q q, Where [B] is the strain-displacement matri: B l e
Constitutive Equation Linear elastic behavior: 800 q E B q 600 STRESS, (MPa) 00 00 Linear = E Non-linear = K( p ) n 0 50 0.0 0. 0. 0. 0. 0.5 0.6 STRESS, (MPa) 00 50 00 E STRAIN, 0 0.0000 0.000 0.000 0.0006 0.0008 0.000 STRAIN,
Tpes of load Bod force, f Distributed force per unit volume or weight densit (N/m ) Eample: self-weight due to gravit Traction force, T Force per unit area (N/m ) (For a -D problem, force/length) Eamples: Frictional forces, viscous drag, and surface shear Point load, P i Concentrated force (N) acting at an point i. 5
Minimum Potential Energ Principle Potential Energ = strain energ work b eternal force Strain energ of linear elastic bod u Uu Wu U dv Work done b eternal forces W T T q B EBAd q u f dv Potential energ functional b S ut ds T T T T q B EBAd q N q f Ad N q b S T ds 6
Theor of Minimum Potential Energ The displacement field (u) which satisfies the equilibrium, and the conditions at the boundar surface is the one that minimizes the potential energ d q e d e q r i q e i dq i 0 Finite element equation: T T B EBAd q N f Ad N K q f b S T T ds 7
Stiffness Matri ( e) T k B EB EA l e EA l e A l e l e d d Recall that: B l d le d d d l e e 8
Force Terms 9 e b e b l e e b b T l Af d d l Af d l N d l N Af Ad f N e e l T Tl Td N e Bod force Traction
Finite Element Equations 50 P P Tl l Af q q l EA e e b e At element level f q k
Assembled global FE equations 5 5 55 5 5 5 5 F F F F F Q Q Q Q Q K K K sm K K K K K K K K K K K K Subjected to boundar condition: Q = 0 Solve for unknown nodal displacements
EXAMPLE A thin steel plate has a uniform thickness t = in., as shown. Its elastic modulus, E = 0 0 6 psi, and weight densit, r = 0.86 lb/in. The plate is subjected to a point load P = 00 lb at its midpoint and a traction force T = 6 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. 5
Suggested solution. Transform the given plate into sections, each having uniform cross-sectional area. Note: Area at midpoint is A mid =.5 in. Average area of section is A = (6 +.5)/ = 5.5 in. Average area of section is A = (.5 + )/ =.75 in.. Model each section using -D (line) element. 5
. Write the element stiffness matri for each element element : element : k k 6 () 5.500 6 ().7500. Assemble global stiffness matri, K 5.5 5.5 0 6 00 5.5 9.00.75 0.75.75 5
5. Write the element force vector for each element a) Due to bod force, f b = 0.86 lb/in element () 5.50.86 f b element ().750.86 f b Assemble global force vector due to bod force, F b 5.5 8.9 0.86 9.00 5..75 6. 55
b) Due to traction force, T = 6 lb/ft 6 element () T 8 element T () 6 8 Assemble global force vector due to traction force, F T 8 8 6 8 56
c) Due to concentrated load, P = 00 lb at node F P 0 00 0 6. Assemble all element force vectors to form the global force vector for the entire structure. F 8.9 8 0 6.9 5. 6 00 5. 6. 8 0. lb 57
7. Write sstem of linear equations (SLEs) for entire model Note: The SLEs can be written in condensed matri form as KQ F Epanding all terms and substituting values, we get 5.5 5.5 0 Q 6.9 6 5.5 9.00.75 Q 5. 0 0 0.75.75 Q.. The global force term includes the unknown reaction force R at the support. But it is ignored for now.. The SLEs have no solutions since the determinant of [K] = 0; Phsicall, the structure moves around as a rigid bod. 58
8. Impose boundar conditions (BCs) on the global SLEs There are tpes of BCs: a) Homogeneous = specified zero displacement; b) Non-homogeneous = specified non-zero displacement. In this eample, homogeneous BC eists at node. How to impose this BC on the global SLEs? DELETE ROW AND COLUMN # OF THE SLEs! 5.5 5.5 0 Q 6.9 6 5.5 9.00.75 Q 5. 0 0 0.75.75 Q. 59
9. Solve the reduced SLEs for the unknown nodal displacements The reduced SLEs are, 00 6 9.00.75 Q 5..75.75 Q. Solve using Gaussian elimination method, ields 5 Q.9 0 Q.5990 5 in 60
0. Estimate stresses in each elements Recall, element ( e) E B q E le q q 0 00.8 psi 5.9 0 6 element 5.9 0 00 6.5 psi 5.599 0 6 6
. Compute the reaction force R at node We now include the reaction force term in the global SLEs. From the st. equation we get, 5.5 5.5 0 0 6.9 R 5.5 9.00.75.90 5. 5 0.75.75.5990. 6 5 0 0 We have, 6 00 5.5 5.5 0.9 0 5 6.9 R 5.5990 R 0.68 lb 0 6
EXAMPLE A concentrated load P = 60 kn is applied at the midpoint of a uniform bar as shown. Initiall, a gap of. mm eists between the right end of the bar and the support there. If the elastic modulus E = 0 0 N/mm, determine the: a) displacements field, 50 mm P. mm b) stresses in the bar, and c) reaction force at the support. 50 mm 50 mm 6
Solution. Write the element stiffness matrices and assemble the global stiffness matri. K 0 00 50 50 0. Write the element force vectors and assemble the global force vector. F 0, 60 0, 0 T. Write the global sstem of linear equations. 0 5 500 500 0 Q 0 500 000 500 Q 0 60 0 500 500 Q 0 6
. Impose the boundar conditions. We have; Q = 0; Q =. mm. Using Gaussian elimination method: a) Delete st row and column. b) Delete rd row and column and modif the force term. 0 5 500 500 0 Q 0 500 000 500 Q 0 60 0 500 500. 0 The reduced SLE becomes, 500. 5 5 0 000 Q 0 60 Modification to force term 65
7. Solve the reduced SLE, we get Q.5 mm 8. Compute stresses in the bar, 0 00 50.5 00 MPa.5 00 50. 0 MPa 9. Compute reaction forces at supports Using the st and rd equations, we obtain, R = -50 0 N; R = -0 0 N. 66
EXAMPLE A composite bar ABC is subjected to aial forces as shown. Given, the elastic moduli, E = 00 GPa and E = 70 GPa. Estimate: a) Displacement of end C; [Answer: d C = 6.60 - mm] b) Stress in section, and c) Reaction force at support A. Verif our results with analtical solution. 67
EXAMPLE Reconsider Eercise -. Suppose a gap of d = 0 - mm eists between end C and a fied support there. Estimate: a) Displacement of point B; b) Stress in section, and c) Reaction forces at both supports. 0 kn 0 - mm 68