1 Lecture 1 Fundamental Concepts. 1.1 Introduction

Size: px

Start display at page:

Download "1 Lecture 1 Fundamental Concepts. 1.1 Introduction"

Russell Jordan Howard
5 years ago
Views:

1 1.1 Introduction 1 ecture 1 Fundamental Concepts The finite element method is a numerical method for solving problems of engineering and physical science. Useful for problems with complicated geometries, loadings, and material properties where analytical solutions can not be obtained. It divides a body it into an equivalent system of smaller bodies (finite elements) interconnected at points common to two or more elements (nodes or nodal points) and/or boundary lines and/or surfaces. The basic idea is based on building the stiffness matri of the system in the form 1 [ K]{ u} {} f { u} [ K] { f }, which can be solved using linear algebra techniques

2 1. Solid Mechanics Variables: Displacement: u { u } T uy uz Surface traction: T Stress: σ { T } T Ty Tz Body weight: f { f } T fy fz Point load: P { P P P} T i y z { σ } T σ y σz τ yz τz τy i

3 1..1 Equilibrium 3 σ τ τ y z y z f τ y σ y τ yz f y 0 y z τ z τ σ y z zy z fz 1.. Boundary conditions Prescribed displacement: 0 if fied u uo if some movement 0 0

4 Prescribed traction/surface forces: σ n + τ n + τ n T y y z z τ n + σ n + τ n T y y y yz z y τ n + τ n + σ n T z zy y z z z Strain-Displacement Relations ε { ε } εy εz γ yz γ z γ y { u v w v w u w u v} y z z y z y y u 0 0 z v 0 z y w 0 z 31 y 0 63 T T

5 1..4 Stress-Strain Relations σ Dε where 1 v v v 0 v 1 v v E v v 1 v D 1 (1 + ν)(1 ν) v 1 v 1 0 v Method of Weighted Residual Galerkin s method Suppose we have a differential equation written in the * Strong form: u P * The abstract operator can take many forms, e.g. * d d EA. The strong form requires us to find d d an eact solution u for every point. In real life, we don t need the solution u for every point. We only need them at some discreet points. If we approimate u u by a trial solution u, we produce a residual * u PR( u ) 0.

6 6 The Galerkin s method argues that it is possible to find a suitable weighting function, which when multiplied with and integrate the equation over the domain, we can make this residual zero. * Weak form: Φi u P dv 0 V This equation is called weak form because we relaed the requirement that u only needs to satisfy the equation approimately at some discreet points. The trial solution u is then approimated by another trial function u u jφ j. We also note that if φ j Bubnov-Galerkin's method Φi : φ j Petrov-Galerkin's method Galerkin s Method for Elastostatic et us define our test function as j { } T i y y Φ Φ Φ Φ, and apply this to the equilibrium equation:

7 V ( σ τ τ ) ( τ σ τ ) y z f y z f y y z y y yz Φ Φ y ( τ τ σ ) y z f z dv 0 z zy z Φ z Integration by part formula: d da db ( ab) b + a d d d da db b d a d + ab d d Integrating the first term yields: σ [ σ ] Φ σ + Φ V V S [ ] 0 Φ dv dv n ds etc which we finally get: + 7 T T T T σεdv fdv TdS P Φ Φ Φ V V S i Elastic 1D Element Strong formulation displacement u b Body force (force per unit length) δ Consider section δ

8 8 bδ P σ A dσ δp σ + δ A d Resolving forces horizontally yields dσ F 0: + b A 0 d du But, σ Eε E, therefore: d d du Strong form: EA + ba 0 d d 1.4. Weak & FE formulation Multiply the governing equation with an arbitrary test function,, and integrate over the length: N i d du Ni EA + b d 0 d d 0 Integrating by parts gives Element stiffness body force traction point load dni du EA d Ni fad NiTd NiP 0 d d i i We now approimate the primary unknown u as

9 + [ u ] where N1 1 shape functions. N 1 u u N1u1 Nu N1 N ujn j u 9 Substituting the test and shape function into the equation gives us the finite element equation: Element stiffness body force traction point load dn dn i j EA d{} u N j i fad NiTd NiP d d i Epanding the FE equation in matri form gives dn1 dn1 dn1 dn d d d d u1 N1 P1 EA d ( ba + T ) d dn 0 dn1 dn dn u + N 0 P d d d d dn1 dn1 d d d d Nd 1 1 d i 0

10 Performing these operations gives finally 10 EA 1 1 u1 ba 1 T 1 P u 1 1 P or symbolically, we write e e [ ] {} { } k u f e. [ ] e k is called the local finite element stiffness. This equation needs to be assembled into the global stiffness e K k, e and so is the local nodal forces as follows: e F f. The final global FE equation is therefore 1 K u F u K F. [ ]{ } {} { } [ ] { } e After computation of the displacement, we can compute the strain using the formula: du u u1 1 1 [ 1 1] u ε Bu. d u

11 To compute the stress, we use the formula E σ Bu Eample: Giving P N. Evaluate the displacements and stresses in the elements. Soln: Element 1: k Element : k Assembly: K Global load vector: F { } Solving the matri yields: T

12 u Element 1 stress: T { } mm σ1 [ 1 1] 54.7 MPa Element stress: σ [ 1 1] MPa N.B. Don t forget to convert the unit into a consistent one

13 1.5 FE Program and Algorithm Overall program structure 13 program TRUSS implicit double precision (a-h,o-z) iin5 iout6 call welcome c...initialise all the arrays to avoid mem. corruption. c...read input data call rdata call stiff call gslod call assemb call greduc call baksub call stress call result write(*,9000) 9000 format(1,/'truss completed successfully'//) stop end program TRUSS

14 1.5. Data Type 14 program TRUSS implicit double precision (a-h,o-z) & parameter(mpoin40, melem0, mdime, mdofn, mnode, mmats10, mprop4, mstif100) parameter(mevabmdofn*mnode, msvabmdofn*mpoin) common/chal/iin,iout common/tape/itp1,itp common/fem1/ & npoin,nelem,nboun,nprop,nnode,nevab,nsvab,ndofn, & ndime,nstre,nplod,nband common/fem/ & props(mmats,mprop),coord(mpoin,mdime),lnods(melem,mnode), & eload(melem,mevab),pload(mpoin,mdofn),treac(mpoin,mdofn), & tdisp(mpoin,mdofn),bdlod(mpoin,mdofn),tload(mpoin,mdofn) common/fem3/ & ifpre(msvab),fied(msvab),matno(melem),disp(msvab), & gload(msvab),react(msvab),tract(melem),stres(melem) common/solver/ & gstif(mstif,mstif) logical bdylod, tractn, pnload common/logic/ bdylod, tractn, pnload The data declaration can be simplified by using an include file statement, e.g. include param.h

15 1.5.3 Element stiffness formation algorithm 15 subroutine stiff implicit double precision (a-h,o-z) c loop over each element do ie1,nelem node1lnods(ie,1) nodelnods(ie,) matidmatno(ie) youngprops(matid,1) dareaprops(matid,) 1coord(node1,1) coord(node,1) y1coord(node1,) ycoord(node,) elengdsqrt( (-1)*(-1) + (y-y1)*(y-y1) ) coeffyoung*darea/eleng if(ndime.eq.1)then estif(1,1) coeff EA 1 1 estif(1,)-coeff K 1 1 estif(,1)-coeff estif(,) coeff else two dimensional stiffness endif write(itp1) estif enddo rewind itp1 return end

16 1.5.4 oad vector assembly algorithm 16 subroutine gslod implicit double precision (a-h,o-z) etc for data declaration c...first, initialise the load vector etc do ie1,nelem etract the information matid, element length etc c c...form body load vector c if(bdylod)then coeffdensi*gravi*eleng*darea/.0d0 if(ndime.eq.1)then bdlod(node1,1)bdlod(node1,1)+coeff bdlod(node,1)bdlod(node,1)+coeff elseif(ndime.eq.)then two dimensional implementation endif f b A 1 e f 1 endif c c...form traction load vector c if(tractn)then tforctract(ie)*eleng/.0d0 if(ndime.eq.1)then tload(node1,1)tload(node1,1)+coeff tload(node,1)tload(node,1)+coeff elseif(ndime.eq.)then two-dimensional implementation endif endif enddo c...finally, we assembly all the load vector into global one f t T 1 1 & do ip1,npoin do id1,ndofn nodei(ip-1)*ndofn+id gload(nodei)gload(nodei)+bdlod(ip,id)+tload(ip,id) +pload(ip,id) point load is assembled directly here! enddo enddo

17 1.5.5 Element assembly algorithm 17 subroutine assemb implicit double precision (a-h,o-z) c initialise etc do isvab1,nsvab do jsvab1,nsvab gstif(isvab,jsvab)0.0d0 enddo enddo & do ie1,nelem read(itp1)estif do in1,nnode nodeilnods(ie,in) do id1,ndofn nrows(nodei-1)*ndofn+id nrowe(in-1)*ndofn+id do jn1,nnode nodejlnods(ie,jn) do jd1,ndofn ncols(nodej-1)*ndofn+jd ncole(jn-1)*ndofn+jd gstif(nrows,ncols)gstif(nrows,ncols)+ estif(nrowe,ncole) enddo enddo enddo enddo enddo return end

18 1.6 Etension to D et the solution of the 1D element (previous section) be denoted: T u' { u' 1 u' }, and let the corresponding displacement in D be denoted: T u { u1 v1 u v}. The transformation is given by u1 ' u m 0 0 v 1 1 ' u 0 0 m u, v u 18 The element now has 4-dofs per node. Notice that only the element stiffness matri in the program needed to be modified according to:

19 T k k '. 19 The stress can be calculated as: E σ [ 1 1] u '. The rest of the program can remain the same. Eample: Soln: Coordinates: NODE X Y

20 0 Connectivity: EEMENT Direction cosine: EEMENT e m Element 1: k E Element 3: k E Element : k E Element 4: k E

21 3 oad vector: F { } T 10 1 So, can assembly and solve the equations.

22 Tutorial 6 1. Consider the 1D bar shown below. Given A e 1. in, E psi. If it is known that u in. and u 0.05 in., calculate a. The displacement at point P b. The strain and stress in the element c. The element stiffness matri. Consider the bar shown below. By hand calculation, evaluate the nodal displacements, element stresses and support reactions. 3. Computer problem: Determine the nodal displacements, element stiffness and support reactions for the sensor shown below when the tip closes the air gap.

23 3 4. Computer problem: Find the deflection at the free end under its own weight. Plot the number of elements vs. deflection. How do you know that your deflection is correct? 5. Consider the truss element shown. The ( y, ) coordinates of the two nodes are T indicated in the figure. If u { } 10 in., determine a. The vector ' u b. The stress in the element c. The k matri

24 4 6. For the pin-jointed configuration shown below, determine the stiffness values of the global stiffness matri. Hence, solve the deflection at node Computer problem: A small railroad bridge is constructed of steel members, all of which have a cross-sectional area of 350 mm. A train stops on the bridge and the vertical loads applied to the truss on one side of the bridge are estimated to be P 1 80 kn, P 10 kn, P The height of the truss is 3.118m, and each section is of equal width of 3.6m. Estimate how much point 5 moves horizontally because of this loading. Also determine the nodal displacements and element stresses.

25 8. Computer problem: Find the deflection at the nodes for the truss configuration shown below. Take A 8 in. Convert all the units into a consistent SI unit. 5

Development of Truss Equations

CIVL 7/87 Chapter 3 - Truss Equations - Part /53 Chapter 3a Development of Truss Equations Learning Objectives To derive the stiffness matri for a bar element. To illustrate how to solve a bar assemblage