Belytschko T. - Introduction (779635), страница 8
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It is an initial-boundary value problem (IBVP). To completethe description of the IBVP, the boundary conditions and initial conditions must be given.The boundary in a one dimensional problem consists of the two points at the ends of thedomain, which in the model problem are the points Xa and Xb . From the linear form ofthe momentum equations, Eq. (2.2.23), it can be seen that the partial differential equation issecond order in X.
Therefore, at each end, either u or u,X must be prescribed as a boundarycondition. In solid mechanics, instead of u,X , the traction t x0 = n0 P is prescribed; n 0 is theunit normal to the body which is given by n0 = 1 at X a, n0 = −1 at X b. Since the stress is afunction of the measure of strain, which in turn depends on the derivative of thedisplacement by Eq.
(2.2.6), prescribing t 0x is equivalent to prescribing u,X ; the superscript"naught" on t indicates that the traction is defined over the undeformed area; the superscriptis always explicitly included on the traction t 0x to distinguish it from the time t. Thereforeeither the traction or the displacement must be prescribed at each boundary.A boundary is called a displacement boundary and denoted by Γu if the displacementis prescribed; it is called a traction boundary and denoted by Γt if the traction is prescribed.The prescribed values are designated by a superposed bar. The boundary conditions areu=uon Γu0n0 P = t x(2.2.27)on Γt(2.2.28)As an example of the boundary conditions in solid mechanics, for the rod in Fig.
2.1, theboundary conditions areu(X a,t) = 0n0 ( X b )P (Xb , t) = P( Xb ,t ) =T (t)A0 ( Xb )(2.2.29)The traction and displacement cannot be prescribed at the same point, but one of thesemust be prescribed at each boundary point; this is indicated byΓu ∩Γ t = 0Γu ∪Γt = Γ(2.2.30)2-8T. Belytschko, Chapter 2, December 16, 1998Thus in a one dimensional solid mechanics problem any boundary is either a tractionboundary or a displacement boundary, but no boundary is both a prescribed traction andprescribed displacement boundary.Initial Conditions. Since the governing equation for the rod is second order in time,two sets of initial conditions are needed.
We will express the initial conditions in terms ofthe displacements and velocities:u( X, 0) = u0 ( X ) for X ∈[ X a ,X b ](2.2.31a)u˙(X, 0) = v0 ( X ) for X ∈ [ Xa ,X b ](2.2.31b)If the body is initially undeformed and at rest, the initial conditions can be written asu( X, 0) = 0u˙ (X , 0) = 0(2.2.32)Jump Conditions. In order for the derivative in Eq.(2.2.12) to exist, the quantity A 0 Pmust be continuous.
However, neither A 0 nor P need be continuous in the entire interval.Therefore momentum balance requires thatA0 P = 0(2.2.33)where 〈f〉 designates the jump in f(X), i.e.f ( X) = f ( X + ε) – f ( X – ε )ε→0(2.2.34)In dynamics, it is possible to have jumps in the stress, known as shocks, which can eitherbe stationary or moving. Moving discontinuities are governed by the Rankine-Hugoniotrelations, but these are not considered in this Chapter.2 . 3 Weak Form for Total Lagrangian FormulationThe momentum equation cannot be discretized directly by the finite element method. Inorder to discretize this equation, a weak form, often called a variational form, is needed.The principle of virtual work, or weak form, which will be developed next, is equivalent tothe momentum equation and the traction boundary conditions (2.2.33).
Collectively, thesetwo equations are called the classical strong form. The weak form can be used toapproximate the strong form by finite elements; solutions obtained by finite elements areapproximate solutions to the strong form.Strong Form to Weak Form. A weak form will now be developed for the momentumequation (2.2.23) and the traction boundary conditions.
For this purpose we define trialfunctions u( X,t ) which satisfy any displacement boundary conditions and are smoothenough so that all derivatives in the momentum equation are well defined. The testfunctions δu( X ) are assumed to be smooth enough so that all of the following steps arewell defined and to vanish on the prescribed displacement boundary. The weak form isobtained by taking the product of the momentum equation expressed in terms of the trialfunction with the test function.
This gives2-9T. Belytschko, Chapter 2, December 16, 1998Xb∫Xa[]δu ( A0 P), X + ρ0 A0 b– ρ 0 A0˙u˙ dX = 0(2.3.1)Using the derivative of the product in the first term in (2.3.1) gives∫Xbδu( A0 P) , X dX =Xa∫ [(δuA P)Xb0Xa,X]– δu, X A0 P dX(2.3.2)Applying the fundamental theorem of calculus to the above gives0∫X δu( A0 P) , X dX =– ∫X δu, X ( A0 P ) dX + (δuA0n P ) ΓXbXbaa= – ∫ δu, X ( A0 P) dX +XbXa(δuA0 t x0)(2.3.3)Γtwhere we obtained the second line using the facts that the test function δu vanishes on theprescribed displacement boundary, the complementarity conditions on the boundaries(2.2.30) and the traction boundary conditions. Substituting (2.3.3) into the first term ofEq. (2.3.1) gives (with a change of sign)∫ [δuXbXa,X(A0 P – δu(ρ 0 A0 b – ρ 0 A0u˙˙)]dX – δuA0 t x0)Γt=0(2.3.4)The above is the weak form of the momentum equation and the traction boundary conditionfor the total Lagrangian formulation.Smoothness of Test and Trial functions; Kinematic Admissibility.
We shallnow investigate the smoothness required to go through the above steps more closely. Forthe momentum equation (2.2.12) to be well defined in a classical sense, the nominal stressand the initial area must be continuously differentiable, i.e. C1 ; otherwise the firstderivative would have discontinuities. If the stress is a smooth function of the derivative ofthe displacement as in (2.2.18), then to obtain this continuity in the stresses requires thatthe trial functions must be C 2 . For Eq. (2.3.2) to hold, the test function δu( X ) must beC1 .However, the weak form is well defined for test and trial functions which are far lesssmooth, and indeed the test and trial functions to be used in finite element methods will berougher. The weak form (2.3.4) involves only the first derivative of the test function andthe trial function appears directly or as a first derivative of the trial function through thenominal stress.
Thus the integral in the weak form is integrable if both functions are C 0 .We will now define the conditions on the test and trial function more precisely. Theweak form is well-defined if the trial functions u(X,t) are continuous functions withpiecewise continuous derivatives, which is stated symbolically by u( X,t ) ∈C0 ( X) , wherethe X in the parenthesis following C 0 indicates that it pertains to the continuity in X; notethat this definition permits discontinuities of the derivatives at discrete points. This is thesame as the continuity of finite element approximations in linear finite element procedures:2-10T.
Belytschko, Chapter 2, December 16, 1998the displacement is continuous and continuously differentiable within elements, but thederivative u,X is discontinuous across element boundaries. In addition, the trial functionu(X,t) must satisfy all displacement boundary conditions. These conditions on the trialdisplacements are indicated symbolically by{u( X,t ) ∈U where U = u( X, t) u( X,t ) ∈ C0 ( X), u = u on Γu}(2.3.5)Displacement fields which satisfy the above conditions, i.e. displacements which are in U,are called kinematically admissible.The test functions are denoted by δu(X); they are not functions of time.
The testfunctions are required to be C0 in X and to vanish on displacement boundaries, i.e.,{δu( X ) ∈U0 where U0 = δu( X ) δ ( X )u ∈C 0( X ), δu = 0 on Γu}(2.3.6)We will use the prefix δ for all variables which are test functions and for variables whichare related to test functions. This convention originates in variational methods, where thetest function emerges naturally as the difference between admissible functions.
Although itis not necessary to know variational methods to understand weak forms, it provides anelegant framework for developing various aspects of the weak form. For example, invariational methods any test function is a variation and defined as the difference betweentwo trial functions, i.e. the variation δu(X) = ua(X) – ub(X), where u a( X ) and ub ( X ) areany two functions in U. Since any function in U satisfies the displacement boundaryconditions, the requirement in (2.3.6) that δu(X) = 0 on Γ u can be seen immediately.Weak Form to Strong Form. We will now develop the equations implied by the weakform with the less smooth trial and test functions, (2.3.5) and (2.3.6), respectively; thestrong form implied with very smooth test and trial functions will also be discussed.
Theweak form is given by0∫X [δu, X A0 P– δu( ρ0 A0 b– ρ0 A0 u˙˙ )]dX – (δuA0 t x )Γt = 0Xba∀δu( X ) ∈U 0(2.3.7)The displacement fields are assumed to be kinematically admissible, i.e. u( X,t ) ∈U . Theabove weak form is expressed in terms of the nominal stress P, but it is assumed that thisstress can always be expressed in terms of the derivatives of the displacement field throughthe strain measure and constitutive equation. Since u(X,t) is C0 and the strain measureinvolves derivatives of u(X,t) with respect to X, we expect P(X,t) to be C–1 in X if theconstitutive equation is continuous: P(X,t) will be discontinuous wherever the derivative ofu(X,t) is discontinuous.To extract the strong form, we need to eliminate the derivative of δu( X ) from theintegrand.
This is accomplished by integration by parts and the fundamental theorem ofcalculus. Taking the derivative of the product δuA0 P we have∫X (δuA0 P), X dX = ∫Xδu, X A0 P dX + ∫X δu (A0 P ), X dXXbXbXbaaa2-11(2.3.8)T. Belytschko, Chapter 2, December 16, 1998The second term on the RHS can be converted to point values by using the fundamentaltheorem of calculus. Let the piecewise continuous function (A 0P) ,X be continuous onintervals X1i, X2i , e = 1 to n, Then by the fundamental theorem of calculus[]0∫X (δuA0 P), X dX = (δuA0 P) X – (δuA0 P ) X ≡ (δuA0 n P) ΓX2ei2e1i1( )(2.3.9)i( )where n0 is the normal to the segments are n X1i =–1, n X 2i =+1, and Γi denotes the twoboundary points of the segment i over which the function is continuously differentiable.Let [ XA , XB ] = ∑ X1i , X i2 ; then applying (2.3.2) over the entire domain givesi[]Xb0∫ ( δuA0 P), X dX =(δuA0n P) Γt – ∑i δu A0 P Γi(2.3.10)Xawhere Γ i are the interfaces between the segments in which the integrand is continuouslydifferentiable.
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