Hutton - Fundamentals of Finite Element Analysis, страница 10

For equilibrium,x A = PxA = A(x ) = A 0 1 −2Land sincethe axial stress variation along the length of the bar is described byx =A0Px1−2LTherefore, the axial strain isεx =x=EE A0Px1−2LSince the bar is fixed at x = 0 , the displacement at x = L is given byL=0=Pε x dx =EA 0L0dxx1−2LL2 PL2 PL2 PLPL[−ln(2L − x )]0 =[ln(2L ) − ln L ] =ln 2 = 1.386E A0E A0E A0A0 E37Hutton: Fundamentals ofFinite Element Analysis382. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsComparison of the results of parts b and c reveals that the two element solutionexhibits an error of only about 1 percent in comparison to the exact solution fromstrength of materials theory.

Figure 2.7e shows the displacement variation along thelength for the three solutions. It is extremely important to note, however, that thecomputed axial stress for the finite element solutions varies significantly from that ofthe exact solution. The axial stress for the two-element solution is shown in Figure 2.7f, along with the calculated stress from the exact solution. Note particularlythe discontinuity of calculated stress values for the two elements at the connectingnode. This is typical of the derived, or secondary, variables, such as stress and strain,as computed in the finite element method.

As more and more smaller elements areused in the model, the values of such discontinuities decrease, indicating solutionconvergence. In structural analyses, the finite element user is most often more interested in stresses than displacements, hence it is essential that convergence of thesecondary variables be monitored.2.4 STRAIN ENERGY, CASTIGLIANO’SFIRST THEOREMWhen external forces are applied to a body, the mechanical work done by thoseforces is converted, in general, into a combination of kinetic and potential energies. In the case of an elastic body constrained to prevent motion, all the workis stored in the body as elastic potential energy, which is also commonlyreferred to as strain energy.

Here, strain energy is denoted U e and mechanicalwork W. From elementary statics, the mechanical work performed by a force Fas its point of application moves along a path from position 1 to position 2 isdefined as2rW = F · d(2.37)1wheredr = dx i + dy j + dz k(2.38)is a differential vector along the path of motion. In Cartesian coordinates, workis given byx2y2z2W = Fx dx + Fy dy + Fz dzx1y1(2.39)z1where Fx , Fy , and Fz are the Cartesian components of the force vector.For linearly elastic deformations, deflection is directly proportional to applied force as, for example, depicted in Figure 2.8 for a linear spring. The slopeof the force-deflection line is the spring constant such that F = k.

Therefore,the work required to deform such a spring by an arbitrary amount 0 from itsHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 2004Force, F2.4 Strain Energy, Castigliano’s First Theoremk1Deflection, ␦Figure 2.8 Force-deflectionrelation for a linear elasticspring.free length is 0 01W = F d = k d = k02 = U e20(2.40)0and we observe that the work and resulting elastic potential energy are quadraticfunctions of displacement and have the units of force-length. This is a generalresult for linearly elastic systems, as will be seen in many examples throughoutthis text.Utilizing Equation 2.28, the strain energy for an axially loaded elastic barfixed at one end can immediately be written asUe =1 21 AE 2k =22 L(2.41)However, for a more general purpose, this result is converted to a different form(applicable to a bar element only) as follows:2 1 21 AE PL1 PP1U e = k ==AL = εV(2.42)22 LAE2 AAE2where V is the total volume of deformed material and the quantity 12 ε is strainenergy per unit volume, also known as strain energy density.

In Equation 2.42,stress and strain values are those corresponding to the final value of appliedforce. The factor 12 arises from the linear relation between stress and strain as theload is applied from zero to the final value P. In general, for uniaxial loading, thestrain energy per unit volume u e is defined byεu e = dε(2.43)0which is extended to more general states of stress in subsequent chapters.

We notethat Equation 2.43 represents the area under the elastic stress-strain diagram.39Hutton: Fundamentals ofFinite Element Analysis402. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsPresently, we will use the work-strain energy relation to obtain the governing equations for the bar element using the following theorem.Castigliano’s First Theorem [1]For an elastic system in equilibrium, the partial derivative of total strain energywith respect to deflection at a point is equal to the applied force in the directionof the deflection at that point.Consider an elastic body subjected to N forces Fj for which the total strainenergy is expressed asjN Ue = W =Fj dj(2.44)j=10where j is the deflection at the point of application of force Fj in the direction ofthe line of action of the force. If all points of load application are fixed exceptone, say, i, and that point is made to deflect an infinitesimal amount i by anincremental infinitesimal force Fi , the change in strain energy isiU e = W = Fi i + Fi di(2.45)0where it is assumed that the original force Fi is constant during the infinitesimalchange.

The integral term in Equation 2.45 involves the product of infinitesimalquantities and can be neglected to obtainU e= Fii(2.46)which in the limit as i approaches zero becomes∂U= Fi∂ i(2.47)The first theorem of Castigliano is a powerful tool for finite element formulation, as is now illustrated for the bar element. Combining Equations 2.30, 2.31,and 2.43, total strain energy for the bar element is given by211u2 − u1U e = x ε x V = EAL(2.48)22LApplying Castigliano’s theorem with respect to each displacement yields∂U eAE=(u 1 − u 2 ) = f 1∂ u1L(2.49)∂U eAE=(u 2 − u 1 ) = f 2∂ u2L(2.50)which are observed to be identical to Equations 2.33 and 2.34.Hutton: Fundamentals ofFinite Element Analysis2.

Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.4 Strain Energy, Castigliano’s First Theorem41The first theorem of Castigliano is also applicable to rotational displacements. In the case of rotation, the partial derivative of strain energy with respectto a rotational displacement is equal to the moment/torque applied at the point ofconcern in the sense of the rotation. The following example illustrates the application in terms of a simple torsional member.EXAMPLE 2.5A solid circular shaft of radius R and length L is subjected to constant torque T. The shaftis fixed at one end, as shown in Figure 2.9. Formulate the elastic strain energy in terms ofthe angle of twist at x = L and show that Castigliano’s first theorem gives the correctexpression for the applied torque.■ SolutionFrom strength of materials theory, the shear stress at any cross section along the length ofthe member is given by =TrJwhere r is radial distance from the axis of the member and J is polar moment of inertia ofthe cross section.

For elastic behavior, we haveTr=GJG =where G is the shear modulus of the material, and the strain energy is then1Ue =21 dV =2V=L TrTrdAdxJJGA0T22J 2 GL r 2 dA dx =T2L2JG0 Awhere we have used the definition of the polar moment of inertiaJ =r2 dAARLTFigure 2.9 Example 2.5:Circular cylinder subjected totorsion.Hutton: Fundamentals ofFinite Element Analysis422. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsAgain invoking the strength of materials results, the angle of twist at the end of the member is known to be=TLJGso the strain energy can be written asUe =1 L JG 2JG 2=2 JG L2LPer Castangliano’s first theorem,∂UeJG=T =∂Lwhich is exactly the relation shown by strength of materials theory. The reader may thinkthat we used circular reasoning in this example, since we utilized many previously knownresults.

However, the formulation of strain energy must be based on known stress andstrain relationships, and the application of Castigliano’s theorem is, indeed, a differentconcept.For linearly elastic systems, formulation of the strain energy function interms of displacements is relatively straightforward. As stated previously, thestrain energy for an elastic system is a quadratic function of displacements. Thequadratic nature is simplistically explained by the facts that, in elastic deformation, stress is proportional to force (or moment or torque), stress is proportionalto strain, and strain is proportional to displacement (or rotation). And, since theelastic strain energy is equal to the mechanical work expended, a quadratic function results.

Therefore, application of Castigliano’s first theorem results in linearalgebraic equations that relate displacements to applied forces. This statementfollows from the fact that a derivative of a quadratic term is linear. The coefficients of the displacements in the resulting equations are the components of thestiffness matrix of the system for which the strain energy function is written.Such an energy-based approach is the simplest, most-straightforward method forestablishing the stiffness matrix of many structural finite elements.EXAMPLE 2.6(a) Apply Castigliano’s first theorem to the system of four spring elements depicted inFigure 2.10 to obtain the system stiffness matrix.

The vertical members at nodes 2and 3 are to be considered rigid.(b) Solve for the displacements and the reaction force at node 1 ifk 1 = 4 N/mmk 2 = 6 N/mmk 3 = 3 N/mmF2 = − 30 NF3 = 0F4 = 50 NHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.4 Strain Energy, Castigliano’s First Theoremk21F2k12k2k3F443Figure 2.10 Example 2.6: Four spring elements.■ Solution(a) The total strain energy of the system of four springs is expressed in terms of thenodal displacements and spring constants asUe =111k 1 (U 2 − U 1 ) 2 + 2k 2 (U 3 − U 2 ) 2 + k 3 (U 4 − U 3 ) 2222Applying Castigliano’s theorem, using each nodal displacement in turn,∂U e= F1 = k 1 (U 2 − U 1 )(−1) = k 1 (U 1 − U 2 )∂U 1∂U e= F2 = k 1 (U 2 − U 1 ) + 2k 2 (U 3 − U 2 )(−1) = −k 1 U 1 + (k 1 + 2k 2 )U 2 − 2k 2 U 3∂U 2∂U e= F3 = 2k 2 (U 3 − U 2 ) + k 3 (U 4 − U 3 )(−1) = −2k 2 U 2 + (2k 2 + k 3 )U 3 − k 3 U 4∂U 3∂U e= F4 = k 3 (U 4 − U 3 ) = −k 3 U 3 + k 3 U 4∂U 4which can be written in matrix form ask1 −k1 00−k1k1 + 2k2−2k200−2k22k2 + k3−k3   U1   F1 0U  F 0 22=−k3  U  F  3  3k3U4F4and the system stiffness matrix is thus obtained via Castigliano’s theorem.(b) Substituting the specified numerical values, the system equations become  4−400  0   F1  −4 16 −12 0  U2−30 = 0 −12 15 −3   U3   0   00−33U450Eliminating the constraint equation, the active displacements are governed by  16 −12 0  U2   −30  −12 15 −3  U3 =0  50U40−33which we solve by manipulating the equations to convert the coefficient matrix (the43Hutton: Fundamentals ofFinite Element Analysis442.