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Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elementsstiffness matrix) to upper-triangular form; that is, all terms below the maindiagonal become zero.Step 1. Multiply the first equation (row) by 12, multiply the second equation (row) by16, add the two and replace the second equation with the resulting equationto obtain  16 −120 U2   −30  096 −48  U3 = −360  50U40 −33Step 2. Multiply the third equation by 32, add it to the second equation, and replacethe third equation with the result.

This gives the triangularized form desired:  16 −120 U2   −30  096 −48  U3 = −360  1240U40048In this form, the equations can now be solved from the “bottom to the top,” and it will befound that, at each step, there is only one unknown. In this case, the sequence isU4 =1240= 25.83 mm48U3 =1[−360 + 48(25.83)] = 9.17 mm96U2 =1[−30 + 12(9.17)] = 5.0 mm16The reaction force at node 1 is obtained from the constraint equationF1 = −4U 2 = −4(5.0) = −20 Nand we observe system equilibrium since the external forces sum to zero as required.2.5 MINIMUM POTENTIAL ENERGYThe first theorem of Castigliano is but a forerunner to the general principle ofminimum potential energy.

There are many ways to state this principle, and it hasbeen proven rigorously [2]. Here, we state the principle without proof but expectthe reader to compare the results with the first theorem of Castigliano. The principle of minimum potential energy is stated as follows:Of all displacement states of a body or structure, subjected to external loading,that satisfy the geometric boundary conditions (imposed displacements), the displacement state that also satisfies the equilibrium equations is such that the totalpotential energy is a minimum for stable equilibrium.Hutton: Fundamentals ofFinite Element Analysis2.

Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 20042.5 Minimum Potential EnergyWe emphasize that the total potential energy must be considered in application of this principle. The total potential energy includes the stored elastic potential energy (the strain energy) as well as the potential energy of applied loads. Asis customary, we use the symbol for total potential energy and divide the totalpotential energy into two parts, that portion associated with strain energy Ue andthe portion associated with external forces UF. The total potential energy is = Ue + U F(2.51)where it is to be noted that the term external forces also includes moments andtorques.In this text, we will deal only with elastic systems subjected to conservativeforces.

A conservative force is defined as one that does mechanical workindependent of the path of motion and such that the work is reversible or recoverable. The most common example of a nonconservative force is the force ofsliding friction. As the friction force always acts to oppose motion, the workdone by friction forces is always negative and results in energy loss. This lossshows itself physically as generated heat.

On the other hand, the mechanicalwork done by a conservative force, Equation 2.37, is reversed, and thereforerecovered, if the force is released. Therefore, the mechanical work of a conservative force is considered to be a loss in potential energy; that is,U F = −W(2.52)where W is the mechanical work defined by the scalar product integral of Equation 2.37. The total potential energy is then given by = Ue − W(2.53)As we show in the following examples and applications to solid mechanicsin Chapter 9, the strain energy term Ue is a quadratic function of system displacements and the work term W is a linear function of displacements.

Rigorously, the minimization of total potential energy is a problem in the calculus ofvariations [5]. We do not suppose that the intended audience of this text isfamiliar with the calculus of variations. Rather, we simply impose the minimization principle of calculus of multiple variable functions.

If we have a total potential energy expression that is a function of, say, N displacements Ui , i = 1, . . . , N;that is, = (U 1 , U 2 , . . . , U N )(2.54)then the total potential energy will be minimized if∂=0∂U ii = 1, . . . , N(2.55)Equation 2.55 will be shown to represent N algebraic equations, which form thefinite element approximation to the solution of the differential equation(s) governing the response of a structural system.45Hutton: Fundamentals ofFinite Element Analysis462.

Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar ElementsEXAMPLE 2.7Repeat the solution to Example 2.6 using the principle of minimum potential energy.■ SolutionPer the previous example solution, the elastic strain energy is11122U e = k 1 (U 2 − U 1 ) + 2 k 2 (U 3 − U 2 ) + k 3 (U 4 − U 3 ) 2222and the potential energy of applied forces isU F = −W = − F1 U 1 − F2 U 2 − F3 U 3 − F4 U 4Hence, the total potential energy is expressed as=11k 1 (U 2 − U 1 ) 2 + 2 k 2 (U 3 − U 2 ) 222+1k 3 (U 4 − U 3 ) 2 − F1 U 1 − F2 U 2 − F3 U 3 − F4 U 42In this example, the principle of minimum potential energy requires that∂=0∂U ii = 1, 4giving in sequence i = 1, 4 , the algebraic equations∂= k 1 (U 2 − U 1 )(−1) − F1 = k 1 (U 1 − U 2 ) − F1 = 0∂U 1∂= k 1 (U 2 − U 1 ) + 2k 2 (U 3 − U 2 )(−1) − F2∂U 2= −k 1 U 1 + (k 1 + 2k 2 )U 2 − 2k 2 U 3 − F2 = 0∂= 2k 2 (U 3 − U 2 ) + k 3 (U 4 − U 3 )(−1) − F3∂U 3= −2k 2 U 2 + (2k 2 + k 3 )U 3 − k 3 U 4 − F3 = 0∂= k 3 (U 4 − U 3 ) − F4 = −k 3 U 3 + k 3 U 4 − F4 = 0∂U 4which, when written in matrix form, arek1 −k1 00−k1k1 + 2k2−2k200−2k22k2 + k3−k3   0  U1   F1    0  U2 = F2−k3 U  F  3  3k3U4F4and can be seen to be identical to the previous result.

Consequently, we do not resolve thesystem numerically, as the results are known.Hutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 2004ReferencesWe now reexamine the energy equation of the Example 2.7 to develop a moregeneral form, which will be of significant value in more complicated systems tobe discussed in later chapters. The system or global displacement vector is U 1U2{U } =(2.56) U3 U4and, as derived, the global stiffness matrix isk−k1 −k[K ] =  1001k1 + 2k2−2k200−2k22k2 + k3−k300 −k3k3(2.57)If we form the matrix triple product11{U } T [K ]{U } = [ U 1 U 2 U 3 U 4 ]22k1−k10−2k2 −k1 k1 + 2k2×0−2k22k2 + k300−k3 0U 10  U2−k3  U3 k3U4(2.58)and carry out the matrix operations, we find that the expression is identical to thestrain energy of the system.

As will be shown, the matrix triple product of Equation 2.58 represents the strain energy of any elastic system. If the strain energycan be expressed in the form of this triple product, the stiffness matrix will havebeen obtained, since the displacements are readily identifiable.2.6 SUMMARYTwo linear mechanical elements, the idealized elastic spring and an elastic tensioncompression member (bar) have been used to introduce the basic concepts involved informulating the equations governing a finite element. The element equations are obtainedby both a straightforward equilibrium approach and a strain energy method using the firsttheorem of Castigliano.

The principle of minimum potential also is introduced. The nextchapter shows how the one-dimensional bar element can be used to demonstrate the finiteelement model assembly procedures in the context of some simple two- and threedimensional structures.REFERENCES1. Budynas, R. Advanced Strength and Applied Stress Analysis. 2d ed. New York:McGraw-Hill, 1998.2. Love, A. E. H.

A Treatise on the Mathematical Theory of Elasticity. New York:Dover Publications, 1944.47Hutton: Fundamentals ofFinite Element Analysis482. Stiffness Matrices,Spring and Bar ElementsCHAPTER 2Text© The McGraw−HillCompanies, 2004Stiffness Matrices, Spring and Bar Elements3. Beer, F. P., E. R.

Johnston, and J. T. DeWolf. Mechanics of Materials. 3d ed.New York: McGraw-Hill, 2002.4. Shigley, J., and R. Mischke. Mechanical Engineering Design. New York:McGraw-Hill, 2001.5. Forray, M. J. Variational Calculus in Science and Engineering. New York:McGraw-Hill, 1968.PROBLEMSFor each assembly of springs shown in the accompanying figures(Figures P2.1–P2.3), determine the global stiffness matrix using the systemassembly procedure of Section 2.2.2.1–2.3k1k21k3234Figure P2.1k3k11k2243k3Figure P2.2k11k2k332…kN24kN1N 1NFigure P2.32.4For the spring assembly of Figure P2.4, determine force F3 required to displacenode 2 an amount = 0.75 in. to the right. Also compute displacement ofnode 3.

Givenk 1 = 50 lb./in.and k 2 = 25 lb./in.␦k11k223F3Figure P2.42.5In the spring assembly of Figure P2.5, forces F2 and F4 are to be applied suchthat the resultant force in element 2 is zero and node 4 displaces an amountHutton: Fundamentals ofFinite Element Analysis2. Stiffness Matrices,Spring and Bar ElementsText© The McGraw−HillCompanies, 2004Problems = 1 in. Determine (a) the required values of forces F2 and F4, (b) displacementof node 2, and (c) the reaction force at node 1.F2k1␦k212k3F443k1 k3 30 lb./in.k2 40 lb./in.Figure P2.52.62.7Verify the global stiffness matrix of Example 2.3 using (a) direct assembly and(b) Castigliano’s first theorem.Two trolleys are connected by the arrangement of springs shown in Figure P2.7.(a) Determine the complete set of equilibrium equations for the system in theform [K ]{U } = {F }.

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