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For a uniform distributed loadq z (x ) = q z , the equivalent nodal load vector is found to beqz L   2 2−qLfz qz1  Mqz112=(4.69)qz L f qz2  2 Mqz22 qLz12The addition of torsion to the general beam element is accomplished withreference to Figure 4.16a, which depicts a circular cylinder subjected to torsionvia twisting moments applied at its ends. A corresponding torsional finite elementL1Mx1T12xxG, JT2(a)Figure 4.16(a) Circular cylinder subjected to torsion. (b) Torsional finite element notation.Mx2(b)Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 20044.8 A General Three-Dimensional Beam Elementis shown in Figure 4.16b, where the nodes are 1 and 2, the axis of the cylinder isthe x axis, and twisting moments are positive according to the right-hand rule.From elementary strength of materials, it is well known that the angle of twist perunit length of a uniform, elastic circular cylinder subjected to torque T is given by=TJG(4.70)where J is polar moment of inertia of the cross-sectional area and G is the shearmodulus of the material.

As the angle of twist per unit length is constant, the totalangle of twist of the element can be expressed in terms of the nodal rotations andtwisting moments asTLx2 − x1 =(4.71)JGorJGT =(x2 − x1 ) = k T (x2 − x1 )(4.72)LComparison of Equation 4.72 with Equation 2.2 for a linearly elastic spring andconsideration of the equilibrium condition M x1 + M x2 = 0 lead directly to theelement equilibrium equations: JG 1 −1x1M x1=(4.73)x2M x2L −1 1so the torsional stiffness matrix isJG 1[k torsion ] =L −1−11(4.74)While this development is, strictly speaking, applicable only to a circular crosssection, an equivalent torsional stiffness Jeq G/L is known for many commonstructural cross sections and can be obtained from standard structural tables orstrength of materials texts.Adding the torsional characteristics to the general beam element, the elementequations become u 1   f x1 u2 f x2  v1   f y1  z1   Mz2 [kaxial ][0][0][0]vf2 y2  [0] z2[k][0][0]Mbendingxyz2=(4.75) [0][0][kbending ]x z[0] w1 f z1  y1   M y1 [0][0][0][ktorsion ]   f z2  w2  My2 y2 M x1   x1 x2Mx2123Hutton: Fundamentals ofFinite Element Analysis1244.

Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elementsand the final stiffness matrix for a general 3-D beam element is observed to be a12 × 12 symmetric matrix composed of the individual stiffness matrices repre-senting axial loading, two-plane bending, and torsion.The general beam element can be utilized in finite element analyses of threedimensional frame structures.

As with most finite elements, it is often necessaryto transform the element matrices from the element coordinate system to theglobal coordinates. The transformation procedure is quite similar to that discussed for the bar and two-dimensional beam elements, except, of course, for theadded algebraic complexity arising from the size of the stiffness matrix andcertain orientation details required.4.9 CLOSING REMARKSIn this chapter, finite elements for beam bending are formulated using elasticflexure theory from elementary strength of materials.

The resulting elements arevery useful in modeling frame structures in two or three dimensions. A generalthree-dimensional beam element including axial, bending, and torsional effectsis developed by, in effect, superposition of a spar element, two flexure elements,and a torsional element.In development of the beam elements, stiffening of the elements owing totensile loading, the possibility of buckling under compressive axial loading, andtransverse shear effects have not been included. In most commercial finiteelement software packages, each of these concerns is an option that can be takeninto account at the user’s discretion.REFERENCES1.

Beer, F. P., E. R. Johnston, and J. T. DeWolf. Mechanics of Materials, 3rd ed.New York: McGraw-Hill, 2002.2. Budynas, R. Advanced Strength and Applied Stress Analysis, 2nd ed. New York:McGraw-Hill, 1999.PROBLEMS4.1Two identical beam elements are connected at a common node as shown inFigure P4.1. Assuming that the nodal displacements vi , i are known, useEquation 4.32 to show that the normal stress x is, in general, discontinuousat the common element boundary (i.e., at node 2).

Under what condition(s)would the stress be continuous?1Figure P4.123Hutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 2004Problems4.2For the beam element loaded as shown in Figure P4.2, construct the shear forceand bending moment diagrams. What is the significance of these diagrams withdMrespect to Equations 4.10, 4.17, and the relation V =from strength ofdxmaterials theory?FFLM1M2Figure P4.24.3For a uniformly loaded beam as shown in Figure P4.3, the strength of materialstheory gives the maximum deflection asvmax = −5qL 4384EI zat x = L /2.

Treat this beam as a single finite element and compute the maximumdeflection. How do the values compare?yqxLFigure P4.34.4The beam element shown in Figure P4.4 is subjected to a linearly varying loadof maximum intensity qo. Using the work-equivalence approach, determine thenodal forces and moments.yqoxLFigure P4.44.5Use the results of Problem 4.4 to calculate the deflection at node 2 of the beamshown in Figure P4.5 if the beam is treated as a single finite element.yqo1Figure P4.5E, Iz, L2x125Hutton: Fundamentals ofFinite Element Analysis1264. Flexure ElementsCHAPTER 44.64.74.8Text© The McGraw−HillCompanies, 2004Flexure ElementsFor the beam element of Figure P4.5, compute the reaction force and moment atnode 1. Compute the maximum bending stress assuming beam height is 2h .

Howdoes the stress value compare to the maximum stress obtained by the strength ofmaterials approach?Repeat Problem 4.5 using two equal length elements. For this problem, letE = 30 × 10 6 psi, I z = 0.1 in.4, L = 10 in., q o = 10 lb/in.Consider the beam shown in Figure P4.8. What is the minimum number ofelements that can be used to model this problem? Construct the global nodal loadvector corresponding to your answer.PqoL3L3L3Figure P4.84.9 What is the justification for writing Equation 4.36 in the form of Equation 4.37?4.10–4.15 For each beam shown in the associated figure, compute the deflection atthe element nodes.

The modulus of elasticity is E = 10 × 10 6 psi and thecross section is as shown in each figure. Also compute the maximum bendingstress. Use the finite element method with the minimum number of elementsfor each case.500 lb10 in.30 in.2 in.1 in.Figure P4.1010 lb/in.0.5 in.0.5 in.10 in.10 in.Figure P4.1145 N/m40 mm30 mm0.3 mFigure P4.120.3 m10 mmHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 2004Problems200 lb18 in.6 in.2 in.1 in.400 lb/in.Figure P4.13500 lb10 in.200 lb8 in.8 in.D400 in.-lbD2 1.0 in.D1 1.5 in.Figure P4.14200 lb2 in.300 lb10 in.20 in.10 in.2 in.0.25 in.Figure P4.154.16The tapered beam element shown in Figure P4.16 has uniform thickness t andvaries linearly in height from 2h to h.

Beginning with Equation 4.37, derive thestrain energy expression for the element in a form similar to Equation 4.39.yh2hxLFigure P4.164.174.18Use the result of Problem 4.16 to derive the value of component k 11 of theelement stiffness matrix.The complete stiffness matrix for the tapered element of Figure P4.16 is given by243156LEth 156L56L 2[k] =60L 3  −243 −156L87L42L 23−243−156L243−87L87L2 42L −87L 45L 2127Hutton: Fundamentals ofFinite Element Analysis1284. Flexure ElementsCHAPTER 4Text© The McGraw−HillCompanies, 2004Flexure Elements10 lb1 in.0.5 in.t 0.25 in.12 in.(a)10 lb0.875 in.6 in.0.625 in.t 0.25 in.6 in.(b)Figure P4.184.19a.

Using the given stiffness matrix with E = 10(106 ) , compute the deflectionof node 2 for the tapered element loaded as shown in Figure P4.18a.b. Approximate the tapered beam using two straight elements, as inFigure P4.18b, and compute the deflection.c. How do the deflection results compare?d. How do the stress computations compare?The six equilibrium equations for a beam-axial element in the element coordinatesystem are expressed in matrix form as[k e ] {} = { f e }with {} as given by Equation 4.61, [k e ] by Equation 4.62, and { f e } as the nodalforce vector{ f e } = [ f 1x4.204.21f1 yM1f 2xf2 yM 2 ]TFor an element oriented at an arbitrary angle relative to the global X axis,convert the equilibrium equations to the global coordinate system and verifyEquation 4.65.Use Equation 4.63 to express the strain energy of a beam-axial element in termsof global displacements.

Apply the principle of minimum potential energy toderive the expression for the element equilibrium equations in the globalcoordinate system. (Warning: This is algebraically tedious.)The two-dimensional frame structure shown in Figure P4.21 is composed oftwo 2 × 4 in. steel members ( E = 10 × 10 6 psi), and the 2-in. dimension isperpendicular to the plane of loading. All connections are treated as weldedjoints. Using two beam-axial elements and the node numbers as shown,determineHutton: Fundamentals ofFinite Element Analysis4. Flexure ElementsText© The McGraw−HillCompanies, 2004Problems30 in.1200 lb20 lb/in.231500 in.ⴢ lb30 in.1Figure P4.214.224.23a. The global stiffness matrix.b.

The global load vector.c. The displacement components of node 2.d. The reaction forces and moments at nodes 1 and 3.e. Maximum stress in each element.Repeat Problem 4.21 for the case in which the connection at node 2 is apin joint.The frame structure shown in Figure P4.23 is the support structure for a hoistlocated at the point of application of load W.

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