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In addition, with modern computer technology, it ispossible to produce color-coded stress contour plots of an entire model, to visually observe the stress distribution, the deformed shape, the strain energy distribution, and many other criteria.9.9 TORSIONTorsion (twisting) of structural members having circular cross sections is a common problem studied in elementary mechanics of materials. (Recall that earlierwe developed a finite element for such cases.) A major assumption (and theassumption is quite valid for elastic deformation) in torsion of circular membersis that plane sections remain plane after twisting. In the case of torsion of a noncircular cross section, this assumption is not valid and the problem is hence morecomplicated. A general structural member subjected to torsion is shown in Figure 9.13a. The member is subjected to torque T acting about the x axis, and it isassumed that the cross section is uniform along the length.

An arbitrary pointlocated on a cross section at position x is shown in Figure 9.13b. If the cross section twists through angle ␪ , the point moves through arc ds and the displacementcomponents in the y and z directions arev = −z␪w = y␪(9.132)respectively. Since the angle of twist varies along the length of the member, weconclude that the displacement components of Equation 9.132 are described byv = v(x , z)w = w(x , y)(9.133)375Hutton: Fundamentals ofFinite Element Analysis3769. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsOwing to the noncircular cross section, plane sections do not remain plane;instead, there is warping, hence displacement, in the x direction described byu = u( y, z)(9.134)Applying the definitions of the normal strain components to Equation 9.133and 9.134, we findεx =∂u=0∂xεy =∂v=0∂yεz =∂w=0∂z(9.135)and it follows that the normal stress components are ␴x = ␴y = ␴z = 0.

To compute the shear strain components, we introduce the angle of twist per unit length␾ such that the rotation of any cross section can be expressed as ␪ = ␾x . The displacement components are then expressed asu = u( y, z)v = −␾x zw = ␾x y(9.136)and the shear strain components are␥x y =∂u∂v∂u+=− ␾z∂y∂x∂y␥x z =∂w∂u∂u+=+ ␾y∂z∂x∂z␥yz =∂v∂w+=0∂z∂y(9.137)It follows from the stress-strain relations that the only nonzero stress componentsare ␶x y and ␶x z and the only equilibrium equation (Appendix B) not identicallysatisfied becomes∂ ␶x y∂ ␶x z+=0(9.138)∂y∂zWe now hypothesize the existence of a scalar function ␺( y, z) such that␶x y =∂␺∂z␶x z = −∂␺∂y(9.139)In the context of the torsion problem, scalar function ␺ is known as Prandtl’sstress function and is generally analogous to the stream function and potentialfunction introduced in Chapter 8 for ideal fluid flow.

If the relations of Equation 9.139 are substituted into Equation 9.138, we find that the equilibrium condition is automatically satisfied. To discover the governing equation for the stressHutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.9 Torsionfunction, we compute the stress components as∂u− ␾z␶x y = G ␥x y = G∂y∂u␶x z = G ␥yz = G+ ␾y∂zand note that 2∂ ␶x y∂ u=G−␾∂z∂y ∂z 2∂ ␶x z∂ u=G+␾∂y∂y ∂z(9.140)(9.141)Combining the last two equations results in∂ 2␺∂ ␶x y∂ ␶x z∂ 2␺−=+= −2G ␾∂z∂y∂ y2∂ z2(9.142)as the governing equation for Prandtl’s stress function. As with the fluid formulations of Chapter 8, note the analogy of Equation 9.142 with the case of heatconduction.

Here the term 2G ␾ is analogous to internal heat generation Q.9.9.1 Boundary ConditionAt the outside surface of the torsion member, no stress acts normal to the surface,so the resultant of the shear stress components must be tangent to the surface.This is illustrated in Figure 9.13c showing a differential element dS of the surface(with the positive sense defined by the right-hand rule). For the normal stress tobe zero, we must have␶x y sin ␣ − ␶x z cos ␣ = 0(9.143)or␶x ydzdy− ␶x z=0dsds(9.144)Substituting the stress function relations, we obtain∂ ␺ dzd␺ dyd␺+==0dz dsdy dsds(9.145)which shows that the value of the stress function is constant on the surface.

Thevalue is arbitrary and most often taken to be zero.9.9.2 TorqueThe stress function formulation of the torsion problem as given previously doesnot explicitly include the applied torque. To obtain an expression relating the377Hutton: Fundamentals ofFinite Element Analysis3789. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsapplied torque and the stress function, we must consider the moment equilibriumcondition.

Referring to the differential element of a cross section shown in Figure 9.13d, the differential torque corresponding to the shear stresses acting on theelement isdT = ( y␶x z − z␶x y ) d A(9.146)and the total torque is computed as ∂␺∂␺+zdy dz = 2T =( y␶x z − z␶x y ) d A = −y␺ dA∂ydzAAA(9.147)The final result in the last equation is obtained by integrating by parts and notingthe condition ␺ = 0 on the surface.9.9.3 Finite Element FormulationSince the governing equation for the stress function is analogous to the heat conduction equation, it is not necessary to repeat the details of element formulation.Instead, we reiterate the analogies and point out one very distinct difference inhow a finite element analysis of the torsion problem is conducted when the stressfunction is used.

First, note that the stress function is discretized asM␺{y, z} =N i ( y, z)␺ i(9.148)iso that the finite element computations result in nodal values analogous to nodaltemperatures (with the conductivity values set to unity). Second, the torsion term2G ␾ is analogous to internal heat generation Q. However, the angle of twist perunit length ␾ is actually the unknown we wish to compute in the first place.Preferably, in such a problem, we specify the geometry, material properties, andapplied torque, then compute the angle of twist per unit length as well as stressvalues.

However, the formulation here is such that we must specify a value forangle of twist per unit length, compute the nodal values of the stress function,then obtain the torque by summing the contributions of all elements to Equation 9.147. Since the governing equation is linear, the angle of twist per unitlength and the computed torque can be scaled in ratio as required. The procedureis illustrated in the following example.EXAMPLE 9.7Figure 9.14a shows a shaft having a square cross section with 50-mm sides. The material has shear modulus 80 Gpa. Shaft length is 1 m.

The shaft is fixed at one end and subjected to torque T at the other end. Determine the total angle of twist if the applied torqueis 100 N-m.Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.9 Torsion4325 mm2125 mm12(b)(a)Figure 9.14 Finite element model of Example 9.7.■ SolutionObserving the symmetry conditions, we model one-fourth of the cross section usingthree-node linear triangular elements, as in Figure 9.14b. For simplicity of illustration, weuse only two elements and note that, at nodes 2, 3, and 4, the value of the stress functionis specified as zero, since these nodes are on the surface.

Also note that the planes of symmetry are such that the partial derivatives of the stress function across those planes arezero. These conditions correspond to zero normal heat flux (perfect insulation) in a conduction problem.The element stiffness matrices are given byk(e) #=∂N∂yT ∂N∂y∂N+∂zT ∂N∂z$dAAand element nodal forces aref(e)=2G ␾ [N ] T d AA(Note the use of y, z coordinates in accord with the coordinate system used in the preceding developments.)These relations are obtained by analogy with Equation 7.35 for heat conduction.

Theinterpolation functions are as defined in Equation 9.28.Element 1N1 =1(625 − 25 y)2A∂ N125=−∂y2A∂ N1=0∂ yzN2 =1(25 y − 25z)2A∂ N225=∂y2A∂ N225=−∂z2AN3 =1(25z)2A∂ N3=0∂y∂ N325=∂z2ASince the partial derivatives are all constant, the stiffness matrix is−25 0  (1) 1 1 k=25−25 [ 0 −25 25 ][ −25 25 0 ] +4A 4A 025379Hutton: Fundamentals ofFinite Element Analysis3809. Applications in SolidMechanicsCHAPTER 9orText© The McGraw−HillCompanies, 2004Applications in Solid Mechanics625 −625 00001 1 0 625 −625 −625 625 0  +125012500000 −625 6250.5 −0.50=  −0.51−0.5 0−0.5 0.5 (1) k=The element nodal forces are readily shown to be given byf(1) 12G␾A  =13  1which we leave in this general form for the time being.Element 2N1 =1(625 − 25z)2A∂ N1=0∂y∂ N125=−∂z2AN2 =1(25y)2A∂ N225=∂y2A∂ N2=0∂zN3 =1(25z − 25y)2A∂ N325=−∂y2A∂ N325=∂z2A0 −25  (2) 1 1 k=250[ 0 25 −25 ] +[ −25 0 25 ]4A 4A −2525000625 0 −62511 0 625 −625  + 0[k (2) ] =00 125012500 −625 625−625 0 6250.50−0.5= 00.5 −0.5 −0.5 −0.5 1.0For element 2, the nodal forces are also given byf(2) 12G␾A  =13  1Noting the element to global nodal correspondences, the assembled system equations are  21−0.50−0.5  ␺ 1 1 −0.5 ␺22G␾A1−0.50= 02−0.51−0.5 3  ␺3  1−0.50−0.51␺4Hutton: Fundamentals ofFinite Element Analysis9.

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