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Element thickness is 0.2 in. and uniform.■ SolutionUsing the nodal coordinates specified, the interpolation functions (with element areaA = 1) areN 1 (x , y) =1[2 − 2x ] = 1 − x2N 2 (x , y) =1[2x − y]2N 3 (x , y) =1y2Hutton: Fundamentals ofFinite Element Analysis3389. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanics(1, 2)3Y300 psi1(0, 0)2(1, 0)X100 psi(a)3Y2110X402010(b)Figure 9.4(a) Distributed loads on a triangular element.(b) Work-equivalent nodal forces.Along edge 1-2, y = 0 , p x = 0 , p y = −100 psi; hence, Equation 9.39 becomes1 − xf( p) = S 0 x0 00 0t dS01−x −1000x 001 − x0 1 = 0.2 0 x00000 0 0  0 0 0dx =lb −10 1 − x  −100x  −10 00Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane StressFor edge 2-3, we have x = 1, px = 150y , p y = 0, so that001 (2 − y)2y  ( p) 2f= 0S 00 150y t dS01(2 − y) 2y200001 (2 − y) 00 220y2 0150y402= 0.2 dy =lb00 10 0(2 − y) 2 0 0y02000Combining the results, the nodal force vector arising from the distributed loads for theelement shown is then f 1x 0 f 2x 20  ( p) f 3x40 f==lbf1 y −10 f2 y  −10  f3 y0as shown in Figure 9.4b.In addition to distributed edge loads on element boundaries, so-called bodyforces may also arise.

In general, a body force is a noncontact force acting on abody on a per unit mass basis. The most commonly encountered body forcesare gravitational attraction (weight), centrifugal force arising from rotationalmotion, and magnetic force. Currently, we consider only the two-dimensionalFB Xin which FB Xcase in which the body force is described by the vectorFBYand FBY are forces per unit mass acting on the body in the respective coordinate directions. As with distributed loads, the body forces are to be replaced byequivalent nodal forces.

Considering a differential mass ␳ t dx dy undergoingdisplacements (u, v) in the coordinate directions, mechanical work done by the339Hutton: Fundamentals ofFinite Element Analysis3409. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsbody forces isdW b = ␳ FB X ut dx dy + ␳ FBY vt dx dy(9.42)Considering the volume of interest to be a CST element in which the displacements are expressed in terms of interpolation functions and nodal displacements as u1   u2  u(x, y)N1 N2 N3 000u3u== [N ](9.43)v(x, y)000 N1 N2 N3 v1 v v2 v3the total work done by the body forces acting on the element is expressed interms of nodal displacement asWb = ␳ tFB X ( N 1 u 1 + N 2 u 2 + N 3 u 3 ) dx dyA+ ␳tFBY ( N 1 v1 + N 2 v2 + N 3 v3 ) dx dy(9.44)AAs desired, Equation 9.44 is in the form(b)(b)(b)(b)(b)(b)W b = f 1x u 1 + f 2x u 2 + f 3x u 3 + f 1y v1 + f 2y v2 + f 3y v3(9.45)in terms of equivalent concentrated nodal forces.

The superscript (b) is used toindicate nodal-equivalent body force. Comparison of the last two equationsyields the nodal force components as(b)f ix = ␳ tN i FB X dx dyi = 1, 3A(b)f iy = ␳ t(9.46)N i FBY dx dyi = 1, 3AThe nodal force components equivalent to the applied body forces can also bewritten in the compact matrix form (b) FB Xf= ␳ t [N ] Tdx dy(9.47)FBYAWhile developed in the specific context of a constant strain triangular element inplane stress, Equation 9.47 proves to be a general result for two-dimensionalelements. A quite similar expression holds for three-dimensional elements.Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane Stress341EXAMPLE 9.2Determine the nodal force components representing the body force for the element ofExample 9.1, if the body force is gravitational attraction in the y direction, so thatFB XFB Y=0−386.4in./sec 2given the density of the element material is ␳ = 7.3 × 10 −4 slug/in.3.■ SolutionAs the x component of the body force is zero, the x components of the nodal force vectorwill be, too, so we need not consider those components. The y components are computedusing the second of Equation 9.46:(b)f iy = ␳ tN i F B Y dx dyi = 1, 3AFrom the previous example, the interpolation functions areN 1 (x , y) =1[2 − 2x ] = 1 − x2N 2 (x , y) =1[2x − y]2N 3 (x , y) =1y2We have, in this instance,(b)f1 y= ␳tF B Y N 1 dx dy = ␳ tF B Y (1 − x ) dx dyAA(b)f2 y = ␳ tF B Y N 2 dx dy = ␳ tAA(b)f3 y = ␳ tF B Y N 3 dx dy = ␳ tAFB Y(2x − y) dx dy2FB Yy dx dy2AThe limits of integration must be determined on the basis of the geometry of the area.

Inthis example, we utilize x as the basic integration variable and compute the y-integrationlimits in terms of x. For the element under consideration, as x varies between zero andone, y is the linear function y = 2x so the integrations become 1 2x(b)f1 y= ␳ t FB Y1(1 − x ) dy dx = ␳ t F B Y00␳ t FB Y== −0.0189 lb3012x 3 2x (1 − x ) dx = ␳ t Y x 2 −3 0Hutton: Fundamentals ofFinite Element Analysis3429. Applications in SolidMechanicsCHAPTER 9TextApplications in Solid Mechanics 1 2x(b)f2 y= ␳ t FB Y0=(b)f3 y© The McGraw−HillCompanies, 2004␳ t FB Y2␳ t FB Y=2␳ t FB Y1(2x − y) dy dx =22012x 2 dx0 2␳ t FB Y== −0.0189 lb33 1 2x0␳ t FB Yy dx dy =2012x 2 dx =␳ t FB Y= −0.0189 lb30showing that the body force is equally distributed to the element nodes.If we now combine the concepts just developed for the CST element in planestress, we have a general element equation that includes directly applied nodalforces, nodal force equivalents for distributed edge loadings, and nodal equivalents for body forces as [k]{␦} = { f } + f ( p) + f (b)(9.48)where the stiffness matrix is given by Equation 9.24 and the load vectors are asjust described.

Equation 9.48 is generally applicable to finite elements used inelastic analysis. As will be learned in studying advanced finite element analysis,Equation 9.48 can be supplemented by addition of force vectors arising from plastic deformation, thermal gradients or temperature-dependent material properties,thermal swelling from radiation effects, and the dynamic effects of acceleration.9.3 PLANE STRAIN: RECTANGULAR ELEMENTA solid body is said to be in a state of plane strain if it satisfies all the assumptions of plane stress theory except that the body’s thickness (length in the zdirection) is large in comparison to the dimension in the xy plane.

Mathematically, plane strain is defined as a state of loading and geometry such thatεz =∂w=0∂z␥x z =∂u∂w+=0∂z∂x␥yz =∂v∂w+=0∂z∂y(9.49)(See Appendix B for a discussion of the general stress-strain relations.)Physically, the interpretation is that the body is so long in the z direction thatthe normal strain, induced by only the Poisson effect, is so small as to be negligible and, as we assume only xy-plane loadings are applied, shearing strainsare also small and neglected.

(One might think of plane strain as in the exampleof a hydroelectric dam—a large, long structure subjected to transverse loadingonly, not unlike a beam.) Under the prescribed conditions for plane strain, theHutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.3 Plane Strain: Rectangular Elementconstitutive equations for the nonzero stress components become␴x =E[(1 − ␯)ε x + ␯e y ](1 + ␯)(1 − 2␯)␴y =E[(1 − ␯)ε y + ␯ε x ](1 + ␯)(1 − 2␯)␶x y =E␥x y = G ␥x y2(1 + ␯)(9.50)and, while not zero, the normal stress in the z direction is considered negligiblein comparison to the other stress components.The elastic strain energy for a body of volume V in plane strain is1Ue =(␴x ε x + ␴y ε y + ␶x y ␥x y ) dV(9.51)2Vwhich can be expressed in matrix notation asεx1Ue =[␴x ␴y ␶x y ] ε y dV2␥(9.52)xyVCombining Equations 9.50 and 9.52 with considerable algebraic manipulation,the elastic strain energy is found to be1EUe =[ εx ε y ␥x y ]2(1 + ␯)(1 − 2␯)V1−␯ ␯×0␯1−␯00 εx 0  εdV1 − 2␯  y ␥x y2(9.53)and is similar to the case of plane stress, in that we can express the energy as1Ue ={ε} T [D]{ε} dV2Vwith the exception that the elastic property matrix for plane strain is defined as1−␯ ␯E[D] =(1 + ␯)(1 − 2␯) 0␯1−␯000 1 − 2␯ 2(9.54)343Hutton: Fundamentals ofFinite Element Analysis3449.

Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsThe nonzero strain components in terms of displacements are∂uεx =∂x∂vεy =∂y␥x y =(9.55)∂u∂v+∂y∂xFor a four-node rectangular element (for example only), the column matrixof strain components is expressed as  u 1 ∂ N1 ∂ N2 ∂ N3 ∂ N4 u2 0000  ∂x∂x∂xu ∂x3 εx  ∂ N1 ∂ N2 ∂ N3 ∂ N4  u 4  0000{ε} = ε y=v ∂y∂y∂y∂ y   1␥x yv2  ∂ N1 ∂ N2 ∂ N3 ∂ N4 ∂ N1 ∂ N2 ∂ N3 ∂ N4 v3∂y∂x∂y∂y∂x∂x∂x∂x  v4(9.56)in terms of the interpolation functions and the nodal displacements. As is customary, Equation 9.56 is written as{ε} = [B]{␦}(9.57)with [B] representing the matrix of derivatives of interpolation functions and {␦}is the column matrix of nodal displacements.

Hence, total strain energy of anelement is11U e = {␦} T[B] T [D][ B] dV {␦} = {␦} T [k]{␦}(9.58)22Vand the element stiffness matrix is again given by (e) k=[B T ][D][ B] dV (e)(9.59)V (e)The interpolation functions for the four-node rectangular element per Equation 6.56 are1N 1 (r, s) = (1 − r )(1 − s)41N 2 (r, s) = (1 + r )(1 − s)4(9.60)1N 3 (r, s) = (1 + r )(1 + s)41N 4 (r, s) = (1 − r )(1 + s)4Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.3 Plane Strain: Rectangular Element(⫺1, 1)(1, 1)43s2br12(⫺1, ⫺1)(1, ⫺1)2aFigure 9.5 A rectangularelement of width 2a andheight 2b.with the natural coordinates defined as in Figure 9.5. To compute the strain components in terms of the natural coordinates, the chain rule is applied to obtain∂∂ ∂r1 ∂==∂x∂r ∂ xa ∂r∂∂ ∂s1 ∂==∂y∂s ∂yb ∂s(9.61)Performing the indicated differentiations, the strain components are found to be εx {ε} = ε y␥x ys−1 4a= 0r −14b1−s4a1+s4a0001+r4b1−r4b−1+r4b−1+s4a0r −14bs−14a001+r4b1−s4a1+r4b1+s4a− u1 u2 0u3u 1−r 4 v1 4b   v2 1 + s −v4a3 v4(9.62)showing that the normal strain ε x varies linearly in the y direction, normal strainε y varies linearly in the x direction, and shear strain ␥x y varies linearly in bothcoordinate directions (realizing that the natural coordinate r corresponds to thex axis and natural coordinate s corresponds to the y axis).From Equation 9.62, the [B] matrix is readily identified as1−s1+s1+ss−1−0000 4a4a4a4ar −11+r 1+r1−r 0000−[B] = 4b4b4b4b r −11+r 1+r1−rs−11−s1+s1+s −−4b4b4b4b4a4a4a4a(9.63)345Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanics346CHAPTER 9 (e) k=V (e)TextApplications in Solid Mechanicshence, the element stiffness matrix is given, formally, by[B T ] [D] [B] dV (e)s−1 4a 1−s 4a 1+s 4a 1+s1 1 −Etab4a=(1 + ␯)(1 − 2␯)0−1 −1 000© The McGraw−HillCompanies, 2004s−14a0×r −14b1−s4a0−1+s4a01+r4b1+r4b0000r −14b1+r−4b1+r4b1−r4b−1+s4a01−r4br −14b1+r−4b1+r4b1−r4bs−14a1−s4a1+s4a1+s−4a 1−␯ ␯000(1 + ␯)(1 − 2␯) 2(1 + ␯)␯1−␯00000r −14bs−14a1+r−4b1−s4a1+r4b1+s4a1−r4b1+s−4a dr ds(9.64)The element stiffness matrix as defined by Equation 9.64 is an 8 × 8 symmetricmatrix, which therefore, contains 36 independent terms.

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