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For the constant straintriangle, we already observed that the strains are constant over the element volume. Assuming that the elastic properties similarly do not vary, Equation 9.15331Hutton: Fundamentals ofFinite Element Analysis3329. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid Mechanicsbecomes simplyU(e)e= 1 (e) T␦[B] T [D][ B] ␦(e)2dV (e)V (e)=12 T (e) ␦(e)V [B] T [D][ B] ␦(e)(9.16)where V (e) is the total volume of the element.Considering the element forces to be as in Figure 9.2b (for this element formulation, we require that forces be applied only at nodes; distributed loads areconsidered subsequently), the work done by the applied forces can be expressedasW = f 1x u 1 + f 2x u 2 + f 3x u 3 + f 1y v1 + f 2y v2 + f 3y v3(9.17)and we note that the subscript notation becomes unwieldy rather quickly in thecase of 2-D stress analysis. To simplify the notation, we use the force notationf 1x f 2x f 3x{f} =(9.18)f 1y f 2y f 3yso that we can express the work of the external forces (using Equation 9.13) asW = {␦} T { f }(9.19)Per Equation 2.53, the total potential energy for an element is then = Ue − W =Ve T{␦} [B] T [D][ B]{␦} − {␦} T { f }2(9.20)If the element is a portion of a larger structure that is in equilibrium, then theelement must be in equilibrium.

Consequently, the total potential energy of theelement must be minimum (we consider only stable equilibrium), and for thisminimum, we must have mathematically∂=0∂ ␦ii = 1, 6(9.21)If the indicated mathematical operations of Equation 9.21 are carried out onEquation 9.20, the result is the matrix relationV e [B] T [D][ B]{␦} = { f }(9.22)and this matrix equation is of the form[k]{␦} = { f }(9.23)Hutton: Fundamentals ofFinite Element Analysis9. Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane Stresswhere [k] is the element stiffness matrix defined by[k] = V e [B] T [D][ B](9.24)and we must keep in mind that we are dealing with only a constant strain triangle at this point.This theoretical development may not be obvious to the reader.

To make theprocess more clear, especially the application of Equation 9.21, we examine theelement stiffness matrix in more detail. First, we represent Equation 9.20 as=1 T{␦} [k]{␦} − {␦} T { f }2(9.25)and expand the relation formally to obtain the quadratic function=1k 11 ␦12 + 2k 12 ␦1 ␦2 + 2k 13 ␦1 ␦3 + 2k 14 ␦1 ␦4 + · · · + 2k 56 ␦5 ␦6 + k 66 ␦622− f 1x ␦1 − f 2x ␦2 − f 3x ␦3 − f 1y ␦4 − f 2y ␦5 − f 3y ␦6(9.26)The quadratic function representation of total potential energy is characteristic oflinearly elastic systems. (Recall the energy expressions for the strain energy ofspring and bar elements of Chapter 2.)The partial derivatives of Equation 9.21 are then in the form∂= k 11 ␦1 + k 12 ␦2 + k 13 ␦3 + k 14 ␦4 + k 15 ␦5 + k 16 ␦6 − f 1x = 0∂ ␦1∂= k 21 ␦1 + k 22 ␦2 + k 23 ␦3 + k 24 ␦4 + k 25 ␦5 + k 26 ␦6 − f 2x = 0∂ ␦2(9.27)for example.

Equations 9.27 are the scalar equations representing equilibriumof nodes 1 and 2 in the x-coordinate direction. The remaining four equationssimilarly represent nodal equilibrium conditions in the respective coordinatedirections.As we are dealing with an elastic element, the stiffness matrix should besymmetric. Examining Equation 9.27, we should have k12 = k21, for example.Whether this is the case may not be obvious in consideration of Equation 9.24,since [D] is a symmetric matrix but [B] is not symmetric. A fundamental property of matrix multiplication (Appendix A) is as follows: If [G ] is a real, symmetric N × N matrix and [F ] is a real N × M matrix, the matrix triple product[F ] T [G ][ F ] is a real, symmetric M × M matrix.

Thus, the stiffness matrix asgiven by Equation 9.24 is a symmetric 6 × 6 matrix, since [D] is 3 × 3 and symmetric and [B] is a 6 × 3 real matrix.9.2.2 Stiffness Matrix EvaluationThe stiffness matrix for the constant strain triangle element given byEquation 9.24 is now evaluated in detail. The interpolation functions per333Hutton: Fundamentals ofFinite Element Analysis3349.

Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsEquation 6.37 are1[(x 2 y3 − x 3 y2 ) + ( y2 − y3 )x + (x 3 − x 2 ) y]2A1=(␣1 + ␤1 x + ␥1 y)2AN 1 (x , y) =1[(x 3 y1 − x 1 y3 ) + ( y3 − y1 )x + (x 1 − x 3 ) y]2A1=(␣2 + ␤2 x + ␥2 y)2AN 2 (x , y) =(9.28)1[(x 1 y2 − x 2 y1 ) + ( y1 − y2 )x + (x 2 − x 1 ) y]2A1=(␣3 + ␤2x + ␥3 y)2AN 3 (x , y) =so the required partial derivatives are∂ N11␤1=( y2 − y3 ) =∂x2A2A∂ N21␤2=( y3 − y1 ) =∂x2A2A1␤3∂ N3=( y1 − y2 ) =∂x2A2A∂ N11␥1=(x 3 − x 2 ) =∂y2A2A∂ N21␥2=(x 1 − x 3 ) =∂y2A2A(9.29)1␥3∂ N3=(x 2 − x 1 ) =∂y2A2AThe [B] (strain-displacement) matrix is then01  y2 − y3 y3 − y1 y1 − y2[B] =000x3 − x22Ax3 − x2 x1 − x3 x2 − x1 y2 − y3␤␤␤0001231 =0 0 0 ␥1 ␥2 ␥3 2A␥1 ␥2 ␥3 ␤1 ␤2 ␤30x1 − x3y3 − y10x2 − x1 y1 − y2(9.30)Noting that, for constant thickness, element volume is tA, substitution into Equation 9.24 results in␤1 0 ␥1  ␤2 0 ␥2  1 ␯0 ␤3 0 ␥3  ␯ 1 ␤1 ␤2 ␤3 0 0 0Et0 0 0 0 ␥ ␥ ␥ [k] =1234A(1 − ␯2 )  0 ␥1 ␤1  0 0 1 − ␯␥␥␥␤␤␤123123 0 ␥2 ␤2 20 ␥3 ␤3(9.31)Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane StressPerforming the matrix multiplications of Equation 9.31 gives the element stiffness matrix as22 ␤1 + C␥ 1Et[k] =4A(1 − ␯2 ) ␤1 ␤2 + C␥1 ␥2␤1 ␤3 + C␥1 ␥31+␯␤1 ␥12␤22 + C␥ 22␤2 ␤3 + C␥2 ␥3␯␤2 ␥1 + C␤1 ␥21+␯␤2 ␥22␤32 + C␥ 23␯␤3 ␥1 + C␤1 ␥3␯␤3 ␥2 + C␤2 ␥3␥ 21SY M+␯␤1 ␥2 + C␤2 ␥1␥1 ␥2 + C␤1 ␤2C␤12␥ 22 + C␤22335␯␤1 ␥3 + C␤3 ␥1 ␯␤2 ␥3 + C␤3 ␥2 1+␯␤3 ␥3 2␥1 ␥3 + C␤1 ␤3 ␥2 ␥3 + C␤2 ␤3 ␥ 23 + C␤32(9.32)where C = (1 − ␯)/2. Equation 9.32 is the explicit representation of the stiffness matrix for a constant strain triangular element in plane stress, presented forillustrative purposes. In finite element software, such explicit representation isnot often used; instead, the matrix triple product of Equation 9.24 is applieddirectly to obtain the stiffness matrix.9.2.3 Distributed Loads and Body ForceFrequently, the boundary conditions for structural problems involve distributedloading on some portion of the geometric boundary.

Such loadings may arisefrom applied pressure (normal stress) or shearing loads. In plane stress, these distributed loads act on element edges that lie on the global boundary. As a generalexample, Figure 9.3a depicts a CST element having normal and tangential loadspn and pt acting along the edge defined by element nodes 2 and 3. Element thickness is denoted t, and the loads are assumed to be expressed in terms of force perunit area.

We seek to replace the distributed loads with equivalent forces actingat nodes 2 and 3. In keeping with the minimum potential energy approach, theconcentrated nodal loads are determined such that the mechanical work is thesame as that of the distributed loads.(p)f 3y33n៝pt3(p)f 3xpypnpx12(a)1(p)f 2y2(b)12(c)Figure 9.3 Conversion of distributed loading to work-equivalentnodal forces.(p)f 2xHutton: Fundamentals ofFinite Element Analysis3369. Applications in SolidMechanicsCHAPTER 9Text© The McGraw−HillCompanies, 2004Applications in Solid MechanicsFirst, the distributed loads are converted to equivalent loadings in the globalcoordinate directions, as in Figure 9.3b, viap x = pn n x − pt n yp y = pn n y + pt n x(9.33)with nx and ny corresponding to the components of the unit outward normal vector to edge 2-3.

Here, we use the notation p for such loadings, as the units arethose of pressure. The mechanical work done by the distributed loads is33Wp = tp x u(x , y) dS + tp y v(x , y) dS(9.34)22where the integrations are performed along the edge defined by nodes 2 and 3.Recalling that interpolation function N 1 (x , y) is zero along edge 2-3, the finiteelement representations of the displacements along the edge areu(x , y) = N 2 (x , y)u 2 + N 3 (x , y)u 3v(x , y) = N 2 (x , y)v2 + N 3 (x , y)v3(9.35)The work expression becomes3Wp = tp x [N 2 (x , y)u 2 + N 3 (x , y)u 3 ] dS23+tp y [N 2 (x , y)v2 + N 3 (x , y)v3 ] dS(9.36)2and is of the form( p)( p)( p)( p)W p = f 2x u 2 + f 3x u 3 + f 2y v2 + f 3y v3(9.37)Comparison of the last two equations yields the equivalent nodal forces as3( p)f 2x=tp x N 2 (x , y) dS23( p)f 3x=tp x N 3 (x , y) dS2(9.38)3( p)f 2y = tp y N 2 (x , y) dS23( p)f 3x = tp y N 3 (x , y) dS2Hutton: Fundamentals ofFinite Element Analysis9.

Applications in SolidMechanicsText© The McGraw−HillCompanies, 20049.2 Plane Stress337as depicted in Figure 9.3c. Recalling again for emphasis that N 1 (x , y) is zeroalong the integration path, Equation 9.38 can be expressed in the compact form ( p) pxt dSf=[N ] T(9.39)pySwithN1 0N 2 0 N3 0 [N ]T =  0 N 1 0 N 20 N3( p)f 1x ( p) f 2x ( p) f 3x  ( p) f=( p)f 1y (p)f2y ( p) f 3y(9.40)(9.41)The reader is urged to write out in detail the matrix multiplication indicated inEquation 9.39 to ensure that the result is correct. Although developed in the context of the three-node triangular element, Equation 9.39 will prove generallyapplicable for two-dimensional elements and require only minor modificationfor application to three-dimensional problems.EXAMPLE 9.1Given the triangular plane stress element shown in Figure 9.4a, determine the nodalforces equivalent to the distributed loads shown via the method of work equivalence discussed previously.

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