Ebook A first course in the finite element method (4th edition) Part 2

(BQ) Part 1 book A first course in the infite element has contents: Structural dynamics and time dependent heat transfer, thermal stress, heat transfer and mass transport, plate bending element, plate bending element, axisymmetric elements,... and other contents.

of the Linear-Strain Triangle Equations

8

C H A P T E R

Introduction

In this chapter, we consider the development of the stiffness matrix and equations for

a higher-order triangular element, called the linear-strain triangle (LST) This element

is available in manycommercial computer programs and has some advantages overthe constant-strain triangle described in Chapter 6

The LST element has six nodes and twelve unknown displacement degrees offreedom The displacement functions for the element are quadratic instead of linear(as in the CST)

The procedures for development of the equations for the LST element follow thesame steps as those used in Chapter 6 for the CST element However, the number ofequations now becomes twelve instead of six, making a longhand solution extremelycumbersome Hence, we will use a computer to perform manyof the mathematicaloperations

After deriving the element equations, we will compare results from problemssolved using the LST element with those solved using the CST element The introduc-tion of the higher-order LST element will illustrate the possible advantages of higher-order elements and should enhance your general understanding of the conceptsinvolved with finite element procedures

Triangular Element Stiffness Matrix

and Equations

d

We will now derive the LST stiffness matrix and element equations The steps usedhere are identical to those used for the CST element, and much of the notation is thesame

398

Step 1 Select Element Type

Consider the triangular element shown in Figure 8–1 with the usual end nodes andthree additional nodes convenientlylocated at the midpoints of the sides Thus, acomputer program can automaticallycompute the midpoint coordinates once thecoordinates of the corner nodes are given as input

The unknown nodal displacements are now given by

We now select a quadratic displacement function in each element as

uðx; yÞ ¼ a1þ a2xþ a3yþ a4x2þ a5xyþ a6y2vðx; yÞ ¼ a7þ a8xþ a9yþ a10x2þ a11xyþ a12y2 ð8:1:2ÞAgain, the number of coefficients aið12Þ equals the total number of degrees of freedomfor the element The displacement compatibilityamong adjoining elements is satisfiedbecause three nodes are located along each side and a parabola is defined bythreepoints on its path Since adjacent elements are connected at common nodes, their dis-placement compatibilityacross the boundaries will be maintained

In general, when considering triangular elements, we can use a complete mial in Cartesian coordinates to describe the displacement field within an element

Using internal nodes as necessaryfor the higher-order cubic and quartic elements, weuse all terms of a truncated Pascal triangle in the displacement field or, equivalently,the shape functions, as shown byFigure 8–2; that is, a complete linear function isused for the CST element considered previouslyin Chapter 6 The complete quadraticfunction is used for the LST of this chapter The complete cubic function is used forthe quadratic-strain triangle (QST), with an internal node necessaryas the tenth node.The general displacement functions, Eqs (8.1.2), expressed in matrix form arenow

Alternatively, we can express Eq (8.1.3) as

where½M is defined to be the first matrix on the right side of Eq (8.1.3) The coefficients

a1through a12 can be obtained bysubstituting the coordinates into u and v as follows:8

377777775

coefficients based on a Pascal triangle

Solving for the ai’s, we have

3777775

u1

u6

v1

where½X is the 12 12 matrix on the right side of Eq (8.1.6) It is best to invert the

½X matrix byusing a digital computer Then the ai’s, in terms of nodal displacements,are substituted into Eq (8.1.4) Note that onlythe 6 6 part of ½X in Eq (8.1.6)reallymust be inverted Finally, using Eq (8.1.7) in Eq (8.1.4), we can obtain the gen-eral displacement expressions in terms of the shape functions and the nodal degrees offreedom as

qxþquqy

35

a1

a2

We observe that Eq (8.1.11) yields a linear strain variation in the element Therefore,the element is called a linear-strain triangle (LST) Rewriting Eq (8.1.11), we have

where½M0 is the first matrix on the right side of Eq (8.1.11) Substituting Eq (8.1.6)for the ai’s into Eq (8.1.12), we havefeg in terms of the nodal displacements as

Step 4 Derive the Element Stiffness Matrix and Equations

We determine the stiffness matrix in a manner similar to that used in Section 6.2 byusing Eq (6.2.50) repeated here as

½k ¼ððð

V

However, the½B matrix is now a function of x and y as given byEq (8.1.14) fore, we must perform the integration in Eq (8.1.16) Finally, the½B matrix is of theform

There-½B ¼ 12A

where the b’s and g’s are now functions of x and y as well as of the nodal coordinates,

as is illustrated for a specific linear-strain triangle in Section 8.2 byEq (8.2.8).The stiffness matrix is then seen to be a 12 12 matrix on multiplying the matrices

in Eq (8.1.16) The stiffness matrix, Eq (8.1.16), is verycumbersome to obtain inexplicit form, so it will not be given here However, if the origin of the coordinates

is considered to be at the centroid of the element, the integrations become amenable[9] Alternatively, area coordinates [3, 8, 9] can be used to obtain an explicit form

of the stiffness matrix However, even the use of area coordinates usuallyinvolvestedious calculations Therefore, the integration is best carried out numerically.(Numerical integration is described in Section 10.4.)

The element bodyforces and surface forces should not be automaticallylumped

at the nodes, but for a consistent formulation (one that is formulated from the sameshape functions used to formulate the stiffness matrix), Eqs (6.3.1) and (6.3.7), respec-tively, should be used (Problems 8.3 and 8.4 illustrate this concept.) These forces can

be added to anyconcentrated nodal forces to obtain the element force matrix Herethe element force matrix is of order 12 1 because, in general, there could be an xand a y component of force at each of the six nodes associated with the element Theelement equations are then given by

f1x

f1y

k11 k1; 12

k21 k2; 12

u1

v1

ð8:1:18Þ

Steps 5–7

Steps 5–7, which involve assembling the global stiffness matrix and equations, mining the unknown global nodal displacements, and calculating the stresses, areidentical to those in Section 6.2 for the CST However, instead of constant stresses ineach element, we now have a linear variation of the stresses in each element Commonpractice was to use the centroidal element stresses Current practice is to use theaverage of the nodal element stresses

To illustrate some of the procedures outlined in Section 8.1 for deriving an LSTstiffness matrix, consider the following example Figure 8–3 shows a specific LSTand its coordinates The triangle is of base dimension b and height h, with midsidenodes

a stiffness matrix

Using the first six equations of Eq (8.1.5), we calculate the coefficients a1

through a6byevaluating the displacement u at each of the six known coordinates ofeach node as follows:

Using Eqs (8.2.3) and (8.2.4), we can express the general displacement expressions interms of the shape functions as

term in Eq (8.2.3) For instance, collecting all terms that multiplybyu1in Eq (8.2.3),

we obtain N1 These shape functions are then given by

ey¼ 12A½g1v1þ g2v2þ g3v3þ g4v4þ g5v5þ g6v6

gxy¼ 12A½g1u1þ b1v1 6v6 The stiffness matrix for a constant-thickness element can now be obtained onsubstituting Eqs (8.2.8) into Eq (8.1.17) to obtain B, then substituting B into

Eq (8.1.16) and using calculus to set up the appropriate integration The explicit pression for the 12 12 stiffness matrix, being extremelycumbersome to obtain, isnot given here Stiffness matrix expressions for higher-order elements are found inReferences [1] and [2].

For a given number of nodes, a better representation of true stress and displacement isgenerallyobtained using the LST element than is obtained with the same number ofnodes using a much finer subdivision into simple CST elements For example, usingone LST yields better results than using four CST elements with the same number ofnodes (Figure 8–4) and hence the same number of degrees of freedom (except for thecase when constant stress exists)

We now present results to compare the CST of Chapter 6 with the LST of thischapter Consider the cantilever beam subjected to a parabolic load variation acting

as shown in Figure 8–5 Let E¼ 30 106psi, n¼ 0:25, and t ¼ 1:0 in

Table 8–1 lists the series of tests run to compare results using the CST and LSTelements Table 8–2 shows comparisons of free-end (tip) deflection and stress sxfor each element type used to model the cantilever beam From Table 8–2, we can ob-serve that the larger the number of degrees of freedom for a given type of triangularelement, the closer the solution converges to the exact one (compare run A-1 to runA-2, and B-1 to B-2) For a given number of nodes, the LST analysis yields some whatbetter results for displacement than the CST analysis (compare run A-1 to run B-1)

However, one of the reasons that the bending stress sx predicted bythe LSTmodel B-1 compared to CST model A-1 is not as accurate is as follows Recall thatthe stress is calculated at the centroid of the element We observe from the table thatthe location of the bending stress is closer to the wall and closer to the top for theCST model A-1 compared to the LST model B-1 As the classical bending stress is alinear function with increasing positive linear stress from the neutral axis for thedownward applied load in this example, we expect the largest stress to be at the verytop of the beam So the model A-1 with more and smaller elements (with eight ele-ments through the beam depth) has its centroid closer to the top (at 0.75 in from thetop) than model B-1 with few elements (two elements through the beam depth) withcentroidal stress located at 1.5 in from the top Similarly, comparing A-2 to B-2 weobserve the same trend in the results—displacement at the top end being more accu-ratelypredicted bythe LST model, but stresses being calculated at the centroid mak-ing the A-2 model appear more accurate than the LST model due to the locationwhere the stress is reported.

Although the CST element is rather poor in modeling bending, we observe fromTable 8–2 that the element can be used to model a beam in bending if a sufficientnumber of elements are used through the depth of the beam In general, both LSTand CST analyses yield results good enough for most plane stress/strain problems,provided a sufficient number of elements are used In fact, most commercial programsincorporate the use of CST and/or LST elements for plane stress/strain problems,

although these elements are used primarilyas transition elements (usuallyduring meshgeneration) The four-sided isoparametric plane stress/strain element is most fre-quentlyused in commercial programs and is described in Chapter 10.

Also, recall that finite element displacements will always be less than (or equalto) the exact ones, because finite element models are normallypredicted to be stifferthan the actual structures when the displacement formulation of the finite elementmethod is used (The reason for the stiffer model was discussed in Sections 3.10 and7.3 Proof of this assertion can be found in References [4–7]

Finally, Figure 8-6 (from Reference [8]) illustrates a comparison of CST andLST models of a plate subjected to parabolicallydistributed edge loads Figure 8–6shows that the LST model converges to the exact solution for horizontal displacement

at point A faster than does the CST model However, the CST model is quite able even for modest numbers of degrees of freedom For example, a CST modelwith 100 nodes (200 degrees of freedom) often yields nearly as accurate a solution asdoes an LST model with the same number of degrees of freedom

accept-In conclusion, the results of Table 8–2 and Figure 8–6 indicate that the LSTmodel might be preferred over the CST model for plane stress applications when rela-tivelysmall numbers of nodes are used However, the use of triangular elements ofhigher order, such as the LST, is not visiblyadvantageous when large numbers ofnodes are used, particularlywhen the cost of formation of the element stiffnesses,equation bandwidth, and overall complexities involved in the computer modeling areconsidered

results for triangular elements (Gallagher, R H Finite Element Analysis: Fundamentals,

d References

[1] Pederson, P., ‘‘Some Properties of Linear Strain Triangles and Optimal Finite ElementModels,’’ International Journal for Numerical Methods in Engineering, Vol 7, pp 415–430,1973

[2] Tocher, J L., and Hartz, B J., ‘‘Higher-Order Finite Element for Plane Stress,’’ Journal ofthe Engineering Mechanics Division, Proceedings of the American Societyof Civil En-gineers, Vol 93, No EM4, pp 149–174, Aug 1967

[3] Bowes, W H., and Russell, L T., Stress Analysis by the Finite Element Method for ticing Engineers, Lexington Books, Toronto, 1975

Prac-[4] Fraeijes de Veubeke, B., ‘‘Upper and Lower Bounds in Matrix Structural Analysis,’’ trix Methods of Structural Analysis, AGAR-Dograph 72, B Fraeijes de Veubeke, ed.,Macmillan, New York, 1964

Ma-[5] McLay, R W., Completeness and Convergence Properties of Finite Element DisplacementFunctions: A General Treatment, American Institute of Aeronautics and Astronautics Paper

No 67–143, AIAA 5th Aerospace Meeting, New York, 1967

[6] Tong, P., and Pian, T H H., ‘‘The Convergence of Finite Element Method in SolvingLinear Elastic Problems,’’ International Journal of Solids and Structures, Vol 3, pp 865–

879, 1967

[7] Cowper, G R., ‘‘Variational Procedures and Convergence of Finite-Element Methods,’’Numerical and Computer Methods in Structural Mechanics, S J Fenves, N Perrone, A R.Robinson, and W C Schnobrich, eds., Academic Press, New York, 1973

[8] Gallagher, R., Finite Element Analysis Fundamentals, Prentice Hall, Englewood Cliffs, NJ,1975

[9] Zienkiewicz, O C., The Finite Element Method, 3rd ed., McGraw-Hill, New York, 1977

8.3 For the element of Figure 8–3 (shown again as Figure P8–3) subjected to the uniformpressure shown acting over the vertical side, determine the nodal force replacementsystem using Eq (6.3.7) Assume an element thickness of t

Figure P8–3

8.4 For the element of Figure 8–3 (shown as Figure P8–4) subjected to the ing line load shown acting over the vertical side, determine the nodal force replace-ment system using Eq (6.3.7) Compare this result to that of Problem 6.9 Are theseresults expected? Explain.

linearlyvary-8.5 For the linear-strain elements shown in Figure P8–5, determine the strains ex;ey, and

gxy Evaluate the stresses sx;sy, and txyat the centroids The coordinates of the nodesare shown in units of inches Let E ¼ 30 106psi, n¼ 0:25, and t ¼ 0:25 in for bothelements Assume plane stress conditions apply The nodal displacements are given as

8.6 For the linear-strain element shown in Figure P8–6, determine the strains ex;ey, and

gxy Evaluate these strains at the centroid of the element; then evaluate the stresses

sx;sy, and txy at the centroid The coordinates of the nodes are shown in units ofmillimeters Let E¼ 210 GPa, n ¼ 0:25, and t ¼ 10 mm Assume plane stress con-ditions apply Use the nodal displacements given in Problem 8.5 (converted to milli-meters) Note that the b’s and g’s from the example in Section 8.2 cannot be used here

as the element in Figure P8–6 is oriented differentlythan the one in Figure 8–3

8.7 Evaluate the shape functions for the linear-strain triangle shown in Figure P8–7 Thenevaluate the B matrix Units are millimeters

8.8 Use the LST element to solve Example 7.2 Compare the results

8.9 Write a computer program to solve plane stress problems using the LST element

Figure P8–7Figure P8–6

is quite useful when symmetry with respect to geometry and loading exists about

an axis of the body being analyzed Problems that involve soil masses subjected tocircular footing loads or thick-walled pressure vessels can often be analyzed using theelement developed in this chapter

We begin with the development of the stiffness matrix for the simplest metric element, the triangular torus, whose vertical cross section is a plane triangle

axisym-We then present the longhand solution of a thick-walled pressure vessel to illustratethe use of the axisymmetric element equations This is followed by a description of sometypical large-scale problems that have been modeled using the axisymmetric element

In this section, we will derive the stiffness matrix and the body and surface force trices for the axisymmetric element However, before the development, we will firstpresent some fundamental concepts prerequisite to the understanding of the deriva-tion Axisymmetric elements are triangular tori such that each element is symmetricwith respect to geometry and loading about an axis such as the z axis in Figure 9–1.Hence, the z axis is called the axis of symmetry or the axis of revolution Each verticalcross section of the element is a plane triangle The nodal points of an axisymmetrictriangular element describe circumferential lines, as indicated in Figure 9–1

ma-In plane stress problems, stresses exist only in the x-y plane ma-In axisymmetricproblems, the radial displacements develop circumferential strains that induce stresses

sr, sy, sz, and trz, where r, y, and z indicate the radial, circumferential, and longitudinal

412

directions, respectively Triangular torus elements are often used to idealize the metric system because they can be used to simulate complex surfaces and are simple towork with For instance, the axisymmetric problem of a semi-infinite half-space loaded

axisym-by a circular area (circular footing) shown in Figure 9–2(a), the domed pressure vesselshown in Figure 9–2(b), and the engine valve stem shown in Figure 9–2(c) can be solvedusing the axisymmetric element developed in this chapter

(a) soil mass

(b) domed vessel (c) engine valve stem

modeled by axisymmetric elements, (b) a domed pressure vessel, and (c) an enginevalve stem

Because of symmetry about the z axis, the stresses are independent of the ycoordinate Therefore, all derivatives with respect to y vanish, and the displacementcomponent v (tangent to the y direction), the shear strains gry and gyz, and the shearstresses tryand tyzare all zero.

Figure 9–3 shows an axisymmetric ring element and its cross section to representthe general state of strain for an axisymmetric problem It is most convenient

to express the displacements of an element ABCD in the plane of a cross section

in cylindrical coordinates We then let u and w denote the displacements in the radialand longitudinal directions, respectively The side AB of the element is displaced anamount u, and side CD is then displaced an amount uþ ðqu=qrÞ dr in the radial direc-tion The normal strain in the radial direction is then given by

er¼qu

In general, the strain in the tangential direction depends on the tangential ment v and on the radial displacement u However, for axisymmetric deformation be-havior, recall that the tangential displacement v is equal to zero Hence, the tangentialstrain is due only to the radial displacement Having only radial displacement u, thenew length of the arc AB_isðr þ uÞ dy, and the tangential strain is then given by

Summarizing the strain/displacement relationships of Eqs (9.1.1a–d) in one equationfor easier reference, we have

37777

An axisymmetric solid is shown discretized in Figure 9–5(a), along with a typical angular element The element has three nodes with two degrees of freedom per node(that is, ui, wi at node i ) The stresses in the axisymmetric problem are shown inFigure 9–5(b)

The element displacement functions are taken to be

uðr; zÞ ¼ a1þ a2rþ a3z

ð9:1:3Þwðr; zÞ ¼ a4þ a5rþ a6z

so that we have the same linear displacement functions as used in the plane stress,constant-strain triangle Again, the total number of a’s (six) introduced in the

rotations of lines of element in the r-zplane

displacement functions is the same as the total number of degrees of freedom for theelement The nodal displacements are

Substituting the coordinates of the nodal points shown in Figure 9–5(a) into

Eq (9.1.6), we can solve for the ai’s in a manner similar to that in Section 6.2 Theresulting expressions are

1

wi

wjw

Performing the inversion operations in Eqs (9.1.7) and (9.1.8), we have

ai aj am

bi bj bm

gi gj gm

26

ai aj am

bi bj bm

gi gj gm

26

Nj¼ 1

2Aðamþ bmrþ gmzÞSubstituting Eqs (9.1.7) and (9.1.8) into Eq (9.1.6), along with the shape func-tion Eqs (9.1.12), we find that the general displacement function is

Rewriting Eq (9.1.15) with the ai’s as a separate column matrix, we have

or, rewriting Eq (9.1.17) in simplified matrix form,

26664

ð9:1:19Þwhere

Similarly, we obtain submatrices Bjand Bm by replacing the subscript i with j andthen with m in Eq (9.1.19) Rewriting Eq (9.1.18) in compact matrix form, we have

where½D is given by the first matrix on the right side of Eq (9.1.2) (As mentioned inChapter 6, for n¼ 0:5, a special formula must be used; see Reference [9].)

Step 4 Derive the Element Stiffness Matrix and Equations

The stiffness matrix is

½k ¼ððð

We can evaluate Eq (9.1.24) for½k by one of three methods:

1 Numerical integration (Gaussian quadrature) as discussed in

Chapter 10

2 Explicit multiplication and term-by-term integration [1]

3 Evaluate½B for a centroidal point ðr; zÞ of the element

If the triangular subdivisions are consistent with the final stress distribution (that

is, small elements in regions of high stress gradients), then acceptable results can beobtained by method 3

Distributed Body Forces

Loads such as gravity (in the direction of the z axis) or centrifugal forces in rotatingmachine parts (in the direction of the r axis) are considered to be body forces (asshown in Figure 9–6) The body forces can be found by

forces per unit volume

where Rb¼ o2rr for a machine part moving with a constant angular velocity o aboutthe z axis, with material mass density r and radial coordinate r, and where Zbis thebody force per unit volume due to the force of gravity.

Considering the body force at node i, we have

Multiplying and integrating in Eq (9.1.28), we obtain

where the origin of the coordinates has been taken as the centroid of the element, and

Rbis the radially directed body force per unit volume evaluated at the centroid of theelement The body forces at nodes j and m are identical to those given by Eq (9.1.30)for node i Hence, for an element, we have

For example, along the vertical face jm of an element, let uniform loads prand pz

be applied, as shown in Figure 9–7 along surface r¼ r We can use Eq (9.1.34)

written for each node separately For instance, for node j, substituting Nj from Eqs.(9.1.12) into Eq (9.1.34), we have

f fsjg ¼

ðz m

z j

12A

fsiand fsm, we obtain the total distribution of surface force to nodes i, j, and m as

of two methods that we use to determine the LST element stresses Either we mine the centroidal element stresses, or we determine the nodal stresses for the ele-ment and then average them The latter method has been shown to be more accurate

deter-in some cases [2]

Example 9.1

For the element of an axisymmetric body rotating with a constant angular velocity

o¼ 100 rev/min as shown in Figure 9–8, evaluate the approximate body forcematrix Include the weight of the material, where the weight density rwis 0.283 lb/in3.The coordinates of the element (in inches) are shown in the figure

We need to evaluate Eq (9.1.31) to obtain the approximate body force matrix.Therefore, the body forces per unit volume evaluated at the centroid of the elementare

Z ¼ 0:283 lb=in3

and by Eq (9.1.32), we have

Rb¼ o2rr¼ 100rev

min

2pradrev

fb1r ¼ ð2:44Þð0:187Þ ¼ 0:457 lb

fb1z¼ ð2:44Þð0:283Þ ¼ 0:691 lb ðdownwardÞBecause we are using the first approximation Eq (9.1.31), all r-directed nodalbody forces are equal, and all z-directed body forces are equal Therefore,

fb2r¼ 0:457 lb fb2z¼ 0:691 lb

To illustrate the use of the equations developed in Section 9.1, we will now solve anaxisymmetric stress problem

To illustrate the finite element solution for the cylinder, we first discretize the cylinderinto four triangular elements, as shown in Figure 9–10 A horizontal slice of the cylin-der represents the total cylinder behavior Because we are performing a longhandsolution, a coarse mesh of elements is used for simplicity’s sake (but without loss ofgenerality of the method) The governing global matrix equation is

Assemblage of the Stiffness Matrix

We assemble the½K matrix in the usual manner by superposition of the individualelement stiffness matrices For simplicity’s sake, we will use the first approximationmethod given by Eq (9.1.26) to evaluate the element matrices Therefore,

For element 1 (Figure 9–11), the coordinates are ri¼ 0:5, zi¼ 0, rj¼ 1:0, zj¼ 0,

rm¼ 0:75, and zm¼ 0:25 (i ¼ 1; j ¼ 2, and m ¼ 5 for element 1) for the coordinate axes as set up in Figure 9–10

We now evaluate½B , where ½B is given by Eq (9.1.19) evaluated at the centroid

of the element r¼ r, z ¼ z, and expanded here as

½B ¼ 12A

37775ð9:2:3Þ

where, using element coordinates in Eqs (9.1.11), we have

Substituting the results from Eqs (9.2.4) into Eq (9.2.3), we obtain

37

7in:1 ð9:2:5Þ

For the axisymmetric stress case, the matrix½D is given in Eq (9.1.2) as

377

6Þð1 þ 0:3Þ½1  2ð0:3Þ

377

37

Using Eqs (9.2.5) and (9.2.8), we obtain

½B T½D ¼57:7ð10

6Þ0:125

377775ð9:2:9Þ

Substituting Eqs (9.2.5) and (9.2.9) into Eq (9.2.2), we obtain the stiffness matrix forelement 1 as

377775

lbin:

ð9:2:10Þwhere the numbers above the columns indicate the nodal orders of degrees of freedom

in the element 1 stiffness matrix

For element 2 (Figure 9–12), the coordinates are ri¼ 1:0, zi¼ 0:0, rj¼ 1:0,

zj¼ 0:5, rm¼ 0:75, and zm¼ 0:25 (i ¼ 2, j ¼ 3, and m ¼ 5 for element 2) Therefore,

377775

lbin:

377775

lbin:

ð9:2:13Þ

lbin:

ð9:2:14ÞUsing superposition of the element stiffness matrices [Eqs (9.2.10) and (9.2.12)–(9.2.14)], where we rearrange the elements of each stiffness matrix in order of increas-ing nodal degrees of freedom, we obtain the global stiffness matrix as

lb in:

ð9:2:15ÞThe applied nodal forces are given by Eq (9.1.36) as

The results for nodal displacements are as expected because radial displacements

at the inner edge are equalðu1¼ u4Þ and those at the outer edge are equal ðu2 ¼ u3Þ

In addition, the axial displacements at the outer nodes and inner nodes are equalbut opposite in sign (w1 ¼ w4and w2¼ w3) as a result of the Poisson effect andsymmetry Finally, the axial displacement at the center node is zeroðw5¼ 0Þ, as itshould be because of symmetry

By using Eq (9.1.22), we now determine the stresses in each element as

fsg ¼ ½D ½B fdg

ð9:2:18ÞFor element 1, we use Eq (9.2.5) for½B , Eq (9.2.8) for ½D , and Eq (9.2.17) for fdg

in Eq (9.2.18) to obtain

sr¼ 0:338 psi sz¼ 0:0126 psi

sy¼ 0:942 psi trz¼ 0:1037 psiSimilarly, for element 2, we obtain

sr¼ 0:105 psi sz¼ 0:0747 psi

sy¼ 0:690 psi trz ¼ 0:000 psiFor element 3, the stresses are

sr¼ 0:337 psi sz¼ 0:0125 psi

sy¼ 0:942 psi trz ¼ 0:1037 psiFor element 4, the stresses are

sr¼ 0:470 psi sz¼ 0:1493 psi

sy¼ 1:426 psi trz¼ 0:000 psiFigure 9–13 shows the exact solution [10] along with the results determined hereand the results from Reference [5] Observe that agreement with the exact solution

is quite good except for the limited results due to the very coarse mesh used inthe longhand example, and in case 1 of Reference [5] In Reference [5], stresses havebeen plotted at the center of the quadrilaterals and were obtained by averaging the

Numerous structural (and nonstructural) systems can be classified as axisymmetric.Some typical structural systems whose behavior is modeled accurately using theaxisymmetric element developed in this chapter are represented in Figures 9–14,9–15, and 9–17

Figure 9–14 illustrates the finite element model of a steel-reinforced concretepressure vessel The vessel is a thick-walled cylinder with flat heads An axis of sym-metry (the z axis) exists such that only one-half of the r-z plane passing throughthe middle of the structure need be modeled The concrete was modeled by using theaxisymmetric triangular element developed in this chapter The steel elements werelaid out along the boundaries of the concrete elements so as to maintain continuity

(or perfect bond assumption) between the concrete and the steel The vessel was thensubjected to an internal pressure as shown in the figure Note that the nodes alongthe axis of symmetry should be supported by rollers preventing motion perpendicular

to the axis of symmetry

Figure 9–15 shows a finite element model of a high-strength steel die used in athin-plastic-film-making process [7] The die is an irregularly shaped disk An axis ofsymmetry with respect to geometry and loading exists as shown The die was modeled

by using simple quadrilateral axisymmetric elements The locations of high stress were

pressure

of primary concern Figure 9–16 shows a plot of the von Mises stress contours for thedie of Figure 9–15 The von Mises (or equivalent, or effective) stress [8] is often used

as a failure criterion in design Notice the artificially high stresses at the location ofload F as explained in Section 7.1

(Recall that the failure criterion based on the maximum distortion energy theoryfor ductile materials subjected to static loading predicts that a material will fail ifthe von Mises stress reaches the yield strength of the material.) Also recall fromEqs (6.5.37) and (6.5.38), the von Mises stress svmis related to the principal stresses

by the expression

svm¼ 1ffiffiffi

2p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðs1 s2Þ2þ ðs2 s3Þ2þ ðs3 s1Þ2

q

ð9:3:1Þ

North Holland Physics Publishing, Amsterdam)

Figure 9–16 von Mises stress contour plot of axisymmetric model of Figure 9–15(also producing a radial inward deflection of about 0.015 in.)

431

where the principal stresses are given by s1, s2, and s3 These results were obtainedfrom the commercial computer code ANSYS [12].

Other dies with modifications in geometry were also studied to evaluate the mostsuitable die before the construction of an expensive prototype Confidence in the ac-ceptability of the prototype was enhanced by doing these comparison studies Finally,Figure 9–17 shows a stepped 4130 steel shaft with a fillet radius subjected to an axialpressure of 1000 psi in tension Fatigue analysis for reversed axial loading required

an accurate stress concentration factor to be applied to the average axial stress of

1000 psi The stress concentration factor for the geometry shown was to be mined Therefore, locations of highest stress were necessary Figure 9–18 shows theresulting maximum principal stress plot using a computer program [11] The largestprincipal stress was 1932.5 psi at the fillet Other examples of the use of the axisym-metric element can be found in References [2]–[6]

deter-In this chapter, we have shown the finite element analysis of axisymmetric tems using a simple three-noded triangular element to be analogous to that of thetwo-dimensional plane stress problem using three-noded triangular elements as

developed in Chapter 6 Therefore, the two-dimensional element in commercial puter programs with the axisymmetric element selected will allow for the analysis ofaxisymmetric structures.

com-Finally, note that other axisymmetric elements, such as a simple quadrilateral(one with four corner nodes and two degrees of freedom per node, as used in thesteel die analysis of Figure 9–15) or higher-order triangular elements, such as in Refer-ence [6], in which a cubic polynomial involving ten terms (ten a’s) for both u and w,could be used for axisymmetric analysis The three-noded triangular element wasdescribed here because of its simplicity and ability to describe geometric boundariesrather easily

[1] Utku, S., ‘‘Explicit Expressions for Triangular Torus Element Stiffness Matrix,’’ Journal ofthe American Institute of Aeronautics and Astronautics, Vol 6, No 6, pp 1174–1176,June 1968

[2] Zienkiewicz, O C., The Finite Element Method, 3rd ed., McGraw-Hill, London, 1977.[3] Clough, R., and Rashid, Y., ‘‘Finite Element Analysis of Axisymmetric Solids,’’ Journal ofthe Engineering Mechanics Division, American Society of Civil Engineers, Vol 91, pp 71–85,Feb 1965

[4] Rashid, Y., ‘‘Analysis of Axisymmetric Composite Structures by the Finite ElementMethod,’’ Nuclear Engineering and Design, Vol 3, pp 163–182, 1966.

[5] Wilson, E., ‘‘Structural Analysis of Axisymmetric Solids,’’ Journal of the American Institute

of Aeronautics and Astronautics, Vol 3, No 12, pp 2269–2274, Dec 1965

[6] Chacour, S., ‘‘A High Precision Axisymmetric Triangular Element Used in the Analysis ofHydraulic Turbine Components,’’ Transactions of the American Society of MechanicalEngineers, Journal of Basic Engineering, Vol 92, pp 819–826, 1973

[7] Greer, R D., The Analysis of a Film Tower Die Utilizing the ANSYS Finite Element age, M.S Thesis, Rose-Hulman Institute of Technology, Terre Haute, IN, May 1989.[8] Gere, J M., Mechanics of Materials, 5th ed., Brooks/Cole Publishers, Pacific Grove, CA,2001

Pack-[9] Cook, R D., Malkus, D S., Plesha, M E., and Witt, R J., Concepts and Applications ofFinite Element Analysis, 4th ed., Wiley, New York, 2002

[10] Cook, R D., and Young, W C., Advanced Mechanics of Materials, Macmillan, NewYork, 1985

[11] Algor Interactive Systems, 150 Beta Drive, Pittsburgh, PA 15238

[12] Swanson, J A ANSYS-Engineering Analysis System’s User’s Manual, Swanson AnalysisSystems, Inc., Johnson Rd., P.O Box 65, Houston, PA 15342

9.1 For the elements shown in Figure P9–1, evaluate the stiffness matrices using Eq

for each element

9.2 Evaluate the nodal forces used to replace the linearly varying surface traction shown

in Figure P9–2 Hint: Use Eq (9.1.34)

(1, 2)

(2, 0) (0, 0)

9.3 For an element of an axisymmetric body rotating with a constant angular velocity

o¼ 20 rpm as shown in Figure P9–3, evaluate the body-force matrix The nates of the element are shown in the figure Let the weight density rwbe 0.283 lb/in3

coordi-9.4 For the axisymmetric elements shown in Figure P9–4, determine the element stresses.Let E 6psi and n¼ 0:25 The coordinates (in inches) are shown in the figures,and the nodal displacements for each element are u1¼ 0:0001 in., w1¼ 0:0002 in.,

u2¼ 0:0005 in., w2¼ 0:0006 in., u3¼ 0, and w3 ¼ 0

9.5 Explicitly show that the integration of Eq (9.1.35) yields the j surface forces given by

Eq (9.1.36)

9.6 For the elements shown in Figure P9–6, evaluate the stiffness matrices using Eq.(9.2.2) The coordinates (in millimeters) are shown in the figures Let E¼ 210 GPaand n¼ 0:25 for each element

(c)

(0, 2) 3

2 (2, 0) (0, 0)

1

Figure P9–4

Figure P9–6

Figure P9–3

9.7 For the axisymmetric elements shown in Figure P9–7, determine the element stresses.Let E ¼ 210 GPa and n ¼ 0:25 The coordinates (in millimeters) are shown in thefigures, and the nodal displacements for each element are

9.8 Can we connect plane stress elements with axisymmetric ones? Explain

9.9 Is the three-noded triangular element considered in Section 9.1 a constant strain ment? Why or why not?

symmetry?

9.11 How would you evaluate the circumferential strain, ey, at r¼ 0? What is this strain interms of the a’s given in Eq (9.1.3) Hint: Elasticity theory tells us that the radialstrain must equal the circumferential strain at r¼ 0

9.12 What will be the stresses srand syat r¼ 0? Hint: Look at Eq(9.1.2) after consideringproblem 9.11

Solve the following axisymmetric problems using a computer program

9.13 The soil mass in Figure P9–13 is loaded by a force transmitted through a circularfooting as shown Determine the stresses in the soil Compare the values of srusing an

Figure P9–7

Figure P9–13

axisymmetric model with the sy values using a plane stress model Let E¼ 3000 psiand n¼ 0:45 for the soil mass.

for the steel liner The steel liner is 2 in thick Let the pressure p equal 500 psi

9.15 Perform a stress analysis of the concrete pressure vessel with the steel liner shown inFigure P9–15 Let E¼ 30 GPa and n ¼ 0:15 for the concrete, and let E ¼ 205 GPaand n¼ 0:25 for the steel liner The steel liner is 50 mm thick Let the pressure p equal

700 kPa

Figure P9–14

1250 mm

Concrete Steel liner

400 mm

325 mm

750 mm