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Applied
FitBute Element
Attalysis
JLarry «/. SegerlindAssociate Professor
Department of Agricultural Engineering
Michigan State University
JOHN Wn.EY & SONS
New York • Chichester • Brisbane • Toronto • Singapore
6Tarsian af
JVancircular
Sections
The preceding four chapters cover the discretiz-ing of the body, the interpolating polynomial fora single element, the interpolation polynomialsfor a discretized body, and the derivation of thefinite element equations. Each of these chapterscontains basic information that is related to the
finite element method. This chapter is a transitionfrom the theory to the implementation. Its objective is to illustrate the steps involved in implementing the finite element method. This objectiveis accomplished by obtaining a numerical solution for a problem that involves the torsion of anoncircular section.
There are two reasons for selecting the torsionof noncircular sections to illustrate the im
plementation of the finite element method. Theelement equations are relatively straightforwardto derive. The [K] matrix is easy to calculate andthe boundary integrals do not have to beevaluated because the boundary values are prescribed. These properties prevent the evaluationof the element equations from detracting fromour main objective. The other reason is that thetorsion of a noncircular section involves conceptsimportant to both mechanics-oriented and field-
87
88 Torsion of Nonclrcular Sections
problem-oriented individuals. The theory will be familiar to an individual with amechanics background while the governing differential equation is similar tothose governing heat transfer and groundwater flow.
6.1 GENERAL THEORY, TORSION OF
NONCIRCULAR SECTIONS
One theory* for calculating the shear stresses in a noncircular shaft subjectedto a twisting moment T about the z axis (Fig. 6.1) states that the shear stresscomponents at any point can be calculated using
9<J)
dy
d(j>
dx(6.1)
where (p is a stress function. The governing differential equation is
1
G dx^
1 9G +261-0 (6.2)
with
</) = 0 (6.3)
on the boundary. The physical parameters in (6.2) are the shear modulus of the
Fig. 6.1 The shear stress components in a noncircularsection subjected to a torque loading.
*There are two theories for calculating the shear stresses that result when a noncircular shaft issubjected to torsion. One of them was developed by St. Venant; the other was proposed by Prantl.Both theories are discussed by Fung (1965). The minimization formulation of Prandtl's theory,which is used in this chapter, is also discussed by Temoshenko and Goodier (1970).
General Theory, Torsion of Nonclrcular Sections 89
material, G, (N/cm^), and the angle of twist per unit length, 6 (rad/cm). Thisformulation does not have the applied torque, T (N ■ m) in the governing equation. Instead, T is calculated once <p is known, using
^=2f <j>dA''Area
(6.4)
The stress function represents a surface covering the cross section of the shaft(Fig. 6.2). The twisting moment is proportional to the volume under this surfacewhile the shear stresses are related to the gradients in the a and y coordinatedirections.
Fig. 6.2 The <f) surface anda related shear
stress component.
Equation 6.2 is usually written
d^cl>^ +—-+269 = 0
dx 9y(6.5)
when the shaft is composed of a single material. The integral formulation of(6.5) is
dV
which according to our discussion in Chapter 5 can be rewritten as
= j\\{gV[D]{g]-(2Ge)<^ dV
(6.6)
(6.7)
90 Torsion of Nonclrcuiar Sections
where
9</)
dx
d<j>
dy
and [/)] =K.. 0
0 K„
The column vector (g) is related to the shear stress components in thisapplication, while [D] becomes the identity matrix since K^^ = K^^ = 1. Theminimization of x with respect to (O) results in the set of linear equations
E E
Sf S f {'^G9)dV (6.8)I/(^) •/ l/(^)
e = 1 e=\
where is defined by (4.1) and the gradient matrix, is defined by(5.39). The evaluation of (6.8) commences as soon as we select a shape anddivide it into elements.
6.2 ASSEMBLAGE OF THE ELEMENT MATRICES
The square shaft (Fig. 6.3a) will be used to illustrate the evaluation and theassemblage of the element matrices into a set of linear equations. This shaft hasfour axes of symmetry, therefore, only one eighth of the total cross section needsto be analyzed. This fractional portion of the cross section is divided into fourelements as shown in Fig. 6.3b. Four elements are not sufficient to obtain anaccurate answer, but they are enough to illustrate the assemblage technique thatis one of our objectives.
Following the methods of Chapter 4, the element interpolating polynomialsare
<^(0 = + 0^)3 + + 005 + 006
C|>(2) = 00, + 7V2^2^02 + + 004 + yVi2^05 + 00^ (6.9)
cj>(3) = GO J -h A^2^^>02 + 003 + A^i^^04 + A^5^^^05 + 00^
= 00, + 002 4- 003 + A^i'^^04 4 A^i^^05 + yV6^^^06
The general equation for the element stiffness matrix is
J y
since it has already been noted that [/)] = [/] for the torsion of a noncircular
Assemblage of the Element Matrices 91
Axes of symmetry
G = 0.8 X 10^ N/cm20 = 1 deg in ICQ cm
Shaft is 1 cm2
(4)
f * i)
(3)
X (1) . (2)
A
0.25 cm
+0.25 cm
V
-0.25 cm - -0.25 cm -
Fig. 6.3 Element subdivision for the torsion
of a square shaft.
section. Evaluation of involves differentiating with respect to x andy. Confining our attention to element one
dx dx0
aM"9x
0 0
12^(0 [M" 0 0 o]
d(j>(1)
dy dy dy dy0 0
2A(1)[cS" 0 4'^ 0 o]
(6.10a, b)
92 Torsion of Noncircuiar Sections
The gradient matrix is
[5<"] =
The area for this element is
0 0 0
(1) .(1)^2 0 cV^ 0 0
32and
2/4">= 16
The b and c coefficients are
b\'^=Y2-Y,= -0.25 c\'^=X^-X2 =
b['^=Y,-Y,= 0.25 c(" = A',-X4 =
M'>= 7,-72= 0 4" = ̂2-'^i =
0
-0.25
0.25
-4
0
Substituting these values into (6.11) yields
[5<"] =
and the product is
4
IS
[fi<'>]^[5">]=
or
0 0 0 0
4 0 4 0 0
4 0
4 -4
0 0
0 4
0 0
0 0
■4 4 0 0 0 00 -4 0 4 0 0
16 - 16 0 0 0 0- 16 32 0 -16 0 0
0 0 0 0 0 00 - 16 0 16 0 00 0 0 0 0 00 0 0 0 0 0
(6.11)
(6.12)
(6.13)
The element stiffness matrix is the integral of (6.13). Since the matrix productconsists of constant values, it may be removed from under the
Assemblage off the Element Matrices 93
integral yielding
(1)
assuming a unit thickness. Recalling that 1/32 and combining this with(6.13) yields
1 -1 0 0 0 0
-1 2 0 -1 0 0
0 0 0 0 0 0
0 -1 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
(6.14)
The evaluation of the volume integral
(/'')= [ 2G<"l9^ ̂ JyO)
0
yvi'*0
0
dV
is straightforward if we employ the area coordinate system discussed in Chapter3. Defining the area coordinates as
L, = Af)'\ T2=A')" and =
the volume integral becomes
^ L,
(6.15)
^ ' JyO)dV (6.16)
Assuming a unit thickness and using the area integral for area coordinates,(3.43), yields
94 Torsion of Nonclrcular Sections
(/")-
Substitution of the values of 0, and yields*
(6.17)
in
29.07
29.07
0
29.07
0
0
The system of equations for element one is
= (/'>}
or
(6.19a )
A similar set of equations can be calculated for each of the other elementsusing an identical procedure. The resulting matrices are given below.
(6.18)
1 -1 0 0 0 0 29.07"
-1 2 0 -1 0 0 29.07
1 0 0 0 0 0 0 $3 0
2 0 -1 0 1 0 0 <I'4 29.07
0 0 0 0 0 0 <^5 0
0 0 0 0 0 0 ^>6 0
= (/«'}
0 0
-1
0
0
0
0
0
0 0
-1 0
0 0
0 0
0 0
-1 0
0 0
1 0
0 0
' 0
29.07
29.07
0
% 29.07
^6 0
(6.19b )
*The units are N/cm^ for G, cm^ for and rad/cm for 0. ̂ = 7r/180x 1/100, since the shafttwists 1 deg in 100 cm.
Assemblage of the Element Matrices 95
[ A:® ]{f } = {/"*}
"0 0 0 0 0 0 ' 0
0 1 0 -1 0 0 <^2 29.07
1 0 0 0 0 0 0 % 0
2 0 1 0 2 -1 0 29.07
0 0 0 -1 1 0 29.07
0 0 0 0 0 0 0
0 0 0 0 0 0 ' 00 0 0 0 0 0 0
1 0 0 0 0 0 0 $3 0
2 0 0 0 1 -11 0 $4 29.07
0 0 0 -1 :! -1 % 29.07
0 0 0 0 -11 1 ^6 29.07
(6.19c )
(6.19d )
The final system of equations is the algebraic sum of the element equations,which is j
1 -1 0 0 0 P $, 1-1 4 -1 -2 0 b
1 0 - 1 2 0 -1 0
2 0 -2 0 4 -2 0
0 0 -1 -2 4 -1
0 0 0 0 - 1 1
C.O-
29.07
87.22
29.07
87.22
87.22
29.07
(6.20 )
The nodal valu^ $3, $5, and are zero,.sin€h these nodes are on the externalboundary. Modifibaftbtrxif (6720) and solution yields
<I), = 218.16 <f>3 = 0
^>2=160 <I>5 = 0 (6.21)
a . ■
-\ o
•- >1 "' i-
$4= 123.63 $6 = 0
The modification of (6.20) is discussed in the next chapter where we consider thecomputer techniques that related to the implementation of the finite element
method. The (j) surface for this set of nodal values is presented in Fig. 6.4.
96 Torsion of Nonclrcular Sections
4), = 218.16
= 123.63
Fig. 6.4 The nodal values for the four-element torsionproblem.
The determination of the nodal values is a major step in the solution of theproblem. In most problems, however, a set of element resultants must becalculated. The shear stress values in each element and the twisting torque Twhich produces the angle of twist 6 are of interest in the present example. Thecalculation of the element resultants is discussed in the next section.
6.3 CONVENTIONAL ELEMENT RESULTANTS
The gradients of the nodal parameter, (f>, are the important quantities in thetorsion problem because the shear stresses are related to these gradients by
d<f)
dyand
dxj)
a;c
The shear stress values are readily calculated because the gradient matrix for
Conventional Element Resultants 97
each element has already been evaluated. The gradient matrix for element one is(6.11).
d(j>
dx
d<j)
dy
2A^'^
6^" 0 0 0
c{" c^" 0 4" 0 0
Combining (6.12) and (6.21)
{0 =-4 4 0 0 0 0
0 -4 0 4 0 0
218.16
160.0
0
123.63
0
0
-232.6
- 145.4
therefore,
4^)= =-145.4 N/cm^dy ■ '
40=-.^dx
232.6 N/cm^
The stress components for the other elements are
Element 2 = ON/cm^. T = 639.4 N/cm^
Element 3 = - 145.4 N/cm^, t = 494.0 N/cm^
Element 4 = ON/cm^. T- =494.0 N/cm^
These values are shown schematically in Fig. 6.5.
98 Torsion of Nonclrcular Sections
Positive shear stress
components
Fig. 6.5 The element shearstress components
for the to u r-
element torsion
problem. All values
are in Newtons persquare centimeter.
The shear stress values calculated for each element are constant over theregion of the element because the element interpolation polynomial was linear inj\: andy. The failure to obtain gradients that vary with x andy within an elementis a definite disadvantage of the simplex elements.
There are three ways to improve the stress values obtained for this example.First, a larger number of elements can be used. As the size of the elementdecreases, the existence of a constant value within the element becomes morerealistic. An alternative approach is to use a triangular element with more nodesand an interpolating polynomial with quadratic and cubic terms. Differentiationwill then yield gradients that are a function of the coordinate directions. A thirdapproach is to utilize conjugate approximation theory. This theory makespossible the determination of the stress values at the nodal points and as afunction of x and y within the element. The implementation of this theory isdiscussed in the next section.
Another resultant of interest is the magnitude of the twisting torque T, whichis given in (6.4) as
^=2f (i>dA''Area
This integral is equivalent to
■=2/.A
dArea
(6.22)
Conventional Element Resultants 99
where the equations for are given in (6.9). Starting with element one
if <l>^'^dA=2f fyvl" /VjO 0 A'i" 0 OlJ^O)
$2
<1.3dA (6.23)
or
if <p('>dA=l[^Vf [N^'^VdAJa(^)
(6.24)
which is identical to the integral in (6.16). We can immediately conclude that
if <#,('>^^ = ̂[<1)]•^(1)
2A(1)
-(0, + (^2 + ̂4) (6.25)
Substituting the nodal values yields
2 I = ̂4—(218.16+ 160+ 123.63)= ̂ 4—3 3
Likewise, for the other elements,
2A^^^2
(501.79)
= ̂($2 + <E'3 + ̂5)= ̂ (160)
if <|>®^(^>2 + <f5+ <i>4) = ̂(283.63)Ja O) J
r (A\ 2A^^^ 2A^^^2jy^dA = ̂(4.4 +$5 + $,)= ̂ (123.63)
Summing these relationships and noting that the area of each element is the
100 Torsion of Nonclrcuiar Sections
same produces
E
T=y f = ̂(501.79+ 160 + 283.63+ 123.63)
r= ̂(1069.05)=^^We must also recall that the region we have subdivided into elements is only oneeighth of the total cross section; therefore, the total twisting moment is
(1069)Moment = 87=8 = 178.16 N-cm
48
It takes a torque of 178 N-cm to produce a twist of 1 deg in a 1-cm square steelshaft that is 100 cm long. The accuracy of this result is highly questionable,however, because of the coarseness of the grid. In fact, the theoretical value* is196.3 N-cm. Our answer is 9.5 percent below this value.
6.4 CONSISTENT ELEMENT RESULTANTS
The failure to obtain gradients that are a function of ;c and y is a definitedisadvantage of the linear interpolation functions. The gradient and any quantity related to the gradient has a constant value within an element. Variousaveraging techniques have been used to obtain a value of the desired quantity ata node. One procedure uses the average of the gradient values in each elementthat surrounds the node. The correct procedure for obtaining nodal values of theelement resultants is to employ conjugate approximation theory (Oden andBrauchli, 1971). This theory yields element resultant values consistent with theapproximating polynomial(s) for a scalar or vector quantity. .The derivation of the conjugate approximation theory is beyond the scope of
this book. The implementation of this theory, however, is straightforward andwill be illustrated relative to the four-element torsion problem that we have beenanalyzing.The nodal values of the element resultants are obtained by solving the system
of equations
[C]{a} = {7?} (6.26)
*The relationship between the applied torque and the angle of twist for a square of dimension 2ais given by 7=0.1406 (la)"^ (Timoshenko and Goodier, 1970, equation 170, page 313). For ourexample, 2a = 1 and 7=0.1406 Gff= 196.3 N-cm.
Consistent Element Resultants 101
where [C] and (/?) are the sum of element matrices defined by
[cW]= rJ y
and
[/^n= fJy
where a is the conventional element resultant. The evaluation of
no problems because it is similar to (6.8), since a is constant within anThe evaluation of is performed most easily using area coordinates,ing our attention to element one, yields
(6.27)
(6.28)
presents
element.
Confin-
[,yV<'>] = [L, L, 0 L,
The product becomes
[;V<')]^[7V<'>] =
Integration over the area using (3.43) yields
0 0]
0 L.Lj 0 0
L,L, Ll 0 ^2^3 0 0
0 0 0 0 0 0
L1L3 L2L3 0 Ll 0 0
0 0 0 0 0 0
0 0 0 0 0 0
['"'1 = 1?
2 1
1 2
0
0
1 0 0
1 0 0
0 0 0 0 0 0
1 1 0 2 0 0
0 0 0 0 0 0
0 0 0 0 0 0
assuming a unit thickness.The column vector for element one is
(6.29)
(1)1 ={,(.)j
r r '77.67"1 1 77.67
0 0. = ̂(0. 0
1 3 1 77.67
0 0 0
0 0 0
(6.30)
102 Torsion of Nonclrcular Sections
The shear stress component was used in the previous calculations becauseit has the largest numerical value within each element. The shear stress component could also be considered. The only quantity that changes is thecolumn vector (/?), which would have to be reevaluated using the numericalvalues for
The element matrices for each element are summarized in (6.29) to (6.33).
[c®] =4(2)
w
.(3)1 = ̂12
[C(3)] =
4(4)
IT
0 0 0 0 0 0 0 '0 2 1 0 1 0 213
0
0
1
0
2
0
0
0
1
0
0
0
213
0■
0 1 1 0 2 0 213
0 0 0 0 0 0 , 0 .
0 0 0 0 0 0" ' 00 2 0 1 1 0 164.67
0
0
0
1
0
0
0
2
0
1
0
0= ̂(3). 0
164.67
0 1 0 1 2 0 164.67
0 0 0 0 0 0 0
0 0 0 0 0 0' ' 00 0 0 0 0 0 0
0
0
0
0
0
0
0
2
0
1
0
1
0
164.67
0 0 0 1 2 1 164.67
0 0 0 1 1 2 164.67
(6.31)
(6.32)
(6.33)
The final system of equations is obtained by summing the element equationswhich yields
J_12
2 1 0 1 0 0
1 6 1 2 2 0
0 1 2 0 1 0
1 2 0 6 2 1
0 2 1 2 6 1
0 0 0 1 1 2
77.67'455.35
213
<34 407.01
<35 542.34
<36 164.67
(6.34)
The element areas cancel since all elements have the same area.
Consistent Element Resultants 103
The nodal resultant values are
(a) ̂ = [ 70.9,436.5,724.1,353.6,671.4,475.5 ]
It is also possible to obtain equations for the element resultants that give thevariation of these values over the element area. There is a very limited use forthis information and the details will not be presented here.The use of conjugate approximation theory requires the solution of another
system of algebraic equations that is identical in size to those used to obtain thenodal values. This is a definite disadvantage when we are solving problemsinvolving a large number of elements. An alternative to solving the completesystem of equations is discussed in Chapter 7 when we reconsider the torsion ofa square shaft with an increased number of elements.The final steps of the finite element method have been illustrated in this
chapter while we solved a specific problem. We have shown how the elementmatrices are evaluated and how the element resultants are determined once the
nodal values are known.
We shall discuss some of the computer techniques used when we implementthe finite element method in the next chapter. We shall consider specificapplication areas in Chapters 8 to 12.
PROBLEMS
41. Verify the element stiffness and force matrix in equation (a) 6.19b, (b)6.19c, and (c) 6.19d.
42. Verify the shear stress values for element (a) two, (b) three, and (c) four ofthe example discussed in Section 6.3.
43. Determine the numerical values in {R} for the consistent element resultants when the values of are desired at the nodes.
References
Fung, Y. C., Foundations of Solid Mechanics, Prentice-Hall, Englewood Cliffs,N.J., 1965, pp. 162-170.
Oden, J. T., and H. J. Brauchli, "On the Calculation of Consistent Stress
Distributions in Finite Element Approximations," International Journal forNumerical Methods in Engineering, 1971, Vol. 3, pp. 317-325.
Timoshenko, S. P., and J. N. Goodier, Theory of Elasticity, McGraw-Hill, NewYork, 1970, pp. 315-316.
104 Torsion of Nonclrcuiar Sections
Uncited Reference
Owen, D. R. J., and O. C. Zienkiewicz, "Torsion of Axi-Symmetric Solids ofVariable Diameter—Including Acceleration Effects," International Journalfor Numerical Methods in Engineering, 1974, Vol. 8, pp. 195-212.
Yamada, Y., T. Kawai, and N. Yoshemura, "Analysis of the Elastic-Plastic-Problems by the Matrix Displacement Method," Proceedings of the SecondConference on Matrix Methods in Structural Mechanics (AFFDL-TR-68-150), Wright-Patterson Air Force Base, Dayton, Ohio, 1968.