Chapter 7 Numerical Methods

7.1 Introduction

In this chapter we discuss the finite difference and finite element approaches to the solution of problems in elasticity and the mechanics of materials. The use of these numerical methods enables the engineer to expand his or her ability to solve practical design problems. The engineer may now treat real shapes as distinct from the somewhat limited variety of shapes amenable to simple analytic solution. Similarly, the engineer need no longer force a complex loading system to fit a more regular load configuration to conform to the dictates of a purely academic situation. Numerical analysis thus provides a tool with which the engineer may feel freer to undertake the solution of problems as they are found in practice.

Analytical solutions of the type discussed in earlier chapters have much to offer beyond the specific cases for which they have been derived. For example, they enable us to gain insight into the variation of stress and deformation with basic shape and property changes. In addition, they provide the basis for rough approximations in preliminary design even though there is only crude similarity between the analytical model and the actual case. In other situations, analytical methods provide a starting point or guide in numerical solutions.

Numerical analyses lead often to a system of linear algebraic equations. The most appropriate method of solution then depends on the nature and the number of such equations, as well as the type of computing equipment available. The techniques introduced in this chapter and applied in the chapters following have clear application to computation by means of electronic digital computer.

7.2 Finite Differences

The numerical solution of a differential equation is essentially a table in which values of the required function are listed next to corresponding values of the independent variable(s). In the case of an ordinary differential equation, the unknown function (y) is listed at specific pivot or nodal points spaced along the x axis. For a two-dimensional partial differential equation, the nodal points will be in the xy plane.

The basic finite difference expressions follow logically from the fundamental rules of calculus. Consider the definition of the first derivative with respect to x of a continuous function y = f(x) (Fig. 7.1):

Figure 7.1 Finite difference approximation of f(x).

The subscript n denotes any point on the curve. If the increment in the independent variable does not become vanishingly small but instead assumes a finite Δx = h, the preceding expression represents an approximation to the derivative:

Here Δy_n is termed the first difference of y at point x_n:

(7.1)

Because relationship (7.1) is expressed in terms of the numerical value of the function at the point in question (n) and a point ahead of it (n + 1), the difference is termed a forward difference. The backward difference at n, denoted ∇y_n, is given by

(7.2)

Central differences involve pivot points symmetrically located with respect to x_n and often result in more accurate approximations than forward or backward differences. The latter are especially useful where, because of geometrical limitations (as near boundaries), central differences cannot be employed. In terms of symmetrical pivot points, the derivative of y at x_n is

(7.3)

The first central difference δy is thus

(7.4)

A procedure similar to that just used will yield the higher-order derivatives. Referring once again to Fig. 7.1, employing Eq. (7.1), we have

The second forward difference at x_n is therefore

(7.5)

The second backward difference is found in the same way:

(7.6)

It is a simple matter to verify the fact that the coefficients of the pivot values in the mth forward and backward differences are the same as the coefficients of the binomial expansion (a – b)^m. Using this scheme, higher-order forward and backward differences are easily written.

The second central difference at x_n is the difference of the first central differences:

This quantity is thus observed to depend on values of the function two intervals ahead of and behind the point under consideration. To improve the approximation, the second difference is expressed in terms of the function only one interval ahead of and behind the point. To do this, consider the following:

δ²y_n = Δ(∇y_n) or δ²y_n = ∇(Δy_n)

The second central difference at x_n is thus

(7.7)

In a like manner, the third and fourth central differences are readily determined:

(7.8)

(7.9)

Examination of Eqs. (7.7) and (7.9) reveals that for even-order derivatives, the coefficients of y_n, y_n+l, are equal to the coefficients in the binomial expansion (a – b)^m.

Unless otherwise specified, we shall use the term finite differences to refer to central differences.

We now discuss a continuous function w(x, y) of two variables. The partial derivatives may be approximated by the following procedures, similar to those discussed in the previous chapter. For purposes of illustration, consider a rectangular boundary as in Fig. 7.2. By taking Δx = Δy = h, a square mesh or net is formed by the horizontal and vertical lines. The intersection points of these lines are the nodal points. Equation (7.7) yields

(7.10)

(7.11)

The subscripts x and y applied to the δ’s indicate the coordinate direction appropriate to the difference being formed. The preceding expressions written for the point 0 are

(7.12)

and

Figure 7.2. Rectangular boundary divided into a square mesh.

(7.13)

(7.14)

Similarly, Eqs. (7.8) and (7.9) lead to expressions for approximating the third- and fourth-order partial derivatives.

7.3 Finite Difference Equations

We are now in a position to transform a differential equation into an algebraic equation. This is accomplished by substituting the appropriate finite difference expressions into the differential equation. At the same time, the boundary conditions must also be converted to finite difference form. The solution of a differential equation thus reduces to the simultaneous solution of a set of linear, algebraic equations, written for every nodal point within the boundary.

Example 7.1

Analyze the torsion of a bar of square section using finite difference techniques.

Solution The governing partial differential equation is (see Sec. 6.3)

(7.15)

where Φ may be assigned the value of zero at the boundary. Referring to Fig. 7.2, the finite difference equation about the point 0, corresponding to Eq. (7.15), is

(7.16)

A similar expression is written for every other nodal point within the section. The solution of the problem then requires the determination of those values of Φ that satisfy the system of algebraic equations.

The domain is now divided into a number of small squares, 16 for example. In labeling nodal points, it is important to take into account any conditions of symmetry that may exist. This has been done in Fig. 7.3. Note that Φ = 0 has been substituted at the boundary. Equation (7.16) is now applied to nodal points b, c, and d, resulting in the following set of expressions:

Figure 7.3. Example 7.1..

4Φ_d – 4Φ_b = –2Gθh²

2Φ_d – 4Φ_c = –2Gθh²

2Φ_c + Φ_b – 4Φ_d = –2Gθh²

Simultaneous solution yields

(a)

The results for points b and d are tabulated in the second column of Table 7.1.

To determine the partial derivatives of the stress function, we shall assume a smooth curve containing the values in Eq. (a) to represent the function Φ. Newton’s interpolation formula [Ref. 7.1], used for fitting such a curve, is

(7.17)

Here the Δ Φ₀’s are the forward differences calculated at x = 0 as follows:

(b)

Table 7.1

The differences are also calculated at x = h, x = 2h, and so on, and are listed in Table 7.1. Note that we can readily obtain the values given in Table 7.1 (for the given Φ’s) by starting at node x = 4h: 0 – 1.75 = –1.75, 1.75 – 2.25 = –0.5, –1.75 – (–0.5) = –1.25, and so on.

The maximum shear stress, which occurs at x = 0, is obtained from (∂Φ/∂x)₀. Thus, differentiating Eq. (7.17) with respect to x and then setting x = 0, the result is

(c)

Substituting the values in the first row of Table 7.1 into Eq. (c), we obtain

The exact value, given in Table 6.2 as τ_max = 0.678Gθa, differs from this approximation by only 4.7%.

By means of a finer network, we expect to improve the result. For example, selecting h = a/6, six nodal equations are obtained. It can be shown that the maximum stress in this case, 0.661Gθa, is within 2.5% of the exact solution. On the basis of results for h = a/4 and h = a/6, a still better approximation can be found by applying extrapolation techniques.

7.4 Curved Boundaries

It has already been mentioned that one important strength of numerical analysis is its adaptability to irregular geometries. We now turn, therefore, from the straight and parallel boundaries of previous problems to situations involving curved or irregular boundaries. Examination of one segment of such a boundary (Fig. 7.4) reveals that the standard five-point operator, in which all arms are of equal length, is not appropriate because of the unequal lengths of arms bc, bd, be, and bf. When at least one arm is of nonstandard length, the pattern is referred to as an irregular star. One method for constructing irregular star operators is discussed next.

Figure 7.4. Curved boundary and irregular star operator.

Assume that, in the vicinity of point b, w(x, y) can be approximated by the second-degree polynomial

(7.18)

Referring to Fig. 7.4, this expression leads to approximations of the function w at points c, d, e, and f:

(a)

At nodal point b(x = y = 0), Eq. (7.18) yields

(b)

Combining Eqs. (a) and (b), we have

Introducing this into the Laplace operator, we obtain

(7.19)

In this expression α₁ = h₁/h and α₂ = h₂//h. It is clear that for irregular stars 0 ≤ α_i ≤ l(i = 1, 2).

The foregoing result may readily be reduced for one-dimensional problems with irregularly spaced nodal points. For example, in the case of a beam, Eq. (7.19), with reference to Fig. 7.4, simplifies to

(7.20)

where x represents the longitudinal direction. This expression, setting α = α₁, may be written

(7.21)

Example 7.2

Find the shearing stresses at the points A and B of the torsional member of the elliptical section shown in Fig. 7.5. Let a = 15 mm, b = 10 mm, and h = 5 mm.

Figure 7.5. Example 7.2. Elliptical cross section of a torsion member.

Solution Because of symmetry, only a quarter of the section need be considered. From the equation of the ellipse with the given values of a, b, and h, it is found that h₁ = 4.4 mm, h₂ = 2.45 mm, h₃= 3 mm. At points b, e, f, and g, the standard finite difference equation (7.18) applies, while at c and d, we use a modified equation found from Eq. (7.15) with reference to Eq. (7.19). We can therefore write six equations presented in the following matrix form:

These equations are solved to yield

Φb = 2.075Gθh2,	Φc = 1.767Gθh2,	Φd = 0.843Gθh2
Φe = 1.536Gθh2,	Φf = 2.459Gθh2,	Φg = 2.767Gθh2

The solution then proceeds as in Example 7.1. The following forward differences at point B are first evaluated:

ΔΦB = 2.075Gθh2,	Δ3ΦB = –0.001Gθh2
Δ2ΦB = –1.383Gθh2,	Δ4ΦB = 0.002Gθh2

Similarly, for point A, we obtain

ΔΦA = 1.536Gθh2,	Δ4ΦA = 0.001Gθh2
Δ2ΦA = –0.613Gθh2,	Δ5ΦA = 0.001Gθh2
Δ3ΦA = –0.002Gθh2,	Δ6ΦA = –0.002Gθh2

Thus,

Note that, according to the exact theory, the maximum stress occurs at y = b and is equal to 1.384Gθb (see Example 6.2), indicating excellent agreement with τ_B.

7.5 Boundary Conditions

Our concern has thus far been limited to problems in which the boundaries have been assumed free of constraint. Many practical situations involve boundary conditions related to the deformation, force, or moment at one or more points. Application of numerical methods under these circumstances may become more complex.

To solve beam deflection problems, the boundary conditions as well as the differential equations must be transformed into central differences. Two types of homogeneous boundary conditions, obtained from υ = 0, dυ/dx = 0, and υ= 0, d²υ/dx² = 0 at a support (n), are depicted in Fig. 7.6a and b, respectively. At a free edge (n), the finite difference boundary conditions are similarly written from d²υ/dx² = 0 and d³υ/dx³ = 0 as follows

(a)

Figure 7.6. Boundary conditions in finite differences: (a) clamped or fixed support; (b) simple support.

Let us consider the deflection of a nonprismatic cantilever beam with depth varying arbitrarily and of constant width. We shall divide the beam into m segments of length h = L/m and replace the variable loading by a load changing linearly between nodes (Fig. 7.7). The finite difference equations at a nodal point n may now be written as follows. Referring to Eq. (7.7), we find the difference equation corresponding to EI(d²υ/dx²) = M to be of the form

Figure 7.7. Finite-difference representation of a nonprismatic cantilever beam with varrying load.

(7.22)

The quantities M_n and (IE)_n represent the moment and the flexural rigidity, respectively, of the beam at n. In a like manner, the equation EI(d⁴υ/dx⁴) = p is expressed as

(7.23)

wherein p_n is the load intensity at n. Equations identical with the foregoing can be established at each remaining point in the beam. There will be m such expressions; the problem involves the solution of m unknowns, υ₁, ..., υ_m.

Applying Eq. (7.4), the slope at any point along the beam is

(7.24)

The following simple examples illustrate the method of solution.

Example 7.3

Use a finite difference approach to determine the deflection and the slope at the midspan of the beam shown in Fig. 7.8a.

Solution For simplicity, take h = L/4 (Fig. 7.8b). The boundary conditions ν(0) = ν(L) = 0 and ν"(0) = ν"(L) = 0 are replaced by finite

Figure 7.8. Example 7.3.

difference conditions by setting υ(0) = υ₀, υ(L) = υ₄ and applying Eq. (7.7):

(b)

When Eq. (7.23) is used at points 1, 2, and 3 the following expressions are obtained:

(c)

Simultaneous solution of Eqs. (b) and (c) yields

Then, from Eq. (7.24), we obtain

Note that, by successive integration of EId⁴υ/dx⁴ = p(x), the result υ₂ = 0.0185pL⁴/EI is obtained. Thus, even a coarse segmentation leads to a satisfactory solution in this case.

Example 7.4

Determine the redundant reaction R for the beam depicted in Fig. 7.9a.

Solution The bending diagrams associated with the applied loads 2P and the redundant reaction R are given in Figs. 7.9b and c, respectively.

Figure 7.9. Example 7.4. Statically indeterminate beam: (a) load diagram; (b) and (c) moment diagrams.

For h = L/2, Eq. (7.22) results in the following expressions at points 1, 2, and 3:

(d)

The number of unknowns in this set of equations is reduced from five to three through application of the conditions of symmetry, υ₁ = υ₃, and the support conditions, υ₀ = υ₂ = υ₄ = 0. Solution of Eq. (d) now yields R = 8P/3.

Example 7.5

Determine the deflection of the free end of the stepped cantilever beam loaded as depicted in Fig. 7.10a. Take h = a/2 (Fig. 7.10b).

Figure 7.10. Example 7.5.

Solution The boundary conditions υ(0) = 0 and υ′(0) = 0, referring to Fig. 7.6a, lead to

(e)

The moments are M₀ = 3Pa, M₁ = 2Pa, M₂= Pa, and M₃ = Pa/2. Applying Eq. (7.22) at nodes 0, 1, 2, and 3, we obtain

(f)

Observe that at point 2, the average flexural rigidity is used. Solving Eqs. (e) and (f) gives υ₁ = 3C/4, υ₂ = 5C/2, υ₃ = 59C/12, and υ₄ = 47C/6, in which C = (Pa/EI)h². Therefore, after setting h = a/2, we have

The foregoing deflection is about 2.2% larger than the exact value.

7.6 Finite Element Method

The powerful finite element method had its beginnings in the 1950s, and with the widespread use of the digital computer it has since gained considerable favor relative to other numerical approaches. The finite element method may be viewed as an approximate Ritz method combined with a variational principle applied to continuum mechanics. It permits the prediction of stress and strain in an engineering structure with unprecedented ease and precision.

The general procedures of the finite element and conventional structural matrix methods are similar. In the latter approach, the structure is idealized as an assembly of structural members connected to one another at joints or nodes at which the resultants of the applied forces are assumed to be concentrated. Employing the mechanics of materials or the elasticity approach, the stiffness properties of each element can be ascertained. Equilibrium and compatibility considered at each node lead to a set of algebraic equations in which the unknowns may be nodal displacements, internal nodal forces, or both, depending on the specific method used. In the displacement or direct stiffness method, the set of algebraic equations involves the nodal displacements. In the force method, the equations are expressed in terms of unknown internal nodal forces. A mixed method is also used in which the equations contain both nodal displacements and internal forces.

In contrast with the foregoing approaches, in the finite element method, the solid continuum is discretized by a finite number of elements, connected not only at their nodes, but along the hypothetical interelement boundaries as well. In addition then to nodal compatibility and equilibrium, as in conventional structural analysis, it is clear that compatibility must also be satisfied along the boundaries between elements. Because of this, the element stiffness can be only approximately determined in the finite element method. Alternative approaches to those of structural analysis are thus employed.

An essential difference exists between the finite element and finite difference analyses. In the latter, the differential equations describing the state of the continuum governing quantities such as stress or deflection are approximated by a set of finite difference equations written for a limited number of nodes. On the other hand, the finite element approach deals with an assembly of elements that replaces the continuous structure, and it is the replaced structure that is then the subject of analysis.

There are a number of finite element methods, as in the structural analyses previously cited. We here treat only the commonly employed finite element displacement approach.^*

*The literature related to this method is extensive. See, for example, Refs. 7.2 through 7.10.

7.7 Properties of a Finite Element

Now we shall define a number of basic quantities relevant to an individual finite element of an isotropic elastic body. In the interest of simple presentation, in this section the relationships are written only for the two-dimensional case. The general formulation of the finite element method applicable to any structure is presented in the next section. Solutions of plane stress and plane strain problems are illustrated in detail in Sec. 7.9. The analyses of axisymmetric structures and thin plates employing the finite element are given in Chapters 8 and 13, respectively.

To begin with, the relatively thin, continuous body shown in Fig. 7.11a is replaced by an assembly of finite elements (triangles, for example) indicated by the dashed lines (Fig. 7.11b). These elements are connected not only at their corners or nodes, but along the interelement boundaries as well. The basic unknowns are the nodal displacements.

Figure 7.11. Plane stress region (a) before and (b) after division into finite elements.

Displacement Matrix

The nodal displacements are related to the internal displacements throughout the entire element by means of a displacement function. Consider the typical element e in Fig. 7.11b, shown isolated in Fig. 7.12a. Designating the nodes i, j, and m, the element nodal displacement matrix is

Figure 7.12. Triangular finite element.

(7.25)

or, for convenience, expressed in terms of submatrices δ_u and δ_υ,

(7.26)

where the braces indicate a column matrix. The displacement function defining the displacement at any point within the element, {f}_e, is given by

(7.27)

which may also be expressed as

(7.28)

where the matrix [N] is a function of position, to be obtained later for a specific element. It is desirable that a displacement function {f}_e be selected such that the true displacement field will be as closely represented as possible. The approximation should result in a finite element solution that converges to the exact solution as the element size is progressively decreased.

Strain, Stress, and Elasticity Matrices

The strain, and hence the stress, are defined uniquely in terms of displacement functions (see Chapter 2). The strain matrix is of the form

(7.29a)

or

(7.29b)

where [B] is also yet to be defined.

Similarly, the state of stress throughout the element is, from Hooke’s law,

(a)

In general,

(7.30)

where [D], an elasticity matrix, contains material properties. If the element is subjected to thermal or initial strain, the stress matrix becomes

(7.31)

The thermal strain matrix, for the case of plane stress, is given by {ε₀} = {αT, αT, 0} (Sec. 3.7). Comparing Eqs. (a) and (7.30), it is clear that

(7.32)

where D₁₁ = E/(1 – ν²), D₁₂ = νE/(1 – ν²).... Equation (7.30) is valid for the case of plane stress. A matrix [D] for plane strain is found in a similar fashion.

7.8 Formulation of the Finite Element Method

A convenient method for executing the finite element procedure relies on the minimization of the total potential energy of the system, expressed in terms of displacement functions. Consider again in this regard an elastic body (Fig. 7.11). The principle of potential energy, from Eq. (10.21), is expressed for the entire body as follows:

(7.33)

Note that variation notation δ has been replaced by Δ to avoid confusing it with nodal displacement. In Eq. (7.33),

n = number of elements comprising the body

V = volume of a discrete element

s = portion of the boundary surface area over which forces are prescribed

F = body forces per unit volume

p = prescribed boundary force or surface traction per unit area

Through the use of Eq. (7.27), Eq. (7.33) may be expressed in the following matrix form:

(a)

where superscript T denotes the transpose of a matrix. Now, using Eqs. (7.28), (7.29), and (7.31), Eq. (a) becomes

(b)

The element stiffness matrix [k]_e and element nodal force matrix {Q}_e (due to body force, initial strain, and surface traction) are

(7.34)

(7.35)

It is clear that the variations in {δ}_e are independent and arbitrary, and from Eq. (b) we may therefore write

(7.36)

We next derive the governing equations appropriate to the entire continuous body. The assembled form of Eq. (b) is

(c)

This expression must be satisfied for arbitrary variations of all nodal displacements {Δδ}. This leads to the following equations of equilibrium for nodal forces for the entire structure, the system equations:

(7.37)

where

(7.38)

It is noted that structural matrix [K] and the total or equivalent nodal force matrix {Q} are found by proper superposition of all element stiffness and nodal force matrices, respectively.

Outline of General Finite Element Analysis

We can now summarize the general procedure for solving a problem by application of the finite element method as follows:

1. Calculate [k]_e from Eq. (7.34) in terms of the given element properties. Generate [K] = Σ[k]_e.

2. Calculate {Q}_e from Eq. (7.35) in terms of the applied loading. Generate {Q} = Σ{Q}_e.

3. Calculate the nodal displacements from Eq. (7.37) by satisfying the boundary conditions: {δ} = [K]^-1{Q}.

4. Calculate the element strain using Eq. (7.29): {ε}_e = [B]{δ}_e.

5. Calculate the element stress using Eq. (7.31): {σ}_e = [D]({ε} – {ε₀})_e.

When the stress found is uniform throughout each element, this result is usually interpreted two ways: the stress obtained for an element is assigned to its centroid; if the material properties of the elements connected at a node are the same, the average of the stresses in the elements is assigned to the common node.

The foregoing outline will be better understood when applied to a triangular element in the next section. Formulation of the properties of a simple one-dimensional element is illustrated in the following example.

Example 7.6

An axial element of constant cross-sectional area A, length L, and modulus of elasticity E is subjected to a distributed load p_x per unit length and a uniform temperature change T (Fig. 7.13a). Determine (a) the stiffness matrix, (b) the total nodal force matrix, and (c) the deflection of the right end u₂ for fixed left end and p_x = 0 (Fig. 7.13b).

Figure 7.13. Example 7.6. Axial finite element.

Solution

a. We shall assume that the displacement u at any point within the element varies linearly with x:

(7.39)

wherein a₁ and a₂ are constants. The axial displacements of nodes 1 and 2 are

u₁ = a₁, u₂ = a₁ + a₂L

from which a₂ = – (u₁ – u₂)/L. Substituting this into Eq. (7.39),

(d)

Applying Eq. (7.29a), the strain in the element is

or in matrix form

(e)

For a one-dimensional element, we have D = E, and the stress from Eq. (7.30) equals

The element stiffness matrix is obtained upon introduction of [B] from Eq. (e) into Eq. (7.34):

(7.40)

b. Referring to Fig. 7.13a,

(7.41)

where p₁ and p₂ are the intensities of the load per unit length at nodes 1 and 2, respectively. Substitution of [N] from Eq. (d) together with Eq. (7.41) into Eq. (7.35) yields

The distributed load effects are obtained by integrating the preceding equations:

(f)

The strain due to the temperature change is ε₀ = αT, where α is the coefficient of thermal expansion. Inserting [B] from Eq. (g) into Eq. (7.35),

The thermal strain effects are then

(g)

The total element nodal matrix is obtained by adding Eqs. (f) and (g):

(7.42)

c. The nodal force-displacement relations (7.37) now take the form

from which the elongation of the bar equals u₂ = α(T)L, a predictable result.

7.9 Triangular Finite Element

Because of the relative ease with which the region within an arbitrary boundary can be approximated, the triangle is used extensively in finite element assemblies. Before deriving the properties of the triangular element, we describe area or triangular coordinates, which are very useful for the simplification of the displacement functions.

In this section, we shall derive the basic constant strain triangular (CST) plane stress and strain element. Note that there are a variety of two-dimensional finite element types that lead to better solutions. Examples are linear strain triangular (LST) elements, triangular elements having additional side and interior nodes, rectangular elements with corner nodes, and rectangular elements having additional side nodes [Refs. 7.4, 7.6]. The LST element has six nodes: the usual corner nodes plus three additional nodes located at the midpoints of the sides. The procedures for the development of the LST element equations follow the identical steps as that of the CST element.

Consider the triangular finite element 1 2 3 (where i = l, j = 2, and m = 3) shown in Fig. 7.12b, in which the counterclockwise numbering convention of nodes and sides is indicated. A point P located within the element, by connection with the corners of the element, forms three subareas denoted A₁, A₂, and A₃. The ratios of these areas to the total area A of the triangle locate P and represent the area coordinates:

(7.43)

It follows from this that

(7.44)

and consequently only two of the three coordinates are independent. A useful property of area coordinates is observed through reference to Fig. 7.12b and Eq. (7.43):

L_n = 0 on side n, n = 1,2,3

and

Li = 1,	Lj = Lm = 0 at node i
Lj = 1,	Li = Lm = 0 at node j
Lm = 1,	Li = Lj = 0 at node m

The area A of the triangle may be expressed in terms of the coordinates of two sides, for example, 2 and 3:

or

(7.45a)

Here i, j, and k are the unit vectors in the x, y, and z directions, respectively. Two additional expressions are similarly found. In general, we have

(7.45b)

where

(7.46)

Note that a_j, a_m, b_j, and b_m can be found from definitions of a_i and b_i with the permutation of the subscripts in the order ijmijm, and so on.

Similar equations are derivable for subareas A₁, A₂, and A₃. The resulting expressions, together with Eq. (7.43), lead to the following relationship between area and Cartesian coordinates:

(7.47)

where

(7.48)

Note again that, given any of these expressions for c_ij, the others may be obtained by permutation of the subscripts.

Now we explore the properties of an ordinary triangular element of a continuous body in a state of plane stress or plane strain (Fig. 7.12). The nodal displacements are

(a)

The displacement throughout the element is provided by

(7.49)

Matrices [N] and [B] of Eqs. (7.28) and (7.29b) are next evaluated, beginning with

(b)

(c)

We observe that Eqs. (7.49) and (b) are equal, provided that

(7.50)

The strain matrix is obtained by substituting Eqs. (7.49) and (7.47) into Eq. (7.29a):

(d)

Here A, a_i, and b_i are defined by Eqs. (7.45) and (7.46). The strain (stress) is observed to be constant throughout, and the element of Fig. 7.12 is thus referred to as a constant strain triangle. Comparing Eqs. (c) and (d), we have

(7.51)

The stiffness of the element can now be obtained through the use of Eq. (7.34):

(e)

Let us now define

(7.52)

where [D] is given by Eq. (7.32) for plane stress. Assembling Eq. (e) together with Eqs. (7.51) and (7.52) and expanding, the stiffness matrix is expressed in the following partitioned form of order 6 × 6:

(7.53)

where the submatrices are

(7.54)

Finally, we consider the determination of the element nodal force matrices. The nodal force owing to a constant body force per unit volume is, from Eqs. (7.35) and (7.50),

(f)

For an element of constant thickness, this expression is readily integrated to yield^*

*For a general function , defined in area coordinates, the integral of f over any triangular area A is given by

where α, β, γ, are constants.

(7.55)

The nodal forces associated with the weight of an element are observed to be equally distributed at the nodes.

The element nodal forces attributable to applied external loading may be determined either by evaluating the static resultants or by application of Eq. (7.35). Nodal force expressions for arbitrary nodes j and m are given next for a number of common cases (Prob. 7.17).

Linear load, p(y) per unit area, Fig. 7.14a:

(7.56a)

Figure 7.14. Nodal forces due to (a) linearly distributed load and (b) shear load.

where t is the thickness of the element.

Uniform load is a special case of the preceding with p_j = p_m = p:

(7.56b)

End shear load, P, the resultant of a parabolic shear stress distribution defined by Eq. (3.21) (see Fig. 7.14b):

(7.57)

Equation (7.53), together with those expressions given for the nodal forces, characterizes the constant strain element. These are substituted into Eq. (7.38) and subsequently into Eq. (7.37) in order to evaluate the nodal displacements by satisfying the boundary conditions.

The basic procedure employed in the finite element method is illustrated in the following simple problems.

Example 7.7

The element e shown in Fig. 7.15 represents a segment of a thin elastic plate having side 23 adjacent to its boundary. The plate is subjected to several loads as well as a uniform temperature rise of 50°C. Determine (a) the stiffness matrix and (b) the equivalent (or total) nodal force matrix for the element if a pressure of p = 14 MPa acts on side 23. Let t = 0.3 cm, E = 200 GPa, ν = 0.3, specific weight γ = 77 kN/m³, and α = 12 × 10^–6/°C.

Solution The origin of the coordinates is located at midlength of side 13, for convenience. However, it may be placed at any point in the x, y plane. Applying Eq. (7.52), we have (in N/cm³)

(g)

a. Stiffness matrix: The nodal points are located at

(h)

Using Eq. (7.46) and referring to Fig. 7.15, we obtain

Figure 7.15. Example 7.7

(i)

Next, the first equation of (7.54), together with Eqs. (g) and (i), yields

The submatrix k_uu is thus

Similarly, from the second and third equations of (7.54), we obtain the following matrices:

Assembling the preceding equations, the stiffness matrix of the element (in newtons per centimeter) is

(j)

b. We next determine the nodal forces of the element owing to various loadings. The components of body force are F_x = 0 and F_y = 0.077 N/cm³.

Body force effects: Through the application of Eq. (7.55), it is found that

Surface traction effects: The total load, , is equally divided between nodes 2 and 3. The nodal forces can therefore be expressed as

Thermal strain effects: The initial strain associated with the 50°C temperature rise is ε₀ = αT = 0.0006. From Eq. (7.36),

Substituting matrix [B], given by Eq. (7.51), into this equation, and the values of the other constants already determined, the nodal force is calculated as follows:

or

Equivalent nodal force matrix: Summation of the nodal matrices due to the several effects yields the total element nodal force matrix:

If, in addition, there are any actual node forces, these must also be added to the value obtained.

Example 7.8

A 0.3-cm-thick cantilever beam is subjected to a parabolically varying end shearing stress resulting in a total load of 5000 N (Fig. 7.16a). Divide the beam into two constant strain triangles and calculate the deflections. Let E = 200 GPa and υ = 0.3.

Solution The discretized beam is shown in Fig. 7.16b.

Suffness matrix: Inasmuch as the dimensions and material properties of element a are the same as that given in the previous example, [k]_a is

Figure 7.16. Example 7.8. Cantilever beam (a) before and (b) after being discretized.

defined by Eq. (j). For element b, assignment of i = 2, j = 4, and m = 3 [Eq. (7.46)] leads to

ai = a2 = 0 – 4 = –4,	bi = b2 = 1 – 1 = 0
aj = a4 = 4 – 0 = 4,	bj = b4 = 1 + 1 = 2
am = a3 = 4 – 4 = 0,	bm = b3 = –1 –1 = –2

Substitution of these and Eq. (g) into Eqs. (7.54) yields

Thus

Similarly, we obtain

The stiffness matrix of element b is therefore

(k)

The displacements u₄, υ₄ and u₁, υ₁ are not involved in elements a and b, respectively. Therefore, prior to the addition of [k]_a and [k]_b to form the system stiffness matrix, rows and columns of zeros must be added to each element matrix to account for the absence of these displacements. In doing so, Eqs. (j) and (k) become

(l)

and

(m)

Then, addition of Eqs. (l) and (m) yields the system matrix (in newtons per centimeter):

(n)

Nodal forces: Referring to Fig. 7.16b and applying Eq. (7.57), we obtain

Because no other external force exists, the system nodal force matrix is

{Q} = {0, 0, 0, 0, 0 –2500, 0 –2500}

Nodal displacements: The boundary conditions are

(o)

The force-displacement relationship of the system is therefore

(p)

Equation (p) is readily reduced to the form

(q)

From this we obtain

(r)

The strains {ε}_a may now be found upon introduction of Eqs. (i) and (r) into (d) as

Finally, the stress is determined by multiplying [D] by {ε}_a:

Element b is treated similarly.

Note that the model employed in the foregoing solution is quite crude. The effect of element size on solution accuracy is illustrated in the following example.

Example 7.9

By means of (a) exact and (b) finite element approaches, investigate the stresses and displacements in a thin beam subjected to end moments applied about the centroidal axis (Fig. 7.17a). Let L = 76.2 mm, h = 50.8 mm, thickness t = 25.4 mm, p = 6895 kPa, E = 207 GPa, and v = 0.15. Neglect the weight of the member.

Solution

a. Exact solution: Replacing the end moments with the statically equivalent load per unit area p = Mh/l (Fig. 7.17b), the stress distribution from Eq. (5.5) is

(s)

Figure 7.17. Example 7.9. Beam in pure bending.

From Hooke’s law and Eq. (2.3), we have

By now following a procedure similar to that of Sec. 5.4, satisfying the conditions u(0, 0) = υ(0, 0) = 0 and u(L,0) = 0, we obtain

(t)

Substituting the data into Eqs. (s) and (t), the results are

(u)

u(0.0762, –0.0254) = 25.4 × 10^–6m, υ(0.0762, 0) = 1.905 × 10^–6m

b. Finite element solution: Considerations of symmetry and antisymmetry indicate that only any one-quarter (shown as shaded portion in the figure) of the beam need be analyzed.

Displacement boundary conditions: Figure 7.18a shows the quarter-plate discretized to contain 12 triangular elements. The origin of coordinates is located at node 3. As no axial deformation occurs along the x and y axes, nodes 1, 2, 6, 9, and 12 are restrained against u deformation; node 3 is restrained against both u and v deformation. The boundary conditions are thus

u₁ = u₂ = u₃ = u₆ = u₉ = u₁₂ = 0, v₃ = 0

Figure 7.18. Example 7.9. Influence of element size and orientation on stress (megapascals) in beam shown in Fig. 7.17.

Nodal forces: For the loading system of Fig. 7.17b, Eq. (7.56a) applies. Upon substitution of numerical values,

The remaining Q’s are zero.

Results: The nodal displacements are determined following a procedure similar to that of Example 7.8. The stresses are then evaluated and representative values (in megapascals) given in Fig. 7.18b. Note that the stress obtained for an element is assigned to the centroid. Observe that there is considerable difference between the exact solution, Eq. (u), and that resulting from the coarse mesh arrangement employed. To demonstrate the influence of element size and orientation, calculations have also been carried out for the grid configurations shown in Figs. 7.18c and d. For purposes of comparison, the deflections corresponding to Figs. 7.18b through d are presented in Table 7.2.

Table 7.2

The effect of element orientation is shown in Figs. 7.18b and c for an equal number of elements and node locations. Figure 7.18d reveals that elements characterized by large differences between their side lengths, weak elements, lead to unfavorable results even though the number of nodes is larger than those of Figs. 7.18b and c. The employment of equilateral or nearly equilateral well-formed elements of finer mesh leads to solutions approaching the exact values.

Note that, owing to the approximate nature of the finite element method, nonzero values are found for σ_y, and τ_xy. These are not listed in the figures. As the mesh becomes finer, these stresses do essentially vanish.

It should be mentioned that, when the three mid-side points as well as the three corners of a triangular element are taken as the nodes, a second-order polynomial can be selected for {f}_e such that it ensures the compatibility of interelement displacement and a linear variation of strain throughout the element. The element is thus referred to as a linear strain triangle and to arrive at the characteristics of this element, the same procedure is followed as for the constant strain triangle (Sec. 7.9). Two other common finite element types are the quadrilateral, formed by four constant strain triangles, and the rectangular element, a special case of the quadrilateral. The factors such as size, arrangement, and shape of the finite element influence directly the accuracy of the finite element solution.

It is clear that we cannot reduce element size to extremely small values, as this would tend to increase to significant magnitudes the computer error incurred. An “exact” solution is thus unattainable, and we therefore seek instead an acceptable solution. The goal is then the establishment of a finite element that ensures convergence to the exact solution in the absence of round-off error. The literature contains many comparisons between the various basic elements. The efficiency of a finite element solution can, in certain situations, be enhanced through the use of a “mix” of elements. For example, a denser mesh within a region of severely changing or localized stress may save much time and effort.

7.10 Use of Digital Computers

The finite difference method, as well as the finite element method, leads to systems of linear algebraic equations. The digital computer is often employed to provide rapid solutions of these simultaneous equations, usually by means of matrix methods. If the number of equations is not great, the computer solution may be expected to be very accurate. Errors associated with round-off and truncation are negligible for matrices as large as 15 × 15, provided that the elements contain eight significant figures. This may also be true for well-conditioned matrices of order higher than 15, that is, those with dominant diagonal terms. In cases involving higher-order matrices, iteration procedures, as well as other techniques of numerical analysis, are used.

Compared with analytical techniques, the finite element method offers a distinct advantage in the treatment of anisotropic situations, three-dimensional and eigenvalue problems, and two-dimensional problems involving irregular geometries and loads. It is evident, however, that the finite element approach, even in the simplest of cases, requires considerable matrix algebra. For any significant problem, the electronic digital computer must be employed.

A variety of techniques is applicable to the efficient execution of a finite element solution. It is worthwhile to mention that many large-scale general purpose system programs and special-purpose programs [Ref. 7.4] have been developed for finite element analysis. Such systems permit structural data to be processed in a general and flexible manner for the purpose of fulfilling the specific requirements of the user. System programs involve almost all known finite element models and can be stored on the disk of a large computer.

The digital computer not only offers extreme speed, but tends to encourage the analyst to retain the complexities of a problem. The latter is in contrast with the tendency to make overly simplifying assumptions in the interest of a tractable hand calculation. The many advantages of digital computation more than outweigh the disadvantage sometimes cited, that digital computation may cause the user to lose the physical grasp of a problem.

Problems

Secs. 7.1 through 7.4

7.1. Referring to Fig. 7.2, demonstrate that the biharmonic equation

takes the following finite difference form:

(P7.1)

7.2. Consider a torsional bar having rectangular cross section of width 4a and depth 2a. Divide the cross section into equal nets with h = a/2. Assume that the origin of coordinates is located at the centroid. Find the shear stresses at points x = ±2a and y = ±a. Use the direct finite difference approach. Note that the exact value of stress at y = ±a is, from Table 6.2, τ_max = 1.860Gθa.

7.3. For the torsional member of cross section shown in Fig. P7.3, find the shear stresses at point B. Take h = 5 mm and h₁ = h₂ = 3.5 mm.

Figure P7.3.

7.4. Redo Prob. 7.3 to find the shear stress at point A. Let h = 4.25 mm; then h₁ = h and h₂ = 2.25 mm.

7.5. Calculate the maximum shear stress in a torsional member of rectangular cross section of sides a and b (a = 1.5b). Employ the finite difference method, taking h = a/4. Compare the results with that given in Table 6.2.

Sec. 7.5

7.6. A force P is applied at the free end of a stepped cantilever beam of length L (Fig. P7.6). Determine the deflection of the free end using the finite difference method, taking n = 3. Compare the result with the exact solution υ(L) = 3PL³/16EI.

Figure P7.6.

7.7. A stepped simple beam is loaded as shown in Fig. P7.7. Apply the finite difference approach, with h = L/4, to determine (a) the slope at point C; (b) the deflection at point C.

Figure P7.7.

7.8. A stepped simple beam carries a uniform loading of intensity p as shown in Fig. P7.8. Use the finite difference method to calculate the deflection at point C. Let h = a/2.

Figure P7.8.

7.9. Cantilever beam AB carries a distributed load that varies linearly as shown in Fig. P7.9. Determine the deflection at the free end by applying the finite difference method. Use n = 4. Compare the result with the “exact” solution w(L) = 11p_oL⁴/120EI.

Figure P7.9.

7.10. Applying Eq. (7.22), determine the deflection at points 1 through 5 for beam and loading shown in Fig. P7.10.

Figure P7.10.

7.11. Employ the finite difference method to obtain the maximum deflection and the slope of the simply supported beam loaded as shown in Fig. P7.11. Let h = L/4.

Figure P7.11.

7.12. Determine the deflection at a point B and the slope at point A of the overhanging beam loaded as shown in Fig. P7.12. Use the finite difference approach, with n = 6. Compare the deflection with its “exact” value υ_B = Pa³/12EI.

Figure P7.12.

7.13. Redo Prob. 7.6 with the beam subjected to a uniform load p per unit length and P = 0. The exact solution is υ(L) = 3pL⁴/32EI.

7.14. A beam is supported and loaded as depicted in Fig. P7.14. Use the finite difference approach, with h = L/4, to compute the maximum deflection and slope.

Figure P7.14.

7.15. A fixed ended beam supports a concentrated load P at its midspan as shown in Fig. P7.15. Apply the finite difference method to determine the reactions. Let h = L/4.

Figure P7.15.

7.16. Use the finite difference method to calculate the maximum deflection and the slope of a fixed-ended beam of length L carrying a uniform load of intensity p (Fig. P7.16). Let h = L/4.

Figure P7.16.

Secs. 7.6 through 7.10

7.17. Verify Eqs. (7.56) and (7.57) by determining the static resultant of the applied loading. [Hint: For Eq. (7.57), apply the principle of virtual work.]

with

to obtain

7.18. Redo Example 7.8 for the beam subjected to a uniform additional load throughout its span, p = – 7 MPa, and a temperature rise of 50°C. Let υ = 77 kN/m³ and α = 12 × 10^–6/°C.

7.19. A 2/7-cm-thick cantilever beam is subjected to a parabolically varying end shear stress resulting in a load of P N and a linearly distributed load p N/cm (Fig. P7.19). Dividing the beam into two triangles as shown, calculate the stresses in the member. The beam is made of a transversely isotropic material, in which a rotational symmetry of properties exists within the xz plane:

Figure P7.19.

E₁ = 210 GPa, v₂ = 0.1

E₂ = 70 GPa, G₂ = 28 GPa

Here E₁ is associated with the behavior in the xz plane, and E₂, G₂, and v₂ with the direction perpendicular to the xz plane. Now the elasticity matrix, Eq. (7.32), becomes

where n = E₁/E₂ and m = G₂/E₂.

7.20. Redo Example 7.8 if the discretized beam consists of triangular elements 143 and 124 (Fig. 7.16b). Assume the remaining data to be unchanged.

7.21. A thin plate containing a small hole of radius a is subjected to uniform tensile load of intensity σ_o at its edges, as shown in Fig. P7.21. Use a computer program with CST (or LST) elements to determine the stress concentration factor k. Compare the result with that obtained in Sec. 3.11. Given: L = 500 mm, a = 40 mm, h = 400 mm, σ_o = 40 MPa, E = 70 GPa, and ν = 0.3.

Figure P7.21.