Chapter 7. Numerical Methods

Search in book...
Toggle Font Controls
Create new playlist

Name your new playlist

Playlist description (optional)
Sign In

Email address

Password

Forgot Password?

or

Continue with Facebook

Continue with Google
Sign Up

Full Name

Email address

Confirm Email Address

Password

or

Continue with Facebook

Continue with Google

Chapter 7. Numerical Methods

7.1 Introduction

This chapter is subdivided into two parts. The finite difference method is treated briefly first. Then, the most commonly employed numerical technique, the finite element method, is discussed. We will apply both of these 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 no longer needs to force a complex loading system to fit a more regular load configuration so as to conform to the dictates of a purely academic situation. Instead, numerical analysis 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 of this text have much to offer beyond the specific cases for which they have been derived. For example, they enable us to gain insights 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 often lead 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 subsequent chapters have clear application to computation by means of electronic digital computers. Formulating and solving a problem (Appendix A) and using tools for computations (Section 7.16) are important in such cases. In these types of computations, fundamentals of matrix algebra—a subset of the finite element method—are extensively used.

Part A: Finite Difference Analysis

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):

${(\frac{d y}{d x})}_{n} = \lim_{Δ x \to 0} \frac{y (x_{n} + Δ x)}{Δ x} = \lim_{Δ x \to 0} \frac{y_{n + 1} - y_{n}}{Δ x}$

Analysis of finite differences for beam deflection is shown. — Figure 7.1. *Finite difference approximation of f(x)*.

The finite difference approximation shows the beam deflection in the form of a curve (y equals f of x), drawn along the five nodal points on the XY plane. Three points X n minus 1, X n, and X n plus 1 are plotted on the x-axis. These three points correspond to the points Y n minus 1, Y n, and Y n plus 1 on the curve. These points are connected using dashed lines. The space between the nodal points is marked, h.

where 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:

${(\frac{d y}{d x})}_{n} \approx \frac{Δ y_{n}}{h} = \frac{y_{n + 1} - y_{n}}{h}$

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

$\begin{matrix} Δ y_{n} = y_{n + 1} - y_{n} \approx h {(\frac{d y}{d x})}_{n} & (7.1) \end{matrix}$

Because the relationship (Fig. 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

$\begin{matrix} \nabla y_{n} = y_{n} - y_{n - 1} & (7.2) \end{matrix}$

7.2.1 Central Differences

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

$\begin{matrix} {(\frac{d y}{d x})}_{n} \approx \frac{y (x_{n} + h) - y (x_{n} - h)}{2 h} = \frac{1}{2 h} (y_{n + 1} - y_{n - 1}) & (7.3) \end{matrix}$

The first central difference δy, then, is

$\begin{matrix} δ y = \frac{1}{2} (y_{n + 1} - y_{n - 1}) \approx h {(\frac{d y}{d x})}_{n} & (7.4) \end{matrix}$

A procedure similar to that just used will yield the higher-order derivatives. The second forward difference at x_n is expressed in the form

$\begin{matrix} Δ^{2} y_{n} = y_{n + 2} - 2 y_{n + 1} + y_{n} \approx h^{2} {(\frac{d^{2} y}{d x^{2}})}_{n} & (7.5) \end{matrix}$

The second backward difference is found in the same way:

$\begin{matrix} \begin{matrix} \nabla^{2} y_{n} & = \nabla (\nabla y_{n}) = \nabla (y_{n} - y_{n - 1}) = \nabla y_{n} - \nabla y_{n - 1} \\ = (y_{n} - y_{n - 1}) - (y_{n - 1} - y_{n - 2}) \\ \nabla^{2} y_{n} & = y_{n} - 2 y_{n - 1} + y_{n - 2} \approx h^{2} {(\frac{d^{2} y}{d x^{2}})}_{n} \end{matrix} & (7.6) \end{matrix}$

It is a simple matter to verify 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. Therefore,

$\begin{matrix} \begin{matrix} δ^{2} y_{n} & = Δ (\nabla y_{n}) = Δ (y_{n} - y_{n - 1}) = Δ y_{n} - Δ y_{n - 1} \\ = (y_{n + 1} - y_{n}) - (y_{n} - y_{n - 1}) \\ = y_{n + 1} - 2 y_{n} + y_{n - 1} \approx h^{2} {(\frac{d^{2} y}{d x^{2}})}_{n} \end{matrix} & (7.7) \end{matrix}$

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

$\begin{matrix} \begin{matrix} δ^{3} y_{n} & = δ (δ^{2} y_{n}) = δ (y_{n + 1} - 2 y_{n} + y_{n - 1}) = δ y_{n + 1} - 2 δ y_{n} + δ y_{n - 1} \\ = \frac{1}{2} (y_{n + 2} - y_{n}) - (y_{n + 1} - y_{n - 1}) + \frac{1}{2} (y_{n} - y_{n} - 2) \\ = \frac{1}{2} (y_{n + 2} - 2 y_{n + 1} + 2 y_{n - 1} - y_{n - 2}) \approx h^{3} {(\frac{d^{3} y}{d x^{3}})}_{n} \end{matrix} & (7.8) \end{matrix}$

$\begin{matrix} δ^{4} y_{n} = y_{n + 2} - 4 y_{n + 1} - 6 y_{n} - 4 y_{n - 1} + y_{n - 2} \approx h^{4} {(\frac{d^{4} y}{d x^{4}})}_{n} & (7.9) \end{matrix}$

Examination of Eqs. (7.7) and (7.9) reveals that for even-order derivatives, the coefficients of y_n, y_n+1 are equal to the coefficients in the binomial expansion (a-b)^m

Unless otherwise specified, we 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 Chapter 6.
For purposes of illustration, consider a rectangular boundary as shown 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

$\begin{matrix} \begin{matrix} \frac{\partial w}{\partial x} \approx \frac{1}{h} δ_{x} w, & \frac{\partial w}{\partial y} \approx \frac{1}{h} δ_{y} w \end{matrix} & (7.10) \end{matrix}$

$\begin{matrix} \begin{matrix} \frac{\partial^{2} w}{\partial x^{2}} \approx \frac{1}{h^{2}} δ_{x}^{2} w, & \frac{\partial^{2} w}{\partial x^{2}} \approx \frac{1}{h^{2}} δ_{y}^{2} w, & \frac{\partial^{2} w}{\partial x \partial y} \approx \frac{1}{h^{2}} δ_{x} (\frac{\partial w}{\partial y}) \end{matrix} & (7.11) \end{matrix}$

A rectangular boundary divided into a square mesh is shown. — Figure 7.2. *Rectangular boundary divided into a square mesh*.

Twelve dots are shown on the rectangular boundary divided into a 6 by 7 square mesh. The positions of the dots are as follows: dot 1 (third-row fourth column); dot 2 (second-row third column); dot 3 (third-row second column); and dot 4 (fourth-row third column). Here dots 1 to 4 form a diamond pattern. Dot 5 (second row fourth column); dot 6 (second row second column); dot 7 (fourth row second column); dot 8 (fourth row fourth column). These dots form a square. Dot 9 (third row fifth column); dot 10 (first row third column); dot 11 (third row first column); and dot 12 (fifth row third column). These dots form a diamond pattern.

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

$\begin{matrix} \begin{matrix} \frac{\partial w}{\partial x} \approx \frac{1}{2 h} [w (x + h, y) - w (x - h, y)] = \frac{1}{2 h} (w_{1} - w_{3}) \\ \frac{\partial w}{\partial y} \approx \frac{1}{2 h} [w (x, y + h) - w (x, y - h)] = \frac{1}{2 h} (w_{2} - w_{4}) \end{matrix} & (7.12) \end{matrix}$

and

$\begin{matrix} \begin{matrix} \frac{\partial^{2} w}{\partial x^{2}} & \approx \frac{1}{h^{2}} [w (x + h, y) - 2 w (x, y) + w (x - h, y)] \\ = \frac{1}{h^{2}} (w_{1} - 2 w_{0} + w_{3}) \end{matrix} & (7.13) \end{matrix}$

$\begin{matrix} \begin{matrix} \frac{\partial^{2} w}{\partial y^{2}} & \approx \frac{1}{h^{2}} (w_{2} - 2 w_{0} + w_{4}) \\ \frac{\partial^{2} w}{\partial x \partial y} & \approx \frac{1}{2 h^{2}} (δ_{x} w_{2} - δ_{x} w_{4}) = \frac{1}{4 h^{2}} (w_{5} - w_{6} + w_{7} - w_{8}) \end{matrix} & (7.14) \end{matrix}$

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 be converted to finite difference form. Through this approach, the solution of a differential equation is reduced to the simultaneous solution of a set of linear, algebraic equations, written for every nodal point within the boundary.

Example 7.1 Torsion Bar with Square Cross Section

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

Solution The governing partial differential equation is (see Section 6.3)

$\begin{matrix} \frac{\partial^{2} Φ}{\partial x^{2}} + \frac{\partial^{2} Φ}{\partial y^{2}} = - 2 G θ & (7.15) \end{matrix}$

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

$\begin{matrix} Φ_{1} + Φ_{2} + Φ_{3} + Φ_{4} - 4 Φ_{0} = - 2 G θ h^{2} & (7.16) \end{matrix}$

A similar expression can be 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, as 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:

$\begin{matrix} 4 Φ_{d} - 4 Φ_{b} & = - 2 G θ h^{2} \\ 2 Φ_{d} - 4 Φ_{c} & = - 2 G θ h^{2} \\ 2 Φ_{c} + Φ_{b} - 4 Φ_{d} & = - 2 G θ h^{2} \end{matrix}$

A square cross-section of a torsion bar is shown. — Figure 7.3. *Example 7.1. A square cross section of a torsion bar*.

A 4 by 4 square grid along the XY plane is shown. The sides of the square grid are labeled, a. The height of each column measures, h. Nine dots are shown on the square grid. The dot positions are as follows: Dots c, d, and c in the first row; dots, d, b, and d in the second row, and dots c, d, and c in the third row, respectively. These dots are plotted in a square pattern. The grid lines along the rows and columns are marked 0. A dashed line is drawn diagonally from the top-left corner to the center point, b.

Simultaneous solution yields

$\begin{matrix} \begin{matrix} Φ_{b} = 2.25 G θ h^{2}, & Φ_{c} = 1.375 G θ h^{2}, & Φ_{d} = 1.75 G θ h^{2} \end{matrix} & (a) \end{matrix}$

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

Table 7.1. Values of the Forward Differences for Given Φ's

x	Φ/Gθh²	Δ/Gθh²	Δ²/Gθh²	Δ³/Gθh²	Δ⁴/Gθh²
0	0	1.75	-1.25	0.25	-0.5
H	1.75	0.5	-1.0	-0.25
2h	2.25	-0.5	-1.25
3h	1.75	-1.75
4h	0

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

$\begin{matrix} \begin{matrix} Φ (x) = Φ_{0} + \frac{x}{h} Δ Φ_{0} + \frac{x (x - h)}{2! h^{2}} Δ^{2} Φ_{0} = \frac{x (x - h) (x - 2 h)}{3! h^{3}} Δ^{3} Φ_{0} \\ + \frac{x (x - h) (x - 2 h) (x - 3 h)}{4! h^{4}} Δ^{4} Φ_{0} \\ + \dots + \frac{x (x - h) \dots (x - n h + h)}{n! h^{n}} Δ^{n} Φ_{0} \end{matrix} & (7.17) \end{matrix}$

Here the ΔΦ₀ values are the forward differences calculated at x = 0 as follows:

$\begin{matrix} \begin{matrix} Δ Φ_{0} & = Φ_{d} - Φ_{0} = (1.75 - 0) G θ h^{2} = 1.75 G θ h^{2} \\ Δ^{2} Φ_{0} & = Φ_{b} - 2 Φ_{d} + Φ = 1.25 G θ h^{2} \\ Δ^{3} Φ_{0} & = Φ_{d} - 3 Φ_{b} + 3 Φ_{d} - Φ_{0} = 0.25 G θ h^{2} \\ Δ^{4} Φ_{0} & = Φ_{0} - 4 Φ_{d} + 6 Φ_{b} - 4 Φ_{d} + Φ_{0} = 0.5 G θ h^{2} \end{matrix} & (b) \end{matrix}$

The differences are also calculated at x = h, x = 2h, and so on, and are listed in Table 7.1. We can readily obtain the values given in Table 7.1 for the given Φ 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, after differentiating Eq. (7.17) with respect to x and then setting x = 0, the result is

$\begin{matrix} {(\frac{\partial Φ}{\partial x})}_{0} = \frac{1}{h} (Δ Φ_{0} - \frac{1}{2} Δ^{2} Φ + \frac{1}{3} Δ^{3} Φ - \frac{1}{4} Δ^{4} Φ_{0} + \dots) & (c) \end{matrix}$

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

${(\frac{\partial Φ}{\partial x})}_{0} = τ_{\max} = \frac{31}{12} G θ h = 0.646 G θ a$

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

Comments By using a finer network, we can improve the result. For example, if we select h = a/6, we obtain six nodal equations. 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

As mentioned earlier, one important strength of numerical analysis is its adaptability to irregular geometries. In this section, we turn 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.

An irregular geometric shape is shown to represent a curved boundary and an irregular star operator. — Figure 7.4. *Curved boundary and irregular star operator*.

An irregular geometric shape shows four arms bc, bd, be, and bf with b as the center. The length of the arms, bc, bd, bf, and ef are labeled, h1, h2, h, h, respectively. A curve joining d and c represent, W equals f of (x, y).

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

$\begin{matrix} w (x, y) = w_{b} + a_{1} x + a_{2} y + a_{3} x^{2} + a_{4} y^{2} + a_{5} x y & (7.18) \end{matrix}$

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

$\begin{matrix} \begin{matrix} w_{c} = w_{b} + a_{1} h_{1} + a_{3} h_{1}^{2} \\ w_{d} = w_{b} + a_{2} h_{2} + a_{4} h_{2}^{2} \\ w_{e} = w_{b} + a_{1} h + a_{3} h^{2} \\ w_{f} = w_{b} + a_{2} h + a_{4} h^{2} \end{matrix} & (a) \end{matrix}$

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

$\begin{matrix} \begin{matrix} {(\frac{\partial^{2} w}{\partial x^{2}})}_{b} = 2 a_{3}, & {(\frac{\partial^{2} w}{\partial y^{2}})}_{b} = 2 a_{4} \end{matrix} & (b) \end{matrix}$

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

$\begin{array}{l} {(\frac{\partial^{2} w}{\partial x^{2}})}_{b} = 2 \frac{h (w_{c} - w_{b}) + h_{1} (w_{e} - w_{b})}{h h_{1} (h + h_{1})} \\ {(\frac{\partial^{2} w}{\partial y^{2}})}_{b} = 2 \frac{h (w_{d} - w_{b}) + h_{2} (w_{f} - w_{b})}{h h_{2} (h + h_{2})} \end{array}$

Introducing this expression into the Laplace operator, we obtain

$\begin{matrix} \begin{matrix} h^{2} {(\frac{\partial^{2} w}{\partial x^{2}})}_{b} = \frac{2 w_{c}}{α_{1} (1 + α_{1})} + \frac{2 w_{d}}{α_{2} (1 + α_{2})} \\ + \frac{2 w_{e}}{1 + α_{1}} + \frac{2 w_{f}}{1 + α_{2}} - (\frac{2}{α_{1}} + \frac{2}{α_{2}}) w_{b} \end{matrix} & (7.19) \end{matrix}$

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

This 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

$\begin{matrix} h^{2} {(\frac{\partial^{2} w}{\partial x^{2}})}_{b} = \frac{2 w_{c}}{α_{1} (1 + α_{1})} + \frac{2 w_{e}}{1 + α_{1}} - \frac{2 w_{b}}{α_{1}} & (7.20) \end{matrix}$

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

$\begin{matrix} h^{2} \frac{d^{2} w}{d x^{2}} = \frac{2}{α (α + 1)} [α w_{n - 1} - (1 + α) w_{n} + w_{n + 1}] & (7.21) \end{matrix}$

Example 7.2 Elliptical Bar Under Torsion

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.

The elliptical cross-section of a torsion member is shown. — Figure 7.5. *Example 7.2. Elliptical cross section of a torsion member*.

An elliptical section is shown along the XY plane. The focal points A and B are marked on the x and y-axes, respectively. The length of the major and minor axes are a and b, appropriately. Six connected points b, c, d, e, f, and g are plotted inside the ellipse. The point g is marked at the origin, at a distance, h from f and e. Other points b, c, and d correspond to the position of the previous dots at a distance h from each other. The distance between the point bc is h1, point cd is h2, and de is h3. The distance between the points bB and eA are h, respectively. The circumference of the ellipse is labeled with an equation x squared over a squared plus y squared over b squared equals 1.

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, and 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, which are presented in the following matrix form:

$[\begin{matrix} 2 & 0 & 0 & 0 & 2 & - 4 \\ 0 & 2 & 0 & 1 & - 4 & 1 \\ 0 & 0 & 2 & - 4 & 1 & 0 \\ - 4 & 2 & 0 & 0 & 0 & 1 \\ 1 & - 4.27 & 1 & 0 & 1.06 & 0 \\ 0 & 1.25 & - 7.41 & 1.34 & 0 & 0 \end{matrix}] {\begin{cases} Φ_{b} \\ Φ_{c} \\ Φ_{d} \\ Φ_{e} \\ Φ_{f} \\ Φ_{g} \end{cases}} = {\begin{cases} - 2 h^{2} G θ \\ - 2 h^{2} G θ \\ - 2 h^{2} G θ \\ - 2 h^{2} G θ \\ - 2 h^{2} G θ \\ - 2 h^{2} G θ \end{cases}}$

These equations are solved to yield

$\begin{matrix} Φ_{b} = 2.075 G θ h^{2}, & Φ_{c} = 1.767 G θ h^{2}, & Φ_{d} = 0.843 G θ h^{2} \\ Φ_{e} = 1.536 G θ h^{2}, & Φ_{f} = 2.459 G θ h^{2}, & Φ_{g} = 2.767 G θ h^{2} \end{matrix}$

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

$\begin{matrix} Δ Φ_{B} & = 2.075 G θ h^{2}, & Δ^{3} Φ_{B} & = - 0.001 G θ h^{2} \\ Δ^{2} Φ_{B} & = - 1.383 G θ h^{2}, & Δ^{4} Φ_{B} & = 0.002 G θ h^{2} \end{matrix}$

Similarly, for point A, we obtain

$\begin{matrix} Δ Φ_{A} & = 1.53 G θ h^{2}, & Δ^{4} Φ_{A} & = 0.001 G θ h^{2} \\ Δ^{2} Φ_{A} & = - 0.613 G θ h^{2}, & Δ^{5} Φ_{A} & = 0.001 G θ h^{2}, \\ Δ^{3} Φ_{A} & = - 0.002 G θ h^{2}, & Δ^{6} Φ_{A} & = - 0.002 G θ h^{2}, \end{matrix}$

Thus,

$\begin{matrix} τ_{B} & = {(\frac{\partial Φ}{\partial y})}_{B} = (2.075 + \frac{1.383}{2} - \frac{0.001}{3} + \frac{0.002}{4}) G θ h \\ = 2.766 G θ h = 1.383 G θ b = 0.01383 G θ \\ τ_{A} & = {(\frac{\partial Φ}{\partial x})}_{A} = (1.536 + \frac{0.613}{2} - \frac{0.002}{3} - \frac{0.001}{4} + \frac{0.001}{5} + \frac{0.002}{6}) G θ h \\ = 1.842 G θ h = 0.614 G θ a = 0.00921 G θ \end{matrix}$

Comment Note that, according to the exact theory, the maximum stress occurs at y = b and is equal to 1.384Gθb (see Example 6.4), 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 to be free of constraints. In the real world, 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:

$\begin{matrix} \begin{matrix} υ_{n + 1} - 2 υ_{n} + υ_{n - 1} & = 0 \\ υ_{n + 2} - 2 υ_{n + 1} + 2 υ_{n - 1} - υ_{n - 2} & = 0 \end{matrix} & (a) \end{matrix}$

Two figures illustrate homogeneous boundary conditions in finite differences. — Figure 7.6. *Boundary conditions in finite differences: (a) clamped or fixed support; (b) simple support*.

Two figures are shown. Figure (a) shows a deflected beam of velocity V subscript n minus 1 on the left edge and V subscript n plus 1 on the right edge supported on a clamped edge (n) at the center of velocity Vn equals 0. Here V subscript n plus 1 equals V subscript n minus 1. Figure (b) shows a simple beam, pinned with a clamped edge at the center. The right-end shows the velocity of V subscript n plus 1 and the left-end shows the velocity of V subscript n minus 1. The pinned edge shows the velocity of V subscript n equals 0. Here, V subscript n plus 1 equals negative v subscript n minus 1.

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

$\begin{matrix} υ_{n + 1} - 2 υ_{n} + υ_{n - 1} = h^{2} {(\frac{M}{E I})}_{n} & (7.22) \end{matrix}$

An illustration of finite difference representation of a non-prismatic cantilever beam with varying load is shown. — Figure 7.7. *Finite difference representation of a nonprismatic cantilever beam with varying load*.

A non-prismatic cantilever beam subjected to various loadings along the XY plane is shown. The deflected beam of length, L, shows m segments of varying depths and constant width and further extended to, m plus 1, and m plus 2 along the x-axis. The nodal point n is shown at the center of the segments. The loads P1, Pn, and Pm alter in terms of the nodes are shown.

where 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) is expressed as

$\begin{matrix} υ_{n + 2} - 4 υ_{n + 1} + 6 υ_{n} - 4 υ_{n - 1} + υ_{n - 2} = h^{4} {(\frac{p}{E I})}_{n} & (7.23) \end{matrix}$

where p_n is the load intensity at n. Equations identical to Eq. (7.23) 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

$\begin{matrix} θ_{n} = {(\frac{\partial υ}{\partial x})}_{n} = \frac{υ_{n + 1} - υ_{n - 1}}{2 h} & (7.24) \end{matrix}$

The following simple examples further illustrate this method of solution.

Example 7.3 Displacements of a Simple Beam

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

An illustration for a simply supported beam with varying load is shown. — Figure 7.8. *Example 7.3. Simply supported beam with varying load*.

A cantilever beam of length L subjected to a trapezoidal loading along the XY plane is shown. The beam is supported by a pin at one end and roller at the other end. The beam is divided (by force) into four parts p (pin), 1, 2, 3, and 2p (roller). The force that acts above the beam is marked 5 over 4 p on 1, 3 over 2 p on 2, and 7 over 4 p on 3. The second figure for the beam shows the pin is marked n equals 0 and roller is marked n equals 4 with 1, 2, and 3 between them. To the left of the beam extends and represents negative 1 and to the right of the beam extends and represents 5.

Solution For simplicity, take h = L/4 (Fig. 7.8b). The boundary conditions υ(0) = υ(L) = 0 and υ"(0) = υ″ = υ″(L) = 0 are replaced by finite difference conditions by setting υ(0) = υ₀ and υ(L)=υ₄, and then applying Eq. (7.7):

$\begin{matrix} \begin{matrix} υ_{0} = 0, & υ_{1} = - υ_{- 1}, & υ_{4} = 0 & υ_{3} = - υ_{5} \end{matrix} & (b) \end{matrix}$

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

$\begin{matrix} \begin{matrix} υ_{3} - 4 υ_{2} + 6 υ_{1} - 4 υ_{0} + υ_{- 1} & = \frac{L^{4}}{256} \frac{5 p}{4 E I} \\ υ_{4} - 4 υ_{2} + 6 υ_{2} - 4 υ_{1} + υ_{0} & = \frac{L^{4}}{256} \frac{3 p}{2 E I} \\ υ_{4} - 4 υ_{3} + 6 υ_{2} - 4 υ_{1} + υ_{0} & = \frac{L^{4}}{256} \frac{7 p}{4 E I} \end{matrix} & (c) \end{matrix}$

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

$υ_{1} = 0.0139 \frac{p L^{4}}{E I}, υ_{2} = 0.0198 \frac{p L^{4}}{E I}, υ_{3} = 0.0144 \frac{p L^{4}}{E I}$

Then, from Eq. (7.24), we obtain

$θ_{2} = \frac{υ_{3} - υ_{1}}{2 h} = 0.001 \frac{p L^{3}}{E I}$

Note that, by successive integration of EId⁴υ/dx⁴ = p(x), the result υ₂ = 0.0185 PL⁴/EI is obtained. Thus, even a coarse segmentation leads to a satisfactory solution in this case.

Example 7.4 Reactions of a Continuous Beam

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

An illustration for the Statically indeterminate beam is shown. — Figure 7.9. *Example 7.4. Statically indeterminate beam: (a) load diagram; (b) and (c) moment diagrams*.

The load diagram shows the beam on XY plane supported by a pin on the left (n equals 0), pin at the center (n equals 2), and pin on the right (n equals 4). The beam marked R at the center. The length of the beam is marked two L over 2 on the left and right of R, equally. Two downward force of 2P is marked above the beam between L over 2 on both sides. The moment diagram for the beam displays a Quadrilateral area labeled PL below the x-axis which marked M subscript P upward force on the left. Another moment diagram for the beam displays a linear slope bend (triangle) labeled RL over 2 above the x-axis. The upward and downward force acts above and below the linear slope. The M subscript P upward force is marked on the left.

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

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

$\begin{matrix} \begin{array}{l} υ_{0} - 2 υ_{1} + υ_{2} = \frac{L^{2}}{4 E I} (- P L + \frac{R L}{4}) \\ υ_{1} - 2 υ_{2} + υ_{3} = \frac{L^{2}}{4 E I} (- P L + \frac{R L}{2}) \\ υ_{2} - 2 υ_{3} + υ_{4} = \frac{L^{2}}{4 E I} (- P L + \frac{R L}{4}) \end{array} & (d) \end{matrix}$

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 Stepped Cantilever Beam

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).

An illustration for Nonprismatic cantilever beam under concentrated loads is shown. — Figure 7.10. *Example 7.5. Nonprismatic cantilever beam under concentrated loads*.

A non-prismatic cantilever beam subjected concentrated loads along the XY plane is shown. The beam is fixed at one end and free at the other end. The beam is marked upper A, C, and B on the left, center, and right end. Two downward force acts above C and B. The distance from C to upper A and C to B are marked a. The second figure shows the beam fixed on the left end (n equals 0) and free on the right end (n equals 4). The non-prismatic cantilever beam is marked 1, 2, and 3. The distance between 1 and 2 is marked h equals a over 2. The fixed end is extended and marks negative 1 and the free end is extended and marks 5.

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

$\begin{matrix} \begin{matrix} υ_{0} = 0, & υ_{1} = υ_{- 1} \end{matrix} & (e) \end{matrix}$

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

$\begin{matrix} \begin{array}{l} υ_{1} - 2 υ_{0} + υ_{- 1} = \frac{3 P a}{2 E I} h^{2} \\ υ_{2} - 2 υ_{1} + υ_{0} = \frac{P a}{E I} h^{2} \\ υ_{3} - 2 υ_{2} + υ_{1} = \frac{P a}{1.5 E I} h^{2} \\ υ_{4} - 2 υ_{3} + υ_{2} = \frac{P a}{2 E I} h^{2} \end{array} & (f) \end{matrix}$

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

$υ_{B} = υ_{4} = \frac{47 P a^{3}}{24 E I}$

Comment The preceding deflection is approximately 2.2% larger than the exact value.

Part B: Finite Element Analysis

7.6 Fundamentals

Structural analysis involves the determination of the forces and deflections within a structure or its members. The earliest demands for structural analysis led to a host of so-called classical methods. Over time, the specialization of these classical methods was replaced by generalities of the modern matrix methods. The presentation of such methods in this chapter is limited to the most widely used of these techniques: the finite element stiffness or displacement method. Unless otherwise specified, we will refer to this approach as the finite element method (FEM). Finite element analysis (FEA) is a numerical approach that is well suited to digital computers. It relies on the formulations of a simultaneous set of algebraic equations relating forces to corresponding displacements at discrete preselected points (called nodes) on the structure. These governing algebraic equations, also called the force-displacement relations, are expressed in matrix notations.

The powerful finite element method had its beginnings in the 1980s. Since then, with the advent of high-speed, large-storage-capacity digital computers, it has gained great prominence throughout industry for the solution of practical analysis and design problems of high complexity. FEA offers many advantages. The system's structural geometry can be readily described, and combined load conditions can be easily handled. This technique also offers the ability to treat discontinuities, to handle composite and anisotropic materials, to handle unlimited numbers and kinds of boundary conditions, to handle dynamic and thermal loadings, and to treat nonlinear structural problems. In addition, it has the capacity for complete automation. The literature related to FEA is extensive (see, for example, Refs. 7.2 through 7.9). Numerous commercial FEA software programs are available, as described in Section 7.16, including some directed at the learning process.

The basic concept of the finite element approach is that the real structure can be divided or discretized by a finite number of elements, connected not only at their nodes but also along the interelement boundaries. Triangular, rectangular, tetrahedral, quadrilateral, and hexagonal forms of elements are often employed in the finite element method. The types of elements that are commonly used in structural idealization are the truss, beam, two-dimensional elements, shell and plate bending, and three-dimensional elements. Figure 7.11a depicts how a multistory hotel building is modeled using bar, beam, column, and plate elements, which can be employed for static, free vibration, earthquake response, and wind response analysis [Ref. 7.2, page 6].

Two figures shows the finite element models of two structure. — Figure 7.11. *Finite element models of two structures: (a) multistory building; (b) pipe connection*.

The multistory building element picks a node from one of the stories and zoomed in. The zoomed in node includes bar element and beam element. Another node picked a node below the story, zoomed in, and seen. It includes the plate element. The figure of the pipe element picks a node, zoomed in, and that includes the triangular element.

A model of a nozzle in a thin-walled cylinder, created using triangular shell elements, is shown in Fig. 7.11b [Ref. 7.3]. Similarly, large structural systems, such as aircraft or ships, are usually analyzed by dividing the structure into smaller units or substructures (e.g., the wing of an airplane). When the stiffness of each unit has been determined, the analysis of the system follows the familiar procedure established by the matrix methods used in structural mechanics.

The network of elements and nodes that discretize the region is called a mesh. The density of a mesh increases as more elements are placed within a given region. In mesh refinement, the mesh is modified through an analysis of a model to give improved solutions. Mesh density is increased in areas of high stress concentrations and when geometric transition zones are meshed smoothly (Fig. 7.11b). Usually, the FEA results converge toward the exact results as the mesh is continuously refined.

To discuss FEA adequately would require a far lengthier presentation than can be justified here. However, the subject is sufficiently important that engineers concerned with the analysis and design of members should have at least an understanding of FEA. The fundamentals presented in this chapter clearly show both the potential of FEA and its complexities. The material provided here can be covered as an option, used as a “teaser” for a student's advanced study of the topic, or used as a professional reference.

7.7 The Bar Element

A bar element, also called a truss bar element or axial element, represents the simplest form of structural finite element. An element of this type of length L, modulus of elasticity E, and cross-sectional area A, is denoted by e (Fig. 7.12). The two ends or joints or nodes are numbered 1 and 2, respectively. A set of two equations in matrix form are required to relate the nodal forces $({\bar{F}}_{1} and {\bar{F}}_{2})$ to the nodal displacements $({\bar{u}}_{1} and {\bar{u}}_{2})$ .

The figure of one-dimensional axial element is shown. — Figure 7.12. *The one-dimensional axial element (e)*.

The one-dimensional axial element of length L marks 1 and 2 on either ends (nodes) with the element representation (encircled 'e') at the center. A vertical dashed line is drawn before the element representation and the distance between the dashed line and the element representation is marked x bar. A rightward arrow from end 1 is marked and labeled F1 bar, u1 bar. And a rightward arrow from the end 2 is marked and labeled F2 bar, u2 bar.

7.7.1 Equilibrium Method

The direct equilibrium approach is simple and physically clear, but it is practically well suited only for truss, beam, and frame elements. The equilibrium of the x-directed forces requires that ${\bar{F}}_{1} = - {\bar{F}}_{2}$ (Fig. 7.12). In terms of the spring rate AE/L of the element, we have

$\begin{matrix} {\bar{F}}_{1} = \frac{A E}{L} ({\bar{u}}_{1} - {\bar{u}}_{2}) & {\bar{F}}_{2} = \frac{A E}{L} ({\bar{u}}_{2} - {\bar{u}}_{1}) \end{matrix}$

This may be written in matrix form,

$\begin{matrix} {\begin{cases} {\bar{F}}_{1} \\ {\bar{F}}_{2} \end{cases}} = \frac{A E}{L} [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] {\begin{cases} {\bar{u}}_{1} \\ {\bar{u}}_{2} \end{cases}} & (7.25 a) \end{matrix}$

or symbolically,

$\begin{matrix} {\bar{F}} = {[\bar{k}]}_{e} {\bar{u}}_{e} & (7.25 b) \end{matrix}$

The quantity ${[\bar{k}]}_{e}$ is called the element stiffness matrix and has dimensions of force per unit displacement. It relates the nodal displacements ${\bar{u}}_{e}$ to the nodal forces ${\bar{F}}_{e}$ on the element.

7.7.2 Energy Method

The energy approach is more general, easier to apply, and more powerful than the direct method just discussed, particularly for complex kinds of finite elements [Ref. 7.2]. Using this technique, it is necessary to first define a displacement function for the element (Fig. 7.12):

$\begin{matrix} \bar{u} = a_{1} + a_{2} \bar{x} & (7.26) \end{matrix}$

where a₁ and a₂ are constants. Clearly, Eq. (7.26) represents a continuous linear displacement variation along the x axis of the element that corresponds to that of engineering formulation for a bar under axial loading. The axial displacements of joints 1 (at x = 0) and 2 (at $\bar{x} = L$ ), respectively, are

${\bar{u}}_{1} = a_{1} {\bar{u}}_{2} = a_{1} + a_{2} L$

Solving these equations, we have $a_{1} = {\bar{u}}_{1}$ and $a_{2} = - ({\bar{u}}_{1} - {\bar{u}}_{2})/ L$ . Carrying these results into Eq. (7.26), we obtain

$\begin{matrix} \bar{u} = (1 - \frac{\bar{x}}{L}) {\bar{u}}_{1} + \frac{\bar{x}}{L} {\bar{u}}_{2} & (7.27) \end{matrix}$

Using Eq. (7.27), we write

$\begin{matrix} ε_{x} = \frac{d \bar{u}}{d x} = \frac{1}{L} (- {\bar{u}}_{1} + {\bar{u}}_{2}) & (7.28) \end{matrix}$

Thus, the axial force in the element is

$\begin{matrix} \bar{F} = (E ε_{x}) A = \frac{A E}{L} (- {\bar{u}}_{1} + {\bar{u}}_{2}) & (7.29) \end{matrix}$

Substituting Eq. (7.29) into Eq. (2.58), the strain energy in the element may be expressed as follows:

$\begin{matrix} U = \int_{0}^{L} \frac{F^{2} d x}{2 A E} = \frac{A E}{2 L} ({\bar{u}}_{1}^{2} - 2 {\bar{u}}_{1} {\bar{u}}_{2} + {\bar{u}}_{2}^{2}) & (7.30) \end{matrix}$

Then, Castigliano's first theorem, Eq. (10.22), gives

$\begin{matrix} {\bar{F}}_{1} & = \frac{\partial U}{\partial {\bar{u}}_{1}} = \frac{A E}{L} ({\bar{u}}_{1} - {\bar{u}}_{2}) \\ {\bar{F}}_{2} & = \frac{\partial U}{\partial {\bar{u}}_{2}} = \frac{A E}{L} (- {\bar{u}}_{1} + {\bar{u}}_{2}) \end{matrix}$

The matrix form of these equations is the same as that given by Eqs. (7.25).

7.8 Arbitrarily Oriented Bar Element

In this section, we develop the global stiffness matrix for an element oriented arbitrarily in a two-dimensional plane. On the one hand, the local coordinates are chosen to conveniently represent the individual element. On the other hand, the reference or global coordinates are chosen to be convenient for the entire structure (Fig. 7.13). The local and global coordinates systems for an axial element are designated by $\bar{x}, \bar{y}$ and x, y, respectively.

The illustration of a truss depicts the global and local coordinates for a bar element. — Figure 7.13. *Global coordinates (x,y) for plane truss and local coordinates $(\bar{x}, \bar{y})$ for a bar element 1–2.*

The truss subjected on an XY plane shows the two members on the left and right connected to two rigid supports 1 and 4 at the bottom and hinge support 3 at the top. Another two members from the rigid support 1 and 4 are connected to the hinge support 2 (above hinge support 3). Finally, the hinge supports 2 and 3 are connected.

7.8.1 Coordinate Transformation

Figure 7.14 depicts a typical axial element e lying along the x axis, which is oriented at an angle θ, measured counterclockwise from the reference axis x. In the local coordinate system, each joint has an axial force ${\bar{F}}_{x}$ , a transverse force ${\bar{F}}_{y}$ , an axial displacement $\bar{u}$ , and a transverse displacement $\bar{υ}$ . Referring to the figure, Eq. (7.25a) may be expanded as follows:

$\begin{matrix} {\begin{cases} {\bar{F}}_{1 x} \\ {\bar{F}}_{1 y} \\ {\bar{F}}_{2 x} \\ {\bar{F}}_{2 y} \end{cases}}_{e} = \frac{A E}{L} [\begin{matrix} 1 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ - 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] {\begin{cases} {\bar{u}}_{1} \\ {\bar{υ}}_{1} \\ {\bar{u}}_{2} \\ {\bar{υ}}_{2} \end{cases}}_{e} & (7.31 a) \end{matrix}$

The two-dimensional bar element with the local and global coordinates are shown. — Figure 7.14. *Local $(\bar{x}, \bar{y})$ and global (x,y) coordinates for a twodimensional bar element (e)*.

The two-dimensional bar element of length L marks 1 and 2 on either end (nodes) with element representation (encircled e) at the center. The angle between the bar and end 1 is marked theta. The end 1 marks the following forces: F bar 1y, nu 1 bar; F1x, u1; F bar 1x, u1 bar and F1y, nu 1. The end 2 marks the following forces: F bar 2y, nu 2 bar; F2y, nu 2; F bar 2x, nu 2 and F2x, u2. Another diagram in XY plane shows the x bar is diagonally drawn from the origin toward right and y bar is diagonally drawn from the origin toward left. The angle between x and x bar is marked theta.

or concisely,

$\begin{matrix} {\bar{F}}_{e} = {[\bar{k}]}_{e} {\bar{δ}}_{e} & (7.31 b) \end{matrix}$

Here the quantities ${[\bar{k}]}_{e}$ and ${[\bar{δ}]}_{e}$ are the stiffness and nodal displacement matrices, respectively, in the local coordinate system.

7.8.2 Force Transformation

From Fig. 7.14, we can see that the two local and global forces at joint 1 may be related by the following expressions:

$\begin{array}{l} {\bar{F}}_{1 x} = F_{1 x} \cos θ + F_{1 y} \sin θ \\ {\bar{F}}_{1 y} = F_{1 x} \sin θ + F_{1 y} \cos θ \end{array}$

Similar equations apply at joint 2. For brevity, let

$\begin{matrix} \begin{matrix} c = \cos θ & and & s = \sin θ \end{matrix} & (a) \end{matrix}$

Hence, the local and global forces are related in the matrix form as

$\begin{matrix} {\begin{cases} {\bar{F}}_{1}_{x} \\ {\bar{F}}_{1}_{y} \\ {\bar{F}}_{2}_{x} \\ {\bar{F}}_{2 y} \end{cases}}_{e} = [\begin{array}{r} c & s & 0 & 0 \\ - s & c & 0 & 0 \\ 0 & 0 & c & s \\ 0 & 0 & - s & c \end{array}] {\begin{cases} F_{1 x} \\ F_{1 y} \\ F_{2 x} \\ F_{2 y} \end{cases}}_{e} & (7.32 a) \end{matrix}$

or symbolically,

$\begin{matrix} {\bar{F}}_{e} = [T] {F}_{e} & (7.32 b) \end{matrix}$

In Eq. (7.32b), [T] represents the coordinate transformation matrix:

$\begin{matrix} [T] = [\begin{array}{r} c & s & 0 & 0 \\ - s & c & 0 & 0 \\ 0 & 0 & c & s \\ 0 & 0 & - s & c \end{array}] & (7.33) \end{matrix}$

{F}_e is the global nodal force matrix:

$\begin{matrix} {F}_{e} = {\begin{cases} F_{1 x} \\ F_{1 y} \\ F_{2 x} \\ F_{2 y} \end{cases}}_{e} & (7.34) \end{matrix}$

We note that the coordinate transformation matrix satisfies the following conditions of orthogonality: c² + s² = 1, (–s)² + c² = 1, and (–s)(c) + (c)(s) = 0. Thus, [T] is an orthogonal matrix.

7.8.3 Displacement Transformation

Inasmuch as the displacement transforms in an identical way as forces, we can write

$\begin{matrix} {\begin{cases} {\bar{u}}_{1} \\ {\bar{υ}}_{1} \\ {\bar{u}}_{2} \\ {\bar{υ}}_{2} \end{cases}}_{e} = [T] {\begin{cases} u_{1} \\ υ_{1} \\ u_{2} \\ υ_{2} \end{cases}}_{e} & (7.35 a) \end{matrix}$

The concise form of this matrix is

$\begin{matrix} {\bar{δ}}_{e} = {T} {δ}_{e} & (7.35 b) \end{matrix}$

where {δ} is the global nodal displacements. Carrying Eqs. (7.35b) and (7.32b) into Eq. (7.31b) results in

$[T] {F}_{e} = {[\bar{k}]}_{e} [T] {δ}_{e}$

from which

${F}_{e} = {[T]}^{- 1} {[\bar{k}]}_{e} [T] {δ}_{e}$

The inverse of the orthogonal matrix [T] is the same as its transpose: [T]^–1 = [T]_T. Here the superscript T denotes the transpose. Recall that the transpose of a matrix is found by interchanging the rows and columns of the matrix.

7.8.4 Governing Equations

The global force-displacement relations or the governing equations for an element e can be expressed in the form

$\begin{matrix} {F}_{e} = {[k]}_{e} {δ}_{e} & (7.36) \end{matrix}$

where

$\begin{matrix} {[k]}_{e} = {[T]}^{T} {[\bar{k}]}_{e} [T & (7.37) \end{matrix}$

The global stiffness matrix for the element, substituting Eq. (7.33) and ${[\bar{k}]}_{e}$ from Eq. (7.31a) into Eq. (7.37), may be written as

$\begin{matrix} \begin{array}{l} {[k]}_{e} = \frac{A E}{L} [\begin{array}{l} c^{2} & c s & - c^{2} & - c s \\ c s & s^{2} & - c s & - s^{2} \\ - c^{2} & - c s & c^{2} & c s \\ - c s & - s^{2} & c s & s^{2} \end{array}] = \frac{A E}{L} [\begin{array}{l} c^{2} & c s & - c^{2} & - c s \\ s^{2} & - c s & - s^{2} \\ c^{2} & c s \\ Symmetric & s^{2} \end{array}] \end{array} & (7.38) \end{matrix}$

This result indicates that the element stiffness matrix depends on its dimensions, orientation, and material property.

7.9 Axial Force Equation

Reconsider the general case of a truss element 1- 2 oriented arbitrarily in a two-dimensional plane, shown in Fig. 7.14. We multiply the third and fourth expressions in Eq. (7.36) to obtain the global forces at node 2 as

$\begin{array}{l} F_{2 x} = \frac{A E}{L} [c^{2} (u_{2} - u_{1}) + c s (υ_{2} - υ_{1})] \\ F_{2 y} = \frac{A E}{L} [c s (u_{2} - u_{1}) + s^{2} (υ_{2} - υ_{1})] \end{array}$

The axial tensile force in the element with nodes 1 and 2, designated by F₁₂, is then F₁₂ = F_2xc+F_2ys, in which cos = c and sin = s. This force is equivalent to the nodal force ${\bar{F}}_{1 x}$ or ${\bar{F}}_{2 x}$ associated with the local coordinates ${\bar{x}}_{1} and {\bar{x}}_{2}$ .

Combining the preceding relationships leads to

$\begin{matrix} \begin{matrix} F_{12} & = \frac{A E}{L} (c^{2} + s^{2}) [c^{2} (u_{2} - u_{1}) + s (υ_{2} - υ_{1})] \\ = \frac{A E}{L} [c (u_{2} - u_{1}) + s (υ_{2} - υ_{1})] \end{matrix} & (a) \end{matrix}$

F₁₂ = (spring rate)(total axial elongation)

Here, the quantities (u₂-u₁) and (υ₂-υ₁) are, respectively, the horizontal and vertical components of the axial elongation. Equation (a) may be expressed in the matrix form

$F_{12} = \frac{A E}{L} [c s] {\begin{matrix} u_{2} - u_{1} \\ υ_{2} - υ_{1} \end{matrix}}$

Thus, the axial force in the bar element with nodes ij is

$\begin{matrix} F_{i j} = {(\frac{A E}{L})}_{i j} {[c s]}_{i j} {\begin{cases} u_{j} - u_{i} \\ υ_{j} - υ_{i} \end{cases}} & (7.39) \end{matrix}$

A positive (negative) value obtained for F_ij indicates that the element is in tension (compression). The axial stress, in the element of cross-sectional area A, then is σ_ij = F_ij/A.

Example 7.6 Properties of a Truss Bar Element

The element 1- 2 of the steel truss shown in Fig. 7.13 with a length L, cross-sectional area A, and modulus of elasticity E, is oriented at angle counterclockwise from the x axis.

Given:

$\begin{matrix} u_{1} & = 0.5 mm & υ_{1} & = 0.625 mm & u_{2} & = - 1.25 mm & υ_{2} & = 0 \\ θ & = 30 ° & A & = 600 {mm}^{2} & L & = 1.5 m & E & = 200 GPa \end{matrix}$

An illustration of an axially loaded bar is shown. — Figure 7.15. *Example 7.6. An axially loaded bar*.

The axially loaded bar diagonally subjected on an XY plane of length L shows the ends (nodes) of the bar marked with 1 and 2. An extended line on either side is labeled F subscript 12. The end 2 is further extended and marked x bar and the end 1 marks y bar, perpendicular to y. The angle between x and the bar is marked theta.

Find: (a) The global stiffness matrix for the element; (b) the local displacements ${\bar{u}}_{1}, {\bar{υ}}_{1}, {\bar{u}}_{2}, and {\bar{υ}}_{2}$ of the element; (c) the axial stress in the element.

Solution The free-body diagram of the element 1- 2 is shown in Fig. 7.15. The spring rate of the element is

$\frac{A E}{L} = \frac{1}{1.5} (600 \times 10^{- 6}) (200 \times 10^{9}) = 80 \times 10^{6} N/m$

and

$\begin{matrix} c = \cos 30 ° = \sqrt{3} / 2, & s = \sin 30 ° = 1 / 2 \end{matrix}$

Element Stiffness Matrix. Applying Eq. (7.38),

${[k]}_{1} = 80 (10^{6}) [\begin{array}{r} \frac{3}{4} & \frac{\sqrt{3}}{4} & - \frac{3}{4} & - \frac{\sqrt{3}}{4} \\ \frac{\sqrt{3}}{4} & \frac{1}{4} & - \frac{\sqrt{3}}{4} & - \frac{1}{4} \\ - \frac{3}{4} & - \frac{\sqrt{3}}{4} & \frac{3}{4} & \frac{\sqrt{3}}{4} \\ - \frac{\sqrt{3}}{4} & - \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{1}{4} \end{array}]$

or

${[k]}_{1} = 80 (10^{6}) [\begin{matrix} 0.75 & 0.433 & - 0.75 & - 0.433 \\ 0.433 & 0.25 & - 0.433 & - 0.25 \\ - 0.75 & - 0.433 & 0.75 & 0.433 \\ - 0.433 & - 0.25 & 0.433 & 0.25 \end{matrix}] N/m$
Element Displacements. Equation (7.35b), ${\bar{δ}}_{e} = [T] {δ}_{e}$ , results in
${\begin{cases} {\bar{u}}_{1} \\ {\bar{υ}}_{1} \\ {\bar{u}}_{2} \\ {\bar{υ}}_{2} \end{cases}} = [\begin{matrix} 0.866 & 0.5 & 0 & 0 \\ - 0.5 & 0.866 & 0 & 0 \\ 0 & 0 & 0.866 & 0.5 \\ 0 & 0 & - 0.5 & 0.866 \end{matrix}] {\begin{matrix} 0.5 \\ 0.625 \\ - 1.25 \\ 0 \end{matrix}} = {\begin{matrix} 0.75 \\ 0.29 \\ - 1.08 \\ 0.63 \end{matrix}} mm$
Axial Force. Substituting the given numerical values into Eq. (7.39) with i= 1 and j = 2, we find

$F_{1} = F_{12} = 80 (10^{6}) [\frac{\sqrt{3}}{1} \frac{1}{2}] {\begin{matrix} - 1.25 - 0.5 \\ 0 - 0.625 \end{matrix}} (10^{- 3}) = - 146.2 kN$

It follows that axial stress in the element is equal to σ₁ = F₁/A = - 244 MPa.

Comment The negative sign indicates a compression.

7.10 Force-Displacement Relations for a Truss

We now develop the finite element method by considering a plane truss. A truss is a structure composed of differently oriented straight bars connected at their joints (or nodes) by means of pins, such as that shown in Fig. 7.13. It is easier to understand the basic procedures of treatment by referring to this simple system. The general derivation of the FEM is discussed in Section 7.13. To derive the truss equations, the global element relations for the axial element given by Eq. (7.36) will be assembled. This gives the following force-displacement relations for the entire truss or the system equations:

$\begin{matrix} {F} = [K] {δ} & (7.40) \end{matrix}$

The global nodal matrix {F} and the global stiffness matrix [K] are expressed as

$\begin{matrix} {F} = \sum_{1}^{n} {F}_{e} & (7.41) \end{matrix}$

and

$\begin{matrix} {K} = \sum_{1}^{n} {k}_{e} & (7.42) \end{matrix}$

where the quantity e designates an element and n is the number of elements found within the truss. Observe that [K] relates the global nodal force {F} to the global displacement {δ} for the entire truss.

7.10.1 The Assembly Process

The element stiffness matrices [k]e in Eq. (7.38) must be properly added together or superimposed. To perform direct addition, a convenient method is to label the columns and rows of each element stiffness matrix in accordance with the displacement components related with it. The truss stiffness matrix [K] is then found simply by summing terms from the individual element stiffness matrix into their corresponding locations in [K]. Alternatively, we can expand [k]e for each element to the order of the truss stiffness matrix by adding rows and columns of zeros. We will employ this process of assemblage of the element stiffness matrices. It is obvious that, for those problems involving a large number of elements, implementing the assembly of [K] requires a digital computer.

The use of the fundamental equations developed are illustrated in the solution of the following example numerical problems.

Example 7.7 Stepped Axially Loaded Bar

A bar consists of two prismatic parts, fixed at the left end, and supports a load P at the right end, as illustrated in Fig. 7.16a. Determine the nodal displacements and nodal forces in the bar.

A stepped bar under an axial load is shown. — Figure 7.16. *Example 7.7. (a) A stepped bar under an axial load; (b) twoelement model.*

The two stepped bar of length L for each step subjected on an XY plane shows one end is fixed and the other end marks force. The bars are marked AE and 2AE. Another diagram for the stepped bar shows the bar marked 1, 2, and 3 from left to right with the co-ordinates F1x, F2x, and load respectively. The distance between 1 and 2 is marked encircled 1 and the distance between 2 and 3 is marked encircled 2.

Solution In this case, the global coordinates (x, y) coincide with local $(\bar{x}, \bar{y})$ coordinates. The bar is discretized into elements with nodes 1, 2, and 3 (Fig. 7.16b). The axial rigidities of the elements 1 and 2 are 2AE and AE, respectively. Therefore, (AE/L)₁ = 2AE/L and (AE/L)₂ = AE/L.

Element Stiffness Matrices. Through the use of Eq. (7.25), the stiffness for the elements 1 and 2 is expressed as

$\begin{array}{l} u_{1} u_{2} u_{2} u_{3} \\ {[k]}_{1} = \frac{A E}{L} [\begin{array}{r} 2 & - 2 \\ - 2 & 2 \end{array}] \begin{matrix} u_{1} \\ u_{2} \end{matrix} {[k]}_{2} = \frac{A E}{L} [\begin{array}{r} 1 & - 1 \\ - 1 & 1 \end{array}] \begin{matrix} u_{2} \\ u_{3} \end{matrix} \end{array}$

The column and row of each stiffness matrix are labeled according to the nodal displacements associated with them. There are three displacement components (u₁,u₂,u₃), so the order of the system matrix must be 3×3. In terms of the bar displacements, we write

A stiffness matrix for elements 1 and 2. — A stiffness matrix of finite element 1 equals (AE over L) times a 3 by 3 matrix. The 3 by 3 matrix is as follows: Row 1: 2, negative 2, 0; Row 2: negative 2, 2, 0; In row 3, all the entries are 0. The 3 rows and 3 columns denote u1, u2, and u3. It is indicated that all the entries of the third row and column corresponding to u3 are 0. A stiffness matrix of finite element 2 equals (AE over L) times a 3 by 3 matrix. The 3 by 3 matrix is as follows: In row 1, all the entries are 0. Row 2: 0, 1, negative 1; Row 3: 0, negative 1, 1;. The 3 rows and 3 columns denote u1, u2, and u3. It is indicated that all the entries of the first row and column corresponding to u1 are 0.

In these matrices, the last row and the first column of zeros are added, respectively (boxed in the dashed lines).

Stiffness Matrix System. The superposition of the terms of each stiffness matrix leads to

$\begin{array}{l} u_{1} u_{2} u_{3} \\ [K] = \frac{A E}{L} [\begin{array}{r} 2 & - 2 & 0 \\ - 2 & 3 & - 1 \\ 0 & - 1 & 1 \end{array}] \begin{matrix} u_{1} \\ u_{2} \\ u_{3} \end{matrix} \end{array}$

Force-Displacement Relations. Equations (7.40) are expressed as

$\begin{matrix} {\begin{cases} F_{1 x} \\ F_{2 x} \\ F_{3 x} \end{cases}} = \frac{A E}{L} [\begin{array}{r} 2 & - 2 & 0 \\ - 2 & 3 & - 1 \\ 0 & - 1 & 1 \end{array}] {\begin{cases} u_{1} \\ u_{2} \\ u_{3} \end{cases}} & (a) \end{matrix}$

The displacement and force boundary conditions are u₁ = 0,F_2x = 0, and F_3x = P. Then, Eqs. (a), referring to Fig. 7.16b, become

${\begin{cases} F_{1 x} \\ 0 \\ F_{3 x} \end{cases}} = \frac{A E}{L} [\begin{array}{r} 2 & - 2 & 0 \\ - 2 & 3 & - 1 \\ 0 & - 1 & 1 \end{array}] {\begin{matrix} 0 \\ u_{2} \\ u_{3} \end{matrix}}$

Displacements. To determine u₂ and u₃, only part of this equation is considered:

${\begin{matrix} 0 \\ P \end{matrix}} = \frac{A E}{L} [\begin{array}{r} 3 & - 2 \\ - 1 & 2 \end{array}] {\begin{cases} u_{2} \\ u_{3} \end{cases}}$

Solution of this equation yields

${\begin{cases} u_{2} \\ u_{3} \end{cases}} = {\begin{cases} P L / 2 A E \\ 3 P L / 2 A E \end{cases}}$

With the displacements available, the axial force and stress in element 1 (or 2) can readily be found as described in Example 7.6.

Nodal Forces. Equations (a) give

${\begin{cases} F_{1 x} \\ F_{2 x} \\ F_{3 x} \end{cases}} = \frac{A E}{L} [\begin{array}{r} 2 & - 2 & 0 \\ - 2 & 3 & - 1 \\ 0 & - 1 & 1 \end{array}] {\begin{matrix} 0 \\ P L / 2 A E \\ 3 P L / 2 A E \end{matrix}} = {\begin{matrix} - P \\ 0 \\ P \end{matrix}}$

Comment The results indicate that the reaction F_1x = -p is equal in magnitude but opposite in direction to the applied force at node 3, F_3x = P. Also, F_2x = 0 shows that no force is applied at node 2. Equilibrium of the bar assembly is thus satisfied.

Case Study 7.1 Analysis of a Three-Bar Truss

A steel plane truss in which all members have the same axial rigidity AE supports a horizontal force P and a load W acting at joint 2, as shown in Fig. 7.17a. Find the nodal displacements, reactions, and stresses in each member.

An illustration shows the forces acting on the members of a plane truss. — Figure 7.17. *Case Study 7.1. (a) Basic plane truss; (b) finite element model*.

An illustration shows two figures. Figure (a) shows a basic plane truss. The plane truss has three members, a member is placed in vertical position. Another member of length 'L' is attached to the top of the vertically placed member by means of a rigid support at the joint 1. Another member connects the base of the vertical member (joint 3) and the joint 2 at the free end of the horizontal member, thus makes an angle 45 degrees. The truss is represented in an xy plane, the horizontal member lies along the x-axis and the vertical member lies along the y-axis. A rightward force 'P' and a load 'W' acts downward at the joint 2. Reaction forces act at the joints 1 and 3 and are represented along the x and the y axes. In both the joints 1 and 2, the respective reaction forces along the x-axis 'R 1x' and 'R 3x' act rightward and the reaction forces along the y-axis 'R 1y' and 'R 3y' act upward. Figure (b) shows a finite element model of the plane truss. This figure represents the forces and velocities (along the x and the y axes) at each joint. At all the three joints, the respective forces along the x-axis: 'F 1x,' 'F 2x,' and 'F 3x' act rightward with velocities 'u1,' 'u2,' and 'u3' respectively. The respective forces along the y-axis at joints 1 and 3: 'F 1y' and 'F 3y,' act upward with velocities 'v1' and 'v3' respectively; and the force at the joint 2: 'F 2y' acts downward with a velocity 'v2.'

Given:

$\begin{matrix} P = 24 kN, W = 36 kN, σ_{yp} = 250 MPa, E = 200 GPa, \\ L_{1} = L_{2} = L = 2 m, L_{3} = 2 \sqrt{2} m, A = 400 {mm}^{2} \end{matrix}$

Solution The reactions are marked and the nodes numbered arbitrarily for elements in Fig. 7.17a.

Input Data. At each node, there are two displacements and two nodal force components (Fig. 7.17b). Recall that θ is measured counterclockwise from the positive x axis to each element (Table 7.2).

Table 7.2. Data for the Truss of Fig. 7.17

Element	θ	c	s	c²	cs	s₂	AE/L
1	0	1	0	1	0	0	4(10⁷)
2	270°	0	-1	0	0	1	4(10⁷)
3	225°	$- 1 / \sqrt{2}$	$- 1 / \sqrt{2}$	0.5	0.5	0.5	$2 \sqrt{2} (10^{7})$

Element Stiffness Matrix. Applying Eq. (7.38) and Table 7.2, we have for the bars 1, 2, and 3, respectively:

$\begin{matrix} {[k]}_{1} = 4 (10^{7}) \begin{matrix} \begin{matrix} u_{1} & {\begin{matrix} υ \end{matrix}}_{1} & u_{2} & υ_{2} \end{matrix} \\ [\begin{matrix} c^{2} & c s & - c^{2} & - c s \\ - c^{2} & s^{2} & - c s & - s^{2} \\ - c^{2} & - c s & c^{2} & c s \\ - c s & - s^{2} & c s & s^{2} \end{matrix}] \begin{matrix} u_{1} \\ υ_{1} \\ u_{2} \\ υ_{2} \end{matrix} \end{matrix} = 4 (10^{7}) \begin{matrix} \begin{matrix} u_{1} & υ_{1} & u_{2} & υ_{2} \end{matrix} \\ [\begin{matrix} 1 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 \\ - 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}] \begin{matrix} u_{1} \\ υ_{1} \\ u_{2} \\ υ_{2} \end{matrix} \end{matrix} \\ {[k]}_{2} = (10^{7}) \begin{matrix} \begin{matrix} u_{1} & υ_{1} & u_{3} & υ_{3} \end{matrix} \\ [\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 \end{matrix}] \begin{matrix} u_{1} \\ υ_{1} \\ u_{3} \\ υ_{3} \end{matrix} \end{matrix} \\ {[k]}_{1} = 2 \sqrt{2} (10^{7}) \begin{matrix} \begin{matrix} u_{2} & υ_{2} & u_{3} & υ_{3} \end{matrix} \\ [\begin{matrix} 0.5 & 0.5 & - 0.5 & - 0.5 \\ 0.5 & 0.5 & - 0.5 & - 0.5 \\ - 0.5 & - 0.5 & 0.5 & 0.5 \\ - 0.5 & - 0.5 & 0.5 & 0.5 \end{matrix}] \begin{matrix} u_{2} \\ υ_{2} \\ u_{3} \\ υ_{3} \end{matrix} \end{matrix} \end{matrix}$

The column and row of each of these stiffness matrices are labeled according to the nodal displacements associated with them. Notice that displacements u₃ and υ₃ are not involved in element 1; u₂ and υ₂ are not involved in element 2; and u₁ and υ₁ are not involved in element 3. Thus, prior to adding [k]₁, [k]₂, and [k]₃ to obtain the system matrix, two rows and columns of zeros must be added (boxed in by the dashed lines) to each of the element matrices to account for the absence of these displacements. Using a common factor 10⁷, the element stiffness matrices then take the following forms:

Three stiffness matrices of finite elements k1, k2, and k3 are shown. — Three stiffness matrices of finite elements are shown. The rows and columns of all the three matrices are labeled as u1, v1, u2, v2, u3, and v3 respectively. The first stiffness matrix k1 equals 10 power 7 times a 6 cross 6 matrix. The matrix values are 4, 0 negative 4, 0, 0, and 0; row 2 reads 0, 0, 0, 0, 0, and 0; row 3 reads negative 4, 0, 4, 0, 0, and 0; row 4 reads 0, 0, 0, 0, 0, and 0; row 5 reads 0, 0, 0, 0, 0, and 0; row 6 reads 0, 0, 0, 0, 0, and 0. It is indicated that all the entries in the fourth and fifth rows and columns that are corresponding to u3 and v3 are highlighted. The second stiffness matrix k2 equals 10 power 7 times a 6 cross 6 matrix. The matrix values are as follows: row 1 reads 0, 0, 0, 0, 0, and 0; row 2 reads 0, 0, 4, 0, 0, and negative 4; row 3 reads 0, 0, 0, 0, 0, and 0; row 4 reads 0, 0, 0, 0, 0, and 0; row 5 reads 0, 0, 0, 0, 0, and 0; row 6 reads 0, negative 4, 0, 0, 0, and 4. It is indicated that all the entries of third and fourth rows and columns are highlighted. The third stiffness matrix k3 equals 10 power 7 times 6 cross 6 matrix. The matrix values are as follows: row 1 reads 0, 0, 0, 0, 0, and 0; row 2 reads 0, 0, 0, 0, 0, and 0; row 3 reads 0, 0, square root of 2, square root of 2, negative square root of 2, and negative square root of 2; row 4 reads 0, 0, square root of 2, square root of 2, negative square root of 2, and negative square root of 2; row 5 reads 0, 0, negative square root of 2, negative square root of 2, square root of 2, and square root of 2; row 6 reads 0, 0, negative square root of 2, negative square root of 2, square root of 2, and square root of 2. It is indicated that all the entries of first and second rows and columns are highlighted. A matrix K equals 10 power 6 times a 8 cross 8 matrix. The matrix values are as follows: row 1 reads 3.97, negative 1.65, negative 2.32, 0, 2.15, negative 1.16, negative 0.99, and 0; row 2 reads negative 1.65, 3.97, 0, negative 2.32, negative 0.99, 0, 2.15, and negative 1.16; row 3 reads negative 2.32, 0, 3.97, negative 1.65, negative 1.16, 2.15, 0, and negative 0.99; row 4 reads 0, negative 2.32, negative 1.65, 3.97, 0, negative 0.99, negative 1.16, and 2.15; row 5 reads: 2.15, negative 0.99, negative 1.16, 0, 7.18, negative 0.58, negative 6.6, and 0; row 6 reads 0, negative 1.16, negative 0.99, 2.15, 0, negative 6.6, negative 0.58, and 7.18. A vertical and horizontal line divides the matrix into 4 equal parts.

System Stiffness Matrix. There are a total of six components of displacement for the truss before boundary constraints are imposed. Hence, the order of the truss stiffness matrix must be 6 × 6. After adding the terms from each element stiffness matrix into their corresponding locations in [K], we readily determine the global stiffness matrix for the truss:

$\begin{matrix} \begin{array}{l} u_{1} υ_{1} u_{2} υ_{2} u_{3} υ_{3} \\ [K] = (10^{7}) [\begin{array}{r} 4 & 0 & - 4 & 0 & 0 & 0 \\ 0 & 4 & 0 & 0 & 0 & - 4 \\ - 4 & 0 & 4 + \sqrt{2} & \sqrt{2} & - \sqrt{2} & - \sqrt{2} \\ 0 & 0 & \sqrt{2} & \sqrt{2} & - \sqrt{2} & - \sqrt{2} \\ 0 & 0 & - \sqrt{2} & - \sqrt{2} & \sqrt{2} & \sqrt{2} \\ 0 & - 4 & - \sqrt{2} & - \sqrt{2} & \sqrt{2} & 4 + \sqrt{2} \end{array}] \begin{matrix} u_{1} \\ υ_{1} \\ u_{2} \\ υ_{2} \\ u_{3} \\ υ_{3} \end{matrix} \end{array} & (b) \end{matrix}$

System Force-Displacement Relationship. With reference to Fig. 7.11, the boundary conditions are u₁ = 0, υ₁ = 0, and u₃ = 0. In addition, we have F_3y = 0.
Then, Eq. (7.40) becomes

$\begin{matrix} {\begin{matrix} R_{1 x} \\ R_{1 y} \\ P \\ W \\ R_{3 x} \\ 0 \end{matrix}} = [K] {\begin{matrix} 0 \\ 0 \\ u_{2} \\ υ_{2} \\ 0 \\ υ_{3} \end{matrix}} & (c) \end{matrix}$

where [K] is given by Eq. (b).

Displacements. To find u₂, υ₂, and υ₃, only the part of Eq. (c) associated with these displacements is considered. Thus, we have

$\begin{matrix} {\begin{matrix} P \\ W \\ 0 \end{matrix}} = {\begin{matrix} 24, 000 \\ - 36, 000 \\ 0 \end{matrix}} = (10^{7}) {\begin{array}{r} (4 + \sqrt{2}) & \sqrt{2} & - \sqrt{2} \\ \sqrt{2} & \sqrt{2} & - \sqrt{2} \\ - \sqrt{2} & - \sqrt{2} & (4 + \sqrt{2}) \end{array}} {\begin{matrix} u_{2} \\ υ_{2} \\ υ_{3} \end{matrix}} & (d) \end{matrix}$

Inversion of this matrix gives

${\begin{matrix} u_{2} \\ υ_{2} \\ υ_{3} \end{matrix}} {= 2.5(10}^{- 8}) {\begin{matrix} 1 & - 1 & 0 \\ - 1 & 4.828 & 1 \\ 0 & 1 & 1 \end{matrix}} {\begin{matrix} 24 \\ - 36 \\ 0 \end{matrix}} {(10}^{3}) = {\begin{matrix} 1.50 \\ - 4.95 \\ - 0.90 \end{matrix}} mm$

Reactions. Carrying these values of u₂, υ₂, and υ₃ into Eq. (c) results in the following reactional forces:

${\begin{matrix} R_{1 x} \\ R_{1 y} \\ R_{3 x} \end{matrix}} {= 10}^{7} {\begin{array}{r} - 4 & 0 & 0 \\ 0 & 0 & - 4 \\ - \sqrt{2} & - \sqrt{2} & \sqrt{2} \end{array}} {\begin{matrix} 1.50 \\ - 4.95 \\ - 0.90 \end{matrix}} {(10}^{- 3}) = {\begin{matrix} - 60 \\ 19.8 \\ 36.1 \end{matrix}} kN$

These results may be verified by applying the equilibrium equations to the free-body diagram of the entire truss (Fig. 7.17a).

Axial Forces in Elements. From Eqs. (7.39) and (d) and Table 7.2, we obtain

$F_{1} = F_{12} = \frac{A E}{L} [1 0] {\begin{matrix} u_{2} \\ υ_{2} \end{matrix}} = 4 (10^{7}) [1 0] {\begin{matrix} 1.5 \\ - 4.95 \end{matrix}} (10^{- 3}) = 60 kN$

$\begin{matrix} F_{2} & = F_{13} & = \frac{A E}{L} [0 - 1] {\begin{matrix} 0 \\ υ_{3} \end{matrix}} = 4 (10^{7}) [0 - 1] {\begin{matrix} 0 \\ - 0.9 \end{matrix}} (10^{- 3}) = 36 kN \\ F_{3} & = F_{23} & = \frac{A E}{\sqrt{2 L}} [- 1 - 1] {\begin{matrix} - u_{2} \\ υ_{2} - υ_{2} \end{matrix}} \\ = 2.83 (10^{7}) [- 1 - 1] {\begin{matrix} - 1.5 \\ - 0.90 + 4.95 \end{matrix}} (10^{- 3}) = - 51 kN \end{matrix}$

Stresses in Elements. By dividing the preceding element forces by the cross-sectional area of each bar, we obtain

$\begin{matrix} σ_{1} = \frac{60 (10^{3})}{400 (10^{- 6})} & = 150 σ_{2} = 150 (\frac{36}{60}) = 90 MPa \\ σ_{3} & = 150 (\frac{- 51}{60}) = - 127.5 MPa \end{matrix}$

The negative sign indicates a compressive stress.

Comments The results found in this case study indicate that member axial stresses are well below the yield strength for the material considered. As seen here, FEA permits the calculation of displacements, forces, and stresses in the truss with unprecedented ease and precision. Even in the simplest cases, however, this analysis requires considerable algebra. For any significant problem, an electronic digital computer must be used to find the solution.

7.11 Beam Element

This section provides only a brief discussion of the development of a stiffness matrix for beam elements. Let us consider an initially straight beam element of uniform flexural rigidity EI and length L (Fig. 7.18a). The element has a transverse deflection υ and a slope θ = dυ/dx at each end or node. Corresponding to these displacements, a transverse shear force F and a bending moment M act at each node. The deflected configuration of the beam element is depicted in Fig. 7.18b.

An illustration shows a beam element before and after deformation. — Figure 7.18. *The beam element with nodal forces and displacements: (a) before deformation; (b) after deformation*.

An illustration shows two figures. Figure (a) represents a beam element before deformation. A beam element of length 'L' is represented along an xy plane. The nodes of the beam are marked, 1 and 2. An upward shear force 'F 1y' acts at node 1 with a transverse deflection 'v1' and an upward shear force 'F 2y' acts at node 2 with a transverse deflection 'v2.' Anticlockwise bending moments M1 and M2 act at nodes 1 and 2 with slopes: theta 1 and theta 2 respectively. Figure (b) represents the beam after deformation and it represented along an xy plane. The beam bends at the middle, node 1 slightly deflects downward at an angle theta 1 from the horizontal; and node 2 slightly deflects upward at an angle theta 2 from the horizontal. Upward forces F 1y and F 2y act at nodes 1 and 2 respectively. The transverse deflection at node 1 is v1 and at node 2 it is v2, v1 is greater than v2. Anticlockwise moments M1 and M2 act at nodes 1 and 2 respectively.

The linearly elastic behavior of a beam element is governed by Eq. (5.32) as d⁴υ/dx⁴ = 0.
Observe that the right-hand side of this equation is zero because in the formulation of the stiffness matrix equations, we assume no loading between nodes. In the elements with a distributed load (see Example 7.9) or a concentrated load between the nodes, the equivalent nodal load components listed in Table D.5 are employed. These equivalent nodal loads correspond to the (oppositely directed) reactions provided by a beam subject to the distributed or concentrated loading under fixed-fixed boundary conditions. For further details, see Ref. 7.4.

The solution of the governing equation is a cubic polynomial function of x:

$\begin{matrix} υ = a_{1} + a_{2} x + a_{3} x^{2} + a_{4} x^{3} & (a) \end{matrix}$

The number of terms in this expression is the same as the number of nodal displacements of the element. Equation (a) satisfies the basic beam equation, the conditions of displacement, and the continuity of interelement nodes. The coefficients a₁, a₂, a₃, and a₄ are obtained from the conditions at both nodes:

$\begin{matrix} \begin{array}{l} υ = υ_{1} and \frac{d υ_{1}}{d x} = θ_{1} (at x = 0) \\ υ = υ_{2} and \frac{d υ_{2}}{d x} = θ_{2} (at x = L) \end{array} & (b) \end{matrix}$

Introducing Eq. (a) into (b) leads to

${\begin{matrix} υ_{1} \\ θ_{1} \\ υ_{2} \\ θ_{2} \end{matrix}} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & L & L^{2} & L^{3} \\ 1 & 1 & 2 L & 3 L^{2} \end{matrix}] {\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix}}$

The inverse of these equations is

${\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix}} = \frac{1}{L^{3}} [\begin{matrix} L^{3} & 0 & 0 & 0 \\ 0 & L^{3} & 0 & 0 \\ - 3 L & - 2 L^{2} & 3 L & - L^{2} \\ 2 & L & - 2 & L \end{matrix}] {\begin{matrix} υ_{1} \\ θ_{1} \\ υ_{2} \\ θ_{2} \end{matrix}}$

Substituting the foregoing expressions into Eq. (a) yields the displacement function in the following form:

$\begin{matrix} \begin{matrix} υ = υ_{1} + x θ_{1} - \frac{3 x^{2}}{L} υ_{1} - \frac{2 x^{2}}{L^{2}} θ_{1} + \frac{3 x^{2}}{L^{2}} υ_{2} - \frac{x^{2}}{L} θ_{2} + \frac{2 x^{3}}{L^{3}} υ_{1} \\ + \frac{x^{3}}{L^{2}} θ_{1} - \frac{2 x^{3}}{L^{3}} υ_{2} + \frac{x^{3}}{L^{2}} θ_{2} \end{matrix} & (7.43) \end{matrix}$

The bending moment M and shear force V in an elastic beam with a cross section that is symmetrical about the plane of loading are related to the displacement function by Eqs. (5.32). Therefore,

$\begin{matrix} \begin{matrix} F_{1 y} = V_{1} = E I \frac{d^{3} υ}{d x^{3}} & and & M_{1} = - E I \frac{d^{2} υ}{d x^{2}} & (at x = 0) \\ F_{2 y} = - V_{2} = E I \frac{d^{3} υ}{d x^{3}} & and & M_{2} = E I \frac{d^{2} υ}{d x^{2}} & (at x = L) \end{matrix} & (7.44) \end{matrix}$

in which the minus signs in the second and third expressions are due to opposite to the sign conventions adopted for V and M in Figs. 5.7b and 7.18.

It can be verified [Ref. 7.2] that inserting Eq. (7.43) into Eqs. (7.44) gives the nodal force (moment)-deflection (slope) relations in the matrix form as

$\begin{matrix} {\begin{matrix} F_{1 y} \\ M_{1} \\ F_{2 y} \\ M_{2} \end{matrix}} = \frac{E I}{L_{3}} [\begin{array}{r} 12 & 6 L & - 12 & 6 L \\ 6 L & 4 L^{2} & - 6 L & 2 L^{2} \\ - 12 & - 6 L & 12 & - 6 L \\ 6 L & 2 L^{2} & - 6 L & 4 L^{2} \end{array}] {\begin{matrix} υ_{1} \\ θ_{1} \\ υ_{2} \\ θ_{2} \end{matrix}} & (7.45 a) \end{matrix}$

or symbolically,

$\begin{matrix} {F} = {[k]}_{e} {δ}_{e} & (7.45 b) \end{matrix}$

where the matrix [k_e] represents the force and moment components. Likewise, {d_e} represents both deflections and slopes. The element stiffness matrix lying along a coordinate x is then

$\begin{matrix} {[k]}_{e} = \frac{E I}{L^{3}} [\begin{array}{r} 12 & 6 L & - 12 & 6 L \\ 6 L & 4 L^{2} & - 6 L & 2 L^{2} \\ - 12 & - 6 L & 12 & - 6 L \\ 6 L & 2 L^{2} & - 6 L & 4 L^{2} \end{array}] & (7.46) \end{matrix}$

With development of the stiffness matrix, formulation and solution of problems involving beam elements proceeds like that of bar elements, as demonstrated in the next few examples.

Example 7.8 Displacements of a Uniformly Loaded Cantilever Beam

A cantilever beam of length L and flexural rigidity EI carries a uniformly distributed load of intensity p, as illustrated in Fig. 7.19a. Find the vertical deflection and rotation at the free end.

The illustration of the beam with a distributed load is shown. — Figure 7.19. *Example 7.8. (a) Beam with distributed load; (b) the equivalent nodal forces.*

The beam of fixed support on node 1 and free support on node 2 of length L is shown. An equal distribution of load p is given to the beam. A downward moment M1 and upward reaction force R1 acts on node 1 of the beam. The nodal forces figure shows the beam of nodes 1 and 2 of length L. An upward force pL superscript 2 over 12 and downward force pL over 2 acts on node 1 and node 2. The element 1 representation, encircled 1 is given at the center.

Solution Only one finite element is used to represent the entire beam. The distributed load is replaced by the equivalent forces and moments, as shown in Fig. 7.19b (see case 3 in Table D.4). The boundary conditions are given by

$\begin{matrix} υ_{1} = 0 & θ_{1} = 0 \end{matrix}$

Since

$F_{1 y} = F_{2 y} = - \frac{p L}{2}, M_{2} = - M_{1} = \frac{p L^{2}}{12}$

the force-displacement relations, Eqs. (7.45a), simplify to

${\begin{matrix} - \frac{p L}{2} \\ \frac{p L^{2}}{12} \end{matrix}} = \frac{E I}{L^{3}} [\begin{matrix} 12 & - 6 L \\ - 6 L & 4 L \end{matrix}] {\begin{matrix} υ_{2} \\ θ_{2} \end{matrix}}$

Solving for the displacements,

${\begin{matrix} υ_{2} \\ θ_{2} \end{matrix}} = \frac{L}{6 E I} [\begin{matrix} 2 L^{2} & 3 L \\ 3 L & 6 \end{matrix}] {\begin{matrix} - \frac{p L}{2} \\ \frac{p L^{2}}{12} \end{matrix}}$

Subsequent to the multiplication, we obtain

${\begin{matrix} υ_{2} \\ θ_{2} \end{matrix}} = {\begin{matrix} - \frac{p L^{4}}{8 E I} \\ - \frac{p L^{3}}{6 E I} \end{matrix}}$

Comment The minus sign indicates a downward deflection and a clockwise rotation at the right end, node 2.

Example 7.9 Analysis of a Statically Indeterminate Stepped Beam

A propped cantilever beam of flexural rigidity EI and 2EI for the parts 1- 2 and 2- 3, respectively, supports a concentrated load P at point 2 (Fig. 7.20a). Calculate (a) the nodal displacements and (b) the nodal forces and moments.

Solution The beam is discretized into elements 1 and 2 with nodes 1, 2, and 3, as illustrated in Fig. 7.20a. There are a total of six displacement components for the beam before the boundary conditions are applied. The order of the system stiffness matrix must therefore be 6 × 6. Applying Eq. (7.46), the stiffness matrix for element 1, with (EI/L³)₁ = EI/L³ and L₁ = L, may be written as

An illustration represents the moments and forces acting at the nodes of a cantilever beam. — Figure 7.20. *Example 7.9. (a) Stepped beam with a load; (b) shear diagram; (c) bending moment diagram*.

An illustration shows three figures. Figure (a) shows a cantilever beam that is partitioned into two parts. A portion of the beam from the left node 1 to a length length 'L' (node2) forms the first part and the rest of the beam (2-3) of length '2L' forms the second part. The second part is of larger dimension than the first part. The flexural rigidity of the first part is EI and for the second part, 2 EI. The node 1 is fixed to a support and the node 3 is supported by a roller. A concentrated load P acts downward at the node 2. Upward forces R1 and R3 acts at the nodes 1 and 3 respectively. Figure (b) shows a shear diagram. The x-axis represents the beam length and the y-axis represents the shear force. The shear force at the first part of the beam (1-2) remains constant at 19 P over 23. At node 2, it suddenly drops to negative 4 P over 23 and remains the same until node , where it drops to 0. Figure (c) represents moment diagram. The moment is represented along the y-axis and the beam length is represented along the x-axis. At node 1, the moment is negative 11 PL over 23, that decreases gradually as the length increases. The moment becomes 0 exactly at the middle of the first part of the beam. The moment then increases gradually with the increase in length and reaches 8 PL over 23 at the node 2. Then it decreases gradually beyond node 2, and becomes 0 at node 3.

A stiffness matrix for element 1. — A stiffness matrix of finite element 1 equals EI over L cube times a 6 by 6 matrix. The 6 by 6 matrix is as follows: Row 1: 12, 6L, negative 12, 6L , 0, 0; Row 2: 6L, 4 L squared, negative 6 L, 2 L squared, 0, 0; Row 3: negative 12, negative 6L, 12, negative 6L, 0, 0; Row 4: 6L, 2 L squared, negative 6 L, 4 L squared, 0, 0; All the entries of row 5 and row 6 are 0. The 6 rows and 6 columns denote v1, theta 1, v2, theta 2, v3, and theta 3. It is indicated that all the entries of the last two rows and columns corresponding to v3 and theta 3 are 0.

Likewise, with (EI/L³)₂ = EI/4L³ and L₂ = 2L, the stiffness matrix for element 2, after rearrangement, is

A stiffness matrix for element 2. — A stiffness matrix of finite element 2 equals (EI over L cube) times a 6 by 6 matrix. The 6 by 6 matrix is as follows: In row 1 and row 2, the entries are 0. Row 3: 0, 0, 3, 3L, negative 3, 3L; Row 4: 0, 0, 3L, 4 L squared, negative 3L, 2 L squared; Row 5: 0, 0, negative 3, negative 3L, 3, negative 3L; Row 6: 0, 0, 3L, negative 2 L squared, negative 3L, 4 L powered 4. The 6 rows and 6 columns denote v1, theta 1, v2, theta 2, v3, and theta 3. It is indicated that all the entries of the first two rows and columns corresponding to v1 and theta 1 are 0.

In these matrices, the nodal displacements are shown to indicate the associativity of the rows and columns of the member stiffness matrices. Hence, rows and columns of zeros are added (boxed in by the dashed lines).

The system stiffness matrix of the beam can now be superimposed [K] = [k]₁+ [k]₂. The beam governing equations, with F_2y = -P (Fig. 7.20b), are therefore

$\begin{matrix} {\begin{array}{r} R_{1} \\ M_{1} \\ - P \\ 0 \\ R_{3} \\ 0 \end{array}} = \frac{E I}{L^{3}} [\begin{matrix} 12 & 6 L & - 12 & 6 L & 0 & 0 \\ 6 L & 4 L^{2} & - 6 L & 2 L^{2} & 0 & 0 \\ - 12 & - 6 L & 15 & - 3 L & 3 & 3 L \\ 6 L & 2 L^{2} & - 3 L & 8 L^{2} & - 3 L & 2 L^{2} \\ 0 & 0 & - 3 & - 3 L & 3 & - 3 L \\ 0 & 0 & 3 L & 2 L^{2} & - 3 L & 4 L^{2} \end{matrix}] {\begin{matrix} υ_{1} \\ θ_{1} \\ υ_{2} \\ θ_{2} \\ υ_{3} \\ θ_{3} \end{matrix}} & (7.47) \end{matrix}$

The boundary conditions are υ₁ = 0,θ₁ = 0, and υ₃ = 0. After multiplying these equations with the corresponding unknown displacements, we obtain

$\begin{matrix} {\begin{matrix} - P \\ 0 \\ 0 \end{matrix}} = \frac{E I}{L^{3}} [\begin{matrix} 15 & - 3 L & 3 L \\ - 3 L & 8 L^{2} & 2 L^{2} \\ 3 L & 2 L^{2} & 4 L^{2} \end{matrix}] [\begin{matrix} υ_{2} \\ θ_{2} \\ θ_{3} \end{matrix}] & (d) \end{matrix}$

and

$\begin{matrix} {\begin{matrix} R_{1} \\ M_{1} \\ R_{3} \end{matrix}} = \frac{E I}{L^{3}} [\begin{matrix} - 12 & 6 L & 0 \\ - 6 L & 2 L^{2} & 0 \\ - 3 & - 3 L & - 3 L \end{matrix}] {\begin{matrix} υ_{2} \\ θ_{2} \\ θ_{3} \end{matrix}} & (e) \end{matrix}$

Solving Eqs. (d), we find the deflection and slopes as follows:
${\begin{matrix} υ_{2} \\ θ_{2} \\ θ_{3} \end{matrix}} = \frac{L^{2}}{276 E I} [\begin{matrix} 28 L & 18 & - 30 \\ 18 & 51/ L & - 39/ L \\ - 30 & - 39/ L & 111/ L \end{matrix}] {\begin{matrix} - P \\ 0 \\ 0 \end{matrix}} = \frac{P L^{2}}{276 E I} {\begin{matrix} - 28 L \\ - 18 \\ 30 \end{matrix}}$

The minus sign means a downward deflection at node 2 and clockwise rotation of the left end 1; the positive sign means a counterclockwise rotation at the right end 3 (Fig. 7.20), as appreciated intuitively.
Substituting the displacements found into Eq. (e), and then multiplying and simplifying, we find the nodal forces and moments to be

${\begin{matrix} R_{1} \\ M_{1} \\ R_{3} \end{matrix}} = \frac{P}{276 L} [\begin{matrix} - 12 & 6 L & 0 \\ - 6 L & 2 L^{2} & 0 \\ - 3 & - 3 L & - 3 L \end{matrix}] {\begin{matrix} - 28 L \\ - 18 \\ 30 \end{matrix}} = \frac{1}{23} {\begin{matrix} 19 P \\ 11 P L \\ 4 P \end{matrix}}$

Comments Usually, it is necessary to obtain the nodal forces and moments associated with each element to analyze the whole structure. For the case considered in this example, it may readily be seen from a free-body diagram of element 2 that M₂ = R₃ (2L) = 8PL/23. So, we have the shear and moment diagrams for the beam, as shown in Figs. 7.20b and c, respectively.

7.12 Properties of Two-Dimensional Elements

Now we 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. Section 7.13 provides the general formulation of the finite element method applicable to any structure. Solutions of plane stress and plane strain problems are then illustrated in detail in Section 7.14. The analyses of axisymmetric structures and thin plates employing the finite element are given in Chapters 8 and 13, respectively.

To begin, the relatively thin, continuous body shown in Fig. 7.21a is replaced or discretized by an assembly of finite elements (triangles, for example) indicated by the dashed lines (Fig. 7.21b). These elements are connected not only at their corners or nodes, but also along the interelement boundaries. The basic unknowns are the nodal displacements.

A figure represents a plane stress region. — Figure 7.21. *Plane stress region (a) before and (b) after division into finite elements*.

A figure (a) shows a irregular body, represented along an xyz plane. The displacement along the x-axis is u, displacement along the y-axis is v, and displacement along the z-axis is w. In figure (b), the irregular body is divided into finite triangular elements. Each element is represented as 'e' and the three nodes of each element are represented as m, i, and j.

7.12.1 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. 7.21b, shown isolated in Fig. 7.22a. Designating the nodes as i, j, and m, the element nodal displacement matrix is

$\begin{matrix} {δ}_{e} = {u_{i}, υ_{i}, u_{j}, υ_{j}, u_{m}, υ_{m}} & (7.48 a) \end{matrix}$

An illustration represents triangular finite element. — Figure 7.22. *Triangular finite element*.

An illustration shows two figures. Figure (a) shows a triangular finite element 'e,' represented along an xy plane. The three nodes of the triangular element are 'i,' 'm,' and 'j.' The node 'i' lies in the xy plane at a horizontal distance 'x subscript i' from the y-axis and at a vertical distance 'y' from the x-axis. The horizontal displacement at the node 'i' is 'u subscript i,' that acts rightward and the vertical displacement is 'v subscript i,' that acts upward. In figure (b), a point P is marked at the middle of the triangular finite element. The nodes are marked as 1, 2, and 3. The nodes at the base of the triangular element are 1 and 2. A dashed line is drawn from each node toward the point 'P,' that separates the triangular element into three sections 'A1,' 'A2,' and 'A3.' The edge 1-3 is labeled '2,' edge 1-2 is labeled '3,' and the edge 2-3 is labeled '1.'

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

$\begin{matrix} {δ}_{e} = {\begin{matrix} δ_{u} \\ - - - \\ δ_{υ} \end{matrix}} = {u_{i}, u_{j}, u_{m}, υ_{i}, υ_{j}, υ_{m}} & (7.48 b) \end{matrix}$

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

$\begin{matrix} {f}_{e} = {u (x, y), υ (x, y)} & (7.49) \end{matrix}$

which may also be expressed as

$\begin{matrix} {f}_{e} = [N]{δ}_{e} & (7.50) \end{matrix}$

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

7.12.2 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

$\begin{matrix} \begin{matrix} {ε}_{e} & = {ε_{x}, ε_{y}, γ_{x y}} \\ = {\frac{\partial u}{\partial x}, \frac{\partial υ}{\partial y}, \frac{\partial u}{\partial y} + \frac{\partial υ}{\partial x}} \end{matrix} & (7.51 a) \end{matrix}$

$\begin{matrix} {ε}_{e} = [B]{δ}_{e} & (7.51 b) \end{matrix}$

where [B] is also yet to be defined.

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

$\begin{matrix} {σ}_{e} = \frac{E}{1 - ν^{2}} [\begin{matrix} 1 & ν & 0 \\ ν & 1 & 0 \\ 0 & 0 & (1 - v) /2 \end{matrix}] {\begin{matrix} ε_{x} \\ ε_{y} \\ γ_{x y} \end{matrix}} & (a) \end{matrix}$

Put succinctly,

$\begin{matrix} {σ}_{e} = [D]{ε}_{e} & (7.52) \end{matrix}$

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

$\begin{matrix} {σ}_{e} = [D]({ε} - {ε_{0} {})}_{e} & (7.53) \end{matrix}$

The thermal strain matrix, for the case of plane stress, is given by {ε₀} = {αT,αT,0} (Section 3.8). Comparing Eqs. (a) and (7.52), it is clear that

$\begin{matrix} [D] = \frac{E}{1 - ν^{2}} [\begin{matrix} 1 & ν & 0 \\ ν & 1 & 0 \\ 0 & 0 & (1 - v) / 2 \end{matrix}] & (7.54) \end{matrix}$

In general, the elasticity matrix may be represented in the form

$\begin{matrix} [D] = λ [\begin{matrix} 1 & D_{12} & 0 \\ D_{12} & 1 & 0 \\ 0 & 0 & D_{33} \end{matrix}] & (7.55) \end{matrix}$

Recall from Section 3.5 that two-dimensional problems are of two classes: plane stress and plane strain. The constants λ,D₁₂, and D₃₃ for a two-dimensional problem are given in Table 7.3.

Table 7.3. Elastic Constants for Two-Dimensional Problems

Quantity	Plane Strain	Plane Stress
λ	$\frac{E}{1 - v^{2}}$	$\frac{E (1 - v)}{(1 + v) (1 - 2 v)}$
D₁₂	ν	$\frac{v}{1 - v}$
D₃₃	$\frac{1 - v}{2}$	$\frac{1 - 2 v}{2 (1 - v)}$

7.13 General 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. To see how this method works, consider again an elastic body (Fig. 7.21). The principle of potential energy, from Eq. (10.21), is expressed for the entire body as follows:

$\begin{matrix} \begin{matrix} Δ Π & = \sum_{1}^{n} \int_{V} (σ_{x} Δ ε_{x} + \dots + σ_{z} Δ ε_{z}) d V \\ - \sum_{1}^{n} \int_{V} (F_{x} Δ u + F_{y} Δ υ + F_{z} Δ w) d V \\ - \sum_{1}^{n} \int_{s} (p_{x} Δ u + p_{y} Δ υ + p_{z} Δ w) d s = 0 \end{matrix} & (7.56) \end{matrix}$

Note that variation notation δ has been replaced by Δ to avoid confusing it with nodal displacement. In Eq. (7.56),

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. (7.49), Eq. (7.56) may be expressed in the following matrix form:

$\begin{matrix} \sum_{1}^{n} \int_{V} ({Δ ε}_{e}^{T} {σ}_{e} - {Δ f}_{e}^{T} {F}_{e}) d V - \sum_{1}^{n} \int_{s} {Δ f}_{e}^{T} {p}_{e} d s = 0 & (a) \end{matrix}$

where superscript T denotes the transpose of a matrix. Now, using Eqs. (7.50), (7.51), and (7.53), Eq. (a) becomes

$\begin{matrix} \sum_{1}^{n} {Δ δ}_{e}^{T} ([k]_{e} {δ}_{e} - {Q}_{e}) = 0 & (b) \end{matrix}$

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

$\begin{matrix} {[k]}_{e} = \int_{V} {[B]}^{T} [D][B] d V & (7.57) \end{matrix}$

$\begin{matrix} {Q}_{e} = \int_{V} {[N]}^{T} {F} d V + \int_{V} {[B]}^{T} [D]{ε_{0}} d V + \int_{s} {[N]}^{T} {p} d s & (7.58) \end{matrix}$

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

$\begin{matrix} {[k]}_{e} {δ}_{e} = {Q}_{e} & (7.59) \end{matrix}$

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

$\begin{matrix} {(Δ δ)}^{T} ([K]{δ} - {Q}) = 0 & (c) \end{matrix}$

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—that is, the system equations:

$\begin{matrix} [K] {δ} = {Q} & (7.60) \end{matrix}$

where

$\begin{matrix} [K] = Σ_{1}^{n} {[k]}_{e}, {Q} = Σ_{1}^{n} {Q}_{e} & (7.61) \end{matrix}$

Note 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, as discussed in Section 7.10 for trusses.

7.13.1 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:

Calculate [K]_e from Eq. (7.57) in terms of the given element properties. Generate [K]=Σk_e.
Calculate [Q]_e from Eq. (7.58) in terms of the applied loading. Generate [Q]=ΣQ_e.
Calculate the nodal displacements from Eq. (7.60) by satisfying the boundary conditions: {δ}=[K]^–1{Q}
Calculate the element strain using Eq. (7.51): {ε}_e = [B]{δ}_e
Calculate the element stress using Eq. (7.53): {σ}_e=[D]({ε}-{ε₀})_e

A simple block diagram of finite element analysis is shown in Fig. 7.23 [Ref. 7.3, page 632].

The block diagram for the finite element analysis is shown. — Figure 7.23. *The finite element analysis block diagram.*

The finite element analysis block diagram shows the following seven blocks flows from top to bottom. DEFINE ANALYSIS PROBLEM - Choose a finite element model Plan the mesh for the model. INPUT DATA - Materials, node and element definition, boundary conditions, loads. FORM ELEMENT [k] subscript e - Calculate element stiffness matrix. FORM SYSTEM [K] - Assemble element [k] subscript es to form the system stiffness matrix. FORM SYSTEM {Q} AND {delta} - Apply boundary displacement and force conditions. COMPUTE DISPLACEMENTS - Solve the system equations {Q} = [K] {delta} for the displacements {delta} = [K] superscript negative 1 {Q}. COMPUTE STRESSES - Calculate stresses (and/or forces) in elements. EVALUATE RESULTS. Which then points to a decision block that reads, Is refined mesh for the model required? If the condition is yes, the loop again points to the Input data via the block Plan refined mesh for the model. And if the condition is no, it points to Present results block.

When the stress found is uniform throughout each element, this result is usually interpreted two ways: (1) The stress obtained for an element is assigned to its centroid; and (2) 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 outline of the finite element method will be better understood when it is applied to a triangular element in the next section. Formulation of the properties of a simple one-dimensional element, using relations developed in this section, is illustrated in the following example.

Example 7.10 Deflection of a Bar Under Combined Loading

A bar 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. 7.24a). Determine (a) the stiffness matrix, (b) the total nodal force matrix, and (c) the deflection of the right end u₂ for a fixed left end and p_x=0 (Fig. 7.24b).

A figure represents the distributed load acting on a bar element. — Figure 7.24. *Example 7.10. (a) Bar element subjected to an axial load px and uniform temperature change T; (b) fixed end bar.*

A bar element of length 'L' is represented along the x-axis. A point 'Q1' is marked at the left end of the beam (at the origin) and a point 'Q2' is marked at the right end. A force 'p subscript x' acts on a small length 'dx' of the beam, that is at a distance 'x' from Q1. The distributed force along Q1 is 'p1' and along Q2, it is 'p2.' The distributed force increases gradually from Q1 toward Q2. Figure (b) represents a fixed end bar along an xy plane. The bar is of length 'L' and that lies along the x-axis.

Solution

a. As before (see Section 7.7), we will assume that the displacement u at any point within the element varies linearly with x:

$\begin{matrix} f = u = a_{1} + a_{2} x & (d) \end{matrix}$

where a₁ and a₂ are constants. The axial displacements of nodes 1 and 2 are u₁=a₁ and u₂=a₁+a₂L, from which a₂ = -(u₁-u₂)/L. Substituting this into Eq. (d),

$\begin{matrix} u = [1 - \frac{x}{L} \frac{x}{L}] {\begin{matrix} u_{1} \\ u_{2} \end{matrix}} = [N] {δ} & (e) \end{matrix}$

Applying Eq. (7.51a), the strain in the element is

$ε_{x} = \frac{u_{2} - u_{1}}{L}$

or, in matrix form,

$\begin{matrix} ε_{x} = \frac{1}{L} [- 1 1] {\begin{matrix} u_{1} \\ u_{2} \end{matrix}} = [B]{δ} & (f) \end{matrix}$

For a one-dimensional element, we have D = E and the stress from Eq. (7.52) equals

$σ_{x} = \frac{E}{L} [- 1 1] {\begin{matrix} u_{1} \\ u_{2} \end{matrix}}$

The element stiffness matrix is obtained by introducing [B] from Eq. (f) into Eq. (7.57):

$\begin{matrix} {[k]}_{e} = {[B]}^{T} E [B] A L = {\begin{array}{r} - 1 \\ 1 \end{array}} [- 1 1] \frac{E A}{L} = \frac{E A}{L} [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] & (7.62) \end{matrix}$

b. Referring to Fig. 7.24a,

$\begin{matrix} p_{x} = p_{1} + \frac{p_{2} - p_{1}}{L} x & (7.63) \end{matrix}$

where p₁ and p₂ are the intensities of the load per unit length at nodes 1 and 2, respectively. Substitution of [N] from Eq. (e) together with Eq. (7.63) into Eq. (7.58) yields

${Q}_{e}^{p} = \int_{0}^{L} {[N]}^{T} {p_{x}} d x = \int_{0}^{L} [\begin{matrix} 1 - x / L \\ x / L \end{matrix}] [p_{1} + \frac{p_{2} - p_{1}}{L} x] d x$

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

$\begin{matrix} {Q}_{e}^{p} = {\begin{cases} Q_{1} \\ Q_{2} \end{cases}}_{e}^{p} = \frac{L}{6} [\begin{matrix} 2 p_{1} + p_{2} \\ p_{1} + 2 p_{2} \end{matrix}] & (g) \end{matrix}$

The strain due to the temperature change is ε₀ =αT, where α is the coefficient of thermal expansion. Inserting [B] from Eq. (f) into Eq. (7.58),

${Q}_{e}^{t} = \int_{0}^{L} \int_{A} {[B]}^{T} [D] {ε_{0}} d A d x = \int_{0}^{L} \frac{1}{L} {\begin{array}{r} - 1 \\ 1 \end{array}} E A α (T) d x$

The thermal strain effects are then

$\begin{matrix} {Q}_{e}^{t} = {\begin{cases} Q_{1} \\ Q_{2} \end{cases}}_{e}^{t} = E A α (T) {\begin{array}{r} - 1 \\ 1 \end{array}} & (h) \end{matrix}$

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

$\begin{matrix} {Q}_{e} = \frac{L}{6} {\begin{matrix} 2 p_{1} + p_{2} \\ p_{1} + 2 p_{2} \end{matrix}} + E A α (T) {\begin{array}{r} - 1 \\ 1 \end{array}} & (7.64) \end{matrix}$

c. The nodal force-displacement relations (7.60) now take the form

$E A α (T) {\begin{array}{r} - 1 \\ 1 \end{array}} = \frac{E A}{L} [\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix}] {\begin{matrix} 0 \\ u_{2} \end{matrix}}$

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

7.14 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 derive the basic constant strain triangular (CST) plane stress and strain element. Note that several two-dimensional finite element types can lead to better solutions, including 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 [Ref. 7.4]. 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 same steps as those for the CST element.

Consider the triangular finite element 1 2 3 (where i = 1, j = 2, and m = 3) shown in Fig. 7.22b, 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 as A₁, A₂, and A₃. The ratios of these areas to the total area A of the triangle locate P and represent the area coordinates:

$\begin{matrix} L_{1} = \frac{A_{1}}{A}, L_{2} = \frac{A_{2}}{A}, L_{3} = \frac{A_{3}}{A} & (7.65) \end{matrix}$

It follows that

$\begin{matrix} L_{1} + L_{2} + L_{3} = 1 & (7.66) \end{matrix}$

Consequently, only two of the three coordinates are independent. A useful property of area coordinates is observed through reference to Fig. 7.22b and Eq. (7.65):

$L_{n} = 0 on side n, n = 1, 2, 3$

and

$\begin{matrix} L_{i} = 1, & L_{j} = L_{m} = 0 & at node i \\ L_{j} = 1, & L_{i} = L_{m} = 0 & at node j \\ L_{m} = 1, & L_{i} = L_{j} = 0 & at node m \end{matrix}$

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

$A = \frac{1}{2} {[(x_{2} - x_{1}) i + (y_{2} - y_{1}) j] \times [(x_{3} - x_{1}) i + (y_{3} - y_{1}) j]} \cdot k$

$\begin{matrix} 2 A = \det | \begin{matrix} 1 & x_{1} & y_{1} \\ 1 & x_{2} & y_{2} \\ 1 & x_{3} & y_{3} \end{matrix} | & (7.67 a) \end{matrix}$

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

$\begin{matrix} 2 A = a_{j} b_{i} - a_{i} b_{j} = a_{m} b_{j} - a_{j} b_{m} = a_{i} b_{m} - a_{m} b_{i} & (7.67 b) \end{matrix}$

where

$\begin{matrix} \begin{matrix} a_{i} = x_{m} - x_{j}, & b_{i} = y_{j} - y_{m} \\ a_{j} = x_{i} - x_{m}, & b_{j} = y_{m} - y_{i} \\ a_{m} = x_{j} - x_{i}, & b_{m} = y_{i} - y_{j} \end{matrix} & (7.68) \end{matrix}$

Note that a_j, a_m, b_j and b_m can be found from the 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. (7.65), lead to the following relationship between area and Cartesian coordinates:

$\begin{matrix} {\begin{matrix} L_{1} \\ L_{2} \\ L_{3} \end{matrix}} = \frac{1}{2 A} [\begin{matrix} 2 c_{23} & b_{1} & a_{1} \\ 2 c_{31} & b_{2} & a_{2} \\ 2 c_{12} & b_{3} & a_{3} \end{matrix}] {\begin{matrix} 1 \\ x \\ y \end{matrix}} & (7.69) \end{matrix}$

in which

$\begin{matrix} \begin{matrix} c_{23} = x_{2} y_{3} - x_{3} y_{2} \\ c_{31} = x_{3} y_{1} - x_{1} y_{3} \\ c_{12} = x_{1} y_{2} - x_{2} y_{1} \end{matrix} & (7.70) \end{matrix}$

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. 7.22). The nodal displacements are

$\begin{matrix} {δ}_{e} = {u_{1}, u_{2}, u_{3}, υ_{1}, υ_{2}, υ_{3}} & (a) \end{matrix}$

The displacement throughout the element is provided by

$\begin{matrix} {f}_{e} = [\begin{matrix} u_{1} & u_{2} & u_{3} \\ υ_{1} & υ_{2} & υ_{3} \end{matrix}] {\begin{matrix} L_{1} \\ L_{2} \\ L_{3} \end{matrix}} & (7.71) \end{matrix}$

Matrices [N] and [B] of Eqs. (7.50) and (7.51b) are next evaluated, beginning with

$\begin{matrix} {f}_{e} = [N]{δ}_{e} & (b) \end{matrix}$

$\begin{matrix} {ε}_{e} = [B]{δ}_{e} & (c) \end{matrix}$

We observe that Eqs. (7.71) and (b) are equal, provided that

$\begin{matrix} [N] = [\begin{matrix} L_{1} & L_{2} & L_{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & L_{1} & L_{2} & L_{3} \end{matrix}] & (7.72) \end{matrix}$

The strain matrix is obtained by substituting Eqs. (7.71) and (7.69) into Eq. (7.51a):

$\begin{matrix} {\begin{matrix} ε_{x} \\ ε_{y} \\ γ_{x y} \end{matrix}} = \frac{1}{2 A} [\begin{matrix} b_{1} & b_{2} & b_{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{1} & a_{2} & a_{3} \\ a_{1} & a_{2} & a_{3} & b_{1} & b_{2} & b_{3} \end{matrix}] {\begin{matrix} u_{1} \\ u_{2} \\ u_{3} \\ υ_{1} \\ υ_{2} \\ υ_{3} \end{matrix}} & (d) \end{matrix}$

Here A, a_i, and b_i are defined by Eqs. (7.67) and (7.68). The strain (stress) is constant throughout, so that the element of Fig. 7.22 is referred to as a CST. Comparing Eqs. (c) and (d), we have

$\begin{matrix} [B] = \frac{1}{2 A} [\begin{matrix} b_{1} & b_{2} & b_{3} & 0 & 0 & 0 \\ 0 & 0 & 0 & a_{1} & a_{2} & a_{3} \\ a_{1} & a_{2} & a_{3} & b_{1} & b_{2} & b_{3} \end{matrix}] & (7.73) \end{matrix}$

The stiffness of the element can now be obtained through the use of Eq. (7.57):

$\begin{matrix} {[k]}_{e} = {[B]}^{T} [D][B] t A & (e) \end{matrix}$

Let us now define

$\begin{matrix} [D^{*}] = \frac{t [D]}{4 A} = [\begin{matrix} D_{11}^{*} & D_{12}^{*} & D_{13}^{*} \\ D_{22}^{*} & D_{23}^{*} \\ Symm. & D_{33}^{*} \end{matrix}] & (7.74) \end{matrix}$

where [D] is given by Eq. (7.54) for plane stress. Combining Eq. (e) ith Eqs. (7.73) and (7.74) and expanding, we can express the stiffness matrix in the following partitioned form of order 6 × 6:

A stiffness matrix for element e. — A stiffness matrix of finite element e is partitioned into four submatrices. The submatrices in the first row are k subscript (uu, ln) and k subscript (uv, ln). The submatrices in the second row are k superscript T, subscript (uu, ln) and k subscript (uv, ln). Here l, n equals 1, 2, 3.

in which the submatrices are

$\begin{matrix} \begin{array}{l} k_{u u, \ln} = D_{11}^{*} b_{l} b_{n} + D_{33}^{*} a_{l} a_{n} + D_{13}^{*} (b_{l} a_{n} + b_{n} a_{l}) \\ k_{υ υ, \ln} = D_{33}^{*} b_{l} b_{n} + D_{22}^{*} a_{l} a_{n} + D_{23}^{*} (b_{l} a_{n} + b_{n} a_{l}) \\ k_{u υ, \ln} = D_{13}^{*} b_{l} b_{n} + D_{23}^{*} a_{l} b_{n} + D_{12}^{*} b_{l} a_{n} + D_{33}^{*} b_{n} a_{l} \end{array} & (7.76) \end{matrix}$

7.14.1 Element Nodal Forces

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. (7.58) and (7.72),

$\begin{matrix} {Q}_{e}^{b} = \int_{V} {[N]}^{T} {F} d V = \int_{V} [\begin{matrix} L_{1} & 0 \\ L_{2} & 0 \\ L_{3} & 0 \\ 0 & L_{1} \\ 0 & L_{2} \\ 0 & L_{3} \end{matrix}] {\begin{matrix} F_{x} \\ F_{y} \end{matrix}} d V & (f) \end{matrix}$

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

^*For a general function $f = L_{1}^{α} L_{2}^{β} L_{3}^{γ}$ , defined in area coordinates, the integral off over any triangular area A is given by

$\int_{A} L_{1}^{α} L_{2}^{β} L_{3}^{γ} d A = 2 A \frac{α! β! γ!}{(α + β + γ)!}$

where α, β, and γ are constants.

$\begin{matrix} {Q}_{e} = \frac{1}{3} A t {F_{x}, F_{x}, F_{x}, F_{y}, F_{y}, F_{y}} & (7.77) \end{matrix}$

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 applying Eq. (7.58). Nodal force expressions for arbitrary nodes j and m are given next for a number of common cases (Prob. 7.41).

Linear load, p(y) per unit area, Fig. 7.25a:

$\begin{matrix} Q_{j} = \frac{h_{1} t}{6} (2 p_{j} + p_{m}), Q_{m} = \frac{h_{1} t}{6} (2 p_{m} + p_{j}) & (7.78 a) \end{matrix}$

Two figures show the nodal forces due to shear load and linearly distributed load. — Figure 7.25. Nodal forces due to (a) linearly distributed load and (b) shear load.

In figure (a), the nodal force on the element due to linearly distributed load is shown. A right triangle section with the nodes i, j, and m is shown. The height of the triangle is h1. The linear loads at nodes m and j are Q subscript m and Q subscript j respectively. A uniformly varying load acts on the right edge of the element. The load is minimum at the bottom of the element and maximum at the top of the element. The load at nodes m and j are p subscript m and p subscript j. In figure (b), the nodal force on the element due to shear load is shown. The element of infinite length and breadth 2h is subjected to a shear force P along the right edge of the element. A right triangle section with the nodes I, j, and m is shown. The shear load at nodes m and j are Q subscript m and Q subscript j respectively. Another figure shows the parabolic shear stress distribution. The parabola is open leftward.

where t is the thickness of the element.

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

$\begin{matrix} Q_{j} = Q_{m} = \frac{1}{2} p h_{1} t & (7.78 b) \end{matrix}$

End shear load, P, is the result of a parabolic shear stress distribution defined by
Eq. (3.24) (see Fig. 7.25b):

$\begin{matrix} \begin{matrix} Q_{m} = \frac{3 P}{4 h} {\frac{1}{2} (y_{m} - y_{j}) + \frac{1}{3 h^{2}} [\frac{y_{i}}{4} (y_{m}^{2} + y_{j} y_{m} + y_{i}^{2}) - \frac{3}{4} y_{m}^{3}]} \\ Q_{j} = \frac{3 P}{4 h} {\frac{1}{2} (y_{m} - y_{j}) - \frac{1}{3 h^{2}} [(y_{m} - \frac{3}{4} y_{j}) (y_{m}^{2} + y_{j} y_{m} + y_{i}^{2}) - \frac{3}{4} y_{m}^{3}]} \end{matrix} & (7.79) \end{matrix}$

Equation (7.75), together with those expressions given for the nodal forces, characterizes the constant strain element. These are substituted first into Eq. (7.61) and subsequently into Eq. (7.60) 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 7.11 Nodal Forces of a Plate Segment Under Combined Loading

The element e shown in Fig. 7.26 represents a segment of a thin elastic plate having side 2–3 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 2–3. Let t = 0.3 cm, E = 200 GPa, v = 0.3, specific weight γ=77kN/m³, and α=12×10^–6/°c.

A triangular plate is represented along an xy plane. — Figure 7.26. *Example 7.11. A triangular plate.*

A right triangular plate 'e' is represented along an xy plane. The adjacent side of the triangular plate (1-2) measures 4 centimeters and that lies parallel to the x-axis in the negative xy plane. The opposite side of the triangular plate (1-3) measures 3 centimeters and that lies along the y-axis. An uniform distributed load 'p' acts on the hypotenuse of the plate (2-3).

Solution The origin of the coordinates is located at midlength of side 1–3, for convenience. However, it may be placed at any point in the x, y plane. Applying Eq. (7.74), we have (in N/cm³)

$\begin{matrix} [D^{*}] = \frac{t [D]}{4 A} = \frac{10^{6}}{8} [\begin{matrix} 3.3 & 0.99 & 0 \\ 0.99 & 3.3 & 0 \\ 0 & 0 & 1.16 \end{matrix}] & (g) \end{matrix}$

Stiffness matrix: The nodal points are located at

$\begin{matrix} \begin{matrix} x_{i} = x_{1} = 0, & y_{i} = y_{1} = - 1 \\ x_{j} = x_{2} = 4, & y_{j} = y_{2} = - 1 \\ x_{m} = x_{3} = 0, & y_{m} = y_{3} = 1 \end{matrix} & (h) \end{matrix}$

Using Eq. (7.68) and referring to Fig. 7.26, we obtain

$\begin{matrix} \begin{matrix} a_{i} = a_{1} = 0 - 4 = - 4, & b_{i} = b_{1} = - 1 - 1 = - 2 \\ a_{j} = a_{2} = 0 - 0 = 0, & b_{j} = b_{2} = 1 + 1 = 2 \\ a_{m} = a_{3} = 4 - 0 = 4, & b_{m} = b_{3} = - 1 + 1 = 0 \end{matrix} & (i) \end{matrix}$

Next, the first equation of Eqs. (7.76), together with Eqs. (g) and (i), yields

$\begin{matrix} k_{u u},_{11} & = \frac{10^{6}}{8} [3.3 (4) + 1.16 (16)] = 3.97 \times 10^{6} \\ k_{u u},_{12} & = k_{u u, 21} = \frac{10^{6}}{8} [3.3 (2) (- 2) + 0] = - 1.65 \times 10^{6} \\ k_{u u},_{13} & = k_{u u, 31} = \frac{10^{6}}{8} [0 + 1.16 (4) (- 4)] = - 2.32 \times 10^{6} \\ k_{u u, 22} & = \frac{10^{6}}{8} [3.3 (4) + 0] = 1.65 \times 10^{6} \\ k_{u u, 23} & = k_{u u, 32} = 0 \\ k_{u u, 33} & = \frac{10^{6}}{8} [0 + 1.16 (16)] = 2.32 \times 10^{6} \end{matrix}$

The submatrix k_uu is thus

$k_{u u} = 10^{6} [\begin{matrix} 3.97 & - 1.65 & - 2.32 \\ - 1.65 & 1.65 & 0 \\ - 2.32 & 0 & 2.32 \end{matrix}]$

Similarly, from the second and third equations of Eqs. (7.76), we obtain the following matrices:

$k_{u υ} = 10^{6} [\begin{matrix} 2.15 & - 1.16 & - 0.99 \\ - 0.99 & 0 & 0.99 \\ - 1.16 & 1.16 & 0 \end{matrix}], k_{υ υ} {= 10}^{6} [\begin{matrix} 7.18 & - 0.58 & - 6.6 \\ - 0.58 & 0.58 & 0 \\ - 6.6 & 0 & 6.6 \end{matrix}]$

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

A stiffness matrix of finite element e equals 10 powered 6 times a 6 by 6 matrix. The entries of the matrix are as follows. Row 1: 3.97, negative 1.65, negative 2.32, 2.15, negative 1.16, negative 0.99; Row 2: negative 1.65, 1.65, 0, negative 0.99, 0, 0.99; Row 3: negative 2.32, 0, 2.32, negative 1.16, 1.16, 0; Row 4: 2.15, negative 0.99, negative 1.16, 7.18, negative 0.58, negative 6.6; Row 5: negative 1.16, 0, 1.16, negative 0.58, 0.58, 0; Row 6: negative 0.99, 0.99, 0, negative 6.6, 0, 6.6. A dotted line in the matrix partitions the matrix into four sub matrices of order 3 by 3.
We next determine the nodal forces of the element owing to various loadings. The components of body force are F_x=0 and _y=0.077 N/cm³.

Body Force Effects. Through the application of Eq. (7.77), it is found that

${Q}_{e}^{b} = {0, 0, 0, - 0.0308, - 0.0308, - 0.0308} N$

Surface Traction Effects. The total load, $s t {p} = \sqrt{20} (0.3) {- 2 \times 1400 / \sqrt{20},$ $- 4 \times 1400 \sqrt{20}} = {- 840, - 16 80},$ , is equally divided between nodes 2 and 3. The nodal forces can therefore be expressed as

${Q}_{e}^{p} = {0, - 420, - 420, 0, - 840, - 840} N$

Thermal Strain Effects. The initial strain associated with the 50°C temperature rise is ε ₀ = αT = 0.0006 0. From Eq. (7.59),

${Q}_{e}^{t} = {[B]}^{T} [D]{ε_{0}}(A t)$

Substituting matrix [B], given by Eq. (7.73), into this equation, and with the values of the other constants already determined, the nodal force is calculated as follows:

${Q}_{e}^{t} = \frac{1}{8} [\begin{array}{r} - 2 & 0 & - 4 \\ 2 & 0 & 0 \\ 0 & 0 & 4 \\ 0 & - 4 & - 2 \\ 0 & 0 & 2 \\ 0 & 4 & 0 \end{array}] \frac{200 \times 10^{5}}{0.91} [\begin{matrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{matrix}] {\begin{matrix} 0.0006 \\ 0.0006 \\ 0 \end{matrix}} (1.2)$

${Q}_{e}^{t} = {- 5142.85, 5142.85, 0, - 10, 285.70, 0, 10, 285.70} N$

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

${Q}_{e} = {\begin{matrix} Q_{x 1} \\ Q_{x 2} \\ Q_{x 3} \\ Q_{y 1} \\ Q_{y 2} \\ Q_{y 2} \end{matrix}} = {\begin{matrix} - {5142.85}_{} \\ {4722.85}_{} \\ - {420.03}_{} \\ - {10,285.73}_{} \\ - {840.03}_{} \\ {9445.67}_{} \end{matrix}} N$

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

7.15 Case Studies in Plane Stress

A good engineering case study includes all data necessary to analyze a problem in detail and may come in many varieties [Refs. 7.5 and 7.6]. Obviously, in solid mechanics, it deals with stress and deformation in load-carrying members or structures. In this section, we briefly present four case studies limited to plane stress situations and CST finite elements: a cantilever beam supporting a concentrated load, a deep beam or plate in pure bending, a plate with a hole subjected to an axial loading, and a disk carrying concentrated diametral compression.

Recall from Chapter 3 that very few elasticity or “exact” solutions to two-dimensional problems exist, especially for any but the simplest shapes. As will become evident in the following discussion, the stress analyst and designer can reach a very accurate solution by employing proper techniques and modeling. Accuracy is often limited by the willingness to model all the significant features of the problem and to pursue the FEA until convergence is reached.

Case Study 7.2 Stresses in a Cantilever Beam Under a Concentrated Load

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. 7.27a). Divide the beam into two constant-strain triangles and calculate the deflections. Let E = 200 GPa and v = 0.3.

Solution The discretized beam is shown in Fig. 7.27b.

A cantilever beam is represented along an xy plane. — Figure 7.27. *Case Study 7.2. Cantilever beam (a) before and (b) after being discretized.*

An illustration shows two figures. A figure (a) represents a cantilever beam before discretized. The length of the beam measures 4 centimeters that is represented along an xy plane. The height of the beam is 3 centimeters in which 1 centimeter lies below the x-axis and the remaining 2 centimeters lie above the x-axis. A downward force 'P' acts along the free edge of the beam. Figure (b) represents the beam after being discretized. The edge along the y-axis is 1-3 and its opposite edge is 2-4. The nodes 1 and 3 are fixed along the y-axis by supports. The beam is diagonally divided from the node 3, forming two triangles. The triangle at the lower portion is (a) and the triangle at the upper portion is (b). A downward force 'Q subscript y 4' acts at the node 4 and a downward force 'Q subscript y 2' acts at the node 2.

Stiffness Matrix. Inasmuch as the dimensions and material properties of element a are the same as those given in Example 7.11, [k]_a is defined by Eq. (j) of that example. For element b, assignment of i = 2, j = 4, and m = 3 [Eq. (7.68)] leads to

$\begin{matrix} a_{i} = a_{2} = 0 - 4 = - 4, & b_{i} = b_{2} = 1 - 1 = 0 \\ a_{j} = a_{4} = 4 - 0 = 4, & b_{j} = b_{4} = 1 + 1 = 2 \\ a_{m} = a_{3} = 4 - 4 = 0, & b_{m} = b_{3} = - 1 - 1 = - 2 \end{matrix}$

Substitution of these and Eq. (g) of Example 7.11 into Eqs. (7.76) yields

$\begin{matrix} k_{u u, 22} & = \frac{10^{6}}{8} [3.3 (0) + 1.16 (16)] = 2.32 \times 10^{6} \\ k_{u u, 44} & = \frac{10^{6}}{8} [3.3 (4) + 1.16 (16)] = 3.97 \times 10^{6} \\ k_{u u, 23} & = k_{u u, 32} = 0 \\ k_{u u},_{24} & = \frac{10^{6}}{8} [3.3 (0) + 1.16 (- 4) (4)] = - 2.32 \times 10^{6} \\ k_{u u, 43} & = \frac{10^{6}}{8} [3.3 (2) (- 2) + 1.16 (0)] = - 1.65 \times 10^{6} \\ k_{u u, 33} & = \frac{10^{6}}{8} [3.3 (4) + 1.16 (0)] = 1.65 \times 10^{6} \end{matrix}$

Thus,

$k_{u u} = 10^{6} [\begin{matrix} 2.32 & 0 & - 2.32 \\ 0 & 1.65 & - 1.65 \\ - 2.32 & - 1.65 & 3.97 \end{matrix}]$

Similarly, we obtain

$\begin{matrix} k_{u υ} = 10^{6} [\begin{matrix} 0 & 1.16 & - 1.16 \\ 0.99 & 0 & - 0.99 \\ - 0.99 & - 1.16 & 2.15 \end{matrix}] \\ k_{υ υ} = 10^{6} [\begin{matrix} 6.6 & 0 & - 6.6 \\ 0 & 0.58 & - 0.58 \\ - 6.6 & - 0.58 & 7.18 \end{matrix}] \end{matrix}$

The stiffness matrix of element b is therefore

A stiffness matrix of finite element “K subscript b” of size 10 power 6 times 6 cross 6 matrix is shown. — A stiffness matrix of finite element K subscript 'a' equals 10 power 6 times a 8 cross 8 matrix. The matrix values are as follows: row 1 reads 2.32, 0, negative 2.32, 0, 1.16, and negative 1.16; row 2 reads 0, 1.65, negative 1.65, 0.99, 0, and negative 0.99; row 3 reads negative 2.32, negative 1.65, 3.97, negative 0.99, negative 1.16, and 2.15; row 4 reads 0, 0.99, negative 0.99, 6.6, 0, and negative 6.6; row 5 reads 1.16, 0, negative 1.16, 0, 0.58, and negative 0.58; row 6 reads negative 1.16, negative 0.99, 2.15, negative 6.6, negative 0.58, and 7.18. A vertical and horizontal line divides the matrix into 4 equal parts.

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) of Example 7.11 and (a) become

A stiffness matrix of finite element “K subscript 'a'” of size 10 power 6 times 8 cross 8 is shown. — A stiffness matrix of finite element K subscript 'a' equals 10 power 6 times a 8 cross 8 matrix. The matrix values are as follows: row 1 reads 3.97, negative 1.65, negative 2.32, 0, 2.15, negative 1.16, negative 0.99, and 0; row 2 reads negative 1.65, 1.65, 0, 0, negative 0.99, 0, 0.99, and 0; row 3 reads negative 2.32, 0, 2.32, 0, negative 1.16, 1.16, 0, and 0; row 4 reads 0, 0, 0, 0, 0, 0, 0, and 0; row 5 reads 2.15, negative 0.99, negative 1.16, 0, 7.18, negative 0.58, negative 6.6, and 0; row 6 reads negative 1.16, 0, 1.16, 0, negative 0.58, 0.58, 0, and 0; row 7 reads negative 0.99, 0.99, 0, 0, negative 6.6, 0, 6.6, and 0; row 8 reads 0, 0, 0, 0, 0, 0, 0, and 0. A vertical and horizontal line divides the matrix into 4 equal parts.

and

A stiffness matrix of finite element “K subscript b” of size 10 power 6 times 8 cross 8 is shown. — A stiffness matrix of finite element K subscript b equals 10 power 6 times a 8 cross 8 matrix. The matrix values are as follows: row 1 reads 0, 0, 0, 0, 0, 0, 0, and 0; row 2 reads 0, 2.32, 0, negative 2.32, 0, 0, negative 1.16; row 3 reads negative 2.32, 0, 3.97, negative 1.65, negative 1.65, negative 1.16, 2.15, 0, and negative 0.99; row 4 reads 0, negative 2.32, negative 1.65, 3.97, 0, negative 0.99, negative 1.16, and 2.15; row 5 reads 2.15, negative 0.99, negative 1.16, 0, 7.18, negative 0.58, negative 6.6, and 0; row 6 reads negative 1.16, 0, 2.15, negative 0.99, negative 0.58, 7.18, 0, and negative 6.6; row 7 reads negative 0.99, 2.15, 0, negative 1.16, negative 6.6, 0, 7.18, and negative 0.58; row 8 reads 0, negative 1.66, negative 0.99, 2.15, 0, negative 6.6, negative 0.58, and 7.18. A vertical and horizontal line divides the matrix into 4 equal parts.

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

A stiffness matrix “K” of size 10 power 6 times 8 cross 8 is shown. — A stiffness matrix K equals 10 power 6 times a 8 cross 8 matrix. The matrix values are as follows: row 1 reads 3.97, negative 1.65, negative 2.32, 0, 2.15, negative 1.16, negative 0.99, and 0; row 2 reads negative 1.65, 3.97, 0, negative 2.32, negative 0.99, 0, 2.15, and negative 1.16; row 3 reads negative 2.32, 0, 3.97, negative 1.65, negative 1.16, 2.15, 0, and negative 0.99; row 4 reads 0, negative 2.32, negative 1.65, 3.97, 0, negative 0.99, negative 1.16, and 2.15; row 5 reads: 2.15, negative 0.99, negative 1.16, 0, 7.18, negative 0.58, negative 6.6, and 0; row 6 reads 0, negative 1.16, negative 0.99, 2.15, 0, negative 6.6, negative 0.58, and 7.18. A vertical and horizontal line divides the matrix into 4 equal parts.

Nodal Forces. Referring to Fig. 7.27b and applying Eq. (7.79), we obtain

$\begin{matrix} Q_{y 4} & = \frac{3 (- 5000)}{4 (1)} {\frac{1}{2} (1 + 1) + \frac{1}{3} [- \frac{1}{4} (1 - 1 + 1) - \frac{3}{4}]} = - 2500 N \\ Q_{y 2} & = \frac{3(- 5000)}{4(1)} {\frac{1}{2} (1 + 1) - \frac{1}{3} [1 + \frac{3}{4} (1 - 1 + 1) - \frac{3}{4}]} = - 2500 N \end{matrix}$

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

$u_{1} = u_{3} = υ_{1} = υ_{3} = 0$

The force-displacement relationship of the system is therefore

$\begin{matrix} {0, 0, 0, 0, 0, - 2 500, 0, - 2500} = [K]{0, u_{2}, 0, u_{4}, 0, υ_{2}, 0, υ_{4}} & (e) \end{matrix}$

Equation (e) is readily reduced to the form

$\begin{matrix} {\begin{matrix} 0 \\ 0 \\ - 2500 \\ - 2500 \end{matrix}} {= 10}^{6} {\begin{matrix} 3.97 & - 2.32 & 0 & - 1.16 \\ - 2.32 & 3.97 & - 0.99 & 2.15 \\ 0 & - 0.99 & 7.18 & - 6.6 \\ - 1.16 & 2.15 & - 6.6 & 7.18 \end{matrix}} {\begin{matrix} u_{2} \\ u_{4} \\ υ_{2} \\ υ_{4} \end{matrix}} & (f) \end{matrix}$

From this, we obtain

$\begin{matrix} {\begin{matrix} u_{2} \\ u_{4} \\ υ_{2} \\ υ_{4} \end{matrix}} {= 10}^{- 6} {\begin{matrix} 0.429 & 0.180 & 0.252 & 0.247 \\ 0.180 & 0.483 & - 0.256 & - 0.351 \\ 0.252 & - 0.256 & 1.366 & 1.373 \\ 0.247 & - 0.351 & 1.373 & 1.546 \end{matrix}} {\begin{array}{r} 0 \\ 0 \\ - 2500 \\ - 2500 \end{array}} = {\begin{matrix} - 0.0012 \\ 0.0015 \\ - 0.0068 \\ - 0.0073 \end{matrix}} cm & (g) \end{matrix}$

The strains {ε}_a may now be found by introducing Eq. (i) of Example 7.11 and Eq. (g) into Eq. (d) of Section 7.14 as follows:

${\begin{matrix} ε_{x} \\ ε_{y} \\ γ_{x y} \end{matrix}} = \frac{1}{8} [\begin{matrix} - 2 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 4 & 0 & 4 \\ - 4 & 0 & 4 & - 2 & 2 & 0 \end{matrix}] {\begin{matrix} 0 \\ - 0.0012 \\ 0 \\ 0 \\ - 0.0068 \\ 0 \end{matrix}} = {\begin{array}{r} - 3 \\ 0 \\ - 17 \end{array}} 10^{- 4}$

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

${\begin{matrix} σ_{x} \\ σ_{y} \\ τ_{x y} \end{matrix}} = \frac{200 \times 10^{9}}{0.91} [\begin{matrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{matrix}] {\begin{matrix} - 3 \\ 0 \\ - 17 \end{matrix}} 10^{-4} = {\begin{matrix} - 65.93 \\ - 19.78 \\ - 130.77 \end{matrix}} MPa$

Element b is treated similarly.

Note that the model employed in this solution is quite crude. The effect of element size on solution accuracy is illustrated in the next case study.

Case Study 7.3 Analysis of a Deep Beam by the Theory of Elasticity and FEM

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. 7.28a). 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.

A beam is represented along an xy plane. — Figure 7.28. *Case Study 7.3. (a) Beam in pure bending; (b) moment is replaced by a statically equivalent load.*

An illustration shows two figures. A figure (a) represents a beam of length '2L' and height '2h' lying on an xy plane, centered at the origin. Half of its length 'L' lies to the left of y-axis and the rest lies to the right of y-axis. Similarly, half of its height lies below x-axis and the rest lies above x-axis. A clockwise moment 'M' acts along the left edge and an anticlockwise moment M acts along the right edge of the beam. A portion of the beam lying in the fourth quadrant is highlighted. Another diagram (b) represents the shear stress due to a statically equivalent load applied about the centroidal axis of the beam. On the upper half region of the beam about the origin, the stress acts leftward and it is maximum along the edge of the beam than in the origin. On the lower half region of the beam, the stress acts rightward, which is maximum along the edge than in the origin.

Solution

Exact Solution. Replacing the end moments with the statically equivalent load per unit area p = Mh/I (Fig. 7.28b), the stress distribution from Eq. (5.5) is

$\begin{matrix} σ_{x} = - \frac{y}{h} p, σ_{y} = τ_{x y} = 0 & (h) \end{matrix}$

From Hooke’s law and Eq. (2.4), we have

$\frac{\partial u}{\partial x} = - \frac{y p}{E h}, \frac{\partial υ}{\partial y} = \frac{ν y p}{E h}, \frac{\partial u}{\partial y} + \frac{\partial υ}{\partial x} = 0$

By now following a procedure similar to that described in Section 5.4, and satisfying the conditions u(0,0)=υ (0,0)=0 and u(L,0)=0, we obtain

$\begin{matrix} u = - \frac{p}{E h} x y, υ = - \frac{p}{2 E h} (x^{2} + ν y^{2}) & (i) \end{matrix}$

Substituting the data into Eqs. (h) and (i), the results are

$\begin{matrix} \begin{matrix} σ_{x} & = - \frac{1}{0.0508} y p, σ_{x, \max} = 6895 kPa \\ u (0.0762, - 0.0254) & = 25.4 \times 10^{- 6} m, υ (0.0762, 0) = 1.905 \times 10^{- 6} m \end{matrix} & (j) \end{matrix}$
Finite Element Solution. Considerations of symmetry and antisymmetry indicate that only any one-quarter (shown as shaded portion in the figure) of the beam needs to be analyzed.

Displacement Boundary Conditions. Figure 7.29a 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 υ deformation. The boundary conditions are thus

Figure 7.29. Case Study 7.3. Influence of element size and orientation on the stress (MPa) in the beam shown in Fig. 7.28.

In figure (a), a two cross three grid divided into twelve right triangular elements is shown. The sides of each square measures 25.4 units. The grid consists of 12 nodes with the roller supports at nodes 1, 2, and 12. The node 3 has hinged support. In figure (b), the two cross three grid with twelve elements is shown. The following elements are labeled as follows: Element 3 - 3.11; Element 4 - 5.76; Element 5 - 5.63; Element 6 - 3.25; Element 7 - 3.14; Element 8 - 0.05; Element 9 - 0.05; Element 10 2.99. In figure (c), the two cross three grid with twelve elements is shown. The following elements are labeled as follows: Element 3 5.60; Element 4 3.01; Element 5 5.51; Element 6 3.26; Element 7 3.23; Element 8 0.10; Element 9 0.03; Element 10 3.04. In figure (d), the quarter plate is divided into 16 elements with 15 nodes. The elements are assumed to be labeled horizontally. The following elements are labeled as follows: Element 1 0.12; Element 2 1.41; Element 3 0.12; Element 4 2.96; Element 5 1.67; Element 6 3.02; Element 7 3.01; Element 8 4.25; Element 9 3.22; Element 10 5.35; Element 11 4.47; Element 12 5.10.

$u_{1} = u_{2} = u_{3} = u_{6} = u_{9} = u_{12} = 0, υ_{3} = 0$

Nodal Forces. For the loading system of Fig. 7.28b, Eq. (7.78a) applies. After substituting numerical values,

$\begin{array}{l} Q_{x 10} = \frac{0.0254 \times 00254}{6} (2 \times 6895 + 3447.5) = 1853.5 N \\ Q_{x 11} = \frac{0.0254 \times 0.0254}{6} (2 \times 3447.5 + 6895) \\ + \frac{0.0254 \times 00254}{6} (2 \times 3447.5 + 0) = 2224 N \\ Q_{x 12} = \frac{0.0254 \times 0.0254}{6} (0 + 3447.5) = 370.5 N \end{array}$

The remaining Q’s are zero.

Results. The nodal displacements are determined by following a procedure similar to that used in Case Study 7.2. The stresses are then evaluated, with representative values (in megapascals) being given in Fig. 7.29b. Note that the stress obtained for an element is assigned to the centroid. Moreover, there is a considerable difference between the exact solution, Eq. (j), 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. 7.29c and d. For purposes of comparison, the deflections corresponding to Figs.7.29b through d are presented in Table 7.4.

Table 7.4. Comparison of Deflections Obtained by FEM and the Elasticity Theory

		Deflection (m)
Case	Number of Nodes	υ ₁₂(10₆)	u ₁₀(10₆)
Fig. 7.28b	12	1.547	2.133
Fig. 7.29c	12	1.745	2.062
Fig. 7.29d	15	1.572	1.976
Exact solution	—	1.905	2.540

Comments The effect of element orientation is shown in Figs. 7.29b and c for an equal number of elements and node locations. Figure 7.29d 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 shown in Figs. 7.29b and c. Using equilateral or nearly equilateral well-formed elements of finer mesh leads to solutions approaching the exact values.

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

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. Since an “exact” solution is unattainable, we seek instead an acceptable solution. The goal is then to find a finite element that ensures convergence to the exact solution in the absence of round-off error. The literature provides 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, using a denser mesh within a region of severely changing or localized stress may save much time and effort.

Case Study 7.4 Stress Concentration in a Plate with a Hole in Uniaxial Tension

A thin aluminum alloy 6061-T6 plate containing a small circular hole of radius a is under uniform tensile stress σ₀ at its edges (Fig. 7.30a). Using finite element analysis, outline the determination of the stress concentration factor K. Compare the result with that obtained by applying the theory of elasticity in Section 3.12.

Given: L = 600 mm, a = 50 mm, h = 500 mm,σ₀ = 42 MPa, v = 0.3. Also, from Table D.1, E = 70 GPa.

Three figures show the stress concentration in a plate. — Figure 7.30. *Case Study 7.4. (a) Circular hole in a plate under uniaxial tension; (b) one-quadrant plate model; (c) uniaxial stress (ó x) distribution.*

In figure (a), a plate of length 2L and breadth 2h is shown. A circular hole of diameter 2a is present at the center of the hole. Axial tensile stress (sigma 0) acts from the left and the right edges of the plate. Figure (b) shows the meshes of one-quadrant of the plate. Tensile stress (sigma 0) acts horizontally from the right edge of the plate. The plate is fixed at the bottom and on the left edge. In figure (c), the enlarged view of the quarter plate is shown. Here, the plate is discretized into 202 CST elements. The tensile stress (sigma 0) acts horizontally from the left and the right edges of the plate. The stress value sigma 0 equals 42 (No Suggestions). The stress is maximum at the bottom left edge of the plate with the value of 3 times sigma 0.

Solution Due to symmetry, only any one-quarter of the plate needs to be analyzed, as illustrated in Fig. 7.30b. In this case, the quarter plate is discretized into 202 CST elements [Ref. 7.2]. The roller boundary conditions are also shown in the figure.

The values of the edge stress σ_x, calculated by the FEM and the theory of elasticity are plotted in Fig. 7.30c for comparison [Ref. 7.7]. Observe that the agreement is reasonably good. The stress concentration factor for σ_x is equal to K ≈ 3σ₀/3σ₀=3.

7.16 Computational Tools

Various computational tools can be employed to carry out analysis calculations. A quality scientific calculator may be the best tool for solving most of the problems in this text. General-purpose analysis tools such as spreadsheets and equation solvers are very useful for certain computational tasks. Mathematical software packages of these types include MATLAB, TK Solver, and MathCAD. Such tools offer a key advantage: The user can both document and save the detailed completed work

As noted earlier and in Appendix E, the website for this text contains MATLAB simulations for advanced mechanics of materials and applied elasticity. Computer-aided drafting or design (CAD) software packages can also produce realistic three-dimensional representations of a member. Most CAD software provides an interface to one or more FEA programs. Such an interface permits direct transfer of the member’s geometry to an FEA package for analysis of stress and vibration as well as fluid and thermal analysis. Computer programs (e.g., NASTRAN, ANSYS, ABAQUS, GT-STRUDL) are used widely for performing the numerical computations required in the analysis and design of structural and mechanical systems.

With the proper use of computer-aided engineering (CAE) software, problems can be solved more quickly and more accurately. Clearly, the results are subject to the accuracy of the various assumptions that must necessarily be made in the analysis and design. This kind of computer-based software may be used as a tool to assist students with lengthy homework assignments. At the same time, it is important to thoroughly understand the basics, and analysts must check the computer solutions for accuracy and feasibility.

References

7.1. KREYSZIG, E. Advanced Engineering Mathematics, 10th ed. Hoboken, NJ: Wiley, 2011.

7.2. YANG, T. Y. Finite Element Structural Analysis. Upper Saddle River, NJ: Prentice Hall, 1986.

7.3. UGURAL, A. C. Mechanics of Materials. Hoboken, NJ: Wiley, 2008.

7.4. LOGAN, D. L. A First Course in the Finite Element Method. Boston: PWS-Kent, 1986.

7.5. KNIGHT, E. The Finite Element Method in Mechanical Design. Boston: PWS-Kent, 1993.

7.6. UGURAL, A. C. Mechanical Design of Machine Components, 2nd ed. Boca Raton, FL: CRC Press, Taylor & Francis Group, 2016.

7.7. BORESI, A. P., and SCHMIDT, R. J. Advanced Mechanics of Materials, 6th ed. New York: Wiley, 2003.

7.8. ZIENKIEWICZ, O. C., and Taylor, R. I. The Finite Element Method, 4th ed., Vol. 2 (Solid and Fluid Mechanics, Dynamics and Nonlinearity). London: McGraw-Hill, 1991.

7.9. BATHE, K. I. Finite Element Procedures in Engineering Analysis. Upper Saddle River, NJ: Prentice Hall, 1996.

Problems

Sections 7.1 through 7.4

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

$\nabla^{4} w = \frac{\partial^{4} w}{\partial x^{4}} + 2 \frac{\partial^{4} w}{\partial x^{2} \partial y^{2}} + \frac{\partial^{4} w}{\partial y^{4}} = 0$

takes the following finite difference form:

$\begin{matrix} \begin{matrix} h^{4} \nabla^{4} w = 20 w_{0} - 8(w_{1} + w_{2} + w_{3} + w_{4}) \\ + 2(w_{5} + w_{6} + w_{7} + w_{8}) + w_{9} + w_{10} + w_{11} + w_{12} \end{matrix} & (P7.1) \end{matrix}$

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 by using 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 whose cross section is shown in Fig. P7.3, find the shear stresses at point B. Given: h = 5 mm and h₁ = h₂ = 3.5 mm.

A figure shows a rectangle placed on an XY plane. — Figure P7.3.

The figure depicts a rectangular structure with curved edges on the right and the left. The rectangle is placed on an XY plane. The center of the rectangle id g and the right curved edge intersecting the x-axis is B. The portion of the rectangle in the first quadrant is divided into 6 sections. Two vertical lines are drawn and a horizontal line intersecting these two lines are also shown. The points are labeled b, c, d, e, f, and g.

7.4. Re-sole 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.

Section 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.

A figure shows a stepped cantilever beam. — Figure P7.6.

A stepped cantilever beam of length L is shown. The diameter of the beam varies after a distance of L over 2. The stiffness of the left portion of the beam is 2EI1, and the stiffness of the right portion of the beam is EI1. A force P acts vertically downward at the free end of the beam.

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 and (b) the deflection at point C.

A figure shows stepped simple beam. — Figure P7.7.

A stepped simple beam of length L is shown. The beam is stepped at point C which is at a distance of L over 2. The beam has hinged support at point A on the left end and roller support at point B on the right end. The diameter of the left portion of the beam is larger than the right portion of the beam. The stiffness of the left portion of the beam is 3EI and the stiffness of the right portion of the beam is EI. A moment M 0 acts in the clockwise direction at point A.

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.

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) = 11P₀L⁴/120EI.

A cantilever beam AB of length L is shown. A uniform varying load acts vertically downward on the beam. The load at the fixed end (A) is zero, and the load at the free end (B) is maximum (p subscript o). — Figure P7.9.

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

A simple beam is shown. — Figure P7.10.

A figure shows a simple beam (AB) of length L. The beam has hinged support at point A and roller support at point B. Seven points (0, 1, 2, 3, 4, 5, and 6) are marked on the beam. The distance between each point is L over 6. A force P acts vertically downward on the beam at a distance L over 3 from A.

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.

A figure shows a simple beam (AB) of length L. The beam has hinged support at point A and roller support at point B. A uniform distributed load (p) acts on the beam for a distance of L over 8 on either side of the center of the beam. — 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.

A figure shows an overhanging beam. — Figure P7.12.

An overhanging beam of length 3a is shown. The beam has hinged support at point A on the left end of the beam, and a roller support at point C, which is at a distance 2a from the left end of the beam. A force P acts at point B, which is at a distance a from the hinged support. Another force P acts on the right end of the beam, which is at a distance a from the roller support.

7.13. Re-solve 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.

A horizontal beam (AB) of length L is fixed to rigid support at point A. The beam has roller support at point B. A uniform distributed load (p) acts on the beam from A to B. — 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.

A horizontal beam (AB) of length L is fixed between two rigid supports at points A and B. A load P acts vertically downward at point C, which is at a distance of L over 2 from A. — 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.

A horizontal beam (AB) of length L is fixed between two rigid supports at points A and B. A uniformly distributed load P acts vertically downward from A to B. — Figure P7.16.

Sections 7.6 through 7.16

7.17. The bar element 4–1 of length L and cross-sectional area A is oriented at an angle a clockwise from the x axis (Fig. P7.17). Calculate (a) the global stiffness matrix of the bar, (b) the axial force in the bar, and (c) the local displacements at the ends of the bar. Given: A = 1350 mm², L = 1.7 m, α = 30°, E = 96 GPa, u₄ = 21.1 mm, υ⁴ = 21.2 mm, u₁ = 2 mm, and υ₁ = 1.5 mm.

A figure shows a structure with 5 nodes. — Figure P7.17.

The ends 1 and 2 are fixed to roller support. A right-angled triangle is formed on the right joining 3, 4, and 5. Also, nodes 2 and 4 are joined by a horizontal member at the bottom and the nodes a and 3 are joined by another horizontal member on the top. The nodes 2 and 3, 1 and 4 are joined using diagonal lines at an angle alpha. A load P acts on node 4 and Q acts on 5.

7.18. The axially loaded bar 1–4 of constant axial rigidity AE is held between two rigid supports and under a concentrated load P at node 3, as illustrated in Fig. P7.18. Find (a) the system stiffness matrix, (b) the displacements at nodes 2 and 3, and (c) the nodal forces and reactions at the supports.

A figure shows a horizontal bar. — Figure P7.18.

An axially loaded bar 1-4 is fixed between two rigid supports at nodes 1 and 4. The bar is divided into three sections. Node 2 is present at a distance L from node 1. Node 3 is present at a distance L over 2 from node 2. The distance between node 3 and 4 is L over 3. A horizontal force P acts leftward from node 3.

7.19. The axially loaded composite bar 1–4 is held between two rigid supports and subjected to a concentrated load P at node 2, as depicted in Fig. P7.19. The steel bar 1–3 has cross-sectional area A and modulus of elasticity E. The brass bar 3–4 has cross-sectional area 2A and elastic modulus E/2. Determine (a) the system stiffness matrix, (b) the displacements of nodes 2 and 3, and (c) the nodal forces and reactions at the supports.

A figure shows a stepped bar fixed between rigid supports. — Figure P7.19.

A stepped bar made up of steel and bronze is fixed between two rigid supports. The steel portion of the bar is on the left. The diameter of the steel bar is smaller than the bronze bar. A horizontal force P acts rightward from node 2 which is at a distance of L over 3 from node 1. The bronze bar is attached to the steel bar at node 3. The other end of the bar is fixed to the rigid support. The distance between each node is L over 3.

7.20. A stepped bar 1–4 is held between rigid supports and carries a concentrated load P at node 3, as illustrated in Fig. P7.20. Find (a) the system stiffness matrix, (b) the displacements of nodes 2 and 3, and (c) the nodal forces and reactions at the supports.

A figure shows a stepped composite bar fixed between rigid supports. — Figure P7.20.

A stepped bar made up of steel and brass is fixed between two rigid supports. The first and the second bars are made up of steel, which is fixed to rigid support at node 1. The two bars are connected at node 2. The area of the steel bar 1-2 is A, and the area of the steel bar 2-3 is 3A over 4. The length of the bar 1-2 is L from the rigid support on the left end. The length of the bar 2-3 is 3L over 4. The brass bar is connected between the steel bar 2-3 and the rigid support on the right end. The area of the brass bar is A over 2. The length of the brass bar is 5L over 8. A horizontal force P acts rightward at node 3.

7.21. A planar truss containing five members with axial rigidity AE is supported at joints 1 and 4, as shown in Fig. P7.21. What is the global stiffness matrix for each element?

A five-member structure is shown forming a triangle. — Figure P7.21.

A figure shows a 5 member structure forming a triangle. The length of the triangle is 4.5 meters and a weight w acts on the top. The base is formed by the second and the fourth member. The length of the second member is 6 meters and the length of the fourth member is 4.5 meters. The right edge is formed by the member 1 and the left edge is formed by the member 5. The third member is drawn joining the apex and the base.

7.22. A plane truss is loaded and supported as illustrated in Fig. P7.22. Each element has an axial rigidity AE. Find (a) the global stiffness matrix for each element, (b) the system stiffness matrix, and (c) the system force-displacement relations.

A five-member structure is shown of length 4.8 meters. — Figure P7.22.

The figure depicts a triangle having member 3 as a base. The members 1 and 2 are joined to the left of the base to form a triangle and the members 4 and 5 are joined to the right of the base to form another triangle. The total length of the structure is 1 meter. The left end of the structure is fixed to a pinned support and the right end of the structure is fixed to roller support.

7.23. A two-bar planar truss is supported by a spring of stiffness k at joint 1, as depicted in Fig. P7.23. Each element has an axial rigidity AE. Calculate (a) the stiffness matrix for bars 1 and 2, and spring 3; (b) the system stiffness matrix; and (c) the force-displacement equations. Given: L₂ = L, (AE)₂ = AE, $L_{1} = \frac{3}{4} L$ , ${(A E)}_{1} \frac{3}{4} A E$ .

A figure shows a two bar planar truss. — Figure P7.23.

In the figure, a horizontal bar and an inclined bar are connected at node 1 on the x-axis. The length of the horizontal bar is L and is fixed to rigid support on the left at node 3. The inclined bar makes an angle 45 degrees with the horizontal bar and is fixed to rigid support at node 2. The inclined bar is labeled 1 and the horizontal bar is labeled 2. A planar truss is supported by a vertical spring at node 1 and is labeled 3. The spring is attached to rigid support at node 4. A force P acts vertically downward at node 1.

7.24. A vertical concentrated load P = 6 kN is applied at joint 2 of the two-bar plane truss supported as shown in Fig. P7.24. Take AE = 20 MN for each member. Find (a) the global stiffness matrix of each bar, (b) the system stiffness matrix, (c) the nodal displacements, (d) the support reactions, and (e) the axial forces in each bar.

7.25. In a two-bar plane truss, its support at joint 1 settles vertically by an amount of u = 15 mm downward when loaded by a horizontal concentrated load P (Fig. P7.25). Calculate (a) the global stiffness matrix of each element, (b) the system stiffness matrix, (c) the nodal displacements, and (d) the support reactions. Given: E = 105 GPa, A = 10 × 10^–4m², P = 10 KN.

7.26. A plane truss is loaded and supported as shown in Fig. P17.26. Find (a) the global stiffness matrix for each member, (b) the system stiffness matrix, (c) the nodal displacements, (d) the reactions, and (e) the axial forces in each member. Given: The axial rigidity AE = 20 MN is the same for each bar.

7.27. A cantilever of constant flexural rigidity EI carries a concentrated load P at its free end, as shown in Fig. P7.27. Find (a) the deflection υ ₁ and angle of rotation u₁ at the free end, and (b) the reactions R₂ and M₂ at the fixed end.

A figure shows a planar truss. — Figure P7.26.

In the figure, the truss consists of five members. The members 2 and 4 represents the base which is supported by hinged and a roller supports. The length of the second member is 6 meters and the length of the fourth member is 4.5 meters. The second and the fourth members are connected by a pinned support. The first and the fifth members are inclined which roughly makes an angle of 45 degrees with the base. The first and the fifth members are fixed to a pinned support at the top. The third member is vertical and is connected between the pinned supports.

A figure shows a cantilever beam 1-2 of length L. The beam is fixed on the right at node 2. A concentrated load P acts vertically downward at node 1. A moment M2 acts at a fixed end in the clockwise direction. — Figure P7.27.

7.28. A simple beam 1–3 of length L and flexural rigidity EI is supports a uniformly distributed load of intensity p, as illustrated in Fig. P7.28. Determine the deflection of the beam at midpoint 2 by replacing the applied load with the equivalent nodal loads (see Table D.5).

A figure shows a horizontal beam 1-3 of length L supported by hinged support (1) and roller support (3). Point 2 is marked at the center of the beam. A uniformly distributed load (p) acts vertically downward on the beam. The left portion of the beam is labeled 1 and the right portion of the beam is labeled 2. — Figure P7.28.

7.29. A beam supported by a pin, a spring of stiffness k, and a roller at points 1, 2, and 3, respectively, is under a concentrated load P at point 2 (Fig. P7.29). Calculate (a) the nodal displacements and (b) the nodal forces and spring force. Given: L = 4 m, P = 12 KN, EI = 14 MN · m², and k = 200 KN/m.

A horizontal beam 1-3 of length 2L is shown. The beam has hinged support at 1 and roller support at 3. A beam is supported by a vertical spring with stiffness k at the center of the beam at point 2. A concentrated load P acts vertically downward at the center of the beam. — Figure P7.29.

7.30. A cantilever beam 1–2 of length L and constant flexural rigidity EI is subjected to a concentrated load P at the midspan, as shown in Fig. P7.30. Find the vertical deflection v₂ and angle of rotation θ ₂ at the free end by replacing the applied load with the equivalent nodal loads acting at each end of the beam (see Table D.5).

A cantilever beam 1-2 of length L is shown. A concentrated load P acts vertically downward at the center of the beam. — Figure P7.30.

7.31. A propped cantilever beam with flexural rigidities EI and EI/2 for the parts 1–2 and 2–3, respectively, carries concentrated load P and moment 3PL at point 2 (Fig. P7.31). Calculate the displacements υ₂, θ ₂, and θ ₃. Given: L = 1.2 m, P = 30 KN, E = 207 GPa, and I = 15×10⁶ mm⁴.

A stepped beam 1-3 of length 2L is shown. — Figure P7.31.

In the figure, the beam is fixed by rigid support at point 1 and roller support at point 3. The beam has two parts 1-2 and 2-3, each of length L. The left end of the first part is fixed to the rigid support. A concentrated load P acts vertically downward at point 2. A moment 3PL acts in the counterclockwise direction at point 2.

7.32. A continuous beam of constant flexural rigidity EI is loaded and supported as seen in Fig. P7.32. Determine (a) the stiffness matrix for each element and (b) the system stiffness matrix and the force-displacement relations.

A horizontal beam 1-4 of length 3L is shown. The beam is fixed on the right at point 4. A concentrated load P acts vertically downward at point 1. The beam is supported by two roller supports at a distance of L and 2L respectively. The sections 1-2, 2-3, and 3-4 are labeled 1, 2, and 3. — Figure P7.32.

7.33. A propped cantilever beam with an overhang is subjected to a concentrated load P, as illustrated in Fig. P7.33. The beam has a constant flexural rigidity EI. Determine (a) the stiffness matrix for each element, (b) the system stiffness matrix, (c) the nodal displacements, (d) the forces and moments at the ends of each member, and (e) the shear and moment diagrams.

A figure shows a horizontal beam fixed at one and free at the other end. — Figure P7.33.

A horizontal beam 1-3 of length 2L is shown. The beam is fixed on the left at point 1. A concentrated load P acts downward on the other end of the beam. The beam is supported by roller support at point 2. The part 1-2 of the beam is labeled 1 and the part 2-3 of the beam is labeled 2. The length of each part is L. A reaction force R2 acts upward at point 2. A moment M1 acts at point 1 in the counterclockwise direction.

7.34. A cantilever aluminum beam is supported at its free end by a spring of stiffness k and supports a concentrated load P (Fig. P7.34). Find (a) the nodal displacements and (b) the nodal forces and spring force. Given: L = 7 m, P= 8 kN, EI= 65 (10⁵) N · m², and k = 200 kN/m.

A figure shows a horizontal beam fixed at one end and supported by a spring at the other end. — Figure P7.34.

A horizontal beam of length L is shown. The beam is fixed to rigid support on its left at point 1. The other end of the beam is supported by a vertical spring, which is fixed to rigid support at point 3. A concentrated load P acts vertically downward on the beam at point 2. The beam portion 1-2 is labeled 1 and the spring portion is labeled 2.

7.35. The planar truss, with the axial rigidity AE the same for each element, is loaded and supported as illustrated in Fig. P7.35. Determine (a) the global stiffness matrix for each element and (b) the system matrix and the system force-displacement equations.

A planar truss structure is shown. — Figure P7.35.

In the figure, the truss consists of five members. The first member represents the base and is supported by hinged and roller supports at points 1 and 4. The second and the third members are inclined and are connected between the points 1-2 and 4-2 respectively. The fourth member is horizontal and is connected between points 2 and 3. The fifth member is connected between the points 4-3. The length of the inclined members is L. A concentrated load P acts vertically downward at point 2. A force Q acts rightward from point 3.

7.36. A vertical load P = 20 kN acts at joint 2 of the two-bar (1–2 and 2–3) truss shown in Fig. P7.35. Find (a) the global stiffness matrix for each member, (b) the system stiffness matrix, (c) the nodal displacements, (d) reactions, and (e) the axial forces in each member; also show the results on a sketch of each member. Assumption: The axial rigidity AE = 60 MN is the same for each bar.

7.37. A plate with a hole is under an axial tension loading P (Fig. P7.37). Dimensions are in millimeters. Given: P = 5 kN and plate thickness t = 12 mm. (a) Analyze the stresses using a computer program with the CST (or LST) elements. (b) Compare the stress concentration factor K obtained in part (a) with that found from Fig. D.9.

A figure shows a two-bar truss. — Figure P7.36.

In the figure, the first member of the truss is inclined and fixed to hinged support at the base at joint 1. This member roughly makes an angle of 45 degrees with the base. The horizontal member is fixed to another rigid support at joint 3 which is at a height of 4 meters from the joint 1. The two bars are connected at joint 2. The length of the horizontal bar is 4 meters, and the horizontal distance of the inclined member is 3 meters. A load P acts vertically downward from the joint 2.

A figure shows a rectangular plate with a hole of diameter 20 millimeters present at the center of the plate. The length and breadth of the rectangular plate are 150 millimeters and 50 millimeters respectively. An axial tension load P acts from the left and the right edges of the plate. — Figure P7.37.

7.38. Re-solve Prob. 7.37 for the plate shown in Fig. P7.38.

A figure shows a filleted plate. — Figure P7.38.

A filleted plate is shown. The diameter of the plate is 50 millimeters at the wider part and 20 millimeters at the narrow part with the fillet radius of 10 millimeters. The length of the wider part is 75 millimeters. The total length of the plate is 150 millimeters.

7.39. A simple beam is under a uniform loading of intensity p (Fig. P7.39). Let L = 10h, t = 1, and v = 0.3. Refine 9 meshes to calculate the stress and deflection within 5% accuracy, by using a computer program with the CST (or LST) elements. Given: The exact solution [Ref. 7.5] is of the following form:

$\begin{matrix} σ_{x, \max} = \frac{3 p L^{2}}{4 t h^{2}}, υ_{\max} = - \frac{5 p L^{4}}{16 E t h^{3}} - \frac{3 p L^{2} (1 + v)}{5 E h t} & (P7.39) \end{matrix}$

A simple horizontal beam of length 2L is supported by hinged and roller supports on the left and the right ends of the beam respectively. A uniformly distributed load (p) acts vertically downward on the beam. The thickness of the beam is t. The height of the beam section is 2h. — Figure P7.39.

where t represents the thickness.

7.40. Re-solve Prob. 7.39 for the case in which a cantilever beam is under a uniform loading of intensity p (Fig. P7.40). Given: The exact solution [Ref. 7.5] is given by

$\begin{matrix} σ_{x, \max} = \frac{3 p L^{2}}{4 t h^{2}}, υ_{\max} = - \frac{3 p L^{4}}{16 E t h^{3}} - \frac{3 p L^{2} (1 + v)}{5 E h t} & (P7.40) \end{matrix}$

A cantilever beam of length L is supported by rigid support on the left. A uniformly distributed load (p) acts vertically downward on the beam. The thickness of the beam is t. The height of the beam section is 2h. — Figure P7.40.

where t is the thickness.

7.41. Verify Eqs. (7.78) and (7.79) by determining the static resultant of the applied loading. [Hint: For Eq. (7.79), apply the principle of virtual work.] Use

$Q_{j} δ υ_{j} + Q_{m} δ υ_{m} - \int_{y_{j}}^{y_{m}} τ_{x y} t δ υ d y = 0$

with

$υ = υ_{j} + \frac{υ_{m} - υ_{i}}{y_{m} - y_{i}} (y - y_{j})$

to obtain

$Q_{j} + Q_{m} = \int_{y_{i}}^{y_{m}} τ_{x y} t d y, Q_{m} = \frac{1}{y_{m} - y_{j}} [\int_{y_{j}}^{y_{m}} τ_{x y} t (y - y_{j}) d y]$

7.42. Re-solve Case Study 7.2 for the beam subjected to a uniform additional load throughout its span, p = −7 MPa, and a temperature rise of 50°C. Let γ = KN/m³ and α = 12×10^–6/°c.

7.43. Re-solve Case Study 7.2 for a discretized beam consisting of triangular elements 1 4 3 and 1 2 4 (Fig. 7.27b). Assume the remaining data are unchanged.

..................Content has been hidden....................

You can't read the all page of ebook, please click here login for view all page.

Table of Contents for Chapter 7. Numerical Methods

Create new playlist

Sign In

Sign Up

Chapter 7. Numerical Methods

7.1 Introduction

7.2 Finite Differences

7.2.1 Central Differences

7.3 Finite Difference Equations

7.4 Curved Boundaries

7.5 Boundary Conditions

7.6 Fundamentals

7.7 The Bar Element

7.7.1 Equilibrium Method

7.7.2 Energy Method

7.8 Arbitrarily Oriented Bar Element

7.8.1 Coordinate Transformation

7.8.2 Force Transformation

7.8.3 Displacement Transformation

7.8.4 Governing Equations

7.9 Axial Force Equation

7.10 Force-Displacement Relations for a Truss

7.10.1 The Assembly Process

7.11 Beam Element

7.12 Properties of Two-Dimensional Elements

7.12.1 Displacement Matrix

7.12.2 Strain, Stress, and Elasticity Matrices

7.13 General Formulation of the Finite Element Method

7.13.1 Outline of General Finite Element Analysis

7.14 Triangular Finite Element

7.14.1 Element Nodal Forces

7.15 Case Studies in Plane Stress

7.16 Computational Tools

References

Problems

Sections 7.1 through 7.4

Section 7.5

Sections 7.6 through 7.16

Table of Contents for
Chapter 7. Numerical Methods