김경종
24 KiB
(b) The effective stress is defined as \overline{\sigma} = \sqrt{\frac{3}{2} S_{ij} S_{ij}} where the S_{ij} are the components of the deviatoric stress tensor, S_{ij} = \tau_{ij} - \tau_m \delta_{ij} and \tau_m is the mean stress \tau_m = \frac{\tau_{ii}}{3} . Prove that \overline{\sigma} is a scalar. Then also show explicitly for the given value of \tau that \overline{\sigma} is the same number in the old and new bases.
2.8. The column \mathbf{q} is defined as
\mathbf {q} = \left[ \begin{array}{c c c} & x _ {1} \\ x _ {1} & + & x _ {2} \end{array} \right]
where (x_{1}, x_{2}) are the coordinates of a point. Prove that q is not a vector.
2.9. The components of the Green-Lagrange strain tensor in the Cartesian coordinate system are defined as (see Section 6.2.2 for details)
\epsilon = \frac {1}{2} (\mathbf {X} ^ {T} \mathbf {X} - \mathbf {I})
where the components of the deformation gradient X are
X _ {i j} = \delta_ {i j} + \frac {\partial u _ {i}}{\partial x _ {j}}
and u_{i}, x_{j} are the displacements and coordinates, respectively. Prove that the Green-Lagrange strain tensor is a second-order tensor.
2.10. The material tensor in (2.66) can be written as [see (6.185)]
C _ {i j r s} = \lambda \delta_ {i j} \delta_ {r s} + \mu (\delta_ {i r} \delta_ {j s} + \delta_ {i s} \delta_ {j r}) \tag {a}
where \lambda and \mu are the Lamé constants,
\lambda = \frac {E \nu}{(1 + \nu) (1 - 2 \nu)}; \quad \mu = \frac {E}{2 (1 + \nu)}
This stress-strain relation can also be written in the matrix form used in Table 4.3, but in the table the use of engineering strain components is implied. (The tensor normal strain components are equal to the engineering normal strain components, but the tensor shear strain components are one-half the engineering components).
(a) Prove that C_{ijrs} is a fourth-order tensor.
(b) Consider the plane stress case and derive from the expression in (a) the expression in Table 4.3.
(c) Consider the plane stress case and write (2.66) in the matrix form \mathbf{C}' = \mathbf{TCT}^T , where \mathbf{C} is given in Table 4.3 and you derive \mathbf{T} . (See also Exercise 4.39.)
2.11. Prove that (2.70) holds.
2.12. The covariant base vectors expressed in a Cartesian coordinate system are
\mathbf {g} _ {1} = \left[ \begin{array}{l} 1 \\ 0 \end{array} \right]; \quad \mathbf {g} _ {2} = \left[ \begin{array}{l} \frac {1}{\sqrt {2}} \\ \frac {1}{\sqrt {2}} \end{array} \right]
The force and displacement vectors in this basis are
\mathbf {R} = 3 \mathbf {g} _ {1} + 4 \mathbf {g} _ {2}; \quad \mathbf {u} = - 2 \mathbf {g} _ {1} + 3 \mathbf {g} _ {2}
(a) Calculate \mathbf{R} \cdot \mathbf{u} using the covariant basis only.
(b) Calculate \mathbf{R} \cdot \mathbf{u} using the covariant basis for \mathbf{R} and the contravariant basis for \mathbf{u} .
2.13. Assume that the covariant basis is given by \mathbf{g}_1 and \mathbf{g}_2 in Exercise 2.12. Let the stress and strain tensor components in the Cartesian basis be
\boldsymbol {\tau} = \left[ \begin{array}{l l} 1 0 0 & 1 0 \\ 1 0 & 2 0 0 \end{array} \right]; \quad \boldsymbol {\epsilon} = \left[ \begin{array}{l l} 0. 0 1 & 0. 0 5 \\ 0. 0 5 & 0. 0 2 \end{array} \right]
Evaluate the components \tilde{\tau}^{mn} and \tilde{\epsilon}_{mn} and show explicitly that the product \tau \cdot \epsilon is the same using on the one side the Cartesian stress and strain components and on the other side the contravariant stress and covariant strain components.
2.14. Let \mathbf{a} and \mathbf{b} be second-order tensors and let \mathbf{A} and \mathbf{B} be transformation matrices. Prove that
\mathbf {a} \cdot (\mathbf {A b B} ^ {T}) = (\mathbf {A} ^ {T} \mathbf {a B}) \cdot \mathbf {b}.
(Hint: This proof is easily achieved by writing the quantities in component forms.)
2.15. Consider the eigenproblem \mathbf{A}\mathbf{v} = \lambda \mathbf{v} with
\mathbf {A} = \left[ \begin{array}{c c} 2 & - 1 \\ - 1 & 1 \end{array} \right]
(a) Solve for the eigenvalues and orthonormalized eigenvectors and write A in the form (2.109).
(b) Calculate A^{6} , A^{-1} and A^{-2} .
2.16. Consider the eigenproblem
\left[ \begin{array}{l l l} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{array} \right] \mathbf {v} = \lambda \mathbf {v}
The smallest eigenvalue and corresponding eigenvector are
\lambda_ {1} = 1; \quad \mathbf {v} _ {1} = \left[ \begin{array}{c} \frac {1}{\sqrt {3}} \\ - \frac {1}{\sqrt {3}} \\ \frac {1}{\sqrt {3}} \end{array} \right]
Also, \lambda_{2} = 2, \lambda_{3} = 4 . Calculate the Rayleigh quotient \rho(\mathbf{v}) with
\mathbf {v} = \mathbf {v} _ {1} + 0. 1 \left[ \begin{array}{l} 1 \\ 1 \\ 0 \end{array} \right]
and show that \rho(\mathbf{v}) is closer to \lambda_{1} than \mathbf{v} is to \mathbf{v}_{1} .
2.17. Consider the eigenproblem
\left[ \begin{array}{r r r} 2 & - 1 & 0 \\ - 1 & 4 & - 1 \\ 0 & - 1 & 8 \end{array} \right] \mathbf {v} = \lambda \mathbf {v}
Evaluate the eigenvalues of the matrices A and \mathbf{A}^{(m)} , m = 1, 2, where \mathbf{A}^{(m)} is obtained by omitting the last m rows and columns in A. Sketch the corresponding characteristic polynomials (see Example 2.38).
2.18. Prove that the 1- and 2-norms of a vector v are equivalent. Then show explicitly this equivalency for the vector
\mathbf {v} = \left[ \begin{array}{c} 1 \\ 4 \\ - 3 \end{array} \right]
2.19. Prove the relation (2.157) for the 1-norm.
2.20. Prove that \| \mathbf{A}\mathbf{v}\| _1\leq \| \mathbf{A}\| _1\| \mathbf{v}\| _1
2.21. Prove that for a symmetric matrix \mathbf{A} we have \| \mathbf{A}\| _2 = \rho (\mathbf{A}) . (Hint: Use (2.108).)
2.22. Prove that (2.177) and (2.178) hold when we use the dual norm of the L -norm for the R -norm.
Some Basic Concepts of Engineering Analysis and an Introduction to the Finite Element Method
3.1 INTRODUCTION
The analysis of an engineering system requires the idealization of the system into a form that can be solved, the formulation of the mathematical model, the solution of this model, and the interpretation of the results (see Section 1.2). The main objective of this chapter is to discuss some classical techniques used for the formulation and solution of mathematical models of engineering systems (see also S. H. Crandall [A]). This discussion will provide a valuable basis for the presentation of finite element procedures in the next chapters. Two categories of mathematical models are considered: lumped-parameter models and continuum-mechanics-based models. We also refer to these as “discrete-system” and “continuous-system” mathematical models.
In a lumped-parameter mathematical model, the actual system response is directly described by the solution of a finite number of state variables. In this chapter we discuss some general procedures that are employed to obtain the governing equations of lumped-parameter models. We consider steady-state, propagation, and eigenvalue problems and also briefly discuss the nature of the solutions of these problems.
For a continuum-mechanics-based mathematical model, the formulation of the governing equations is achieved as for a lumped-parameter model, but instead of a set of algebraic equations for the unknown state variables, differential equations govern the response. The exact solution of the differential equations satisfying all boundary conditions is possible only for relatively simple mathematical models, and numerical procedures must in general be employed. These procedures, in essence, reduce the continuous-system mathematical model to a discrete idealization that can be solved in the same manner as a lumped-parameter model. In this chapter we summarize some important classical procedures that are employed to reduce continuous-system mathematical models to lumped-parameter numerical models and briefly show how these classical procedures provide the basis for modern finite element methods.
In practice, the analyst must decide whether an engineering system should be represented by a lumped-parameter or a continuous-system mathematical model and must choose all specifics of the model. Furthermore, if a certain mathematical model is chosen, the analyst must decide how to solve numerically for the response. This is where much of the value of finite element procedures can be found; that is, finite element techniques used in conjunction with the digital computer have enabled the numerical solution of continuous-system mathematical models in a systematic manner and in effect have made possible the practical extension and application of the classical procedures presented in this chapter to very complex engineering systems.
3.2 SOLUTION OF DISCRETE-SYSTEM MATHEMATICAL MODELS
In this section we deal with discrete or lumped-parameter mathematical models. The essence of a lumped-parameter mathematical model is that the state of the system can be described directly with adequate precision by the magnitudes of a finite (and usually small) number of state variables. The solution requires the following steps:
- System idealization: the actual system is idealized as an assemblage of elements
- Element equilibrium: the equilibrium requirements of each element are established in terms of state variables
- Element assemblage: the element interconnection requirements are invoked to establish a set of simultaneous equations for the unknown state variables
- Calculation of response: the simultaneous equations are solved for the state variables, and using the element equilibrium requirements, the response of each element is calculated.
These steps of solution are followed in the analyses of the different types of problems that we consider: steady-state problems, propagation problems, and eigenvalue problems. The objective in this section is to provide an introduction showing how problems in these particular areas are analyzed and to briefly discuss the nature of the solutions. It should be realized that not all types of analysis problems in engineering are considered; however, a large majority of problems do fall naturally into these problem areas. In the examples in this section we consider structural, electrical, fluid flow, and heat transfer problems, and we emphasize that in each of these analyses the same basic steps of solution are followed.
3.2.1 Steady-State Problems
The main characteristic of a steady-state problem is that the response of the system does not change with time. Thus, the state variables describing the response of the system under consideration can be obtained from the solution of a set of equations that do not involve time as a variable. In the following examples we illustrate the procedure of analysis in the solution of some problems. Five sample problems are presented:
- Elastic spring system
- Heat transfer system
- Hydraulic network
- Dc network
- Nonlinear elastic spring system.
The analysis of each problem illustrates the application of the general steps of analysis summarized in Section 3.2. The first four problems involve the analysis of linear systems, whereas the nonlinear elastic spring system responds nonlinearly to the applied loads. All the problems are well defined, and a unique solution exists for each system response.
EXAMPLE 3.1: Figure E3.1 shows a system of three rigid carts on a horizontal plane that are interconnected by a system of linear elastic springs. Calculate the displacements of the carts and the forces in the springs for the loading shown.
text_image
U1, R1 k4 k3 U2, R2 k1 1 2 k2 k5 3 U3, R3
(a) Physical layout
text_image
k₁ U₁ F₁⁽¹⁾
k_{1} U_{1} = F_{1}^{(1)}
text_image
F₁⁽²⁾ U₁ k₂ U₂ F₂⁽²⁾
text_image
F₁⁽⁴⁾ U₁ k₄ U₃ F₃⁽⁴⁾
text_image
F₁⁽³⁾ U₁ k₃ U₂ F₂⁽³⁾
text_image
F₂⁽⁵⁾ U₂ k₅ U₃ F₃⁽⁵⁾
(b) Element equilibrium relations
Figure E3.1 System of rigid carts interconnected by linear springs
We perform the analysis by following steps 1 to 4 in Section 3.2. As state variables that characterize the response of the system, we choose the displacements U_{1} , U_{2} , and U_{3} . These displacements are measured from the initial positions of the carts, in which the springs are unstretched. The individual spring elements and their equilibrium requirements are shown in Fig. E3.1(b).
To generate the governing equations for the state variables we invoke the element interconnection requirements, which correspond to the static equilibrium of the three carts:
F _ {1} ^ {(1)} + F _ {1} ^ {(2)} + F _ {1} ^ {(3)} + F _ {1} ^ {(4)} = R _ {1}
F _ {2} ^ {(2)} + F _ {2} ^ {(3)} + F _ {2} ^ {(5)} = R _ {2} \tag {a}
F _ {3} ^ {(4)} + F _ {3} ^ {(5)} = R _ {3}
We can now substitute for the element end forces F_{i}^{(j)} ; i = 1, 2, 3; j = 1, \ldots, 5 ; using the element equilibrium requirements given in Fig. E3.1(b). Here we recognize that corresponding to the displacement components U_{1}, U_{2} , and U_{3} we can write for element 1,
\left[ \begin{array}{c c c} k _ {1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \end{array} \right] = \left[ \begin{array}{c} F _ {1} ^ {(1)} \\ 0 \\ 0 \end{array} \right]
or
\mathbf {K} ^ {(1)} \mathbf {U} = \mathbf {F} ^ {(1)}
for element 2,
\left[ \begin{array}{c c c} k _ {2} & - k _ {2} & 0 \\ - k _ {2} & k _ {2} & 0 \\ 0 & 0 & 0 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \\ U _ {3} \end{array} \right] = \left[ \begin{array}{c} F _ {1} ^ {(2)} \\ F _ {2} ^ {(2)} \\ 0 \end{array} \right]
or \mathbf{K}^{(2)}\mathbf{U} = \mathbf{F}^{(2)} , and so on. Hence the element interconnection requirements in (a) reduce to
\mathbf {K} \mathbf {U} = \mathbf {R} \tag {b}
where \mathbf{U}^T = [U_1\quad U_2\quad U_3]
\mathbf {K} = \left[ \begin{array}{c c c} \left(k _ {1} + k _ {2} + k _ {3} + k _ {4}\right) & - \left(k _ {2} + k _ {3}\right) & - k _ {4} \\ - \left(k _ {2} + k _ {3}\right) & \left(k _ {2} + k _ {3} + k _ {5}\right) & - k _ {5} \\ - k _ {4} & - k _ {5} & \left(k _ {4} + k _ {5}\right) \end{array} \right]
and \mathbf{R}^T = [R_1\quad R_2\quad R_3]
Here it is noted that the coefficient matrix K can be obtained using
\mathbf {K} = \sum_ {i = 1} ^ {5} \mathbf {K} ^ {(i)} \tag {c}
where the \mathbf{K}^{(i)} are the element stiffness matrices. The summation process for obtaining in (c) the total structure stiffness matrix by direct summation of the element stiffness matrices is referred to as the direct stiffness method.
The analysis of the system is completed by solving (b) for the state variables U_{1} , U_{2} , and U_{3} and then calculating the element forces from the element equilibrium relationships in Fig. E3.1.
EXAMPLE 3.2: A wall is constructed of two homogeneous slabs in contact as shown in Fig. E3.2. In steady-state conditions the temperatures in the wall are characterized by the external surface temperatures \theta_{1} and \theta_{3} and the interface temperature \theta_{2} . Establish the equi-
text_image
Surface coefficient 3k θ₀ θ₁ θ₂ θ₃ Surface coefficient 2k θ₄ Conductance 2k Conductance 3k
Figure E3.2 Slab subjected to temperature boundary conditions
librium equations of the problem in terms of these temperatures when the ambient temperatures \theta_{0} and \theta_{4} are known.
The conductance per unit area for the individual slabs and the surface coefficients are given in Fig. E3.2. The heat conduction law is q/A = \bar{k} \Delta \theta , where q is the total heat flow, A is the area, \Delta \theta is the temperature drop in the direction of heat flow, and \bar{k} is the conductance or surface coefficient. The state variables in this analysis are \theta_{1} , \theta_{2} , and \theta_{3} . Using the heat conduction law, the element equilibrium equations are
for the left surface, per unit area:
q _ {1} = 3 k \left(\theta_ {0} - \theta_ {1}\right)
for the left slab: q_{2} = 2k(\theta_{1} - \theta_{2})
for the right slab: q_{3} = 3k(\theta_{2} - \theta_{3})
for the right surface: q_{4} = 2k(\theta_{3} - \theta_{4})
To obtain the governing equations for the state variables, we invoke the heat flow equilibrium requirement q_{1} = q_{2} = q_{3} = q_{4} . Thus,
3 k \left(\theta_ {0} - \theta_ {1}\right) = 2 k \left(\theta_ {1} - \theta_ {2}\right)
2 k \left(\theta_ {1} - \theta_ {2}\right) = 3 k \left(\theta_ {2} - \theta_ {3}\right)
3 k \left(\theta_ {2} - \theta_ {3}\right) = 2 k \left(\theta_ {3} - \theta_ {4}\right)
Writing these equations in matrix form we obtain
\left[ \begin{array}{c c c} 5 k & - 2 k & 0 \\ - 2 k & 5 k & - 3 k \\ 0 & - 3 k & 5 k \end{array} \right] \left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \end{array} \right] = \left[ \begin{array}{c} 3 k \theta_ {0} \\ 0 \\ 2 k \theta_ {4} \end{array} \right] \tag {a}
These equilibrium equations can be also derived in a systematic manner using a direct stiffness procedure. Using this technique, we proceed as in Example 3.1 with the typical element equilibrium relations
\bar {k} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{c} \theta_ {i} \\ \theta_ {j} \end{array} \right] = \left[ \begin{array}{c} q _ {i} \\ q _ {j} \end{array} \right]
where q_{i}, q_{j} are the heat flows into the element and \theta_{i}, \theta_{j} are the element-end temperatures. For the system in Fig. E3.2 we have two conduction elements (each slab being one element), hence we obtain
\left[ \begin{array}{c c c} 2 k & - 2 k & 0 \\ - 2 k & 5 k & - 3 k \\ 0 & - 3 k & 3 k \end{array} \right] \left[ \begin{array}{l} \theta_ {1} \\ \theta_ {2} \\ \theta_ {3} \end{array} \right] = \left[ \begin{array}{c} 3 k \left(\theta_ {0} - \theta_ {1}\right) \\ 0 \\ 2 k \left(\theta_ {4} - \theta_ {3}\right) \end{array} \right] \tag {b}
Since \theta_{1} and \theta_{3} are unknown, the equilibrium relations in (b) are rearranged for solution to obtain the relations in (a).
It is interesting to note the analogy between the displacement and force analysis of the spring system in Example 3.1 and the temperature and heat transfer analysis in Example 3.2. The coefficient matrices are very similar in both analyses, and they can both be obtained in a very systematic manner. To emphasize the analogy we give in Fig. 3.1 a spring model that is governed by the coefficient matrix of the heat transfer problem.
text_image
3k 2k 3k 2k U₁, R₁ U₂, R₂ U₃, R₃
Figure 3.1 Assemblage of springs governed by same coefficient matrix as the heat transfer problem in Fig. E3.2
We next consider the analyses of a simple flow problem and a simple electrical system, both of which are again analyzed in much the same manner as the spring and heat transfer problems.
EXAMPLE 3.3: Establish the equations that govern the steady-state pressure and flow distributions in the hydraulic network shown in Fig. E3.3. Assume the fluid to be incompressible and the pressure drop in a branch to be proportional to the flow q through that branch, \Delta p = Rq , where R is the branch resistance coefficient.
In this analysis the elements are the individual branches of the pipe network. As unknown state variables that characterize the flow and pressure distributions in the system we select the
flowchart
```mermaid graph TD A["A"] -->|q1| B["B"] B -->|q2| C["C"] C -->|q3| D["D"] D -->|q4| C C -->|R = 5b| E["E"] E -->|R = 10b| A style A fill:#f9f,stroke:#333 style B fill:#f9f,stroke:#333 style C fill:#f9f,stroke:#333 style D fill:#f9f,stroke:#333 style E fill:#f9f,stroke:#333 ```Figure E3.3 Pipe network
pressures at A, C , and D , which we denote as p_A, p_C , and p_D , and we assume that the pressure at B is zero. Thus, we have for the elements
q _ {1} = \frac {p _ {A}}{1 0 b}; \quad q _ {3} = \frac {p _ {C} - p _ {D}}{2 b}
q _ {2} \mid_ {A C} = \frac {p _ {A} - p _ {C}}{5 b}; \quad q _ {2} \mid_ {D B} = \frac {p _ {D}}{5 b}; \quad q _ {4} = \frac {p _ {C} - p _ {D}}{3 b} \tag {a}
The element interconnectivity requirements require continuity of flow, hence
Q = q _ {1} + q _ {2}
q _ {2} \mid_ {A C} = q _ {3} + q _ {4}; \quad q _ {2} \mid_ {D B} = q _ {3} + q _ {4} \tag {b}
Substituting from (a) into (b) and writing the resulting equations in matrix form, we obtain
\left[ \begin{array}{r r r} 3 & - 2 & 0 \\ - 6 & 3 1 & - 2 5 \\ - 1 & 1 & 1 \end{array} \right] \left[ \begin{array}{l} p _ {A} \\ p _ {C} \\ p _ {D} \end{array} \right] = \left[ \begin{array}{c} 1 0 b Q \\ 0 \\ 0 \end{array} \right]
or
\left[ \begin{array}{r r r} 9 & - 6 & 0 \\ - 6 & 3 1 & - 2 5 \\ 0 & - 2 5 & 3 1 \end{array} \right] \left[ \begin{array}{l} p _ {A} \\ p _ {C} \\ p _ {D} \end{array} \right] = \left[ \begin{array}{c} 3 0 b Q \\ 0 \\ 0 \end{array} \right] \tag {c}
The analysis of the pipe network is completed by solving from (c) for the pressures p_{A} , p_{C} , and p_{D} , and then the element equilibrium relations in (a) can be employed to obtain the flow distribution.
The equilibrium relations in (c) can also be derived—as in the preceding spring and heat transfer examples—using a direct stiffness procedure. Using this technique, we proceed as in Example 3.1 with the typical element equilibrium relations
\frac {1}{R} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \left[ \begin{array}{l} p _ {i} \\ p _ {j} \end{array} \right] = \left[ \begin{array}{l} q _ {i} \\ q _ {j} \end{array} \right]
where q_{i} , q_{j} are the fluid flows into the element and p_{i} , p_{j} are the element-end pressures.
EXAMPLE 3.4: Consider the dc network shown in Fig. E3.4. The network with the resistances shown is subjected to the constant-voltage inputs E and 2E at A and B, respectively. We are to determine the steady-state current distribution in the network.
In this analysis we use as unknown state variables the currents I_{1} , I_{2} , and I_{3} . The system elements are the resistors, and the element equilibrium requirements are obtained by applying Ohm's law to the resistors. For a resistor \overline{R} , carrying current I, we have Ohm's law
\Delta E = \overline {{{R}}} I
where \Delta E is the voltage drop across the resistor.
The element interconnection law to be satisfied is Kirchhoff's voltage law for each closed loop in the network,
2 E = 2 R I _ {1} + 2 R \left(I _ {1} - I _ {3}\right)
E = 4 R (I _ {2} - I _ {3})
0 = 6 R I _ {3} + 4 R (I _ {3} - I _ {2}) + 2 R (I _ {3} - I _ {1})