김경종
624 lines
22 KiB
Markdown
624 lines
22 KiB
Markdown
<!-- source-page: 111 -->
|
||
|
||
As in the derivation of steady-state equilibrium equations (see Section 3.2.1), we can employ the direct procedure or a variational approach to establish the problem-governing equations, and we describe both techniques in this problem solution.
|
||
|
||
In the direct approach we establish the governing equilibrium equations directly by considering the equilibrium of the structure in its deformed configuration. Referring to Fig. E3.11, the moment equilibrium of bar AB requires that
|
||
|
||
$$
|
||
P L \sin (\alpha + \beta) = k U _ {1} L \cos (\alpha + \beta) + k _ {r} \alpha \tag {a}
|
||
$$
|
||
|
||
Similarly, for bars CBA we need
|
||
|
||
$$
|
||
P L [ \sin (\alpha + \beta) + \sin \beta ] = k U _ {1} L [ \cos (\alpha + \beta) + \cos \beta ] + k U _ {2} L \cos \beta \tag {b}
|
||
$$
|
||
|
||
We select $U_{1}$ and $U_{2}$ as state variables that completely describe the structural response. We also assume small displacements, for which
|
||
|
||
$$
|
||
L \sin (\alpha + \beta) = U _ {1} - U _ {2}; \quad L \sin \beta = U _ {2}
|
||
$$
|
||
|
||
$$
|
||
L \cos (\alpha + \beta) \doteq L; \quad L \cos \beta \doteq L; \quad \alpha \doteq \frac {U _ {1} - 2 U _ {2}}{L}
|
||
$$
|
||
|
||
Substituting into (a) and (b) and writing the resulting equations in matrix form, we obtain
|
||
|
||
$$
|
||
\left[ \begin{array}{c c} k L + \frac {k _ {r}}{L} & - 2 \frac {k _ {r}}{L} \\ 2 k L & k L \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right] = P \left[ \begin{array}{c c} 1 & - 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right]
|
||
$$
|
||
|
||
We can symmetrize the coefficient matrices by multiplying the first row by -2 and adding the result to row 2, which gives the eigenvalue problem
|
||
|
||
$$
|
||
\left[ \begin{array}{c c} k L + \frac {k _ {r}}{L} & - \frac {2 k _ {r}}{L} \\ - \frac {2 k _ {r}}{L} & k L + \frac {4 k _ {r}}{L} \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] = P \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 2 \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] \tag {c}
|
||
$$
|
||
|
||
It may be noted that the second equation in (c) can also be obtained by considering the moment equilibrium of bar $CB$ .
|
||
|
||
Considering next the variational approach, we need to determine the total potential $\Pi$ of the system in the deformed configuration. Here we have
|
||
|
||
$$
|
||
\Pi = \frac {1}{2} k U _ {1} ^ {2} + \frac {1}{2} k U _ {2} ^ {2} + \frac {1}{2} k _ {r} \alpha^ {2} - P L [ 1 - \cos (\alpha + \beta) + 1 - \cos \beta ] \tag {d}
|
||
$$
|
||
|
||
As in the direct approach, we now assume small displacement conditions. Since we want to derive, using $(3.1)$ , an eigenvalue problem of form $(3.5)$ in which the coefficient matrices are independent of the state variables, we approximate the trigonometric expressions to second order in the state variables. Thus, we use
|
||
|
||
$$
|
||
\cos (\alpha + \beta) \doteq 1 - \frac {(\alpha + \beta) ^ {2}}{2}
|
||
$$
|
||
|
||
$$
|
||
\cos \beta \doteq 1 - \frac {\beta^ {2}}{2} \tag {e}
|
||
$$
|
||
|
||
$$
|
||
\alpha + \beta \doteq \frac {U _ {1} - U _ {2}}{L}; \quad \alpha \doteq \frac {U _ {1} - 2 U _ {2}}{L}; \quad \beta \doteq \frac {U _ {2}}{L} \tag {f}
|
||
$$
|
||
|
||
<!-- source-page: 112 -->
|
||
|
||
Substituting from (e) and (f) into (d) we obtain
|
||
|
||
$$
|
||
\Pi = \frac {1}{2} k U _ {1} ^ {2} + \frac {1}{2} k U _ {2} ^ {2} + \frac {1}{2} k _ {r} \left(\frac {U _ {1} - 2 U _ {2}}{L}\right) ^ {2} - \frac {P}{2 L} (U _ {1} - U _ {2}) ^ {2} - \frac {P}{2 L} U _ {2} ^ {2}
|
||
$$
|
||
|
||
Applying the stationarity principle,
|
||
|
||
$$
|
||
\frac {\partial \Pi}{\partial U _ {1}} = 0; \quad \frac {\partial \Pi}{\partial U _ {2}} = 0
|
||
$$
|
||
|
||
the equations in (c) are obtained.
|
||
|
||
Considering now dynamic analysis, an eigenvalue problem may need to be formulated in the solution of the dynamic equilibrium equations. In essence, the objective is then to find a mathematical transformation on the state variables that is employed effectively in the solution of the dynamic response (see Section 9.3). In the analysis of physical problems, it is then most valuable to identify the eigenvalues and vectors with physical quantities (see Section 9.3).
|
||
|
||
To illustrate how eigenvalue problems are formulated in dynamic analysis, we present the following examples.
|
||
|
||
EXAMPLE 3.12: Consider the dynamic analysis of the system of rigid carts discussed in Example 3.8. Assume free vibration conditions and that
|
||
|
||
$$
|
||
\mathbf {U} = \boldsymbol {\phi} \sin (\omega t - \psi) \tag {a}
|
||
$$
|
||
|
||
where $\phi$ is a vector with components independent of time, $\omega$ is a circular frequency, and $\psi$ is a phase angle. Show that with this assumption an eigenvalue problem of the form given in (3.5) is obtained when searching for a solution of $\phi$ and $\omega$ .
|
||
|
||
The equilibrium equations of the system when considering free-vibration conditions are
|
||
|
||
$$
|
||
\mathbf {M} \ddot {\mathbf {U}} + \mathbf {K} \mathbf {U} = \mathbf {0} \tag {b}
|
||
$$
|
||
|
||
where the matrices M and K and vector U have been defined in Examples 3.1 and 3.8. If U given in (a) is to be a solution of the equations in (b), these equations must be satisfied when substituting for U,
|
||
|
||
$$
|
||
- \omega^ {2} \mathbf {M} \phi \sin (\omega t - \psi) + \mathbf {K} \phi \sin (\omega t - \psi) = \mathbf {0}
|
||
$$
|
||
|
||
Thus, for (a) to be a solution of (b) we obtain the condition
|
||
|
||
$$
|
||
\mathbf {K} \boldsymbol {\phi} = \omega^ {2} \mathbf {M} \boldsymbol {\phi} \tag {c}
|
||
$$
|
||
|
||
which is an eigenvalue problem of form (3.5). We discuss in Section 9.3 the physical characteristics of a solution, $\omega_{i}^{2}$ and $\Phi_{i}$ , to the problem in (c).
|
||
|
||
EXAMPLE 3.13: Consider the electric circuit in Fig. E3.13. Determine the eigenvalue problem from which the resonant frequencies and modes can be calculated when $L_{1} = L_{2} = L$ and $C_{1} = C_{2} = C$ .
|
||
|
||
Our first objective is to derive the dynamic equilibrium equations of the system. The element equilibrium equation for an inductor is
|
||
|
||
$$
|
||
L \frac {d I}{d t} = V \tag {a}
|
||
$$
|
||
|
||
<!-- source-page: 113 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
L₁
|
||
I₂
|
||
C₁
|
||
I₁
|
||
L₂
|
||
I₁ + I₂
|
||
I₁
|
||
C₂
|
||
</details>
|
||
|
||
Figure E3.13 Electric circuit
|
||
|
||
where L is the inductance, I is the current through the inductor, and V is the voltage drop across the inductor. For a capacitor of capacitance C the equilibrium equation is
|
||
|
||
$$
|
||
I = C \frac {d V}{d t} \tag {b}
|
||
$$
|
||
|
||
As state variables we select the currents $I_{1}$ and $I_{2}$ shown in Fig. E3.13. The governing equilibrium equations are obtained by invoking the element interconnectivity requirements contained in Kirchhoff's voltage law:
|
||
|
||
$$
|
||
V _ {C _ {1}} + V _ {L _ {2}} + V _ {C _ {2}} = 0
|
||
$$
|
||
|
||
$$
|
||
V _ {L _ {1}} + V _ {L _ {2}} + V _ {C _ {2}} = 0 \tag {c}
|
||
$$
|
||
|
||
Differentiating (a) and (c) with respect to time and substituting into (c) with $L_{1} = L_{2} = L$ and $C_{1} = C_{2} = C$ , we obtain
|
||
|
||
$$
|
||
L \left[ \begin{array}{l l} 1 & 1 \\ 1 & 2 \end{array} \right] \left[ \begin{array}{l} \ddot {I} _ {1} \\ \ddot {I} _ {2} \end{array} \right] + \frac {1}{C} \left[ \begin{array}{l l} 2 & 1 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{l} I _ {1} \\ I _ {2} \end{array} \right] = \left[ \begin{array}{l} 0 \\ 0 \end{array} \right] \tag {d}
|
||
$$
|
||
|
||
We note that these equations are quite analogous to the free-vibration equilibrium equations of a structural system. Indeed, recognizing the analogy
|
||
|
||
$$
|
||
I \rightarrow \text { displacement }; \quad \frac {1}{C} \rightarrow \text { stiffness }; \quad L \rightarrow \text { mass }
|
||
$$
|
||
|
||
the eigenproblem for the resonant frequencies is established as in Example 3.12 (and an equivalent structural system could be constructed).
|
||
|
||
# 3.2.4 On the Nature of Solutions
|
||
|
||
In the preceding sections we discussed the formulation of steady-state, propagation, and eigenvalue problems, and we gave a number of simple examples. In all cases a system of equations for the unknown state variables was formulated but not solved. For the solution of the equations we refer to the techniques presented in Chapters 8 to 11. The objective in this section is to discuss briefly the nature of the solutions that are calculated when steady-state, propagation, or eigenvalue problems are considered.
|
||
|
||
For steady-state and propagation problems, it is convenient to distinguish between linear and nonlinear problems. In simple terms, a linear problem is characterized by the fact that the response of the system varies in proportion to the magnitude of the applied loads. All other problems are nonlinear, as discussed in more detail in Section 6.1. To demonstrate in an introductory way some basic response characteristics that are predicted in linear steady-state, propagation, and eigenvalue analyses we consider the following example.
|
||
|
||
<!-- source-page: 114 -->
|
||
|
||
EXAMPLE 3.14: Consider the simple structural system consisting of an assemblage of rigid weightless bars, springs, and concentrated masses shown in Fig. E3.14. The elements are connected at A, B, and C using frictionless pins. It is required to analyze the discrete system for the loading indicated, when the initial displacements and velocities are zero.
|
||
|
||
The response of the system is described by the two state variables $U_{1}$ and $U_{2}$ shown in Fig. E3.14(c). To decide what kind of analysis is appropriate we need to have sufficient information on the characteristics of the structure and the applied forces F and P. Let us assume that the structural characteristics and the applied forces are such that the displacements of the element assemblage are relatively small,
|
||
|
||
$$
|
||
\frac {U _ {1}}{L} < \frac {1}{1 0} \quad \text { and } \quad \frac {U _ {2}}{L} < \frac {1}{1 0}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
k_{r1} = kL^2
|
||
m
|
||
2F
|
||
k_{r2} = kL^2
|
||
m/2
|
||
F
|
||
Rigid bar
|
||
Rigid bar
|
||
A
|
||
B
|
||
C
|
||
P
|
||
k
|
||
L
|
||
L
|
||
Smooth hinges at
|
||
A, B, and C
|
||
</details>
|
||
|
||
(a) Discrete system
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
F
|
||
F = sin πt/Td
|
||
Td
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
P
|
||
t
|
||
</details>
|
||
|
||
(b) Loading conditions
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
H
|
||
A
|
||
V
|
||
kr1α
|
||
α
|
||
2F
|
||
mÜ1
|
||
β
|
||
U1
|
||
F
|
||
mÜ2/2
|
||
kU2
|
||
U2
|
||
P
|
||
</details>
|
||
|
||
(c) External forces in deformed configuration
|
||
Figure E3.14 A two degree of freedom system
|
||
|
||
<!-- source-page: 115 -->
|
||
|
||
We can then assume that
|
||
|
||
$$
|
||
\cos \alpha = \cos \beta = \cos (\beta - \alpha) = 1
|
||
$$
|
||
|
||
$$
|
||
\sin \alpha = \alpha ; \quad \sin \beta = \beta \tag {a}
|
||
$$
|
||
|
||
$$
|
||
\alpha = \frac {U _ {1}}{L}; \quad \beta = \frac {U _ {2} - U _ {1}}{L}
|
||
$$
|
||
|
||
The governing equilibrium equations are derived as in Example 3.11 but we include inertia forces (see Example 3.8); thus we obtain
|
||
|
||
$$
|
||
\left[ \begin{array}{l l} m & 0 \\ 0 & \frac {m}{2} \end{array} \right] \left[ \begin{array}{l} \ddot {U} _ {1} \\ \ddot {U} _ {2} \end{array} \right] + \left[ \begin{array}{c c} \left(5 k + \frac {2 P}{L}\right) & - \left(2 k + \frac {P}{L}\right) \\ - \left(2 k + \frac {P}{L}\right) & \left(2 k + \frac {P}{L}\right) \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] = \left[ \begin{array}{l} 2 F \\ F \end{array} \right] \tag {b}
|
||
$$
|
||
|
||
The response of the system must depend on the relative values of k, m, and P/L. In order to obtain a measure of whether a static or dynamic analysis must be performed we calculate the natural frequencies of the system. These frequencies are obtained by solving the eigenvalue problem
|
||
|
||
$$
|
||
\left[ \begin{array}{l l} \left(5 k + \frac {2 P}{L}\right) & - \left(2 k + \frac {P}{L}\right) \\ - \left(2 k + \frac {P}{L}\right) & \left(2 k + \frac {P}{L}\right) \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] = \omega^ {2} \left[ \begin{array}{l l} m & 0 \\ 0 & \frac {m}{2} \end{array} \right] \left[ \begin{array}{l} U _ {1} \\ U _ {2} \end{array} \right] \tag {c}
|
||
$$
|
||
|
||
The solution of (c) gives (see Section 2.5)
|
||
|
||
$$
|
||
\omega_ {1} = \left(\frac {9 k}{2 m} + \frac {2 P}{m L} - \sqrt {\frac {3 3 k ^ {2}}{4 m ^ {2}} + \frac {8 P k}{m ^ {2} L} + \frac {2 P ^ {2}}{m ^ {2} L ^ {2}}}\right) ^ {1 / 2}
|
||
$$
|
||
|
||
$$
|
||
\omega_ {2} = \left(\frac {9 k}{2 m} + \frac {2 P}{m L} + \sqrt {\frac {3 3 k ^ {2}}{4 m ^ {2}} + \frac {8 P k}{m ^ {2} L} + \frac {2 P ^ {2}}{m ^ {2} L ^ {2}}}\right) ^ {1 / 2}
|
||
$$
|
||
|
||
We note that for constant k and m the natural frequencies (radians per unit time) are a function of P/L and increase with P/L as shown in Fig. E3.14(d). The ith natural period, $T_{i}$ , of the system is given by $T_{i} = 2\pi/\omega_{i}$ , hence
|
||
|
||
$$
|
||
T _ {1} = \frac {2 \pi}{\omega_ {1}}; \quad T _ {2} = \frac {2 \pi}{\omega_ {2}}
|
||
$$
|
||
|
||
The response of the system depends to a large degree on the duration of load application when measured on the natural periods of the system. Since P is constant, the duration of load application is measured by $T_{d}$ . To illustrate the response characteristics of the system, we assume a specific case $k = m = P/L = 1$ and three different values of $T_{d}$ .
|
||
|
||
<!-- source-page: 116 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| P/kL | ω₁/(k/m)¹/² | ω₂/(k/m)¹/² |
|
||
|------|-------------|-------------|
|
||
| -2 | 1.0 | 0.5 |
|
||
| 0 | 2.5 | 1.5 |
|
||
| 2 | 3.5 | 2.0 |
|
||
| 4 | 4.5 | 2.5 |
|
||
| 6 | 5.5 | 2.8 |
|
||
| 8 | 6.0 | 3.0 |
|
||
| 10 | 6.5 | 3.2 |
|
||
</details>
|
||
|
||
(d) Frequencies of the system
|
||
|
||
|
||
(e) Analysis of the system: Case i
|
||
Figure E3.14 (continued)
|
||
|
||
Case (i) $T_{d} = 4T_{1}$ : The response of the system is shown for this case of load application in Fig. E3.14(e), Case i. We note that the dynamic response solution is somewhat close to the static response of the system and would be very close if $T_{1} \ll T_{d}$ .
|
||
|
||
<!-- source-page: 117 -->
|
||
|
||
Case (ii) $T_{d} = (T_{1} + T_{2})/2$ : The response of the system is truly dynamic as shown in Fig. E3.14(e), Case ii. It would be completely inappropriate to neglect the inertia effects.
|
||
|
||
Case (iii) $T_{d} = 1/4 T_{2}$ : In this case the duration of the loading is relatively short compared to the natural periods of the system. The response of the system is truly dynamic, and inertia effects must be included in the analysis as shown in Fig. E3.14(e), Case iii. The response of the system is somewhat close to the response generated assuming impulse conditions and would be very close if $T_{2} \gg T_{d}$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| t/Td | U1 | U2 | U2_static | U1_static |
|
||
|------|----|----|----------|----------|
|
||
| 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
|
||
| 0.1 | 0.2 | 0.4 | 0.2 | 0.1 |
|
||
| 0.2 | 0.4 | 0.6 | 0.4 | 0.2 |
|
||
| 0.3 | 0.6 | 0.8 | 0.6 | 0.3 |
|
||
| 0.4 | 0.8 | 1.0 | 0.8 | 0.4 |
|
||
| 0.5 | 1.0 | 1.2 | 1.0 | 0.5 |
|
||
| 0.6 | 1.2 | 1.4 | 1.2 | 0.6 |
|
||
| 0.7 | 1.4 | 1.6 | 1.4 | 0.7 |
|
||
| 0.8 | 1.6 | 1.8 | 1.6 | 0.8 |
|
||
| 0.9 | 1.4 | 1.6 | 1.4 | 0.7 |
|
||
| 1.0 | 1.2 | 1.4 | 1.2 | 0.6 |
|
||
</details>
|
||
|
||
(e) Analysis of the system: Case ii
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| t/T₁ | U₁ | U₂ | Impulse response |
|
||
|------|--------|--------|------------------|
|
||
| 0.0 | 0.0 | 0.0 | 0.0 |
|
||
| 0.1 | 0.8 | 0.9 | 0.7 |
|
||
| 0.2 | 1.0 | 1.2 | 0.9 |
|
||
| 0.3 | 1.1 | 1.7 | 1.0 |
|
||
| 0.4 | 0.9 | 1.6 | 0.8 |
|
||
| 0.5 | 0.5 | 1.0 | 0.5 |
|
||
| 0.6 | 0.0 | 0.0 | 0.0 |
|
||
| 0.7 | -0.5 | -0.8 | -0.5 |
|
||
| 0.8 | -1.0 | -1.2 | -1.0 |
|
||
| 0.9 | -0.5 | -0.8 | -0.5 |
|
||
| 1.0 | 0.0 | 0.0 | 0.0 |
|
||
</details>
|
||
|
||
(e) Analysis of the system: Case iii (here the actual displacements are obtained by multiplying the given values by $2T_{d}/\pi$ ; the impulse response was calculated using $^{0}U_{1}=^{0}U_{2}=0$ , $^{0}\dot{U}_{1}=^{0}\dot{U}_{2}=4T_{d}/\pi$ and setting the external loads to zero).
|
||
Figure E3.14 (continued)
|
||
|
||
<!-- source-page: 118 -->
|
||
|
||
To identify some conditions for which the structure becomes unstable, we note from (b) that the stiffness of the structure increases with increasing values of P/L (which is why the frequencies increase as P/L increases). Therefore, for the structure to become unstable, we need a negative value of P; i.e., P must be compressive. Let us now assume that P is decreased very slowly (P increases in compression) and that F is very small. In this case a static analysis is appropriate, and we can neglect the force F to obtain from (b) the governing equilibrium equations
|
||
|
||
$$
|
||
\left[ \begin{array}{c c} 5 k & - 2 k \\ - 2 k & 2 k \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right] = \frac {P}{L} \left[ \begin{array}{c c} - 2 & 1 \\ 1 & - 1 \end{array} \right] \left[ \begin{array}{c} U _ {1} \\ U _ {2} \end{array} \right]
|
||
$$
|
||
|
||
The solution of this eigenvalue problem gives two values for P/L. Because of the sign convention for P, the larger eigenvalue gives the critical load
|
||
|
||
$$
|
||
P _ {\mathrm{crit}} = - 2 k L
|
||
$$
|
||
|
||
It may be noted that this is the load at which the smallest frequency of the system is zero [see Fig. E3.14(d)].
|
||
|
||
Although we considered a structural system in this example, most of the solution characteristics presented are also directly observed in the analysis of other types of problems. As shown in an introductory manner in the example, it is important that an analyst be able to decide what kind of analysis is required: whether a steady-state analysis is sufficient or whether a dynamic analysis should be performed, and whether the system may become unstable. We discuss some important factors that influence this decision in Chapters 6 and 9.
|
||
|
||
In addition to deciding what kind of analysis should be performed, the analyst must select an appropriate lumped-parameter mathematical model of the actual physical system. The characteristics of this model depend on the analysis to be carried out, but in complex engineering analyses, a simple lumped-parameter model is in many cases not sufficient, and it is necessary to idealize the system by a continuum-mechanics-based mathematical model. We introduce the use of such models in the next section.
|
||
|
||
# 3.2.5 Exercises
|
||
|
||
3.1. Consider the simple cart system in static (steady-state) conditions shown. Establish the governing equilibrium equations.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Spring stiffness k
|
||
Cart 1
|
||
R1 = 10
|
||
3k
|
||
k
|
||
R2 = 20
|
||
4k
|
||
Cart 2
|
||
Cart 3
|
||
R3 = 0
|
||
2k
|
||
</details>
|
||
|
||
<!-- source-page: 119 -->
|
||
|
||
3.2. Consider the wall of three homogeneous slabs in contact as shown. Establish the steady-state heat transfer equilibrium equations of the analysis problem.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Conductance 4k
|
||
Conductance 2k
|
||
Prescribed
|
||
temperature θ₁
|
||
θ₂
|
||
θ₃
|
||
θ₄
|
||
Conductance 3k
|
||
Surface coefficient 3k
|
||
Environmental
|
||
temperature θ₅
|
||
</details>
|
||
|
||
3.3. The hydraulic network shown is to be analyzed. Establish the equilibrium equations of the system when $\Delta p = Rq$ and R is the branch resistance coefficient.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
R = 6b
|
||
R = 3b
|
||
R = 2b
|
||
R = 4b
|
||
R = 10b
|
||
R = 8b
|
||
Q
|
||
Q
|
||
</details>
|
||
|
||
3.4. The dc network shown is to be analyzed. Using Ohm's law, establish the current-voltage drop equilibrium equations of the system.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
R
|
||
2R
|
||
3R
|
||
E
|
||
E
|
||
</details>
|
||
|
||
<!-- source-page: 120 -->
|
||
|
||
3.5. Consider the spring-cart system in Exercise 3.1. Determine the variational indicator $\Pi$ of the total potential of this system.
|
||
3.6. Consider the slab in Example 3.2. Find a variational indicator $\Pi$ that has the property that $\delta \Pi = 0$ generates the governing equilibrium equations.
|
||
3.7. Establish the dynamic equilibrium equations of the system of carts in Exercise 3.1 when the carts have masses $m_1, m_2,$ and $m_3$ .
|
||
3.8. Consider the simple spring-cart system shown initially at rest. Establish the equations governing the dynamic response of the system.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| Time | Ri |
|
||
|---|---|
|
||
| 0 | 1 |
|
||
| 1 | 1 |
|
||
| 2 | 1 |
|
||
| 3 | 1 |
|
||
| 4 | 1 |
|
||
| 5 | 1 |
|
||
| 6 | 1 |
|
||
| 7 | 1 |
|
||
| 8 | 1 |
|
||
| 9 | 1 |
|
||
| 10 | 1 |
|
||
| 11 | 1 |
|
||
| 12 | 1 |
|
||
| 13 | 1 |
|
||
| 14 | 1 |
|
||
| 15 | 1 |
|
||
| 16 | 1 |
|
||
| 17 | 1 |
|
||
| 18 | 1 |
|
||
| 19 | 1 |
|
||
| 20 | 1 |
|
||
| 21 | 1 |
|
||
| 22 | 1 |
|
||
| 23 | 1 |
|
||
| 24 | 1 |
|
||
| 25 | 1 |
|
||
| 26 | 1 |
|
||
| 27 | 1 |
|
||
| 28 | 1 |
|
||
| 29 | 1 |
|
||
| 30 | 1 |
|
||
| 31 | 1 |
|
||
| 32 | 1 |
|
||
| 33 | 1 |
|
||
| 34 | 1 |
|
||
| 35 | 1 |
|
||
| 36 | 1 |
|
||
| 37 | 1 |
|
||
| 38 | 1 |
|
||
| 39 | 1 |
|
||
| 40 | 1 |
|
||
| 41 | 1 |
|
||
| 42 | 1 |
|
||
| 43 | 1 |
|
||
| 44 | 1 |
|
||
| 45 | 1 |
|
||
| 46 | 1 |
|
||
| 47 | 1 |
|
||
| 48 | 1 |
|
||
| 49 | 1 |
|
||
| 50 | 1 |
|
||
| 51 | 1 |
|
||
| 52 | 1 |
|
||
| 53 | 1 |
|
||
| 54 | 1 |
|
||
| 55 | 1 |
|
||
| 56 | 1 |
|
||
| 57 | 1 |
|
||
| 58 | 1 |
|
||
| 59 | 1 |
|
||
| 60 | 1 |
|
||
| 61 | 1 |
|
||
| 62 | 1 |
|
||
| 63 | 1 |
|
||
| 64 | 1 |
|
||
| 65 | 1 |
|
||
| 66 | 1 |
|
||
| 67 | 1 |
|
||
| 68 | 1 |
|
||
| 69 | 1 |
|
||
| 70 | 1 |
|
||
| 71 | 1 |
|
||
| 72 | 1 |
|
||
| 73 | 1 |
|
||
| 74 | 1 |
|
||
| 75 | 1 |
|
||
| 76 | 1 |
|
||
| 77 | 1 |
|
||
| 78 | 1 |
|
||
| 79 | 1 |
|
||
| 80 | 1 |
|
||
| 81 | 1 |
|
||
| 82 | 1 |
|
||
| 83 | 1 |
|
||
| 84 | 1 |
|
||
| 85 | 1 |
|
||
| 86 | 1 |
|
||
| 87 | 1 |
|
||
| 88 | 1 |
|
||
| 89 | 1 |
|
||
| 90 | 1 |
|
||
| 91 | 1 |
|
||
| 92 | 1 |
|
||
| 93 | 1 |
|
||
| 94 | 1 |
|
||
| 95 | 1 |
|
||
| 96 | 1 |
|
||
| 97 | 1 |
|
||
| 98 | 1 |
|
||
| 99 | 1 |
|
||
| 100 | 1 |
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
R1 = 50
|
||
Spring stiffness k
|
||
R2 = 0
|
||
R3 = 10
|
||
Mass m1
|
||
2k
|
||
2k
|
||
3k
|
||
m2
|
||
m3
|
||
Rigid carts
|
||
</details>
|
||
|
||
3.9. The rigid bar and cable structure shown is to be analyzed for its dynamic response. Formulate the equilibrium equations of motion.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Massless cable
|
||
Tension T
|
||
Length L
|
||
R
|
||
L
|
||
Mass M
|
||
M
|
||
Rigid, massless bars
|
||
k
|
||
Spring stiffness k
|
||
</details>
|