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x A = A(x).
(a)
(b)
Figure 16.8-1. (a) A pipe of variable cross-sectional area A. (b) Possible finite elements for acoustic mode analysis.
16.9 FLUID-STRUCTURE INTERACTION
Remarks. A structure may be surrounded by fluid or may contain fluid. A phenomenon of fluid–structure interaction exists when fluid and structure both move and exert forces upon one another associated with the motion. Different categories of the phenomenon may be identified [16.10].
- Problems having large relative motion; for example, aerodynamic flutter. Behavior is dominated by flow characteristics of the fluid.
- Problems having limited fluid displacement; for example, a reservoir behind a dam under earthquake load, or wave action on an offshore structure.
In this section we are concerned with the second category.
One analytical approach is to model the structure by finite elements as usual, then load its wetted surface by the pressures described by Eq. 16.8-11. This approach results in a system of equations that uses displacement d.o.f. of the structure and pressure d.o.f. at nodes of the wetted surface. The coefficient matrices are unsymmetric. With the simplifying assumptions of no surface waves and incompressibility, the pressure d.o.f. may be eliminated, leaving a smaller system of equations that includes only the structural d.o.f. and has symmetric coefficient matrices. In this smaller system the fluid is represented by a “virtual” mass matrix that is added to the mass matrix of the structure. However, the simplifying assumption of incompressibility may be crude, as seems to be the case for a relatively stiff submerged structure [16.9] .
Another analytical approach is to model the fluid by displacement-based elements, as if the fluid were a structure with special material properties. This “mock-fluid” approach is described in some detail in the following discussion. As compared with the pressure-load approach, the mock-fluid approach typically requires more d.o.f. but produces symmetric matrices with smaller bandwidth or wave front.
Other approaches are also available, for example, the boundary element method.
Mock-Fluid Elements. We will model the fluid by elements that have displacement d.o.f., as do solid elements. Displacements are assumed to be small. Our needs are for a compressibility stiffness matrix [k_{B}] , a wave or slosh stiffness matrix [k_{S}] , and a mass matrix [m]. This formulation appears to be reliable for static analysis and for calculation of natural frequencies of fluid in rigid or flexible containers, but does not always accurately predict natural frequencies of elastic structures surrounded by fluid [16.15].
The strain energy per unit volume associated with compressibility is B\epsilon_V^2 / 2 , where B is the bulk modulus, and \epsilon_V is the volumetric strain, \epsilon_V = dV / V . The expression B\epsilon_V^2 / 2 can be obtained from the standard expression \{\epsilon\}^T [\mathbf{E}]\{\epsilon\} / 2 by recognizing that \epsilon_V = \epsilon_x + \epsilon_y + \epsilon_z and using [E] from Eq. 9.4-7 with G = 0 . If displacements \{\mathbf{u}\} = [u \ v \ w]^T are interpolated over an element in standard fashion—that is, \{\mathbf{u}\} = [\mathbf{N}]\{\mathbf{d}\} —then
\epsilon_ {V} = \left\lfloor \mathbf {B} _ {V} \right\rfloor \{\mathbf {d} \} \quad \text { where } \quad \left\lfloor \mathbf {B} _ {V} \right\rfloor = \left\lfloor \frac {\partial}{\partial x} \quad \frac {\partial}{\partial y} \quad \frac {\partial}{\partial z} \right\rfloor [ \mathbf {N} ] \tag {16.9-1}
Let us now restrict attention to linear elements, such as the four-node plane element shown in Fig. 16.9-1. We elect to evaluate \epsilon_{V} at only the center of the element and regard its value there as representative of the element as a whole. Thus, with \epsilon_{V0} = [B_{V0}]\{d\} the center value of \epsilon_{V} , the strain energy of compressibility in an element of volume V_{e} is
U = \frac {1}{2} \left\{\mathbf {d} \right\} ^ {T} \left[ \mathbf {k} _ {B} \right] \left\{\mathbf {d} \right\} \quad \text { where } \quad \left[ \mathbf {k} _ {B} \right] = B V _ {e} \left\lfloor \mathbf {B} _ {V 0} \right\rfloor^ {T} \left\lfloor \mathbf {B} _ {V 0} \right\rfloor \tag {16.9-2}
It is not necessary to use Eq. 16.9-2 to generate [k_{B}] . One can use instead the formula for [k] of a structural element (Eq. 4.1-5), integrate using a single Gauss point at the element center, and take [E] from Eq. 9.4-7 but with G = 0. Thus, with reduced integration, bulk modulus B provides the element with resistance only to volume change of the element as a whole, and [k_{B}] of Eq. 16.9-2 is produced.
Note that the constraint of near-incompressibility is enforced at each Gauss point used to evaluate [k_{B}] . Accordingly, if [k_{B}] were formed by full integration of B[B_{V}]^{T}[B_{V}] , [k_{B}] would exhibit undesirable stiffnesses. For example, the plane four-node element of Fig. 16.9-1a would yield zero strain energy only when \{d\} represents rigid-body motion or a pure shear deformation. Thus the slosh mode in Fig. 16.9-2b would be resisted by bulk modulus B, which is clearly unreasonable.
One may contemplate the use of elements of higher order than the linear elements discussed thus far. One would then probably use more than one Gauss point per element, yet underintegrate so as to avoid locking due to too many penalty constraints, as discussed in Sections 9.4 and 9.5. However, higher-order elements seem not to be used in practice.
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a x w4 w3 u4 4 3 u3 u1 1 2 u2 w1 w2 Fluid surface for slosh stiffness (a)
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The only mode resisted εv0 > 0 εv0 = 0 εv0 = 0 εv0 = 0 εv0 = 0 (b)
Figure 16.9-1. (a) Plane rectangular element. (b) Deformation modes of a plane element, showing volumetric strain at the element center. Rigid-body modes are not shown.
A slosh stiffness matrix [k_{S}] is needed only if there is a free surface with waves or an oscillating boundary between fluids of different density. The energy associated with small-amplitude waves on a surface S is
U = \int_ {S} \frac {1}{2} \rho g w _ {s} ^ {2} d S \quad \text { where } \quad w _ {s} = \left[ \mathbf {N} _ {S} \right] \{\mathbf {d} \} \tag {16.9-3}
and \rho = \cdot mass density of the fluid, g = acceleration of gravity, w_{s} = surface displacement in the upward direction, and \lfloor \mathbf{N}_S\rfloor operates only on element d.o.f. in surface S . For example, for the plane element of Fig. 16.9-1a, w_{s} = \lfloor \mathbf{N}_{S}\rfloor \{\mathbf{d}\} = w_{3}x / a + w_{4}(a - x) / a . Equation 16.9-3 becomes
U = \frac {1}{2} \left\{\mathbf {d} \right\} ^ {T} \left[ \mathbf {k} _ {S} \right] \left\{\mathbf {d} \right\} \quad \text { where } \quad \left[ \mathbf {k} _ {S} \right] = \int_ {S} \left[ \mathbf {N} _ {S} \right] ^ {T} \left[ \mathbf {N} _ {S} \right] \rho g d S \tag {16.9-4}
The element mass matrix [m] is formulated in standard fashion, as though the element were elastic (e.g., see Eq. 13.2-5).
If desired, a viscous damping matrix [c] can be included [16.13]
[ \mathbf {c} ] = \int_ {V _ {e}} [ \mathbf {B} ] ^ {T} [ \mathbf {E} _ {G} ] [ \mathbf {B} ] \mu d V \tag {16.9-5}
where [B] is the standard strain-displacement matrix of Eq. 4.1-3, [E_{G}] is the first rectangular matrix on the right-hand side of Eq. 9.4-7 (exclusive of G), and \mu is the dynamic coefficient of viscosity [13.3].
As is evident from Eq. 16.9-2, [k_{B}] is a matrix of rank one. Accordingly, the assembled structure may have several zero-energy modes, called circulation modes. They involve no volume change of the fluid and no surface waves. One such mode is shown in Fig. 16.9-2a. A way to slightly restrain circulation modes is to make G a small positive number in Eq. 9.4-7 and use the structural formulation to obtain [k_{B}] , as explained in the paragraph that follows Eq. 16.9-2. (This is not the same as attributing viscosity to the fluid.) Another way to inhibit circulation is to augment [k_{B}] by the penalty matrix
[ \mathbf {k} _ {c} ] = \alpha \int_ {V _ {e}} [ \mathbf {B} _ {c} ] ^ {T} [ \mathbf {B} _ {c} ] d V \tag {16.9-6}
where \alpha is a penalty number and \{d\}^{T}[k_{c}]\{d\} is the integral over the element of the square of the curl of the displacement field [16.14]. For a plane element in
Figure 16.9-2 (a) A circulation mode in a plane three-element model. (b) A one-element model for slosh in a plane rectangular tank. (c) A two-element model for acoustic analysis of a pipe with an elastic end.
the xz plane, (\operatorname{curl} \mathbf{u})^2 = (u_{,z} - w_{,x})^2 and [\mathbf{B}_c] = \left\lfloor \frac{\partial}{\partial z} - \frac{\partial}{\partial x} \right\rfloor [\mathbf{N}] . If \alpha = B , circulation modes disappear and triangular elements yield good results [16.14]. If circulation modes are unrestrained, they may appear with frequencies indistinguishable from other low frequencies of greater interest [16.14]. Note also that [\mathbf{k}_B] >> [\mathbf{k}_S] (Eqs. 16.9-2 and 16.9-4). Accordingly, if both are used in an analysis, there is a risk that [\mathbf{k}_S] will "fall off the end" of computer words, as discussed in Section 18.2.
Example: Plane Slosh Analysis. Consider the one-element model shown in Fig. 16.9-2b. Displacements in the y (out of plane) direction are assumed to be zero. The element has eight d.o.f., but because of the rigid walls, only w_{3} and w_{4} are unrestrained. Rather than invoke all eight d.o.f. and then eliminate six, we elect to use only w_{3} and w_{4} at the outset. Now u = 0, and
w = \frac {x z}{a b} w _ {3} + \frac {(a - x) z}{a b} w _ {4} \tag {16.9-7}
Using [\mathbf{k}_B], [\mathbf{k}_S] , and the consistent mass matrix [\mathbf{m}] , we obtain, for a unit thickness,
\left(\frac {B a}{4 b} \left[ \begin{array}{l l} 1 & 1 \\ 1 & 1 \end{array} \right] + \frac {\rho g a}{6} \left[ \begin{array}{l l} 2 & 1 \\ 1 & 2 \end{array} \right]\right) \left\{ \begin{array}{l} w _ {3} \\ w _ {4} \end{array} \right\} + \frac {\rho a b}{1 8} \left[ \begin{array}{l l} 2 & 1 \\ 1 & 2 \end{array} \right] \left\{ \begin{array}{l} \ddot {w} _ {3} \\ \ddot {w} _ {4} \end{array} \right\} = \left\{ \begin{array}{l} R _ {3} \\ R _ {4} \end{array} \right\} \tag {16.9-8}
where R_{3} and R_{4} are time-varying loads. Typically (Ba/4b) >> (\rho ga/6) , which implies the condition w_{3} = -w_{4} . This is a constraint of zero volume change. Imposing this constraint, and setting R_{3} = R_{4} = 0 and \left[w_{3} \quad w_{4}\right] = \left[\overline{w}_{3} \quad \overline{w}_{4}\right] \sin \omega t , we obtain an eigenvalue problem that yields \omega^{2} = 3g/b . The theoretical result is \omega^{2} = (\pi g/a) \tanh(\pi b/a) , which yields \omega^{2} = 3.13g/b for a = b.
Example: Acoustic Modes. Air in a uniform pipe is modeled by two two-node elements (Fig. 16.9-2c). At the left end there is a massless plug connected to an elastic spring of stiffness k. For this problem, there is no slosh stiffness, \epsilon_{V} = \epsilon_{x} , and [k_{B}] has the same form as the stiffness matrix of a two-node elastic bar. To [k_{B}] we add the structure (spring) stiffness k. Thus, with [m] the consistent mass matrix and the boundary condition u_{3} = 0 already imposed, the structure equations are
\left(\frac {B A}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 2 \end{array} \right] + \left[ \begin{array}{l l} k & 0 \\ 0 & 0 \end{array} \right]\right) \left\{ \begin{array}{l} u _ {1} \\ u _ {2} \end{array} \right\} + \frac {\rho A L}{6} \left[ \begin{array}{l l} 2 & 1 \\ 1 & 4 \end{array} \right] \left\{ \begin{array}{l} \ddot {u} _ {1} \\ \ddot {u} _ {2} \end{array} \right\} = \left\{ \begin{array}{l} R _ {1} \\ R _ {2} \end{array} \right\} \tag {16.9-9}
where A is the constant cross-sectional area. The fundamental acoustic frequency in a pipe with rigid ends is obtained by setting u_{1} = 0 , R_{2} = 0 , and u_{2} = \bar{u}_{2} \sin \omega t . Thus
\omega^ {2} = \frac {3 B}{\rho L ^ {2}} = \frac {3 c ^ {2}}{L ^ {2}} = \frac {1 2 c ^ {2}}{L _ {T} ^ {2}} \quad \text { and } \quad \omega = 3. 4 6 \frac {c}{L _ {T}} \tag {16.9-10}
The exact result (for infinite k) is \omega = \pi c / L_{T} .
PROBLEMS
Section 16.2
16.1 Obtain Eq. 16.2-7, with matrices defined as in Eq. 16.2-6, by use of the Galerkin method.
16.2 The three-node bar shown is uniform. Temperature T_{1} is prescribed, Q = 0 , node 3 is insulated ( q_{3} = 0 ), and heat is transferred across the lateral surface by convection. In the units used in Section 16.1, let k = 180 , h = 12 , A = 0.1 , p = 1 , and L = 2 .
(a) Use two two-node elements to determine T_{2} and T_{3} in terms of T_{1} and T_{f} .
(b) Repeat part (a), but use a single three-node element of length 2L. Note that [h_{ls}] has the form of a mass matrix, and use an “optimally lumped” form (see Eq. 13.3-9).
Problem 16.2
Section 16.3
16.3 Verify that [\kappa] is as defined by Eq. 16.3-3 (resolve q_x and q_y into components q_r and q_s ).
16.4 Imagine that an essentially two-dimensional body has a gradual variation in thickness \tau —that is, \tau = \tau(x, y) . What changes must be made in the equations developed in Section 16.3?
16.5 Verify that Eq. 16.3-10 yields the correct governing equation and nonessential boundary conditions from the condition \delta \Pi = 0 . Include the term 2h(T_fT - T^2 /2) in \Pi , so as to account for convection on two identical lateral surfaces.
Section 16.4
16.6 Devise an example that shows why \theta must be a principal material direction if T in a solid of revolution is to be symmetric with respect to the \theta = 0 plane.
16.7 Using conventional scalar notation like that in Eq. 16.3-5, write Eqs. 16.4-1 and 16.4-2 for the following special cases.
(a) Solid of revolution with an axially symmetric temperature field and isotropic material.
(b) The plane problem in polar coordinates (T, z = 0) . Let [\kappa] be full, as for an anisotropic material.
16.8 Derive the governing equation of Problem 16.7(b) from first principles, analogous to the Cartesian coordinate derivation in Eqs. 16.3-4 and 16.3-5.
Section 16.5
16.9 The three-node triangle shown is to be used for heat conduction analysis. The body in question is homogeneous, plane, isotropic, and of unit thickness.
(a) Evaluate [k] in terms of k and nodal coordinates.
(a) Evaluate [k] in terms of h and nodal coordinates if only side 1–3 transfers
(b) Evaluate [11] in terms of heat by convection.
(c) Write the “lumped” forms of [h] and [c].
(d) Write the lumped \{r_Q\} in terms of Q and nodal coordinates if Q is constant over the element.
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y 3 1 2 x
Problem 16.9
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z 1 2 r r₁ r₂
Problem 16.11
16.10 Repeat Problem 16.9, but regard the body as a solid of revolution, so that r replaces x and axis of revolution z replaces y .
16.11 The solid of revolution element shown is homogeneous, isotropic, flat, and of unit thickness. Surfaces z = constant are insulated so that heat flows only radially. Assume that T varies linearly with r in the element and let Q be constant. Evaluate the matrices defined by Eqs. 16.5-3 in terms of r_1 , r_2 , and material properties k, c , and so on.
16.12 The element shown is a quadratic triangle with straight sides and midside nodes. Evaluate \{\mathbf{r}_Q\} due to heat input Q_P (units J/s), if Q_P is
(a) Concentrated at node 1.
(b) Concentrated at node 4.
(c) Concentrated at the centroid of the triangle.
(d) Uniformly distributed along a line between nodes 5 and 6. Use a simple approximation.
(e) Improve your answer to part (d) by devising a better approximation.
16.13 The uniform bar element shown is described by Eqs. 16.2-6 and 16.2-7. Conductivity k is a function of temperature. Let T_{a} be the average temperature in the element. In the units used in Section 16.1, let k = 100 - 0.4T_{a} , h = 30 , A = 0.1 , p = 1 , L = 2 , and T_{f} = 200 . Heat flows across the lateral surface. If T = 400^{\circ}\mathrm{C} at node 1 and the right end is insulated ( q_{R} = 0 in Fig. 16.2-1), what is the steady-state value of T_{2} ? Use an iterative solution method.
Problem 16.12
Problem 16.13
Problem 16.14
16.14 The uniform bar shown is modeled by two linear elements. Lateral surfaces are insulated. Heat flows into the bar at node 3 at the prescribed rate 20A, where A is the cross-sectional area of the bar and the units are those used in Section 16.1. The temperature at node 1 is kept at zero. Let conductivity in an element be given by k = 2 + 0.04T_{a} , where T_{a} is the average temperature in an element. Perform two steps of an iterative solution for T_{2} and T_{3} , starting with T_{a} = 0 in each element.
Section 16.6
16.15 Prove Eqs. 16.6-3.
16.16 The sketch shows the actual variation of temperature with time at a certain node. Imagine that we start at point A and use Eq. 16.6-6 to predict T at time t_{n+1} . Consider \beta = 0 , \beta = 0.5 , and \beta = 1.0 . For each of these three values make a sketch that shows how the predicted temperature compares with T_{B} .
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| Point | Time (t) | Temperature (T) |
|---|---|---|
| A | t_n | T_A |
| B | t_{n+1} | T_B |
Problem 16.16
16.17 Show that Eq. 16.6-8 follows from Eqs. 16.6-6 and 16.6-7.
16.18 Imagine that the bar of Fig. 16.2-1 is initially at zero temperature and then is subjected to a heat flux at the right end. The flux appears at time t = 0 and thereafter remains constant. The left end is kept at zero temperature. For a one-element model, physical constants are such that Eq. 16.6-1 becomes 6T + 2\dot{T} = 3 , where T is the temperature of the node where heat flux is imposed. We are to compute T = T(t) by use of Eq. 16.6-8.
(a) To what T should T(t) converge as t becomes large?
(b) What is the exact solution for T = T(t) ?
(c) What is \Delta t_{\mathrm{cr}} for Euler's method?
In what follows, use Eq. 16.6-8 with the values of \beta and \Delta t given. Take five time steps in each case. Compare results with exact values and with the other direct integration results.
(d-g) With \Delta t = 0.1 , take \beta as (d) 0, (e) \frac{1}{2} , (f) \frac{2}{3} , (g) 1.0.
(h-k) With \Delta t = 1.0 , take \beta as (h) 0, (i) \frac{1}{2} , (j) \frac{2}{3} , (k) 1.0.
Section 16.7
16.19 Equations 16.5-3 apply to the problem of groundwater flow if T = hydraulic head (meters of water), [\kappa] = permeability coefficients, q = seepage velocity, and Q = flow rate per unit volume. The sketch represents the plan view of a square array of four wells in a homogeneous, horizontal aquifer of infinite extent and constant thickness \tau between impermeable strata above and below it. The hydraulic head at a large distance from the group of wells is h_0 . The hydraulic head in the neighborhood of a typical well is
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Simple abstract shape with four small dots inside, no text or symbols present
Problem 16.19
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Free surface (zero pressure)
Problem 16.21
desired. Each well pumps at constant flow rate Q_{P} m ^{3} /s. Outline how to set up a finite element solution for steady-state T = T(x, y) in the aquifer, with attention to mesh layout, prescribed d.o.f., and load terms.
16.20 Derive Eqs. 16.7-2.
16.21 The sketch depicts a steady flow that emerges from an enclosure, thereafter to flow with a “free surface” (open to the air). Bernoulli’s equation is \frac{1}{2}(u^{2} + v^{2}) + (p/\rho) + gy = \text{constant} , where p = pressure, \rho = mass density , g = acceleration of gravity, and y = elevation above a datum. Locating the free surface requires an iterative solution process. Outline the major steps of this process [16.8].
Section 16.8
16.22 (a) Derive Eqs. 16.8-1.
(b) Show that dV / V = \epsilon_x + \epsilon_y + \epsilon_z in Eq. 16.8-3.
(b) Show that \alpha \in \mathbb{R}_{\mathbb{X}} is a finite set of \mathbb{R}_{\mathbb{X}} for \mathbb{R}_{\mathbb{X}} = 0 (c) Show that with \Pi defined by Eq. 16.8-7, the condition \delta \Pi = 0 yields Eqs. 16.8-4, 16.8-5, and 16.8-6.
16.23 The formulation of Eqs. 16.8-9 is to be applied with a plane rectangular element of unit thickness (see sketch).
(a) Generate element matrices [k], [c], and [h], in terms of a, b, \rho, B , and g .
(b) Imagine that the sketch represents a one-element model of an incompressible liquid in a rectangular tank that is open at the top. Write the appropriate form of Eq. 16.8-13 for the case a = b = h .
(c) To reduce the number of equations, set p_1 = -p_2 and p_4 = -p_3 , then solve for the frequency of the slosh mode, again with a = b = h . (Do not expect high accuracy. The theoretical result is stated below Eq. 16.9-8.)
16.24 (a) Show that the functional \Pi = \int [\overline{p}_{;x}^2 +\overline{p}_{;y}^2 +\overline{p}_{;z}^2 -(\omega \overline{p} /c)^2 ]dV yields the latter form of Eq. 16.8-12 from the stationary condition \delta \Pi = 0
(b) Derive Eq. 16.8-13 (with [\mathbf{H}] = [\mathbf{0}]) from the functional stated in part (a).
16.25 (a) Specialize Eq. 16.8-7 to the one-dimensional acoustic mode problem, Fig. 16.8-1. Use this result to show that the condition \delta \Pi = 0 yields the governing differential equation d(Ap_{xx}) / dx - (A / c^2)\ddot{p} = 0 .
(b) Using Eq. 16.8-14, obtain the equation given by the condition \delta \Pi = 0 . Check your result by specializing the differential equation given in part (a).
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a 4 3 1 2 b
Problem 16.23
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1 2 3 L/2 L/2
Problem 16.27
16.26 Consider a uniform pipe of length L and having closed ends. Model the pipe by a single element. Determine the frequency of the fundamental acoustic mode by use of the following element formulations. The exact result is \omega = \pi c / L .
(a) Use an element with d.o.f. p_1 and p_2 at ends 1 and 2.
(b) Use a four d.o.f. element. The d.o.f. at each end are p and p_{,x} .
16.27 The sketch represents a three-node model of air in a uniform pipe. The end at the left is open. Determine \omega for the acoustic mode of lowest frequency. The exact result is \omega = \pi c/2L .
(a) Use a single quadratic element.
(b) Use two two-node (linear) elements.
Section 16.9
16.28 The sketch represents a rectangular tank of liquid. Beam EFGH is fixed at E and H and is in contact with the liquid surface. Imagine that motion is confined to the xz plane.
(a) If the liquid were absent, what would be the appearance of the first two vibration modes of the beam? Sketch them. Which has the higher frequency?
(b) If the liquid is restored, what now are the appearances of the first two vibration modes that involve bending of the beam? Sketch them.
(c) If beam and liquid are each divided into three elements, with displacement d.o.f. and nodes at points A–H, how many d.o.f. are nonzero after boundary conditions are imposed? Identify these d.o.f.
(d) Are beam and liquid elements compatible? If no, is this acceptable? If yes, how is it accomplished?
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z E F G H A B C D x
Problem 16.28
16.29 Imagine that a very small amount of liquid is to be analyzed, and that the effects of surface tension \sigma (force per unit length) are to be included. Determine an expression for the appropriate stiffness matrix that operates on displacement d.o.f. {d}.
16.30 For the three-element fluid model shown in Fig. 16.9-2a, sketch all possible circulation modes. Show the direction of the circulation in each case. Note: A simple reversal of the direction of all fluid displacements is not considered a different mode.
16.31 (a) Write an expression for [\mathbf{B}_c] in Eq. 16.9-6, in the form of an operator matrix times shape function matrix [N]. Assume that the problem is three-dimensional.
(b) Generate [\mathbf{k}_c] of Eq. 16.9-6 for the two-d.o.f. element described by Eq. 16.9-7.
16.32 Verify the correctness of the rectangular matrices in Eq. 16.9-8.
16.33 Use a three-node, mock-fluid model to solve for the lowest acoustic frequency in a uniform pipe with one open end (see Problem 16.27).
(a) Use a single quadratic element.
(b) Use two two-node (linear) elements.