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Section 9.5
9.28 Give a detailed explanation of why the constraint ratio r is the same for a one-element test mesh as it is for a square test mesh with N_{es} elements per side.
9.29 The sketch shows a mesh of plane constant-strain triangular elements. The material is incompressible, and plane strain conditions are enforced. All nodes i in contact with the supports are fixed ( u_{i} = v_{i} = 0 ).
(a) What is the constraint ratio? Will the mesh lock?
(b) Can the constraint ratio be increased by use of reduced integration?
(c) What is the constraint ratio if the number of elements is greatly increased?
(d) Consider elements 1 and 2, and show that the incompressibility constraint implies u_{j} = v_{j} = 0 at node j . Extend this argument to show that the mesh is completely locked.
(e) Try some other arrangements of constant-strain triangles to see whether the same conclusion holds.
(f) Revise the support conditions so that part (a) of this problem will yield the constraint ratio 1/1 .
9.30 What are the constraint ratios r (in a mesh of many elements) for the plane and solid Q6 elements of Section 8.3 if integration is done using an order 2 Gauss rule and the material is nearly incompressible?
9.31 The elements shown are (a) a cubic triangle, and (b) a quartic triangle. The stiffness matrix of the cubic triangle can be exactly integrated by the six-point formula of Table 6.8-1 (when material properties and element thickness are constant). Similarly, there is a ten-point formula that integrates the quartic element stiffness matrix exactly. In a refined mesh, for an incompressible material, what is the constraint ratio for each of these exactly integrated elements?
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y,v j ① ② x,u
Problem 9.29
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Simple geometric diagram of a triangle with internal dots (no text or labels)
(a)
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Simple geometric diagram of a triangle with internal dots (no text or symbols)
(b)
Problem 9.31
Section 9.6
9.32 (a) Show that \lambda = (\sigma_x + \sigma_y + \sigma_z) / 3 , as stated below Eq. 9.6-2.
(b) Show that \beta \alpha H in Eq. 9.4-10 can be expressed as B\epsilon_V^2 / 2 .
9.33 Consider the element shown, which is proposed as a higher-order solid element for incompressible media. It has 20 nodes on the edges and a “bubble” shape function added at \xi = \eta = \zeta = 0 . (Shape functions N_{1} through N_{20} are not needed in this problem but may be found in Ref. 2.1, p. 201.)
(a) Give an expression for the shape function N_{21}(\xi, \eta, \zeta) that assures inter-element compatibility.
(b) What is the constraint ratio when four internal pressure points (at the nodes of a tetrahedron) are used? (Note: This is an attempt to generalize
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η ξ ξ
Problem 9.33
Fig. 9.6-1 to a solid element. Little is currently known of the behavior of this element.)
SOLIDS OF REVOLUTION
Analysis methods for axially symmetric bodies are described. Loads may be with or without axial symmetry. Loads without axial symmetry are treated by superposition, using Fourier series.
10.1 INTRODUCTION
A solid of revolution is generated by revolving a plane figure about an axis, and is most easily described in cylindrical coordinates r, \theta , and z (Fig. 10.1-1). The geometry is axially symmetric, and if material properties and loads are also axially symmetric, the problem is mathematically two-dimensional. That is, if geometry, support conditions, loads \{R\} , and material property matrix [E] are all independent of \theta , and if the material either is isotropic or has \theta as a principal material direction, then static displacements and stresses are independent of \theta : circumferential displacement v is zero, material points have only u (radial) and w (axial) displacement components, and the nonzero stresses are those shown in Fig. 10.1-1a. The analysis procedure for static problems having axial symmetry is very similar to the procedure used for static problems of plane stress or plane strain. (In a vibration or buckling problem, symmetric and unsymmetric modes should be expected, even if geometry, support conditions, and elastic properties are all axially symmetric. A vibration or buckling analysis that assumes \theta independence would miss all modes that are not axially symmetric.)
If the solid is axially symmetric but the loading is not, displacements and stresses are three-dimensional rather than axially symmetric. A Fourier series method can then be used. The given loading is expressed as the sum of several component loadings, and an analysis is done for each load component. According to the principle of superposition, the original problem is solved by adding the solutions of the component problems. Each component analysis remains mathematically two-dimensional. Thus the original three-dimensional problem is exchanged for a series of two-dimensional problems. The exchange is usually worthwhile because three-dimensional problems are expensive to set up and run.
A finite element model of a solid of revolution has nodal circles rather than nodal points (Fig. 10.1-1). So does a shell of revolution, which is an effective model if the body is thin-walled (Section 12.4). If a body of revolution (having nodal circles) must be attached to a solid body (having nodal points), there is some difficulty in making the connection.
Finite element analysis for axially symmetric solids was first published in 1965 [10.1]. Computer programs are readily available [10.2]. Indeed, minor additions
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r,u z,w θ,v σz τzr σθ (a) β y' x' 3 4 z,w y' β 4 3 x' 2 1 Axis r,u Nodal circle 2 (b)
Figure 10.1-1. An axially symmetric finite element of rectangular cross section. (a) Isometric view, showing stresses produced by axially symmetric loading. (b) Cross section containing the z axis. Hatching suggests an orthotropic material whose principal directions are x' , y' , and \theta .
to a program for plane problems makes the program capable of analyzing solids of revolution as well.
10.2 ELASTICITY RELATIONS FOR AXIAL SYMMETRY
If the analysis problem is axially symmetric, then (see Fig. 10.1-1)
v = 0 \text { and } \tau_ {r \theta} = \tau_ {\theta z} = \gamma_ {r \theta} = \gamma_ {\theta z} = 0 \tag {10.2-1}
Equations 10.2-1 prevail if geometry, support conditions, and loading are all axially symmetric, \theta is a principal material direction, and \beta in Fig. 10.1-1b is independent of \theta . Thus the material may be isotropic. Or, if orthotropic, principal material axes x' and y' must not change direction with \theta , and the third principal material axis must not form a helix. Accordingly, the most general stress–strain relation allowed has the form
\left\{ \begin{array}{l} \sigma_ {r} \\ \sigma_ {\theta} \\ \sigma_ {z} \\ \tau_ {z r} \end{array} \right\} = \left[ \begin{array}{c c c c} E _ {1 1} & E _ {1 2} & E _ {1 3} & E _ {1 4} \\ \varepsilon_ {2 1} & E _ {2 2} & E _ {2 3} & E _ {2 4} \\ E _ {3 1} & \varepsilon_ {3 2} & E _ {3 3} & E _ {3 4} \\ \text {symm} _ {4 1} & \varepsilon_ {4 2} & \varepsilon_ {4 3} & E _ {4 4} \end{array} \right] \left(\left\{ \begin{array}{l} \epsilon_ {r} \\ \epsilon_ {\theta} \\ \epsilon_ {z} \\ \gamma_ {z r} \end{array} \right\} - \{\epsilon_ {0} \}\right) \tag {10.2-2}
in which \{\epsilon_{0}\} represents initial strains, and the trivial relations \tau_{r\theta}=0 and \tau_{\theta z}=0 are simply not written. If, in addition to \theta , r and z are also principal material directions ( \beta=0 in Fig. 10.1-1), then E_{14}=E_{24}=E_{34}=0 . For the special case of isotropy and thermal loading, Eq. 10.2-2 becomes
\left\{ \begin{array}{l} \sigma_ {r} \\ \sigma_ {\theta} \\ \sigma_ {z} \\ \tau_ {z r} \end{array} \right\} = \frac {(1 - \nu) E}{(1 + \nu) (1 - 2 \nu)} \left[ \begin{array}{c c c c} 1 & f & f & 0 \\ & 1 & f & 0 \\ & & 1 & 0 \\ \text {symm.} & & & \mathrm{g} \end{array} \right] \left(\left\{ \begin{array}{l} \epsilon_ {r} \\ \epsilon_ {\theta} \\ \epsilon_ {z} \\ \gamma_ {z r} \end{array} \right\} - \left\{ \begin{array}{l} \alpha T \\ \alpha T \\ \alpha T \\ 0 \end{array} \right\}\right) \tag {10.2-3a}
in which
f = \frac {\nu}{1 - \nu} \quad \text { and } \quad g = \frac {1 - 2 \nu}{2 (1 - \nu)} \tag {10.2-3b}
and \alpha = coefficient of thermal expansion and T = temperature relative to a reference temperature at which the body is free of stress. Thus E_{44} = G , the shear modulus.
The strain-displacement relations are
\epsilon_ {r} = u _ {, r} \quad \epsilon_ {\theta} = \frac {2 \pi (r + u) - 2 \pi r}{2 \pi r} = \frac {u}{r} \tag {10.2-4}
\epsilon_ {z} = w _ {, z} \quad \gamma_ {z r} = u _ {, z} + w _ {, r}
In matrix format, Eqs. 10.2-4 are
\left\{ \begin{array}{l} \epsilon_ {r} \\ \epsilon_ {\theta} \\ \epsilon_ {z} \\ \gamma_ {z r} \end{array} \right\} = [ \partial ] \left\{ \begin{array}{l} u \\ w \end{array} \right\}, \quad \text { where } \quad [ \partial ] = \left[ \begin{array}{c c} \partial / \partial r & 0 \\ 1 / r & 0 \\ 0 & \partial / \partial z \\ \partial / \partial z & \partial / \partial r \end{array} \right] \tag {10.2-5}
Or, in alternative format, the same relations are
\left\{ \begin{array}{l} \epsilon_ {r} \\ \epsilon_ {\theta} \\ \epsilon_ {z} \\ \gamma_ {z r} \end{array} \right\} = [ \mathbf {H} ] \left\{ \begin{array}{l} u _ {, r} \\ u _ {, z} \\ w _ {, r} \\ w _ {, z} \\ u \end{array} \right\}, \quad \text {where} \quad [ \mathbf {H} ] = \left[ \begin{array}{c c c c c} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 / r \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 \end{array} \right] \tag {10.2-6}
10.3 FINITE ELEMENTS FOR AXIAL SYMMETRY
One may follow the standard formulation procedure, which is contained in Eqs. 4.1-5 and 4.1-6. Consider, for example, an eight-d.o.f. element of rectangular cross section, shown in Fig. 10.3-1. Its displacement field \{u\} = [N]\{d\} is
\left\{ \begin{array}{l} u \\ w \end{array} \right\} = \left[ \begin{array}{c c c c c c c c} N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} & 0 \\ 0 & N _ {1} & 0 & N _ {2} & 0 & N _ {3} & 0 & N _ {4} \end{array} \right] \left\{ \begin{array}{l} u _ {1} \\ w _ {1} \\ u _ {2} \\ \vdots \\ w _ {4} \end{array} \right\} \tag {10.3-1}
where shape functions N_{1} through N_{4} are stated in Eq. 3.12-10, except that z replaces y and r - r_{m} replaces x, where r_{m} is the mean radius (r_{1} + r_{2} + r_{3} + r_{4})/4 . Thus
N _ {i} = \frac {[ a \pm (r - r _ {m}) ] (b \pm z)}{4 a b} \tag {10.3-2}
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Axis of revolution z y a a 4 3 b x 1 2 b r_m x = r - r_m r
Figure 10.3-1. Geometry of an element of rectangular cross section.
in which z = 0 at the element center. The element stiffness matrix is
[ \mathbf {k} ] _ {8 \times 8} = \int_ {A} \int_ {- \pi} ^ {\pi} [ \mathbf {B} ] ^ {T} [ \mathbf {E} ] [ \mathbf {B} ] r d \theta d A \tag {10.3-3}
where A = cross-sectional area of the element and dA = drdz . From Eqs. 10.2-5 and 10.3-1, [B] = [∂][N]. However, since \partial/\partial r = \partial/\partial x and r = r_m + x , we can also write
[ \mathbf {B} ] = \left[ \begin{array}{c c} \partial / \partial x & 0 \\ 1 / (r _ {m} + x) & 0 \\ 0 & \partial / \partial z \\ \partial / \partial z & \partial / \partial x \end{array} \right] [ \mathbf {N} ], \quad \text { where } \quad N _ {i} = \frac {(a \pm x) (b \pm z)}{4 a b} \tag {10.3-4}
and where x = 0 at r = r_m , the element center.
Equation 10.3-4 yields the same [B] as does [\mathbf{B}] = [\partial][\mathbf{N}] with the N_{i} given by Eq. 10.3-2. Equation 10.3-4 shows that, as compared with a plane problem, the only change in [B] is the added row that computes \epsilon_{\theta} = u / r (compare Eqs. 10.3-4 and 4.2-14). Moreover, [k] is the same size; for example, it is 8 by 8 for the foregoing four-node bilinear element, whether the problem is plane or axially symmetric.
symmetric. If the element is isoparametric, shape functions N_{i} are functions of \xi and \eta , and we must use the usual coordinate transformation to relate derivatives,
\left\{ \begin{array}{l} u, _ {r} \\ u, _ {z} \\ w, _ {r} \\ w, _ {z} \\ u \end{array} \right\} = [ \Gamma ] _ {5 \times 5} \left\{ \begin{array}{l} u, _ {\xi} \\ u, _ {\eta} \\ w, _ {\xi} \\ w, _ {\eta} \\ u \end{array} \right\}, \quad \text { where } \quad [ \Gamma ] = \left[ \begin{array}{c c c} \mathrm{J} ^ {- 1} & 0 & 0 \\ 0 & \mathrm{J} ^ {- 1} & 0 \\ 0 & 0 & 1 \end{array} \right] \tag {10.3-5}
Jacobian matrix [J] is unchanged: it is still 2 by 2 and is as described in Section 6.3. The \xi and \eta derivatives in Eq. 10.3-5 are related to nodal d.o.f. by the equation
\left[ \begin{array}{l l l l l} u _ {, \xi} & u _ {, \eta} & w _ {, \xi} & w _ {, \eta} & u \end{array} \right] ^ {T} = \left[ \begin{array}{l} \mathbf {Q} \end{array} \right] \left[ \begin{array}{l l l l} u _ {1} & w _ {1} & u _ {2} \dots w _ {4} \end{array} \right] ^ {T} \tag {10.3-6}
The first four rows of [Q] appear in Eq. 6.3-19. The fifth row is [N_{1} , 0, N_{2} , 0, N_{3} , 0, N_{4} , 0]. Shape functions N_{i} of a four-node isoparametric element are given by Eqs. 6.3-2. From Eqs. 10.2-6, 10.3-5, and 10.3-6, [B] = [H][ \Gamma ][Q]. The element stiffness matrix of an isoparametric element is
[ \mathbf {k} ] = \int_ {- 1} ^ {1} \int_ {- 1} ^ {1} \int_ {- \pi} ^ {\pi} [ \mathbf {B} ] ^ {T} [ \mathbf {E} ] [ \mathbf {B} ] r d \theta J d \xi d \eta \tag {10.3-7}
where
r = \sum N _ {i} r _ {i} \quad \text { or } \quad r = r _ {m} + \sum N _ {i} x _ {i} \tag {10.3-8}
In numerical integration, either of Eqs. 10.3-8 may be used to determine r at quadrature points.
Element nodal loads \{r_{e}\} —for example, from heating or from centrifugal force—are calculated in straightforward fashion from Eq. 4.1-6. Here dV = r \, d\theta \, dA or dV = r \, d\theta \, J \, d\xi \, d\eta , and dS = r \, d\theta \, d\ell , where d\ell is an increment of meridional length.
A uniform line load q (units N/m) on a nodal circle of radius r_{i} produces the nodal load 2\pi r_{i}q . The net static force is also 2\pi r_{i}q if q acts axially, but is zero if q acts radially. Nevertheless, radial load q produces deformations and stresses.
Remarks. The preceding formulation produces 2\pi as a multiplier of every K_{ij} and every R_{i} in the structural equations [K]\{D\} = \{R\} . This superfluous multiplier can be avoided by letting integrals for [k] and \{r_{e}\} have theta limits of from zero to one radian. With this approach, externally applied loads must also pertain to a one-radian segment.
Some coefficients in the integrands of Eqs. 10.3-3 and 10.3-7 have 1/r as a multiplier. With Gauss quadrature these terms remain finite because there are no Gauss points at r = 0 . If [\mathbf{k}] is formed explicitly, we can produce “core” elements by using a displacement field for which u = 0 for r = 0 and evaluating indeterminate forms 0/0 that appear in the formulation by the L'Hôpital rule. If numerical integration is used instead, for acceptable accuracy core elements may require more integration points in the radial direction than are used for elements distant from the axis of revolution.
During stress computation, the indeterminate form \epsilon_{\theta}=u/r=0/0 arises for points on the z axis. We can avoid this trouble by calculating \epsilon_{\theta} slightly alway from the axis or by extrapolating strains at Gauss points to the axis. Another option is to exploit the theoretical requirement that \epsilon_{r}=\epsilon_{\theta} at r=0. Thus, for stress computation at r=0, we merely replace the \epsilon_{\theta} row of [B] by the \epsilon_{r} row.
Because of axial symmetry, z-direction translation is the only possible rigid-body motion. It can be restrained by prescribing w on a single nodal circle. The radial displacement u = 0 should be prescribed at all nodes that lie on the z axis.
Valid elements for solids of revolution must pass a weak patch test (see Section 4.6). Consider, for example, elements Q6 and QM6 (Section 8.3). These elements develop a spurious radial bulge because internal d.o.f. are activated. The bulge creates spurious shear strain \gamma_{zr} except at \xi = \eta = 0 . Nevertheless, the element is valid because the bulge tends to vanish as element cross-sectional dimensions become small in comparison with the mean radius of the element. In general use,
stresses predicted by the QM6 element may be more reliable if \gamma_{zr} is evaluated only at \xi = \eta = 0 .
10.4 FOURIER SERIES
The response of an axially symmetric body to asymmetric loads can be analyzed by superposing component analyses, each of which represents the response attributable to one component of the total load. The method relies on Fourier series, which is summarized as follows without reference to bodies of revolution.
Fourier series represent functions that are periodic. A Fourier series for a dependent variable \phi = \phi(\theta) can be written
\phi = \sum_ {n = 0} ^ {\infty} p _ {n} \cos n \theta + \sum_ {n = 1} ^ {\infty} q _ {n} \sin n \theta \tag {10.4-1}
where n is an integer. The period of \phi is 2\pi , for example, from \theta = -\pi to \theta = \pi .
Sine terms are called odd or antisymmetric, as \phi(\theta) = -\phi(-\theta) . Cosine terms are called even or symmetric, as \phi(\theta) = \phi(-\theta) . Coefficients p_n and q_n are functions of n but not of \theta . The following integrals, where m and n are integers, are useful:
\int_ {- \pi} ^ {\pi} \sin m \theta \sin n \theta d \theta = \left\{ \begin{array}{l l} \pi & \text { for } m = n \neq 0 \\ 0 & \text { for } m \neq n \text { and for } m = n = 0 \end{array} \right. \tag {10.4-2a}
\int_ {- \pi} ^ {\pi} \cos m \theta \cos n \theta d \theta = \left\{ \begin{array}{l l} 2 \pi & \text { for } m = n = 0 \\ \pi & \text { for } m = n \neq 0 \\ 0 & \text { for } m \neq n \end{array} \right. \tag {10.4-2b}
\int_ {- \pi} ^ {\pi} \sin m \theta \cos n \theta d \theta = 0 \quad \text { for all } m \text { and } n \tag {10.4-2c}
Imagine that a certain periodic function \phi = \phi(\theta) is known, but not expressed as a Fourier series. An equivalent Fourier series representation of \phi requires that p_{n} and q_{n} in Eq. 10.4-1 be determined. To do so, we integrate the function, then multiply it by the single term \cos n\theta and integrate, then multiply it by the single term \sin n\theta and integrate. Equation 10.4-1 is similarly integrated, making use of Eqs. 10.4-2. Thus equations that determine p_{0}, p_{n} , and q_{n} are
\int_ {- \pi} ^ {\pi} \phi d \theta = 2 \pi p _ {0} \quad \int_ {- \pi} ^ {\pi} \phi \cos n \theta d \theta = \pi p _ {n} \quad \int_ {- \pi} ^ {\pi} \phi \sin n \theta d \theta = \pi q _ {n} \tag {10.4-3}
Integrals in Eq. 10.4-3 may be evaluated analytically, numerically, or even graphically.
Example. The square wave in Fig. 10.4-1 is to be represented by a Fourier series. Here \theta = \pi x / L , and \phi = \phi_0 can be regarded as a uniformly distributed load of intensity \phi_0 on a span of length L . In Eqs. 10.4-3 we use \phi = -\phi_0 for -L < x < 0 and \phi = +\phi_0
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φ φ₀ -L 0 L -φ₀
(a)
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φ φ₀ L x
(b)
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φ φ₀ L x
{c}
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φ φ₀ x L
{d}
Figure 10.4-1. (a) Square wave and its representation by truncated Fourier series, using terms (b) n = 1, (c) n = 1 and 3, and (d) n = 1, 3, and 5.
for 0 < x < L . Results of the respective integrations in Eq. 10.4-3 are 0, 0, and 0 ( n even) or 4\phi_0 / n ( n odd). Hence, p_0 = 0 , p_n = 0 , and q_n = 0 ( n even). For n odd,
q _ {n} = \frac {4 \phi_ {0}}{n \pi} \quad \text { and } \quad \phi = \sum_ {n = 1, 3, \dots} \frac {4 \phi_ {0}}{n \pi} \sin \frac {n \pi x}{L} \tag {10.4-4}
As suggested by Fig. 10.4-1, the square wave can be modeled arbitrarily closely by using enough series terms.
Example. Concentrated loads P produce an interesting result. In the coordinates of Fig. 10.4-1a, consider a load P downward at x = -L/2 and a load P upward at x = +L/2. Here \theta = \pi x / L , and P can be regarded as a concentrated center load on a beam that extends from x = 0 to x = L. Equations 10.4-3 yield p_{0} = 0 , p_{n} = 0 , and
q _ {n} = \frac {2 P}{L} \sin \frac {n \pi}{2} \quad \phi = \sum_ {n = 1, 2, 3, \dots} \left(\frac {2 P}{L} \sin \frac {n \pi}{2}\right) \sin \frac {n \pi x}{L} \tag {10.4-5}
This series for \phi does not converge. However, when used to load the aforementioned beam, convergent results are obtained for displacement and stress.
A Simple Application in Stress Analysis. The beam problem depicted in Fig. 10.4-2 illustrates features of series analysis that also appear in series analysis of solids (and shells) of revolution. For a beam, the equilibrium and moment-curvature relations are
V _ {, x} = \phi \quad M _ {, x} = V \quad E I w _ {, x x} = M \tag {10.4-6}
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z, w x L
{a}
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φ qₙ qₙ L x
{b}
Figure 10.4-2. (a) Simply supported beam. (b) The sine wave loading \phi = q_{n} \sin(n\pi x/L) , where q_{n} is the amplitude of \phi . The case n = 5 is depicted.
where V is transverse shear force, M is bending moment, and \phi is distributed load. When combined, Eqs. 10.4-6 yield, for constant bending stiffness EI,
E I w _ {, x x x x} = \phi \tag {10.4-7}
Consider the loading of Fig. 10.4-2b, which corresponds to one term of the second summation in the Fourier series of Eq. 10.4-1:
\phi = q _ {n} \sin \frac {n \pi x}{L} \tag {10.4-8}
Here q_{n} does not depend on x. Assume that the displacement is the admissible function
w = w _ {n} \sin \frac {n \pi x}{L} \tag {10.4-9}
where w_{n} does not depend on x . We substitute w and \phi into Eq. 10.4-7 and obtain
\left[ E I \left(\frac {n \pi}{L}\right) ^ {4} w _ {n} - q _ {n} \right] \sin \frac {n \pi x}{L} = 0 \tag {10.4-10}
This can be true for all x only if the bracketed expression vanishes. Hence
w _ {n} = \frac {q _ {n}}{E I} \left(\frac {L}{n \pi}\right) ^ {4} \tag {10.4-11}
Substitution of Eq. 10.4-11 into Eq. 10.4-9 defines the correct and unique solution for w, since it satisfies all requirements of equilibrium, compatibility, and boundary conditions (Section 1.6). Note that a sine wave of loading produces corresponding sine waves of deflection and bending moment, regardless of n. That is, the various harmonics are uncoupled—the nth wave does not interact with the mth wave.
Now that w_{n} is known, Eq. 10.4-9 defines w for any x and for any n. If two or more sine wave loadings act simultaneously, each associated with a different n, the net deflection and the net bending moment are determined by superposition; that is,
w = \sum_ {n} q _ {n} \frac {L ^ {4}}{E I n ^ {4} \pi^ {4}} \sin \frac {n \pi x}{L} \quad \text { and } \quad M = - \sum_ {n} q _ {n} \frac {L ^ {2}}{n ^ {2} \pi^ {2}} \sin \frac {n \pi x}{L} \tag {10.4-12}
where M = EIw_{,xx} . A particular loading requires a particular q_{n} . For example, to analyze the effect of a uniformly distributed loading of intensity \phi_{0} , we substitute q_{n} = 4\phi_{0}/n\pi from Eq. 10.4-4 into Eqs. 10.4-12. For a concentrated load at midspan, we substitute q_{n} from Eq. 10.4-5.
What we have done is find the response of the beam to a single load component from a single equation (Eq. 10.4-11) that says nothing about how w varies with x . We pay for this simplicity by having to solve the equation several times, once for each Fourier component of loading.
In dealing with bodies of revolution, we replace Eq. 10.4-11 by a set of simultaneous algebraic equations, which must be solved for each Fourier component of loading.