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김경종
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<!-- source-page: 91 -->
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<details>
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<summary>text_image</summary>
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z,w
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P
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x
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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z,w
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A
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B
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P
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x
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</details>
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(b)
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Figure 3.2-2. (a) A cantilever beam. (b) An inadmissible configuration (upper dashed line) and two admissible configurations (lower dashed lines).
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at $x = 0$ . Nonessential boundary conditions are that $w_{,xx} = 0$ and $w_{,xxx} = 0$ at $x = L$ , since bending moment $M = EIw_{,xx}$ and transverse shear force $V = EIw_{,xxx}$ are both zero at $x = L$ .
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An admissible configuration is any configuration that satisfies internal compatibility and essential boundary conditions. Examples appear in Fig. 3.2-2b. The uppermost curve, which is inadmissible, has four faults: it violates the two essential boundary conditions w = 0 and $w_{,x} = 0$ at x = 0, and it violates compatibility because of the jump at A and the cusp at B. The lower two curves are both admissible, even though only the lower one seems physically reasonable. An admissible configuration need not satisfy nonessential boundary conditions. Thus, at x = L, neither of the two lower curves need display $w_{,xx} = 0$ or $w_{,xxx} = 0$ .
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A conservative mechanical system has a potential energy. That is, one can express the energy content of the system in terms of its configuration, without reference to whatever deformation history or path may have led to that configuration [3.1]. Potential energy, also called total potential energy, includes (a) the strain energy of elastic distortion, and (b) the potential possessed by applied loads, by virtue of their having the capacity to do work if displaced through a distance. The principle of stationary potential energy states that
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Among all admissible configurations of a conservative system, those that satisfy the equations of equilibrium make the potential energy stationary with respect to small admissible variations of displacement. $\mathrm{d}T = 0$ (dy) This principle is applicable whether or not the load versus deformation relation is linear. If the stationary condition is a relative minimum, the equilibrium state is stable. Note that loads are kept constant while displacements are varied.
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Example. Linear Spring with Axial Load. A very simple system is shown in Fig. 3.2-3. Its (total) potential energy $\Pi_{p}$ has two parts;
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<details>
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<summary>text_image</summary>
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k
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P
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x
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L
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</details>
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(a)
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<details>
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<summary>text_image</summary>
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k
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D
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P
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L + D
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x
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</details>
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(b)
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Figure 3.2-3. (a) Unstretched (reference) configuration of a linear spring of stiffness k. (b) Configuration after force P is applied, stretching the spring D units.
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<!-- source-page: 92 -->
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$$
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\Pi_ {p} = U + \Omega \tag {3.2-1}
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$$
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where U is the strain energy of the system, which in the present example is given by
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$$
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U = \frac {1}{2} k D ^ {2} \tag {3.2-2}
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$$
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The potential of loads, called $\Omega$ , is here given by
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$$
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\Omega = - P D \tag {3.2-3}
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$$
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The load is regarded as always acting at its full value P. In moving through displacement D it does work in the amount PD, thereby losing potential of equal amount; hence the negative sign in the expression $\Omega = -PD$ . The potential energy
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$$
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\left[ \Pi_ {p} = \frac {1}{2} k D ^ {2} - P D\right) \tag {3.2-4}
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$$
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can be regarded as the total internal and external work done in changing the configuration from the reference state $D = 0$ to the displaced state $D \neq 0$ . Note that if $P$ were directed toward the left, while $D$ remains positive toward the right, then $\Omega$ would become $+PD$ . This is in essence the same as increasing potential energy by increasing the elevation of a weight.
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If only displacements along the $x$ axis are allowed, then the single d.o.f. $D$ defines all admissible configurations. The equilibrium configuration $D_{\mathrm{eq}}$ is found from the stationary value of $\Pi_p$ :
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$$
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\left\{ \begin{array}{l} d \Pi_ {p} = (k D _ {\mathrm{eq}} - P) d D = 0, \quad \text { hence } \quad D _ {\mathrm{eq}} = \frac {P}{k} \end{array} \right. \tag {3.2-5}
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$$
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The equation $(kD_{\mathrm{eq}} - P)dD = 0$ is an instance of the virtual work principle: zero network is done by all forces during a small admissible displacement $dD$ from the equilibrium configuration. This is graphically apparent in Fig. 3.2-4. We see also that $\Pi_p$ is a relative minimum, which means that the equilibrium state is stable.
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The reference datum for $\Omega$ can be arbitrarily changed by a constant. For example, if we say that $\Omega$ is zero at the equilibrium configuration, then $\Omega = P(D_{\mathrm{eq}} - D)$ . The added constant $PD_{\mathrm{eq}}$ disappears in the process of writing $d\Pi_p = 0$ , and the same value of $D_{\mathrm{eq}}$ is again obtained.
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<details>
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<summary>text_image</summary>
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Potential
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U=½kD²
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Πp=U+Ω
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D
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D_eq
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Ω=-PD
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</details>
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Figure 3.2-4. Graphical interpretation of the potential energy relations for the problem of Fig. 3.2-3.
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In Eq. 3.2-2 we write $\Omega = -PD$ rather than $\Omega = -PD/2$ . This is because P is regarded as always acting at full intensity. True, we could bypass the stationary potential energy principle and imagine that D is produced by a gradually increasing load whose final value is P. Thus, equating work done against a linear spring to strain energy stored, we have
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$$
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\frac {1}{2} P D _ {\mathrm{eq}} = \frac {1}{2} k D _ {\mathrm{eq}} ^ {2} \quad \text { from which } \quad D _ {\mathrm{eq}} = \frac {P}{k} \tag {3.2-6}
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$$
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This energy balance argument is valid but rarely helpful. It yields but one equation, even if there are a great many d.o.f. that must be determined.
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# 3.3 PROBLEMS HAVING MANY D.O.F.
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A finite element analysis typically uses hundreds of d.o.f. They may be the x and y displacements of nodes (as in a plane stress problem), or lateral displacement w and its first derivative $w_{xx}$ at nodes (as in a beam problem), and so on. Let n be the number of d.o.f. that must be calculated, and let them be collected in the structure displacement vector $\{D\} = \left[D_{1}, D_{2}, \ldots, D_{n}\right]^{T}$ . We assume here that support conditions are already imposed, so that arbitrary values of the $D_{i}$ always create admissible configurations.
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Potential $\Pi_p$ is a function of the $D_i$ . Symbolically, $\Pi_p = \Pi_p (D_1, D_2, \ldots, D_n)$ . Applying the principle of stationary potential energy, we write
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$$
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d \Pi_ {p} = \frac {\partial \Pi_ {p}}{\partial D _ {1}} d D _ {1} + \frac {\partial \Pi_ {p}}{\partial D _ {2}} d D _ {2} + \dots + \frac {\partial \Pi_ {p}}{\partial D _ {n}} d D _ {n} = 0 \tag {3.3-1}
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$$
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$$
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\text { S I A ( } \quad \text { d } \text { 1 } \quad \text { d } \text { 1 } \quad \text { d } \text { 1 } \quad \text { d } \text { 1 } \quad \text { d } \text { 1 } \quad \text { d } \text { 1 } \quad \text { d } \text { p } \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 } \quad \text { 2 }
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$$
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The stationary principle states that equilibrium prevails when $d\Pi_{p} = 0$ for any small admissible variation of the configuration. We can imagine that only $dD_{1}$ is nonzero, or that only $dD_{2}$ and $dD_{3}$ are nonzero, and so on. For any and all such choices, $d\Pi_{p}$ must vanish. This is possible only if coefficients of the $dD_{i}$ vanish separately. Thus, for $i = 1, 2, 3, \ldots, n$ ,
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$$
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\boxed {\frac {\partial \Pi_ {p}}{\partial D _ {i}} = 0} \text { or, in alternative notation, } \left\{\frac {\partial \Pi_ {p}}{\partial \mathbf {D}} \right\} = \{\mathbf {0} \} \tag {3.3-2}
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$$
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These are n equations to be solved for the n values of d.o.f. $D_{i}$ that define the static equilibrium configuration.
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Example. Springs in Series. The structure shown in Fig. 3.3-1 has the potential
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$$
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\Pi_ {p} = \frac {1}{2} k _ {1} D _ {1} ^ {2} + \frac {1}{2} k _ {2} \left(D _ {2} - D _ {1}\right) ^ {2} + \frac {1}{2} k _ {3} \left(D _ {3} - D _ {2}\right) ^ {2} - P _ {1} D _ {1} - P _ {2} D _ {2} - P _ {3} D _ {3} \tag {3.3-3}
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$$
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Equations 3.3-2 and 3.3-3 yield, for $i = 1, 2, 3$ ,
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$$
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k _ {1} D _ {1} - k _ {2} (D _ {2} - D _ {1}) - P _ {1} = 0
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$$
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$$
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k _ {2} (D _ {2} - D _ {1}) - k _ {3} (D _ {3} - D _ {2}) - P _ {2} = 0 \tag {3.3-4}
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$$
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$$
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k _ {3} (D _ {3} - D _ {2}) - P _ {3} = 0
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$$
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<!-- source-page: 94 -->
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<details>
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<summary>text_image</summary>
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k₁
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D₁
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k₂
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D₂
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k₃
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D₃
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P₁
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P₂
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P₃
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</details>
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Figure 3.3-1. A three d.o.f. system of three linear springs and three axial loads $P_{1}, P_{2}$ , and $P_{3}$ . D.o.f. $D_{i}$ are axial displacements relative to a fixed point, such as the left support. The springs are unstretched when $D_{1} = D_{2} = D_{3} = 0$ .
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In the matrix form $[\mathbf{K}]\{\mathbf{D}\} = \{\mathbf{R}\}$ , Eqs. 3.3-4 are
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$$
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\left[ \begin{array}{c c c} k _ {1} + k _ {2} & - k _ {2} & 0 \\ - k _ {2} & k _ {2} + k _ {3} & - k _ {3} \\ 0 & - k _ {3} & k _ {3} \end{array} \right] \left\{ \begin{array}{l} D _ {1} \\ D _ {2} \\ D _ {3} \end{array} \right\} = \left\{ \begin{array}{l} P _ {1} \\ P _ {2} \\ P _ {3} \end{array} \right\} \tag {3.3-5}
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$$
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The correctness of stiffness matrix [K] in Eq. 3.3-5 can be checked by the procedure of activating one d.o.f. at a time, as described in Section 2.2.
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Example. Plane Truss Problems. Consider the plane truss element of Figs. 2.4-1 and 2.4-2. Its potential expression is
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$$
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\Pi_ {p} = \frac {1}{2} k e ^ {2} - p _ {i} u _ {i} - q _ {i} v _ {i} - p _ {j} u _ {j} - q _ {j} v _ {j} \tag {3.3-6}
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$$
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where $k = AE / L$ and elongation $e$ is given by
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$$
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e = (u _ {j} - u _ {i}) \cos \beta + (v _ {j} - v _ {i}) \sin \beta \tag {3.3-7}
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$$
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Setting to zero the four derivatives of $\Pi_p$ with respect to d.o.f. $u_i, v_i, u_j$ , and $v_j$ , we obtain the four rows of Eq. 2.4-3.
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The same procedure can be applied to an entire structure. Consider the three-bar truss of Fig. 2.2-1, with support conditions $u_{2} = v_{2} = u_{3} = 0$ already imposed. Its potential energy is
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$$
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\Pi_ {p} = \frac {1}{2} k _ {3} u _ {1} ^ {2} + \frac {1}{2} k _ {1} v _ {3} ^ {2} + \frac {1}{2} k _ {2} \left[ (0 - u _ {1}) (- 0. 6) + (v _ {3} - v _ {1}) (0. 8) \right] ^ {2} + P v _ {1} \tag {3.3-8}
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$$
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The $Pv_{1}$ term bears a positive sign because the potential of the (downward) load P is increased by a positive (upward) displacement $v_{1}$ . The derivatives of $\Pi_{p}$ with respect to $u_{1}$ , $v_{1}$ and $v_{3}$ , when equated to zero, are found to yield Eqs. 2.3-8.
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From the foregoing examples we draw the following conclusions, which are true in general.
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A system that has linear load versus displacement characteristics has a symmetric stiffness matrix; that is, $K_{ij} = K_{ji}$ . This happens because each symmetrically located pair of off-diagonal coefficients comes from a single term in $\Pi_p$ whose form is a constant times $D_i D_j$ . Thus $K_{ij} = \partial^2 \Pi_p / \partial D_i \partial D_j = \overline{\partial^2 \Pi_p / \partial D_j} \partial D_i = K_{ji}$ .
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2. If $D_{i}$ is a nodal displacement (or rotation), the equation $\sqrt{\partial\Pi_{p}/\partial D_{i}} = 0$ is a nodal equilibrium equation stating that forces (or moments) applied to the
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<!-- source-page: 95 -->
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node sum to zero in the direction of $D_{i}$ . Included in the sum are (a) loads applied externally, and (b) loads applied internally, owing to deformation of structural components (and perhaps also to thermal load, body force, etc.).
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3. Static indeterminacy does not alter the procedure or make a problem more difficult. For example, in Fig. 3.3-1 we could connect a fourth spring between the fixed support and node 3. Then $\Pi_p$ is augmented by $k_4D_3^2/2$ , and the last stiffness coefficient in Eq. 3.3-5 is changed from $k_3$ to $k_3 + k_4$ , but the same three d.o.f. still suffice.
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√ 4. The potential energy of a structure can be written in the form
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$$
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\left| \Pi_ {p} = U + \Omega , \right. \text { where } \left. U = \frac {1}{2} \{\mathbf {D} \} ^ {T} [ \mathbf {K} ] \{\mathbf {D} \} \right| \text { and } \left. \Omega = - \{\mathbf {D} \} ^ {T} \{\mathbf {R} \} \right. \tag {3.3-9}
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$$
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If $U = 0$ , then either $\{\mathbf{D}\} = \{\mathbf{0}\}$ or $\{\mathbf{D}\}$ expresses a rigid-body motion. If the structure is stable and is supported so that rigid-body motion is not possible (as in Eqs. 3.3-3 and 3.3-5), then $\frac{1}{2}\{\mathbf{D}\}^T [\mathbf{K}]\{\mathbf{D}\} > 0$ for any nonzero $\{\mathbf{D}\}$ ; that is, $[\mathbf{K}]$ is said to be positive definite.
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# 3.4 POTENTIAL ENERGY OF AN ELASTIC BODY
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The potential energy of an elastic body consists of the strain energy contained in elastic distortions and the potential of loads that act within the body or on its surface. The potential energy expression can be used to formulate element stiffness matrices and element load vectors. Simple finite element formulations appear later in this chapter. Additional formulations appear in subsequent chapters.
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In this section we present formulas, argue their validity, show that special cases yield correct results, and consider examples. Derivations and detailed arguments may be found elsewhere [3.1,3.2].
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Consider a linearly elastic body that carries conservative loads. Let its volume be V and its surface area be S. The expression for its potential energy is
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$$
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\begin{array}{l} \Pi_ {p} = \int_ {V} \left(\frac {1}{2} \{\epsilon \} ^ {T} [ \mathbf {E} ] \{\epsilon \} - \{\epsilon \} ^ {T} [ \mathbf {E} ] \{\epsilon_ {0} \} + \{\epsilon \} ^ {T} \{\sigma_ {0} \}\right) d V \\ - \int_ {V} \{\mathbf {u} \} ^ {T} \{\mathbf {F} \} d V - \int_ {S} \{\mathbf {u} \} ^ {T} \{\boldsymbol {\Phi} \} d S - \{\mathbf {D} \} ^ {T} \{\mathbf {P} \} \tag {3.4-1} \\ \end{array}
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$$
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The notation is explained in Section 1.6 and in the following.
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Explanation and Justification. In the first integral of Eq. 3.4-1, the expression within parentheses represents $U_{0}$ , the strain energy per unit volume. The expression is derived as follows. Consider a unit cube (i.e., a cube of unit length along each edge). Stresses that act on faces of the cube do work during infinitesimal straining of the cube. This work is stored as an increment of strain energy $dU_{0}$ . On a unit cube, stress and force have the same magnitude on each face, and strain and elongation have the same magnitude along each edge. Therefore, since work is equal to force times displacement,
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$$
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\left| d U _ {0} = \sigma_ {x} d \epsilon_ {x} + \sigma_ {y} d \epsilon_ {y} + \sigma_ {z} d \epsilon_ {z} + \tau_ {x y} d \gamma_ {x y} + \tau_ {y z} d \gamma_ {y z} + \tau_ {z x} d \gamma_ {z x} \right. \tag {3.4-2}
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$$
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<!-- source-page: 96 -->
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Changes in stress produced by infinitesimal strain increments have been discarded from $dU_0$ because they produce terms of higher order. For example, $(\sigma_x + d\sigma_x)$ $d\epsilon_x \approx \sigma_x d\epsilon_x$ . From Eq. 3.4-2 we conclude that $\partial U_0 / \partial \epsilon_x = \sigma_x$ , $\overline{\partial U_0 / \partial \epsilon_y} = \overline{\sigma_y, \ldots, \partial U_0 / \partial \gamma_{zx}} = \tau_{zx}$ . Expressing these six derivatives in matrix format and using the stress-strain relation (Eq. 1.7-8), we obtain
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$$
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\left\{\frac {\partial U _ {0}}{\partial \epsilon} \right\} = \{\boldsymbol {\sigma} \} \quad \text { or } \quad \left\{\frac {\partial U _ {0}}{\partial \epsilon} \right\} = [ \mathrm{E} ] \{\boldsymbol {\epsilon} \} - [ \mathrm{E} ] \{\boldsymbol {\epsilon} _ {0} \} + \{\boldsymbol {\sigma} _ {0} \} \tag {3.4-3}
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$$
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Integration of the latter equation with respect to the strains yields the parenthetic expression in Eq. 3.4-1. That the integration is correct may be shown by applying differentiation rules given in Appendix A. A constant of integration has been discarded. It is superfluous because it disappears during the differentiation process that makes $\Pi_{n}$ stationary.
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Integrals in Eq. 3.4-1 that contain body forces $\{F\}$ and surface tractions $\{\Phi\}$ represent work done (hence potential lost) by $\{F\}$ and $\{\Phi\}$ as the body is deformed. Displacements in the x, y, and z coordinate directions are
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$$
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\{\mathbf {u} \} ^ {T} = \left[ \begin{array}{l l l} u & v & w \end{array} \right] \tag {3.4-4}
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$$
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Thus potential changes associated with $\{F\}$ and $\{\Phi\}$ , per unit volume and per unit area, respectively, are
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$$
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- F _ {x} u - F _ {y} v - F _ {z} w \quad \text { and } \quad - \Phi_ {x} u - \Phi_ {y} v - \Phi_ {z} w \tag {3.4-5}
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$$
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These are the products $-\{\mathbf{u}\}^T\{\mathbf{F}\}$ and $-\{\mathbf{u}\}^T\{\Phi\}$ . It is assumed that positive senses correspond—for example, that $F_x$ , $\Phi_x$ , and $u$ are all considered positive when acting in the $+x$ direction. Integrals that contain $\{\mathbf{F}\}$ and $\{\Phi\}$ are evaluated only over the portions of $V$ and $S$ where $\{\mathbf{F}\}$ and $\{\Phi\}$ are prescribed.
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The final term in Eq. 3.4-1, $-\{\mathbf{D}\}^{T}\{\mathbf{P}\} = -D_{1}P_{1} - D_{2}P_{2} - \cdots - D_{n}P_{n}$ , accounts for work done (hence potential lost) by concentrated forces and/or moments applied to the body. The potential of these loads could be included in the surface integral by imagining large tractions to be applied over small areas, but is easier to regard the potential of such loads as $-\{\mathbf{D}\}^{T}\{\mathbf{P}\}$ . As usual, the same sense is considered positive for a displacement or rotation $D_{i}$ and for its corresponding force or moment $P_{i}$ .
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In a typical problem, many of the load terms $\{\epsilon_0\}$ , $\{\sigma_0\}$ , $\{\mathbf{F}\}$ , $\{\Phi\}$ , and $\{\mathbf{P}\}$ are zero. For example, if heating is localized $\{\epsilon_0\}$ is zero over most of the body. If gravity or acceleration loads are considered unimportant, $\{\mathbf{F}\} = \{\mathbf{0}\}$ . Surface tractions $\{\Phi\}$ usually act on only a portion of surface $S$ and may be absent altogether. Indeed, all these load terms might be zero if nonzero values of one or more d.o.f. are prescribed instead.
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Equation 3.4-1 is not restricted to rectangular coordinates. It requires only that x, y, and z refer to three mutually perpendicular directions at each material point.
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Particular Cases. In its most general form, Eq. 3.4-1 includes all six strains in $\{\epsilon\}$ . Material property matrix [E] is 6 by 6. Its coefficients are stated in Eq. 1.7-3 for the case of isotropy.
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If the problem is one of plane stress or plane strain, then $\{\epsilon\}^T = \left[\epsilon_x \quad \epsilon_y \quad \gamma_{xy}\right]$ and [E] is a 3 by 3 array whose coefficients are given by Eq. 1.7-4 or Eq. 1.7-5 for the case of isotropy.
|
||||
|
||||
<!-- source-page: 97 -->
|
||||
|
||||
The simplest special case is that of uniaxial stress. For this particular case, and perhaps for others as well, it may be easiest to derive the strain energy expression afresh, as follows, rather than specialize Eq. 3.4-1. Equations 3.4-2 and 3.4-3 become $dU_{0} = \sigma_{x} d\epsilon_{x}$ and $dU_{0}/d\epsilon_{x} = \sigma_{x} = E\epsilon_{x} - E\epsilon_{x0} + \sigma_{x0}$ . After integration and inclusion of the load terms, we obtain in place of Eq. 3.4-1
|
||||
|
||||
$$
|
||||
\Pi_ {p} = \int_ {0} ^ {L} \left(\frac {1}{2} E \epsilon_ {x} ^ {2} - \epsilon_ {x} E \epsilon_ {0} + \epsilon_ {x} \sigma_ {0}\right) A d x - \int_ {0} ^ {L} u F _ {x} A d x - \{\mathbf {D} \} ^ {T} \{\mathbf {P} \} \tag {3.4-6}
|
||||
$$
|
||||
|
||||
where $E =$ elastic modulus, $dV = A dx$ , $A =$ cross-sectional area, and $L =$ length. In the second integral, the integrand could be regarded as $uF_x dV$ or as $uq dx$ , where $q = F_x A$ : axial body force $F_x$ and axial line load $q$ have the same effect when the body is mathematically one-dimensional.
|
||||
|
||||
In beam bending, Fig. 3.4-1, each z = constant layer of the beam is regarded as being in a state of uniaxial stress if, as is common, transverse shear deformation is neglected. Accordingly, the expression for strain energy in a beam can be derived from Eq. 3.4-6. Let b represent the width of the beam. Since $\epsilon_{x} = u_{,x}$ and $u = -zw_{,x}$ , the first term in Eq. 3.4-6 yields
|
||||
|
||||
$$
|
||||
\int \frac {1}{2} E \epsilon_ {x} ^ {2} d V = \iint \frac {1}{2} E (- z w _ {, x x}) ^ {2} b d z d x = \int \frac {1}{2} E I w _ {, x x} ^ {2} d x \tag {3.4-7}
|
||||
$$
|
||||
|
||||
where, if the cross section is rectangular, $I = bt^{3}/12$ is the moment of inertia of cross-sectional area A. If terms analogous to $\epsilon_{0}$ and $\sigma_{0}$ are omitted (but which the reader may add as an exercise), the expression for potential of a straight beam without transverse shear deformation is
|
||||
|
||||
$$
|
||||
\Pi_ {p} = \int_ {0} ^ {L} \frac {1}{2} E I w _ {, x x} ^ {2} d x - \int_ {0} ^ {L} w q d x - \{\mathbf {w} \} ^ {T} \{\mathbf {F} \} - \{\boldsymbol {\theta} \} ^ {T} \{\mathbf {M} \} \tag {3.4-8}
|
||||
$$
|
||||
|
||||
where $\{\mathbf{w}\}^T = \lfloor w_1 \quad w_2 \quad \ldots \rfloor$ and $\{\theta\}^T = \lfloor \theta_1 \quad \theta_2 \quad \ldots \rfloor$ are lateral deflections and rotations $(\theta = w_{,x})$ at locations where lateral forces $\{\mathbf{F}\}$ and moments $\{\mathbf{M}\}$ are applied as loads. The second integral in Eq. 3.4-8 accounts for work done by lateral force increments $q dx$ during lateral displacement $w$ . Axial stress in the beam is $\sigma_x = E\epsilon_x = Eu_x = -Ezw_{,xx}$ .
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
z,w
|
||||
F₁ F₂ q
|
||||
M₁ M₂
|
||||
x,u
|
||||
L
|
||||
(a)
|
||||
</details>
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
z,w
|
||||
w_x
|
||||
w_x
|
||||
w
|
||||
x,u
|
||||
u = -zw_x
|
||||
t/2
|
||||
z
|
||||
dx
|
||||
t/2
|
||||
(b)
|
||||
</details>
|
||||
|
||||
Figure 3.4-1. (a) A beam loaded by lateral forces $F_{i}$ , moments $M_{i}$ , and distributed lateral load q (force per unit length). (b) A slice cut from the beam, shown after it has undergone lateral deflection w and small rotation $w_{xx}$ . Transverse shear deformation is ignored.
|
||||
|
||||
<!-- source-page: 98 -->
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
y
|
||||
D
|
||||
P
|
||||
x,u
|
||||
L
|
||||
</details>
|
||||
|
||||
Figure 3.4-2. Uniform bar under axial load P.
|
||||
|
||||
Plates in bending are analogous to beams in that strain energy is conveniently expressed in terms of curvatures rather than strains. Specifically, in a thin plate having lateral deflection w, strain energy can be expressed in terms of $w_{,xx}, w_{,xy}$ , and $w_{,yy}$ instead of $\epsilon_{x}, \epsilon_{y}$ , and $\gamma_{xy}$ . Potential expressions for problems of plates, shells, and other special cases are introduced in subsequent chapters where they are used.
|
||||
|
||||
Example. Bar Under Axial Load. The correctness of Eq. 3.4-6 can be established by solving simple test problems. For example, let the uniform bar of Fig. 3.4-2 carry an end load P and be uniformly heated T degrees. D designates the end displacement produced by P and T. With $\epsilon_{0} = \alpha T$ , $\sigma_{0} = 0$ , and $\epsilon_{x} = D/L$ , Eq. 3.4-6 becomes
|
||||
|
||||
$$
|
||||
\Pi_ {p} = \int_ {0} ^ {L} \left(\frac {1}{2} E \frac {D ^ {2}}{L ^ {2}} - \frac {D}{L} E \alpha T\right) A d x - D P = \frac {E A D ^ {2}}{2 L} - D E A \alpha T - D P \tag {3.4-9}
|
||||
$$
|
||||
|
||||
End displacement $D$ is found from the equation $d\Pi_{\rho} / dD = 0$ :
|
||||
|
||||
$$
|
||||
D = \frac {P L}{A E} + \alpha T L \tag {3.4-10}
|
||||
$$
|
||||
|
||||
Finally, the axial stress $\sigma_{x}$ is, with $\epsilon_{x} = D / L$ ,
|
||||
|
||||
$$
|
||||
\sigma_ {x} = E \epsilon_ {x} - E \epsilon_ {0} = E \left(\frac {P}{A E} + \alpha T\right) - E \alpha T = \frac {P}{A} \tag {3.4-11}
|
||||
$$
|
||||
|
||||
which is the result expected.
|
||||
|
||||
# 3.5 THE RAYLEIGH-RITZ METHOD
|
||||
|
||||
A structure composed of discrete members, such as a truss or a frame, can be represented exactly by a finite number of d.o.f.; these d.o.f. are motions of the joints. A continuum, such as an elastic solid, has infinitely many d.o.f.; these d.o.f. are the displacements of every material point. The behavior of a continuum is described by partial differential equations. For all but the simplest problems there is little hope of discovering a stress field or a displacement field that solves the differential equations and satisfies boundary conditions. The need to solve differential equations can be avoided by applying the Rayleigh–Ritz method to a functional such as $\Pi_{p}$ that describes the problem. The result is a substitute problem that has a finite number of d.o.f. and is described by algebraic equations rather than by differential equations. A Rayleigh–Ritz solution is rarely exact but becomes more accurate as more d.o.f. are used.
|
||||
|
||||
<!-- source-page: 99 -->
|
||||
|
||||
The Rayleigh–Ritz method began in 1870 with studies of vibration problems by Lord Rayleigh. He used an approximating field that contained a single d.o.f. In 1909, Ritz generalized the method by building an approximating field from several functions, each satisfying essential (i.e., kinematic) boundary conditions, and each associated with a separate d.o.f. Ritz applied the method to equilibrium problems and to eigenvalue problems. The procedure for an equilibrium (static) problem is as follows.
|
||||
|
||||
Consider an elastic solid. Displacements and stresses produced by applied loads are required. The displacement of a point is described by the displacement components $u, v,$ and $w$ . A Rayleigh-Ritz solution begins with approximating fields for $u, v,$ and $w$ . Each field is a series, whose typical term is a function of the coordinates, $f_{i} = f_{i}(x,y,z)$ , times an amplitude $a_{i}$ whose value is yet to be determined. The $a_{i}$ may be called generalized coordinates. We write
|
||||
|
||||
$$
|
||||
u = \sum_ {i = 1} ^ {\ell} a _ {i} f _ {i} \quad v = \sum_ {i = \ell + 1} ^ {m} a _ {i} f _ {i} \quad w = \sum_ {i = m + 1} ^ {n} a _ {i} f _ {i} \tag {3.5-1}
|
||||
$$
|
||||
|
||||
Each of the functions $f_{i} = f_{i}(x,y,z)$ must be admissible; that is, each must satisfy compatibility conditions and essential boundary conditions. It is not required that any of the $f_{i}$ satisfy nonessential boundary conditions (but doing so yields a more accurate approximation for a given number of d.o.f.). Usually, but not necessarily, the $f_{i}$ are polynomials. The analyst must estimate how many terms are needed in each series in order to achieve the accuracy required. Thus the series are truncated rather than infinite, having, respectively, $\ell, m - \ell$ , and $n - m$ terms, for a total of $n$ terms.
|
||||
|
||||
The d.o.f. of the problem are the $n$ amplitudes $a_{i}$ . They are determined as follows. Substitute Eqs. 3.5-1 into the strain-displacement relations (Eqs. 1.5-6) to find strains $\{\epsilon\}$ , then use Eq. 3.4-1 to evaluate $\Pi_{p}$ . Thus $\Pi_{p}$ becomes a function of d.o.f. $a_{i}$ , just as $\Pi_{p}$ is a function of d.o.f. $D_{i}$ in Eq. 3.3-1. According to the principle of stationary potential energy, the equilibrium configuration is defined by the $n$ algebraic equations
|
||||
|
||||
$$
|
||||
\frac {\partial \Pi_ {p}}{\partial a _ {i}} = 0 \quad \text { for } \quad i = 1, 2, \dots , n \tag {3.5-2}
|
||||
$$
|
||||
|
||||
After Eqs. 3.5-2 are solved for numerical values of the $a_{i}$ , the displacement fields of Eqs. 3.5-1 are completely defined. Differentiation of the displacement fields yields strains, which enter the stress-strain relations to produce stresses.
|
||||
|
||||
The foregoing procedure has two principal steps. First, establish a trial family of admissible solutions. Second, apply a criterion to select the best form of the family. Here the criterion is that $\Pi_{p}$ be stationary. Alternative criteria are available, such as methods of weighted residuals (Chapter 15).
|
||||
|
||||
Equations 3.5-1 create a substitute problem because the infinitely many d.o.f. of the real structure are replaced by the finite number of d.o.f. in the mathematical model. A Rayleigh–Ritz solution is usually approximate because the functions $f_{i}$ are usually incapable of exactly representing the actual displacements. The solution process selects amplitudes $a_{i}$ so as to combine the functions $f_{i}$ to best advantage. When $\Pi_{p}$ is the functional, “best” means tending to satisfy differential equations of equilibrium and stress boundary conditions more and more closely as more and more terms $a_{i}f_{i}$ are added to the series.
|
||||
|
||||
<!-- source-page: 100 -->
|
||||
|
||||
Equations 3.5-2 are found to be stiffness equations. They can be written in the usual form $[K]\{D\} = \{R\}$ , where $\{D\} = \left[a_{1} - a_{2} - \ldots - a_{n}\right]^{T}$ . Not all $D_{i}$ have units of displacement and not all $R_{i}$ have units of force, but each product $R_{i}D_{i}$ has units of work or energy.
|
||||
|
||||
Example. Bar Under Axial Load. Consider the uniform bar of Fig. 3.5-1a. The load is distributed along the length of the bar in linearly varying fashion: $q = cx$ , where $c$ is a constant that has units of force divided by the square of length. Axial displacement $u$ and axial stress $\sigma_x$ are to be computed by the Rayleigh-Ritz method.
|
||||
|
||||
Axial strain is $\epsilon_{x} = u_{,x}$ . Thus eq. 3.4-6 becomes
|
||||
|
||||
$$
|
||||
\Pi_ {\rho} = \int_ {0} ^ {L _ {T}} \frac {1}{2} E u _ {, x} ^ {2} A d x - \int_ {0} ^ {L _ {T}} u (c x) d x \tag {3.5-3}
|
||||
$$
|
||||
|
||||
Equation 3.5-1 becomes, with $f_{i} = f_{i}(x)$ and only polynomial functions considered,
|
||||
|
||||
$$
|
||||
u = \sum_ {i = 1} ^ {n} a _ {i} f _ {i} = a _ {1} x + a _ {2} x ^ {2} + a _ {3} x ^ {3} + \dots + a _ {n} x ^ {n} \tag {3.5-4}
|
||||
$$
|
||||
|
||||
Note that there is no initial term $a_0$ : the displacement mode $u = a_0$ is inadmissible because it violates the essential boundary condition—namely $u = 0$ at $x = 0$ .
|
||||
|
||||
The simplest approximation results from using only the first term of the series, $u = a_{1}x$ . From Eqs. 3.5-2, 3.5-3, and 3.5-4,
|
||||
|
||||
$$
|
||||
\Pi_ {p} = \frac {A E L _ {T}}{2} a _ {1} ^ {2} - \frac {c L _ {T} ^ {3}}{3} a _ {1} \tag {3.5-5a}
|
||||
$$
|
||||
|
||||
$$
|
||||
\frac {d \Pi_ {p}}{d a _ {1}} = 0 \quad \text { yields } \quad a _ {1} = \frac {c L _ {T} ^ {2}}{3 A E} \tag {3.5-5b}
|
||||
$$
|
||||
|
||||
$$
|
||||
\text { hence } \quad u = \frac {c L _ {T} ^ {2}}{3 A E} x ^ {\nu} \quad \text { and } \quad \sigma_ {x} = E u _ {, x} = \frac {c L _ {T} ^ {2}}{3 A} \tag {3.5-5c}
|
||||
$$
|
||||
|
||||
Before commenting on these results we consider a two-term solution, using the field
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
y
|
||||
q = cx
|
||||
x,u
|
||||
L_T
|
||||
(a)
|
||||
</details>
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>text_image</summary>
|
||||
|
||||
Exact and two
|
||||
terms (almost
|
||||
coincident)
|
||||
One term
|
||||
0
|
||||
x
|
||||
L_T
|
||||
(b)
|
||||
</details>
|
||||
|
||||
|
||||
|
||||
<details>
|
||||
<summary>line</summary>
|
||||
| x | One Term | Exact | Two Terms |
|
||||
| ---- | -------- | ----- | --------- |
|
||||
| 0 | 0 | 0 | 0 |
|
||||
| L_T | 0 | 0 | 0 |
|
||||
</details>
|
||||
|
||||
Figure 3.5-1. (a) Uniform bar under linearly varying distributed axial load of intensity q = cx, where c is a constant. (b) Exact and approximate axial displacements. (c) Exact and approximate axial stresses.
|
||||
Reference in New Issue
Block a user