김경종
450 lines
31 KiB
Markdown
450 lines
31 KiB
Markdown
<!-- source-page: 451 -->
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$$
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U _ {m} = P \int_ {0} ^ {L} \epsilon_ {m} d x = \frac {1}{2} \int_ {0} ^ {L} P w _ {, x} ^ {2} d x \tag {14.1-4}
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$$
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Let us assume that $w$ varies as half a sine wave, which happens to be the exact shape for Euler column buckling. Thus
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$$
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w = w _ {c} \sin \frac {\pi x}{L} \quad \text { yields } \quad \left\{ \begin{array}{l} U _ {b} = \frac {\pi^ {4} E I}{4 L ^ {3}} w _ {c} ^ {2} \\ U _ {m} = \frac {\pi^ {2} P}{4 L} w _ {c} ^ {2} \end{array} \right. \tag {14.1-5a}
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$$
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where $w_{c}$ is the center deflection of the beam.
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To do a buckling analysis, we presume that lateral load q is zero. Thus, during buckling, membrane energy is exchanged for bending energy without any input of external work. Therefore
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$$
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U _ {b} + U _ {m} = 0 \quad \text { yields } \quad P = - \frac {\pi^ {2} E I}{L ^ {2}} \tag {14.1-6}
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$$
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which is the classical Euler buckling load, independent of $w_{c}$ so long as $w_{c}$ is small.
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Now consider a deflection problem rather than a buckling problem. Imagine that distributed lateral load q, in the form of a half sine wave with amplitude $q_{c}$ , is applied to the beam in the positive z direction. With $w = w_{c} \sin(\pi x/L)$ and $q = q_{c} \sin(\pi x/L)$ , load q has potential
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$$
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\Omega = - \int_ {0} ^ {L} q w d x = - \frac {q _ {c} L}{2} w _ {c} \tag {14.1-7}
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$$
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The total potential is $\Pi_p = U_b + U_m + \Omega$ , and the equilibrium value of $w_c$ is given by $\partial \Pi_p / \partial w_c = 0$ . Thus, from Eqs. 14.1-5 and 14.1-7,
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$$
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(k + k _ {\sigma}) w _ {c} = \frac {q _ {c} L}{2} \quad \text { where } \quad \left\{ \begin{array}{l} k = \frac {\pi^ {4} E I}{2 L ^ {3}} \\ k _ {\sigma} = \frac {\pi^ {2} P}{2 L} \end{array} \right. \tag {14.1-8}
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$$
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The stiffness coefficient $(k + k_{\sigma})$ is the sum of conventional stiffness k and stress stiffness $k_{\sigma}$ . If P = 0, we obtain $w_{c} = q_{c}L^{4}/\pi^{4}EI$ . This result is exact (see Eq. 10.4-11). A tensile load $(P > 0)$ decreases the lateral deflection $w_{c}$ produced by transverse load q. This is the “stress stiffening” effect. If P is compressive $(P < 0)$ , then $w_{c}$ is increased, becoming infinite when $P = -\pi^{2}EI/L^{2}$ , which again defines the buckling load $P = P_{cr}$ . When $P = P_{cr}$ , the net stiffness $(k + k_{\sigma})$ is zero.
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Stress stiffness $k_{\sigma}$ can be written in terms of displacement rather than force. Imagine that end x = L of the bar is roller-supported and can have axial displacement $u_{L}$ . Expressing load P in terms of displacement $u_{L}$ , we have
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<details>
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<summary>text_image</summary>
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z, w
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w_c
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F
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F
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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>line</summary>
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| w_c | F (Primary path) | F (Bifurcation point) | F (Secondary path (e = 0)) | F (Increasing e) |
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|-----|------------------|------------------------|-----------------------------|------------------|
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| 0 | 0 | 0 | 0 | 0 |
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| >0 | >0 | >0 | >0 | >0 |
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</details>
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(b)
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Figure 14.1-2. Bar under compressive load F. In (b), e indicates the magnitude of imperfection.
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$$
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P = \frac {A E}{L} u _ {L} \quad \text { and } \quad k _ {\sigma} = \frac {\pi^ {2}}{2 L} \frac {A E}{L} u _ {L} = \frac {\pi^ {2} A E}{2 L ^ {2}} u _ {L} \tag {14.1-9}
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$$
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Equation 14.1-9 suggests that the following two-stage analysis is possible (although unnecessary in this simple example). In the first stage, one does a conventional static analysis (without $k_{\sigma}$ ) to determine $u_{L}$ produced by load P. Hence, from Eq. 14.1-9, $k_{\sigma}$ becomes known. One can now use the net stiffness ( $k + k_{\sigma}$ ) in Eq. 14.1-8 to determine the lateral displacement $w_{c}$ produced by lateral load q. Specifically, in this example we obtain $u_{L} = PL/AE$ and $k_{\sigma} = \pi^{2}P/2L$ , exactly as in Eq. 14.1-8. A two-stage analysis is accurate if displacements associated with the first stage are not coupled to displacements associated with the second stage. (If coupling is significant, a multistage analysis is required; see Sections 14.5 and 17.7.) The motivation for using two stages rather than one is that in most structures the distribution of membrane forces is not known a priori and must be determined by the first-stage calculation before the effect of an additional loading can be determined or a buckling analysis performed.
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Caution. Buckling theory presumes the existence of a bifurcation point. Consider, for example, Fig. 14.1-2. At the bifurcation (buckling) load, two equilibrium configurations are possible: the column could remain straight (primary path) or it could buckle (secondary path). A bifurcation point exists if the column is perfectly straight, perfectly uniform, perfectly free of end moments and lateral loads, and forces F are perfectly centered and perfectly axial. In reality there are always imperfections, whose magnitude we denote by e. If $e \neq 0$ , the column displays no bifurcation point and structures in general display “limit points.” A computed buckling load is then only an approximation of how much load a structure will carry. The approximation may be quite wrong, and may err on the unconservative (unsafe) side. Most “buckling” problems should be approached as nonlinear problems in which prebuckling deformations are taken into account. These concepts are discussed further in Sections 14.4, 14.5, and 14.7.
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These cautionary remarks do not obviate the usefulness of $[k_{\sigma}]$ in stress-stiffening and nonlinear analyses.
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# 14.2 STRESS STIFFNESS MATRICES FOR BEAMS AND BARS
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In this section, stress stiffness matrices $[k_{\sigma}]$ for prismatic members are derived from Eq. 14.1-4 by use of an assumed lateral displacement field $w = w(x)$ . Axial
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force P in the member is presumed known in terms of loads applied to the structure, either a priori or by elastic analysis, depending on whether the structure in which the member resides is statically determinate or not. Attention is restricted to displacements in a plane. Beam and bar elements are placed on a local x axis. Matrices $[k_{\sigma}]$ for elements arbitrarily oriented in global coordinates can be obtained by straightforward use of the transformation $[T]^{T}[k_{\sigma}][T]$ (see Eq. 7.4-4). Matrices $[k_{\sigma}]$ for space frames and space trusses are derivable by use of expressions discussed in Section 14.4.
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To begin, we include conventional strain terms as well as $w_{,x}^{2}/2$ from Eq. 14.1-3 in the strain expression. In this way we show clearly which terms lead to the conventional stiffness matrix [k] and which to $[k_{\sigma}]$ .
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Plane Beam. The beam in Fig. 14.2-1 can have axial displacement $u = u(x)$ and lateral displacement $w = w(x)$ . Membrane strain is $\epsilon_{m} = u_{,x} + \frac{1}{2} w_{,x}^{2}$ , where the latter term comes from Eq. 14.1-3. At a distance z from the centroidal axis, the contribution of bending to the axial strain is $\epsilon = -z w_{,xx}$ , as derived in Eq. 11.1-3. The total axial strain of an arbitrarily located fiber is therefore
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$$
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\epsilon_ {x} = u _ {, x} + \frac {1}{2} w _ {, x} ^ {2} - z w _ {, x x} \tag {14.2-1}
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$$
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Each fiber carries uniaxial stress. Strain energy in the element is therefore
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$$
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U = \int_ {V _ {e}} \frac {1}{2} E \epsilon_ {x} ^ {2} d V = \int_ {0} ^ {L} \int_ {A} \frac {1}{2} E \epsilon_ {x} ^ {2} d A d x \tag {14.2-2}
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$$
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We substitute Eq. 14.2-1 into Eq. 14.2-2 and note that
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$$
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\int_ {A} d A = A \quad \int_ {A} z d A = 0 \quad \int_ {A} z ^ {2} d A = I \quad \int_ {A} E u _ {, x} d A = P \tag {14.2-3}
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$$
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where P is the axial force, positive in tension. If a term dependent on $w_{,x}^{4}$ is discarded as negligible in comparison with other terms, we obtain
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$$
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U = \int_ {0} ^ {L} \frac {A E}{2} u _ {, x} ^ {2} d x + \int_ {0} ^ {L} \frac {P}{2} w _ {, x} ^ {2} d x + \int_ {0} ^ {L} \frac {E I}{2} w _ {, x x} ^ {2} d x \tag {14.2-4}
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$$
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The first integral yields [k] for a bar element; it contains coefficients AE/L and is associated with d.o.f. $u_{1}$ and $u_{2}$ . The third integral yields [k] for a standard beam element; it contains coefficients such as $12EI/L^{3}$ and is associated with d.o.f. $w_{1}$ , $\theta_{1}$ , $w_{2}$ , and $\theta_{2}$ . The second integral yields $[k_{\sigma}]$ . This integral was previously seen
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<details>
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<summary>text_image</summary>
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z, w
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w₁
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θ₁
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A, E, I
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w₂
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θ₂
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u₁
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u₂
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x, u
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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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w₁
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A, E
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w₂
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u₁
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u₂
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x, u
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L
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</details>
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(b)
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Figure 14.2-1. Plane elements and their d.o.f. (a) Beam. (b) Bar. A = cross-sectional area, E = elastic modulus, I = moment of inertia of A.
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as Eq. 14.1-4. It describes work done, and strain energy stored, when lateral displacement w causes differential elements to stretch an amount $w_{,x}^{2} dx/2$ in the presence of a constant axial force P. The standard $[k_{\sigma}]$ for a beam is developed from the integral expression as follows.
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With nodal d.o.f. $\{\mathbf{d}\} = [w_1 \theta_1 w_2 \theta_2]^T$ and shape functions $[\mathbf{N}]$ , lateral displacement $w$ and its first derivative $w_{xx}$ are
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$$
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w = [ \mathbf {N} ] \{\mathbf {d} \} \quad \text { where } \quad [ \mathbf {N} ] = \left[ \begin{array}{l l l l} N _ {1} & N _ {2} & N _ {3} & N _ {4} \end{array} \right] \tag {14.2-5}
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$$
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$$
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w _ {, x} = [ \mathbf {G} ] \{\mathbf {d} \} \quad \text { where } \quad [ \mathbf {G} ] = \left\lfloor N _ {1, x} \quad N _ {2, x} \quad N _ {3, x} \quad N _ {4, x} \right\rfloor \tag {14.2-6}
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$$
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The second integral in Eq. 14.2-4 yields
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$$
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\int_ {0} ^ {L} \frac {P}{2} w _ {, x} ^ {2} d x = \frac {1}{2} \int_ {0} ^ {L} w _ {, x} ^ {T} P w _ {, x} d x = \frac {1}{2} \left\{\mathbf {d} \right\} ^ {T} [ \mathbf {k} _ {\sigma} ] \left\{\mathbf {d} \right\} \tag {14.2-7}
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$$
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where
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$$
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[ \mathbf {k} _ {\sigma} ] = \int_ {0} ^ {L} [ \mathbf {G} ] ^ {T} P [ \mathbf {G} ] d x \tag {14.2-8}
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$$
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Force P is constant in this member and can be removed from the integral. Using the standard $N_{i}$ given in Fig. 3.13-2, we obtain [14.2]
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$$
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\left[ \mathbf {k} _ {\sigma} \right] = \frac {P}{3 0 L} \left[ \begin{array}{c c c c} 3 6 & 3 L & - 3 6 & 3 L \\ 3 L & 4 L ^ {2} & - 3 L & - L ^ {2} \\ - 3 6 & - 3 L & 3 6 & - 3 L \\ 3 L & - L ^ {2} & - 3 L & 4 L ^ {2} \end{array} \right] \tag {14.2-9}
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$$
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where P is positive in tension. (By inserting rows and columns of zeros, $[k_{\sigma}]$ could be written as a 6 by 6 matrix that operates on the d.o.f. $\{d\} = [u_{1} w_{1} \theta_{1} u_{2} w_{2} \theta_{2}]^{T}$ . Coordinate transformation could follow; the resulting $[k_{\sigma}]$ could then be used for an arbitrarily oriented member of a plane frame.)
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Plane Bar. The foregoing arguments can be repeated, but with curvature $w_{,xx}$ and d.o.f. $\theta_{1}$ and $\theta_{2}$ omitted from Eq. 14.2-4. Nonzero terms in $[k_{\sigma}]$ are then associated with d.o.f. $w_{1}$ and $w_{2}$ , and matrix $[G]$ describes a rotation of the bar that is independent of x,
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$$
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w _ {, x} = \left\lfloor \mathbf {G} \right] \left\{ \begin{array}{l} w _ {1} \\ w _ {2} \end{array} \right\} \quad \text { where } \quad \left\lfloor \mathbf {G} \right\rfloor = \left\lfloor - \frac {1}{L} \quad \frac {1}{L} \right\rfloor \tag {14.2-10}
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$$
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Hence, Eq. 14.2-8 yields
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$$
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[ \mathbf {k} _ {\sigma} ] = \frac {P}{L} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right] \quad \text { for } \quad \{\mathbf {d} \} = \left\lfloor w _ {1} \quad w _ {2} \right\rfloor^ {T} \tag {14.2-11a}
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$$
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<!-- source-page: 455 -->
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or
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$$
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\left[ \mathbf {k} _ {\sigma} \right] = \frac {P}{L} \left[ \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right] \quad \text { for } \quad \{\mathbf {d} \} = \left[ \begin{array}{c c c c} u _ {1} & w _ {1} & u _ {2} & w _ {2} \end{array} \right] ^ {T} \tag {14.2-11b}
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$$
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Remarks. Equations 14.2-11 are exact for small deflections of a bar that may rotate but does not bend. Equation 14.2-9, which allows bending, is approximate because a cubic lateral-displacement field is not exact when a beam carries axial load as well as loads that produce bending. As usual, accuracy is gained by dividing a given beam into two or more elements. A single beam-column element can be exact if the element formulation is based on the exact displacement field. Formulation of such an element yields a combined matrix $[\mathbf{k} + \mathbf{k}_{\sigma}]$ , which remains 4 by 4 but has coefficients that are more complicated than coefficients in Eq. 14.2-9.
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Note that if $\{d\}$ represents a small rigid-body rotation, the conventional stiffness matrix [k] yields zero forces; that is, $[k]\{d\} = \{0\}$ . Such is not the case for the stress stiffness matrix; that is, $[k_{\sigma}]\{d\} \neq \{0\}$ . This result does not imply that $[k_{\sigma}]$ is in error. Consider, for example, $[k_{\sigma}]$ of Eq. 14.2-11. If the element is given a small rotation $\theta$ , then axial strain $\epsilon_{x} = \theta^{2}/2$ appears and transverse “kickoff” forces of magnitude $P\theta$ appear at the nodes. These forces can be regarded as inseparable from buckling; that is, in the buckled state of a structure, the loading provided by kickoff forces from the various elements is exactly resisted by a deformation state whose associated rotations create the kickoff forces. If all higher-order terms were retained in the expression for $\epsilon_{x}$ , the rigid-body rotation of an element would not create nodal forces (see Section 14.4).
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When conventional stiffness and stress stiffness are both taken into account, the total or effective stiffness matrix is $[k] + [k_{\sigma}]$ for an element and $[K] + [K_{\sigma}]$ for a structure. Thus, in direct analogy to Eq. 14.1-8, one accounts for the stiffening or weakening effect of axial load on bending stiffness. The inclusion of $[K_{\sigma}]$ does not require the inclusion of extra d.o.f. in $\{D\}$ when the equation $([K] + [K_{\sigma}])\{D\} = \{R\}$ is used in place of $[K]\{D\} = \{R\}$ . Use of $[K] + [K_{\sigma}]$ to solve buckling problems is discussed in Section 14.5.
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# 14.3 STRESS STIFFNESS MATRIX OF A PLATE ELEMENT
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For a flat plate, just as for a bar or a beam, an expression for $[k_{\sigma}]$ can be obtained by examination of the work done by constant membrane forces as they act through displacements associated with small lateral deflections. Membrane forces, Fig. 14.3-1, are defined by
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$$
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N _ {x} = \int_ {- t / 2} ^ {t / 2} \sigma_ {x} d z \quad N _ {y} = \int_ {- t / 2} ^ {t / 2} \sigma_ {y} d z \quad N _ {x y} = \int_ {- t / 2} ^ {t / 2} \tau_ {x y} d z \tag {14.3-1}
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$$
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<!-- source-page: 456 -->
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<details>
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<summary>text_image</summary>
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z, w
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y
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Nx
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Ny
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Nx y
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</details>
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Figure 14.3-1. Differential element of a flat plate, showing membrane forces $N_{x}$ , $N_{y}$ , and $N_{xy}$ .
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where membrane stresses $\sigma_{x}, \sigma_{y}$ , and $\tau_{xy}$ are either known a priori or calculated by standard static stress analysis, using, for example, plane bilinear isoparametric elements.
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Membrane strains associated with small rotations $w_{,x}$ and $w_{,y}$ of the plate mid-surface are [11.1,14.4]
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$$
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\epsilon_ {x} = \frac {1}{2} w _ {, x} ^ {2} \quad \epsilon_ {y} = \frac {1}{2} w _ {, y} ^ {2} \quad \gamma_ {x y} = w _ {, x} w _ {, y} \tag {14.3-2}
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$$
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If membrane forces $N_{x}$ , $N_{y}$ , and $N_{xy}$ are assumed to be independent of the small lateral deflection $w = w(x, y)$ , then the work associated with the membrane forces and the strains of Eqs. 14.3-2 is
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$$
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U _ {\sigma} = \int_ {A} \left(\frac {1}{2} w _ {, x} ^ {2} N _ {x} + \frac {1}{2} w _ {, y} ^ {2} N _ {y} + w _ {, x} w _ {, y} N _ {x y}\right) d A \tag {14.3-3a}
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$$
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$$
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U _ {\sigma} = \frac {1}{2} \iint \left\{ \begin{array}{l} w _ {, x} \\ w _ {, y} \end{array} \right\} ^ {T} \left[ \begin{array}{l l} N _ {x} & N _ {x y} \\ N _ {x y} & N _ {y} \end{array} \right] \left\{ \begin{array}{l} w _ {, x} \\ w _ {, y} \end{array} \right\} d x d y = \frac {1}{2} \{\mathbf {d} \} ^ {T} [ \mathbf {k} _ {\sigma} ] \{\mathbf {d} \} \tag {14.3-3b}
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$$
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One must choose a displacement field $w = w(x, y)$ whose form is appropriate to the element shape and its d.o.f., for example, Eq. 11.2-5. From the displacement field one obtains rotations, that is,
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$$
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w \doteq \lfloor \mathbf {N} \rfloor \{\mathbf {d} \} \quad \text { yields } \quad \left\{ \begin{array}{l} w _ {, x} \\ w _ {, y} \end{array} \right\} = \left[ \begin{array}{l} \mathbf {G} \end{array} \right] \left\{ \begin{array}{l} \mathbf {d} \end{array} \right\} \tag {14.3-4}
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$$
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where n is the number of d.o.f. per element. Matrix [G] is in general a function of x and y. Equations 14.3-3b and 14.3-4 yield
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$$
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[ \mathbf {k} _ {\sigma} ] = \iint [ \mathbf {G} ] ^ {T} \left[ \begin{array}{l l} N _ {x} & N _ {x y} \\ N _ {x \bar {y}} & N _ {y} \end{array} \right] [ \mathbf {G} ] d x d y \tag {14.3-5}
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$$
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where integration extends over the element area. If the element is of the isoparametric family, shape functions [N] are expressed in terms of dimensionless coordinates $\xi$ and $\eta$ . Therefore, one must invoke [J], the Jacobian matrix of Eq. 6.3-11. Thus
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$$
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\left\{ \begin{array}{l} w, \xi \\ w, \eta \end{array} \right\} = [ \mathbf {G} _ {I} ] \{\mathbf {d} \} \quad \text { and } \quad \left\{ \begin{array}{l} w, x \\ w, y \end{array} \right\} = [ \mathbf {J} ] ^ {- 1} \left\{ \begin{array}{l} w, \xi \\ w, \eta \end{array} \right\} \tag {14.3-6}
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$$
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<!-- source-page: 457 -->
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where $[G_{I}]$ contains derivatives of the shape functions with respect to $\xi$ and $\eta$ . Equations 14.3-3b and 14.3-6 yield
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$$
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\left[ \mathbf {k} _ {\sigma} \right] = \int_ {- 1} ^ {1} \int_ {- 1} ^ {1} \left[ \mathbf {G} _ {I} \right] ^ {T} [ \mathbf {J} ] ^ {- T} \left[ \begin{array}{l l} N _ {x} & N _ {x y} \\ N _ {x y} & N _ {y} \end{array} \right] [ \mathbf {J} ] ^ {- 1} \left[ \mathbf {G} _ {I} \right] J d \xi d \eta \tag {14.3-7}
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$$
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where $J$ is the Jacobian determinant.
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Membrane forces $N_{x}$ , $N_{y}$ , and $N_{xy}$ may vary over an element. Then, in numerical integration to evaluate $[k_{\sigma}]$ , different membrane forces would be used at different sampling points.
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Note that $[k_{\sigma}]$ is determined independently of material properties, except to the extent that material properties may influence computed values of $N_{x}$ , $N_{y}$ , and $N_{xy}$ . Accordingly, a given $[k_{\sigma}]$ is equally applicable to both isotropic and anisotropic structures.
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# 14.4 A GENERAL FORMULATION FOR $[\mathbf{k}_{\sigma}]$
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In Sections 14.2 and 14.3, each type of element is approached as a special case. It is desirable to also have a general formula for $[k_{\sigma}]$ , analogous to the formula for the conventional [k] (Eq. 4.1-5), that may be specialized to particular geometries, and requires only that a specific displacement field be chosen. Such a formula is derived in the present section. $^{2}$
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The formula for $[k_{\sigma}]$ , Eq. 14.4-7, is “linearized” and is limited to small displacements. To demonstrate in a general way that this is so requires comparatively lengthy and complicated arguments. We will omit these arguments [2.1] and instead illustrate the nature of the approximation by means of a simple particular case (Eqs. 14.4-10 to 14.4-16).
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Green-Lagrange Strain. Various advanced texts [e.g., 2.1, 3.1, 9.9] discuss the expressions for stress and strain appropriate to problems that involve large deformations. The following equations define a strain measure commonly known as Green-Lagrange strain:
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$$
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\epsilon_ {x} = u _ {, x} + \frac {1}{2} \left(u _ {, x} ^ {2} + v _ {, x} ^ {2} + w _ {, x} ^ {2}\right) \tag {14.4-1a}
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$$
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$$
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\epsilon_ {y} = v _ {, y} + \frac {1}{2} \left(u _ {, y} ^ {2} + v _ {, y} ^ {2} + w _ {, y} ^ {2}\right) \tag {14.4-1b}
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$$
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$$
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\epsilon_ {z} = w _ {, z} + \frac {1}{2} \left(u _ {, z} ^ {2} + v _ {, z} ^ {2} + w _ {, z} ^ {2}\right) \tag {14.4-1c}
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$$
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$$
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\gamma_ {x y} = u _ {, y} + v _ {, x} + \left(u _ {, x} u _ {, y} + v _ {, x} v _ {, y} + w _ {, x} w _ {, y}\right) \tag {14.4-1d}
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$$
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$$
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\gamma_ {y z} = v _ {, z} + w _ {, y} + \left(u _ {, y} u _ {, z} + v _ {, y} v _ {, z} + w _ {, y} w _ {, z}\right) \tag {14.4-1e}
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$$
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$$
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\gamma_ {z x} = w _ {, x} + u _ {, z} + (u _ {, z} u _ {, x} + v _ {, z} v _ {, x} + w _ {, z} w _ {, x}) \tag {14.4-1f}
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$$
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<!-- source-page: 458 -->
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The initial terms in Eqs. 14.4-1 are the customary engineering definitions of normal and shear strain ( $\epsilon_{x} = u_{,x}$ , etc.). The added terms, in parentheses, become significant if displacement gradients are not small. Green–Lagrange strains are zero for a rigid-body rotation of any magnitude. In Eqs. 14.4-1, all displacement derivatives are computed in the original coordinate system, regardless of how large a rigid-body rotation may be superposed on the deformations. This is the “total Lagrangian” approach, in which all displacements are measured in a reference frame that is stationary rather than attached to the deforming structure. The stationary coordinates may also be called “material coordinates” and may be denoted in some papers by uppercase labels X, Y, and Z.
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Green-Lagrange normal strains correspond to defining the strain of a line segment by the equation
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$$
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\epsilon = \frac {1}{2} \left[ \left(\frac {d s ^ {*}}{d s}\right) ^ {2} - 1 \right] \tag {14.4-2}
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$$
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where $ds$ and $ds^*$ are respectively the initial and final lengths of the line segment. If $ds \approx ds^*$ , Eq. 14.4-2 reduces to the usual small-strain approximation, $\epsilon = (ds^* - ds)/ds$ .
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Formula for $[k_{\sigma}]$ . Imagine that initial stresses $\{\sigma_{0}\}$ prevail. If these stresses are assumed to remain constant as strains $\{\epsilon\}$ occur, the associated work is $^{3}$
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$$
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\int_ {V} \{\boldsymbol {\epsilon} \} ^ {T} \left\{\boldsymbol {\sigma} _ {0} \right\} d V \quad \text {where} \quad \left\{ \begin{array}{l l} \left\{\boldsymbol {\epsilon} \right\} ^ {T} = \left\lfloor \epsilon_ {x} \quad \epsilon_ {y} \dots \gamma_ {z x} \right\rfloor & (1 4. 4 - 3 a) \\ \left\{\boldsymbol {\sigma} _ {0} \right\} = \left\lfloor \sigma_ {x 0} \quad \sigma_ {y 0} \dots \tau_ {z x 0} \right\rfloor^ {T} & (1 4. 4 - 3 b) \end{array} \right.
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$$
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With $\{\epsilon\}$ given by Eqs. 14.4-1, the integrand $\{\epsilon\}^{T}\{\sigma_{0}\}$ first displays the terms $u_{,x}\sigma_{x0} + v_{,y}\sigma_{y0} + \cdots$ . These terms lead to nodal loads associated with $\{\sigma_{0}\}$ , as given by Eq. 4.1-6. What remains is
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$$
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U _ {\sigma} = \int_ {V} \left[ \frac {1}{2} \left(u _ {, x} ^ {2} + v _ {, x} ^ {2} + w _ {, x} ^ {2}\right) \sigma_ {x 0} + \dots + \left(u _ {, z} u _ {, x} + v _ {, z} v _ {, x} + w _ {, z} w _ {, x}\right) \tau_ {z x 0} \right] d V \tag {14.4-4}
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$$
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If we define
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$$
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\{\delta \} = \left[ \begin{array}{l l l l l l l l l} u _ {, x} & u _ {, y} & u _ {, z} & v _ {, x} & v _ {, y} & v _ {, z} & w _ {, x} & w _ {, y} & w _ {, z} \end{array} \right] ^ {T} \tag {14.4-5}
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$$
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then Eq. 14.4-4 can be written in the form
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$$
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U _ {\sigma} = \frac {1}{2} \int_ {V} \left\{\delta \right\} ^ {T} \left[ \begin{array}{l l l} \mathrm{s} & \mathbf {0} & \mathbf {0} \\ \mathbf {0} & \mathrm{s} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} & \mathrm{s} \end{array} \right] \left\{\delta \right\} d V \quad \text {where} \quad [ \mathrm{s} ] = \left[ \begin{array}{l l l} \sigma_ {x 0} & \tau_ {x y 0} & \tau_ {z x 0} \\ \tau_ {x y 0} & \sigma_ {y 0} & \tau_ {y z 0} \\ \tau_ {z x 0} & \tau_ {y z 0} & \sigma_ {z 0} \end{array} \right] \tag {14.4-6}
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$$
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$^{3}$ This assumption restricts the subsequent development, Eqs. 14.4-3 to 14.4-7, to small strains and small rotations. More advanced arguments [2.1] show that a more elaborate definition of stress than engineering stresses $\{\sigma_{0}\}$ is required if one is to write a strain energy expression that is meaningful in analyses of large deformations.
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<!-- source-page: 459 -->
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This expression is analogous to Eqs. 14.2-7 and 14.3-3b, and yields $[k_{\sigma}]$ in an analogous way. Let the element displacement field be given by $\{u\} = [N]\{d\}$ , as usual, where $\{u\} = [u \quad v \quad w]^{T}$ and $\{d\}$ contains nodal d.o.f. Also let $\{\delta\} = [G]\{d\}$ , where [G] is obtained from shape functions [N] by appropriate differentiation and ordering of terms. Equation 14.4-6 becomes $U_{\sigma} = \{d\}^{T}[k_{\sigma}]\{d\}/2$ , where
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$$
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\left[ \mathbf {k} _ {\sigma} \right] = \int_ {V _ {e}} [ \mathbf {G} ] ^ {T} \left[ \begin{array}{l l l} \mathbf {s} & \mathbf {0} & \mathbf {0} \\ \mathbf {0} & \mathbf {s} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} & \mathbf {s} \end{array} \right] [ \mathbf {G} ] d V \tag {14.4-7}
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$$
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As an example, consider the bar of Fig. 14.2-1b, again with motion restricted to the xz plane. For this case all initial stresses are zero except for axial stress $\sigma_{x0}$ . We assume that u and w are linear in x and require that v = 0. Accordingly, with $N_{1} = (L - x)/L$ and $N_{2} = x/L$ , we write
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$$
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\begin{array}{l} u = N _ {1} u _ {1} + N _ {2} u _ {2} \\ w = N _ {1} w _ {1} + N _ {2} w _ {2} \end{array} \quad [ \mathbf {G} ] = \frac {1}{L} \left[ \begin{array}{c c c c} - 1 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right] \tag {14.4-8}
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$$
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Nonzero d.o.f. are $\{\mathbf{d}\} = \left[u_1 w_1 u_2 w_2\right]^T$ . Also, $\{\delta\} = \left[u_{,x} w_{,x}\right]^T$ . Equation 14.4-7 reduces to
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$$
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\left[ \mathbf {k} _ {\sigma} \right] = \int_ {0} ^ {L} \left[ \mathbf {G} \right] ^ {T} \left[ \begin{array}{c c} \sigma_ {x 0} & 0 \\ 0 & \sigma_ {x 0} \end{array} \right] [ \mathbf {G} ] \mathrm{A} d x = \frac {P}{L} \left[ \begin{array}{c c c c} 1 & 0 & - 1 & 0 \\ 0 & 1 & 0 & - 1 \\ - 1 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \end{array} \right] \tag {14.4-9}
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$$
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where $P = \sigma_{x0}A$ . This $[k_{\sigma}]$ is almost the same as that in Eq. 14.2-11, but contains four more nonzero terms. However, note that the additional nonzero terms occupy the same positions as the nonzero terms in the conventional stiffness matrix (see Eq. 2.5-3). Thus, in the net stiffness matrix $[k] + [k_{\sigma}]$ , we see coefficients $\pm(AE + P)/L$ (corresponding to d.o.f. $u_{1}$ and $u_{2}$ ) and $\pm P/L$ (corresponding to d.o.f. $w_{1}$ and $w_{2}$ ). Since AE >> P in any practical problem, the “extra” P/L terms in $[k_{\sigma}]$ can be discarded. In this way Eq. 14.4-9 reduces to Eq. 14.2-11.
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Full Nonlinearity: An Example. In the foregoing development it is not obvious what approximations are contained in $[k_{\sigma}]$ . In what follows we allow large rotations, and show by example that use of $[k_{\sigma}]$ implies complete linearization of the problem and negligible rotations prior to buckling.
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Consider the one-element elastic bar in Fig. 14.4-1. If forces are applied only at the ends, displacements vary linearly.
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$$
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u = \frac {x}{L} u _ {2} \quad \text { and } \quad w = \frac {x}{L} w _ {2} \tag {14.4-10}
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$$
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Measurements are made in the original coordinate system $xz$ ; that is, $x$ is not regarded as an axial coordinate that rotates as the bar rotates. For example, the bar becomes vertical if $u_2 = -L$ and $w_2 = \pm L$ ; nevertheless, one uses the horizontal coordinate $x$ in the fields $u = u_2x/L$ and $w = w_2x/L$ . Thus $u_2$ is the $x$ -direction component of the displacement of node 2; it is not the stretch of the bar unless $w_2 = 0$ . End 2 of the bar is located by coordinates $x = L$ and $z = 0$ , regardless of the values of $u_2$ and $w_2$ .
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<!-- source-page: 460 -->
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<details>
|
||
<summary>text_image</summary>
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|
||
z, w
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1 A, E 2 x, u
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||
L
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||
</details>
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|
||
{a}
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||
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||
|
||
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||
<details>
|
||
<summary>text_image</summary>
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|
||
z, w
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||
1
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||
θ
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||
L
|
||
u₂
|
||
Rₓ
|
||
2
|
||
Rₓ
|
||
w₂
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||
x, u
|
||
</details>
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(b)
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Figure 14.4-1. (a) A bar element, hinged at node 1, prior to loading. (b) The displaced and deformed bar after forces $R_{x}$ and $R_{z}$ are applied to node 2.
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The bar moves in the $xz$ plane and carries uniaxial stress. If strains are small, the total potential of the bar is
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|
||
$$
|
||
\Pi_ {p} = \frac {1}{2} \int_ {0} ^ {L} A E \epsilon_ {x} ^ {2} d x - R _ {x} u _ {2} - R _ {z} w _ {2} \tag {14.4-11}
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||
$$
|
||
|
||
where $\epsilon_{x}$ is axial strain, as if the bar occupies its original x-parallel orientation. From Eqs. 14.4-1a and 14.4-10,
|
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|
||
$$
|
||
\epsilon_ {x} = \frac {u _ {2}}{L} + \frac {1}{2} \left(\frac {u _ {2} ^ {2}}{L ^ {2}} + \frac {w _ {2} ^ {2}}{L ^ {2}}\right) \tag {14.4-12}
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||
$$
|
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|
||
The resulting expression for $\Pi_p$ still allows large rotation of the bar. Static equilibrium prevails when $\partial \Pi_p / \partial u_2 = 0$ and $\partial \Pi_p / \partial w_2 = 0$ . Results of these calculations can be written in the form
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|
||
$$
|
||
\frac {A E}{L} \left(\left[ \begin{array}{l l} 1 & 0 \\ 0 & 0 \end{array} \right] + \frac {1}{2 L} \left[ \begin{array}{l l} 3 u _ {2} & w _ {2} \\ w _ {2} & u _ {2} \end{array} \right] + \frac {1}{2 L ^ {2}} \left[ \begin{array}{l l} u _ {2} ^ {2} & u _ {2} w _ {2} \\ u _ {2} w _ {2} & w _ {2} ^ {2} \end{array} \right]\right) \left\{ \begin{array}{l} u _ {2} \\ w _ {2} \end{array} \right\} = \left\{ \begin{array}{l} R _ {x} \\ R _ {z} \end{array} \right\} \tag {14.4-13}
|
||
$$
|
||
|
||
One finds that if $u_{2}$ and $w_{2}$ represent rigid-body rotation about node 1—that is, if $u_{2} = -L(1 - \cos \theta)$ and $w_{2} = L \sin \theta$ —then $R_{x} = 0$ and $R_{z} = 0$ for any rotation $\theta$ , no matter how large.
|
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|
||
In Eq. 14.4-13, $u_{2}$ and $w_{2}$ are total displacements. An analogous incremental form can be obtained from Eq. 14.4-13 by writing $R_{x} = R_{x}(u_{2}, w_{2})$ and $R_{z} = R_{z}(u_{2}, w_{2})$ , then forming expressions for $dR_{x}$ and $dR_{z}$ by differentiation. Thus
|
||
|
||
$$
|
||
\underbrace {\left(\frac {A E}{L} \left[ \begin{array}{l l} 1 & 0 \\ 0 & 0 \end{array} \right] \right.} _ {[ \mathrm{K} ]} + \underbrace {\frac {A E}{L ^ {2}} \left[ \begin{array}{l l} 3 u _ {2} & w _ {2} \\ w _ {2} & u _ {2} \end{array} \right]} _ {[ \mathrm{N} _ {1} ]} + \underbrace {\frac {A E}{2 L ^ {3}} \left[ \begin{array}{c c} 3 u _ {2} ^ {2} + w _ {2} ^ {2} & 2 u _ {2} w _ {2} \\ 2 u _ {2} w _ {2} & u _ {2} ^ {2} + 3 w _ {2} ^ {2} \end{array} \right]} _ {[ \mathrm{N} _ {2} ]} \Bigg) \left\{ \begin{array}{l} d u _ {2} \\ d w _ {2} \end{array} \right\} = \left\{ \begin{array}{l} d R _ {x} \\ d R _ {z} \end{array} \right\} \tag {14.4-14}
|
||
$$
|
||
|
||
This equation describes nodal force increments $dR_{x}$ and $dR_{z}$ associated with small nodal displacements $du_{2}$ and $dw_{2}$ , where $du_{2}$ and $dw_{2}$ are measured from a current
|