김경종
488 lines
27 KiB
Markdown
488 lines
27 KiB
Markdown
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in several stages, with convergence required in each stage before going on to the next. Indeed, the latter alternative, or underrelaxation, may be mandatory in some problems that are reluctant to converge (or tend to converge to the wrong result, which is possible if there is more than one equilibrium state that is mathematically possible).
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In view of the many physical and computational alternatives that can be tested, coding the foregoing algorithm is a good educational device. Numerous test cases are available [17.32, 17.33]. A particularly simple one is that of Fig. 17.7-1a under tip moment $M_{L}$ alone. The y-direction deflection of the tip is
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$$
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v _ {\text { tip }} = \frac {E I}{M _ {L}} \left(1 - \cos \frac {M _ {L} L _ {T}}{E I}\right) \tag {17.7-2}
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$$
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which is valid for all values of $M_{L}$ provided that the material remains linearly elastic.
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Some Computational Details. Figure 17.7-2 shows a typical element. For the structures of Fig. 17.7-1, initial angle $\alpha_0$ is zero for all elements. Global coordinates $xy$ are fixed in space. A local system $x'y'$ is attached to each element and moves with it: we attach the origin $x' = y' = 0$ to node 1 and direct the $x'$ axis through node 2. Thus, in a local system $x'y'$ , three nodal d.o.f. are always zero: $u_1 = v_1 = v_2 = 0$ . All element d.o.f. in global directions, $D_1$ through $D_6$ , are in general nonzero. We presume that elements are small enough that local rotations are small—that is, that $|\theta_1| << 1$ and $|\theta_2| << 1$ .
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In the deformed and displaced configuration, element length projections $x_{L}$ and $y_{L}$ on global axes xy and the orientation $\alpha$ of the local $x'$ axis are
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$$
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x _ {L} = x _ {0} + D _ {4 1}, \quad y _ {L} = y _ {0} + D _ {5 2} \quad \alpha = \arctan (y _ {L} / x _ {L}) \tag {17.7-3a}
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$$
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where
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$$
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D _ {4 1} = D _ {4} - D _ {1} \quad \text { and } \quad D _ {5 2} = D _ {5} - D _ {2} \tag {17.7-3b}
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$$
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If $\alpha_{0}=0$ and $\alpha$ is computed in Fortran as DATAN2(YL,XL), then $\alpha$ can reach $\pm\pi$ before it becomes ambiguously defined.
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<details>
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<summary>text_image</summary>
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y
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y'
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L0
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D6
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x'
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2
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D4
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y0
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D3
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1
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α0
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D5
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D1
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D2
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x0
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(a)
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x
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</details>
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<details>
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<summary>text_image</summary>
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y
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L
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u2
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θ2
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x'
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L0
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u1 = v1 = 0
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θ1
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v2 = 0
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2
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yL
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α
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1
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xL
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(b)
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x
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</details>
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Figure 17.7-2. (a) Plane frame element, shown before any deformation or motion, but identifying global d.o.f. $D_{1}$ through $D_{6}$ . (b) The same element after deformation and motion, showing d.o.f. in local coordinates $x'y'$ .
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Element d.o.f. in the local system $x'y'$ are $\{\mathbf{d}'\} = \begin{bmatrix} 0 & 0 & \theta_1 & u_2 & 0 & \theta_2 \end{bmatrix}^T$ , where
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$$
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\theta_ {1} = D _ {3} - (\alpha - \alpha_ {0}) \quad \theta_ {2.} = D _ {6} - (\alpha - \alpha_ {0}) \tag {17.7-4a}
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$$
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$$
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u _ {2} = \frac {1}{L + L _ {0}} \left[ \left(2 x _ {0} + D _ {4 1}\right) D _ {4 1} + \left(2 y _ {0} + D _ {5 2}\right) D _ {5 2} \right] \tag {17.7-4b}
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$$
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where again $D_{41} = D_{4} - D_{1}$ and $D_{52} = D_{5} - D_{2}$ . The expression for $u_{2}$ in Eq. 17.7-4b is more accurate than the expression $u_{2} = L - L_{0}$ (in which $L^{2} = x_{L}^{2} + y_{L}^{2}$ and $L_{0}^{2} = x_{0}^{2} + y_{0}^{2}$ ) because Eq. 17.7-4b avoids finding the small difference between large numbers. Equation 17.7-4b is obtained by writing $L^{2} - L_{0}^{2}$ , substituting for $x_{L}^{2}$ and $y_{L}^{2}$ from Eq. 17.7-3, factoring $L^{2} - L_{0}^{2}$ , and solving for $u_{2} = L - L_{0}$ [17.30]. In the denominator, $L + L_{0} \approx 2L_{0}$ , as we presume that strains are small.
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In the local system $x'y'$ , loads $\{\mathbf{r}'\}$ applied to nodes by the distorted element can be obtained from Eq. 4.1-6, using for $\{\sigma_0\}$ the stresses produced by local d.o.f. $\{\mathbf{d}'\}$ , that is, $\{\sigma_0\} = [\mathbf{E}][\mathbf{B}]\{\mathbf{d}'\}$ . An alternative calculation is
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$$
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\left\{\mathbf {r} ^ {\prime} \right\} = - \left[ \mathbf {k} ^ {\prime} \right] \left\{\mathbf {d} ^ {\prime} \right\} \tag {17.7-5}
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$$
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in which the element stiffness matrix $[k']$ in the local system does not change as the local system moves and the element deforms. Here $[k']$ is given by Fig. 7.5-2a. If axial forces are significant, particularly in compression, one should add to this matrix a stress stiffness matrix $[k_{\sigma}]$ , for example, Eq. 14.2-11b (with zeros added to expand the matrix to size 6 by 6). In $[k_{\sigma}]$ , the element axial force P is given by $P = (AE/L)u_{2}$ . Thus P is deformation-dependent, and may be considered unknown at the outset of the iterative process.
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Referred to global coordinates $xy$ , the element stiffness matrix and element nodal load vector are
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$$
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[ \mathbf {k} ] = [ \mathbf {T} ] ^ {T} [ \mathbf {k} ^ {\prime} ] [ \mathbf {T} ] \quad \text { and } \quad \{\mathbf {r} \} = [ \mathbf {T} ] ^ {T} \{\mathbf {r} ^ {\prime} \} \tag {17.7-6}
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$$
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where, for the frame element, transformation matrix [T] is given by Eq. 7.5-7 (with $\beta$ replaced by $\alpha$ ). In general, each element has a different $\alpha$ and therefore requires a different [T]. The structure tangent-stiffness matrix and resisting load vector in global coordinates are formed by the usual assembly process, which is symbolized by
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$$
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[ \mathbf {K} _ {t} ] = \sum [ \mathbf {k} ] \quad \text { and } \quad \{\mathbf {R} _ {R} \} = \sum \{\mathbf {r} \} \tag {17.7-7}
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$$
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Note that coordinate transformation is not the essence of an analysis procedure for geometric nonlinearity, but only the vehicle adopted here to establish what is essential: the properties of the system in its current configuration.
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# 17.8 OTHER NONLINEAR PROBLEMS
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Nonlinear computational mechanics continues to be an active area of research. Some of the nonlinear structural problems we have not discussed in this book are as follows.
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Materials that creep are analyzed by use of constitutive relations and computational algorithms very similar to those used for analysis of plastic deformations. Indeed, creep and plasticity analyses are often combined [17.12, 17.34].
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Forming processes such as extrusion and rolling involve large plastic deformations. Elastic response may be considered negligible, and the material analyzed as if it were a viscous fluid. “Springback” after the forming process is complete can be modeled if a viscoelastic material is invoked. Casting processes involve considerable heat transfer calculations and changes of phase $[17.35]$ .
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Problems of moving contact fields, such as the rolling contact of a tire on pavement, have been addressed by special algorithms $[17.36]$ . Other contact problems, either static or dynamic, include bearings, joints in rock, and gaps that may open or close. Special elements for such problems have been devised $[17.13, 17.37, 17.38]$ .
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Membranes may be deflected by pressure, as when a balloon is inflated. Such a problem typically involves large strains, large deflections, and large rotations. Pressure loads change in direction, as they continue to act normal to the membrane. Element nodal loads $\{r_{e}\}$ also change in magnitude as pressure increases and as the element surface area subjected to pressure increases [17.39]. If a membrane is initially flat and initially unstressed, it has no resistance to lateral load. To avoid a singular structure stiffness matrix in the first iterative cycle, one can add a fictitious initial stress for the first cycle only, thereafter to be replaced by the computed stress and corresponding stress stiffness matrix.
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Cable problems are somewhat similar to membrane problems. One can follow the standard approach of dividing each cable of a cable network into many elements (e.g., two-node bar elements). However, this approach is inefficient if deflections are large. A method that treats an entire cable as a single element appears to be much more economical [17.40, 17.41].
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A nonlinear vibration problem leads to an eigenvalue problem of the standard symbolic form—that is, $([K] - \omega^{2}[M])\{\overline{D}\} = \{0\}$ ; however, stiffness matrix [K] is a function of $\{\overline{D}\}$ , the nodal amplitudes [17.42].
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# PROBLEMS
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# Section 17.2
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17.1 Sketch the progress of the two forms of direct substitution, as in Fig. 17.2-2, but let the curves be concave up, as for a hardening structure.
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17.2 Load P acts on a nonlinear spring, as shown. Let $k = 0.2 - u$ and P = 0.006. Apply three cycles of direct substitution according to Eq. 17.2-3. Also, apply three cycles of modified Newton–Raphson iteration, using k at u = 0. Thus, show that both methods yield the same results.
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Problem 17.2
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17.3 In Problem 17.2, what is the expression for the tangent stiffness $k_{t}$ in terms of displacement $u$ ?
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<details>
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<summary>text_image</summary>
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k = 24 N/m
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P
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0.02 m
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Rigid
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k = 10 N/m
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D
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100 mm
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200 mm
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</details>
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Problem 17.4
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17.4 The bar shown in the sketch is rigid. Contact with the right-hand spring is made only when the $0.02\mathrm{-m}$ gap closes. Set up an equation for displacement $D$ of load $P$ , then apply three cycles of a direct-substitution solution for $D$ when $P = 0.24\mathrm{N}$ , as follows:
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(a) Use the procedure in which stiffness $K$ is repeatedly updated but the right-hand side (i.e., load $P$ ) is unchanged.
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(b) Use the procedure in which $K$ is unchanged from its initial value, but the effective load is repeatedly updated.
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17.5 The bar shown is of length $L$ when unstressed, where $L^2 = a^2 + c^2$ . When load $P$ is zero, displacement $D$ is also zero. The bar has axial stiffness $AE / L$ , rolls without friction at $B$ , and does not buckle as a column. Assume that the roller is constrained to remain in contact with the wall.
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(a) For $a >> c$ , show that $\Pi_p = U - PD = (AE/8a^3)(-2cD + D^2)^2 - PD$ .
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(b) Show that the equilibrium values of $D$ are given by roots of the equation $P = (AE / 2a^3)(2cD - D^2)(c - D)$ .
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(c) Determine expressions for the secant stiffness and the tangent stiffness.
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(d) Show that limit points are at $D = c(1 \pm \sqrt{3}/3)$ .
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(d) Show that limit points are at $D = 0$ , $F = 241$ , and $F_{\mathrm{D}} = 241$ . Let constants of this system be such that points $A$ and $F$ on the $P$ versus $D$ curve are at $P_A = 241$ , $D_A = 0.211$ , $P_F = 250$ , and $D_F = 1.080$ . After convergence at $P = 200$ , $P$ is increased to 250, and the following sequence of displacements $D$ is generated by a Newton-Raphson algorithm: 0.173, 0.219, 0.071, 0.143, 0.190, 0.249, 0.199, 0.294, 0.235, 0.175, 0.222, 0.108, 0.166, 0.210, 1.178, 1.096, 1.080, 1.080. Explain this path to convergence by sketching it on the $P$ versus $D$ plot.
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(f) Similarly, sketch the path that would be taken by a modified Newton-Raphson algorithm.
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<details>
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<summary>text_image</summary>
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a
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B
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A
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P
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c
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D
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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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P
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A
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0
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B
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C
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E
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D
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F
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</details>
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(b)
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Problem 17.5
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17.6 Let loads $P_{1}$ and $P_{2}$ be functions of displacements $D_{1}$ and $D_{2}$ , that is, $P_{1} = f_{1}(D_{1}, D_{2})$ and $P_{2} = f_{2}(D_{1}, D_{2})$ . Let $D_{A}$ and $D_{B}$ be exact values of $D_{1}$
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and $D_{2}$ produced by loads $P_{A}$ and $P_{B}$ . Let $D_{A}^{*}$ and $D_{B}^{*}$ be approximations of $D_{A}$ and $D_{B}$ . Assume that $D_{A} = D_{A}^{*} + \Delta D_{A}$ and $D_{B} = D_{B}^{*} + \Delta D_{B}$ . Derive the following equations (analogous to Eq. 17.2-9):
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$$
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\left[ \begin{array}{c c} \partial P _ {1} / \partial D _ {1} & \partial P _ {1} / \partial D _ {2} \\ \partial P _ {2} / \partial D _ {1} & \partial P _ {2} / \partial D _ {2} \end{array} \right] _ {D _ {A} ^ {*}, D _ {B} ^ {*}} \left\{ \begin{array}{c} \Delta D _ {A} \\ \Delta D _ {B} \end{array} \right\} = \left\{ \begin{array}{c} P _ {A} - f _ {1} (D _ {A} ^ {*}, D _ {B} ^ {*}) \\ P _ {B} - f _ {2} (D _ {A} ^ {*}, D _ {B} ^ {*}) \end{array} \right\}
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$$
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17.7 The sketch shows a nonlinear load versus deflection curve. In the exercises of this problem we pretend that $P$ and $dP / dD$ can be found when $D$ is known, but that an explicit expression for $D$ in terms of $P$ is not available. Solve for $D$ , using the situations and methods indicated. Sketch the progress of each solution on a plot of $P$ versus $D$ .
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(a) What $D$ is predicted by five cycles of Newton-Raphson iteration if $P = 8$ , starting from $P = D = 0$ ?
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(b) What $D$ is predicted by five cycles of modified Newton-Raphson iteration if $P = 8$ , starting from $P = 7.5$ , $D = 3$ ?
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(c) Repeat part (b) but use three cycles and update the tangent stiffness after the first cycle.
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(d) What $D$ is predicted by four purely incremental (Euler's method) steps of $\Delta P = 2$ , starting from $P = 1$ and going to $P = 9$ ? Given: $D = 0.11111$ at $P = 1$ .
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(e) Repeat part (d) but include a force imbalance correction at every step.
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(f) What $D$ is predicted by five cycles of the direct-substitution algorithm of Eq. 17.2-2 if $P = 8$ , starting from $P = D = 0$ ?
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(g) What $D$ is predicted by five cycles of the secant method depicted in Fig. 17.2-7 if $P = 8$ , starting from $P = D = 0$ ?
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17.8 The introductory remarks of Problem 17.7 again apply, but now to the hardening curve sketched.
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(a) What $D$ is predicted by five cycles of Newton-Raphson iteration if $P = 0.8$ , starting from $P = D = 0$ ?
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(b) What $D$ is predicted by five cycles of modified Newton-Raphson iteration if $P = 3$ , starting from $P = 1.5$ , $D = 6$ ?
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(c) Repeat part (b) but apply an underrelaxation factor $\beta = 0.6$ to the increments $\Delta D$ .
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(d) Repeat part (b) but use four cycles and update the tangent stiffness after the first cycle.
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(e) What $D$ is predicted by three purely incremental (Euler's method) steps
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<details>
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<summary>line</summary>
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| D | P |
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|---------|----------|
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| 0 | 0 |
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| >0 | 10D/D + 1|
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</details>
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Problem 17.7
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<details>
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<summary>line</summary>
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| D | P |
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|---|---|
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| 0 | 0 |
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| 10 | D / (10 - D)² |
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| 20 | D / (10 - D)² |
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</details>
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Problem 17.8
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of $\Delta P = 1$ , starting from $P = 1.5$ and going to $P = 4.5$ ? Given: $D = 6$ at $P = 1.5$ .
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(f) Repeat part (e) but include a force imbalance correction at every step.
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(g) What $D$ is predicted by four cycles of the direct-substitution algorithm of Eq. 17.2-2 if $P = 4$ , starting from $P = D = 0$ ?
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(h) Repeat part (g) but apply an underrelaxation factor $\beta = 0.3$ in Eq. 17.2-5b.
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(i) What $D$ is predicted by five cycles of the secant method in Fig. 17.2-7 if $P = 0.8$ , starting from $P = D = 0$ ?
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(j) Repeat part (i), but apply an underrelaxation factor $\beta = 0.3$ to the increments $\Delta D$ .
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17.9 Use four cycles of the secant method depicted in Fig. 17.2-7 to calculate iterates $u_{i}$ for the spring problem posed in Problem 17.2.
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17.10 The Newton-Raphson method to solve $f(x) = 0$ can be formulated by defining an iteration function
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$$
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g (x) = x - \frac {f (x)}{f ^ {\prime} (x)}
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$$
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and seeking a solution $x^{*}$ that satisfies $x^{*} = g(x^{*})$ by taking a given $x_{0}$ and iterating: $x_{i+1} = g(x_{i})$ .
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(a) Verify that $x^{*}$ satisfies $f(x^{*}) = 0$ provided that $f'(x^{*}) \neq 0$ .
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(b) Use a three-term exact Taylor series for $g(x)$ , that is,
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$$
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g (x) = g \left(x ^ {*}\right) + g ^ {\prime} \left(x ^ {*}\right) \left(x - x ^ {*}\right) + \frac {1}{2} g ^ {\prime \prime} (\bar {x}) \left(x - x ^ {*}\right) ^ {2}
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$$
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where $\overline{x}$ lies between $x$ and $x^{*}$ , to verify that the Newton-Raphson method terminates quadratically.
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17.11 Use Eq. 17.2-1, $(k_0 + k_N)u = P$ , to show that the principle of superposition does not apply to a nonlinear problem.
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# Section 17.3
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17.12 The bar shown is to be modeled by a single element. Thus there is one d.o.f.—namely, the displacement $u_{2}$ of load P. Apply the tangent-stiffness method to determine $u_{2}$ for P = 3.0 kN and for P = 6.0 kN. Use two load increments of $\Delta P = 3.0$ kN. Start from P = 0. Use two cycles of step 3 of the algorithm. Plot $u_{2}$ versus P, showing the exact solution and the progress of the incremental solution.
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<details>
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<summary>line</summary>
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| ε | σ, MPa |
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| ------- | ------ |
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| 0.001 | 20 |
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| 0.001 | 0 |
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</details>
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Problem 17.12
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<!-- source-page: 557 -->
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17.13 Repeat Problem 17.12 but apply only the 3.0-kN load. Start from $P = 2.0$ kN, apply $\Delta P = 1.0$ kN, and carry out five cycles of the initial-stiffness method. Compute the percentage error in the resulting $u_{2}$ .
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17.14 The two bars shown are fixed to rigid walls at their outer ends and are welded together where load $P$ is applied. The material behavior is shown in the sketch. Use the tangent-stiffness method and the successive load increments $\Delta P = 20$ , $\Delta P = 10$ , and $\Delta P = -30$ . Determine the corresponding values of displacement $D$ . Show results on a plot of $P$ versus $D$ .
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<details>
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<summary>text_image</summary>
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10
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10
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D
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P
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A = 1.0 throughout
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</details>
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Problem 17.14
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17.15 For the bar sketched in Problem 17.14, apply the single load increment $\Delta P = 10$ after the yield point value of load $P$ is reached. Determine the value of displacement $D$ predicted by five cycles of the initial-stiffness algorithm.
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17.16 The horizontal bar in the sketch is perfectly rigid and is constrained to remain horizontal as load $P$ and displacement $v$ increase. The three vertical bars are elastic-perfectly plastic with $A = 1$ , $E = 1$ and $L = 2$ . These bars have the respective yield point loads $F_{1} = 2$ , $F_{2} = 4$ , and $F_{3} = 6$ . Use the tangent-stiffness method to generate the $P$ versus $v$ relation. Use three steps. Scale each step so that one bar begins to yield at the end of the step.
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<details>
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<summary>text_image</summary>
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①
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②
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③
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L
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P, v
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</details>
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Problem 17.16
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17.17 A weightless and rigid block $B$ is pushed down in a frictionless guide by force $P$ . The force versus deflection plot for each of two supporting bars is given in the sketch. Solve for displacement $D$ of block $B$ under a force $P = 24 \mathrm{~N}$ , as follows.
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(a) Determine the exact solution.
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(b) Use the tangent-stiffness method. Let the first load increment be $\Delta P = 19\mathrm{N}$ .
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(c) Use the initial-stiffness method. Apply one load increment $\Delta P = 5$ N starting from the exact solution at P = 19 N.
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<!-- source-page: 558 -->
|
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|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
P
|
||
B
|
||
100 mm
|
||
D
|
||
Force, N
|
||
10
|
||
9
|
||
1
|
||
2
|
||
Deflection, mm
|
||
0
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||
1
|
||
4
|
||
</details>
|
||
|
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Problem 17.17
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17.18 Assume that members of a truss carry only uniaxial stress, and that members in compression will buckle elastically at their critical loads without yielding. Tensile members are elastic-perfectly plastic. Outline a tangent-stiffness algorithm for computation of the displacements produced by applied loads. For simplicity, assume that load reversal does not occur in any member.
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17.19 In both the initial-stiffness algorithm and the tangent-stiffness algorithm, factor m of Eq. 17.3-4 is used only to correct a trial solution after it is computed. Can the correction be anticipated instead? Explain.
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|
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# Section 17.4
|
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|
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17.20 Verify Eq. 17.4-7.
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17.21 Show that Eq. 17.4-13 becomes $F = \sigma_{a} - \sigma_{Y}$ when a state of uniaxial stress $\sigma_{a}$ is defined in the following ways. Let $\{\alpha\} = \{\mathbf{0}\}$ .
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|
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(a) $\sigma_{x} = \sigma_{a}, \sigma_{y} = \sigma_{z} = \tau_{xy} = \tau_{yz} = \tau_{zx} = 0.$
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(b) $\sigma_{x} = \sigma_{y} = \tau_{xy} = \sigma_{a} / 2, \sigma_{z} = \tau_{yz} = \tau_{zx} = 0.$
|
||
(c) $\sigma_{x} = 0.8\sigma_{a}, \sigma_{y} = 0.2\sigma_{a}, \tau_{xy} = 0.4\sigma_{a}, \sigma_{z} = \tau_{yz} = \tau_{zx} = 0.$
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|
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17.22 Verify that Eq. 17.4-14 follows from Eq. 17.4-13.
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17.23 Imagine that plastic action in bending is to be modeled and that several sampling points are used in the thickness direction (d in the sketch).
|
||
|
||
(a) Why should the sampling points pertain to a trapezoidal or Simpson quadrature rule rather than to a Gauss–Legendre quadrature rule?
|
||
(b) Imagine that the stress distribution shown prevails across the depth of a beam of rectangular cross section. What is the percentage error of the computed bending moment $M_c$ , if $M_c$ is integrated from the stress distribution using a two-point Gauss rule? Repeat the calculation using a three-point Gauss rule.
|
||
(c) Repeat part (b), but use trapezoidal rules instead of Gauss rules. Try three, five, seven, and then nine sampling points.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
d
|
||
σₐ
|
||
σₐ
|
||
d/4
|
||
d/4
|
||
d/4
|
||
d/4
|
||
</details>
|
||
|
||
Problem 17.23
|
||
|
||
<!-- source-page: 559 -->
|
||
|
||
17.24 Imagine that a state of uniaxial stress causes plastic strains. Show that Eq. 17.4-15 yields the correct value of $\epsilon_{\mathrm{ef}}^{p}$ if
|
||
|
||
(a) The stress is parallel to the $x$ axis.
|
||
|
||
(b) The stress acts at 45 degrees to the $x$ and $y$ axes.
|
||
|
||
# Section 17.5
|
||
|
||
17.25 Describe the steps of a tangent-stiffness solution algorithm in which each load increment causes a single sampling point to be brought to the initiation of yielding.
|
||
|
||
17.26 Consider a plane structure modeled by finite elements. The material is isotropic but brittle: it cracks when the tensile stress in any direction exceeds a value $\sigma_{t}$ . Outline a tangent-stiffness algorithm for predicting deformations caused by increasing load. How will the collapse load be detected by this algorithm?
|
||
|
||
17.27 Imagine that corrective loads $\{\Delta R_{c}\}$ are to be computed for a mesh of elements having internal d.o.f. Should internal d.o.f. carry loads that result from $\{\sigma\}$ , or should these loads be omitted from internal d.o.f.? If carried, should they be distributed to remaining d.o.f. (that is, condensed) by means of elastic element stiffness equations?
|
||
|
||
17.28 Imagine that a plane beam of rectangular cross section is modeled by plane finite elements. The material is linearly elastic, but elastic moduli in tension and compression are different. Outline an algorithm that will calculate the stresses produced by a pure bending load. What are the comparative merits of tangent-stiffness and initial-stiffness solutions?
|
||
|
||
# Section 17.6
|
||
|
||
17.29 Consider an elastic-perfectly plastic material with an associated flow rule. The constitutive law for such a material is given by Eqs. 17.4-7 through 17.4-9 with $C = 0$ , $\partial F / \partial W_p = 0$ , and $F = Q$ . Equation 17.4-9 can be written as $[\mathbf{E}_{\mathrm{ep}}] = [\mathbf{E}] + [\mathbf{E}_p]$ , where $[\mathbf{E}_p] = -[\mathbf{E}]\{\partial F / \partial \sigma\} \{\mathbf{C}_\lambda\}^T$ . Using Eq. 17.4-7, show that $[\mathbf{E}_p]$ is negative semidefinite.
|
||
|
||
17.30 Starting with the basic definition of the rate of internal work for an element $e$ as
|
||
|
||
$$
|
||
\dot {W} _ {e} ^ {\mathrm{int}} = \int_ {V _ {e}} \{\dot {\epsilon} \} ^ {T} \{\sigma \} d V
|
||
$$
|
||
|
||
show that the rate of internal work for the entire structure is $\dot{W}^{\mathrm{int}} = \{\dot{\mathbf{D}}\}^T\{\mathbf{R}^{\mathrm{int}}\}$ .
|
||
|
||
17.31 Starting with $\dot{W}_n^{\mathrm{int}} = \{\dot{\mathbf{D}}\}_n^T\{\mathbf{R}^{\mathrm{int}}\}_n$ , show that for linearly elastic material behavior, $W_n^{\mathrm{int}} = \frac{1}{2}\{\mathbf{D}\}_n^T[\mathbf{K}]\{\mathbf{D}\}_n$ . Note: $\frac{d}{dt} (\{\mathbf{D}\}^T[\mathbf{K}]\{\mathbf{D}\}) = 2\{\dot{\mathbf{D}}\}^T[\mathbf{K}]\{\mathbf{D}\}$ if [K] is symmetric.
|
||
|
||
17.32 Verify that Eq. 17.6-3 results from Eq. 17.6-2.
|
||
|
||
17.33 Of the energies $W^{int}$ , $W^{ext}$ , and T, and the energy rates $\dot{W}^{int}$ and $\dot{W}^{ext}$ , which are always nonnegative and which can be positive, zero, or negative? Assume that the material is linearly elastic.
|
||
|
||
<!-- source-page: 560 -->
|
||
|
||
17.34 Repeat the example of Section 17.6 using different time steps and longer analysis duration.
|
||
17.35 Repeat the example of Section 17.6 using the nonlinearly elastic stiffening material shown (in which loading and unloading are both on the same path). It will be necessary to write a subroutine similar to Fig. 17.6-1 for this constitutive law. Note that the transition between slopes E and $E_{t}$ may occur during both loading and unloading. Analyze response for approximately one wave traversal along the bar. Carefully address: (a) the largest stable time step; (b) the correctness of the stress wave shape, and (c) the wave arrival time(s).
|
||
|
||
|
||
|
||
<details>
|
||
<summary>line</summary>
|
||
|
||
| ε | σ |
|
||
| ------- | ----- |
|
||
| 0 | 0 |
|
||
| E | 30(10)⁶ psi |
|
||
| E_t | 4E |
|
||
</details>
|
||
|
||
Problem 17.35
|
||
|
||
17.36 Derive Eqs. 17.6-11 through 17.6-13.
|
||
|
||
# Section 17.7
|
||
|
||
17.37 (a) If $P = 1.884P_{\mathrm{cr}}$ in Fig. 17.7-1b, the tip of the column rotates $120^{\circ}$ from its unloaded position. To what other equilibrium configuration might the numerical process converge? Answer qualitatively, without calculation.
|
||
|
||
(b) Repeat part (a) if force $P$ is supplemented by a vertically directed tip force $Q = 0.001P_{\mathrm{cr}}$ . Sketch your answer.
|
||
|
||
17.38 Verify Eq. 17.7-2, and show that it agrees with the formula $M_L L_T^2 / 2EI$ of elementary beam theory if $M_L$ is small.
|
||
|
||
17.39 (a) Derive the expression for $u_{2}$ , Eq. 17.7-4b.
|
||
|
||
(b) Show that $u_{2} = 0$ for a rigid-body rotation $\alpha$ about node 1, starting from $\alpha_0 = 0$ .
|
||
|
||
17.40 Imagine that $M_L = 0$ in Fig. 17.7-1a and that the right end of the cantilever beam has rotated about $45^\circ$ under the action of force $P$ . Let the beam be divided into several elements. For the rightmost element, qualitatively sketch loads $[\mathbf{k}']\{\mathbf{d}'\}$ of Eq. 17.7-5 in the local system $x'y'$ and loads $\{\mathbf{r}\}$ of Eq. 17.7-6 in the global system $xy$ . Show these loads in the directions they actually act. Which of these loads will be zero after the iterative numerical process has converged?
|
||
|
||
17.41 A long uniform beam has bending stiffness $EI$ , weight $q$ per unit length, and rests on a flat horizontal surface, as shown. A vertical force $F$ , where $F < qL_T / 2$ , is applied to one end.
|
||
|
||
(a) Analytically determine the length $L_{s}$ that lifts off the surface $(L_{s} < L_{T})$ . (b) Imagine that the beam is modeled by several elements. Outline a numerical algorithm for calculating length $L_{s}$ .
|