김경종
656 lines
19 KiB
Markdown
656 lines
19 KiB
Markdown
<!-- source-page: 231 -->
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When using the load lumping procedure it should be recognized that the nodal point loads are, in general, calculated only approximately, and if a coarse finite element mesh is employed, the resulting solution may be very inaccurate. Indeed, in some cases when higher-order finite elements are used, surprising results are obtained. Figure 4.7 demonstrates such a case (see also Example 5.12).
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<details>
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<summary>text_image</summary>
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Thickness = 1 cm
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2 cm
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x
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6 cm
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y
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p
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</details>
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(a) Problem
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$$
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\begin{array}{l} p = 3 0 0 \mathrm{N} / \mathrm{cm} ^ {2} \\ E = 3 \times 1 0 ^ {7} \mathrm{N} / \mathrm{cm} ^ {2} \\ \nu = 0. 3 \end{array}
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$$
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<details>
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<summary>text_image</summary>
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y
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x
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A
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B
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C
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x
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p/3
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4p/3
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p/3
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</details>
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(b) Finite element model with consistent loading
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<table><tr><td>Integration point</td><td> $\tau_{xx}$ </td><td> $\tau_{yy}$ </td><td> $\tau_{xy}$ </td></tr><tr><td>A</td><td>300.00</td><td>0.0</td><td>0.0</td></tr><tr><td>B</td><td>300.00</td><td>0.0</td><td>0.0</td></tr><tr><td>C</td><td>300.00</td><td>0.0</td><td>0.0</td></tr></table>
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(All stresses have units of N/cm²)
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<details>
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<summary>text_image</summary>
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y
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x
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A
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B
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C
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x
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p/2
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p
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p/2
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</details>
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(c) Finite element model with lumped loading
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<table><tr><td>Integration point</td><td> $\tau_{xx}$ </td><td> $\tau_{yy}$ </td><td> $\tau_{xy}$ </td></tr><tr><td>A</td><td>301.41</td><td>-7.85</td><td>-24.72</td></tr><tr><td>B</td><td>295.74</td><td>-9.55</td><td>0.0</td></tr><tr><td>C</td><td>301.41</td><td>-7.85</td><td>24.72</td></tr></table>
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(All stresses have units of N/cm²)
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(3 × 3 Gauss points are used, see Table 5.7)
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Figure 4.7 Some sample analysis results with and without consistent loading
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Considering dynamic analysis, the inertia effects can be thought of as body forces. Therefore, if a lumped mass matrix is employed, little might be gained by using a consistent load vector, whereas consistent nodal point loads should be used if a consistent mass matrix is employed in the analysis.
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# 4.2.5 Exercises
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4.1. Use the procedure in Example 4.2 to formally derive the principle of virtual work for the one-dimensional bar shown.
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<!-- source-page: 232 -->
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<details>
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<summary>text_image</summary>
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A(x)
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x, u
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f^B_x
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R
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L
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E = Young's modulus
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</details>
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The differential equations of equilibrium are
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$$
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E \frac {\partial}{\partial x} \left(A \frac {\partial u}{\partial x}\right) + f _ {x} ^ {B} = 0
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$$
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$$
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E A \left. \frac {\partial u}{\partial x} \right| _ {x = L} = R
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$$
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4.2. Consider the structure shown.
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(a) Write down the principle of virtual displacements by specializing the general equation (4.7) to this case.
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(b) Use the principle of virtual work to check whether the exact solution is
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$$
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\tau (x) = \left(\frac {7 2}{7 3} + \frac {2 4 x}{7 3 L}\right) \frac {F}{A _ {0}}
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$$
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Use the following three virtual displacements: (i) $\overline{u}(x) = a_0x$ , (ii) $\overline{u}(x) = a_0x^2$ , (iii) $\overline{u}(x) = a_0x^3$ .
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(c) Solve the governing differential equations of equilibrium,
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$$
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E \frac {\partial}{\partial x} \left(A \frac {\partial u}{\partial x}\right) = 0
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$$
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$$
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E A \left. \frac {\partial u}{\partial x} \right| _ {x = L} = F
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$$
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(d) Use the three different virtual displacement patterns given in part (b), substitute into the principle of virtual work using the exact solution for the stress [from part (c)], and explicitly show that the principle holds.
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<details>
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<summary>text_image</summary>
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x
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A₀
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L
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</details>
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F = total force exerted on right end
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$E =$ Young's modulus
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$A(x) = A_0(1 - x / 4L)$
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<!-- source-page: 233 -->
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# 4.3. Consider the bar shown.
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(a) Solve for the exact displacement response of the structure.
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(b) Show explicitly that the principle of virtual work is satisfied with the displacement patterns (i) $\overline{u} = ax$ and (ii) $\overline{u} = ax^2$ .
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(c) Identify a stress $\tau_{xx}$ for which the principle of virtual work is satisfied with pattern (ii) but not with pattern (i).
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<details>
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<summary>text_image</summary>
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A = A₀(4 - 3x/L)
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f^D
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x, u
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L
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</details>
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$f^{D}=$ constant force per unit length
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Young's modulus E
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4.4. For the two-dimensional body shown, use the principle of virtual work to show that the body forces are in equilibrium with the applied concentrated nodal loads.
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$$
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f _ {x} ^ {B} = 1 0 (1 + 2 x) \mathrm{N} / \mathrm{m} ^ {3}
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$$
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$$
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f _ {y} ^ {B} = 2 0 (1 + y) \mathrm{N} / \mathrm{m} ^ {3}
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$$
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$$
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R _ {1} = 6 0 \mathrm{N}
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$$
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$$
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R _ {2} = 4 5 \mathrm{N}
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$$
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$$
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R _ {3} = 1 5 \mathrm{N}
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$$
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<details>
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<summary>text_image</summary>
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Unit thickness
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2 m
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f_y^B
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f_x^B
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1 m
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y
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x
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R_1
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R_2
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R_3
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</details>
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<!-- source-page: 234 -->
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4.5. Idealize the bar structure shown as an assemblage of 2 two-node bar elements.
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(a) Calculate the equilibrium equations $\mathbf{K}\mathbf{U} = \mathbf{R}$ .
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(b) Calculate the mass matrix of the element assemblage.
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<details>
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<summary>text_image</summary>
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60
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80
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η
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x
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A = A₀(1 - η/120); η ≤ 60
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f^B(x)
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→ 20 A₀f₁
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f^B(x) = 0.1f₁ force/unit volume
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E = Young's modulus
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ρ = mass density
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f₁ₓ
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10.0
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t₁
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Time
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</details>
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4.6. Consider the disk with a centerline hole of radius 20 shown spinning at a rotational velocity of $\omega$ radians/second.
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<details>
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<summary>text_image</summary>
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20
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60
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80
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ω
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3.0
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x, u
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1.0
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E = Young's modulus
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ρ = mass density
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ν = Poisson's ratio
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</details>
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Idealize the structure as an assemblage of 2 two-node elements and calculate the steady-state (pseudostatic) equilibrium equations. (Note that the strains are now $\partial u/\partial x$ and u/x, where u/x is the hoop strain.)
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4.7. Consider Example 4.5 and the state at time t = 2.0 with $U_{1}(t) = 0$ at all times.
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(a) Use the finite element formulation given in the example to calculate the static nodal point displacements and the element stresses.
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(b) Calculate the reaction at the support.
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<!-- source-page: 235 -->
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(c) Let the calculated finite element solution be $u^{FE}$ . Calculate and plot the error r measured in satisfying the differential equation of equilibrium, i.e.,
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$$
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r = E \left[ \frac {\partial}{\partial x} \left(A \frac {\partial u ^ {\mathrm{FE}}}{\partial x}\right) \right] + f _ {x} ^ {B} A
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$$
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(d) Calculate the strain energy of the structure as evaluated in the finite element solution and compare this strain energy with the exact strain energy of the mathematical model.
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4.8. The two-node truss element shown, originally at a uniform temperature, $20^{\circ}$ C, is subjected to a temperature variation
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$$
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\theta = (1 0 x + 2 0) ^ {\circ} \mathrm{C}
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$$
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Calculate the resulting stress and nodal point displacement. Also obtain the analytical solution, assuming a continuum, and briefly discuss your results.
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<details>
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<summary>text_image</summary>
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E = 200,000
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A = 1
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α = 1 × 10⁻⁶ (per °C)
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</details>
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4.9. Consider the finite element analysis illustrated.
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<details>
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<summary>other</summary>
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| Position | U1 (in) | U2 (in) | U3 (in) | U4 (in) | U5 (in) | U6 (in) | U7 (in) | U8 (in) | U9 (in) | U10 (in) | U11 (in) | U12 (in) | U13 (in) |
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|----------|---------|---------|---------|---------|---------|---------|---------|---------|---------|----------|----------|----------|----------|
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| U1 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U2 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U3 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U4 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U5 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U6 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U7 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U8 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U9 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U10 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U11 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U12 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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| U13 | 3 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 | 4 |
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</details>
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Plane stress condition (thickness t). All elements are 4-node elements
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(a) Begin by establishing the typical matrix $\mathbf{B}$ of an element for the vector $\hat{\mathbf{u}}^T = [u_1 v_1 u_2 v_2 u_3 v_3 u_4 v_4]$ .
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(b) Calculate the elements of the K matrix, $K_{U_{2}U_{2}}$ , $K_{U_{6}U_{7}}$ , $K_{U_{7}U_{6}}$ , and $K_{U_{5}U_{12}}$ of the structural assemblage.
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(c) Calculate the nodal load $R_{9}$ due to the linearly varying surface pressure distribution.
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<!-- source-page: 236 -->
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<details>
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<summary>text_image</summary>
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v2
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u2
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2
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1
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3 in
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v3
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u3
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3
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4 in
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v1
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u1
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v4
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u4
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</details>
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# 4.10. Consider the simply supported beam shown.
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(a) Assume that usual beam theory is employed and use the principle of virtual work to evaluate the reactions $R_{1}$ and $R_{2}$ .
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<details>
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<summary>text_image</summary>
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P
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a b
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R₁ R₂ h
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</details>
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(b) Now assume that the beam is modeled by a four-node finite element. Show that to be able to evaluate $R_{1}$ and $R_{2}$ as in part (a) it is necessary that the finite element displacement functions can represent the rigid body mode displacements.
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<details>
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<summary>text_image</summary>
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P
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a b
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h
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R₁ R₂
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</details>
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4.11. The four-node plane stress element shown carries the initial stresses
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$$
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\tau_ {x x} ^ {I} = 0 \mathrm{MPa}
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$$
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$$
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\tau_ {y y} ^ {I} = 1 0 \mathrm{MPa}
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$$
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$$
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\tau_ {x y} ^ {I} = 2 0 \mathrm{MPa}
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$$
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<!-- source-page: 237 -->
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(a) Calculate the corresponding nodal point forces $R_{i}$ .
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(b) Evaluate the nodal point forces $\mathbf{R}_s$ equivalent to the surface tractions that correspond to the element stresses. Check your results using elementary statics and show that $\mathbf{R}_s$ is equal to $\mathbf{R}_l$ evaluated in part (a). Explain why this result makes sense.
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(c) Derive a general result: Assume that any stresses are given, and $R_{l}$ and $R_{s}$ are calculated. What conditions must the given stresses satisfy in order that $R_{l} = R_{s}$ , where the surface tractions in $R_{s}$ are obtained from equation (b) in Example 4.2?
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<details>
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<summary>text_image</summary>
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60 mm
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30 mm
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y
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x
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</details>
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Young's modulus E Poisson's ratio v Thickness = 0.5 mm
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4.12. The four-node plane strain element shown is subjected to the constant stresses
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$$
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\tau_ {x x} = 2 0 \mathrm{psi}
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$$
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$$
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\tau_ {y y} = 1 0 \mathrm{psi}
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$$
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$$
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\tau_ {x y} = 1 0 \mathrm{psi}
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$$
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Calculate the nodal point displacements of the element.
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<details>
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<summary>text_image</summary>
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3 in
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2
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1
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2 in
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3
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4
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y
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x
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</details>
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Young's modulus $E = 30 \times 10^{6}$ psi
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Poisson's ratio $\nu = 0.30$
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<!-- source-page: 238 -->
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4.13. Consider element 2 in Fig. E4.9.
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(a) Show explicitly that
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$$
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\mathbf {F} ^ {(2)} = \int_ {V ^ {(2)}} \mathbf {B} ^ {(2) T} \boldsymbol {\tau} ^ {(2)} d V ^ {(2)}
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$$
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(b) Show that the element nodal point forces $\mathbf{F}^{(2)}$ are in equilibrium.
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4.14. Assume that the element stiffness matrices $K_{A}$ and $K_{B}$ corresponding to the element displacements shown have been calculated. Assemble these element matrices directly into the global structure stiffness matrix with the displacement boundary conditions shown.
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<details>
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<summary>text_image</summary>
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U₁
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U₄
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U₃
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U₂
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A
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B
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Structural assemblage
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</details>
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Structural assemblage and degrees of freedom
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<details>
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<summary>text_image</summary>
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v2
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u2
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v1
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u1
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A
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v3
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u3
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θ2
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v2
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u2
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θ1
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v1
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u1
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B
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Individual elements
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</details>
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$$
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\mathbf {K} _ {A} = \left[ \begin{array}{l l l l l l} a _ {1 1} & a _ {1 2} & a _ {1 3} & a _ {1 4} & a _ {1 5} & a _ {1 6} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} & a _ {2 4} & a _ {2 5} & a _ {2 6} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} & a _ {3 4} & a _ {3 5} & a _ {3 6} \\ a _ {4 1} & a _ {4 2} & a _ {4 3} & a _ {4 4} & a _ {4 5} & a _ {4 6} \\ a _ {5 1} & a _ {5 2} & a _ {5 3} & a _ {5 4} & a _ {5 5} & a _ {5 6} \\ a _ {6 1} & a _ {6 2} & a _ {6 3} & a _ {6 4} & a _ {6 5} & a _ {6 6} \end{array} \right] \left[ \begin{array}{l} u _ {1} \\ v _ {1} \\ u _ {2} \\ v _ {2} \\ u _ {3} \\ v _ {3} \end{array} \right]
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$$
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$$
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\mathbf {K} _ {B} = \left[ \begin{array}{l l l l l l} b _ {1 1} & b _ {1 2} & b _ {1 3} & b _ {1 4} & b _ {1 5} & b _ {1 6} \\ b _ {2 1} & b _ {2 2} & b _ {2 3} & b _ {2 4} & b _ {2 5} & b _ {2 6} \\ b _ {3 1} & b _ {3 2} & b _ {3 3} & b _ {3 4} & b _ {3 5} & b _ {3 6} \\ b _ {4 1} & b _ {4 2} & b _ {4 3} & b _ {4 4} & b _ {4 5} & b _ {4 6} \\ b _ {5 1} & b _ {5 2} & b _ {5 3} & b _ {5 4} & b _ {5 5} & b _ {5 6} \\ b _ {6 1} & b _ {6 2} & b _ {6 3} & b _ {6 4} & b _ {6 5} & b _ {6 6} \end{array} \right] \begin{array}{l} u _ {1} \\ v _ {1} \\ \theta_ {1} \\ u _ {2} \\ v _ {2} \\ \theta_ {2} \end{array}
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$$
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4.15. Assume that the element stiffness matrices $K_{A}$ and $K_{B}$ corresponding to the element displacements shown have been calculated. Assemble these element matrices directly into the global structure stiffness matrix with the displacement boundary conditions shown.
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<details>
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<summary>text_image</summary>
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U2
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U1
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A
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U4
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U5
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U3
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B
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</details>
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$$
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\mathsf {K} _ {A} = \left[ \begin{array}{c c c c} a _ {1 1} & \ldots & \ldots & a _ {1 8} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\ a _ {8 1} & \ldots & \ldots & a _ {8 8} \end{array} \right] \begin{array}{l} u _ {1} \\ v _ {1} \\ u _ {2} \\ \vdots \\ v _ {4} \end{array}
|
||
$$
|
||
|
||
$$
|
||
\mathbf {K} _ {B} = \left[ \begin{array}{c c c c} b _ {1 1} & \dots & \dots & b _ {1 6} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\ b _ {6 1} & \dots & \dots & b _ {6 6} \end{array} \right] \begin{array}{l} u _ {1} \\ v _ {1} \\ \theta_ {1} \\ \vdots \\ \theta_ {2} \end{array}
|
||
$$
|
||
|
||
<!-- source-page: 239 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
v₂
|
||
u₂
|
||
A
|
||
v₁
|
||
u₁
|
||
v₃
|
||
u₃
|
||
v₄
|
||
u₄
|
||
</details>
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
θ₁
|
||
v₁
|
||
u₁
|
||
B
|
||
θ₂
|
||
v₂
|
||
u₂
|
||
</details>
|
||
|
||
4.16. Consider Example 4.11. Assume that at the support $A$ , the roller allows a displacement only along a slope of 30 degrees to the horizontal direction. Determine the modifications necessary in the solution in Example 4.11 to obtain the structure matrix $\mathbf{K}$ for this situation.
|
||
|
||
(a) Consider imposing the zero displacement condition exactly.
|
||
(b) Consider imposing the zero displacement condition using the penalty method.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Quadrilateral plane
|
||
stress element
|
||
U₁
|
||
A
|
||
30°
|
||
U₀
|
||
</details>
|
||
|
||
4.17. Consider the beam element shown. Evaluate the stiffness coefficients $k_{11}$ and $k_{12}$ .
|
||
|
||
(a) Obtain the exact coefficients from the solution of the differential equation of equilibrium (using the mathematical model of Bernoulli beam theory).
|
||
(b) Obtain the coefficients using the principle of virtual work with the Hermitian beam functions (see Example 4.16).
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
h(x) = h₀ (1 + x/L)
|
||
u₁
|
||
u₃
|
||
u₂
|
||
x
|
||
h₀
|
||
u₄
|
||
u₅
|
||
u₆
|
||
L
|
||
</details>
|
||
|
||
Young's modulus E Unit thickness
|
||
|
||
<!-- source-page: 240 -->
|
||
|
||
4.18. Consider the two-element assemblage shown.
|
||
|
||
(a) Evaluate the stiffness coefficients $K_{11}$ , $K_{14}$ for the finite element idealization.
|
||
(b) Calculate the load vector of the element assemblage.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
4.0
|
||
4.0
|
||
3
|
||
U2
|
||
U1
|
||
1
|
||
Element 1
|
||
Element 2
|
||
U5
|
||
4
|
||
U4
|
||
2
|
||
U3
|
||
x
|
||
y
|
||
1.0
|
||
p
|
||
E = 200,000
|
||
v = 0.3
|
||
</details>
|
||
|
||
Plane stress, thickness = 0.1
|
||
|
||
4.19. Consider the two-element assemblage in Exercise 4.18 but now assume axisymmetric conditions. The $y$ -axis is the axis of revolution.
|
||
|
||
(a) Evaluate the stiffness coefficients $K_{11}$ , $K_{14}$ for the finite element idealization.
|
||
(b) Evaluate the corresponding load vector.
|
||
|
||
4.20. Consider Example 4.20 and let the loading on the structure be $R_r = f_1(t) \cos \theta$ .
|
||
|
||
(a) Establish the stiffness matrix, mass matrix, and load vector of the three-node element
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y, v
|
||
f₁(t)
|
||
1
|
||
z, w
|
||
10
|
||
2
|
||
1
|
||
3
|
||
x (or r), u
|
||
</details>
|
||
|
||
$$
|
||
E = 2 0 0, 0 0 0
|
||
$$
|
||
|
||
$$
|
||
\nu = 0. 3
|
||
$$
|
||
|
||
$$
|
||
\rho = \text { mass density }
|
||
$$
|