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2.2: Розширена тема - Локальна рівновага з принципу віртуальної роботи
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$\frac{1}{2}\sigma_{ij} \delta u_{i,j} + \frac{1}{2}\sigma_{ji}\delta u_{j,i} = \sigma_{ij} \delta u_{i,j}$ \[\sigma_{ij} (\delta u_{i})_{,j} = (\sigma_{ij}\delta u_{i})_{,j} - (\sigma_{ij})_{,j} \... $\frac{1}{2}\sigma_{ij} \delta u_{i,j} + \frac{1}{2}\sigma_{ji}\delta u_{j,i} = \sigma_{ij} \delta u_{i,j}$ $\sigma_{ij} (\delta u_{i})_{,j} = (\sigma_{ij}\delta u_{i})_{,j} - (\sigma_{ij})_{,j} \delta u_{i} \label{2.2.7}$ $\int_{V} \sigma_{ij}\delta \epsilon_{ij}dV = \int_{V} \underbrace{(\sigma_{ij}\delta u_i)}_{A_i} \left._{,j} \right. \[\int_{S} \sigma_{ij}\delta u_i n_j dS - \int_{V} \sigma_{ij,j}\delta u_i dV = \int_{S} T_i \delta u_i dS$
2.6: Поглиблена тема - Математична теорія пучків
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$\int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{xx}}{\partial x} z (b dz) + b \int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{xz}}{\partial z} z (bdz) = 0$ \[\int_{-\frac{h}{2}}^{... $\int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{xx}}{\partial x} z (b dz) + b \int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{xz}}{\partial z} z (bdz) = 0$ $\int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{xz}}{\partial x} (b dz) + \int_{-\frac{h}{2}}^{\frac{h}{2}} \frac{\partial \sigma_{zz}}{\partial z} (bdz) = 0$
2.7: Рівновага в теорії помірно великих прогинів балок
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На малюнку (2.5.3) променевий елемент показаний в теорії малих прогинів (нескінченно мале обертання) і помірно великих прогинів (кінцеве обертання). \[\frac{dV}{dx} + \frac{d}{dx} \left(N\frac{dw}{dx}...На малюнку (2.5.3) променевий елемент показаний в теорії малих прогинів (нескінченно мале обертання) і помірно великих прогинів (кінцеве обертання). $\frac{dV}{dx} + \frac{d}{dx} \left(N\frac{dw}{dx}\right) + q = \frac{dV}{dx} + \frac{dN}{dx} \frac{dw}{dx} + N \frac{d^2w}{dx^2} + q = 0$ Усі рівняння рівноваги для нескінченно малих деформацій 3-D тіл і малих прогинів пучків задіяні лише статичні величини та їх градієнти $(M, V, N)$ .
11.8: Умова врожайності Tresca
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$\tau_1 = \frac{|\sigma_1 − \sigma_2|}{2}, \quad \tau_2 = \frac{|\sigma_2 − \sigma_1|}{2}, \quad \tau_3 = \frac{|\sigma_3 − \sigma_1|}{2}$ \[\tau_o = \text{ max } \left\{ \frac{|\sigma_1 − \sigma_2... $\tau_1 = \frac{|\sigma_1 − \sigma_2|}{2}, \quad \tau_2 = \frac{|\sigma_2 − \sigma_1|}{2}, \quad \tau_3 = \frac{|\sigma_3 − \sigma_1|}{2}$ $\tau_o = \text{ max } \left\{ \frac{|\sigma_1 − \sigma_2|}{2}, \frac{|\sigma_2 − \sigma_3|}{2}, \frac{|\sigma_3 − \sigma_1|}{2} \right\}$ $\text{ max } \{|\sigma_1 − \sigma_2|, |\sigma_2 − \sigma_3|, |\sigma_3 − \sigma_1|\} = \sigma_y$ Вплив гідростатичного тиску на врожайність можна легко оцінити, розглянувши $\sigma_1 = \sigma_2 = \sigma_3 = p$ .
3.4: Закон Гука в узагальнених кількостях для пучків
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\[N = \int_{A} \sigma_{xx} dA = \int_{A} E \epsilon_{xx}dA = E\int_{A} (\epsilon^{\circ} + z\kappa) dA \\ = E \int_{A} \epsilon^{\circ} dA + E \int_{A} \kappa z dA = E\epsilon^{\circ} \int_{A} dA + E\... $N = \int_{A} \sigma_{xx} dA = \int_{A} E \epsilon_{xx}dA = E\int_{A} (\epsilon^{\circ} + z\kappa) dA \\ = E \int_{A} \epsilon^{\circ} dA + E \int_{A} \kappa z dA = E\epsilon^{\circ} \int_{A} dA + E\kappa \int_{A} z dA$ $N = \int_{A} \sigma_{xx}z dA = \int_{A} E(\epsilon^{\circ} + z\kappa)z dA \\ = E\epsilon^{\circ} \int_{A} z dA + E\kappa \int_{A} z^2 dA$
1.10: Відносини деформації-переміщення для циркулюючих пластин
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Багато простих розчинів закритої форми можна отримати для кругових і кільцевих пластин при різних граничних і навантажувальних умовах. $\epsilon_{\theta \theta}^{\circ} = \frac{u}{r}$ \[\epsilon_{rr...Багато простих розчинів закритої форми можна отримати для кругових і кільцевих пластин при різних граничних і навантажувальних умовах. $\epsilon_{\theta \theta}^{\circ} = \frac{u}{r}$ $\epsilon_{rr}(r,z) = \epsilon_{rr}^{\circ}(r) + z\kappa_{rr}$ $\epsilon_{\theta \theta}(r,z) = \epsilon_{\theta \theta}^{\circ}(r) + z\kappa_{\theta \theta}$ Можна помітити, що вираз для радіальних деформацій і кривизни ідентичні виразу балки $r$ при заміні на $x$ .
8.5: Стабільність у напрузі
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$\delta V = \delta (lA) = \delta lA + l \delta A = 0$ $\delta \epsilon = \frac{\delta l}{l}= − \frac{\delta A}{A}$ $\delta \prod = \int_{V} \sigma \delta \epsilon dv + Pl \frac{\delta A}{A}$ ... $\delta V = \delta (lA) = \delta lA + l \delta A = 0$ $\delta \epsilon = \frac{\delta l}{l}= − \frac{\delta A}{A}$ $\delta \prod = \int_{V} \sigma \delta \epsilon dv + Pl \frac{\delta A}{A}$ $\delta^2 \prod = \int_{V} \delta \sigma \delta \epsilon dv − Pl \frac{\delta A \delta A}{A^2}$ $lA \delta \sigma \delta \epsilon = Pl \delta \epsilon \delta \epsilon$ $\delta \sigma = \frac{P}{A} \delta \epsilon = \sigma \delta \epsilon$
6.6: Жорсткі пластини
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$\bar{w} = \frac{w(r)}{w^p_o} = \left[ 2 \left( \frac{r}{a} \right)^2 \ln \frac{r}{a} + \frac{3 + \nu}{1+ \nu} \left( 1- \left( \frac{r}{a}\right)^2 \right) \right] \label{7.57a}$ \[P = \frac{1}{2}... $\bar{w} = \frac{w(r)}{w^p_o} = \left[ 2 \left( \frac{r}{a} \right)^2 \ln \frac{r}{a} + \frac{3 + \nu}{1+ \nu} \left( 1- \left( \frac{r}{a}\right)^2 \right) \right] \label{7.57a}$ $P = \frac{1}{2} Ehw_o \left[ \frac{8\pi}{3(1-\nu^2)} \left( \frac{h}{a}\right)^2 + \left(\frac{H}{a}\right)^3\right]$ є одна лінія прогину і ребра жорсткості сприяють жорсткості на вигин плити.
6: Відповідь на вигин пластин та оптимальна конструкція
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8.1: Прелюдія до стійкості пружних структур
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$\delta^2 \prod = \delta (\delta \prod) = 2C \delta u \delta u$ \[\prod (u) = \prod (u_o) + \left. \frac{d\prod}{du} \right|_{u=u_o} (u − u_o) + \frac{1}{2} \left. \frac{d^2\prod}{du^2}\right|_{u=u... $\delta^2 \prod = \delta (\delta \prod) = 2C \delta u \delta u$ $\prod (u) = \prod (u_o) + \left. \frac{d\prod}{du} \right|_{u=u_o} (u − u_o) + \frac{1}{2} \left. \frac{d^2\prod}{du^2}\right|_{u=u_o} (u − u_o)^2 + \dots$ Інкрементна зміна потенційної енергії $\Delta \prod = \prod (u) − \prod (u_o)$ при малій варіації $\delta u = u − u_o$ аргументу $\Delta \prod = \frac{d\prod}{du} \delta u + \frac{1}{2} \frac{d^2 u}{du^2} (\delta u)^2 = \delta \prod + \delta^2 \prod + \dots$
11.3: нестисливість
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$\dot{\epsilon}_{11} + \dot{\epsilon}_{22} + \dot{\epsilon}_{33} = 0, \quad \dot{\epsilon}_{kk} = 0$ Тому $\dot{\epsilon}_{11} + 2\dot{\epsilon}_{22} = 0$ або\(\dot{\epsilon}_{11} + 2\dot{\epsilon... $\dot{\epsilon}_{11} + \dot{\epsilon}_{22} + \dot{\epsilon}_{33} = 0, \quad \dot{\epsilon}_{kk} = 0$ Тому $\dot{\epsilon}_{11} + 2\dot{\epsilon}_{22} = 0$ або $\dot{\epsilon}_{11} + 2\dot{\epsilon}_{33} = 0$ . $\dot{\epsilon}_{22} = −0.5\dot{\epsilon}_{11}, \quad \dot{\epsilon}_{33} = −0.5\dot{\epsilon}_{11}$ $\nu = − \frac{\dot{\epsilon}_{22}}{\dot{\epsilon}_{11}} = − \frac{\dot{\epsilon}_{33}}{\dot{\epsilon}_{11}} = 0.5$