Zadanie 9 – Pałasz Sebastian, Spychała Jakub, M2

Przedstaw sposób znakowania wewnętrznego momentu gnącego w ramach płaskich. Dla ramy przedstawionej na rysunku wyznaczyć za pomocą twierdzenia Castigliano przemieszczenie poziome u_A^poz oraz kąt obrotu θ_A.

Znakowanie momentów wewnętrznych:

Rozwiązanie zadania:

Równania statyki:

$\sum_{}^{}{F_{x} = 0}$ : $R_{\text{Dx}} - q*l + P_{f} = 0\ \overset{\Rightarrow}{}\ R_{\text{Dx}} = q*l - P_{f}$

$\sum_{}^{}{F_{x} = 0}$ : $R_{\text{Dy}} + R_{A} = 0\ \overset{\Rightarrow}{}\ R_{\text{Dy}} = - R_{A}$

$\sum_{}^{}{M_{D} = 0}$ : $- R_{A}*l - q*l*\frac{3}{2}l + M_{f} + P_{f}*l = 0$

$R_{A} = - \frac{3}{2}q*l + \frac{M_{f}}{l} + P_{f}$

Momenty gnące:

$$M\left( x_{1} \right) = - \frac{q*x_{1}^{2}}{2} + P_{f}*x_{1} - M_{f}$$

$$M\left( x_{2} \right) = \left( - \frac{3}{2}*q*l + \frac{M_{f}}{l} + P_{f} \right)*x_{2} - \frac{q*l^{2}}{2} + P_{f}*l - M_{f}$$

M(x₃) = (−q*l+P_f) * x₃

$$\frac{\partial M(x_{1})}{P_{f}} = x_{1}$$

$$\frac{\partial M(x_{2})}{P_{f}} = x_{2} + l$$

$$\frac{\partial M(x_{3})}{P_{f}} = x_{3}$$

$$\frac{\partial M\left( x_{1} \right)}{M_{f}} = - x_{1}$$

$$\frac{\partial M\left( x_{2} \right)}{M_{f}} = \frac{M_{f}*x_{2}^{2}}{2l} - x_{2}$$

$$\frac{\partial M\left( x_{3} \right)}{M_{f}} = 0$$

Przemieszczenie poziome i kąt obrotu:

$$\frac{\partial V}{\partial P_{f}} = u_{A}^{\text{poz}} = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{l}{\left( - \frac{q*x_{1}^{2}}{2} + P_{f}*x_{1} - M_{f} \right)*\left( x_{1} \right)dx_{1} + \int_{0}^{l}{\left( \left( - \frac{3}{2}*q*l + \frac{M_{f}}{l} + P_{f} \right)*x_{2} - \frac{q*l^{2}}{2} + P_{f}*l - M_{f} \right)*\left( x_{2} + l \right)dx_{2}}} + \int_{0}^{2l}{\left( \left( - q*l + P_{f} \right)*x_{3} \right)*\left( x_{3} \right)dx_{3}} \right\rbrack$$

$$\frac{\partial V}{\partial P_{f}} = u_{A}^{\text{poz}} = \frac{1}{\text{EI}}$$

$$*\left\lbrack - \frac{q*x^{4}}{8}\left|_{0}^{l}{\ - \frac{3}{6}*q*l*x_{2}^{3}\left|_{0}^{l}{\ - \frac{3}{4}*q*l^{2}*x_{2}^{2}\left|_{0}^{l}{- \frac{q*l^{2}*x_{2}^{2}}{4}\left|_{0}^{l}{- \frac{q*l^{3}*x_{2}}{2}} \right.\ } \right.\ } \right.\ } \right.\ \left|_{0}^{\text{l\ }}{\frac{q*l*x_{3}^{3}}{3}\left|_{0\ }^{2l}\ \right.\ } \right.\ \right\rbrack$$

$$\frac{\partial V}{\partial P_{f}} = \frac{1}{\text{EI}}\left\lbrack - \frac{3}{24}*q*l^{4} - \frac{12}{24}*q*l^{4} - \frac{18}{24}*q*l^{4} - \frac{6}{24}*q*l^{4} - \frac{12}{24}*q*l^{4} - \frac{64}{24}*q*l^{4} \right\rbrack$$

$$\frac{\partial V}{\partial P_{f}} = - \frac{115}{24}*\frac{q*l^{4}}{\text{EI}}$$

$$\frac{\partial V}{\partial M_{f}} = \theta_{A} = \int_{0}^{l}\frac{M(x)}{\text{EI}}*\frac{\partial M(x)}{\partial M_{f}}\text{dx}$$

$$\frac{\partial V}{\partial M_{f}} = \theta_{A} = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{l}{\left( - \frac{q*x_{1}^{2}}{2} + P_{f}*x_{1} - M_{f} \right)*\left( - x_{1} \right)dx_{1} + \int_{0}^{l}{\left( \left( - \frac{3}{2}*q*l + \frac{M_{f}}{l} + P_{f} \right)*x_{2} - \frac{q*l^{2}}{2} + P_{f}*l - M_{f} \right)*\left( \frac{M_{f}*x_{2}^{2}}{2} - x_{2} \right)dx_{2} + \int_{0}^{2l}{\left( \left( - q*l + P_{f} \right)*x_{3} \right)*\left( 0 \right)dx_{3}}}} \right\rbrack$$

$$\frac{\partial V}{\partial M_{f}} = \theta_{A} = \frac{1}{\text{EI}}\left\lbrack \frac{q*x_{1}^{4}}{8}\left|_{0}^{l}{+ \frac{3}{6}*q*l*x_{2}^{3}\left|_{0}^{l}{+ \frac{q*l^{2}*x_{2}^{2}}{4}} \right.\ } \right.\ \left|_{0}^{l}{+ 0} \right.\ \right\rbrack$$

$$\frac{\partial V}{\partial M_{f}} = \theta_{A} = \frac{1}{\text{EI}}\left\lbrack \frac{3}{24}*q*l^{4} + \frac{12}{24}*q*l^{4} + \frac{6}{24}*q*l^{4} \right\rbrack$$

$$\theta_{A} = \frac{7}{8}*\frac{q*l^{4}}{\text{EI}}$$