Zadanie 11

Dla ramy przedstawionej na rysunku wyznaczyć za pomocą twierdzenia Castigliano przemieszczenie węzła A oraz obrót węzła B.

W punkcie A przykładamy siłę fikcyjną P_f o wartości P_f = 0, a w punkcie B moment fikcyjny M_f = 0 ∖ n

Z warunków równowagi określamy wartości reakcji podpór

$$\sum_{}^{}{F_{x} = 0\ \ \ \ \ \ \ R_{\text{Dx}} - R_{A} = 0}$$

R_Dx = R_A

$$\sum_{}^{}{F_{Y} = 0\ \ \ {- R}_{\text{Dy}} - qa - P_{f} = 0}$$

R_Dy= − qa − P_f

$$\sum_{}^{}{M_{D} = 0\ \ \ \ \ \ \ {- qa*\frac{1}{2}a + M}_{f} + R_{A}a - P_{f}(a + 2a\operatorname{tg}\alpha) = 0}$$

${R_{A}a = \frac{1}{2}qa^{2} - M}_{f} + P_{f}a(1 + 2\operatorname{tg}\alpha)$

$R_{A} = \frac{1}{2}qa - \frac{M_{f}}{a} + P_{f}(1 + 2\operatorname{tg}\alpha)$

$R_{\text{Dx}} = \frac{1}{2}qa - \frac{M_{f}}{a} + P_{f}(1 + 2\operatorname{tg}\alpha)$

Funkcje momentów zginających i ich pochodne cząstkowe w poszczególnych przedziałach

0≤x₁≤a

$$M(x_{1}) = R_{\text{Dx}}{*x}_{1} = \lbrack\frac{1}{2}qa - \frac{M_{f}}{a} + P_{f}(1 + 2\operatorname{tg}\alpha)\rbrack{*x}_{1}$$

$$\frac{\partial M(x_{1})}{\partial P_{f}} = (1 + 2\operatorname{tg}\alpha){*x}_{1}\text{\ \ \ \ \ \ \ \ }\frac{\partial M(x_{1})}{\partial M_{f}} = - \frac{x_{1}}{a}$$

0≤x₂≤a

$$M(x_{2}) = \ R_{\text{Dx}}*a - R_{\text{Dy}}*x_{2} - qx_{2}*\frac{1}{2}x_{2} = \lbrack\frac{1}{2}qa - \frac{M_{f}}{a} + P_{f}(1 + 2\operatorname{tg}\alpha)\rbrack*a + (qa + P_{f})*x_{2} - \frac{1}{2}gx_{2}^{2} = {\frac{1}{2}qa^{2} - M}_{f} + P_{f}a(1 + 2\operatorname{tg}\alpha) + (qa + P_{f})*x_{2} - \frac{1}{2}gx_{2}^{2}$$

$$\frac{\partial M(x_{2})}{\partial P_{f}} = (1 + 2\operatorname{tg}\alpha)*a + x_{2}\text{\ \ \ \ \ }\frac{\partial M(x_{2})}{\partial M_{f}} = - 1$$

$$\mathbf{0 \leq}\mathbf{x}_{\mathbf{3}}\mathbf{\leq}\frac{\mathbf{2}\mathbf{a}}{\cos\mathbf{\alpha}}$$

$$M(x_{3}) = R_{A}\cos\alpha*x_{3} - P_{f}\sin\alpha*x_{3} = \lbrack\frac{1}{2}qa - \frac{M_{f}}{a} + P_{f}(1 + 2\operatorname{tg}\alpha)\rbrack\operatorname{*cos}\alpha x_{3} - P_{f}\sin\alpha*x_{3}$$

$$\frac{\partial M(x_{3})}{\partial P_{f}} = (1 + 2\operatorname{tg}\alpha)\operatorname{*cos}\alpha x_{3} - \sin\alpha x_{3}$$

$$\frac{\partial M(x_{3})}{\partial M_{f}} = - \frac{\cos\alpha}{a}x_{3}$$

Kąt obrotu w węzła B

$$Q_{B} = \frac{\partial V}{\partial M_{f}} = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{a}{(\frac{1}{2}}\text{qa}x_{1})*( - \frac{x_{1}}{a})dx_{1} + \int_{0}^{a}{(\frac{1}{2}}qa^{2} + qax_{2} - \frac{1}{2}qx_{2}^{2})*( - 1)dx_{2} + \int_{0}^{\frac{2a}{\cos\alpha}}{(\frac{1}{2}}\text{qa}\cos\alpha*x_{3})*( - \frac{\cos\alpha}{a}x_{3})dx_{3} \right\rbrack = = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{a}{( - \frac{1}{2}}qx_{1}^{2})dx_{1} + \int_{0}^{a}{( - \frac{1}{2}}qa^{2} - qax_{2} + \frac{1}{2}qx_{2}^{2})dx_{2} + \int_{0}^{\frac{2a}{\cos\alpha}}{( - \frac{1}{2}}q{\cos\alpha}^{2}x_{3}^{2})dx_{3} \right\rbrack = = \frac{1}{\text{EI}}\left\lbrack \left( - \frac{1}{2}g\frac{x_{1}^{3}}{3} \right)\left| \frac{a}{0} + \left( - \frac{1}{2}qa^{2}x_{2} \right)\left| \frac{a}{0} + \left( - qa\frac{x_{2}^{2}}{2} \right)\left| \frac{a}{0} + \left( \frac{1}{2}q\frac{x_{2}^{3}}{3} \right)\left| \frac{a}{0}\ ( - \frac{1}{2}q\cos{\alpha^{2}\frac{x_{3}^{3}}{3})\left| \frac{\frac{2a}{\cos\alpha}}{0} \right.\ } \right.\ \right.\ \right.\ \right.\ \right\rbrack = = \frac{1}{\text{EI}}\left( - \frac{qa^{3}}{6} - \frac{qa^{3}}{2} - \frac{qa^{3}}{2} + \frac{qa^{3}}{6} - \frac{q\cos\alpha^{2}{*8a}^{3}}{6\cos\alpha^{3}} \right) = \ \frac{1}{\text{EI}}\left( - qa^{3} - \frac{4qa^{3}}{3\cos\alpha} \right) = = - \frac{qa^{3}}{\text{EI}}\left( 1 + \frac{4}{3\cos\alpha} \right)$$

Przemieszczenie pionowe węzła A

$U_{A}^{\text{pion}} = \frac{\partial V}{\partial P_{f}} = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{a}{(\frac{1}{2}\text{qa}x_{1})(1 + 2\operatorname{tg}\alpha)*x_{1}dx_{1} + \int_{0}^{a}{(\frac{1}{2}}qa^{2} + qax_{2} - \frac{1}{2}qx_{2}^{2})(a(1 + 2\operatorname{tg}\alpha) + x_{2})dx_{2} + \int_{0}^{\frac{2a}{\cos\alpha}}{(\frac{1}{2}}\text{qa}x_{3}cos\alpha)((1 + 2\operatorname{tg}\alpha)\cos\alpha*x_{3} - sin\alpha*x_{3})dx_{3}} \right\rbrack = \frac{1}{\text{EI}}\left\lbrack \int_{0}^{a}{\frac{1}{2}qa(1 + 2\operatorname{tg}\alpha)x_{1}^{2}}dx_{1} + \int_{0}^{a}{(\frac{1}{2}}qa^{3}(1 + 2\operatorname{tg}\alpha) + qa^{2}(1 + 2\operatorname{tg}\alpha)x_{2} - \frac{1}{2}qa(1 + 2\operatorname{tg}\alpha)*x_{2}^{2}) + \frac{1}{2}qa^{2}x_{2} + qax_{2}^{2} - \frac{1}{2}qx_{2}^{3})dx_{2} + \int_{0}^{\frac{2a}{\cos\alpha}}{(\frac{1}{2}}qa(1 + 2\operatorname{tg}\alpha)\operatorname{}{\cos^{2}\alpha}x_{3}^{2} - \frac{1}{2}\text{qasinαcosα}x_{3}^{2})dx_{3} \right\rbrack = = \frac{1}{\text{EI}}\left\lbrack \left( \frac{1}{2}\text{qa}\left( 1 + 2\operatorname{tg}\alpha \right)\frac{x_{1}^{3}}{3} \right)\left| \frac{a}{0} \right.\ + \left( \frac{1}{2}qa^{3}\left( 1 + 2\operatorname{tg}\alpha \right)x_{2} \right)\left| \frac{a}{0} \right.\ + \left( qa^{2}\left( 1 + 2\operatorname{tg}\alpha \right)\frac{x_{2}^{2}}{2} \right)\left| \frac{a}{0} + \right.\ \left( - \frac{1}{2}\text{qa}\left( 1 + 2\operatorname{tg}\alpha \right)\frac{x_{2}^{3}}{3} \right)\left| \frac{a}{0} \right.\ + \left( \frac{1}{2}qa^{2}\frac{x_{2}^{2}}{2} \right)\left| \frac{a}{0} \right.\ + \left( \text{qa}\frac{x_{2}^{3}}{3} \right)\left| \frac{a}{0} \right.\ + \left( - \frac{1}{2}q\frac{x_{2}^{4}}{4} \right)\left| \frac{a}{0} \right.\ + \left( \frac{1}{2}\text{qa}\cos^{2}\alpha\left( 1 + 2\operatorname{tg}\alpha \right)\frac{x_{3}^{3}}{3} \right)\left| \frac{\frac{2a}{\cos\alpha}}{0} \right.\ + ( - \frac{1}{2}\text{qasinαcosα}\frac{x_{3}^{3}}{3})\left| \frac{\frac{2a}{\cos\alpha}}{0} \right.\ \right\rbrack = \frac{1}{\text{EI}}\left\lbrack \frac{(1 + 2\operatorname{tg}\alpha)qa^{4}}{6} + \frac{(1 + 2\operatorname{tg}\alpha)qa^{4}}{2} + \frac{(1 + 2\operatorname{tg}\alpha)qa^{4}}{2} - \frac{(1 + 2\operatorname{tg}\alpha)qa^{4}}{6} + \frac{qa^{4}}{4} + \frac{qa^{4}}{3} - \frac{qa^{4}}{8} + \frac{\text{qa}\cos^{2}\alpha(1 + 2tg\alpha)\frac{{8a}^{3}}{\cos^{3}\alpha}}{6} - \frac{\text{qasinαcosα}\frac{{8a}^{3}}{\cos^{3}\alpha}}{6} \right\rbrack = \frac{\text{qa}^{4}}{\text{EI}}$($\frac{35}{24} + 2tg\ \alpha + \frac{4cos\alpha\left( 1 + 2tg\alpha \right) - 4sin\alpha}{{3cos}^{2}\alpha}$