Dane dla kształtowników
| C400E : | A=61,5cm2 | C120 : | A=17cm2 |
|---|---|---|---|
| Iy=15220cm4 | Iy=364cm4 | ||
| Iz=642cm4 | Iz=43,2cm4 | ||
| ey=2,75cm | ey=1,6cm | ||
Współrzędne środków ciężkości kształtowników w początkowym układzie współrzędnych y1z1
$${y_{1}^{C1} = 2,75\text{cm}\backslash n}{z_{1}^{C1} = 20\text{cm}\backslash n}{y_{2}^{C2} = - 6\text{cm}\backslash n}{z_{2}^{C2} = 1,6\text{cm}\backslash n}{y_{0} = \frac{A^{C1}*y_{1}^{C1} + A^{C2}*y_{2}^{C2}}{A^{C1} + A^{C2}} = \frac{61,5*\left( 2,75 \right) + 17*( - 6)}{61,5 + 17} = 0,86\text{cm}\backslash n}{z_{0} = \frac{A^{C1}*z_{1}^{C1} + A^{C2}*z_{2}^{C2}}{A^{C1} + A^{C2}} = \frac{61,5*20 + 17*(1,6)}{61,5 + 17} = 16,02\text{cm}}$$
Iy0 = 15220 + 61, 5 * (20−16,2)2 + 364 + 17 * (16,2−1,6)2 = 20139, 1cm4 ∖ nIz0 = 642 + 61, 5 * (2,75−0,86)2 + 43, 2 + 17 * (0,86+6)2 = 1704, 9cm4 ∖ nIy0z0 = −490 + 33, 3 * (9−8,87) * (0,475+1,08) − 184 + 23, 2 * (8,69−8,87) * (−3,31+1,08) = −658, 0cm4
$${\tan\left( 2\varphi_{0} \right) = - \frac{2*I_{y0z0}}{I_{y0} - I_{z0}} = - \frac{2*\left( - 658,0 \right)}{1914 - 778,9} = 1,159\backslash n}{2\varphi_{0} = \arctan\left( 1,159 \right) = 49,21\backslash n}{\varphi_{0} = 24,61}$$
$${I_{1} = \frac{1}{2}\left( I_{y0} + I_{z0} \right) + \frac{1}{2}\sqrt{\left( I_{y0} - I_{z0} \right)^{2} + 4*{I_{y0z0}}^{2}} = \frac{1}{2}\left( 1914 + 778,9 \right) + \frac{1}{2}\sqrt{\left( 1914 - 778,9 \right)^{2} + 4*\left( - 658,0 \right)^{2}} = 1346 + 869,0 = 2215\text{cm}^{4}\backslash n}{I_{2} = \frac{1}{2}\left( I_{y0} + I_{z0} \right) - \frac{1}{2}\sqrt{\left( I_{y0} - I_{z0} \right)^{2} + 4*{I_{y0z0}}^{2}} = \frac{1}{2}\left( 1914 + 778,9 \right) - \frac{1}{2}\sqrt{\left( 1914 - 778,9 \right)^{2} + 4*\left( - 658,0 \right)^{2}} = 1346 - 869,0 = 477\text{cm}^{4}\ \backslash n}{I_{y0} > I_{z0}I_{y} = I_{1}\ ;\ I_{1} = I_{2}}$$
My0max = 12, 625q m2 ∖nMy = My0 * cosφ0 ∖nMz = −My0 * sinφ0
$$z = y*\frac{M_{z}}{M_{y}}*\frac{J_{y}}{J_{z}} = - tan\varphi_{0}*\frac{J_{y}}{J_{z}}*y = - tan\ 24,61*\frac{2215}{477}*y = - 2,127*y$$
$${y_{1} = - 1cm\backslash n}{z_{1} = 0\backslash n}{y_{0} = 0,08cm\backslash n}{z_{0} = - 8,87cm\backslash n}{y = y_{0}*cos\varphi_{0} + z_{0}*sin\varphi_{0} = 0,07 - 3,69 = - 3,62cm\backslash n}{z = - y_{0}*sin\varphi_{0} + z_{0}*cos\varphi_{0} = - 0,03 - 8,06 = - 8,09cm\backslash n}{\sigma_{x}^{1} = \frac{M_{y}}{I_{y}}*z - \frac{M_{z}}{I_{z}}*y = M_{y0}*\left( \frac{\cos\varphi_{0}}{I_{y}}*z + \frac{\sin\varphi_{0}}{I_{z}}*y \right) = - M_{y0}*0,006481\frac{1}{\text{cm}^{3}}}$$
$${y_{1} = - 12cm\backslash n}{z_{1} = 11cm\backslash n}{y_{0} = - 10,92cm\backslash n}{z_{0} = 2,13cm\backslash n}{y = y_{0}*cos\varphi_{0} + z_{0}*sin\varphi_{0} = - 9,93 + 0,89 = - 9,04cm\backslash n}{z = - y_{0}*sin\varphi_{0} + z_{0}*cos\varphi_{0} = 4,55 + 1,94 = 6,49cm\backslash n}{\sigma_{x}^{1} = \frac{M_{y}}{I_{y}}*z - \frac{M_{z}}{I_{z}}*y = M_{y0}*\left( \frac{\cos\varphi_{0}}{I_{y}}*z + \frac{\sin\varphi_{0}}{I_{z}}*y \right) = - M_{y0}*0,005228\frac{1}{cm^{3}}}$$
$${y_{1} = 0,95cm\backslash n}{z_{1} = 18cm\backslash n}{y_{0} = 2,03cm\backslash n}{z_{0} = 9,13\text{cm}\backslash n}{y = y_{0}*cos\varphi_{0} + z_{0}*sin\varphi_{0} = 1,85 + 3,80 = 5,65cm\backslash n}{z = y_{0}*sin\varphi_{0} + z_{0}*cos\varphi_{0} = - 0,85 + 8,30 = 7,45cm\backslash n}{\sigma_{x}^{1} = \frac{M_{y}}{I_{y}}*z - \frac{M_{z}}{I_{z}}*y = M_{y0}*\left( \frac{\cos\varphi_{0}}{I_{y}}*z + \frac{\sin\varphi_{0}}{I_{z}}*y \right) = M_{y0}*0,007991\frac{1}{\text{cm}^{3}}}$$
Największe naprężenia występują w punkcie 3
$${\left| \sigma_{x} \right| \leq K_{g}\backslash n}{\sigma_{x}^{\max} = M_{y0}*0,007991\frac{1}{cm^{3}} = 126250q\text{\ cm}^{2}*0,007991\frac{1}{\text{cm}^{3}} = 1008q\frac{1}{\text{cm}}\backslash n}{q_{\text{dop}} = \frac{21,5\frac{\text{kN}}{\text{cm}^{2}}}{1008\frac{1}{\text{cm}}} = 0,0213\frac{\text{kN}}{\text{cm}} = 2,13\frac{\text{kN}}{m}}$$
Schemat statyczny belki wtórnej
Obciążenie belki wtórnej
Sposób rozkładu obciążenia o zmienności
parabolicznej na składowe figury proste
$${F_{1}^{*} = \frac{2}{3}*5,5*\frac{q\left( 5,5 \right)^{2}}{8} = 13,864q\ m^{3}\backslash n}{F_{2}^{*} = \frac{1}{2}*5,5*12,625q = 34,719q\ m^{3}\backslash n}{F_{3}^{*} = \frac{1}{2}*5,5*2,5q = 6,875q\ m^{3}\backslash n}{F_{4}^{*} = \frac{1}{2}*2,5*5q = 6,250q\ m^{3}\backslash n}{T_{B}^{*} = - F_{1}^{*} - F_{2}^{*} + F_{3}^{*} = - 41,708q\ m^{3}}$$
$${M_{A}^{*} = T_{B}^{*}*2,5 + F_{4}^{*}*\frac{2}{3}*2,5 = - 41,708q*2,5 + 6,250q*\frac{5}{3} = - 104,27q + 10,417q = - 93,853q\ m^{4}\backslash n}{\varphi_{B} = \frac{T_{B}^{*}}{\text{EI}}\backslash n}{\omega_{A} = \frac{M_{A}^{*}}{\text{EI}}}$$
qz = q * cosφ0 ∖ nqy = q * sinφ0
$${{\omega'}_{B} = \frac{- 41,708*2,13\frac{\text{kN}}{m}*\ m^{3}*cos\varphi_{0}}{205*10^{6}\frac{\text{kN}}{m^{2}}*2215*10^{- 8}m^{4}} = - 0,017\ rad\backslash n}{\omega_{A} = \frac{- 93,853*2,13\frac{\text{kN}}{m}*\ m^{4}*cos\varphi_{0}}{205*10^{6}\frac{\text{kN}}{m^{2}}*2215*10^{- 8}m^{4}} = - 0,040m}$$
$${{v'}_{B} = \frac{- 41,708*2,13\frac{\text{kN}}{m}*\ m^{3}*sin\varphi_{0}}{205*10^{6}\frac{\text{kN}}{m^{2}}*477*10^{- 8}m^{4}} = - 0,038\ rad\backslash n}{v_{A} = \frac{- 93,853*2,13\frac{\text{kN}}{m}*\ m^{4}*sin\varphi_{0}}{205*10^{6}\frac{\text{kN}}{m^{2}}*477*10^{- 8}m^{4}} = - 0,085m}$$
ω0 = ωA * cosφ0 + vA * sinφ0 = −0, 071m
ω0 = ω′A * cosφ0 + v′A * sinφ0 = −0, 031rad
TzB − B = 52, 5kN
∖nMyB − B = −31, 5kNm
$$z_{c} = \frac{2*2,3*7*3,5 + 18*4,8*9,4 + 3,8*7,8*15,7}{2*2,3*7 + 18*4,8 + 3,8*7,8} = 9,378cm$$
$$I_{y} = \frac{2,3*7^{3}}{12} + 2,3*7*\left( 9,378 - 3,5 \right)^{2} + \frac{{18*4,8}^{3}}{12} + 18*4,8*\left( 9,378 - 9,4 \right)^{2} + \frac{3,8*{7,8}^{3}}{12} + 3,8*7,8*\left( 9,378 - 15,7 \right)^{2} = 2123cm^{4}$$
Naprężenia normalne będziemy wyznaczać w punktach (1), (6) oraz (5) – aby wyznaczyć σzred. Naprężenia styczne będziemy wyznaczać w punktach (2), (3), (4) – ze względu na zmieniającą się szerokość przekroju we włóknach dolnych i górnych; oraz (5) – aby wyznaczyć σzred.
$${\overset{\overline{}}{S}}_{y}^{\left( 2 \right)} = 2*2,3*7*5,878 = 189,3cm^{3}$$
$${{\overset{\overline{}}{S}}_{y}^{\left( 3 \right)} = {\overset{\overline{}}{S}}_{y}^{\left( 2 \right)} + 18*2,378*1,189 = 240,2cm^{3}\backslash n}{{\overset{\overline{}}{S}}_{y}^{\left( 4 \right)} = 3,8*7,8*6,322 = 187,4cm^{3}\backslash n}{{\overset{\overline{}}{S}}_{y}^{\left( 5 \right)} = 3,8*6,24*7,102 = 168,4cm^{3}\backslash n}{z^{\left( 1 \right)} = - 9,378cm\backslash n}{z^{\left( 5 \right)} = 3,982cm\backslash n}{z^{\left( 6 \right)} = 10,222cm\backslash n}{b^{\left( 1 \right)} = b^{\left( 2A \right)} = 4,6cm\backslash n}{b^{\left( 2B \right)} = b^{\left( 3 \right)} = b^{\left( 4A \right)} = 18cm\backslash n}{b^{\left( 4B \right)} = b^{\left( 5 \right)} = b^{\left( 6 \right)} = 3,8cm}$$
$${\sigma_{x} = \frac{M_{y}}{I_{y}}*z\backslash n}{\sigma_{x}^{\left( 1 \right)} = \frac{- 3150\ kNcm}{2123\ cm^{4}}*\left( - 9,378cm \right) = 13,91\frac{\text{kN}}{cm^{2}} = 139,1MPa\backslash n}{\sigma_{x}^{\left( 3 \right)} = 0\backslash n}{\sigma_{x}^{\left( 5 \right)} = \frac{- 3150\ kNcm}{2123\ cm^{4}}*3,982cm = - 5,908\frac{\text{kN}}{cm^{2}} = - 59,08MPa\backslash n}{\sigma_{x}^{(6)} = \frac{- 3150\ kNcm}{2123\ cm^{4}}*10,222 = - 15,17\frac{\text{kN}}{cm^{2}} = - 151,7MPa}$$
$${T_{\text{xz}} = \frac{T_{z}*{\overset{\overline{}}{S}}_{y}}{I_{y}*b}\backslash n}{T_{\text{xz}}^{\left( 1 \right)} = 0\backslash n}{T_{\text{xz}}^{\left( 2A \right)} = \frac{52,5kN*189,3cm^{3}}{2123cm^{4}*4,6cm} = 0,494\frac{\text{kN}}{cm^{2}} = 4,94MPa\backslash n}{T_{\text{xz}}^{\left( 2B \right)} = \frac{52,5kN*189,3cm^{3}}{2123cm^{4}*18cm} = 0,126\frac{\text{kN}}{cm^{2}} = 1,26MPa\backslash n}{T_{\text{xz}}^{\left( 3 \right)} = \frac{52,5kN*240,2cm^{3}}{2123cm^{4}*18cm} = 0,160\frac{\text{kN}}{cm^{2}} = 1,60MPa\backslash n}{T_{\text{xz}}^{\left( 4A \right)} = \frac{52,5kN*187,4cm^{3}}{2123cm^{4}*18cm} = 0,125\frac{\text{kN}}{cm^{2}} = 1,25MPa\backslash n}{T_{\text{xz}}^{\left( 4B \right)} = \frac{52,5kN*187,4cm^{3}}{2123cm^{4}*3,8cm} = 0,592\frac{\text{kN}}{cm^{2}} = 5,92MPa\backslash n}{T_{\text{xz}}^{\left( 5 \right)} = \frac{52,5kN*168,4cm^{3}}{2123cm^{4}*4,6cm} = 0,532\frac{\text{kN}}{cm^{2}} = 5,32MPa\backslash n}{T_{\text{xz}}^{(6)} = 0}$$
Naprężenia zredukowane wyznaczone zgodnie z hipotezą Hubera-Misesa w punkcie A przekroju B-B
$${\sigma_{\text{zred}} = \sqrt{\sigma_{x}^{2} + 3*T_{\text{xz}}^{2}}\backslash n}{\sigma_{\text{zred}}^{5} = \sqrt{{( - 59,08)}^{2} + 3*{5,92}^{2}} = 59,96\text{MPa}}$$