Zadanie 4

Dane:

γ = 168   β = 26   α = 50

Położenia osi tensometrów a, b oraz c można przedstawić jako osie powstałe w efekcie obrotów osi x układu współrzędnych odpowiednio o kąty:

φ_a = γ = 168

φ_b = γ − β = 168 − 50 = 118

φ_c = γ + α = 168 + 26 = 194(14)

Wyznaczone na drodze eksperymentu wielkości ε_a, ε_b oraz ε_c można traktować jako składowe stanu odkształcenia określone w układach obróconych względem osi układu xy. Muszą one więc spełniać następujące równania zapisane zgodnie z wzorami transformacyjnymi:

$$\varepsilon_{a} = \frac{1}{2}*\left( \varepsilon_{x} + \varepsilon_{y} \right) + \frac{1}{2}*\left( \varepsilon_{x} - \varepsilon_{y} \right)*\cos\left( 2\varphi_{a} \right) + \frac{1}{2}*\gamma_{\text{xy}}*sin(2\varphi_{a})$$

$$\varepsilon_{b} = \frac{1}{2}*\left( \varepsilon_{x} + \varepsilon_{y} \right) + \frac{1}{2}*\left( \varepsilon_{x} - \varepsilon_{y} \right)*\cos\left( 2\varphi_{b} \right) + \frac{1}{2}*\gamma_{\text{xy}}*sin(2\varphi_{b})$$

$$\varepsilon_{c} = \frac{1}{2}*\left( \varepsilon_{x} + \varepsilon_{y} \right) + \frac{1}{2}*\left( \varepsilon_{x} - \varepsilon_{y} \right)*\cos\left( 2\varphi_{c} \right) + \frac{1}{2}*\gamma_{\text{xy}}*sin(2\varphi_{c})$$

cos(2φ_a) = 0, 9135 sin(2φ_a) = −0, 4067

cos(2φ_b) = −0, 5592 sin(2φ_b) = −0, 8290

cos(2φ_c) = 0, 8829 sin(2φ_c) = 0, 4695

−455 * 10⁻⁶ = 0, 95675ε_x + 0, 04325ε_y − 0, 20335γ_xy

−461 * 10⁻⁶ = 0, 2204ε_x + 0, 7796ε_y − 0, 4145γ_xy

91 * 10⁻⁶ = 0, 94145ε_x + 0, 05855ε_y + 0, 23475γ_xy

Równanie rozwiązałem korzystając z programu Excel, wykorzystując obliczenia macierzowe

A * X = B

0,95675	0,04325	-20335		εx		-455*10^-6
0,2204	0,7796	-0,4145	x	εy	=	-461*10^-6
0,94145	0,05855	0,23475		γxy		91*10^-6

X = A⁻¹ * B

εx		0,000014	-0,081201	1,081187		-455*10^-6
εy	=	-0,000030	1,305665	-0,305635	x	-461*10^-6
γxy		-0,000049	-0,000001	0,000050		91*10^-6

Wyniki :  ε_x = 135, 82 * 10⁻⁶

ε_y = −629, 71 * 10⁻⁶

γ_xy = 0, 03 * 10⁻⁶

Wyznaczenie kierunków odkształceń głównych:

$$\tan\left( 2\alpha \right) = \frac{\gamma_{\text{xy}}}{\varepsilon_{x} - \varepsilon_{y}} = \frac{0,03*10^{- 6}}{135,82*10^{- 6} + 629,71*10^{- 6}} = 39,19*10^{- 6}$$

2α = 0, 22  →   α = 0, 11

Wyznaczenie wartości odkształceń głównych:

$$\varepsilon_{1,2} = \frac{1}{2}*\left( \varepsilon_{x} + \varepsilon_{y} \right) \pm \frac{1}{2}\sqrt{\left( \varepsilon_{x} - \varepsilon_{y} \right)^{2} + \gamma_{\text{xy}}^{2}} = \frac{1}{2}*\left( 135,82*10^{- 6} - 629,71*10^{- 6} \right) \pm \frac{1}{2}\sqrt{\left( 135,82*10^{- 6} + 629,71*10^{- 6} \right)^{2} + \left( 0,03*10^{- 6} \right)^{2}} =$$

ε₁ = 518, 56 * 10⁻⁶ 

ε₂ = −1012, 46 * 10⁻⁶

Wyznaczenie wartości naprężeń σ_x, σ_y oraz τ_xy:

$$\sigma_{x} = \frac{E}{1 - v^{2}}*\left( \varepsilon_{x} + v*\varepsilon_{y} \right) = \frac{61,68*10^{3}\text{MPa}}{1 - {0,15}^{2}}*\left( 135,82 - 0,15*629,71 \right)*10^{- 6} = 2,61M\text{Pa}$$

$$\sigma_{y} = \frac{E}{1 - v^{2}}*\left( \varepsilon_{y} + v*\varepsilon_{x} \right) = \frac{61,68*10^{3}\text{MPa}}{1 - {0,15}^{2}}*\left( - 629,71 + 0,15*135,82 \right)*10^{- 6} = - 38,45\text{MPa}$$

$$\tau_{\text{xy}} = G*\gamma_{\text{xy}} = \frac{E}{2*\left( 1 + v \right)}*\gamma_{\text{xy}} = \frac{61,68*10^{3}\text{MPa}}{2*\left( 1 + 0,15 \right)}*0,03*10^{- 6} = 804,52*10^{- 6}\text{MPa}$$

Zadanie 5

1. Charakterystyki geometryczne

1. Środek ciężkości

$\delta_{1} = \frac{L_{1}}{d} = \frac{8,50cm}{20} = 0,425cm$ $\delta_{2} = \frac{19,60cm}{20} = 0,98cm$

$\delta_{3} = \frac{10,4cm}{20} = 0,52cm$ $\delta_{4} = \frac{16,8cm}{20} = 0,84cm$

A = L₁ * δ₁ + L₂ * δ₂ + L₃ * δ₃ + L₄ * δ₄ = 8, 5cm * 0, 425cm + 19, 6cm * 0, 98cm + 10, 4cm * 0, 52cm + 16, 8cm * 0, 84cm = 42, 34cm²

S_y1 = L₁ * δ₁ * z₁ + L₂ * δ₂ * z₂ + L₃ * δ₃ * z₃ + L₄ * δ₄ * z₄ = 8, 5cm * 0, 425cm * 0 + 19, 6cm * 0, 98cm * 9, 8cm + 10, 4cm * 0, 52cm * 19, 6cm + 16, 8cm * 0, 84cm * 11, 2cm = 452, 29cm³

S_z1 = L₁ * δ₁ * y₁ + L₂ * δ₂ * y₂ + L₃ * δ₃ * y₃ + L₄ * δ₄ * y₄ = 8, 5cm * 0, 425cm * (−4,25cm) + 19, 6cm * 0, 98cm * 0 + 10, 4cm * 0, 52cm * (−5,2cm) + 16, 8cm * 0, 84cm * (−10,4cm) = −190, 24cm³

$$y_{c} = \frac{S_{z1}}{A} = \frac{- 190,24cm^{3}}{42,34cm^{2}} = - 4,49cm$$

$$z_{c} = \frac{S_{y1}}{A} = \frac{452,29cm^{3}}{42,34cm^{2}} = 10,68cm$$

1. Centralne momenty bezwładności

$$I_{y0} = 8,5cm*0,425cm*\left( - 10,68cm \right)^{2} + \frac{0,98cm*\left( 19,6cm \right)^{3}}{12} + 19,6cm*0,98cm*\left( - 0,88cm \right)^{2} + 10,4cm*0,52cm*(8,92c{m)}^{2} + \frac{0,84cm*\left( 16,8cm \right)^{3}\ }{12} + 16,8cm*0,84cm*\left( 0,52cm \right)^{2} = 1807,86cm^{4}$$

$$I_{z0} = \frac{0,425cm*\left( 8,5cm \right)^{3}}{12} + 8,5cm*0,425cm*\left( 0,24cm \right)^{2} + 19,6cm*0,98cm*\left( 4,49cm \right)^{2} + \frac{0,52cm*\left( 10,4cm \right)^{3}}{12} + 10,4cm*0,52cm*\left( - 0,71cm \right)^{2} + 16,8cm*0,84cm*\left( - 5,91cm \right)^{2} = 953,57cm^{4}$$

I_y0z0 = 8, 5cm * 0, 425cm * ( − 10, 68cm)*0, 24cm + 19, 6cm * 0, 98cm * (−0,88cm) * 4, 49cm + 10, 4cm * 0, 52cm * 8, 92cm * (−071cm) + 16, 8cm * 0, 84cm * 0, 52cm * (−5,91cm) = −162, 77cm⁴

1. Osie główne

$$\tan\left( 2\varphi_{0} \right) = - \frac{{2I}_{y0z0}}{I_{y0} - I_{\text{zo}}} = - \frac{2*( - 162,77\text{cm}^{4})}{1807,86\text{cm}^{4} - 953,57\text{cm}^{4}} = 0,381$$

2φ₀ = arctan(0,381) = 20, 86

φ₀ = 10, 43

1. Wykresy współrzędnych

y⁽¹⁾ = (−8,5cm+4,49cm) * cos(10,43) + (0−10,68cm) * sin(10,43) = −5, 87cm

y⁽²⁾ = (0+4,49cm) * cos(10,43) + (0−10,68cm) * sin(10,43) = 2, 48cm

y⁽³⁾ = (0+4,49cm) * cos(10,43) + (19,6cm−10,68cm) * sin(10,43) = 6, 03cm

y⁽⁴⁾ = (−10,4cm+4,49cm) * cos(10,43) + (19,6cm−10,68cm) * sin(10,43) = −4, 19cm

y⁽⁵⁾ = (−10,4cm+4,49cm) * cos(10,43) + (19,6cm−16,8cm−10,68cm) * sin(10,43) = −7, 24cm

z⁽¹⁾ = −(−8,5cm+4,49cm) * sin(10,43) + (0−10,68cm) * cos(10,43) = −9, 78cm

z⁽²⁾ = −(0+4,49cm) * sin(10,43) + (0−10,68cm) * cos(10,43) = −11, 32cm

z⁽³⁾ = −(0+4,49cm) * sin(10,43) + (19,6cm−10,68cm) * cos(10,43) = 7, 96cm

z⁽⁴⁾ = −(−10,4cm+4,49cm) * sin(10,43) + (19,6cm−10,68cm) * cos(10,43) = 9, 84cm

z⁽⁵⁾ = −(−10,4cm+4,49cm) * sin(10,43) + (19,6cm−16,8cm−10,68cm) * cos(10,43) = −6, 68cm

ω⁽¹⁾ = −8, 5cm * 19, 6cm = −166, 6cm² 

ω⁽²⁾ = 0     ω⁽³⁾ = 0     ω⁽⁴⁾ = 0

ω⁽⁵⁾ = 10, 4cm * 16, 8cm = 174, 72cm²

1. Momenty bezwładności

$$I_{y} = \frac{8,5cm}{6}*\left\lbrack 2*\left( \left( - 9,78cm \right)^{2} + \left( - 11,32cm \right)^{2} \right) + 2*\left( - 9,78cm \right)*\left( - 11,32cm \right) \right\rbrack*0,425cm$$

$$+ \frac{19,6cm}{6}*\left\lbrack 2*\left( \left( - 11,32cm \right)^{2} + \left( 7,96cm \right)^{2} \right) + 2*\left( - 11,32cm \right)*\left( 7,96cm \right) \right\rbrack*0,98cm$$

$$+ \frac{10,4cm}{6}*\left\lbrack 2*\left( \left( 7,96cm \right)^{2} + \left( 9,84cm \right)^{2} \right) + 2*\left( 7,96cm \right)*\left( 9,84cm \right) \right\rbrack*0,52cm$$

$$+ \frac{16,8cm}{6}*\left\lbrack 2*\left( \left( 9,84cm \right)^{2} + \left( - 6,68cm \right)^{2} \right) + 2*\left( 9,84cm \right)*\left( - 6,68cm \right) \right\rbrack*0,84cm = 1837,8cm^{4}$$

$$I_{z} = \frac{8,5cm}{6}*\left\lbrack 2*\left( \left( - 5,87cm \right)^{2} + \left( 2,48cm \right)^{2} \right) + 2*\left( - 5,87cm \right)*\left( 2,48cm \right) \right\rbrack*0,425cm$$

$$+ \frac{19,6cm}{6}*\left\lbrack 2*\left( \left( 2,48cm \right)^{2} + \left( 6,03cm \right)^{2} \right) + 2*\left( 2,48cm \right)*\left( 6,03cm \right) \right\rbrack*0,98cm$$

$$+ \frac{10,4cm}{6}*\left\lbrack 2*\left( \left( 6,03cm \right)^{2} + \left( - 4,19cm \right)^{2} \right) + 2*\left( 6,03cm \right)*\left( - 4,19cm \right) \right\rbrack*0,52cm$$

$$+ \frac{16,8cm}{6}*\left\lbrack 2*\left( \left( - 4,19cm \right)^{2} + \left( - 7,24cm \right)^{2} \right) + 2*\left( - 4,19cm \right)*\left( - 7,24cm \right) \right\rbrack*0,84cm = 923,6cm^{4}$$

$$K_{s} = \frac{1}{3}\sum_{}^{}{l_{i}\delta_{i}^{3} = \frac{1}{3}\left( 8,5cm*\left( 0,425cm \right)^{3} + \ 19,6cm*\left( 0,98cm \right)^{3} + \ 10,4cm*\left( 0,52cm \right)^{3}\ + 16,8cm*\left( 0,84cm \right)^{3} \right)} = 10,17cm^{4}$$

$$I_{\text{ωBy}} = \frac{8,5cm}{6}*0,425cm*\left\lbrack 2*\left( - 5,87cm \right)*\left( - 166,6cm^{2} \right) + 2,48cm*\left( - 166,6cm^{2} \right) \right\rbrack$$

$$+ \frac{16,8cm}{6}*0,84cm*\left\lbrack 2*\left( - 7,24cm \right)*\left( 174,72cm^{2} \right) + \left( - 4,19cm \right)*\left( 174,72cm^{2} \right) \right\rbrack = - 6741,95cm^{5}$$

$$I_{\text{ωBz}} = \frac{8,5cm}{6}*0,425cm*\left\lbrack 2*\left( - 9,78cm \right)*\left( - 166,6cm^{2} \right) + ( - 11,32cm)*\left( - 166,6cm^{2} \right) \right\rbrack$$

$$+ \frac{16,8cm}{6}*0,84cm*\left\lbrack 2*\left( - 6,68cm \right)*\left( 174,72cm^{2} \right) + 9,84cm*\left( 174,72cm^{2} \right) \right\rbrack = 1648,74cm^{5}$$

1. Główny biegun wycinkowy

$$y^{(A)} = y^{(B)} + \alpha_{y} = 6,03cm + \frac{1648,74cm^{5}}{1837,8cm^{4}} = 6,03cm + 0,9cm = 6,93cm$$

$$z^{(A)} = z^{(B)} + \alpha_{z} = 7,96cm - \frac{- 6741,95cm^{5}}{923,6cm^{4}} = 7,96cm + 7,3cm = 15,26cm$$

1. Główna współrzędna wycinkowa

1. Współrzędne głównego bieguna wycinkowego w układzie osi centralnych

y₀^(A) = y^(A) * cos(−φ) + z^(A) * sin(−φ) = 6, 93cm * cos(−10,43) + 15, 26cm * sin(−10,43) = 4, 05cm

z₀^(A) = −y^(A) * cos(−φ) + z^(A) * sin(−φ) = −6, 93cm * sin(−10,43) + 15, 26cm * cos(−10,43) = 16, 26cm

1. Współrzędne głównego bieguna wycinkowego w układzie osi początkowych

y₁^(A) = y₀^(A) + y_C = 4, 05cm − 4, 49cm = −0, 44cm

z₁^(A) = z₀^(A) + z_c = 16, 26cm + 10, 68cm = 26, 94cm

1. Współrzędna ω_A

ω_A⁽³⁾ = 0

ω_A⁽²⁾ = ω_A⁽³⁾ − |y₁^(A)| * L₂ = 0 − 0, 44cm * 19, 6cm = −8, 61cm²

ω_A⁽¹⁾ = ω_A⁽²⁾ − z₁^(A) * L₁ = −8, 61cm² − 26, 94cm * 8, 5cm = −237, 61cm²

ω_A⁽⁴⁾ = ω_A⁽³⁾ − (z₁^(A)−L₂) * L₃ = 0 − (26,94cm−19,6cm) * 10, 4cm = −76, 35cm²

ω_A⁽⁵⁾ = ω_A⁽⁴⁾ + (L₃−|y₁^(A)|) * L₄ = −76, 35 + (10,4cm−0,44cm) * 16, 8cm = 90, 99cm²

1. Moment statyczny współrzędnej ω_A

$$S_{\text{ωA}} = L_{1}*\delta_{1}*\frac{\omega_{A}^{\left( 1 \right)} + \omega_{A}^{\left( 2 \right)}}{2} + L_{2}*\delta_{2}*\frac{\omega_{A}^{\left( 2 \right)} + \omega_{A}^{\left( 3 \right)}}{2} + L_{3}*\delta_{3}*\frac{\omega_{A}^{\left( 3 \right)} + \omega_{A}^{\left( 4 \right)}}{2} + L_{4}*\delta_{4}*\frac{\omega_{A}^{\left( 4 \right)} + \omega_{A}^{\left( 5 \right)}}{2} = 8,5cm*0,425cm*\frac{- 273,61cm^{2} - 8,61cm^{2}}{2} + 19,6cm*0,98cm*\frac{- 8,61cm^{2} + 0}{2} + 10,4cm*0,52cm*\frac{0 - 76,35cm^{2}}{2} + 16,8cm*0,84cm*\frac{- 76,35cm^{2} + 90,99cm^{2}}{2} = - 630,56cm^{4}$$

1. Współrzędna ω – główna współrzędna wycinkowa

$$\omega = - \frac{S_{\text{ωA}}}{A} = - \frac{- 630,56cm^{4}}{42,34cm^{2}} = 14,89cm^{2}$$

ω⁽ⁱ⁾ = ω_A⁽ⁱ⁾ + ω

ω⁽¹⁾ = −237, 61cm² + 14, 89cm² = −222, 72cm²

ω⁽²⁾ = −8, 61cm² + 14, 89cm² = 6, 28cm²

ω⁽³⁾ = 0 + 14, 89cm² = 14, 89cm²

ω⁽⁴⁾ = −76, 35cm² + 14, 89cm² = −61, 46cm²

ω⁽⁵⁾ = 90, 99cm² + 14, 89cm² = 105, 88cm²

1. Główny wycinkowy moment bezwładności

$$I_{\omega} = \frac{8,5cm}{6}*\left\lbrack 2*\left( \left( - 222,72cm^{2} \right)^{2} + \left( 6,28cm^{2} \right)^{2} \right) + 2*\left( - 222,72cm^{2} \right)*6,28cm^{2} \right\rbrack*0,425cm$$

$$+ \frac{19,6cm}{6}*\left\lbrack 2*\left( \left( 6,28cm^{2} \right)^{2} + \left( 14,89cm^{2} \right)^{2} \right) + 2*6,28cm^{2}*14,89cm^{2} \right\rbrack*0,98cm$$

$$+ \frac{10,4cm}{6}*\left\lbrack 2*\left( \left( 14,89cm^{2} \right)^{2} + \left( - 61,46cm^{2} \right)^{2} \right) + 2*14,89cm^{2}*\left( - 61,46cm^{2} \right) \right\rbrack*0,52cm$$

$$+ \frac{16,8cm}{6}*\left\lbrack 2*\left( \left( - 61,46cm^{2} \right)^{2} + \left( 105,88cm^{2} \right)^{2} \right) + 2*\left( - 61,46cm^{2} \right)*105,88cm^{2} \right\rbrack*0,84cm = 105818,12cm^{6}$$

1. Współczynnik giętno-skrętny

$$\alpha = \sqrt{\frac{GK_{s}}{EI_{\omega}}} = \sqrt{0,4*\frac{10,17cm^{4}}{105818,12cm^{6}}} = 0,006201\frac{1}{\text{cm}} = 0,6201\frac{1}{m}$$

1. Momenty statyczne odciętej części przekroju w punkcie C

y^(c) = y⁽²⁾ + ξ * (y⁽³⁾−y⁽²⁾) = 2, 48cm + 0, 84 * (6,03cm−2,48cm) = 5, 46cm

z^(c) = z⁽²⁾ + ξ * (z⁽³⁾−z⁽²⁾) = −11, 32cm + 0, 84 * (7,96cm+11,32cm) = 4, 87cm

ω^(c) = ω⁽²⁾ + ξ * (ω⁽³⁾−ω⁽²⁾) = 6, 28cm² + 0, 84 * (14,89cm²−6,28cm²) = 13, 51cm²

${\overset{\overline{}}{S}}_{y}^{(c)} = L_{1}*\delta_{1}*\frac{z^{\left( 1 \right)} + z^{\left( 2 \right)}}{2} + \xi*L_{2}*\delta_{2}*\frac{z^{\left( 2 \right)} + z^{\left( c \right)}}{2} = 8,5cm*0,425cm*\frac{- 9,78cm - 11,32cm}{2} + 0,84*19,6cm*0,98cm*\frac{- 11,32cm + 4,87cm}{2} = - 90,14cm^{3}$

${\overset{\overline{}}{S}}_{z}^{(c)} = L_{1}*\delta_{1}*\frac{y^{\left( 1 \right)} + y^{\left( 2 \right)}}{2} + \xi*L_{2}*\delta_{2}*\frac{y^{\left( 2 \right)} + y^{\left( c \right)}}{2} = 8,5cm*0,425cm*\frac{- 5,87cm + 2,48cm}{2} + 0,84*19,6cm*0,98cm*\frac{2,48cm + 5,46cm}{2} = 57,97cm^{3}$

${\overset{\overline{}}{S}}_{\omega}^{(c)} = L_{1}*\delta_{1}*\frac{\omega^{\left( 1 \right)} + \omega^{\left( 2 \right)}}{2} + \xi*L_{2}*\delta_{2}*\frac{\omega^{\left( 2 \right)} + \omega^{\left( c \right)}}{2} = 8,5cm*0,425cm*\frac{- 222,72cm^{2} + 6,28cm^{2}}{2} + 0,84*19,6cm*0,98cm*\frac{6,28cm^{2} + 13,51cm^{2}}{2} = - 231,25cm^{4}$

1. Siły wewnętrzne

1. Pochodzące od zginania: siły tnące i momenty zginające

P_y = P * cosφ₀ = 38kN * cos(10,43) = 37, 37kN

P_z = −P * sinφ₀ = −38 * sin(10,43) = −6, 88kN

1. Siły pochodzące od skręcania pręta cienkościennego: B, M_ω, M_s

M = −P_y * (z⁽¹⁾−z^(A)) + P_z * (y⁽¹⁾−y^(A)) = −37, 37kN * (−9,78cm−15,26cm) + ( − 6, 88kN)*(−5,87cm−6,93cm) = 10, 2381kNm

Przedział I:

$$M_{s} = \frac{- M}{2sinh(\alpha l)}\left\lbrack 2\left( \gamma - 1 \right)\sinh\left( \text{αl} \right) - \sinh\left( \alpha\left( \left( \gamma - 1 \right)l + x \right) \right) + sinh(\alpha\left( \left( 1 - \gamma \right)l + x \right)) \right\rbrack$$

$$B = \frac{- M}{\alpha sinh(\alpha l)}\sinh\left( \alpha\left( \gamma - 1 \right)l \right)sinh(\alpha x)$$

$$M_{\omega} = \frac{- M}{sinh(\alpha l)}\sinh\left( \alpha\left( \gamma - 1 \right)l \right)cosh(\alpha x)$$

Przedział II:

$$M_{s} = \frac{M}{2}\left\lbrack - 2\gamma + \frac{1}{sinh(\alpha l)}\left\lbrack \sinh\left( \alpha\left( \left( 1 + \gamma \right)l - x \right) \right) + sinh(\alpha\left( \left( \gamma - 1 \right)l + x \right)) \right\rbrack \right\rbrack$$

$$B = \frac{M}{\alpha\sinh\left( \text{αl} \right)}\sinh\left( \text{αγl} \right)\sinh\left( \alpha\left( l - x \right) \right)$$

$$M_{\omega} = \frac{- M}{sinh(\alpha l)}\sinh\left( \text{αγl} \right)cosh(\alpha\left( l - x \right))$$

Ms	x
0	0
3,1178	0
3,0367	0,3552
2,7892	0,7104
2,3635	1,0656
1,7387	1,4208
1,2608	1,632
0,8844	1,776
0,8844	1,776
-0,4945	2,3808
-1,4046	2,9856
-1,9755	3,5904
-2,2886	4,1952
-2,3882	4,8
0	4,8

B	x
0,0000	0
1,1932	0,3552
2,4445	0,7104
3,8149	1,0656
5,3711	1,4208
6,4151	1,632
7,1889	1,776
7,1889	1,776
4,8078	2,3808
3,1109	2,9856
1,8567	3,5904
0,8667	4,1952
0	4,8

Mω	x
0	0
3,3322	0
3,4133	0,3552
3,6608	0,7104
4,0865	1,0656
4,7113	1,4208
5,1892	1,632
5,5656	1,776
-4,6725	1,776
-3,2936	2,3808
-2,3835	2,9856
-1,8125	3,5904
-1,4995	4,1952
-1,3999	4,8
0	4,8

1. Naprężenia

1. Naprężenia normalne w przekroju α-α

$$\sigma_{x}^{(i)} = \frac{M_{y}}{I_{y}}*z^{\left( i \right)} - \frac{M_{z}}{I_{z}}*y^{\left( i \right)} + \frac{B}{I_{\omega}}*\omega^{(i)}$$

$$\sigma_{x}^{\left( 1 \right)} = \frac{- 7,07MN*10^{- 3}}{1837,8m^{4}*10^{- 8}}*\left( - 9,78m*10^{- 2} \right) - \frac{- 38,42MN*10^{- 3}}{923,6m^{4}*10^{- 8}}*\left( - 5,87m*10^{- 2} \right) + \frac{6,4151MNm*10^{- 3}}{105818,12m^{6}*10^{- 12}}*\left( - 222,72m^{2}*10^{- 4} \right) = - 1556,95MPa$$

$$\sigma_{x}^{(2)} = \frac{- 7,07MN*10^{- 3}}{1837,8m^{4}*10^{- 8}}*\left( - 11,32m*10^{- 2} \right) - \frac{- 38,42MN*10^{- 3}}{923,6m^{4}*10^{- 8}}*\left( 2,48m*10^{- 2} \right) + \frac{6,4151MNm*10^{- 3}}{105818,12m^{6}*10^{- 12}}*\left( 6,28m^{2}*10^{- 4} \right) = 185,01MPa$$

$$\sigma_{x}^{(3)} = \frac{- 7,07MN*10^{- 3}}{1837,8m^{4}*10^{- 8}}*\left( 7,96m*10^{- 2} \right) - \frac{- 38,42MN*10^{- 3}}{923,6m^{4}*10^{- 8}}*\left( 6,03m*10^{- 2} \right) + \frac{6,4151MNm*10^{- 3}}{105818,12m^{6}*10^{- 12}}*\left( 14,89m^{2}*10^{- 4} \right) = 310,65MPa$$

$$\sigma_{x}^{(4)} = \frac{- 7,07MN*10^{- 3}}{1837,8m^{4}*10^{- 8}}*\left( 9,84m*10^{- 2} \right) - \frac{- 38,42MN*10^{- 3}}{923,6m^{4}*10^{- 8}}*\left( - 4,19m*10^{- 2} \right) + \frac{6,4151MNm*10^{- 3}}{105818,12m^{6}*10^{- 12}}*\left( - 61,46m^{2}*10^{- 4} \right) = - 584,93MPa$$

$$\sigma_{x}^{(5)} = \frac{- 7,07MN*10^{- 3}}{1837,8m^{4}*10^{- 8}}*\left( - 6,68m*10^{- 2} \right) - \frac{- 38,42MN*10^{- 3}}{923,6m^{4}*10^{- 8}}*\left( - 7,24m*10^{- 2} \right) + \frac{6,4151MNm*10^{- 3}}{105818,12m^{6}*10^{- 12}}*\left( 105,88m^{2}*10^{- 4} \right) = 366,60MPa$$

punkt	y^(i)	z^(i)	ω^(i)	σM	σω	σx
1	-5,87	-9,78	-222,72	-206,75	-1350,20	-1556,95
2	2,48	-11,32	6,28	146,91	38,10	185,01
3	6,03	7,96	14,89	220,36	90,28	310,65
4	-4,19	9,84	-61,46	-212,35	-372,58	-584,93
5	-7,24	-6,68	105,88	-275,31	641,91	366,60

1. Naprężenia styczne w punkcie C

$$\tau_{s} = \frac{M_{s}}{K_{s}}*\delta = \frac{1,1434}{10,17m^{4}*10^{- 8}}*0,98m*10^{- 2} = 121,493MPa$$

$$\tau_{M + \omega} = - \left\lbrack \frac{T_{z}*\overset{\overline{}}{S_{y}}}{I_{y}*\delta} + \frac{T_{y}*\overset{\overline{}}{S_{z}}}{I_{z}*\delta} + \frac{M_{\omega}*\overset{\overline{}}{S_{\omega}}}{I_{\omega}*\delta} \right\rbrack = - \left\lbrack \frac{- 4,33kN*\left( - 90,14m^{3}*10^{- 6} \right)}{1837,8m^{4}*10^{- 8}*0,98m*10^{- 2}} + \frac{23,54kN*57,97m^{3}*10^{- 6}}{923,6m^{4}*10^{- 8}*0,98m*10^{- 2}} + \frac{5,1892kNm*\left( - 231,25m^{4}*10^{- 8} \right)}{105818,12m^{6}*10^{- 12}*0,98m*10^{- 2}} \right\rbrack = - \left\lbrack 2167,10kPa + 15076,47kPa - 11571,67kPa \right\rbrack = - 5671,9kPa = - 5,67MPa$$

Zadanie 6

1. Postać wyboczenia

1. Równanie różniczkowe osi odkształconej

przedzial I) 0 ≤ x ≤ 0, 48L

$$M_{y}\left( x \right) = M_{A} + V_{A}*x - \frac{1}{2}*qx^{2}$$

$$\text{EI}{w_{I}}^{''}\left( x \right) = - M_{A} - V_{A}*x + \frac{1}{2}*qx^{2}$$

$$\text{EI}{w_{I}}^{'}\left( x \right) = - M_{A}x - \frac{1}{2}*V_{A}*x^{2} + \frac{1}{6}*qx^{3} + C_{1}$$

$$\text{EI}w_{I}\left( x \right) = - {\frac{1}{2}M}_{A}x^{2} - \frac{1}{6}*V_{A}*x^{3} + \frac{1}{24}*qx^{4} + C_{1}x + D_{1}$$

przedzial II) 0, 48L ≤ x ≤ L

$$M_{y}\left( x \right) = M_{A} + V_{A}*x - \frac{1}{2}*qx^{2}$$

$$0,56EI{w_{\text{II}}}^{''}\left( x \right) = - M_{A} - V_{A}*x + \frac{1}{2}*qx^{2}$$

$$0,56EI{w_{\text{II}}}^{'}\left( x \right) = - M_{A}x - \frac{1}{2}*V_{A}*x^{2} + \frac{1}{6}*qx^{3} + C_{2}$$

$$0,56EIw_{\text{II}}\left( x \right) = - {\frac{1}{2}M}_{A}x^{2} - \frac{1}{6}*V_{A}*x^{3} + \frac{1}{24}*qx^{4} + C_{2}x + D_{2}$$

1. Wyznaczenie stałych z warunków brzegowych

w_I(0) = 0 -> D₁ = 0

w_I^′(0) = 0 -> C₁ = 0

w_II^′(L) = 0 -> $- M_{A}L - \frac{1}{2}*V_{A}*L^{2} + \frac{1}{6}*qL^{3} + C_{2} = 0$

w_I(0,48L) = w_II(0, 48L) -> $- {\frac{1}{2}M}_{A}\left( 0,48L \right)^{2} - \frac{1}{6}*V_{A}*\left( 0,48L \right)^{3} + \frac{1}{24}*q\left( 0,48L \right)^{4} = \frac{1}{0,56}*\left( - {\frac{1}{2}M}_{A}\left( 0,48L \right)^{2} - \frac{1}{6}*V_{A}*\left( 0,48L \right)^{3} + \frac{1}{24}*q\left( 0,48L \right)^{4} + C_{2}*\left( 0,48L \right) + D_{2} \right)$

w_I′(0,48L) = w_II′(0, 48L) -> $- M_{A}(0,48L) - \frac{1}{2}*V_{A}*\left( 0,48L \right)^{2} + \frac{1}{6}*q\left( 0,48L \right)^{3} = \frac{1}{0,56}* - M_{A}(0,48L) - \frac{1}{2}*V_{A}*\left( 0,48L \right)^{2} + \frac{1}{6}*q\left( 0,48L \right)^{3} + C_{2}$

$\sum_{}^{}Y = qL - V_{A} = 0$ -> V_A = qL

$$1) - M_{A}L - \frac{1}{2}qL^{3} + \frac{1}{6}qL^{3} + C_{2} = 0$$

−M_A + C₂ = 0, 33qL²

2)−0, 1152M_AL² − 0, 0184qL⁴ + 0, 0022qL⁴ = −0, 20571M_AL² − 0, 0329qL⁴ − 0, 0039qL⁴ + 0, 8571C₂ + 1, 7857D₂

0, 0905M_AL² − 0, 8571C₂L − 1, 7857D₂ = −0, 0127qL⁴

3)−0, 48M_AL − 0, 1152qL³ + 0, 0184qL⁴ = −0, 8571M_AL − 0, 2057qL³ + 0, 0329qL³ + 1, 7857C₂

0, 3771M_AL − 1, 786C₂ = −0, 0760qL³

Równanie rozwiązałem korzystając z programu Excel, wykorzystując obliczenia macierzowe

A * X = B

-1	1	0		MAL^2		0,3333
0,0905	-0,8571	-1,7857	x	C2L	=	-0,0127
0,3771	-1,7857	0		D2		-0,0760

A⁻¹ * B = X

-1,2677	0	-0,7099		0,3333		-0,3686
-0,2677	0	-0,7099	x	-0,0127	=	-0,0353
0,0642	-0,56	0,3048		-0,076		0,0054

Wyniki:

M_a = −0, 3686qL²

C₂ = −0, 0353qL³

D₂ = 0, 0054qL⁴

1. Obciążenie krytyczne

1. Kryterium energetyczne Timoshenki

∫_LEI(w^″)²dx + ∫_LN(w^′)²dx = 0

$$w_{I}^{'}\left( x \right) = \frac{1}{\text{EI}}\left( - 0,3686qL^{2}x - 0,5qL*x^{2} + \frac{1}{6}*qx^{3} \right)$$

$${w_{I}}^{''}\left( x \right) = \frac{1}{\text{EI}}\left( - 0,3686qL^{2} - qL*x + \frac{1}{2}*qx^{2} \right)$$

N_I(x) = −2, 28P

$$w_{\text{II}}^{'}\left( x \right) = \frac{1}{0,56EI}\left( - 0,3686qL^{2}x - 0,5qL*x^{2} + \frac{1}{6}*qx^{3} - 0,0353qL^{3} \right)$$

$$w_{\text{II}}^{''}\left( x \right) = \frac{1}{0,56EI}*\left( - 0,3686qL^{2} - qL*x + \frac{1}{2}*qx^{2} \right)$$

N_II(x) = −P

∫₀^0, 48LEI(w_I^″)²dx + ∫_0, 48L^L0, 56EI(w_II^″)²dx − P∫₀^0, 48L2, 28(w_I^′)²dx − P∫_0, 48L^L(w_II^′)²dx = 0

1. Wartość krytyczna siły P

$$P_{\text{kr}} = \frac{I_{1} + I_{2}}{I_{3} + I_{4}}$$

I₁ = 0, 161 * q² * L⁵/EI

I₂ = 0, 631*q² * L⁵/EI

I₃ = 0, 023*q² * L⁷/EI²

I₄ = 0, 469*q² * L⁷/EI²

$$P_{\text{kr}} = \frac{I_{1} + I_{2}}{I_{3} + I_{4}} = \frac{\frac{0,161q^{2}L^{5}}{\text{EI}} + \frac{{0,631q^{2}L}^{5}}{\text{EI}}}{\frac{0,023q^{2}L^{7}}{EI^{2}} + \frac{0,469q^{2}L^{7}}{EI^{2}}} = 1,61\frac{\text{EI}}{L^{2}} = 1,61*\frac{1608}{{6,42}^{2}} = 62,80kN$$

$$\mu = \sqrt{\frac{\pi^{2}\text{EI}}{P_{\text{kr}}*L^{2}}} = \sqrt{\frac{\pi^{2}*1608}{62,8*{6,42}^{2}}} = 2,476$$