• Styk montażowy podciagu:

$$X = \ \frac{1}{5} \bullet 10,8 = 2,16\ m$$

M_α = 321, 0 • 8, 64 − 94, 0 • 8, 64 • 4, 32 = −735, 0 kNm    T_α = 94, 0 − 321, 0 = 227, 0 kN

$$In = \frac{1 \bullet 100^{3}}{12} = 83333\ \text{cm}^{4}\ \ \ \ If = \frac{25 \bullet 2^{3}}{12} + 25 \bullet 2 \bullet 51^{2} = 260133,0\ \text{cm}^{4}\text{\ \ \ \ }$$

Ix = In + If = 343466cm⁴

$$Mn = M \bullet \frac{\text{In}}{\text{Ix}} = 735,0 \bullet \frac{83333}{343466} = 735,0 \bullet 0,24 = 176,40\text{\ kN}m$$

M_n^k = M • V • e = 176, 0 • 227, 0 • 0, 12 = 5193, 05 kNm

Maksymalna wysokość przekładki:

$$h_{p}^{\max} = 112 - 2 \bullet y = 112 - 2 \bullet \left( \frac{50 - 1,0}{2} \bullet tg30 + 1,0 \bullet \frac{1}{cos30} \right) = 112,0 - 2 \bullet 16,0 = 80cm$$

a₁ = min(1,5•d) = min(1,5•20) = 30mm     przyjmuje a₁ = 40mm

a = min(2,5•d) = min(2,5•20) = 50mm     przyjmuje a = 80mm

a₂ = min(1,5•d) = min(1,5•20) = 30mm     przyjmuje a₂ = 40mm

$$S_{i,F} = \frac{F}{n}\text{\ \ \ \ \ \ }S_{i,M} = M_{o}\frac{\text{ri}}{\sum\text{ri}^{2}}$$

$$F_{r} = \frac{227,0}{15} = \ \ \ 15,1\ kN\ \ \ $$

∑r = 10 • 8² + 6 • 16² + 6 • 8² = 2560 cm²

$$F_{m}^{x} = 51,93 \bullet \frac{30}{2560} = 0,608\text{kN\ \ \ \ \ }F_{m}^{y} = 51,93 \bullet \frac{8}{7390} = 0,160\ kN\ \ $$

F_y = 15, 1 + 0, 160 = 15, 26 kN

$$F = \ \sqrt{\left( 15,26 \right)^{2}{+ \left( 0,608 \right)}^{2}} = 15,27\text{\ kN\ }$$

Przekładki dla śruby M20 klasy 5.8

S_Rv = 73, 5 • 2 = 147 kN 

$$\alpha = \left( \frac{a_{1}}{d};\frac{a}{d} - \frac{3}{4};2,5 \right) = \ \left( \frac{40}{20};\frac{80}{20} - \frac{3}{4};2,5 \right) = \left( 2;3,25;2,5 \right) = 2,0$$

$$S_{\text{Rb}} = \alpha \bullet f_{d} \bullet d \bullet \sum_{}^{}t\ \ \ = 2,0\ \bullet 20,5 \bullet 2,0 \bullet 1,0 = 82,0\ kN\ \ \ $$

$$S_{R} = min\begin{Bmatrix} S_{\text{Rv}} \\ S_{\text{Rb}} \\ \end{Bmatrix} = min\begin{Bmatrix} 147\ kN \\ 82\ kN \\ \end{Bmatrix} = 82\ kN$$

$$\frac{F}{S_{R}} = \frac{15,27}{82} = 0,182 < 1\ \ \ \ \ Warunek\ spelniony$$

Podkładki dla śruby M20 klasy 6,6

M − M_N = 735, 0 − 176, 4 = 558, 6 kNm

$$F_{t} = \frac{M}{h} = \frac{558,6}{1,14} = 481,7\ kN\ \ \ \ \ \ \ \ \ \ \ \ \frac{F_{t}}{b} = \frac{481,7}{6} = 81,6\text{\ kN}$$

S_Rv = 84, 8 kN

$$\alpha = \left( \frac{a_{1}}{d};\frac{a}{d} - \frac{3}{4};2,5 \right) = \ \left( \frac{50}{20};\frac{80}{20} - \frac{3}{4};2,5 \right) = \left( 2,5;3,25;2,5 \right) = 2,0$$

$$S_{\text{Rb}} = \alpha \bullet f_{d} \bullet d \bullet \sum_{}^{}t\ \ \ = 2,5\ \bullet 20,5 \bullet 2,0 \bullet 1,0 = 102,5\text{\ kN\ \ \ }$$

$$S_{R} = min\begin{Bmatrix} S_{\text{Rv}} \\ S_{\text{Rb}} \\ \end{Bmatrix} = min\begin{Bmatrix} 84,8\text{\ kN} \\ 102,5\text{\ kN} \\ \end{Bmatrix} = 82,0\text{\ kN}$$

$$\frac{F}{S_{R}} = \frac{81,6}{102,5} = 0,78 < 1\ \ \ \ \ Warunek\ spelniony$$

• SŁUP

1. zebranie obciążeń

zebranie obciążeń
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1. wstępne określenie przekroju

$$\sigma = \frac{N}{\varphi \bullet A} = fd\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A = \frac{N}{\varphi \bullet fd} = \frac{691,0}{0,75 \bullet 20,5} = 44,9\ \text{cm}^{2}$$

dane dla ceownika 200p A = 2 • A1 =  2 • 28, 7 = 57, 4cm²

I_x1 = 1821cm⁴    I_x = 2 • 1821cm⁴     = 3642  cm⁴    i_x = 7, 97cm²

 I_y1 = 142cm⁴    i_y = 2, 22 cm² 

1. klasa przekroju

środnik:

$$\frac{\text{hw}}{\text{tw}} = \frac{200 - 2 \bullet \left( 11,5 + 11,5 \right)}{6,5} = 23,9\ \ \ < 33\varepsilon = 33 \bullet \sqrt{\frac{215}{205}} = 33 \bullet 1,02 = 32,26\ $$

Spełnia warunek klasy 1

Półka:

$$\frac{\text{bf}}{\text{tf}} = \frac{75 - 11,5 - 6,5}{11,5} = 8,76\ \ \ \ < 9 \bullet \varepsilon = 8,8\ \ $$

Spełnia warunek klasy 1

1. Wyboczenie dla osi materiałowej x

μ = 1, 0         l_w = μ • l = 1, 0 • 6, 3 = 6, 3 m = 630cm

$$\lambda_{x} = \frac{630}{7,97} = 79,0\ \ \ \ \ \ \ \ \ \lambda_{p} = \frac{\pi}{1,15} \bullet \sqrt{\frac{215}{\text{fd}}} = 84,0 \bullet \sqrt{\frac{215}{205}} = 86,0\ \ \ \ \ $$

$\text{\ \ }\overset{\overline{}}{\lambda_{x}} = \frac{\lambda_{x}}{\lambda_{p}}\ = \frac{79,0}{86,0} = 0,91\ \ \ \ \varphi x = 0,620\ $

N_RC = ψ • A • fd = 1, 0 • 57, 4 • 20, 5 = 1176, 0 kN    

$$\frac{N}{\varphi x \bullet N_{\text{RC}}} = \frac{691,0}{0,620 \bullet 1176} = \frac{691,0}{729,12} = 0,94\ \ < 1\ \ \ warunek\ spelniony$$

Wyboczenie dla osi niemateriałowej y

l1 =  λ_v • i_y = 79, 0 • 2, 22 = 175, 3 cm  ≅ 175 cm 

$$n = \frac{l}{l1} = \frac{630}{175} = 3,6\ przyjeto\ n^{'} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ l1 = \frac{l}{n^{'}} = \frac{630}{5} = 126,0\ cm$$

$$1,1 \bullet 2 \bullet 1821 = 2 \bullet \left\lbrack 142 + 28,7 \bullet \frac{e^{2}}{2} \right\rbrack\ \ \ \ \ \ \ e = 2 \bullet \sqrt{\frac{1,1 \bullet 1821 - 142}{28,7}} = 16,10\ \ \ cm\ \ \ \ \ $$

e = d − e1 • 2 = 24, 0 + 2 • 2, 01 ≅ 20 cm

$$Iy = 2 \bullet \left\lbrack 142 + 28,7 \bullet \frac{20^{2}}{2} \right\rbrack = 11740\ \text{cm}^{4}\text{\ \ \ \ \ \ \ \ \ }i_{y} = \sqrt{\frac{11740}{79,0}} = 12,2\ cm\ \ \ $$

$\ \lambda_{y} = \frac{1,0 \bullet 630}{12,2\ } = 51,6\ \ \ \ \ \ \ \ \ \lambda_{1} = \frac{1,0 \bullet 126}{2,22\ } = 56,7\ \ \ \ \ \ \ \ \ \ \overset{\overline{}}{\lambda_{1}} = \frac{56,7}{84} = 0,675\ \rightarrow \varphi 1 = 0,776\ $

$$\lambda_{\text{my}} = \sqrt{{51,6}^{2} + \frac{2}{2} \bullet 56,7} = \ \ \sqrt{2719} = 52,1\ \ \ \ \begin{matrix} \text{\ \ \ \ \ \ \ } \\ \text{\ \ \ \ \ λ} \\ \end{matrix}_{p} = 84 \bullet \sqrt{\frac{21,5}{20,5}} = \ 85,68\ \ \ \ \ $$

$\text{\ \ \ }\overset{\overline{}}{\lambda_{p}} = \frac{52,1}{84} \bullet \sqrt{0,776} = 0,620 \bullet 0,880 = 0,545\ \ \rightarrow \ \varphi y = 0,919$

N_RC = ψ • A • fd = 0, 776 • 57, 4 • 20, 5 = 913, 1 kN    

$$\frac{N}{\varphi x \bullet N_{\text{RC}}} = \frac{691,0}{0,919 \bullet 913,1} = \frac{691,0}{839,1} = 0,82\ \ < 1\ \ \ warunek\ spelniony$$

Przewiązka pośrednia

Q = 0, 012 • 57, 4 • 20, 5 = 14, 12 kN

$$V_{Q} = \frac{Q \bullet l1}{n(m - 1) \bullet e} = \frac{14,12 \bullet 126}{2(2 - 1) \bullet 20} = \frac{1779}{40} = 44,5\ kN\text{\ \ M}_{Q} = \frac{Q \bullet l1}{m \bullet n} = \frac{1779}{4} = 441,0kNcm$$

Nośność na ścinanie

V_r = 0, 58 • Av • fd = 0, 58 • 0, 9 • 1, 0 • 20, 0 • 20, 5 = 214, 0 kN

$$\frac{44,5}{214} = 0,20 < 1\ warunek\ spelniony$$

Nośność na zgnanie

$$M_{R} = \psi \bullet W_{x} \bullet fd = 1,0 \bullet \frac{1,0 \bullet 20^{2}}{6} \bullet 20,5 = 1365,3\ kNcm$$

$$\frac{441,0}{1356,3} = 0,325 < 1\ \ warunek\ spelniony\ $$

Zakładamy klasę 3 przekroju.

Spoina:

Grubość spoiny przyjmuję a= 6mm

$$\tau_{Q} = \frac{V_{Q}}{\text{Asp}} = \frac{44,5}{0,6 \bullet 20} = 3,70\frac{\text{kN}}{\text{cm}^{2\ }}\text{\ \ \ \ \ \ \ }\tau_{M} = \frac{M_{Q}}{\text{Wsp}} = \frac{441,0}{\frac{0,6 \bullet 20^{2}}{6}} = 6,62\frac{\text{kN}}{\text{cm}^{2\ }} = 66,2\ MPa$$

$$\tau = \sqrt{{\tau_{Q}}^{2} + {\tau_{M}}^{2}} = \sqrt{{3,70}^{2} + {6,62}^{2}} = 7,58\frac{\text{kN}}{\text{cm}^{2\ }} = 75,8\ MPa\ \leq \ \ \alpha_{\bot} \bullet fd = 0,9 \bullet 205 = 184,5MPa$$

Warunek spełniony

Sprawdzenie naprężeń w spoinach przewiązki

Ap = 20, 0 • 1, 0 = 20cm² $W_{p} = \frac{1,0 \bullet 20^{2}}{6} = 66,6\text{cm}^{2}$

$$M_{\text{sp}} = V_{Q} \bullet \frac{d}{2} = 44,5 \bullet \frac{20}{2} = 445,0kNcm$$

$$\tau = \frac{M_{\text{sp}}}{W_{p}} = \frac{445,0}{66,6} = 6,68\frac{\text{kN}}{\text{cm}^{2}} = 66,8MPa\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha_{\bot} \bullet f_{d} = 0,85 \bullet 205 = 174,5\ MPa\ \ \ $$

τ ≤ α_⊥ • f_d       66, 8 MPa ≤ 174, 5MPa      warunek spelniony      

$$\tau_{\text{sp}} = \ \frac{V_{Q}}{A_{p}} = \frac{445,0}{54,0} = 8,25\frac{\text{kN}}{\text{cm}^{2}} = 82,5MPa\ \ \ \ \ \ \ \ \ \ \alpha_{\parallel} \bullet f_{d} = 0,6 \bullet 205 = 123,0\ MPa\ \ \ $$

τ_sp ≤ α_∥ • f_d       82, 5 MPa ≤ 123, 0 MPa      warunek spelniony  

$$\sqrt{\left( \frac{\tau}{\alpha_{\bot}} \right)^{2} + \left( \frac{\tau_{\text{sp}}}{\alpha_{\parallel}} \right)^{2}} \leq f_{\text{d\ }}\text{\ \ \ \ \ \ \ \ \ }\sqrt{\left( \frac{66,8}{0,85} \right)^{2} + \left( \frac{82,5}{0,6} \right)^{2}} \leq 205,0MPa\ \ \ \ \ \ 1580\ MPa \leq 205\ MPa\ \ $$

warunek spelniony           

Blacha trapezowa( przewiązka dolna):

$\sigma_{c} = \frac{N}{A} = \frac{691,0}{52 \bullet 35} = 0,379\frac{\text{kN}}{\text{cm}^{2\ }} = 3,79\ MPa\ \ < fd = 8MPa\ \ $

Grubość blachy postawy $t_{p} = \omega \bullet \sqrt{\frac{\sigma_{c}}{\text{fd}}}$

A → ω • l = 1, 73 • 6, 3 = 10, 89 cm

$$B \rightarrow \frac{b}{l} = \frac{24}{20} = 1,2\ \ \ \ \ \ \ \ \frac{\omega}{l} = 0,56\ \ \ \ \ \omega = 0,56 \bullet 24 = 12,3cm$$

$$C \rightarrow \frac{b}{l} = \frac{14}{20} = 0,70\ \ \ \ \ \frac{\omega}{l} = 0,70\ \ \ \ \ \omega = 0,70 \bullet 20 = 14,0cm\ $$

$$t_{p} = 15,5 \bullet \sqrt{\frac{3,79}{205}} = 15,5 \bullet 0,13 = 2,01\ cm \cong 2cm\ \ \ $$

Naprężenia w przekroju L-L

$$V_{\alpha} = \sigma_{c} \bullet 35 \bullet 14 = 0,379 \bullet 35 \bullet 14 = 185,7kN\ \ \ \ \ \ M_{\alpha} = \ V_{\alpha} \bullet \frac{\text{lz}^{2}}{2} = 185,7 \bullet \frac{14^{2}}{2} = 18130\frac{\text{kN}}{\text{cm}^{2\ \ }}$$

Nośność na ścinanie

V_R = 0, 58 • Av • fd = 0, 58 • (2•1,0•28) • 20, 5 = 665, 0kN

$$\frac{V_{\alpha}}{V_{R}} = \frac{185,7}{665} = 0,27 < 1\ \ \ warunek\ spelniony\ $$

Nośność na zginanie

$$M_{R} = \omega_{x} \bullet fd = \frac{2 \bullet 52^{2}}{6} \bullet 20,5 = 901,0 \bullet 20,5 = 18470,5\frac{\text{kN}}{\text{cm}^{2}}$$

$$\frac{M_{\alpha}}{M_{R}} = \frac{18130}{18470,5} = 0,98 < 1\ \ \ warunek\ spelniony\ $$

Spoina przewiązki dolnej

τ_gr = α_∥ • f_d τ_gr = 0, 6 • 205 = 123, 0 MPa

$$\ \tau_{\parallel} = \frac{P}{4 \bullet a \bullet h} \leq \alpha_{\parallel} \bullet f_{d}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\tau_{\parallel} \leq \ \ \tau_{\text{gr}}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\tau_{\text{gr}} \geq \frac{P}{4 \bullet a \bullet h}\ \ \ \ \ \ \ \ \ \ \ V = \tau_{\text{gr}} \bullet 4 \bullet a \bullet h\ \ \ $$

$$h \geq \frac{P}{4 \bullet \tau_{\text{gr}} \bullet h}\ \geq \ \frac{185,7}{4 \bullet 123,0 \bullet 2,0} \geq 175,0mm\ \ \ \ \ przyjeto\ h = 300mm = 30cm$$

0, 2 • t₁ ≤ a ≤ 0, 7 • t₂  

0, 2 • 20, 0 ≤ a ≤ 0, 7 • 12, 5             4mm ≤ a ≤ 9mm               przyjmuje a = 7mm