KATEDRA PRZERÓBKI PLASTYCZNEJ

PRZERÓBKA PLASTYCZNA – PROJEKT WYTŁACZANIA

3MM-DI, P1

Mateusz Baran
Wojciech Bonar
Paweł Chuchla

1. Dane:

materiał	Re [MPa]	Rm [MPa]	n	C [MPa]	r	Dk [MPa]	d [mm]	h [mm]	go [mm]
Al	45	80	0,22	160	1,1	180	150	420	1,5

1. Obliczenia geometrii wsadu:

$${V_{o} = \ V_{w} = \ \frac{\left\lbrack \pi{D_{k}}^{2}g_{o} - \ \pi d^{2}g_{o} \right\rbrack + \left\lbrack \pi\left( d + 2g_{o} \right)^{2}h - \pi d^{2}\left( h - g_{o} \right) \right\rbrack}{4}\backslash n}{V_{o} = \ \frac{\left\lbrack \pi 180^{2} 1,5 - \pi 150^{2} 1,5 \right\rbrack + \left\lbrack \pi \left( 150 + 2 1,5 \right)^{2} 420 - \pi 150^{2} \left( 420 - 1,5 \right) \right\rbrack}{4}}$$

V_o = 338019, 66 mm³

$$V_{o} = \frac{\pi{D_{o}}^{2}}{4}g_{o}\ \rightarrow \ D_{o} = \ \sqrt{\frac{4V_{o}}{\pi g_{o}}}$$

$$D_{o} = \ \sqrt{\frac{4 338019,66}{\pi 1,5}} = 535,65\ mm$$

Do obliczonej średnicy D_o należy dodać 7% jako naddatek technologiczny

D_o = 535, 651, 07 = 573, 14 mm

1. Obliczenia granicznej wartości współczynnika odkształcenia β_gr­:

$\beta = \frac{D_{o}}{d} = \frac{573,14}{150} = 3,82096$ – współczynnik odkształcenia

D_o – średnica krążka wyjściowego

d – średnica wytłoczki

β_gr = β_materialβ_{geometria wsadu}β_{geometria nadzedzi}β_tarcie

gdzie:

• β_m – czynnik uwzględniający właściwości materiału

$$\beta_{m} = (1,56 + 0,12r)(1 + 0,08n)(1,05 - 0,09\frac{R_{e}}{R_{m}})$$

$$\mathbf{\beta}_{\mathbf{m}} = \left( 1,56 + 0,12 1,1 \right)\left( 1 + 0,08 0,22 \right)\left( 1,05 - 0,09\frac{45}{80} \right) = 1,72$$

• β_gw – czynnik uwzględniający geometrię wsadu

$$\beta_{\text{gw}} = 0,65 + 18\left( \frac{g_{o}}{D_{o}} \right) + 0,15{(\frac{D_{k}}{d + 2g_{o}})}^{\frac{1}{2}}$$

$$\mathbf{\beta}_{\mathbf{\text{gw}}} = 0,65 + 18 \frac{1,5}{573,14} + 0,15 {(\frac{180}{150 + 2 1,5})}^{\frac{1}{2}} = 0,86$$

• β_gn – czynnik uwzględniający geometrię narzędzi

$$\beta_{\text{gn}} = \left( 1,08 - \frac{0,57}{\rho_{m}} \right)\left( 1,02 - \frac{0,19}{\rho_{s}} \right)$$

ρ_m = (2÷6)g_o - promień zaokrąglenia matrycy

Przyjmujemy ρ_m = 6g_o

ρ_m = 5g_o = 51, 5 =  7, 5 mm 

ρ_s = (5÷10)g_o - promień zaokrąglenia stempla

ρ_s = 7g_o = 71, 5 =  10, 5 mm

$$\mathbf{\beta}_{\mathbf{\text{gn}}} = \left( 1,08 - \frac{0,57}{7,5} \right)\left( 1,02 - \frac{0,19}{10,5} \right) = \ 1,02$$

• β_t – czynnik uwzględniający tarcie

β_t = 2exp(0,1−μ₁) − 1

gdzie:

µ₁ = 0,08 – dla smaru – Mo₂s + olej

µ₁ = 0,10 – dla smaru – olej maszynowy

µ₁ = 0,15 – dla smaru – folia PE

µ₁ = 0,20 – bez smarowania

przyjmujemy µ₁ = 0,08 – dla smaru – Mo₂s + olej

β_t = 2exp(0,1−0,08) − 1 =  1, 04

β_gr = β_mβ_gwβ_gnβ_t = 1, 72 0, 841, 021, 04 = 1, 56

Musi być spełniona zależność:

β < β(1) β(2)β(3)β(4)

przyjmujemy:

β(1) = β_gr = 1,56

β(2) = 0,64 · 1,56 = 1,0004

β(3) = 0,67 · β_gr = 1,56

β(4) = β_gr = 1,56

sprawdzamy:

3,82096 < 1,56 · 1,56 · 1,0004 · 1,56

3,82096 < 3,82116 – zależność prawdziwa

Wykonamy w tym procesie wytłaczania 4 operacje tłoczenia

Obliczamy geometrię wytłoczek pośrednich:

$$d_{1} = \ \frac{D_{o}}{\beta_{1}} = \frac{573,14}{1,56} = 366,657\ mm$$

$$d_{2} = \ \frac{d_{1}}{\beta_{2}} = \frac{366,567}{1,0004} = 366,502\ mm$$

$$d_{3} = \ \frac{d_{2}}{\beta_{3}} = \frac{239,07}{1,04} = 234,46\ mm$$

$$d_{4} = \ \frac{d_{3}}{\beta_{4}} = \frac{230,45}{1,55} = 149,99\ mm$$

$$h_{1} = \ \frac{{D_{o}}^{2} - {d_{1}}^{2}}{4(d_{1} + g_{o})} = \ \frac{{573,14}^{2} - {370,17}^{2}}{4 (370,17 + 1,5)} = 131,78\ mm$$

$$h_{2} = \ \frac{{D_{o}}^{2} - {d_{2}}^{2}}{4(d_{2} + g_{o})} = \ \frac{{573,14}^{2} - {239,07}^{2}}{4 (239,07 + 1,5)} = 131,91\ mm$$

$$h_{3} = \ \frac{{D_{o}}^{2} - {d_{3}}^{2}}{4(d_{3} + g_{o})} = \ \frac{{573,14}^{2} - {230,45}^{2}}{4 (230,45 + 1,5)} = 289,79\ mm$$

$$h_{4} = \ \frac{{D_{o}}^{2} - {d_{4}}^{2}}{4(d_{4} + g_{o})} = \ \frac{{573,14}^{2} - {148,84}^{2}}{4 (148,84 + 1,5)} = 504,97\ mm$$

1. Obliczenia wielkości docisku:

Nacisk jednostkowy q (wzór Siebla):

$$q_{1} = 0,00225\left\lbrack \left( \beta_{1} - 1 \right)^{2} + 0,01\ \frac{\beta_{1}d_{1}}{2g_{o}} \right\rbrack C\left( \frac{n}{e} \right)^{n}$$

$$q_{1} = 0,00225\left\lbrack \left( 1,55 - 1 \right)^{2} + 0,01\ \frac{1,55 370,17}{2 1,5} \right\rbrack C{(\frac{0,22}{e})}^{0,22} = 0,396\ MPa$$

Całkowita siła docisku Q:

$$Q_{1} = \ \frac{\pi}{4}\lbrack{D_{o}}^{2} - \left( d + 2\rho_{m} \right)^{2}\rbrack q_{1}$$

$$Q_{1} = \ \frac{\pi}{4}\left\lbrack {573,14}^{2} - \left( 150 + 2 7,5 \right)^{2} \right\rbrack 0,396 = 93458,59N = 93,459\ kN\ $$

1. Obliczenia siły wytłaczania:
   
   P = (P_id+P_gn+P_tk+P_tm)sinα

gdzie:

P_id – siła niezbędna do uplastycznienia materiału
P_gn – siły gięcia na promieniu matrycy

P_tk – siły tarcia na powierzchni kontaktu kołnierza z matrycą i dociskaczem

P­_tm – siła tarcia na promieniu matrycy

α – kąt jaki tworzy ścianka wytłoczki z kierunkiem działania siły

$$sin\alpha = \ \frac{h}{\rho_{m} + \rho_{s} + g_{o}}$$

dla h > ρ_m + ρ_s + g_o     sinα = 1

9 + 9 + 1, 5 = 19, 5 < 420 stad sinα = 1

Wartości poszczególnych sił cząstkowych oblicza się według następujących zależności:

$$R_{s} = \frac{d}{2} + g_{o} = \frac{366}{2} + 1,5 = 186,cos\text{\ mm}$$

$$R_{0} = \frac{D_{0}}{2} = \ \frac{573,14}{2} = 286,57\ mm\backslash n$$

R₁ = 77

R₂ = 100 tu się zaczyna od 185 mniej więcej tu był błąd

R₃ = 123

R₄ = 146

R₅ = 169

R₆ = 192

R₇ = 215

R₈ = 238

R₉ = 261

R₁₀ = 284

$$\mathbf{P}_{\mathbf{\text{id}}}\mathbf{= 2}\mathbf{\text{πC}}\mathbf{R}_{\mathbf{s}}\mathbf{g}\left\lbrack \ln\left( \frac{\mathbf{R}_{\mathbf{o}}}{\mathbf{R}} \right) \right\rbrack^{\mathbf{n}}\ln\left( \frac{\mathbf{R}}{\mathbf{R}_{\mathbf{s}}} \right)$$

gdzie:

$${g = g_{o}exp\lbrack 0,5\ln\left( \frac{R_{o}}{R} \right)\rbrack\backslash n}{g_{1} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{1}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{77})} = 2,894\ mm}$$

$$g_{2} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{2}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{100})} = 2,593\ mm$$

$$g_{3} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{3}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{123})} = 2,289\ mm$$

$$g_{4} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{4}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{146})} = 2,102\ mm$$

$$g_{5} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{5}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{169})} = 1,953\ mm$$

$$g_{6} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{6}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{192})} = 1,833\ mm$$

$$g_{7} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{7}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{215})} = 1,732\ mm$$

$$g_{8} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{8}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{238})} = 1,646\ mm$$

$$g_{9} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{9}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{261})} = 1,572\ mm$$

$$g_{10} = g_{o}\exp\left\lbrack 0,5\ln\left( \frac{R_{o}}{R_{10}} \right) \right\rbrack = 1,5 e^{0,5 ln(\frac{286,57}{284})} = 1,507\ mm$$

$$P_{id1} = 2\pi CR_{s}g_{1}\left\lbrack \ln\left( \frac{R_{o}}{R_{1}} \right) \right\rbrack^{n}\ln\left( \frac{R_{1}}{R_{s}} \right) = 2 \pi 160 76,5 2,894 \left\lbrack \ln\left( \frac{286,57}{77} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{77}{76,5} \right) = 1539,64\ N$$

$$P_{id2} = 2\pi CR_{s}g_{2}\left\lbrack \ln\left( \frac{R_{o}}{R_{2}} \right) \right\rbrack^{n}\ln\left( \frac{R_{2}}{R_{s}} \right) = 2 \pi 160 76,5 2,539 \left\lbrack \ln\left( \frac{286,57}{100} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{100}{76,5} \right) = 52908,39\ N$$

$$P_{id3} = 2\pi CR_{s}g_{3}\left\lbrack \ln\left( \frac{R_{o}}{R_{3}} \right) \right\rbrack^{n}\ln\left( \frac{R_{3}}{R_{s}} \right) = 2 \pi 160 76,5 2,289 \left\lbrack \ln\left( \frac{286,57}{123} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{123}{76,5} \right) = 80595,43\ N$$

$$P_{id4} = 2\pi CR_{s}g_{4}\left\lbrack \ln\left( \frac{R_{o}}{R_{4}} \right) \right\rbrack^{n}\ln\left( \frac{R_{4}}{R_{s}} \right) = 2 \pi 160 76,5 2,102 \left\lbrack \ln\left( \frac{286,57}{146} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{146}{76,5} \right) = 95784,4\ N$$

$$P_{id5} = 2\pi CR_{s}g_{5}\left\lbrack \ln\left( \frac{R_{o}}{R_{5}} \right) \right\rbrack^{n}\ln\left( \frac{R_{5}}{R_{s}} \right) = 2 \pi 160 76,5 1,953 \left\lbrack \ln\left( \frac{286,57}{169} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{169}{76,5} \right) = 103461,2\ N$$

$$P_{id6} = 2\pi CR_{s}g_{6}\left\lbrack \ln\left( \frac{R_{o}}{R_{6}} \right) \right\rbrack^{n}\ln\left( \frac{R_{6}}{R_{s}} \right) = 2 \pi 160 76,5 1,833 \left\lbrack \ln\left( \frac{286,57}{192} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{192}{76,5} \right) = 106040,4\ N$$

$$P_{id7} = 2\pi CR_{s}g_{7}\left\lbrack \ln\left( \frac{R_{o}}{R_{7}} \right) \right\rbrack^{n}\ln\left( \frac{R_{7}}{R_{s}} \right) = 2 \pi 160 76,5 1,732 \left\lbrack \ln\left( \frac{286,57}{215} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{215}{76,5} \right) = 104602,9\ N$$

$$P_{id8} = 2\pi CR_{s}g_{8}\left\lbrack \ln\left( \frac{R_{o}}{R_{8}} \right) \right\rbrack^{n}\ln\left( \frac{R_{8}}{R_{s}} \right) = 2 \pi 160 76,5 1,646 \left\lbrack \ln\left( \frac{286,57}{238} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{238}{76,5} \right) = 99200,19\ N$$

$$P_{id9} = 2\pi CR_{s}g_{9}\left\lbrack \ln\left( \frac{R_{o}}{R_{9}} \right) \right\rbrack^{n}\ln\left( \frac{R_{9}}{R_{s}} \right) = 2 \pi 160 76,5 1,572 \left\lbrack \ln\left( \frac{286,57}{261} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{261}{76,5} \right) = 88067,09\ N$$

$$P_{id10} = 2\pi CR_{s}g_{10}\left\lbrack \ln\left( \frac{R_{o}}{R_{10}} \right) \right\rbrack^{n}\ln\left( \frac{R_{10}}{R_{s}} \right) = 2 \pi 160 76,5 1,507 \left\lbrack \ln\left( \frac{286,57}{284} \right) \right\rbrack^{0,22}\operatorname{ln}\left( \frac{284}{76,5} \right) = 53933,97\ N$$

$$\mathbf{P}_{\mathbf{\text{gn}}}\mathbf{= \pi C}\mathbf{g}^{\mathbf{2}}\frac{\mathbf{R}_{\mathbf{s}}}{\mathbf{\rho}_{\mathbf{m}}}{\mathbf{\lbrack}\frac{\mathbf{g}}{\mathbf{2}\mathbf{\rho}_{\mathbf{m}}}\mathbf{+ ln}\frac{\mathbf{R}_{\mathbf{0}}}{\mathbf{R}}\mathbf{\rbrack}}^{\mathbf{n}}$$

$$P_{gn1} = \pi C{g_{1}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{1}}{2\rho_{m}} + ln\frac{R_{0}}{R_{1}}\rbrack}^{n} = \pi 160 {2,894}^{2} \frac{76,5}{9} {\lbrack\frac{2,894}{2 9} + ln\frac{286,57}{77}\rbrack}^{0,22} = 38971,02\text{\ N}$$

$$P_{gn2} = \pi C{g_{2}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{2}}{2\rho_{m}} + ln\frac{R_{0}}{R_{2}}\rbrack}^{n} = \pi 160 {2,593}^{2} \frac{76,5}{9} {\lbrack\frac{2,593}{2 9} + ln\frac{286,57}{100}\rbrack}^{0,22} = 28643,99\ N$$

$$P_{gn3} = \pi C{g_{3}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{3}}{2\rho_{m}} + ln\frac{R_{0}}{R_{3}}\rbrack}^{n} = \pi 160 {2,289}^{2} \frac{76,5}{9} {\lbrack\frac{2,289}{2 9} + ln\frac{286,57}{123}\rbrack}^{0,22} = 22262,88\ N$$

$$P_{gn4} = \pi C{g_{4}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{4}}{2\rho_{m}} + ln\frac{R_{0}}{R_{4}}\rbrack}^{n} = \pi 160 {2,102}^{2} \frac{76,5}{9} {\lbrack\frac{2,102}{2 9} + ln\frac{286,57}{146}\rbrack}^{0,22} = 17921,04\ N$$

$$P_{gn5} = \pi C{g_{5}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{5}}{2\rho_{m}} + ln\frac{R_{0}}{R_{5}}\rbrack}^{n} = \pi 160 {1,953}^{2} \frac{76,5}{9} {\lbrack\frac{1,953}{2 9} + ln\frac{286,57}{169}\rbrack}^{0,22} = 14759,3\ N$$

$$P_{gn6} = \pi C{g_{6}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{6}}{2\rho_{m}} + ln\frac{R_{0}}{R_{6}}\rbrack}^{n} = \pi 160 {1,833}^{2} \frac{76,5}{9} {\lbrack\frac{1,833}{2 9} + ln\frac{286,57}{192}\rbrack}^{0,22} = 12331,39\ N$$

$$P_{gn7} = \pi C{g_{7}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{7}}{2\rho_{m}} + ln\frac{R_{0}}{R_{7}}\rbrack}^{n} = \pi 160 {1,732}^{2} \frac{76,5}{9} {\lbrack\frac{1,732}{2 9} + ln\frac{286,57}{215}\rbrack}^{0,22} = 10377,8\ N$$

$$P_{gn8} = \pi C{g_{8}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{8}}{2\rho_{m}} + ln\frac{R_{0}}{R_{8}}\rbrack}^{n} = \pi 160 {1,646}^{2} \frac{76,5}{9} {\lbrack\frac{1,646}{2 9} + ln\frac{286,57}{238}\rbrack}^{0,22} = 8728,181\ N$$

$$P_{gn9} = \pi C{g_{9}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{9}}{2\rho_{m}} + ln\frac{R_{0}}{R_{9}}\rbrack}^{n} = \pi 160 {1,572}^{2} \frac{76,5}{9} {\lbrack\frac{1,572}{2 9} + ln\frac{286,57}{261}\rbrack}^{0,22} = 7244,958\ N$$

$$P_{gn10} = \pi C{g_{10}}^{2}\frac{R_{s}}{\rho_{m}}{\lbrack\frac{g_{10}}{2\rho_{m}} + ln\frac{R_{0}}{R_{10}}\rbrack}^{n} = \pi 160 {1,507}^{2} \frac{76,5}{9} {\lbrack\frac{1,507}{2 9} + ln\frac{286,57}{284}\rbrack}^{0,22} = 5748,571\ N$$

P_tk=Qμ₁

P_tk = Qμ₁ = 93458, 590, 08 = 7476, 687 N

$$\mathbf{P}_{\mathbf{\text{tm}}}\mathbf{= \lbrack exp(}\frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{\mu}_{\mathbf{2}}\mathbf{- 1\rbrack(}\mathbf{P}_{\mathbf{\text{id}}}\mathbf{+}\mathbf{P}_{\mathbf{\text{tk}}}\mathbf{)}$$

μ₂ = 2μ₁ = 20, 08 = 0, 16

$$P_{tm1} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id1} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 1539,64\ + 7476,687 \right) = 4264,669\ N$$

$$P_{tm2} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id2} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 52908,39 + 7476,687 \right) = 28561,78\ N$$

$$P_{tm3} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id3} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 80595,43 + 7476,687 \right) = 41657,58\ N$$

$$P_{tm4} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id4} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 95784,4\ + 7476,687 \right) = 48841,88\ N$$

$$P_{tm5} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id5} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 103461,2 + 7476,687 \right) = 52472,97\ N$$

$$P_{tm6} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id6} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 106040,4 + 7476,687 \right) = 53692,89\ N$$

$$P_{tm7} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id7} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 99200,19 + 7476,687 \right) = 53012,97\ N$$

$$P_{tm8} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id8} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 88067,09 + 7476,687 \right) = 45191,63\ N$$

$$P_{tm2} = \left\lbrack \exp\left( \frac{\pi}{2}\mu_{2} - 1 \right) \right\rbrack\left( P_{id2} + P_{\text{tk}} \right) = \left\lbrack \exp\left( \frac{\pi}{2} 0,16 - 1 \right) \right\rbrack\left( 53933,97 + 7476,687 \right) = 29046,87\ N$$

P=(P_id+P_gn+P_tk+P_tm)sinα

sinα=1, stąd:

P=(P_id+P_gn+P_tk+P_tm)

P₁ = (P_id1+P_gn1+P_tk+P_tm1) = 1539, 641 + 38971, 02 + 7476, 687 + 4264, 699 = 52252, 02 N

P₂ = (P_id2+P_gn2+P_tk+P_tm2) = 52908, 39 + 28643, 99 + 7476, 687 + 28561, 78 = 117590, 8 N

P₃ = (P_id3+P_gn3+P_tk+P_tm3) = 80595, 43 + 22262, 88 + 7476, 687 + 41657, 58 = 151992, 6 N

P₄ = (P_id4+P_gn4+P_tk+P_tm4) = 95784, 4 + 17921, 04 + 7476, 687 + 48841, 88 = 170024 N

P₅ = (P_id5+P_gn5+P_tk+P_tm5) = 103461, 2 + 14759, 3 + 7476, 687 + 52472, 97 = 178170, 2 N

P₆ = (P_id6+P_gn6+P_tk+P_tm6) = 106040, 4 + 12331, 39 + 7476, 687 + 53692, 89 = 179542, 3 N

P₇ = (P_id7+P_gn7+P_tk+P_tm7) = 104602, 9 + 10377, 8 + 7476, 687 + 53012, 97 = 175470, 4 N

P₈ = (P_id8+P_gn8+P_tk+P_tm8) = 99200, 19 + 8728, 181 + 7476, 687 + 50457, 52 = 165862, 6 N

P₉ = (P_id9+P_gn9+P_tk+P_tm9) = 88067, 09 + 7244, 958 + 7476, 687 + 45191, 63 = 147980, 4 N

P₂ = (P_id2+P_gn2+P_tk+P_tm2) = 53933, 97 + 5748, 571 + 7476, 687 + 29046, 87 = 96206, 09 N

1. Obliczenia siły zrywającej:

$$P_{\text{zr}} = 2\pi R_{s}g_{o}\left\lbrack 1 - \frac{g_{0}}{2\rho_{s}} + \mu_{s}\frac{h}{2R_{s}} \right\rbrack R_{m} = 2 \pi 76,5 1,5 \left\lbrack 1 - \frac{1,5}{2 9} + 0,1 \frac{420}{2 76,5} \right\rbrack 80 = 68706,63\ $$

P_w = klg₀R_t

gdzie:

k – współczynnik wykrawania (k=1,25)
l – długość cięcia ( l = 2πR₀ )

R_t – wytrzymałość na cięcie ( R_t = 0, 8R_m )

P_w = klg₀R_t = 1, 252π286, 571, 50, 880 = 216068, 69 N