τ_max==$\ \frac{\text{Ms}}{\text{Wo}}$<ks

Rs=0,7*Rm

Ks=$\ \frac{\text{Rs}}{n}$

Ms=9550N/n

Wo=π(D⁴-d⁴)/16D

d₁=$\sqrt[3]{\frac{16Ms}{\pi*f*ks}}$ [mm]

ϕ=$\frac{Ms*l}{G*Jo}$ < ϕ_dop

ΣM(x)=0 => M_E+M₄+M₃-M₂+M₁=0 =>M_E=-3kNm

MS1=M1 MS2=M1-M2 MS3=M1-M2+M3

MS4=M1-M2+M3+M4

τ₁=$\frac{Ms1}{Wo1}$ τ₂=$\frac{Ms2}{Wo2}\ $ τ₃=$\frac{Ms3}{Wo2}\ $ τ₄=$\frac{Ms4}{Wo3}$

W_O1=$\frac{\pi(\text{dz}^{4} - \text{dw}^{4})}{16dz}$ W_O2=$\frac{\pi{d2}^{3}}{16}$ W_O3=$\frac{\pi{d1}^{3}}{16}$

ϕ_ΑΒ=$\frac{Ms*2a}{G*Jo1}$ ϕ_ΒC=$\frac{Ms*3a}{G*Jo2}\ $ ϕ_CD=$\frac{Ms*a}{G*Jo2}$ ϕ_DE=$\frac{Ms*4a}{G*Jo3}$

J_O1=$\frac{\pi(\text{dz}^{4} - \text{dw}^{4})}{32}$ J_O2=$\frac{\pi{d2}^{4}}{32}$ J_O3=$\frac{\pi{d1}^{4}}{32}$

ϕ_ΕC = ϕ_DE + ϕ_CD ϕ_ΕB = ϕ_EC + ϕ_CB ϕ_ΕA = ϕ_EB + ϕ_BA

D=25δ W₁=2Fo*δmin=2*$\frac{\text{πD}}{4}^{2}$*δ

J₁=$\frac{{4Fo}^{2}}{\oint_{}^{}\frac{\text{ds}}{\delta}}$=$\frac{4*{(\frac{\pi D^{2}}{4})}^{2}}{\frac{1}{\delta}*\pi D}$=$\frac{\pi D^{3}*\delta}{4}$

W₂=$\frac{J_{2}}{\text{δmax}}$=$\frac{\pi\text{Dδ}^{3}}{3\delta}$=$\frac{\pi\text{Dδ}^{2}}{3}$

J₂=κ*$\frac{1}{3}$Σδi³bi=$\frac{1}{3}$bi*δ³=$\frac{\pi\text{Dδ}^{3}}{3}$

k₁₌$\frac{W1}{W2}$₌$\frac{\pi D^{2}\delta*3}{2*\pi\text{Dδ}^{2}}$=$\frac{3D}{2\delta}$=$\frac{3*25\delta}{2\delta}$=37,5razy

k2=$\frac{J1}{J2}$₌$\frac{\pi D^{3}\delta*3}{4*\pi\text{Dδ}^{3}}$=$\frac{3D^{2}}{{4\delta}^{2}}$=$\frac{3*25^{2}\delta^{2}}{4\delta^{2}}$=486,75razy

J=κ*$\frac{1}{3}$Σδi³bi=1,2$\frac{1}{3}$[2*90*11,3³+177,4*7,5³]=

133825mm⁴ W=$\frac{J}{\text{δmax}}$=$\frac{133825}{11,3}$=11843mm³

τ_max==$\ \frac{\text{Ms}}{W}$=$\frac{400*10^{3}}{11843}$≈33,8MPa

ϕ=$\frac{\text{Ms}*l}{G*J}$=$\frac{400*10^{3}*10^{3}}{8**10^{4}*133825}$=0,0374rad=>2⁰08’

F=$\frac{\pi d^{2}}{16}$ =>d=$\sqrt{\frac{4F}{\pi}}$=$\sqrt{\frac{4*3350}{\pi}}$=65,3mm

W_O2=$\frac{\pi d^{3}}{16}$ J_O2=$\frac{\pi d^{4}}{32}$

τ_1max==$\ \frac{\text{Ms}}{\text{Wo}}$=$\frac{400*10^{3}*16}{\pi*{65,3}^{3}}$=7,3MPa

ϕ₁=$\frac{\text{Ms}*l}{G*\text{Jo}}$=$\frac{400*10^{3}*10^{3}*32}{8**10^{4}*\pi*{65,3}^{4}}$=0,0028rad=>0^o10’

k₁=$\frac{\text{Tmax}}{T1}$≈4,63 razy wytrzymalszy

k₂=$\frac{\varphi}{\varphi 1}$=$\frac{2^{o}08'}{10'}$=13,6 razy sztywniejszy

1-1zerwanie śruby

σr=$\frac{N}{\text{Fs}}$ ≤kr kr=$\frac{\text{Rm}}{n}$ Fs=$\frac{\pi\text{dr}^{2}}{4}$

$\frac{4P}{\pi\text{dr}^{2}}$ <kr => dr>$\sqrt{\frac{4P}{\text{πkr}}}$

2-2 ścięcie łba śruby

τ₁=$\frac{T}{F}$ <kt F=πDh Kt=0,6kr

$\frac{P}{\text{πDh}}$ <kt => h>$\frac{P}{\text{πDkt}}$

3-3 przekroczenie nacisków dopuszczalnych

P=$\frac{T}{\text{Fn}}$ <p_dop p_dop=2kc kc=$\frac{\text{Rc}}{n}$

Fn=$\frac{\pi(D^{2} - d^{2})}{4}$

$\frac{4P}{\pi(D^{2} - d^{2})}$ <p_dop=>D>$\sqrt{\frac{4P}{\pi\left( - d^{2} \right)*\text{pdop}}}$

i>$\frac{4P}{\pi d^{2}\text{kt}}$ t=2,2do do=d+1 t=2,2do

e₁=1,2do bk=t+2e₁ e=1,6do a=2do

l=2a+2e σ=$\frac{P}{\text{Fo}}$ <kr Fo=g*b-2$\frac{\pi\text{do}^{2}}{4}$

$\frac{P}{b*g - \frac{\text{πdo}^{2}}{2}}$ <kr=>b=$\frac{P}{kr*g - \frac{\text{πdo}^{2}}{2}}$

Po=$\frac{P}{n}$ M=P*L=30(0,08+0,18)=7,8kNm

Pi=l*ri M=$\sum_{1}^{n}{\text{Pi}*\text{ri} = k\sum_{1}^{n}\text{ri}^{2}} = \frac{\text{Pi}}{\text{ri}}\sum_{1}^{n}\text{ri}^{2}$

Pi=$\frac{\text{Mri}}{\sum_{1}^{n}\text{ri}^{2}}$

ρ₁²= ρ₃²=ρ₄²=ρ₆²=80²+100²=16400mm²

ρ₂²= ρ₅²=80²=6400mm²

$\sum_{1}^{6}\text{ri}^{2}$=4*16400+2*6400=78400mm²

P₁=P₃=P₄=P₆=$\frac{7,8*10^{6}\sqrt{16400}}{78400}$=12741N

P₂=P₅=$\frac{7,8*10^{6}*80}{78400}$=7959N

T₁=$\sqrt{P_{1}^{2} + P_{0}^{2} + 2\text{Po}P_{1}\text{sinα}}$=

$\sqrt{12741^{2} + 5000^{2} + 2*12741*5000*0,625}$=16339N

T₂=Po+P₂=5000+7959=12959N

τ=$\frac{4T_{1}}{\pi d^{2}}$ <kt => d>$\sqrt{\frac{4T_{1}}{\text{πkt}}}$=$\sqrt{\frac{4*16339}{\pi*70}}$=17,2

Przyjmujemy d=18mm

τ₌=$\ \frac{P}{\text{Fs}}$<kts Fs=2(l₁+l₂)*gs*0,7

$\frac{P}{2*\left( l1 + l2 \right)gs*0,7}$ <kts Kts=0,65kr

(1)2(l₁+l₂)=$\ \frac{P}{gs*0,7*\text{kts}}$ $\frac{l1}{l2}$ =$\frac{e2}{e1}$ (2)

L₁=l₂$\frac{e2}{e1}$ Wst. do (1) 2$(\frac{l2e2}{e1} + l2)$

τ₌=$\ \frac{P}{\text{Fs}}$<kt $\frac{4P}{\pi d^{2}}$ <kt=>d=$\sqrt{\frac{4P}{\text{πkt}}}$

Ms=Mo=P*$\frac{D}{2}$ 2Mo=P*D

P=$\frac{2Mo}{D}$=$\frac{2*6800000}{215}$=63225N

Dla i=6 P=$\frac{63225}{6}$=10542,5

D=$\sqrt{\frac{4*10542,5}{\pi*60}}$≈15mm

Fs=B*l τ₌=$\ \frac{P}{\text{Fs}}$<kt $\frac{P}{b*l}$ <kt=>l> $\frac{P}{b*\text{kt}}$

l> $\frac{24000}{8*80}$=37,5 Mo= $\frac{P*d}{2}$=>P= $\frac{2Mo}{d}$=24kN

P<F*kr=g*b*kr=30*60*120=216000N

P<F*kr=(D-d)g*kr=60*30*120=216000N

P<F*kd=g*d*kd=30*40*180=216000

P<F*kt=2$\frac{\pi d^{2}}{4}$kt=2$\frac{\pi 40^{2}}{4}$*70=175930

Decyduje najmniejsza wartość siły P

ΣM(x)=0=>P₁*$\frac{D1}{2}$+P₂*$\frac{D2}{2}$=0=>P₂=…

Ms= P₁*$\frac{D1}{2}$= P₂*$\frac{D2}{2}$

σ_z1=$\sqrt{\sigma{g_{1}}^{2} + 3{\tau_{1}}^{2}}$ <kg W_O1=$\frac{\pi{d_{1}}^{3}}{16}$

τ₁=$\frac{\text{Ms}}{Wo_{1}}$ <ks=>d₁<$\sqrt{\frac{16\text{Ms}}{\text{πks}}}$

σ_z2=$\sqrt{\sigma{g_{A}}^{2} + 3{\tau_{A}}^{2}}$ <kg W_Z2=$\frac{\pi{d_{2}}^{3}}{32}$

σg_A=$\frac{\text{Mg}_{A}}{\text{Wz}2}$=$\frac{\text{Ra}*a*32}{\pi{d_{2}}^{3}}$ W_O2=$\frac{\pi{d_{2}}^{3}}{16}$

τ_Α=$\frac{\text{Ms}}{Wo_{2}}$ σ_z2=>d₃

σ_z3=$\sqrt{\sigma{g_{3}}^{2} + 3{\tau_{3}}^{2}}$ <kg W_Z3=$\frac{\pi{d_{3}}^{3}}{32}$

σg₃=$\frac{\text{Mg}_{3}}{Wz3}$=$\frac{RB*3a*32}{\pi{d_{3}}^{3}}$ W_O3=$\frac{\pi{d_{3}}^{3}}{16}$

τ₃=$\frac{\text{Ms}}{Wo_{3}}$ σ_z3=>d3

W_Z4=$\frac{\pi{d_{4}}^{3}}{32}$

σg₄=$\frac{\text{Mg}_{3}}{Wz4}$<kg => d₄<$\sqrt{\frac{32\text{Mg}3}{\text{πkg}}}$

ΣM(x)=0=>P*r-Q*2r=>Q=$\frac{1}{2}$P Ms=P*r=Q*2r

Μ_g1=$\sqrt{M{g_{z1}}^{2} + {\text{Mg}_{y1}}^{2}}$=$\sqrt{{(\frac{1}{4}pl)}^{2} + {(\frac{3}{8}pl)}^{2}}$

Μ_g2=$\sqrt{M{g_{z2}}^{2} + {\text{Mg}_{y2}}^{2}}$=$\sqrt{{(\frac{3}{4}pl)}^{2} + {(\frac{1}{8}pl)}^{2}}$

Μ_z1=$\sqrt{M{g_{1}}^{2} + {0,75\text{Ms}}^{2}}$

Μ_z2=$\sqrt{M{g_{2}}^{2} + {0,75\text{Ms}}^{2}}$

σz=$\frac{\text{Mz}}{\text{Wz}}$ <kg $\frac{32\text{Mz}_{2}}{\pi d^{3}}$<kg = d>$\sqrt{\frac{32\text{Mz}2}{\text{πkg}}}$

Jz=$\frac{30*200^{3}}{12}$+30*200(a1)²+$\frac{120*40^{3}}{12}$+120*40(a2)²

σ_A=$\frac{\text{Mg}}{\text{Iz}}$*ya=$\frac{Q*L}{\text{Iz}}$*ya τ_Α=$\frac{\text{Tmax}*\text{Syz}}{\text{Iz}*b}$=0

σ_B=$\frac{Q*L}{\text{Iz}}$*yb τ_Β=$\frac{\text{Tmax}*\text{Syz}}{\text{Iz}*b}$=$\frac{\frac{Q}{2}*120*40*a2}{\text{Iz}*30}$

σ_C =$\frac{Q*L}{\text{Iz}}$*0=0 τ_C=$\frac{\frac{Q}{2}*(120*40*a2 + 30*\text{yb}*\frac{\text{yb}}{2})}{\text{Iz}*30}$

σ_D=$\frac{Q*L}{\text{Iz}}$*yo τ_D=$\frac{\text{Tmax}*\text{Syz}}{\text{Iz}*b}$=0

σ_zA=$\sqrt{{\sigma_{A}}^{2} + 3{\tau_{A}}^{2}}$ σ_zB=$\sqrt{{\sigma_{B}}^{2} + 3{\tau_{B}}^{2}}$

σ_zC=$\sqrt{{\sigma_{C}}^{2} + 3{\tau_{C}}^{2}}$ σ_zD=$\sqrt{{\sigma_{D}}^{2} + 3{\tau_{D}}^{2}}$

σ_zmax=kg=kr Kt=0,6kr(kg) τ_S= $\frac{Q}{\text{Fs}}$ <kt=>gs

Fs=ls*gs*0,7 Ls=2(2l+3a)

Kw=β*kc

σ_z=σ_c+σ_g<kr(kc) Mg=P*e

E=$\frac{4r}{3\pi}$ Wz1=$\frac{\text{Iz}}{r - e}$

Iz=$\frac{\pi d^{4}}{128}$dla koła Iz=$\frac{a^{4}}{72}$dla kwadratu

σ_z^r=-$\frac{P}{F}$+$\frac{\text{Mg}}{\text{Wz}1}$ <kr=>P1<…

Wz2= $\frac{\text{Iz}}{e}$

σ_z^c=-$\frac{P}{F}$-$\frac{\text{Mg}}{\text{Wz}2}$ <kc=>P2<…

S=$\frac{\text{Lw}}{i}$ i=Jmin/Fmin Sgr=π$\sqrt{\frac{E}{\text{Rs}}}$

S<Sgr=>wzór Tetmajera Jaśińskiego=>P3

σ=a-b*s=>σ=Re-$\frac{\text{Re} - \text{Rs}}{\text{Sgr}}$ σ=$\frac{P}{F}$ <kw

S>Sgr =>Wzór Eulera=>P_kr=$\frac{\pi^{2}*E*\text{Jmin}}{\text{Lw}^{2}}$

σkr=$\frac{\text{Pkr}}{F}$