Respuesta :
Answer:
k = 2463 N/m
Fs = 81.12 N
Lo = 47.7 mm
Explanation:
given data:
d =2 mm
OD = 22 mm
N_2 = 8.5
from data book
A = 1783 MPa
n = 0.19
[tex]S_[ut] = \frac{1783}{2^{0.19}} = 1563 MPa[/tex]
[tex]S_{sy] = 0.45 S_{ut} = 70.3.4 MPa[/tex]
D = OD - d = 20 mm
c = 2D/2 = 10
[tex]K_b =\frac{4c +2}{4c -3} = \frac{4(10) +2}{4(10) + 3} = 1.135[/tex]
Na = 8.5 -1 = 7/5 turns
[tex]Ls = 2\times 6.5 = 17 mm[/tex]
for solid safe use ns = 1.2
spring rate [tex]k = \frac{Gd^4}{8D^3 Na} = \frac{79.3\times 10^9 (2^4 \times 10^{-12}}{8\times (20^3 \times 10^{-9} \times (7.5)}[/tex]
k = 2463 N/m
solid force
[tex]Fs = \frac{\pi d^3 (S_{sy}/ns) }{8 Kb D}[/tex]
[tex]Fs =\frac{\pi (2^3 \times 10^{-19} (703.4\times 10^6)/1.2}{8\times 1.135\times 20\times 10^{-3}}[/tex]
Fs = 81.12 N
Free length
[tex]Lo = y + Ls = \frac{Fs}{k} + Ls = \frac{81.12}{2463\times 10^{-3}} + 17 = 47.7 mm[/tex]
[tex]Lo_{et} = \frac{2.63 D}{\alpha} = \frac{2.63\times 20}{0.5} = 105.2[/tex]
[tex]\frac{LO_{et}}{Lo} = \frac{105.2}{47.7} = 2.2[/tex]
As Lo is less than 105.2 the spring wiill not be buckle
