Answer
given,
D = 50 mm = 0.05 m
d = 10 mm = 0.01 m
Force to compress the spring
[tex]F = \dfrac{d^4G\delta}{8D^3N}[/tex]
[tex]\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm[/tex]
[tex]F = \dfrac{d^4G}{8D^3}\times 0.004[/tex]
[tex]F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004[/tex]
F = 3160 N
stress correction factor from stress correction curve is equal to 1.1
now, calculation of corrected stress
[tex]\tau = \dfrac{8FDk_s}{\pi d^3}[/tex]
[tex]\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}[/tex]
= 442.6 Mpa
The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa
now,
[tex]\tau_s \leq 0.45 S_u[/tex]
[tex]\tau_s \leq 0.45 \times 1300[/tex]
[tex]\tau_s \leq 585\ Mpa[/tex]
since corrected stress is less than the [tex]\tau_s [/tex]
hence, spring will return to its original shape.
