Question

3. Use oil-tempered 0.060” diameter wire to make a spring with squared and closed ends, a
spring index of 4, and solid length of 1.25”. Use a load factor of safety of 1.05 and make the free
length 1.75 inches.

e) What is the force to close to solid, Fs?
f) What is the critical frequency?
g) What is the factor of safety, solid, ns?

Answer 1

Given

Oil tempered wire

G =11.6 x 10⁶ psi

Wire diameter = 0.06 inches

Given that squared and grounded

Solid length = 1.25 inches

1.25 = (n+2) d

1.25 = (n+2) * 0.06

n+2 = 1.25 / 0.06 = 20. 8333

n = 20.8 - 2 = 18.833

n is the number of active coils

Free length = p* n + 2* d

1.75 = 18.833 * p + 2* 0.06

1.75 - 0.12 = 18.333p

1.63 = 18.333 p

Pitch , p = 1.63 / 18.333 = 0.08655 inches

Given spring index = 4

D/ d = 4

D = 4 * 0.06 = 0.24 inches

The Spring rate

= 11.6 x 10⁶ * (0.06⁴) / 8* 0.24³ * 18.8333 = 72.18 lbs/ inches

Maximum deflection = Free length - Solid length = 0.5 inches

Max load possible = 72.18 * 0.5 = 36.09 lbs

Given a load FOS = 1.05

Then max load allowed = 36.09 / 1.05 = 34.37 lbs.

Critical frequency

W is the weight of the spring

= density of material = 0.282 lb/in³

W = (² * d² * D * n* ) / 4

W = (² * 0.06² * 0.24 * 18.8333 * 0.282) / 4 = 0.011322 lbs

Now the Critical frequency

f= 1/4 k*g / W

f = 1/4 72.18 * 386.0892 / 0.011322 = 392.22 cycles /seconds

We have F max = 34.37 lbs

F min = 0

F average = (Fmax-Fmin)/2 = 34.37/2 = 17.19 lbs

K = (4C + 2) / (4C-3)

C=D/d=4

K= (4*4 +2) / (4*4-3) = 1.3846

Tavg = K * 8*Fa*D / d³

T avg = 1.3846 * ( 8* 17.19 * 0.24) / * 0.06³ = 67327.04 psi

T max = K * 8*Fm*D / d³

T max = 1.3846 * ( 8* 34.37 * 0.24) / * 0.06³ = 134654.1 psi

We have to fins Sut and Suy

Sut = A/ d^m = 147 / 0.06^0.187 = 248.7729 kpsi

Suy = 0.67 * Sut = 166.6778 Kpsi

For unpeened spring

Ssa = 35kpsi

Ssm = 55kpsi

Sse = Ssa / ( 1-(Ssm/Su)) = 35 / ( 1- (55/166.6778)) = 52.23708 ksi

Ta / Sse + Tm / Ssu = 1/ nf

( Sse * Ssu) / (Ta Ssu + Tm Se) = nf

(52.24*166.68) / (67.32 * 166.68 + 134.65* 52.23) = 0.47693