Question

2. Consider the special steel spring in the illustration. (partial answers: F. = 81.9 N, T = 271 MPa) a) Find the pitch, solid height, and number of active turns. b) Find the spring rate. Assume the material is A227 HD steel. c) Find the force F, required to close the spring solid. Nts 12.5 Tums d: 3.4 mm D.50 -3.4-46-6 is = 120mm (squared or close a riant hand d) Find the shear stress in the spring due to the force Fs. - 120 mm 28401338 G = 76.6 50 mm WWW BMW18 13.4 mm

please solve it step by step to understand all step please

Answer 1

a)

The pitch of the spring (P) :

p-L-d..Eq. (1)

where L = length, d = wire diameter, N_a = active coils

Substitute L = 120 mm, d = 3.4 mm, N_a = 12 in Eq. (1)

Here, total no. of coils = no. of active coils

120 – 3.4 P = 12 P=9.717 mm

Solid height = (No. of coils + 1)*wire diameter = (12 + 1)*3.4 = 44.2 mm

No. of active turns = 12 no.

(b)

Spring rate, (k):

. Ga = 803 y ... Eq. (2)

Where,

d = wire size (inches)
D = Mean Diameter (inches)
N = Number of active coils
G = Shear Modulus of Material
k = Spring rate

Here, d = 3.4 mm = 0.1338 in.

D = 50 -3.4 = 46.6 mm = 1.8346 in.

N = 12

G = 11.5*10⁶ psi (For the material A227 HD steel)

substitute values of d, D, N, and G in Eq. (2)

11.5 + 106 * (0.1338) 18+ (1.8346) * 12

k = 6.217 lbf/in. = 1.0887 N/mm

(c)

Force required to close the spring, (F_s)

F_s = k(N/mm)*(free length in mm - solid length in mm)

=1.0887(120 - 44.2)

F_s = 82.5 N

(d)

Shear stress in the spring due to the force F_s

8FD T=k* d3 ..... Eq. (3)

Where,

k = spring rate

F_s = Force required to close the spring

D = Mean Diameter (mm)

d = wire size (mm)

substitute f, F_s, d, and d values in Eq. (3)

8 * 82.5 * 46.6 T= 1.0887 * 7 * 3.43 T = 271.25 N/mm² T= 271.25 MPa