a)
The pitch of the spring (P) :
where L = length, d = wire diameter, Na = active coils
Substitute L = 120 mm, d = 3.4 mm, Na = 12 in Eq. (1)
Here, total no. of coils = no. of active coils
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):
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)
k = 6.217 lbf/in. = 1.0887 N/mm
(c)
Force required to close the spring, (Fs)
Fs = k(N/mm)*(free length in mm - solid length in mm)
=1.0887(120 - 44.2)
Fs = 82.5 N
(d)
Shear stress in the spring due to the force Fs
Where,
k = spring rate
Fs = Force required to close the spring
D = Mean Diameter (mm)
d = wire size (mm)
substitute f, Fs, d, and d values in Eq. (3)
please solve it step by step to understand all step please 2. Consider the special steel...
please show all work
2. (12 pts) A shaft w ith a step in diameter is made of SAE 1045 steel. It is required to withstand 107 cycles kN. The shaft has 2(a), di 25 mm, d2-30 mhm, and ρ :0.625 mm. Use Peterson's fitted on (equation below) for a value of a. Modification factors shall be applied for the following of an axial force amplitude P, 16kN applied along with a mean force of Pm dimensions, as in Figu...