Figure 2 Shows the schematic arrangement of a springcontroll

Figure 2 Shows the schematic arrangement of a spring-controlled lever. The spring is to be inserted with an initial compression to produce n force of 23 lb between the right hand end of the lever \'A\' and the stop. When a force of 42 lb is applied at A, the end of the lever moves downward 1.00 in. The spring is to be wound from chrome-silicon steel wire with squared and ground ends. Consider static loading of the spring and a factor of safety of 1.4. Determine (a) the maximum force in the spring, (b) spring stiffness, (c) wire diameter and gage #. (d) N_a, (e) pitch of coils. and (f) force needed to make the spring solid. Use C = 7.25 and epsilon = 0.15

Solution

Solution:

(a) Initial compressive force in the spring = 23*(16/8) = 46 lb

Additional compressive force in the spring when a load of 42 lb is applied at A = 42*(16/8) = 84 lb

So Maximum force in the spring = 84+46 = 130 lb

(b) Deflection of spring due to 42 lb force at A = 0.5 in

So spring stiffness = 84/0.5 = 168 lb/in

(c) For Chrome-silicon wire: (Ref: Mechanical engineering design by Shigley)

m = 0.108

A = 202 kpsi.in^m

E = 29.5 Mpsi

G = 11.2 Mpsi

Maximum shear stress Ssy = 0.45*(A/d^m) = 0.45*(202/d^0.108) kpsi

Wahl’s correction factor K = (4C-1)/(4C-4) + 0.615/C = 1.205

Ssy = 1.4*K * (8WC/d²) = 1.4*1.205*(8*130*7.25/d²) = 0.45*(202/d^0.108)*1000

d^1.892 = 1.4*1.205*8*130*7.25/(*0.45*202*1000) = 0.0445

Gauge # 6, and corresponding diameter = 0.192 in

(d) Number of active turn:

k = Gd/(8C³N)

N = 11.2*10⁶*0.192/(8*7.25³ * 168) = 4.2 5 .

For square and ground end, total turn = 5+2 = 7

(e) Maximum compression = 130/168 = 0.774 in

Free length = 7*0.192 + 0.774*(1+0.15) = 2.234

Pitch = 2.234/(7-1) = 0.372

(f) Force needed to make the spring solid = 0.774*(1+0.15)*168 = 150 lb