Page 436 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
414
Example Problem 14.3
Design a torsion spring with an initial torque of T = 1000 N mm and maximum torque
1
∘
of T = 3000 N mm. The minimum angular deflection is = 25 and the maximum
2 1
∘
∘
angular deflection is = 75 . The free angular deflection is = 120 . The length of
2 f
spring arm is 40 mm. The spring is to be installed in a steadily operated machine.
Solution
Steps Computation Results Units
1. Select materials Select oil tempered carbon steel for spring wire. [ ] = 1280 MPa
b
and determine the Initially select the diameter of spring wire as
allowable bending d = 3 mm.
stress From Figure 14.16, select = 1150 MPa,
e e
= =(1277 ∼ 1916)MPa
b
0.6 ∼ 0.9
Select = 1600 MPa and from Table 14.3,
b
we have
[ ]=0.8× = 1280MPa
b
b
2. Select spring Initially select C = 8 C = 8
2
2
index C and K = 4C − C − 1 = 4 × 8 − 8 − 1 = 1.1 K wT = 1.1
calculate the wT 4C(C − 1) 4 × 8 ×(8 − 1)
curvature factor
K wT
3. Calculate the Therefore, from Eq. (14.35), the spring’s wire d = 3.5 mm
spring wire diameter is
√ √
diameter d ≥ 3 K wT T = 3 1.1×3000 = 2.95
0.1[ b ] 0.1×1280
Select spring wire diameter as d = 3.5 mm.
4. Calculate spring D = Cd = 8 × 3.5 = 28 mm D = 28 mm
diameters D = D + d = 28 + 3.5 = 31.5 mm D = 31.5 mm
o o
D = D − d = 28 − 3.5 = 24.5 mm D = 24.5 mm
i i
4. Calculate the From n = 7.5
number of active Table 14.2, select E = 206 000 MPa. Since the
coils moment of inertia of spring wire is
d 4 × 3.5 4 4
I = = = 7.36mm
64 64
From Eq. (14.39), the number of active coils is
EI 206000 × 7.36 × 50
n = =
180TD 180 ×(3000 − 1000)× 28
= 7.52
Select n = 7.5
5. Calculate From Eq. (14.2) k = 40.11 N

∘
spring rate dT 3000 − 1000 ∘ mm/( )
k = = = 40N ⋅ mm∕

d 75 − 25
From Eq. (14.38)
EI 206000 × 7.36
k = =

180Dn 180 × 28 × 7.5
∘
= 40.11N ⋅ mm∕( )
The spring rate of the designed spring is close
to the requirement.
(continued)