Page 433 - Analysis and Design of Machine Elements

Steps Computation Results Springs 411
Units
5. Compute spring From Eq. (14.13) d = 3 mm
wire diameter √ 8K w CF √ K w CF √ 1.2525×6×250 D = 18 mm
d ≥ = 1.6 = 1.6
[ ] [ ] 880
= 2.33 mm
Select d = 3 mm.
6. Compute the From n = 7
number of active Table 14.2, select G = 80 000 MPa.
coils From Eq. (14.21)
8nFC 3
=
Gd
We have
8n(F − F )C 3
= − = max min
max min Gd
Gd
n =
8(F − F )C 3
max min
80000 × 3
= × 5 = 6.95
8 ×(250 − 150)× 6 3
Select n = 7
7. Check the The required spring rate is
rigidity, load and ΔF F max − F min 250 − 150
deflection k = Δ = max − min = 5
= 20N∕mm

The spring rate of designed spring is calculated
from Eq. (14.22) as
Gd 80000 × 3
k = = = 19.84 N∕mm
8nC 3 8 × 7 × 6 3
Close to the required spring rate.
The deflection at F min
F 150
= min = = 7.56 mm
min k 19.84
The deflection at F max is
max = 5 + 7.56 = 12.56 mm
The maximum load is
F = k =19.84 × 12.56 = 249.2N
max max
Close to the maximum design load.
8. Decide the (1) Outside diameter of spring D = 21 mm
o
dimensions of the D = D + d = 18 + 3 = 21 mm
o
spring (2) Inside diameter of spring
D = D − d = 18 − 3 = 15 mm mm
i
i
(3) Total number of coils D = 15
The number of end coils is 2. Therefore,
n total = n +2 = 7 +2 = 9
(4) The pitch n total = 9
p = (0.28 ∼ 0.5) D = 5.04 ∼ 9mm
(5) Free length p = 5.04 ∼ 9 mm
H = pn + 3d = (5.04 ∼ 9) × 7 + 3 × 3
f
= 44.28 ∼ 72.0 mm H = 44.28 mm
f
∼ 72.0
(continued)