Homework 10 Springs
Problem 17.23
An overflow valve, shown in sketch b, has a piston diameter of 15 mm and a slit length of 5 mm.
The spring has mean coil diameter D = 10 mm and wire diameter d = 2 mm. The valve should
open at 1 bar pressure and be totally open at 3 bar pressure when the spring is fully compressed.
Calculate the number of active coils, the free length, and the pitch of the spring. The shear
modulus for the spring material G = 80 GPa. The spring ends are squared and ground.
Determine the maximum shear stress for this geometry.
Notes: The number of active coils is found from Eq. (17.15). Equations from Table 17.3 are used
for solid length, free length, pitch, and number of total coils. The maximum shear stress is
calculated from Eq. (17.10).
Solution:
The piston area is
When the pressure is one bar (0.1 MPa), the valve starts to open. The force associated with this
pressure is (1.767 × 10−4 m2)(0.1 MPa) = 17.7 N. When the pressure is 3 bar (0.3 MPa), the
force is 53.0 N, and the spring deflects a further 5 mm = 0.005 m. Note that from Eq. (17.7), C =
D/d = 10/2 = 5. Therefore, from Eq. 17.15,
If you use Eq. (17.17), you will get Na = 22.19, a more accurate number.
The stiffness of the spring, from Eq. (17.18) is
Therefore, the spring From Table 17.3, for squared and ground ends,
We know that at 53.0 N, the spring is at the solid length of 0.0492 m. When this load is removed,
the deflection is δ = P/k = 53 N/7060 N/m = 0.00751 m. Therefore the free length is
Lf = Ls + ds = (0.0492 m) + 0.0075 m = 0.0567 m
From Table 17.3, the pitch is
The maximum shear stress is obtained from Eq. (17.11) after the curvature shear correction
factor is calculated as Kw = (4C − 1)/(4C − 4) + 0.615/C = 1.31,
Problem 17.29
A compression spring made of music wire is used for static loading. Wire diameter d = 1.4 mm,
coil outside diameter Do = 12.1 mm, and there are eight active coils. Also, assume that the spring
ends are squared and ground. Find the following:
(a) Spring rate and solid length
(b) Greatest load that can be applied without causing a permanent set in excess of 2%
(c) Spring free length with load determined in part b that causes the spring to be solid
(d) Whether buckling is a problem. If it is, recommend what you would change in the redesign.
Notes: The spring rate is obtained from Eq. (17.18), and other equations are taken from Table
17.3. The strength of the wire is obtained from Eq. (17.2) and data from Table 17.2. Whether or
not buckling is a problem is assessed from Fig. 17.8.
Solution:
Note from Fig. 17.2(b), that the spring diameter D = D0 − d = 10.7 mm = 0.0107 m.
Therefore, C is, from Eq. 17.7, C = D/d = 10.7/1.4 = 7.64. From Table 17.1, G = 11.5 Mpsi = 79.3
GPa. The spring rate is obtained from Eq. (17.18):
From Table 17.3, using squared and ground ends, the solid length is
From Table 17.2 for music wire, Ap = 2170 MPa and m = 0.146. Therefore, from Eq. (17.2),
From Eq. (17.3),
Ssy = 0.4Sut = 0.4(2066 MPa) = 826.4 MPa
The transverse shear factor is Kd = (C + 0.5)/C = 8.14/7.64 = 1.065, so that Eq. (17.8) gives
The deflection caused by this load is
Therefore, the free length is lf = ls + δ = 14 mm + 20.3 mm = 34.3 mm. For this spring, δ/lf =
20.3/32.1 = 0.63, and lf/D = 34.3/10.7 = 3.21. Notice from Fig. 17.7 that buckling is not a problem
for this spring.
Problem 17.36
An 18-mm mean diameter helical compression spring has 22 coils, has 2-mm wire diameter, and
is made of chromium vanadium. Determine the following:
(a) The maximum load-carrying capacity for a safety factor of 1.5 guarding against yielding
(b) The maximum deflection of the spring
(c) The free length for squared and ground ends
Notes: The spring wire diameter is found from the stress requirements, and Eq. (17.8) is used
for the shear stress. The strength of the wire is obtained from Eq. (17.2) and data from Table
17.2. The solution assumes that 22 is the number of active coils, rather than total coils.
Solution:
Chrome vanadium refers to the alloys for steel, so use the stiffness for steel wire, or G =
11.5 Mpsi = 79.3 GPa. From Table 17.2 for chrome vanadium wire, Ap = 2000 MPa and m =
0.167. Therefore the strength of the wire in shear is, from Eqs. (17.2) and (17.3),
From (17.7), C = D/d = 18 mm/2 mm = 9. The transverse shear factor is Kd = 9.5/9 = 1.056.
Therefore, from Eq. (17.8), the maximum shear stress is:
Since the safety factor is 1.5,
From Eq. (17.18),
From Table 17.3 for squared and ground ends,
ls = d(Na + 2) = (0.002 m)(24) = 0.048 m
lf = ls + δt = 0.048 m + 0.0635 m = 0.111 m .
Problem 17.51
A helical torsion spring, shown in sketch k, is made from hard-drawn steel with a wire diameter of
2.2 mm and 8.5 turns. Dimensions are in millimeters.
(a) Using a safety factor of 2, find the maximum force and the corresponding angular
displacement.
(b) What would the coil inside diameter be when the maximum load is applied?
(c) We are skipping part c.
Notes: The force is obtained from Eq. (17.40) after the allowable stress in the wire is
determined. The angular displacement is calculated from Eq. (17.43) once the number of active
coils is found from Eq. (17.45). The inside diameter under maximum load is obtained from Eq.
(17.46).
Solution:
From Table 17.1, for hard drawn wire, G = 11.5 Mpsi = 79.3 GPa, and take E = 207 GPa
as for carbon steel from the inside front cover. From Table 17.2 for hard drawn wire, Ap = 1750
MPa, m = 0.192. Therefore, the ultimate strength of the wire is obtained from Eq. (17.2) as:
The allowable shear stress is calculated from Eq. (17.3)as:
Also, for this spring, from Eq. (17.7), C = D/d = 22 mm/2.2 mm = 10, so from Eq. (17.40),
Recognizing that M = P(0.035 m), Eq. (17.39) gives
From Eq. (17.44),
Na = Nb + Ne = Nb +
l1 + l2
0.035m + 0.035m
= 8.5 +
= 8.5 + 0.338 = 8.84
3πD
3π (0.022m)
So that from Eq. (17.42),
θ rev =
10.18MDN a 10.18(12.5 N )(0.035m)(0.022m)(8.84)
=
= 0.178
Ed 4
(207 ×109 Pa )(0.0022m) 4
Therefore, under the maximum load, the number of coils is N0a = Na+θrev = 8.84+0.18 = 9.02.
From Eq. (17.45),