Download Helical Compression Spring Design Calculator and more Exams Machine Design in PDF only on Docsity!
F = 3.0 kN
do di
Di Do
Do = 45 mm do = 8 mm No = 5
Di = 25 mm di = 5 mm Ni = 10
MECH 344 Problem Set 6-Chapter 12_Helical Compression
Spring-Selected Problems Concordia University
SOLUTION (12.21)
Known: A machine uses a pair of concentric helical compression springs to support a known static load. Both springs are made of steel and have the same length when loaded and when unloaded.
Find: Calculate the deflection and the maximum stress in each spring.
Schematic and Given Data:
Assumptions:
- There are no unfavorable residual stresses.
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
Analysis:
- From Eq. (12.8), k =^
d^4 G 3 8D N where G = 79 ✕ 109 Pa for steel. (Appendix C-1)
k = (8 mm)^4 (79, 000 N/mm^2 ) o = 88. 77 N/mm 8(45 mm)^3 (5)
k = (5 mm)^4 (79, 000 N/mm^2 ) i = 39. 50 N/mm 8(25 mm)^3 (10)
Assumptions:
- There are no unfavorable residual stresses.
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
Analysis:
- From Eq. (12.8), k =^ d 4 G 3 8D N where G = 79 ✕ 109 Pa for steel. (Appendix C-1)
k = (9 mm)^4 (79, 000 N/mm^2 ) o = 103. 66 N/mm 8(50 mm)^3 (5)
k = (5 mm)^4 (79, 000 N/mm^2 ) i = 22. 86 N/mm 8(30 mm)^3 (10)
- From Eq. (12.8), k = F or = F
k
=
= 23. 71 mm (22. 86 + 103. 66) (^) ■
- Using F = k, we can calculate the force on each spring. Fo = ko = (103.66 N/mm)(23.71 mm) = 2458.0 N Fi = ki = (22.86 N/mm)(23.71 mm) = 542 N
- Using Fig. 12.4, we can find Ks values for each spring. For outer spring, C = 50/9 = 5. Ks = 1. For inner spring, C = 30/5 = 6 Ks = 1.
- From Eq. (12.6), = 8FD Ks d
o =
(1. 09) = 467. 94 MPa (^) ■ (9)
i =
(1. 08) = 357. 75 MPa (^) ■ (5) SOLUTION (12.23) Known: A helical coil spring with given D and d is wound with a known pitch value. The material is ASTM B197 beryllium copper spring wire.
Find: If the spring is compressed solid, would you expect it to return to its original free-length when the force is removed?
ASTM B Beryllium copper spring wire
p = 14 mm
d = 10 mm D = 50 mm
Schematic and Given Data:
Assumptions:
- There are no unfavorable residual stresses.
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
Analysis:
- Force to compress spring solid can be calculated by using Eq. (12.7).
F = d
4 G
8D^3 N
where /N = p - d = 14 - 10 = 4 mm G = 50 ✕ 109 Pa (Appendix C-1)
(10 ^10
- ) 4 (50 ^10 9 )(4 ^10 - ) F = 8(50 ^10
) 3
= 2000 N
- The corresponding stress can be calculated by using Eq. (12.6).
= 8FD Ks d^3 for C = D/d = 50/10 = 5 Ks = 1.1 (Fig. 12.4) 8(2000)(50 ^10 -3) (^) = (10 10 -3)^3
(1.1) = 280. 1 MPa
- For beryllium copper spring wire without presetting, s ≤ 0.35 Su. Also, for beryllium copper spring wire without presetting, Su = 750 MPa and 0.35 Su = 262.5 MPa
- Since 280.1 MPa > 262.5 MPa, set should occur; therefore, spring would not return to original length. ■
Comment: By considering the curvature (stress concentration) factor of the inner surface by using Kw = 1.3, the inner surface stress is (1.3)(280.1) = 364.1 MPa which is clearly even larger than 262.5 MPa.
k =
= 3. 22 N/mm 8(17)^3 (10) ■
- The amount of deflection when Fmax = 122 N is applied to the spring is
s = Fmax/k =^122 = 37. 89 mm
- 22 From Fig. 12.8(d), Ls = Ntd Ls = (N + 2)d = (10 + 2)2 = 24 mm Therefore, Lf = Ls + s Lf = 24 + 37.89 = 61.89 mm ■
- Lf/D = 61.89/17 = 3. s/Lf = 37.89/61.89 = 0. From Fig. 12.10, for the case B, no buckling should occur. ■
500 N 1000 N (^) F s
60 mm Lf
Music wire with d = 5 mm
Clash allowance
SOLUTION (12.27)
Known: A helical compression spring with squared and ground ends is to be designed with given force and deflection requirement. Presetting is to be used. The loading is static.
Find: Determine appropriate values for D, N, and Lf. Check for possible buckling.
Schematic and Given Data:
Assumptions:
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
- The clash allowance is 10% of the maximum deflection.
Analysis:
- From Eq. (12.8), k =
F
k = F^ 1000 - 500 = 8. 33 N/mm 60
- Since class allowance is 10% of maximum deflection, Fs = 110% of 1000 N = 1100 N
- From Eq.(12.9), s = 0.65 Su
- From Fig. 12.7, Su 1650 MPa for music wire with d = 5 mm.
- Thus, s = 0.65(1650) = 1073 MPa.
- From Eq. (12.6), ^ = 8F CK^ or CK (^) = d^2
CKs
d^2
(1073)()(5)^2
s (^) 8F
Therefore, C = 9.0 (Fig. 12.4) and D = Cd = 9.0(5) = 45.0 mm ■
- From Eq. (12.8), N = dG
8kC 3 where G = 79 GPa (Appendix C-1)
s
40 lb 90 lb (^) F s
1.5 in. Lf
Su = 200 ksi^ Clash^ allowance C = 8
SOLUTION (12.28)
Known: A helical compression spring with squared and ground ends is to be made of steel, and presetting is to be used. The loading is static.
Find: Determine D, d, N, and Lf.
Schematic and Given Data:
Assumptions:
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
- The clash allowance is 10% of the maximum deflection.
Analysis:
- From Eq. (12.9), s = 0.65 Su = 0.65(200) = 130 ksi.
- With 10% clash allowance, Fs = 1.10(90) = 99 lb.
- k =^
F = 50
= 33. 3 lb/in. 1.^5
- From Eq. (12.6), s = 8F CKs where Ks = 1.06 (Fig. 12.4) d^2
d = 8Fs CK 1/ s s
d =
1/ = 0. 128 in. (^) ■
- D = Cd = 8(0.128) = 1.02 in. ■
- From Eq. (12.8), N = d
4 G
8D^3 k (0. 128)^4 (11. 5 106 )
where G = 11.5 ✕ 106 psi (Appendix C-1)
N =
8(1. 02)^3 (33. 3)
k
- From Fig. 12.8, for squared and ground ends,
Ls = Ntd or Ls = (N + 2)d
Ls = (10.92 + 2)(0.128) = 1.65 in.
Lf = Ls + s = Ls +
Fs
Lf = 1.65 +
= 4.62 in. ■
Comment:
Lf/D =
s/Lf =
From Fig. 12.10, we can see that the end plates should be constrained parallel to avoid buckling.
F
F
D = 3 in. F = 500 lb = 80 ksi k = 200 lb/in.
Thus, F =
((0. 1)^3
= 56. 4 lb ■
- Ls = (N + 2)d = (10)(0.1) = 1.0 in. Lf = Ls + Fs/k = 1.0 + 56.4/73.6 = 1.77 in.
SOLUTION (12.30)
Known: A particular machine requires a helical compression spring, having ends squared and ground, to support a known essentially static load. The spring constant, the stress at the design load, and D are known. The clash allowance is to be 0.10 in.
Find: Determine N, d, and Lf.
Schematic and Given Data:
Assumptions:
- Both end plates are in contact with nearly a full turn of wire.
- The end plate loads coincide with the spring axis.
Analysis:
- From Eq. (12.6), = 8FD Ks assume Ks = 1.06, check later d^3
80, 000 =
d^3
d^3 = 0.
d = 0.370 in.
- C = D/d = 3/0.370 = 8.
From Fig. 12.4, Ks = 1.
Therefore, d = 0.370 in. ■
- From Eq. (12.8), k = d
4 G
or N = 8D^3 N
d^4 G 8kD^3
where G = 11.5 ✕ 106 psi (Appendix C-1)
(0. 370)^4 (11. 5 106 )
N =
8(200)(3)^3
- Ls = (N + 2)d = (5.0 + 2)(0.370) = 2.59 in.
Expansion from solid = 0.1 + 500/200 = 2.6 in.
Lf = 2.59 + 2.6 = 5.19 in. ■
(^3 )
solid
(^3 )
d = 0.164 in.
- C = D/d = 12.19, Kw 1.
From Eq. (12.5), d = 0.164 in. ■ 4
- From Eq. (12.8), k = d G
8D N
4 or N = d G 8D k where G = 11.5 ✕ 106 psi (Appendix C-1) (0. 164)^4 (11. 5 106 ) N = 8(2)^3 (90)
- Clash allowance = 10%
= (1.^ 1) 90 lb 90 lb/in.
= 1. 1 in.
Solid height = (N + 2)d = (1.44 + 2)(0.164) = 0.565 in. Lf = 0.565+ 1.1 = 1.665 in. ■
Analysis--Case B--without shot peening:
- See Fig. 12.16: For max/min = 2, max = 640 MPa = 93 ksi
- k = F/ = 45/0.5 = 90 lb/in.
- From Eq. (12.5), max =
8FmaxD Kw d^3 Assume Kw = 1.
93, 000 =
d^3 d = 0.181 in.
- D/d = 2/0.181 = 11. Kw = 1.13 (Fig. 12.4)
- Assume Kw = 1.
93, 000 =
d^3 d = 0.177 in.
- D/d = 11.3, Kw 1.
Thus, d = 0.177 in. ■ 4
- From Eq. (12.8), k =
d G
8D N
4 or N = d G 8D k where G = 11.5 ✕ 106 psi (Appendix C-1) (0. 177)^4 (11. 5 106 ) N = 8(2)^3 (90)
- Clash allowance = 10 %
solid
F = 100 to 250 N F = 100 to 250 N
F F Original spring failed in service after about 10 5 cycles
Replacement spring stretched to the same length as the original spring
(^) max^ Design points^ overload Without residual stress Residual stress
(^0) min
= (1. 1) 90 lb 90 lb/in.
= 1. 1 in.
Solid height = (N + 2)d = (1.96 + 2)(0.186) = 0.737 in. Lf = 0.737 + 1.1 = 1.837 in.
SOLUTION (12.40) Known: A helical compression spring is subjected to a load fluctuating between 100 and 250. Fatigue properties of the spring wire correspond to the curve for shot-peened wire given in Fig. 12.16. The spring failed in service after about 10^5 cycles. A replacement spring was found which was identical in all respect except that its free- length was slightly shorter. To correct for this, the spring was stretched slightly to increase its free-length to exactly that of the original spring.
Find: Show by means of a max - min plot, whether you would expect the life of the replacement spring to be the same as, less than, or greater than that of the original.
Schematic and Given Data:
Analysis:
max
3 3
- See Fig. 12.
- From above graph, max = 133 ksi
For SF = 1.3, use = (^133) = 102 ksi
- 3
- From Eq. (12.5), max = 8Fmax CKw d^2 102, 000 =
CKw (0. 192)^2 CKw = 7.
- From Fig. 12.4, C = 6. D = Cd = (6.0)(0.192) = 1.15 in. ■
- k =
F = 25 lb (^) = 250 lb/in. 0.^ 1 in.
From Eq. (12.8), k =^ dG
8C N
or N =^ dG
8C k where G = 11.5 ✕ 106 ksi (Appendix C-1)
(0. 192)(11. 5 106 ) N = 8(6. 0)^3 (250)
Assuming squared and ground ends, Ls = (N + 2)d = (7.1)(0.192) = 1.36 in. ■
Lf = Ls + clash allowance +
Fmax
k
= 1. 36 + 0. 05 +^195 = 2. 19 in. (^) ■ 250
150 life, shot peened
max = 133
100 max (^) = 1. min
50
0 (^0 50 100 ) min (ksi)
max
(ksi)
52 mm
4.45 kN 9
D 0 40 mm Engaged Position
4.45 kN (^) + (3)(k)^ mm 9 9
Disengaged Position
3 mm
See Fig. 18.2 where total clamping force = 4.45 kN
SOLUTION (12.42D)
Known: A force of 4.45 kN is required to engage a clutch similar to the one shown in Fig. 18.2. This force is to be provided by nine identical springs equally spaced around the pressure plate of the clutch. The outside diameter of the coils can be no more than 40 mm, and the length of the springs when the clutch is engaged cannot exceed 52 mm. The pressure plate must move 3 mm to disengage the friction surfaces, and the lowest spring rate reasonably possible is desired.
Find: Design the springs, determining a satisfactory combination of D, d, N, wire material, type of ends, Ls, and Lf.
Schematic and Given Data:
Decisions/Assumptions:
- The clash allowance is 2 mm.
- The safety factor is 1.3.
- The ends are squared and ground.
- Presetting is to be used but not taken into account in the calculations.
- The spring is made of shot-peened steel wire having fatigue properties as indicated in Fig. 12.16.
Analysis:
- Design constraints are: (a) Since the lowest spring rate reasonably possible is desired, from Eq. 12.
k = d
4 G
8D^3 N
the largest D, largest N, and smallest d is desired.
(b) D + d ≤ 40 mm (c) Ls ≥ Engaged length Working deflection Clash allowance Assuming the clash allowance to be 2 mm, Ls ≤ (52 3 2) = 47 mm
