Spring Solution-CH 16

6-1

Chapter 16 • Springs

16.1 A guitar string is made of ASTM A228 steel music wire. To get the correct tone from thestring, a tension of 200N has to be applied. When the string breaks, it is replaced by astring made of ASTM B159 phosphor bronze. What tension will the new string need toget the same tone if the string diameter is the same?

Solution:

If the mass per length is aρ, and there is negligible bending stiffness, vertical force equilibriumgives (see free body diagram above)

F∂ 2y

∂x2 dx = aρdx∂2y

∂t2

This has the solution

y = Asin ωtsinπxL

where

ω = πL

Faρ

= π FmaL

Therefore, for the two strings,

ω2L

π 2 = F1

m1= F2

m2; F1 = m1

m2F2 = ρ1

ρ2F2

From Table 16.1 on page 738, the density of phosphor bronze is 0.320lb/in3. The density of steel

wire is 0.283lb/in3. Therefore the tension needed for phosphor bronze is:

F = 0.32lb / in3

0.283 lb / in3

200N( )= 226N

16.2 A tuning fork is made of ASTM A227 steel and vibrates at an eigenfrequency of 400 Hz.To avoid corrosion, the material is changed to phosphor bronze. At what frequency willthe tuning fork vibrate?

Solution:

From the figure, summing forces:

2

T + ∂T∂x

dx − T − aρdx∂2 y

dt2

;

∂T∂x

dx = aρdx∂2y

dt2

From moment equilibrium:

M + ∂M∂x

dx − M − Tdx = 0;∂M∂x

= T

Taking the derivative with respect to x, then substituting for ∂T/∂x,

∂ 2M

∂x 2 = ∂T∂x

= aρ ∂ 2y

∂t2

This has the solution

y = y0 sinπx2L

sinω t

In addition, from Equation (5.3),

M = −EI∂ 2y

∂x2 ;∂ 2M

∂x2 = −EI∂4 y

∂x4 = aρ ∂ 2y

∂t2

Substituting for y and solving for ω,

ω2 = EIaρ

π2L

4

; ω = π2

4L2EIaρ

For the two materials, the ratio of the frequencies isω1

ω2= E1

E2

ρ2

ρ1Therefore, if a b subscript denotes bronze and an s subscript is steel,

ωb = ωsEb

Es

ρs

ρb= 400Hz( ) 15MPa

30MPa0.283lb / in3

0.32 lb / in3 = 266Hz

16.3 An overflow valve, shown in sketch a, has a piston diameter of 15mm and a slit length of5 mm. The spring has a mean coil diameter D=10mm and wire diameter d=2mm. Thevalve should open at 1 bar pressure and be totally open at 3 bar pressure when the springis fully compressed. Calculate the number of active coils, the free length and the pitch ofthe spring. The shear modulus for the spring material G=80GPa. The spring ends aresquared and ground. Determine the maximum shear stress for this geometry.

Notes: The number of active coils is found from Equation (16.15). Equations from Table 16.3 areused for solid length, free length, pitch, and number of total coils. The maximum shear stress iscalculated from Equation (16.9).

Solution:The piston area is

3

A =πd p

2

4= π 0.015m( )2

4= 1.767× 10−4 m2

When the pressure is one bar (0.1MPa), the valve starts to open. The force associated with this

pressure is (1.767x10-4m2)(0.1MPa)=17.7N. When the pressure is 3 bar (0.3MPa), the force is

53.0N, and the spring deflects a further 5mm=0.005m. Note that from Equation (16.6),C=D/d=10/2=5. Therefore, from Equation 16.15,

δ t = 8PC3Na

Gd; Na = Gdδt

8PC3 = 80GPa( ) 0.002m( ) 0.005m( )8 53.0 N −17.7N( ) 5( )3 = 22.6

The stiffness of the spring, from Equation (16.19) is

k = Pδ t

= 53.0 N −17.7N0.005m

= 7060N / m

Therefore, the spring From Table 16.3, for squared and ground ends,Nt=Na+2=24.6

Ls=d(Nt)=(0.002m)(24.6)=0.0492mWe know that at 53.0N, the spring is at the solid length of 0.0492m. When this load is removed,the deflection is δ=P/k=53.N/7060N/m=0.00751m. Therefore the free length is

Lf=Ls+ds=(0.0492m)+0.0075m=0.0567mFrom Table 16.3, the pitch is

p =l f − 2d

Na= 0.0567m − 2(0.002m)

24.6= 0.00233m = 2.33mm

The maximum shear stress is obtained from Equation (16.11) after the curvature shear correctionfactor is calculated as Kw=(4C-1)/(4C-4)+0.615/C=1.31,

τmax =8DKwP

πd3 =8 0.010m( ) 1.31( ) 53.1N( )

π 0.002m( )3 = 221MPa

16.4 A helical compression spring has mean coil diameter D=90mm and wire diameterd=15mm. The spring has 12 active coils. Derive the equation for the spring deflection andcalculate it for P=4000N. Also, find the shear stress in the wire if the stress concentrationfactor Ki=1.26. The shear modulus of elasticity is G=80GPa.

Notes: The derivation uses equilibrium for a portion of the wire combined with Equation (4.31).The shear stress is calculated from Equation (16.5) and using the stress concentration factorgiven.

Solution:

Consider a small portion of a wire, shown in the sketch. Taking momentequilibrium,

M=PD/2

From Equation (4.31),

θ = MlGJ

= P D / 2( )LGJ

; dθ = P D / 2( )dLGJ

= P D / 2( ) D / 2( )dγGJ

= PD2dγ4GJ

For a circular cross section, J=πd4/32, so

4

dθ = PD2dγ4GJ

= 8PD2

πd4Gdγ

Notice from the sketch that the change in length is dδ=(D/2)dθ, so that

dδ = D2

dθ = 4PD3

πd 4Gdγ

The total deflection is

δ = dδ∫ = 4PD3

πd4Gdγ =

0

2πNa

∫ 4PD3

πd 4G2πNa( ) = 8PD3Na

d4G= 8PC3Na

dG

where the substitution C=D/d has been used. For P=4000N, D=0.09m, d=0.015m, Na=12 andG=80GPa, C=D/d=6 and the deflection is

δ = 8PC3Na

dG=

8 4000N( ) 63( ) 12( )0.015m( ) 80GPa( ) = 0.0691m = 69.1mm

The maximum shear stress is obtained from Equation (16.5) combined with the stressconcentration factor as:

τmax = Ki 8PD

πd3 = 1.26( ) 8( ) 4000N( ) 0.090m( )π 0.015m( )3 = 342MPa

16.5 A vehicle has individual wheel suspension in the form of helical springs. The free lengthof the spring lf=360mm and the solid length ls=160mm at a compressive force of 5000N.The shear modulus G=80GPa. Use D/d=9 and calculate the shear stress for pure torsionof the spring wire. The spring ends are squared and ground. Find Na, p, d, D, and τmax.

Notes: An equation for the number of active coils as a function of wire diameter is obtained fromTable 16.3. The deflection equation then gives another equation containing number of active coilsand wire diameter, which can then be solved. The remainder of the terms are then calculated instraightforward fashion similar to Problem 16.3. The maximum shear stress is obtained fromEquation (16.5).

Solution:From Equation (16.15), and going from the solid to free length so that d=0.200m, P=5000N, C=9and G=80GPa gives

δ t = 8PC3Na

Gd;d = 8PC3

Gδ tNa =

8 5000N( ) 93( )80GPa( ) 0.20m( ) Na = 0.00182m( )Na

From Table 16.3 for squared and ground ends, the free length and solid length equations givels=dNt=d(Na+2)

Substituting d=(0.00182m)Na givesLs = 0.00182m( )Na Na + 2( ) = 0.160m; Na = 8.43

Therefore, d=0.0153m=15.3mm. Therefore, since C=9, Equation (16.6) gives D/d=9;D=9d=0.138m=138mm. From Table 16.3, the pitch is obtained as

p =l f − 2d( )

Na= 0.36m − 2 0.0153m( )

8.43= 0.0391m = 39.1mm

The transverse shear factor is Kd=1, since pure torsion is specified. Therefore the maximum shearstress is obtained from Equation (16.5) as

5

τmax =8DP

πd3 =8 0.138m( ) 5000N( )

π 0.0153m( )3 = 486MPa

16.6 Two equally long cylindrical helical compressionsprings are placed one inside the other (see sketch b) andloaded in compression. How should the springs bedimensioned to get the same shear stress in both springs?

Notes: The dimensions which can be varied are thenumber of active coils, the spring diameters and the wirediameters. Equations (16.5) and (16.8) are needed to solvethis problem.

Solution:Recognizing that the deflection is the same in both springs, it can be written from Equation(16.18) that

δ = P1

k1= P2

k2;

P1

P2= k1

k2=

Gd1

8C13Na1

Gd2

8C23Na2

= d14D2

3Na2

d24 D1

3Na1

If we can use Equation (16.5) for the maximum stress in the spring, then since the stresses are thesame in both springs, we can write,

τ1max = τ2 m a x;P1D1

d13 = P2 D2

d23 ;

P1

P2= D2d1

3

D1d23

Using these two equations gives

P1

P2= D2d1

3

D1d23 = d1

4D23Na2

d24 D1

3Na1;

Na1D12

d1= Na2D2

2

d2If we include a stress concentration factor, such as from Equation (16.9), then the relationship is:

Na1Kd1D12

d1= Na2 Kd 2D2

2

d2Where Kd is a transverse shear factor given on page 743 and is a function of D and d.

16.7 A mechanism is used to press as hard as possible against the moving horizontal surfaceshown in sketch c. The mechanism consists of a stiff central beam and two flexiblebending springs made of circular rods with length l, diameter d, modulus of elasticity E,and allowable stress σall. Wheels are mounted on these rods and can roll over a bump withheight f. Calculate the diameter of the springs so that the prestress of the wheels against

6

the moving surface is as high as possible without plastically deforming the springs whenthe bump is rolled over. The deflection of a spring is shown in sketch d.

Notes: This problem shows that cantilevered beams can be considered as springs. Therelationship between force and deflection can de derived using the approach in Chapter 5, but thisproblem uses the result given in the first row of Table 5.1 on page 193. The bending stress givesthe requirement for P in terms of δ, which is differentiated, set to zero and solved for d to obtainthe optimum diameter.

Solution:We want to maximize the force with respect to cantilever diameter. Because of symmetry, theloads and deflections for the two wheels are equal, or P1=P2=P. There is an extra force caused bythe bump, ∆P=∆P1=∆P2. Also, the deflections are δ1=δ2=f/2. For a cantilever with a load at theend, the relationship between deflection at the end and the load is obtained from the first row ofTable 5.1 on page 193, using a=l,

δ = − P6EI

l − l( )3 − l3 + 3l3[ ] = − Pl3

3EITherefore,

f2

= ∆Pl3

3EI;∆P = 3EIf

2l3 = 3Efπd4

128l3

The bending moment at the end of the cantilever is M=(P+∆P)l, but also since σ=Mc/I,

l P + ∆P( )=σI

lc; P =

σI

lc− ∆P =

σ πd4 / 6 4( )ld / 2

−3Efπd4

128l3=

σπd 3

32l−

3Efπd4

128l3

Therefore, taking the derivative and setting equal to zero, then solving for d:

∂P∂d

= ∂∂d

σπd3

32 l− 3Efπd4

128l3

=

3σπd2

32l− 3Efπd3

32l3= 0 ; 3σd2 = 3Efd3

l2 ; d = σ l2

Ef

The maximum force can be found by substituting this into the expression for P:

Pmax = σπd3

32 l− 3Efπd4

128l3=

σπ σ l2 / Ef( )3

32 l−

3Efπ σl2 / Ef( )4

128l3= πσ4 l5

128E3 f3

16.8 A helical compression spring is used as a catapult. Calculate the maximum speed of abody weighing 10kg being thrown by the catapult, given that τmax=500MPa, D=50mm,d=8mm, Na=20 and G=80GPa.

7

Notes: Equation (16.11) will allow for the calculation of the maximum force in the compressionof the spring. The stored energy can then be calculated and equated to the kinetic energy of thebody after launch.

Solution:For this spring, Equation (16.6) gives C=50/8=6.25. Therefore, the Transverse shear factor is

Kd=(C+0.5)/C=6.75/6.25=1.08.Equation (16.9) then gives:

τmax = 8DKdP

πd3 ; P = τmaxπd3

8DKd= 500MPa( )π 0.008m( )3

8 0.050m( ) 1.08( ) = 1860N

From Equation (16.18), the deflection under this load is

δ t = 8PC3Na

Gd= 8 1 8 6 0N( ) 6.25( )3 20( )

80GPa( ) 0.008m( ) = 0.114m = 114mm

The energy stored in the spring is U=Pδ/2=(1860N)(0.114m)/2=106 Nm. Since the kinetic energyof the object is mv

2/2, the maximum velocity is found as

U = K ; 1

2mv2 = 106Nm; v =

212Nm

m=

212Nm

10kg= 4.60m / s

16.9 A spring is preloaded with force Pi and is then exposed to a force increase whereby theshear stress increases to a certain value τ. Choose the mean coil diameter D to maximizethe energy adsorption caused by the force increase.

Notes: The approach is to obtain an equation of energy as a function of spring diameter, then setthe derivative of this expression equal to zero. This solution uses Equation (16.5), althoughEquation (16.9) or (16.12) could be used with an increase in complexity.

Solution:The energy stored in a spring at a load P is

U =Pδ2

=P P / k( )

2=

P2

2k=

4P 2C3Na

Gd=

4P2D3Na

Gd 4

If the shear stress is τ, then from Equation (16.5),

τ = 8DP

πd3 ; P = τπd3

8DTherefore, the energy stored between the two conditions described is

U =

4τπd3

8D

2

D3Na

Gd 4 − 4Pi2D3Na

Gd4 = π 2τ 2d2NaD16G

− 4Pi2D3Na

Gd4

Taking the derivative, setting equal to zero and solving for D yields:

dUdD

= π 2τ 2d2Na

16G− 12Pi

2D2Na

Gd 4 = 0;D = π8 3

τd3

Pi

16.10 A compression spring made of music wire is used for static loading. Wire diameterd=1.4mm, coil outside diameter D0=12.1mm, and there are eight active coils. Also,assume that the spring ends are squared and ground. Find the following:

8

a) Spring rate and solid lengthb) 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 theredesign.

Notes: The spring rate is obtained from Equation (16.19), and other equations are taken fromTable 16.3. The strength of the wire is obtained from Equation (16.2) and data from Table 16.2.Whether or not buckling is a problem is assessed from Figure 16.7.

Solution:Note from Figure 16.2(b), that the spring diameter D=D0-d=10.7mm=0.0107m. Therefore, C is,from Equation 16.6, C=D/d=10.7/1.4=7.64. From Table 16.1 on page 738,G=11.5Mpsi=79.3GPa. The spring rate is obtained from Equation (16.19):

k = Gd

8C3Na 1+ 0.5

C2

= 79.3GPa( ) 0.0014m( )8 7.64( )3 8( ) 1+ 0.5

7.642

= 3850N / m

From Table 16.3 on page 746, using squared and ground ends, the solid length isLs=dNt=d(Na+2)=0.0014m(8+2)=0.014m

From Table 16.2 on page 739 for music wire, Ap=2170MPa and m=0.146. Therefore, fromEquation (16.2),

Sut =Ap

d m = 2170MPa

1.4( )0.146 = 2066MPa

From Equation (16.3),Ssy=0.4Sut=0.4(2066MPa)=826.4MPa

The transverse shear factor is Kd=(C+0.5)/C=8.14/7.64=1.065, so that Equation (16.9) gives

τmax = 8DKdP

πd3 ; P = τmaxπd3

8DKd= 826.4MPa( )π 0.0014m( )3

8 0.0107m( ) 1.065( ) = 78.1N

The deflection caused by this load is

δ = Pk

= 78.1N3850N / m

= 0.0203m = 20.3mm

Therefore, the free length is lf=ls+δ=14mm+20.3mm=34.3mm.For this spring, δ/lf=20.3/32.1=0.63, and lf/D=34.3/10.7=3.21 Notice from Figure 16.7 on

page 748 that buckling is not a problem for this spring.

16.11 Design a helical compression spring made of music wire with squared and parallel ends.Assume a spring index C=12, a spring rate k=300N/m, and a force to solid length of 60N.Find the wire diameter and the mean coil diameter while assuming a safety factor ns=1.5.Also, check whether buckling is a problem. Assume steady loading.

Notes: The spring wire diameter is found from the stress requirements, and Equation (16.11) isused for the shear stress. The strength of the wire is obtained from Equation (16.2) and data fromTable 16.2. Whether or not buckling is a problem is assessed from Figure 16.7.

Solution:

9

From Table 16.1 on page 738 for music wire, G=11.5Mpsi=79.3GPa. From Table 16.2 on page739 for music wire, Ap=2170MPa and m=0.146. Therefore the strength of the wire in shear is,from Equation (16.2) and (16.3),

Ssy = τall = 0.4 Sut = 0.4Ap

d m = 0.42170MPa

d0.146 = 868MPa

d 0.146

where d is in millimeters. The transverse shear factor is Kd=(C+0.5)/C=12.5/12=1.04. Therefore,the maximum shear stress is, from Equation (16.11),

τmax = 8DKdP

πd3 = 8CKdP

πd2 = 8 1 2( ) 1.04( ) 60N( )πd2 = 1907N

d2

The safety factor is 1.5, so that (note the care to keep consistent units)

ns = τall

τmax= 868MPa( )

d 1000mm / m( )( )0.146d2

1907N=1.5; d = 0.00190 m =1.90mm

Therefore D=1.90mm(12)=22.8mm. From Equation (16.19),

k =Gd

8C3Na 1 + 0.5

C2

; Na =Gd

8C3k 1 + 0.5

C2

=79.3GPa( ) 0.00190m( )

8 123( ) 300N / m( ) 1 + 0.5

12

= 36.2

From Table 16.3, Nt=Na+2=38.2 coils, ls=Ntd=(38.2)(0.00190m)=0.0726m. The deflection fromfree to solid length is

δ = Pk

= 60N300N / m

= 0.2m = 200mm

Therefore the free length is lf=ls+δ=0.0726m+0.2m=0.2726m. Note thatlf/D=0.2726/0.0228=11.96, and ls/lf=0.2/0.2726=0.73. From Figure 16.7, it is clear that the springwill buckle unless constrained or supported.

16.12 Two helical springs with spring ratesof k1 and k2 are mounted one inside the other asshown in sketch e. The difference in unstressedlength is ∆ and the second spring is longer. Aloose clamping device is mounted at the top ofthe springs so that they are deformed andbecome equal in length. Calculate the forces inthe two springs. An external force P is the loadapplied thereafter to the spring. In a diagramshow how the spring forces vary with P.

Notes: The shorter spring elongates, the longer spring compresses, and the total deflection is ∆.Each spring has the same force. This is enough to solve the problem.

Solution:From equilibrium on the clamping device, each spring has the load P0. Also, the compressivedisplacement in the longer spring and the elongation in the shorter add up to ∆. Therefore,

δ1+δ2=∆From Equation (16.18),

10

P0

k1+ P0

k2= ∆; P0 = ∆ k1k2

k1 + k2If there is an external force P, equilibrium requires

P1-P2=PBut the deflections must be the same. Since there is a prestress P0, we can write

P1 = P0 + k1Pk1 + k2

P2 = P0 −k2P

k1 + k2The forces in each spring as a function of external load are as follows:

16.13 Using Eqs (16.5) and (16.18) describe how the spring diameter can be chosen to give aslow a maximum shear stress as possible.

Notes: It is helpful to consider the circumstance where there is a load P, and a change in load ∆Pleads to a deflection δ. Then use the suggested equations to obtain an expression for D.

Solution:A deformation increase of δ gives a force increase of ∆P=kδ and a total force of P+∆P.Therefore, from Equation (16.5),

τmax = 8 P + ∆P( )D

πd3 = 8DP

πd3 + 8D∆P

πd3


∆P = kδ = Gdδ8C3Na

= Gd4δ8D3Na

Therefore, the maximum shear stress is

τmax = 8DP

πd3 + 8GdδπNa D2

Taking derivative with respect to D, setting equal to zero and solving,

∂τ∂D

=8P

πd3 −2Gdδ

πNaD3 = 0; 8P

πd3 =2Gdδ

πNaD3 ; D3 =Gd4δ4PNa

; D =Gd 4δ4PNa

3

16.14 Consider a helical compression spring with plain ends made of hard drawn wire with aspring index of 12 and a stiffness of 0.3kN/m. The applied load on the spring is 60N. For

11

a safety factor of 1.5 guarding against yielding find the spring diameter, the wirediameter, the number of coils, and the free and solid lengths of the spring. Does thespring have buckling and/or dynamic instability problems?

Notes: This is the same problem as 16.11, but the wire material is different and additionalinformation is requested. Equation (16.21) gives the critical frequency of the spring.

Solution:From Table 16.1 on page 738, G=11.5Mpsi=79.3GPa. From Table 16.2 on page 739 for musicwire, Ap=1750MPa and m=0.192. Therefore the strength of the wire in shear is, from Equation(16.2) and (16.3),

Ssy = τall = 0.4 Sut = 0.4Ap

d m = 0.41750MPa

d0.192 = 700MPa

d0.192

where d is in millimeters. From Equation (16.12) the transverse shear factor with C=12 isKd=(C+0.5)/C=12.5/12=1.04 Therefore, the maximum shear stress is, from Equation (16.9),

τmax = 8DKwP

πd3 = 8CKwP

πd 2 = 8 12( ) 1.04( ) 60N( )πd2 = 1907N

d2

The safety factor is 1.5, so that (note the care to keep consistent units)

ns = τall

τmax= 700MPa( )

d 1000mm / m( )( )0.192d2

1907N=1.5;d = 0.00218 m = 2.18mm

Therefore D=2.18mm(12)=26.16mm. From Equation (16.19),

k = Gd

8C3Na 1+ 0.5

C2

; Na = Gd

8C3k 1+ 0.5

C2

= 79.3GPa( ) 0.00218m( )8 123( ) 300N / m( ) 1+ 0.5

122

= 41.5

From Table 16.3, Nt=Na=41.5 coils, ls=(Nt+1)d=(42.5)(0.00218m)=0.0927m. The deflection fromfree to solid length is

δ = Pk

= 60N300N / m

= 0.2m = 200mm

Therefore the free length is lf=ls+δ=0.093m+0.2m=0.293m. Note that lf/D=0.293/0.02616=11.2,and δ/lf=0.2/0.3=0.66. From Figure 16.7, it is clear that the spring will buckle unless constrainedor supported. The density of the cold drawn wire is, from Table 16.1, ρ=0.283lb/in

3=7810kg/m

3.

The critical frequency for the spring is, from Equation (16.21),

16.15 An 18 mm mean diameter helical compression spring has 22 coils, has 2 mm wirediameter and is made of chromium vanadium. Determine the following:a) The maximum load-carrying capacity for a safety factor of 1.5 guarding againstyielding.b) The maximum deflection of the spring.c) The free length for squared and ground ends.

fn = 2πNa

d

D2G

32ρ= 2

π 41.5( )0.00218m

0.02616m( )279.3GPa( )

32 7810kg / m3( ) = 27.5Hz

12

Notes: The spring wire diameter is found from the stress requirements, and Equation (16.9) isused for the shear stress. The strength of the wire is obtained from Equation (16.2) and data fromTable 16.2.

Solution:Chrome vanadium refers to the alloys for steel, so use the stiffness for steel wire, orG=11.5Mpsi=79.3GPa. From Table 16.2 on page 739 for chrome vanadium wire, Ap=2000MPaand m=0.167. Therefore the strength of the wire in shear is, from Equation (16.2) and (16.3),

Ssy = τall = 0.4Sut = 0.4Ap

d m = 0.42000MPa

20.167 = 712.6MPa

From (16.6), C=D/d=18mm/2mm=9. The transverse shear factor is Kd=9.5/9=1.056. Therefore,from Equation (16.9), the maximum shear stress is:

τmax = 8DKdP

πd3 = 8CKdP

πd2 = 8 9( ) 1.056( )P

π 0.002m( )2 = 6.05 ×106m−2( )PSince the safety factor is 1.5,

ns =Ssy

τmax= 712.6MPa

6.05 ×106m−2( )P = 1.5; P = 78.5 N


Pδ t

= Gd

8C3Na;δ t = 8PC3Na

Gd=

8 78.5N( ) 93( ) 22( )79.3GPa( ) 0.002m( ) = 0.0635m

From Table 16.3 for squared and ground ends,ls=d(Na+2)=(0.002m)(24)=0.048mlf=ls+δt=0.048m+0.0635m=0.111m

16.16 A 60mm mean diameter helical compression spring with plain ends is made of hard-drawn steel and has a wire diameter of 2.5mm. The shear strength of the wire material is750MPa and the spring has 20 coils. The free length is 500mm. Find the following:a) The required load needed to compress the spring to its solid length.b) By applying the load found in part (a) and then unloading it, whether the spring willreturn to its free length.

Notes: Part (a) is similar to previous problems. Part (b) really asks whether or not the wire’s yieldstrength is exceeded by the loading.

Solution:From Table 16.1, for hard drawn wire, G=11.5Mpsi=79.3GPa. From Table 16.2 for hard drawnwire, Ap=1750MPa, m=0.192. Therefore, the allowable shear stress is calculated from Equations(16.3) and (16.2) as:


d m = 0.41750MPa

2.50.192 = 587MPa

From Table 16.3 on page 746, Nt=Na=20, and ls=d(Nt+1)=(0.0025m)(21)=0.0525m. Since the freelength is 500mm=0.5m, then the change in length in going from free to solid is δ=0.5-0.0525=0.4475m. C=60/2.5=24. From Equation (16.18),

Pδ t

= Gd

8C3Na; P = Gdδt

8C3Na= 79.3GPa( ) 0.0025m( ) 0.4475m( )

8 243( ) 20( )= 40.11N

The transverse shear factor is Kd=(C+0.5)/C=24.5/24 =1.02. Therefore, from Equation (16.9),

13

τmax = 8DKdP

πd3 = 8 0.06m( ) 1.02( ) 40.11N( )π 0.0025m( )3 = 400MPa

Since τmax<Ssy, no plastic deformation occurs and the spring will return to its original free lengthwhen released.

16.17 A desk lamp has four helical extension springs to make it possible to position it overdifferent parts of the desk. Each spring is preloaded to 15N, so that no deflection takesplace for forces below 15N. Above 15N force the spring rate is 100N/m. The mean coildiameter is 10mm and the wire diameter is 1mm. Calculate the torsion needed on the wireduring manufacturing to get the correct preload, and calculate how many coils are neededto get the correct spring rate.

Notes: This problem needs Equations (4.31) and (16.35) to obtain a solution.

Solution:Assuming the wire is high-carbon steel with G=11.5ksi=79.3GPa. The torque on each spring dueto the preload is

T=PiR=(15N)(0.005m)=0.075NmFrom Equation (4.31),

θl

= TGJ

= 0.075Nm

79.3GPa( ) π 0.001m( )4 / 3 2( ) = 9.63rad / m

This is the required preload; Equation (16.35) gives the number of required active coils:

k = dG

8NaC3 ; Na = dG

8kC3 = 0.001m( ) 79.3GPa( )8 100N / m( ) 103( ) = 99.125

16.18 A spring balance for weighing fish needs to be dimensioned. The weighing mechanism isa sharp hook hanging in a helical extension spring. To make it easy to read the weight ofthe fish in the range from zero to 10kg, the length of the scale should be 100mm. Thespring material is music wire.

Notes: This is an open-ended problem in that there are more unknowns than equations. Thissolution demonstrates the design approach used for C=10 and d=2mm.

Solution:Take C=10, d=2mm, so D=20mm. From Table 16.1, for music wire, G=11.5Mpsi=79.3GPa.From Table 16.2 for music wire, Ap=2170MPa, m=0.146. Therefore, the allowable shear stress iscalculated from Equations (16.3) and (16.2) as:


d m = 0.42170MPa

2.0.146 = 784MPa

The spring has a load ranging from 0 to (10kg)(9.81m/s2)=98.1N. The deflection over this load is

100mm, so the spring rate is k=P/δ=(98.1N)/(0.1m)=981N/m. From Equation (16.18), the numberof active coils is

k = Gd

8C3Na; Na = Gd

8C3k= 79.3GPa( ) 0.002m( )

8 103( ) 981N / m( )= 20.2

14

This spring will be loaded in a cyclic manner; therefore Equation (16.11) should be used for theshear stress (see the text on page 744). The Wahl correction factor is given by Equaton (16.12) as

Kw = 4C −14C− 4

+ 0.615C

= 3936

+ 0.61510

=1.14

The maximum shear stress is then, from Equation (16.11),

τmax = 8DKwP

πd3 = 8 0.02m( ) 1.14( ) 98.1N( )π 0.002 m( )3

= 712MPa

which is, fortunately, less than the allowable shear stress. The unloaded spring length, without thehook ends, is Nad=(20.2)(0.002m)=0.0404m. The extended length is then 0.1404m.

16.19 A muscle-training device consists of two handles with three parallel 500-m-long springsin between. The springs are tightly wound but without prestress. When the springs arefully extended to 1600mm, the force in each spring is 100N. The springs are made ofmusic wire. Dimension the springs.

Notes: This is an open-ended problem in that there are more unknowns than equations. Thissolution demonstrates the design approach used for C=10.

Solution:Assume C=10. Initially, take d=2.mm, so that D=20mm. From Table 16.1, for music wire,G=11.5Mpsi=79.3GPa. From Table 16.2 for music wire, Ap=2170MPa, m=0.146. Therefore, theallowable shear stress is calculated from Equations (16.3) and (16.2) as:


d m = 0.42170MPa

2.0.146 = 784MPa

The spring has a load ranging from 0 to 100N. The deflection over this load is 1100mm, so thespring rate is k=P/δ=(100N)/(1.1m)=90.9N/m. From Equation (16.18), the number of active coilsis

k = Gd

8C3Na; Na = Gd

8C3k= 79.3GPa( ) 0.002m( )

8 103( ) 90.9N / m( )= 218

This spring will be loaded in a cyclic manner; therefore Equation (16.11) should be used for theshear stress (see the text on page 744). The Wahl correction factor is given by Equaton (16.12) as

Kw = 4C −14C− 4

+ 0.615C

= 3936

+ 0.61510

=1.14

The maximum shear stress is then, from Equation (16.11),

τmax = 8DKwP

πd3 = 8 0.02m( ) 1.14( ) 100N( )π 0.002m( )3 = 726MPa

The safety factor for this case is ns=784/726=1.08. If this analysis is repeated, the larger thediameter, the larger the safety factor, but also the larger the spring diameter. Ultimately, there willbe constraints on the size of the spring from the application, but this approach is the one whichwould allow for dimensioning of the spring.

16.20 A 45mm diameter extension spring (similar to that shown in Figure 16.8(a) and (b)) has102 coils and is made of 4 mm diameter music wire. The stress due to preload isequivalent to 10% of the yield shear strength. The hook radius is 5 mm and the bendradius is 2.5 mm. Determine the following:a) The solid length of the spring and the spring stiffness.

15

b) The preload and the load that causes failure.

Notes: This is similar to the previous problems. The preload is calculated from Equation (16.36),and failure must be analyzed with respect to shear (Equation (16.38)) and tensile failure(Equation 16.37)).

Solution:For this spring C is given by Equation (16.6) as C=D/d=45/4=11.25. The solid length of the bodyis given by Equation (16.32) as

lb=dNt=d(Na+1)=(0.004m)(102+1)=0.412mThe spring stiffness is obtained from Equation (16.35):

k = dG

8NaC3 = 0.004m( ) 79.3GPa( )8 102( ) 11.253( ) = 273N / m

From Table 16.1, for music wire, G=11.5Mpsi=79.3GPa. From Table 16.2 for music wire,Ap=2170MPa, m=0.146. Therefore, the ultimate strength of the wire is

Sut =Ap

d m = 2170MPa

4.0.146 = 1772MPa

The allowable stress in tension is given by Equation (16.39) as Sty=0.60Sut= 0.60(1772MPa)=1063MPa. The allowable shear stress is calculated from Equations (16.3) and (16.2) as:


d m = 0.42170MPa

4.0.146 = 709MPa

It is given that the shear stress due to the preload is 10% of the shear yield strength, orti=70.9MPa. Therefore, from Equation (16.36), the preload is

Pi = πτ id2

8C= π 70.9MPa( ) 0.004m( )2

8 11.25( ) = 39.6 N

To analyze failure, one must consider both tensile and shear possibilities. The hook radius isr1=0.005m, and the bend radius is r2=0.0025m. From Figure 16.8(a), r3=r1-d/2=0.005m-0.002m=0.003m. From Figure 16.8(b), r4=r2-d/2=0.0005m. The tensile stress is given byEquation (16.37) as

σ = 32PAr1πd3

r1r3

+ 4PA

πd2 = 32 0.005m( )π 0.004m( )3

0.005m0.003m

+

4

π 0.004m( )2

PA = 1.41 ×106m−2( )PA

Or, if σ=Sty, P=754N. The shear stress in the spring is given by Equation (16.38) as:

τB = 8PBC

πd2r2r4

=

8 11.25( )π 0.004m( )2

2.5mm0.5mm

PA = 8.95 ×106m−2( )PA

If τB=Ssy, then PA=79N. Therefore, failure will be by shear at a load of 79N.

16.21 The extension spring shown in sketch f is used in a cyclic motion in turning on an off apower switch. The spring has a 15mm outer diameter and is made of a 1.5 mm diameterwire of hard-drawn steel. The spring has no preload. In a full stroke of the spring theforce varies between 25 and 33 N. Determine the following:a) The number of coils, the free length, the maximum and minimum lengths during cyclicloading, and the spring rate.b) For infinite life with 99% reliability, the safety factors guarding against static andfatigue failure.

-16

Notes: This is similar to problem 16.20, but now introduces fatigue failure. The shear stress mustbe now calculated with Equation (16.19). Alternating and mean shear stress components must becalculated, and a kc value obtained for the 99% reliability. This solution only shows the results forshear stresses. The normal stresses are less critical in this case and are not shown, but normallyshould be analyzed to ensure this is not the failure mode.

Solution:For this spring, the outer diameter is 15mm, so D=D0-d=15mm-1.5mm=13.5mm=0.0135m.Therefore, C=D/d=13.5/1.5=9. From Table 16.1, for hard drawn wire, G=11.5Mpsi=79.3GPa.Note that the elongation of the spring can be seen from the sketch to be d=(30mm)sin30°=15mm= 0.015m. Therefore, from Equation (16.19),

Pδ

= Gd

8C3Na 1 + 0.5

C2

;Na = Gdδ

8C3P 1 + 0.5

C2

= 79.3GPa( ) 0.0015m( ) 0.015m( )8 93( ) 33N −25N( ) 1+ 0.5

92

= 38.0

The length of the body and the free lengths are, from Equations (16.32) and (16.33):lb=dNt=d(Na+1)=(0.0015m)(39)=0.0585m

lf=lb+δ=0.0585m+0.015m=0.0735mThis loading gives Pa=(33-25)/2=4N, and Pm=(33+25)/2=29N. The Wahl correction factor isgiven by Equation (16.12) as

Kw = 4C −14C− 4

+ 0.615C

= 3532

+ 0.6159

=1.162

so that, from Equation (16.11),

τa = 8CKwPa

πd 2 = 8 9( ) 1.162( ) 4N( )π 0.0015m( )2 = 47.3MPa

τm = 8CKwPm

πd 2 = 8 9( ) 1.162( ) 29N( )π 0.0015m( )2 = 343MPa

From Table 16.2 for hard drawn wire, Ap=1750MPa, m=0.191. Therefore, the ultimate strength ofthe wire is obtained from Equation (16.2) as:

Sut =Ap

d m = 1750MPa

1.50.191 = 1619MPa

The allowable shear stress is calculated from Equations (16.3)as:Ssy = τall = 0.4Sut = 0.4 1619MPa( ) = 648MPa

The static safety factor is then, from Equation (16.27),

ns =Ssy

τa +τm= 648MPa

47.3MPa+ 343MPa= 1.66

From Equation (7.7) and for torsion, Se’=0.29Su=0.29(1619MPa)=469MPa. From Table 7.4 on

page 275, kr=0.82. The manufacturing process is drawing, so from Table 7.3, e=4.51MPa and f=-0.265 so that kf is given by Equation (7.21) as

17

k f = eSutf = 4.51MPa( ) 1619( )−0.265 = 0.64

The size factor is, from Equation 7.22, ks=1. Therefore, the shear endurance strength isSse=kfkrSe

’=(0.64)(0.82)(469MPa)=246MPaTherefore, the safety factor against endurance limit fatigue is, from Equation (16.26),

ns = Sse

τa= 246MPa

47.3MPa= 5.2

16.22 Calculate the safety factor guarding against fatigue failure if the spring given in Problem16.21 is designed for 50,000 strokes of motion with 50% reliability.

Notes: Results from Problem 16.21 will be used here. Equation (7.14) is used to get the fatiguestrength at this life.

Solution:For spring steels, Equation (16.30) gives

Ssu=0.6Sut=0.6(1619MPa)=971.4MPaThe fatigue strength will be calculated from Equation (7.14) on page 268. First we need, fromEquation (7.8) that Sl

’=0.72Su=0.72(971.4MPa)=699MPa. From Equation (7.12),

bs = −13

logSl

′

Se′

= − 1

3log

699MPa469 MPa

= −0.0578


C = logSl

′( )2

Se′

= log6992

469

= 3.02

Therefore, from Equation (7.14),

S f′ = 10C Nt

′( )bs=103.02 50,000( )−0.0578 = 560MPa

Therefore, the safety factor is, from Equation (16.28),

ns =Ssf

τ a= 560MPa

47.3MPa= 11.8

16.23 The extension spring shown in sketch g is used in a car braking system. The spring ismade of 2-mm-diameter wire of hard drawn steel, has a mean coil diameter of 10mm, andhas a free length of 106mm. The spring has 18 active coils and a deflection under brakingforce of 11mm. Determine the following:a) The safety factor of the springb) The torsional and normal stresses at the hook.

18

Notes: This is similar to problem 16.20. The key equations needed to obtain a solution to thisproblem are (16.2), (16.11), (16.35), (16.37) and (16.38). It will be assumed that there is nopreload on the spring.

Solution:From Table 16.1, for hard drawn wire, G=11.5Mpsi=79.3GPa. From Table 16.2 for hard drawnwire, Ap=1750MPa, m=0.191. Therefore, the ultimate strength of the wire is obtained fromEquation (16.2) as:

Sut =Ap

d m = 1750MPa

20.191 = 1533MPa

The allowable shear stress is calculated from Equations (16.3)as:Ssy = τall = 0.4Sut = 0.4 1533MPa( ) = 613MPa

From Equation (16.39), the allowable stress that will be needed in Equation (16.37) isSty=0.60Sut=0.60(1533MPa)=920MPa

For this spring C=D/d=10mm/2mm=5. From Equation (16.35),

P = dG

8NaC3 δ = 0.002m( ) 79.3GPa( )8 1 8( ) 53( ) 0.011m( ) = 96.9N

A brake involves a cyclic loading, so Equation (16.11) should be used. The Wahl factor, given byEquation (16.12), is

Kw = 4C −14C− 4

+ 0.615C

= 1916

+ 0.6155

= 1.31

So that the maximum shear stress is:

τmax = 8DKwP

πd3 = 8 0.010m( ) 1.31( ) 96.9( )π 0.002 m( )3 = 404MPa

The safety factor against static shear stress is

ns =Ssy

τmax= 613MPa

404MPa= 1.52

From the sketch, r1=D/4=2.5mm. Also, r3=r1-d/2=2.5mm-2mm/2=1.5mm. r2=D/2=5mm, andr4=r2-d/2=4mm. Therefore, from Equation (16.37),

σ = 32PAr1πd3

r1r3

+ 4PA

πd2 = 32 96.9N( ) 0.0025m( )π 0.002m( )3

2.5mm1.5mm

+ 4 96.9N( )

π 0.002m( )2 = 545MPa

Therefore, the safety factor for tensile stress at A is

ns = Sst

σ A= 920MPa

545MPa= 1.69

From Equation (16.38), the shear stress at B is:

τB = 8PBC

πd2r2r4

=

8 96.9N( ) 5( )π 0.002m( )2

5mm4mm

= 386MPa

with an associated safety factor of

ns =Ssy

τB= 613MPa

386 MPa=1.59

The lowest safety factor is for shear stress inside the spring, and this is the likely failure mode(although shear at the hook is essentially the same safety factor and failures at that location arepossible).

19

16.24 Derive a simple expression for the spring rate of a spring in pure shear. The force P isevenly distributed over the surface. All deflections are small.

Notes: This merely uses elasticity equations applied to pure shear, as opposed to torsion orbending mechanisms as previously in this chapter.

Solution:

A small volume of material is sketched above under a shear force P causing a shear strain. Theshear stress is τ=P/A, and for small deflections, the shear strain is γ=δ/e. Since τ=Gγ, we canwrite

PA

= δe

G

recalling that a spring rate is P/δ, we can reorganize this to:

k = Pδ

= GAe

which is the desired relationship.

16.25 A torsion rod is used as a vehicle spring. The torque on the rod is created by a forceP=1500N acting on a radius R=200mm. The maximum torsion angle is 45°. Calculate thediameter and length of the rod if the maximum shear stress is 500MPa.

Notes: This problem uses Equations (4.31) and (4.28). The material used in this solution is carbonsteel.

Solution:No material has been specified, so use steel, where G=80GPa. For a round torsion member (Note:J=πr

4/2), Equation (4.31) gives

θ = TLGJ

= PRLGJ

Equation (4.28) gives the shear stress as:

τ = GrθL

= GrPRLLGJ

= rPRJ

= 2PR

πr3

Since the maximum shear stress is 500MPa, the radius is:

τ = 2PR

πr3 ;r = 2PRπτ

3 = 2 1 5 0 0N( ) 0.2m( )π 500MPa( )

3 = 0.00725m = 7.25mm

Therefore, since the maximum torsion angle is θ=45°=π/4,

20

θ = PRLGJ

; L = θGπr4

2 PR= π / 4( ) 80GPa( )π 0.00725m( )4

2 1 5 0 0N( ) 0.2m( ) = 0.909m

16.26 A torsional spring, shown in sketch h, consists of a steel cylinder with a rubber ring gluedto it. The dimensions of the ring are D=45mm, d=15mm, and h=20mm. Calculate thetorque as a function of the angular deflection. The shear modulus of elasticity for therubber is 150N/cm

2.

Notes: This is solved through equilibrium and by considering an incremental deformation of thering and the associated torque.

Solution:For each cylinder, the torque M is constant. Also,

τ = Gγ = M

2πhr2

The shear deformation of the ring g gives an angular rotation of

dφ = γdrr

;φ = γdrrri

ry

∫ = M

2πhr2G

drrri

ry

∫ = MπGh

1

d 2 − 1

D2

Therefore the spring rate M/φ is

Mφ

= πhG1

d2 − 1

D2

=π 0.02m( ) 150 ×104N / m( )

1

0.015m( )2 − 1

0.045m( )2

= 23.9Nm / rad

16.27 To keep a sauna door shut, helical torsion springs are mounted at each of the two doorhinges. The friction torque in each hinge is 0.2Nm. It should be possible to open the door180° without plastically deforming the spring. Dimension the springs so that a 10kg door700 mm wide will shut itself in 2s. The length of the wire ends can be neglected. The airdrag on the door is also neglected. The springs are manufactured from music wire andhave a diameter of 2.5mm.

Notes: The solution requires one to investigate the equations of motion of the door. The springdimensions which need to be specified are the number of coils, wire length, and spring diameter.

21

Solution:The motion of the door is governed by

M − M f = −J ˙ ̇ φ where M is the moment of the spring, Mf is the friction moment, J is the door inertia and φ is theangular position of the door. Using Equation (16.42), M can be expressed as

M = M0 + EIlw

φ

where M0 is the prestress moment. Substituting this into the equation of motion gives

M0 − M f +EI

lwφ = −J ˙ ̇ φ ; ˙ ̇ φ +

EI

Jlwφ +

−M0 + M f

J= 0

This differential equation has the solution

φ = φ0 cosω t + lwEI

−M0 + M f( )where ω is given by

ω = EIJlw

Note that with this solution for φ,˙ ̇ φ = −ω2φ0 cosωt

M0 should be chosen to be as small as possible to close the door from any position, so that M0=Mf.The prestress angle for the spring is

φpEI

lw= M f = 0.2Nm; φp =

M f lwEI

In two seconds, the door should have shut if at time t=0 the door is fully open. Therefore at t=t0,φ=φ0, at t=2s, φ=φ0cos2ω=0. This requires that cos2ω=0 or that ω=π/4rad/s. Substituting in theexpression for ω,

ω = EIJlw

= π4

; lw = 16EI

π2 J=

16 207GPa( ) π 0.00125m( )4 / 4( )π2 0 . 5 1 0kg( ) 0.7m( )2

3

= 0.7878m

For half the door, since there are two hinges, At t=0, φ=φ0+π, or

φ =M f lw

EI+ π = 0.2Nm( ) 0.7878 m( )

207GPa( ) π 0.00125m( )4 / 4( ) + π = 3.539rad = 202.7°

The maximum bending moment in the spring wire is

Mmax = M0 + EIlw

φmax = 0.2Nm +207GPa( ) π 0.00125m( )4 / 4( )

0.7878m3.539rad( ) =1.983Nm

If C=10, Equation (16.41) gives

Ki =4C2 − C − 1

4C C − 1( )=

400 − 10 −1

40 9( )= 1.08

So that from Equation (16.40), the bending stress is

σ = 32K iM

πd3 = 32 1.08( ) 1.983 Nm( )π 0.0025m( )3 = 1397MPa

From Table 16.2, Ap=2170MPa and m=0.146, so that from Equation (16.2),

Sut = A

dm = 2170MPa

2.5( )0.146 = 1898MPa

22

so that the safety factor is ns=1898/1397=1.359. The number of coils in the spring is

Na = lwπD

= 0.7878mπ 10( ) 0.0025m( ) =10.03

16.28 A helical torsion spring used in a door handle is made of hard-drawn steel wire with adiameter of 3mm. The spring has 10 active coils and a coil diameter of 21mm. After 5years of use the spring cracks due to fatigue. To get longer fatigue life, the stresses shouldbe decreased by 5%, but the only available cylindrical space is used by the present spring,which has an outside diameter of 24mm, an inside diameter of 18mm, and a length of30mm. The spring rate for the new spring would be the same as the old one. Find asolution.

Notes: There are numerous possibilities, such as choosing a new material or changing the crosssection. Note however that a new material would not be able to reduce the stresses, but a newmaterial may still be acceptable if its fatigue life is superior. This solution uses a rectangular crosssection to achieve the new spring design.

Solution:Note that for this spring, d=3mm. The spring rate needs to be the same, so from Equation (16.42),

Mθ

= EIlw

Also, Equation (16.40) gives the stress, which must be reduced by 5%. If the spring has to take upthe same space, C is constant and Ki will not be changed. Since M is also constant, then c/I has todecrease by 5%. This will be achieved by changing the cross section from circular to rectangular.Therefore, while keeping the same dimensions so that the rectangle’s width is 3mm,

0.95( ) cI

circle= c

I

rect; 0.95( ) d / 2

π64

d4

= h / 2

bh3 / 1 2( ) ;h = 2.36mm

So that the new spring will be of rectangular cross section with the same inner and outer diameter,but the cross section will be of width 3mm and height 2.36mm.

16.29 The oven door on a kitchen stove is kept shut by a helical torsion spring. The springtorque is 1Nm when the door is shut. When the oven door is fully open, it must stay openby gravitational force. The door height is 450mm and it weighs 4kg. Dimension thehelical torsion spring using music wire as the spring material and a wire diameter of4.5mm. Is it also possible to use 3mm wire?

Notes: This is an open ended problem in that one can use any reasonable combination of outerdiameter and number of coils can be used to get an answer. This solution uses C=D/d=10.Equation (16.42) is used in this problem.

Solution:This is an open-ended problem in that any reasonable value of D/d will give a good answer. Thissolution will be based on D/d=10. Recognizing that there needs to be a 1Nm moment at the doorclosed state, the torsion springs needs to be prewound a certain angular deflection θ. At θ+π/2(door open state), then the spring has to just barely support the door weight, which if uniformly

23

distributed leads to a moment of M=(4kg)(0.450m/2)(9.81m/s2)=8.83Nm. We then have twoconditions, which allow us to write Eq. 16.42 twice:

θ = MlwEI

= 1Nm( )lwEI

;lwEI

= θ1Nm

θ + π2

= MlwEI

= (8.83Nm)lwEI

Substituting for lw/EI in the second equation gives θ=0.20rad=11.5°. This is the minimum angulardeflection from the free state, and in practice may be more than this so that the door stays openand requires a certain force to close. However, the design analysis will progress using this value.If d=0.0045m, then Equation (16.42) can be solved for the wire length:

θ = MlwEI

; lw = θEIM

=0.20rad( ) 207GPa( ) π

640.0045m( )4

1Nm= 0.83m

The number of turns needed is obtained from this wire length:

lw = πDNa; Na = lwπD

= 0.83mπ 0.045m( ) = 5.89

If d=0.003m, so that D=0.03m, then Equation (16.42) can be solved for the wire length:

θ =MlwEI

; lw =θEI

M=

0.20rad( ) 207GPa( ) π64

0.003m( )4

1Nm= 0.165m

The number of turns needed is obtained from this wire length:

lw = πDNa; Na =lwπD

=0.165m

π 0.03m( )=1.74

16.30 A helical torsion spring, shown in sketch i, is made from hard-drawn steel with a wirediameter 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 angulardisplacement.b) What would the coil inside diameter be when the maximum load is applied?c) For 100,000 loading cycles calculate the maximum moment and the correspondingangular displacement for a safety factor of 2.5 guarding against fatigue failure.

Notes: The force is obtained from Equation (16.40) after the allowable stress in the wire isdetermined. The angular displacement is calculated from Equation (16.43) once the number ofactive coils is found from Equation (16.45). The inside diameter under maximum load is obtainedfrom Equation (16.46). The fatigue strength of the wire is calculated from Equations (7.9)through (7.13), although the Zimmerli equations (16.29) could alternatively be used.

Solution:

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From Table 16.1, for hard drawn wire, G=11.5Mpsi=79.3GPa, and take E=207GPa as for carbonsteel from the inside front cover. From Table 16.2 for hard drawn wire, Ap=1750MPa, m=0.192.Therefore, the ultimate strength of the wire is obtained from Equation (16.2) as:

Sut =Ap

d m = 1750MPa

2.20.192 = 1505MPa

The allowable shear stress is calculated from Equations (16.3)as:Ssy = σall = 0.6Sut = 0.6 1505MPa( ) = 903MPa

Also, for this spring, from Equation (16.6), C=D/d=22mm/2.2mm=10, so from Equation (16.41),

Ki = 4C2 − C −14C C −1( ) = 400 −10 −1

40 9( ) = 1.08

Recognizing that M=P(0.035m), Equation (16.40) gives

σ =Sy

ns=

32K iP 0.035m( )πd3 ; P =

Syπd3

32ns Ki 0.035m( )=

903MPa( )π 0.0022m( )332 2( ) 1.08( ) 0.035m( )

= 12.5N

From Equation (16.45) on page 762,

Na = Nb + Ne = Nb + l1 + l23πD

= 8.5+ 0.035m + 0.035m3π 0.022m( ) = 8.84

So that from Equation (16.43),

θrev =10.18MDNa

Ed4 =10.18 12.5N( ) 0.035m( ) 0.022 m( ) 8.84( )

207GPa( ) 0.0022m( )4= 0.18

Therefore, under the maximum load, the number of coils is Na’=Na+θrev=8.84+0.18=9.02. From

Equation (16.46),

Di′ =

Na Di

Na′ =

8.84 0.0198m( )9.02

= 0.0194m

Note from Equation (7.7) that for bending, Se’=0.5Su=0.5(1505MPa)=753MPa. From Equation

(7.8), Sl’=0.9Su=0.9(1505MPa)=1355MPa. From Equation (7.12),

bs = −13

logSl

′

Se′

= − 1

3log

0.9Sut

0.5Sut

= −0.085


C = logSl

′( )2

Se′

= log13552

753

= 3.387

Therefore, the fatigue strength is, from Equation (7.14) on page 269,

S f′ = 10C Nt

′( )bs=103.387 100000( )−0.085 = 916MPa

For this spring, σmin=0, σmax=σ, so σm=σa=σ/2. Using a Goodman relationship (see Equation(7.28)),

σa

S f′ + σm

Sut= σ

2 916MPa( ) + σ2 1505MPa( ) = 1

ns= 1

2.5;σ = 455MPa

Therefore, from Equation (16.40), the maximum force is

σ = 32K iP 0.035m( )πd3 ; P = σπd3

32Ki 0.035m( ) = 455MPa( )π 0.0022m( )3

32 1.08( ) 0.035m( ) =12.6 N

Therefore, Mmax=P(0.035m)=12.6N(0.035m)=0.441Nm. The corresponding angular displacementis, from Equation (16.43),

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θrev = 10.18 MDNa

Ed4 = 10.18 0.44Nm( ) 0.022m( ) 8.84( )207GPa( ) 0.0022m( )4 = 0.180rev

16.31 The helical torsion spring shown in sketch j has a coil outside diameter of 22mm, 8.25turns, and a wire diameter of 2 mm. The spring material is hard-drawn steel. What wouldthe applied moment be if the maximum stress equaled the yield limit? Calculate theinside diameter of the loaded spring and the corresponding angular displacement.Dimensions are in millimeters.

Notes: This is similar to the first part of problem 16.30. Equations (16.2) and (16.3) are needed toget the strength of the wire. Equation (16.40) is used to get the normal stress, and Equation(16.43) gives the angular deflection.

Solution:From Table 16.1, for hard drawn wire, G=11.5Mpsi=79.3GPa, and take E=207GPa as for carbonsteel from the inside front cover. From Table 16.2 for hard drawn wire, Ap=1750MPa, m=0.192.Therefore, the ultimate strength of the wire is obtained from Equation (16.2) as:

Sut =Ap

d m = 1750MPa

20.191 = 1532MPa

The yield stress in tension is approximately given by Equation (16.39):Sy =0.6Sut = 0.6 1532MPa( ) = 919MPa

For this spring, D=D0-d=22mm-2mm=20mm=0.020m. From Equation (16.6),C=D/d=20mm/2mm=10. Therefore, from Equation (16.41),

Ki = 4C2 − C −14C C −1( ) = 400 −10 −1

40 9( ) = 1.08

The normal stress is given by Equation (16.40) and set equal to the yield strength, and solved forthe applied moment as follows:

σ = 32K iM

πd3 = Sy ;M =Syπd3

32Ki= 919MPa( )π 0.002m( )3

32 1.08( ) = 0.668Nm

The angular deflection in revolutions is given by Equation (16.43) as

θrev = 10.18MDNa

Ed4 = 10.18 0.668Nm( ) 0.020m( ) 8.25( )207GPa( ) 0.002m( )4 = 0.339

The new number of active coils is given by Equation (16.45) asNa

’=Na+θrev=8.25+0.339=8.589Therefore, from Equation (16.46), the new inner diameter is

Di′ = Na Di

Na′ = 8.25( ) 20mm − 2mm( )

8.589( ) = 17.29mm

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16.32 A helical torsion spring made of music wire has a coil diameter of 17.5mm and a wirediameter of 1.5mm while supporting a 0.15Nm moment with 20% fluctuation. Themaximum number of turns is 12, and the load is 22 mm from the center of the spring. Forinfinite life with 99% reliability, find the safety factors guarding against static and fatiguefailure. Also, determine the inside diameter of the spring when the load has been applied.

Notes: The wire strength is determined as in the previous problems. A mean and alternatingmoment is calculated, which allows calculation of mean and alternating stresses. A Goodmanrelationship is then used for fatigue.

Solution:From Table 16.1, for music wire, G=11.5Mpsi=79.3GPa, and take E=207GPa as for carbon steelfrom the inside front cover. From Table 16.2 for music wire, Ap=2170MPa, m=0.146. Therefore,the ultimate strength of the wire is obtained from Equation (16.2) as:

Sut =Ap

d m = 2170MPa

1.50.146 = 2045MPa

From Equation (7.7) on page 265, Se’=0.5Sut=0.5(2045)=1022MPa. If the wire is drawn, from

Table 7.3, e=4.51 and f=-0.265, so kf is from Equation (7.21):

k f = eSutf = 4.511022( )−0.265 = 0.72

From Table 7.2, kr=0.82 for 99% reliability. The size factor is, from Equation (7.22), ks=1.Therefore, from Equation (7.16),

Se=kfkskrSe’=(0.72)(0.82)(1)(1022MPa)=603MPa

The maximum moment is Mmax=1.2(0.15Nm)=0.18Nm. The minimum moment isMmin=0.8(0.l5Nm)=0.12Nm. Therefore, Ma=(Mmax-Mmin)/2=0.03Nm, and Mm=0.15Nm. For thisspring, C=D/d=17.5/1.5=11.67. Therefore, from Equation (16.41),

Ki = 4C2 − C −14C C −1( ) = 4 11.67( ) −11.67 −1

4 11.67( ) 10.67( ) = 1.07

From Equation (16.40), the alternating stress is:

σa = 32Ki Ma

πd3 = 32 1.07( ) 0.03Nm( )π 0.0015m( )3

= 96.9MPa

Similarly the mean stress is:

σm = 32Ki Mm

πd 3 = 32 1.07( ) 0.15Nm( )π 0.0015m( )3

= 484.4MPa

The static safety factor is

ns = Sut

σ max= Sut

σ a +σ m= 2045MPa

96.9MPa + 484.4MPa= 3.52

The safety factor against fatigue is calculated using a Goodman relationship (Equation (7.28)):1ns

= σa

Se+ σm

Sut= 96.9MPa

603MPa+ 484.4MPa

2045MPa= 0.398; ns = 2.51

16.33 A fishing rod is made like an ideal leaf spring with rectangular cross sections. It is madeof carbon-fiber reinforced plastic with a 150GPa modulus of elasticity. The thickness isconstant at 8mm and the length is 2m. Find how large the cross section must be at the

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handle to carry a 0.3 kg fish by the hook without bending the top of the rod more than200mm. Neglect the weight of the lure. Also, calculate the bending stress.

Notes: Equation (16.51) is needed to obtain the width. Equation (16.49) gives the stress.

Solution:The load is P=(0.3kg)(9.81m/s

2)=2.94N. From Equation (16.51), with d=0.2m and n=1,

δ = 6Pl3

Enbt3;b = 6Pl3

Enδt3 = 6 2.94N( ) 2m( )3

150GPa( ) 1( ) 0.2m( ) 0.008m( )3= 0.00919m = 9.19mm

The stress is obtained from Equation (16.49) as

σ = 6Pl

bt2 = 6 2.94N( ) 2m( )0.00919m( ) 0.008m( )2 = 60.0MPa

16.34 A trampoline is made like a leaf spring with variable width so that the maximum bendingstress in each section of the trampoline is constant. The material is glass-fiber reinforcedplastic with a modulus of elasticity of 28GPa and a bending strength of 300MPa.Calculate the spring rate at the tip of the trampoline and the corresponding safety factor ifa swimmer weighing 80kg jumps onto the trampoline from a height of 2 m. The activelength of the trampoline is 3m, its width is 1.2 m, and its thickness is 38mm.

Notes: The change in potential energy equals the energy stored in bending, and this allowscalculation of the deflection. Using Equation (16.52) gives the stiffness of the spring, so theapplied force can be calculated from the deflection. Equation (16.49) then gives the stress.

Solution:The stiffness of the spring is calculated from Equation (16.52) using n=1:

k = Enbt3

6l3= 28GPa( ) 1( ) 1.2m( ) 0.038m( )3

6 3m( )3 = 11.38kN / m

From the first law of thermodynamics, the energy stored in the spring must be the same as thechange in potential energy of the swimmer. Therefore,

mag h + δ( ) = kδ2

2; δ 2 − 2mag

kδ − 2magh

k= 0; δ = 2mag

2k± 1

22mag

k

2

+ 42magh

k

substituting ma=80kg, g=9.81m/s2, h=2m and k=11.38kN/m gives δ=0.5987m. Therefore, the

maximum force is P=kδ=(11.38kN/m)(0.5987m)=6.81kN. Therefore, from Equation (16.49), thestress is

σmax = 6Pl

bt2= 6 6810N( ) 3m( )

1.2m( ) 0.038m( )2 = 70.7MPa

Therefore, the safety factor is ns=300MPa/70.7MPa=4.24

16.35 The leaf spring of a truck should be able to accommodate 55 mm deflections (up anddown) of the wheels from an equilibrium position when the truck is driven on a roughroad. The static load in the middle of the leaf spring is 50,000N. The modulus ofelasticity for the spring is 207GPa.Assume an allowable stress of 1050 MPa, a safetyfactor of 3, leaf length of 0.8 m, a thickness of 0.02 m, and that they are split into 10layers. Determine the width of the spring.

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Notes: Equation (16.49) gives the stress in the spring, which gives an equation for the springwidth. The deflection is calculated from Equation (16.52).

Solution:With a safety factor of 3, the allowable stress is Using Equation (16.49) to obtain the maximumstress,

σmax =6Pl

bt2=

σall

3;b =

18Pl

σallt2 =

18( ) 50,000 N( ) 0.8m( )1050MPa( ) 0.02m( )2 = 1.714m

From Equation (16.52), the deflection is

δ =6Pl3

Enbt3=

6 50 ,000N( ) 0.8 m( )3

207GPa( ) 10( ) 1.714 m( ) 0.02 m( )3 = 0.0512m = 54.12mm

Spring Solution-CH 16

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Transcript of Spring Solution-CH 16