Springs
Types of Springs
Learning Objectives:
- Understand the various types of springs that are available and their function.
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Solved Example: 92-1-01
A spring used to absorb shocks and vibrations is:
A. Torsion spring
B. Conical spring
C. Leaf spring
D. Disc spring
Correct Answer: C
Solved Example: 92-1-02
In a close-coiled helical spring:
A. Plane of the coil and axis of the spring are closely attached
B. Angle of helix is large
C. Plane of the coil is normal to the axis of the spring
D. Deflection is small
Correct Answer: C
Equivalent Spring Constant
Learning Objectives:
- Get acquainted with spring terminology.
- Solid length, \(L_S\) is that length of the spring, when pressed, all the spring coils will clash with each other and will appear as a solid cylindrical body.
- N is the number of active coils, i.e., only these coils take part in the spring action.
- However, few other coils may be present due to manufacturing consideration, thus total number of coils, $N_T$ may vary from total number of active coils.
- Spring index is the ratio of spring diamter to coil diameter.
The deflection and force are related by $F = kx$ where the spring rate (spring constant) k is given by:
where G is the shear modulus of elasticity and N is the number of active coils.
Spring in Series:
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Spring in Parallel:
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Solved Example: 92-2-01
In spring equation F = k.x, what is name of the parameter ‘k’?
A. Linear constant
B. Spring index
C. Spring Constant
D. Spring quality factor
The spring constant (k in the Hooke's law equation) is the ratio of the F/x.
Correct Answer: C
Solved Example: 92-2-02
When spring index increases, the value of Wahl's stress factor:
A. Decreases linearly
B. Increases linearly
C. Increases exponentially
D. Remains same
Wahl's Correction Factor accounts for shear stress resulting from spring curvature, \[W = \dfrac{4C-1}{4C-4} + \dfrac{0.615}{C}\]
Correct Answer: C
Solved Example: 92-2-03
Spring index is:
A. Ratio of coil diameter to wire diameter
B. Load required to produce unit deflection
C. Its capability of storing energy
D. Indication of quality of spring
The spring index is the ratio between the mean diameter and wire diameter. \[C = \dfrac{D}{d}\] Where, C = Spring Index, D = Spring diameter, d = wire diameter
Correct Answer: A
Solved Example: 92-2-04
Spring stiffness is:
A. Ratio of coil diameter to wire diameter
B. Load required to produce unit deflection
C. Its capability of storing energy
D. Its ability to absorb shocks
Spring stiffness is defined as the load per unit deflection.
Correct Answer: B
Solved Example: 92-2-05
Spring index is:
A. D - d
B. $\dfrac{D}{d}$
C. $D^2 - d^2$
D. None of these
The spring index is the ratio between the mean diameter and wire diameter. \[C = \dfrac{D}{d}\] Where, C = Spring Index, D = Spring diameter, d = wire diameter
Correct Answer: B
Solved Example: 92-2-06
Wahl's stress concentration factor is:
A. $\dfrac{(4C - 1)}{(4C - 3)} + \dfrac{0.615}{C}$
B. $\dfrac{(4C - 1)}{(4C - 2)} + \dfrac{0.615}{C}$
C. $\dfrac{(4C - 1)}{(4C - 4)} + \dfrac{0.615}{C}$
D. None
In order to take into account the effect of direct shear and change in coil curvature a stress factor is defined, which is known as Wahl's factor.
Correct Answer: C
Solved Example: 92-2-07
Resilience of spring is:
A. Strain energy per unit length
B. Strain energy per unit area
C. Strain energy per unit mass
D. None
Resilience is the ability of a material to absorb energy under elastic deformation and to recover this energy upon removal of load. The modulus of resilience is the maximum elastic energy absorbed by a material when a load is applied. The modulus of resilience $E_r$ is the area contained under the elastic portion of the stress-strain curve. It is the elastic energy that a material absorbs during loading and subsequently releases when the load is removed. For linear elastic behavior:
$E_r$ = $\dfrac{1}{2}$ (yield strength) $\times$ (strain at yielding)
Correct Answer: D
Solved Example: 92-2-08
Form coefficient of spring is:
A. Ratio of coil diameter to wire diameter
B. Load required to produce unit deflection
C. Its capability of storing energy
D. Concerned With strength of wire of spring
Correct Answer: C
Solved Example: 92-2-09
The spring constant of a helical compression spring does not depend on:
A. Coil diameter
B. Material strength
C. Number of active turns
D. Wire diameter
Correct Answer: B
Helical Linear Springs
Learning Objectives:
- Design and analyze helical compression springs, including compatibility with allowable stresses.
Springs are subjected to two types of stresses:
- Direct force: F which causes a stress of \(\dfrac{F}{A}\)
- Torque: \(\dfrac{FD}{2}\) which causes a stress of \(\dfrac{Tr}{J}\)
Each causes shear stress in the spring. The maximum shear stress will be in the inside surface of the wire. \[\tau _{max} = \dfrac{Tr}{J} + \dfrac{F}{A}\] Substituting, \[T = \dfrac{FD}{2}\] \[r = \dfrac{D}{2}\] \[J = \dfrac{\pi }{32}d^{4}\] \[A = \frac{\pi }{4}d^{2}\] We get, \[\tau _{max} = \dfrac{8FD}{\pi d^{3}} + \dfrac{4F}{\pi d^{2}}\] Spring index is a measure of coil curvature. It is defined as the ratio of spring diameter to wire diameter.
Where, C = Spring Index, D = Spring diameter, d = wire diameter
Also, by using K\(_s\) = Shear stress correction factor, which is:
\[\tau = \dfrac{2C+1}{2C} \left ( \dfrac{8FD}{\pi d^{3}} \right )\]
In order to take into account the effect of direct shear and change in coil curvature a stress factor is defined, which is known as Wahl’s factor.
Solved Example: 92-4-01
In the calculation of induced shear stress in helical springs, the Wahl's stress (correction) factor is used to take care of:
A. Combined effect of transverse shear stress and bending stress in the wire
B. Combined effect of bending stress and curvature of the wire
C. Combined effect of transverse shear stress and curvature of wire
D. Combined effect of torsional shear stress and transverse shear stress in the wire
Correct Answer: B
Solved Example: 92-4-02
Determine the maximum shearing stress in a helical steel spring composed of 20 turns of 20 mm diameter wire on mean radius of 80 mm when the spring is supporting a load of 2.0 kN?
A. 529 GPa
B. 370 Pa
C. 25 KPa
D. 121 MPa
\[\tau_{max} = \dfrac{16PR}{\pi d^3} \left( \dfrac{4C - 1}{4C - 4} + \dfrac{0.615}{C} \right)\] Where: P = 2 kN = 2000 N, R = 80 mm, d = 20 mm, n = 20 turns, \[C = \dfrac{D}{d} = \dfrac{2 \times 80}{20} = 8\] \[\tau_{max} = \dfrac{16(2000)(80)}{\pi(20^3)} \left( \dfrac{4(8) - 1}{4(8) - 4} + \dfrac{0.615}{8} \right) = 121\ \text{MPa}\]
Correct Answer: D
Solved Example: 92-4-03
The shearing stresses in the inner face as compared to outer face of the wire in a heavy close coiled spring is:
A. Larger
B. Smaller
C. Equal
D. Larger/smaller depending on diameter of spring coil
Correct Answer: A
Solved Example: 92-4-04
When a close coiled helical spring is compressed, its wire is subjected to:
A. Tension
B. Shear
C. Compression
D. All of the above.
Correct Answer: A
Solved Example: 92-4-05
Compute the maximum shearing stress in MPa, developed in a phosphor bronze spring having mean diameter of 200 mm and consisting of 24 turns of 200-mm diameter wire when the spring is stretched 100 mm. G = 42 GPa.
A. 81.10
B. 90.24
C. 21.09
D. 31.89
\[\delta = \dfrac{64PR^3n}{Gd^4}\] Where: $\delta$ = 100 mm, R = 100 mm, d = 20 mm, n = 24 turns, G = 42000 MPa \[100 = \dfrac{64P(100^3)24}{42\,000(20^4)}\] \[P = 437.5 \, \text{N}\] \[\tau_{max} = \dfrac{16PR}{\pi d^3} \left( \dfrac{4C - 1}{4C - 4} + \dfrac{0.615}{C} \right)\] Where $C = \dfrac{D}{d} = \left(\dfrac{2 \times 100}{20}\right) = 10$ \[ \tau_{max} = \dfrac{16(437.5)(100)}{\pi (20^3)} \left[ \dfrac{4(10) - 1}{4(10) - 4} + \dfrac{0.615}{10} \right] = 31.89 \ \text{MPa}\]
Correct Answer: D
Solved Example: 92-4-06
Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B' The ratio of deflections in spring 'A' to spring 'B' is:
A. $\dfrac{1}{8}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{2}$
D. 2
Correct Answer: C
Solved Example: 92-4-07
When two springs are in series (having stiffness K), the equivalent stiffness will be:
A. K
B. $\dfrac{K}{2}$
C. 2K
D. $\dfrac{K}{A}$
Correct Answer: B
Solved Example: 92-4-08
If a spring is cut down into two springs, the stiffness of cut springs will be:
A. Half
B. Same
C. Double
D. Unpredictable
Correct Answer: C
Solved Example: 92-4-09
If two springs are in parallel then their overall stiffness will be:
A. Half
B. Same
C. Double
D. Unpredictable
Correct Answer: C
Solved Example: 92-4-10
Which of the following statement is wrong?
A. The solid length of a spring is the product of total number of coils and the diameter of the wire
B. The spring index is the ratio of mean diameter of the coil to the diameter of the wire
C. The spring stiffness is the load required per unit deflection of the spring
D. The pitch of the coil is the axial distance between adjacent coils in the compressed state
Correct Answer: A
Solved Example: 92-4-11
If a closed coiled helical spring absorbs 30 N-mm of energy while extending by 5 mm, its stiffness will be:
A. 2 N/mm
B. 4 N/mm
C. 2.4 N/mm
D. 10 N/mm
\begin{align*} E &= \dfrac{1}{2}kx^2\\ 30 &= \dfrac{1}{2}k(5)^2\\ \dfrac{60}{25} &= k\\ 2.4\ \mathrm{N/mm} &= k \end{align*}
Correct Answer: C
End Conditions of Compression Springs
Learning Objectives:
- Analyze and design mechanical springs.
- Acquaintance with spring terminology and different types of springs.
- Square (or Closed) end helical springs are the least expensive and most widely used. However, they tend to buckle under the load.
- Square end springs can be ground to improve their utility. Square and Ground springs can be placed accurately so that they don’t buckle due to misalignment.
- Plain (or Open) end helical springs can be used in applications where less solid height is needed or height tolerances are not very stringent.
- Plain end springs when not ground tend to entangle when stored in bulk and hence, at times, they are ground for easy storage and avoid buckling as mentioned above.
-
Plain ends:
Total coils, \(N_T\): N
Solid length, \(L_S\): d ( \(N_T\) + 1 )
Free length, L : L + \(\delta_{max}\) + \(\delta_{allowance}\)
Pitch, p : (L - d) / N -
Plain and Ground ends:
Total coils, \(N_T\): N + 1
Solid length, \(L_S\): d ( \(N_T\))
Free length, L : \(L_s\) + \(\delta_{max}\) + \(\delta_{allowance}\)
Pitch, p : L / (N + 1) -
Squared or closed ends:
Total coils, \(N_T\): N + 2
Solid length, \(L_S\): d ( \(N_T\) + 1 )
Free length, \(L_S\) + \(\delta_{max}\) + \(\delta_{allowance}\)
Pitch, p : (L - 3d) / N -
Squared and ground ends:
Total coils, \(N_T\): N + 2
Solid length, \(L_S\): d ( \(N_T\))
Pitch, p : (L - 2d) / N
Solved Example: 92-3-01
The most commonly used spring for truck, trailer, and railway carriage is:
A. Helical spring
B. Extension spring
C. Compression spring
D. Leaf spring
Correct Answer: D
Solved Example: 92-3-02
A spring used to absorb shocks and vibrations is:
A. Torsion spring
B. Conical spring
C. Leaf spring
D. Disc spring
Correct Answer: C
Helical Torsion Springs
Learning Objectives:
- To calculate bending stress in helical torsion springs.
- To understand the concept of correction factor.
- To calculate angular deflection $\theta$ in helical torsion springs.
- To calculate allowable stress in helical torsion springs.
The bending stress
The correction factor
Angular displacement $\theta$
The spring rate
Strength of spring material, \begin{align*} S_y &= \sigma = 0.78 S_{ut}\ \ \ \ \mathrm{for\ (A227,\ A228,\ A229)}\\ S_y &= \sigma = 0.87 S_{ut}\ \ \ \ \mathrm{for\ (A232,\ A401)} \end{align*}
Spring Material
Learning Objectives:
- To calculate minimum tensile strength of common spring steels.
The minimum tensile strength of common spring steels may be determined from:
where A and m are factors depending upon the spring material and are to be selected from table on Page 433 of FE Reference Handbook.