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UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Unit I: Design of springs Leaf Spring The term leaf spring or flat spring applies to a wide variety of shapes made out of flat strip. An advantage of leaf spring over helical spring is that the ends of the leaf spring may be guided along a definite path as it deflects. Thus the spring may act as a structural member in addition to energy absorbing device. Thus the leaf spring may carry lateral loads, brake torque, driving torque etc., in addition to shocks. Leaf springs can be designed to have progressive spring rates. This “non-linear spring constant” is useful for vehicles which must operate with widely varying loads, such as trucks. The leaf spring may be arranged in two fashions as follows: 1. Cantilever member and 2. Simply supported member

1

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

1. Cantilever member: Consider a rectangular flat strip of uniform width as shown in Figure 18(a) (Ref: DDHB: Fig. 11.10 (a)). It is noticed that in beams of uniform width, the stresses act heavily at one specific point (i.e., at fixed end point) rather than equally along the length of the plate. In order to achieve the uniform strength either width or thickness of the beam made uniformly taper, while other parameter constant. The general procedure adopted is to keep the thickness constant and vary the width as shown in Figure 18 (b) (Ref: DDHB: Figure 11.10 (b)). F

F

h

h l

l b

b

(a) Constant cross-section (b) Constant thickness variable width (uniform strength) Fig. 18 Cantilever member Let, F = Applied load

b = Width of flat spring

h = Depth or thickness of spring and

l = Length of the plate

Bending stress is given by,

𝑀 𝐹𝑙 6𝐹𝑙 𝐶1 𝐹𝑙 = = 2= 2 𝑏ℎ 𝑍 𝑏ℎ 𝑏ℎ2 ( ) 6 Deflection is given by, 𝜎=

(11.27(𝑎))

(𝐶1 = 6)

4𝐹𝑙 3 𝐶2 𝐹𝑙3 𝐹𝑙3 𝐹𝑙3 (11.27(𝑏)) = (𝐶2 = 4) = = 𝐸𝑏ℎ3 3𝐸𝐼 3𝐸 (𝑏ℎ 3 ) 𝐸𝑏ℎ3 12 Using Eqn. 11.27 (a) and (b), the thickness of the spring plate is obtained in terms of ‘b’, 𝑦=

𝐶2 𝜎𝑙2 ℎ= 𝐶1 𝐸𝑦 The width of the plate is obtained as follows:

(11.27(𝑐))

𝑏=

(11.27(𝑑))

𝐶1 𝐹𝑙 𝜎ℎ 2

2

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Constants 𝐶1 𝑎𝑛𝑑 𝐶2 for various types of beams refer Figure 11.10 (a) and (b) in Page No. 180 and Table 11.9 in Page No. 198 DDHB. 2. Simply supported member Consider a rectangular flat strip of uniform width as shown in Figure 19(a) (Ref: DDHB: Fig. 11.10 (d)). It is noticed that in beams of uniform width, the stresses act heavily at one specific point (i.e., at fixed end point) rather than equally along the length of the plate. In order to achieve the uniform strength either width or thickness of the beam made uniformly taper, while other parameter constant. The general procedure adopted is to keep the thickness constant and vary the width as shown in Figure 19 (b) (Ref: DDHB: Figure 11.10 (e)). F

F h

h

2l

2l b

b

(a) Constant cross-section (b) Constant thickness variable width (uniform strength) Fig. 19 Simply supported member Let, F = Applied load h = Depth or thickness of spring and Bending stress is given by, 𝐹2𝑙 𝑀 ( 4 ) 3𝐹𝑙 𝐶1 𝐹𝑙 = 2= 𝜎= = 𝑍 𝑏ℎ2 𝑏ℎ2 𝑏ℎ ( 6 ) Deflection is given by,

b = Width of flat spring 2l = Length of the plate

(11.27(𝑎))

(𝐶1 = 3)

2𝐹𝑙3 8𝐹𝑙 3 𝐶2 𝐹𝑙 3 (11 .27(𝑏)) = (𝐶2 = 2) = 3 𝐸𝑏ℎ3 𝑏ℎ3 𝐸𝑏ℎ 48𝐸 ( 12 ) Using Eqn. 11.27 (a) and (b), the thickness of the spring plate is obtained in terms of ‘b’ as follows: 𝐹(2𝑙)3 = 48𝐸𝐼

𝑦=

𝐶2 𝜎𝑙2 𝐶1 𝐸𝑦 The width of the plate is obtained as follows:

(11.27(𝑐))

𝑏=

(11.27(𝑑))

ℎ=

𝐶1 𝐹𝑙 𝜎ℎ2

3

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Constants 𝐶1 𝑎𝑛𝑑 𝐶2 for various types of beams refer Figure 11.10 in Page No. 180 and Table 11.9 in Page No. 198 DDHB. Laminated springs Beams of uniform strength permits a considerable saving in material and at the same time provides a greater deflection. Thus their resilience and capacity for absorbing shocks is greater. In order to make the spring of uniform strength more compact, so as to minimize the width space, the width of the spring is cut into several strips and are placed one below the other as a laminates. Referring to Figure 20 (Figure 11.11(b) in DDHB), a triangular cantilever strip is cut longitudinally as 1, 2-2 and 3-3. The central strip marked 1 is the master leaf which place at the top. The two strips marked 2-2 are put together, side by side to form another leaf and placed below the top leaf. In the similar manner other pairs of strips marked 3-3 is placed in the decreasing order of strip length to form a laminated spring.

F

F

l

l h

3

2

b

1 3

h

b’

2

Fig. 20 Laminated Springs – Cantilever

The width of each leaf is,

(11.28(𝑎))

𝑏 ′ = 𝑏/𝑖

𝑤ℎ𝑒𝑟𝑒, 𝑖 is the number of leaves

The maximum bending stress and deflection values for these springs are same as that of the original plate. Referring to Figure 21 (Figure 11.11(c) in DDHB (title is missing)), a diamond shaped simply supported strip is cut longitudinally as 1, 2-2, 3-3 and 4-4. The central strip marked 1 is the master leaf which place at the top. The two strips marked 2-2 are put together, side by side to form another leaf and placed below

4

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

the top leaf. In the similar manner other pairs of strips marked 3-3 and 4-4 is placed in the decreasing order of strip length to form a laminated spring. F

F

h

h 2l 2l 3 2 1 2 3

b’

b

(a) Equivalent flat bar

(b) Multi leaf spring

Fig. 21 Laminated Springs – simply supported The width of each leaf is,

(11.28(𝑎))

𝑏 ′ = 𝑏/𝑖

𝑤ℎ𝑒𝑟𝑒, 𝑖 is the number of leaves

The maximum bending stress and deflection values for these springs are same as that of the original plate. The expression for bending stress and deflection, for both cantilever and simply supported laminated spring are same as follows, except the 𝐶1 𝑎𝑛𝑑 𝐶2 values. Bending stress is given by, 𝜎=

𝐶1 𝐹𝑙 𝑖𝑏 ′ ℎ2

(11.28(𝑏))

Deflection is given by, 𝐶2 𝐹𝑙3 𝑦= 𝐸𝑖𝑏 ′ ℎ3

(11.28(𝑐))

Note:

For cantilever laminated spring 𝐶1 = 𝐶2 = 6 and

For simply supported laminated spring 𝐶1 = 𝐶2 = 3

5

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Semi-Elliptical Laminated Spring: Semi-elliptical laminated (multi-leaf) springs are widely used for the suspension of cars, trucks and railway wagons. A multi-leaf spring consists of a series of flat plates, usually of semi-elliptical shape, as shown in Figure 22. The fl at plates are called leaves of the spring. The leaves have graduated lengths. The leaf at the top has maximum length. The length gradually decreases from the top leaf to the bottom leaf. The longest leaf at the top is called master leaf. It is bent at both ends to from the spring eyes. Two bolts are inserted through these eyes to fix the leaf spring to the automobile body.

Fig. 22 Semi-elliptic Leaf Spring The leaves are held together by means of two U-bolts and a centre clip. Rebound clips are provided to keep the leaves in alignment and prevent lateral shifting of the leaves during operation. At the centre, the leaf spring is supported on the axle. Multi-leaf springs are provided with one or two extra full length leaves in addition to master leaf. The extra full-length leaves are stacked between the master leaf and the graduated length leaves. The extra full-length leaves are provided to support the transverse shear force. For the purpose of analysis, the leaves are divided into two groups namely, master leaf along with graduated-length leaves forming one group and extra full-length leaves forming the other. Analysis: The following notations are used in the analysis: if = Number of extra full-length leaves ig = Number of graduated-length leaves including master leaf i = Total number of leaves (if + ig) b = Width of each leaf (mm) t = Thickness of each leaf (mm) l = Length of the cantilever or half the length of semi-elliptic spring (mm)

6

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

2l = Total length of the of semi-elliptic spring (mm) Ff = Portion of F taken by the extra full-length leaves (N) Fg = Portion of F taken by the graduated-length leaves (N) F = Total force applied on the spring (N)

f = The stress in full length leaves g = The stress in graduated leaves b’= The width of laminated spring h = The thickness of spring For analysis purpose half portion of the spring is considered as cantilever. The graduated leaves along with master leaf are treated as a triangular plate of thickness h and maximum width at the supports as igb’ as shown in Figure 23.

(a) Only front view (b) front and top view Fig. 23 Graduated-length leaves as Triangular Plate Bending stress is given by, 𝜎𝑔 =

𝐶1 𝐹𝑔 𝑙 𝑖𝑔 𝑏 ′ ℎ2

𝐶1 = 6

(11.28 (𝑏))

𝐶2 = 6

(11.28 (𝑐))

Deflection is given by, 𝑦𝑔 =

𝐶2 𝐹𝑔 𝑙 3 𝐸𝑖𝑔 𝑏 ′ ℎ3

7

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

The extra full length leaves can be treated as a rectangular plate of thickness h and uniform width ifb’ as shown in Figure 24.

(a) Only front view (b) front and top view Fig. 24 Extra full-length leaves as rectangular plate Bending stress is given by, 𝜎𝑓 =

𝐶1 𝐹𝑓 𝑙 𝑖𝑓 𝑏 ′ ℎ2

𝐶1 = 6

(11.28(𝑏))

𝐶2 = 4

(11.28(𝑐))

Deflection is given by, 𝑦𝑓 =

𝐶2 𝐹𝑓 𝑙 3 𝐸𝑖𝑓 𝑏 ′ ℎ3

8

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Since the graduated leave and full length leaves are clamped together, their deflection is same, ∴ 𝑦𝑓 = 𝑦𝑔 4𝐹𝑓 𝑙 3

6𝐹𝑔 𝑙 3 = ′ 3 𝐸𝑖𝑓 𝑏 ′ ℎ3 𝐸𝑖𝑔 𝑏 ℎ 2𝐹𝑓 3𝐹𝑔 = 𝑖𝑔 𝑖𝑓 𝐹𝑓 3𝑖𝑓 = 𝐹𝑔 2𝑖𝑔 But,

(11.29(𝑑))

𝐹 = 𝐹𝑔 + 𝐹𝑓

𝐹𝑓 ] 𝐹𝑔 3𝑖𝑓 𝐹 = 𝐹𝑔 [1 + ] 2𝑖𝑔 2𝑖𝑔 + 3𝑖𝑓 ] 𝐹 = 𝐹𝑔 [ 2𝑖𝑔 2𝑖𝑔 ∴ 𝐹𝑔 = [ ]𝐹 2𝑖𝑔 + 3𝑖𝑓 Similarly, 3𝑖𝑓 ]𝐹 𝐹𝑓 = [ 2𝑖𝑔 + 3𝑖𝑓 Substituting Eqn. (11.29(c)) in Eqn. (11.28(b)) 𝐹 = 𝐹𝑔 [1 +

(11.29(𝑏)) (11.29(𝑐))

6𝐹𝑓 𝑙 𝑖𝑓 𝑏 ′ ℎ2 6𝑙 3𝑖𝑓 𝜎𝑓 = ′ 2 [ ]𝐹 𝑖𝑓 𝑏 ℎ 2𝑖𝑔 + 3𝑖𝑓 18𝐹𝑙 𝜎𝑓 = ′ 2 𝑏 ℎ (2𝑖𝑔 + 3𝑖𝑓 ) Similarly, substituting Eqn. (11.29(b)) in Eqn. (11.28(b)) 𝜎𝑓 =

(11.30(𝑎))

6𝐹𝑔 𝑙 𝑖𝑔 𝑏 ′ ℎ2 2𝑖𝑔 6𝑙 [ ]𝐹 𝜎𝑔 = ′ 2 𝑖𝑔 𝑏 ℎ 2𝑖𝑔 + 3𝑖𝑓 12𝐹𝑙 𝜎𝑔 = ′ 2 𝑏 ℎ (2𝑖𝑔 + 3𝑖𝑓 ) 𝜎𝑔 =

(11.30(𝑏))

Dividing Eqn. (11.30(c)) by Eqn. (11.30(c)),

9

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

𝜎𝑓

Unit – I: Springs

3 𝑂𝑅 =2 𝜎𝑔 3 (11.29(𝑎)) 𝜎𝑓 = 𝜎𝑔 2 It is seen that the stress developed in full length leaves is 50% greater than that developed in graduated leaves. Substituting Eqn. (11.29(b)) in Eqn. (11.28(c)) 𝑦𝑔 = 𝑦𝑔 = 𝑦𝑔 =

6𝐹𝑔 𝑙 3 𝐸𝑖𝑔 𝑏 ′ ℎ3

2𝑖𝑔 6𝑙 3 [ ]𝐹 ′ 3 𝐸𝑖𝑔 𝑏 ℎ 2𝑖𝑔 + 3𝑖𝑓 12𝐹𝑙3

(𝑎)

𝐸𝑏 ′ ℎ3 (2𝑖𝑔 + 3𝑖𝑓 )

Substituting Eqn. (11.29(c)) in Eqn. (11.28(c)) 𝑦𝑓 = 𝑦𝑓 = 𝑦𝑓 =

4𝐹𝑓 𝑙 3 𝐸𝑖𝑓 𝑏 ′ ℎ3

4𝑙 3 3𝑖𝑓 [ ]𝐹 ′ 3 𝐸𝑖𝑓 𝑏 ℎ 2𝑖𝑔 + 3𝑖𝑓 12𝐹𝑙3

(𝑏)

𝐸𝑏 ′ ℎ3 (2𝑖𝑔 + 3𝑖𝑓 )

Comparing expression (a) and (b), we found that, deflection in graduated leaves and full length leaves are the same. ∴ 𝑦𝑔 = 𝑦𝑓 = 𝑦 =

12𝐹𝑙3

(11 .30(𝑐))

𝐸𝑏 ′ ℎ3 (2𝑖𝑔 + 3𝑖𝑓 )

Equalized stress in springs - Nipping: (Nipping of Leaf Springs) In the previous section it is seen that the stress developed in full length leaves is 50% greater than the stresses developed in graduated leaves. One of the methods of equalising the stresses in different leaves

10

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

is to pre-stress the spring. The pre-stressing is achieved by bending the leaves to different radii of curvature, before they are assembled with the centre clip as shown in Figure 25. The full-length leaf is given a greater radius of curvature than the adjacent leaf. The radius of curvature decreases with shorter leaves. The initial gap c between the extra full-length leaf and the graduated-length leaf before the assembly is called a ‘nip’. Such pre-stressing, achieved by a difference in radii of curvature, is known as ‘nipping’. Nipping is common in automobile suspension springs.

Fig. 25 Nipping of Leaf Spring When the central bolt is tightened, the upper leaf will bend back and have an initial stress in a direction opposite to that of the normal load, while the lower leaf will have an initial stress in the same direction as that of the normal load. Thus on loading, the upper leaf will have lower value of resultant stress than that of lower leaf. The initial gap between the leaves may be adjusted so that under maximum load condition the stress in all the leave is equal. This is desirable in automobile springs in which full length leaves are designed for lower stress because the full length leave carry additional loads caused by the swaying of the vehicle, twisting and in some cases due to driving the vehicle through the rear springs. For equal stresses between the graduated and full length leaves at maximum load, the total deflection of the graduated leaves will exceed the deflection of the full length leaves by an amount equal to the initial gap 'c'. 6 IS 1135–1966: Specifications for leaf-springs for automobile suspension. Following are equations used to design laminated semi-elliptic springs: (Ref: Figure 11.12(b) DDHB) The stress in the full length leaves with no pre-stress 9𝐹𝑙 𝜎𝑓 = ′ 2 (11.31(𝑎)) 𝑏 ℎ (2𝑖𝑔 + 3𝑖𝑓 )

The stress in the graduated leaves with no pre-stress 6𝐹𝑙

𝜎𝑔 =

(11.31(𝑏))

𝑏 ′ ℎ2 (2𝑖𝑔 + 3𝑖𝑓 ) The deflection of the spring

11

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

𝑦=

6𝐹𝑙3 𝑏 ′ ℎ3 𝐸(2𝑖𝑔 + 3𝑖𝑓 ) The initial gap between the full length and graduated leave with pre-stress,

(11.31(𝑐))

𝑐=

𝐹𝑙3 𝑖𝑏 ′ ℎ3 𝐸 The load on the clip bolts to close the initial gap,

(11.32(𝑎))

𝐹𝑏 =

𝑖𝑔 𝑖𝑓 𝐹 𝑖(2𝑖𝑔 + 3𝑖𝑓 ) The maximum stress in the spring with the full-length leaf pre-stressed,

(11.32(𝑏))

3𝐹𝑙 (11.32(𝑐)) 𝑖𝑏 ′ ℎ2 Multi-leaf springs are designed using load stress and load-deflection equations. The standard dimensions

𝜎𝑓 = 𝜎𝑔 =

for the width and thickness of the leaf section are as follows: Nominal thickness (mm): 3.2, 4.5, 5, 6, 6.5, 7, 7.5, 8, 9, 10, 11, 12, 14, and 16 Nominal width (mm): 32, 40, 45, 50, 55, 60, 65, 70, 75, 80, 90, 100 and 125 The leaves are usually made of steels, 55Si2Mn90, 50Cr1 or 50CrlV23. The plates are hardened and tempered. The factor of safety based on the yield strength is from 2 to 2.5 for the automobile suspension.

Problems on Multi-leaf Spring Example 18 A six leaf cantilever spring of 420 mm long to carry a load of 3 kN and having a deflection of 42 mm. The allowable stress in the spring material is 320 MPa and the Young’s modulus is 200 GPa. Determine the width and thickness of leaves.

12

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Example 19 Design a cantilever spring to absorb 600 N-m energy without exceeding the deflection of 150 mm and an allowable stress of 750 MPa. The length of the spring is 600 mm and the Young’s modulus is 200 GPa.

13

Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

Unit – I: Springs

Example 20 A 1 m long cantilever spring is composed of 8 graduated leaves and one extra full length leaves. The leaves are 45 mm wide. A load of 2,000 N at the end spring causes a deflection of 75 mm. Determine the thickness of the leaves and the maximum bending stress in the full length leaf assuming first that the extra full length has been pre-stressed to give the same stress in all the leaves and then determine the stress in the full extra length leaf assuming no pre-stress. 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:

𝑙 = 1,000 mm, 𝑎). ℎ =? ,

𝑖𝑔 = 8,

𝑖𝑓 = 1,

𝑏). 𝜎𝑓 =? −Pre − stressed

𝑏 ′ = 45 𝑚𝑚,

𝐹 = 2,000 𝑁,

𝑐). 𝜎𝑓 =? −No Pre − stress

𝑦 = 75 𝑚𝑚.

Refer Figure 11.12 (a),

a. Thickness: Using Eqn. 11.30(c), y=

14

12𝐹𝑙 3 𝑏 ′ ℎ3 𝐸(2𝑖𝑔 + 3𝑖𝑓 ) Department of Mechanical Engineering, PES University, Bangalore, Chandrashekara

Dr. C V

UE17ME 351Design of Machine Elements – II (Jan-May 2020 - 6th Sem)

12 × 2,000 × 1,0003 75 = 45 × ℎ3 × 2 × 105 (2 × 8 + 3 × 1) ; Using Table 11.10(a), we have

Unit – I: Springs

∴ ℎ = 12.32 𝑚𝑚

Thus, the standard thickness h = 14 mm

b. Stress in full length leaf − Pre-stressed: We know that, 𝜎𝑔 = 𝜎𝑓 = =

3𝐹𝑙 𝑖𝑏 ′ ℎ2

3 × 2000 × 1000 (𝑖 = 𝑖𝑓 + 𝑖𝑔 ) 9 × 45 × 142

𝜎𝑓 = 75.59 𝑀𝑃𝑎

c. Stress in full length leaf – No pre-stress: We know that, 𝜎𝑓 = =

9𝐹𝑙 + 2𝑖𝑔 )

𝑏 ′ ℎ2 (3𝑖𝑓

9 × 2000 × 1000 (𝑖 = 𝑖𝑓 + 𝑖𝑔 ) 45 × 142 [3 × 1 + 2 × 8]

𝜎𝑓 = 107.41 𝑀𝑃𝑎

Example 21 Determine the thickness and deflection of a cantilever leaf spring for the following...