Stress distribution in interference joints

Pergamon

0045-7949(93)EOO42-M

STRESS DISTRIBUTION N. SIVA F%A~AD,~

P.

Computers & Srucwes Vol. 51, No. 5, pp. 53S540, 1994 Copyright 0 1994 Elsetier Science Ltd

Printed in Great Britain. All rights reserved 0045-x349/94 s7.00 + 0.00

IN INTERFERENCE

SASHIKANTH~

and V.

JOINTS

RAMAMURTIS

Departments of tMechanica1 Engineering and SApplied Mechanics, Indian Institute of Technology, Madras-600 036, India (Received 1 October 1992)

Abstract-The use of the finite element method in predicting the stress distribution in an interference joint is illustrated and the results of a study of the influence of various parameters on these stresses is reported.

INTRODUCTION

Interference

joints

The vectors {u,,} and {us} are related as

assemblies are engineering. The most

and shrink-fitted

widely used in mechanical

common examples are axle-bearing and flywheelcrankshaft assemblies. In the design of these assem-

blies, dimensions are normally arrived at by empirical relations. Hence, the study of the stress distribution in these assemblies becomes important for better designs. It would be very useful to the designer if it could be possible to estimate the deformations and stresses in realistic assembly conditions. In shrinkfitted joints, experimental methods have been limited to photo-elastic analysis for the estimation of stresses. This is applicable in only a limited range of materials such as Araldite. It may not be possible to exactly correlate this analysis for materials such as steel, which are often used in practice. The empirical relations available in the literature have the limitation of being applicable only to a few standard shapes such as a cylinder-on-cylinder. The most well known of these is Lam& solution applied to a shrink-fitted jacket on a thick cylinder [l]. The finite element approach is a powerful tool in such situations which has the advantage that it can be applied to any arbitrary shape and to any material or combinations of materials.

MATHEMATICAL

It is assumed that there is sufficient friction to prevent slip at the contact nodes. Hence, the elements of {1(0}corresponding to the normal displacements equal the prescribed machined interference, and those corresponding to displacement in the plane of the contact surface equals zero. For the hub, listing displacements on contact surface last, we have

(2) where [K,,] is the stiffness matrix of the hub, {Fhi} and {Fhc} are any external forces or body forces on the hub, and {P} is a vector of nodal forces acting along the contact plane by virtue of the interference. A similar equation can be written for the shaft, listing contact displacements first

(3) Now, eqn (2) can be rewritten, using eqn (l), as

FORMULATION

For convenience, in the following description the outer member is referred to as the hub and the inner member as the shaft. Daniel [2] has formulated a method for modelling a shrink fit. In the present paper a modified form of this method has been used to carry out the parametric study on interference joints. The formulations are briefly described below. Let the displacements of the shaft be written as a vector of displacements on the contact surface, {us), and a vector of remaining displacements, {uSi}. Similarly, hub displacements consists of {uh} and {Q}.

or

~~~l{~}={~~}+{~}-~~~l{~}.

(4)

Equations (3) and (4) can be assembled into one system as =F+F,,

535

(5)

N. SIvA PRAsnD

536

et al.

Table I d (mm)

Did

Lid

IO0

2.0 1.5

0.5,0‘75, 1.0 0.5,0.75, I .o

150

2.0 1.333

0.5,0.75, 1.0 0.5,0.75, 1.0

ANALYSIS

The stiffness matrix was formed using the method of isoparametric formulation for axisymmetric bodies [3]. A computer program was written incorporating the formulation given in eqns (1x5) and tested using previously known results. Then hubshaft systems with different combinations D/d, L/d, and interferences were considered (see Fig. 1 for nomenclature). The study was carried out for shaft diameters of 100 and 150 mm. The shaft was assumed to be hollow with an inner diameter of 20 mm. Analysis was carried out for the cases listed in Table 1. The interference values taken were 52, 86 and 139pm corresponding to H6p5, H&s5 and H6u5, respectively. Both hub and shaft are considered to be steel members with Young’s modulus of 2.1 x lo5 MPa. The discretization for one of the cases is shown in Fig. 2. It was similar for all other cases. The system was assumed to be only under the influence of interference forces since it was found that they were very high compared to normal magnitudes of external loads or body forces. However, various types of external loads or body forces, in the form of pressure forces or concentrated nodal loads, can be incorporated either before or after assembly of the hub and shaft global stiffness matrices. The boundary condition that was specified was zero axial displacement at the two ends of the shaft.

Fig. 1. Nomenclature of hubshaft system.

where [ZC]is the stiffness matrix assembled from [iu,] and [KS], I; is a vector of external or body forces assembled from Fhi, Fhhc,Fsi and F,, and F, is the vector of contact forces which represents the interference fit. Equation (5) can be solved by any of the known solution techniques to obtain the vector of displacements

Equation (1) can be used to obtain {ah}. Also contact forces P and stresses can be obtained using element stiffness matrices.

I

1,

L--

2

Ll

16

4

d = 1OOmm Ail dimensions

Fig. 2. Discretization

ore in -mm

of hub and

shaft.

Stress distribution in interference joints

-Lld.l;

.\4

Did=2

4

c

.

2 Node

nos.(rrf

L Fig.21

6

Fig. 3. Distribution of radial displacement along interference plane. RESULTS AND DISCUSSION

Figure 3 shows the variation of different parameters; that is L/d, D/d and interference. Figures 4-6 show the variation of radial, axial and circumfer-

537

ential stresses with different interferences, L/d and D/d, respectively. Figure 7 is a plot of radial, axial and circumferential stresses along a radial line passing through the geometric centre of the hub-shaft system. As can be seen from Fig. 7, the radial stress attains the maximum value in the region of the interference plane. The rate of decay of radial stress is fairly high. Hence, the value of the inner diameter of the shaft is of no consequence as long as the ratio of shaft outer diameter to shaft inner diameter is more than 2.5 to 3. The shaft inner diameter in the cases studied was 20 mm. Another prominent feature of Fig. 7 is the reversal of sign of axial and circumferential stresses across the interference plane. The results of the study on shafts with d = 100 and 150 mm are presented in Tables 2 and 3, respectively. Table 4 shows the percentage change in displacement, stress and torque capacity for a known variation in each parameter. It is observed from Table 4 that percentage change in maximum radial displacement, when L/d is varied at a given shaft, is comparatively small (8-12%). The order of variation as seen from Tables 2 and 3 is about 0.001 mm. The change in stress is almost insignificant, i.e. (about 1% for 50% increase in L/d). But it should be noted that the value of the stress is still significantly lower than that computed from Lame’s formula. Lame’s solution assumes that both hub and shaft are of same length. This suggests that radial stress would increase significantly as the hub length approaches the shaft length. Since in normal practice the L/d ratio does not exceed 1.0, it would be an advantage to have interference joints with lower ratio of hub to shaft length.

RADIAL

--

-

--

AXIAL

STRESS

STRESS

CIRCUM-STRESS

0;

O-= 0;

Fig. 4. Variation of stresses with interference (L/d = 1.O, D/d = 2.0, d = 100 mm).

N. SIVAhASAD et al.

538

I

0

P

I 1

I

0.2

1

0.4

I

0.6

0.6

1.

zi -lO-

Fig. 5. Variation of stresses with L/d (D/d = 2, interference = 139 pm, d = 100 mm).

When there is a decrease of about 25% in D/d, radial stresses decrease by 14-16% while axial stresses increase by 10-l 1% and circumferential stresses increase by 18%. This suggests that it would be favourable to decrease D/d to decrease radial

c

40

stresses, as long as the axial and circumferential stresses do not exceed the tensile strength of the material. As expected, the stresses were found to be directly proportional to the interference.

2.0

t 30-

20-

: E

lo-

2505 1.5

=

=

=-

-

---

-

-

----=:

0 "0 : z

I

I

I

I

0.2

0.4

0.6

0.6

1.1

ci -107

-20 --

-30

--e---

RADIAL -

AXIAL CIRCUM

STRESS STRESS STRESS

cr cz q

1 -401

Fig. 6. Variation of stresses with D/d (L/d = 1.0, interference = 139 pm, d = 100 mm).

-30

539

in interference joints

Stress distribution

-___

AXIAL

-+

CIRCUM.STRESS

STRESS

rz re

t

Fig.

7. Variation

of stresses

along

central

radial

line (D/d = 1.5, L/d = 0.75, interference

= 139 pm,

d = 1oOmm).

CONCLUSIONS

To minimize stresses in the interference plane, the ratio of hub outer diameter to inner diameter should not be too low. The hub length has a significant influence on the stresses in the joint and therefore the designer should take advantage of the decrease in

Table 2. Results

SlabNo.

L/d

of a parametric

D/d

Interference brn)

with d = 100

study

mm

stresses occurring when L/d is increased. It is essential to arrive at an optimal L/D ratio to have minimum stresses in the contact plane. The finite element method has been applied here to a cylinder-on-cylinder case taking advantage of the axisymmetric nature of the system. But the methodology presented above can be extended to other

Table 3. Results

of a parametric

Stress (MPa) Radial

Axial

Circum.

-46.07 -76.19 - 123.14

34.54 57.13 92.33

146.03 241.50 390.34

Slab No. L/d 1

0.5

Interference @m)

study with d = 150 mm Stress (MPa) Radial

Axial

Circum.

2.0

52 86 139

- 32.24 - 53.32 -86.18

21.94 36.28 58.64

97.00 160.42 259.28

D/d

1

0.5

2.0

52 86 139

2

0.5

1.5

52 86 139

-40.89 -67.63 - 109.31

37.33 61.73 99.76

168.52 278.71 450.48

2

0.5

1.5

52 86 139

- 28.19 -46.62 -75.36

24.06 39.79 64.32

112.30 185.73 300.19

3

0.75

2.0

52 86 139

-49.63 -82.09 - 132.67

35.59 58.85 95.12

140.13 231.76 374.58

3

0.75

2.0

52 86 139

- 34.00 - 56.24 - 90.90

23.71 39.21 63.37

94.45 156.21 252.48

4

0.75

1.5

52 86 139

-46.22 -76.44 - 123.55

35.91 59.38 95.98

161.36 266.87 431.34

4

0.75

1.5

52 86 139

-31.42 -51.96 -83.99

23.98 39.66 64.10

108.61 179.63 290.33

5

1.0

2.0

52 86 139

-53.33 - 88.20 - 142.55

35.78 59.18 95.66

137.13 226.79 366.57

5

1.0

2.0

52 86 139

- 36.67 - 60.64 -98.02

23.52 38.90 62.87

92.22 152.53 246.53

6

1.0

1.5

52 86 139

-49.10 -81.20 -131.25

34.54 57.12 92.33

158.76 262.56 424.37

6

1.0

1.5

52 86 139

- 33.31 - 55.09 -89.04

23.05 38.12 61.62

106.92 176.84 285.82

N. SIVAPRAsADef al.

540

Table 4. Percentage changes in displacements, stresses and torque capacity with changing parameters: d, = 100 mm and d2= 150mm Average percentage change in Varying Darameter Interference Lid Dld

Percentage change 65.38 50.0 -25.0

Max. disp. d, d, 65.5 65.4 8.8 8.2 -23.0

-23.6

S,, d, 65.38

1.29 - 13.92

regular and irregular shapes with appropriate modifications to the method of stiffness matrix formulation. REFERENCES

1. F. B. Seely and .I. 0. Smith, Advanced Mechanics of Materials, 2nd Edn, pp. 321-327. John Wiley, London (1963).

&; dz 65.32 0.88 -15.88

d, 65.40 5.46 10.22

& d2 65.41 9.92 11.11

d, 65.38 -2.31 17.92

dr 65.38 - 1.10 17.93

Torque capacity d, d, 65.5 65.4 -49.0 -49.0 -27.0 -27.0

2. W. J. T. Daniel, Flywheel design by the finite element method. Mech. Engng Tram, Inst. Engrs, Australia ME7, 75-79 (1982).

3. L. J. Segerlind, Applied Finite Element Analysis, 2nd Edn. John Wiley, New York (1984). Standards Recommendations for limits 4. Indian for engineering, IS 919: 1963 and tolerances (1963).