Fatigue Analysis of Composite Drive Shaft

Fatigue Analysis of Composite Drive Shaft

Prof. Mahesh. N. Pradhan Hrishikesh .T. Gaikwad

Mechanical engg dept. M.I.T, Kothrud ,Pune- 52. Student Mech. Engg Dept. M.I.T, Kothrud ,Pune- 52.

Pune, India. Pune, India.

Abstract – The projects aims at replacing the conventional drive shaft with composite drive shaft which will provide us the better mechanical properties i.e (Torque transmitting capacity, and fatigue life of the shaft).The research paper will include the comparison of these properties of conventional steel shaft with the composite drive shaft .The drive shaft selected is applicable to TATA 407 pick up vehicle which is at present using the conventional steel drive shaft. Our main aim is to show that the fatigue life of composite drive shaft is much better than conventional steel drive shaft.

Keywords- Composite shaft, conventional shaft, TATA 407 pick up, fatigue analysis, mass saving.

1. INTRODUCTON

   Power transmitted from the engine to the final drive where useful work is applied through a system consists of a gearbox, clutch, universal joint, drive shaft and a differential in the rear-drive automotives. In conventional drive shafts there is problem of instant crack propagation in case of heavy loads unlike the crack arrest property of composite materials. As compare to conventional metallic drive shaft, the composite drive shafts have many parameters to be altered, namely the fiber orientation angles, stacking sequences, layers thicknesses and number of layers. These parameters, due to the tailorability of elastic constants, could provide a large number of possible designs, which must satisfy optimally the performance characteristics of the composite drive shaft (critical speed, fatigue life and load carrying capacity). It is well-known that the steel drive shaft is usually manufactured in two pieces. There are many design studies but the information about the design variables and their effect on the performance characteristics is not comprehensive.

   Generally, all accessed design studies were not including the fatigue consideration, which may be needed to be explored in relation to composite shafts design. Therefore, the aim of this work is to investigate numerically the effect of stacking sequence and fiber orientation angle on the performance of drive shaft. The numerical results Will be validated by results obtained from analytical solutions. The specimens used will be filament wounded. Till now very less work is done on fatigue analysis of shaft made up of composite material.

   1.1 Methodology

   The composite shaft will be manufactured using the process known as filament winding. The process involves winding filaments under tension over a male mandrel. The mandrel rotates while a wind eye on a carriage moves horizontally, laying down fibers in the desired pattern .Filament winding is well suited to automation, and there are many applications, such as pipe and small pressure vessel that are wound and cured without any human intervention.

   Many sample specimens will be tested for fatigue failure and the optimum one would be selected. Detail stress analysis will be performed on ansys software .The matlab software will be used to verify the analytical values and to obtain quick results of the same.

2. DESIGN SPECIFICATION

   a. Conventional drive shaft of TATA 407 pick up.

   The drive shaft outer diameter should not exceed 100 mm due to space limitations. Here outer diameter of the shaft is taken as 75 mm.

   Based on standards available specifications of drive shaft are

   1. The torque transmission capacity of the drive shaft(T) = 2058.75 N-m

   2. Speed of drive shaft = 2800 rpm

   3. Outside diameter of drive shaft = 75 mm

   4. Length of drive shaft = 1.3 m

   The steel drive shaft should satisfy three design specifications such as torque transmission capability, and bending natural frequency. Steel (SM45C) used for automotive drive shaft applications. The material properties of the steel (SM45C) are given in following table.

   Properties of steel (SM45C)
   S.r No.	Mechanical properties	Sym bol	Unit s	Steel
   1	Youngs Modulus	E	GPa	207
   2	Shear modulus	G	GPa	80
   3	Poissons ratio	µ	–	0.3
   4	Density		Kg/ m3	7600
   5	Yield Strength	Sy	MPa	370

   Table 1

   Gear Ratio for Tata 407 pickup

   No.of Gears are 5 forward and 1 reverse. Table.2

   Gear ratios for TATA 407 Pick up
   S.r No.	Gear Position	Gear ratio
   1	1st Gear	6.01:1
   2	2nd Gear	3.46:1
   3	3rd gear	1.97:1
   4	4th Gear	1.37:1
   5	Reverse Gear	5.69:1

   So, the torque is maximum at 1st gear, when the speed of vehicle is low. Therefore the maximum torque will be,

   Ma = 64.5× 650 – 1300× By = 0 By = 32.25N

   similarly

   Ay = 32.25N

   M = A y × 650 M = 32.25× 650

   M = 20962.5N.mm

   Tmax Tmax

   = 225 × 6.01 × 1.5

   = 2058.75 × 103 N – mm

   3

   Fig.2 Bending moment diagram

   M tmax

   = 2058.75 ×10 N – m

   2.1 Torsional Strength:

   The primary load in the drive shaft is torsion. The maximum shear stress, max in the drive shaft is at the outer radius, and is given as

   Fig.3 Shear diagram

   2.3 Fatigue life prediction

   (Mb)Max=20.963 x 103 N-mm.

   = 16Mt =

   do3(1 – C4 )

   16(2058.75 ×103)

   (75)3(1 – C4 )

   (Mb)Min= 0 N-mm Then,

   a. To find Mean & Amplitude Stresses.

   C = 0.9223

   We know,

   C = di / do 0.9223 = di

   0.075

   di = 69.182mm

   Mass of steel drive shaft

   (Mb)m= 1/2(Mb)max + (Mb)min ) (Mb)m= 10.48 X 103 N-mm.

   (Mb) a = 1/2(Mb)max – (Mb)min ) (Mb) a = 10.481 x 103 n-mm

   (Mt) max = 225 x 6.01 x 1.5 = 2058.75 x 103 N-mm

   (given)

   (Mt) min =225 x 1.37 x 1.5 = 4.49 X 106 N-mm (given)

   2 2

   32(Mb )m

   m = × 4 (do – di

   ) × L

   xm d 3

   m = 6.575kg

   xm = 0.253 N/mm2

   2.2 Design of shaft against fatigue loading

   Bending moment in shaft

   We assume that only the force due to self weight of the

   xa

   32(Mb )a

   d 3

   shaft is acting on the shaft .It is acting at the centered the shaft.

   xa = 0.253 N/mm2

   P w g

   (M ) 1 (M )

   (M )

   p 6.575 9.81

   t m 2

   t max

   t min

   p 64.5N

   (M )

   1 2058.75 462.38103

   t m 2

   t m

   (M ) 1260.565103 N mm

   (M ) 1 (M )

   (M )

   t a 2

   t max

   t min

   Fig.1 loading diagram

   (Mt )a

   1 2058.75 462.38103 2

   Acoording to modified goodman method

   t m

   (M ) 798.185103 N mm

   Sa Sm 1

   1 6(Mt )m

   Se Syt

   <>xym

   d 3

   Put above values ,

   xym

   15.2177 N / mm2

   0.632Sa Sm 1

   xya

   1 6(Mt )a

   a

   d 3

   86.88 370

   m

   S 100.125N / mm2

   xya

   9.6358 N / mm2

   S 63.36N / mm2

   2 2

   FOS Sa 63.36

   m ( xm Kb ) 3( xym Kt )

   a 25.03

   2 2

   m (0.253 2) 3(15.2177 1.5)

   6

   m 39.54 x 10 Pa

   log(m ) 7.597Pa

   Fatigue factor of safety = 2.53

   c. To find number of cycles

   S a Sut

   ( K )2 3( K )2

   f S

   a xa b xa t

   ut m

   a

   a

   (0.253 2)2 3(15.2177 1.5)2

   25.0375 x 106 Pa

   S f

   25.03 630

   630 39.54

   log(a ) 7.39Pa

   S f 26.706Mpa

   Since ,

   0.9 Sut =0.9(630) = 567 Mpa

   a eq 39.54N / mm

   2

   log10(0.9 Sut ) = 2.5670

   So, von mises stresses are 39.54 N / mm2

   tan

   a & tan

   sa s

   log10( Se ) = 1.6474 log10( S f ) = 1.3090

   m m

   tan 0.633

   32.33 33

   Log10

   (N ) (6 3)(2.75358 1.4266)

   (2.75358 1.9389)

   b. To find endurance strength Se

   Se Kload Ksize Ksurf Ktemp

   Log10 (N ) 4.889 3 7.889

   N = 77.446 x 106 cycles

3. DESIGN OF COMPOSITE DRIVE SHAFT

   K K K S '

   reliability d stressconcentration e

   1. Selection of Material

      Kload 1

      Ksize 0.75

      Carbon Fiber(Panex 35):

      Ksurf A.Sut

      0.265 0.82

      Ktemp 1

      Panex® 35 continuous carbon fiber is manufactured from polyacrylonitrile (PAN) precursor. The consistency in yield and mechanical properties that are

      for T 450

      Kreliability 90% 0.897

      K 1 0.5

      provided by large filament count strands gives the user the ability to design and manufacture composite materials with

      2

      d greater confidence and allows for efficient and fast buildup of carbon fiber reinforced composite structures.Material

      Se 1 0.751 0.82 0.897 0.5 315

      e

      S 86.88N / mm2

      properties of Carbo fiber(panex 35) is as follows.

      Table .3

      Properties of panex 35
      Sr.no	Parameter	SI	US
      1	Tensile Strength	4137 MPa	600 ksi
      2	Tensile Modulus	242 GPa	35 msi
      3	Electrical conductivity	0.00155 ohm- cm	0.00061 ohm-in
      4	Density	1.81 g/cc	0.065 lb/in3
      5	Fiber Diameter	7.2 microns	0.283 mils
      6	Carbon content	95%	95%
      7	Yield	270 m/kg	400 ft/lb
      8	Spool weight	5.5 kg, 11 kg	12 lb, 24 lb
      9	Spool Length	1,500 m , 3,000 m	1,640 yd, 3,280 yd

   2. Epoxy resin

Epoxy resins are polyether resins containing more than one epoxy group capable of being converted into the thermoset form.

Table . 5

MECHANICAL PROPERTIES EPOXY RESIN (HEXCEL HEXPLY EH04 EPOXY RESIN)

Table.4

1. Mass Saving

   1. Mass of steel drive shaft = 6.575 kg

   2. Mass of Composite drive shaft

      MECHANICAL PROPERTIES EPOXY RESIN (HEXCEL HEXPLY EH04 EPOXY RESIN)
      Sr. No	Mechanical Property	Value	Units
      1	Tensile Strength	81	MPa
      2	Tensile Modulus	3.45	GPa
      3	Specific Gravity	1.34	g/cc
      4	Poissons ratio	0.3	–
      5	Shear Strength	1.34	MPa
      6	Shear Modulus	1.3269	GPa

      m (r 2 r 2 ) L

      0 i

      m (0.03752 0.03602 )(1.3)(1622) m 0.7303Kg

   3. Percentage of mass saving over steel is

      6.575 0.7303

      Mechanical Analysis of Lamina

      6.575

      88.89%

      Bending moment in shaft

      100

      1. Volume fraction of fiber 0.7 (70%)

      2. Volume fraction of matrix 0.3 (30%)

      3. Volume of composites 1 (100%)

      we know = 1810 kg/m3, =1340 kg/m3

      We assume that only the force due to self weight of the shaft is acting on the shaft .It is acting at the centered the shaft.

      Force P is given as ,

      f m

      c f Vf mVm

      P w g

      p 1.772 9.81

      c 1810 0.6 1340 0.4

      p 17.384N

      c

      1622kg / m3

      Fig.4 Loading diagram

      M a 17.384 650 1300 By 0

      By 8.69N

      similarly, Ay 8.69N

      Maximum bending moment is given as ,

      (M )

      1 (M )

      (M )

      t a 2

      t max

      t min

      (Mt )a

      (Mt )a

      1 2058.75 462.38103 2

      798.185103 N mm

      Fig.5 Bending Moment diagram

      xym

      16(Mt )m

      d 3

      161260.565103

      (75)3

      xym 15.2177

      N

      mm2

      Fig.6 Shear dagram Maximum bending moment is given as ,

      xya

      16(Mt )a

      d 3

      16 798.185103

      (75)3

      N

      M = A y × 650 M = 8.69 × 650

      M = 5648.5N.mm M = 5.6483N.m

      xya = 9.6358

      mm2

      ( K )2 3( K )2

      m xm b xym t

2. Fatigue life prediction (Mb)Max=5.648 x 103 N-mm. (Mb)Min= 0 N-mm

   Then

   6.1 To find Mean & Amplitude Stresses.

   m

   m

   (0.02704 2)2 3(15.2177 1.5)2

   39.53 x 106 Pa

   Log( m )=7.596 Pa

   ( K )2 3( K )2

   (Mb)m= 1/2(Mb)max + (Mb)min )

   a xa b xa t

   (Mb)m= 2.834X 103 N-mm.

   m

   (0.253 2)2 3(15.21771.5)2

   (Mb) a = 1/2(Mb)max – (Mb)min )

   (Mb) a = 2.824 x 103 n-mm

   a = 25.032 x 106 Pa

   Log( m )=7.4 Pa

   (Mt) max = 225 x 6.01 x 1.5 = 2058.75 x 103 N-mm

   Since ,

   a eq

   39.54N / mm2

   (given)

   (Mt) min =225 x 1.37 x 1.5 = 4.49 X 106 N-mm (given)

   32(M )m

   So, von mises stresses are 39.54 N / mm2

   tan a & tan sa

   xm =

   b

   d3

   m sm

   xm = 0.068224 N / mm2

   = 32(Mb )m

   tan 0.633

   32.33 33

   1. To fing endurance strength Se

      xa d3

      S ' 0.5 S

      xa = 0.068224 N / mm2

      e

      e

      S ' 0.5

      e

      1169.58

      e

      S ' 584.5Mpa

      (M ) 1 (M )

      (M )

      S K K K K

      t m 2

      t max

      t min

      e load size surf temp

      '

      (Mt )m

      (Mt )m

      1 2058.75 462.38103 2

      1260.565103 N mm

      Kreliability Kd Kstressconcentration S e

      K 1 K 0.75

      load

      Ksurf

      T 450

      size

      ut temp

      A.S 0.265 0.82 K 1for

      K 90% 0.897 K 1 0.5

      reliability d 2

      Se 1 0.751 0.82 0.897 0.5 584.5

3. RESULTS AND DISCUSSION

   COMPARISON BETWEEN STEEL AND COMPOSITE DRIVE SHAFT.
   Sr.No	Parameter	Steel shaft	Composite shaft
   1	Applied Torque (T)	2058.75 N-m	2058.75 N-m
   2	Fatigue factor of safety	2.53	5.28
   3	Number of cycles (N)	77.446 x106	834.02 x 106
   4	Mass (m)	6.575 Kg	0.7303 kg
   5	Percentage of mass saving	–	88.89%

   Table .6

   Se 161.46

   N

   mm2

   Acoording to modified goodman method

   Sa Sm 1

   Se S yt

   Put above values ,

   0.624Sm Sm 1

   174.46 1169

   Sm 208.91

   S 132.29

   N

   mm2

   N

4. CONCLUSION

1. From the above results we Come to the conclusion that the fatigue factor of safety of composite drive shaft is much

   a mm2

   S 132.14

   higher than steel drive shaft.

2. The number of cycles sustained by the composite drive shaft is considerably high .

   FOS a

   a

   25.03

3. We have also achieved a high percentage of mass saving.

Fatigue factor of safety = 5.28

1. To find number of cycles

S a Sut

REFERENCES

• M.A. Badie, E. Mahdi , A.M.S. Hamouda. An investigation into hybrid carbon/glass fiber reinforced epoxy composite automotive drive shaft. Materials and Design 32 (2011) 14851500 Contents lists.

  f

  Sut m

  S 25.032 1169

  f 1169 39.53

  S f 25.92Mpa

  R. A. Gujar, S. V. Bhaskar, Shaft Design under Fatigue Loading By Using Modified Goodman Method, (IJERA), Vol. 3, Issue 4, Jul-Aug 2013, pp.1061-1066.

  • M. A. Badie, A. Mahdi. AUTOMOTIVE COMPOSITEDRIVESHAFTS:INVESTIGATION OF THE DESIGN VARIABLES EFFECTS. International Journal of Engineering and Technology, Vol. 3, No.2, 2006, pp. 227-237 227

    0.9 Sut

    =0.9(1265) = 1138.5Mpa

  • RITA ROY, B K SARKAR, A K RANA and N R BOSE. Impact fatigue behaviour of carbon fibre-reinforced vinylester resin

    log10(0.9 Sut ) = 3.0263 log10( Se ) = 2.24 log10( S f ) = 1.413

    (6 3)(3.0263 1.413)

    composites. Bull. Mater. Sci., Vol. 24, No. 1, February 2001, pp. 7986. © Indian Academy of Sciences.

  • Ali Movaghghar*. Theoretical and Experimental Study of Fatigue Strength of Plain Woven Glass/Epoxy Composite. Strojniki vestnik – Journal of Mechanical Engineering 58(2012)3, 175-182

    • Books

    1. Bryan Harris, Fatigue in composites Science and technology of the fatigue response of fibre-reinforced plastics.Pg no. 224-238.

       Log10 (N )

       (3.0563 2.207)

    2. P.K.Mallick,Fiber-reinforced composite,Material ,manufacturing

    Log10 (N ) 6.043 3 9.043

    N = 834 x 106 cycles

    and design. Pg no. 269 -288.

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  • http://www.maintenancetechnology.com/2012/07/failure-analysis-of- machine-shafts/