ME308_Tutorial5_2011

8/6/2019 ME308_Tutorial5_2011

1/9

8/6/2019 ME308_Tutorial5_2011

2/9

METU ME 308 Machine Elements II Spring 2011 2

o Determine the viscosity, From Figure 12.13 in 8 Ed. or from the

Notes to be used in the Examinations and the density of oil.o Choose a23 close the upper boundary of the normal operating

condition region using Fig.4 of STEYR catalogue

i) Minimum possible P,

P=Fr=3.756 kN (X=1, Y=0), assuminga

r

Fe

F

ii) Most accurate maximum value of a23

Take outer diameter , D=90mm

d = 50 mm D = 90 mm2

md Dd md = 70 mm

From Fig.2 of STEYR catalogue with nm =2310 rpm & dm =70 mm, the required

minimum viscosity of lubricant is 1= 12 mm2/s.

From Figure 12.13 in 8 Ed. or from the Notes to be used in the Examinations :

with SAE 30 and T=40

o

C, = 78 mPa.s. With = 860 kg/m3

for oil,36 27.8 10 10 90.7 /

860mm s

Using Fig.4 of STEYR catalogue, with /1= 7.55 and normal operating conditions,since we are sure about normal operating conditions a23can be taken very close to itsmaximum value 3,9.

Assuming a23 = 3.8

Now, Cso can be determined by using life equation1

3

6

.60.

3.9 10

h mISO

L nC P

33.44

ISOC kN

With CISO> 33.44 kN, a bearing with bore diameter of 50 mm is to be selected andchecked whether it actually satisfies the life equation. Lets select light series bearing

63 10 (on page 153 of STEYR catalogue) since CISO = 53.6 > 33.44 kN.

d = 50 mm D = 110 mm2

m

d Dd

md = 80 mm

COISO= 41.6 kN

0.0148a

OISO

F

C e = 0.1917 (by interpolation table 2.2/2 on page 37)

0.1648a

r

Fe

F then X=1 & Y=0 that P = Fr , P =3.756 kN

Now, a23 assumption should be checked for specified bearing.

From Fig.2 of STEYR catalogue with nm =2310 rpm & dm =80 mm, the required

minimum viscosity of lubricant is 1= 11 mm2/s.

For/1= 8.24 : a23=3.8 is valid. Therefore, assumption is correct and no need to toupdate a23 value.

Since, P and a23 do not updated, CISO will not change

1

3

6

.60.

3.9 10

h mISO

L n

C P

33.44ISOC kN

So selected bearing is suitable, since CISO= 53.6 kN > 33.44 kN

Notes:

- Ifa

r

Fe

F condition is not satisfied, minimum P should be updated for

a

r

FF

>e (P=0.56*Fr+1*Fa)

- If /1 for the specified bearing is not be satisfied with the assumeda23=3.8, all the steps should be revisited for new a23.

8/6/2019 ME308_Tutorial5_2011

3/9

8/6/2019 ME308_Tutorial5_2011

4/9

8/6/2019 ME308_Tutorial5_2011

5/9

8/6/2019 ME308_Tutorial5_2011

6/9


Da= 130 mm (outer diameter of housing, given)

In tolerance specification, the aim is to specify tolerances such that there will be nointerference between raceway and rotating elements creating pre-stresses which will

reduce the life of bearing considerably. Hence, the expansion of inner ring andcontraction of outer ring will be calculated and the sum should not exceed theminimum clearance of the unmounted bearing so that after mounting no interferenceexists.

Expansion of the inner ring,For d = 50 mm and precision ground shaft Gi=6 m (page 66).Effective interference is evaluated using deff= dth-Gi

From Fig. 3.2/2 (page 68), with 1 50 0.7765a

dCd

& 2 0idCd

gives, 0.77i

eff

L

d

So, 0.77 ( 6)i thL d Contraction of the outer ring,

For D = 110 mm and precision turned housing surface Go=10 m.Effective interference is evaluated using Deff= Dth- Go

From Fig. 3.2/3 (page 69) with,3

970.75

130a

DC

D &

4

1100.85

130a

DC

D gives, 0.35a

eff

L

D

So, 0.35 ( 10) A thL D

Total reduction in clearance is the sum.

i A R L L 0.77 ( 6) 0.35 ( 10)th th R d D

Standard clearance ranges are given for cylindrical roller bearings in Table 3.2/2 onpage 62. for d = 50 mm with paired rings, Rmin= 30 m and Rmax =45 m. So

R < 30 should be satisfied.

Looking at recommended tolerance fields for solid shafts (Table 4.1/1 on page 76):

10.9

0.073150

P

C P / C < 0.08 with d = 50 mm,j6 is recommended for shaft.

For housing, J7 is recommended; point load on outer ring, outer ring not displaceableand split housing (page 78).

Table 4.1/2 (page 77): d = 50 mm (j6) dmin= -5 m dmax= 23 m

Table 4.1/5 (page 79): D = 110 mm (J7) Dmin= -37 m Dmax= 13 m

Checking

0.77 ( 6) 0.35 ( 6)th th R d D 30 0.77 (23 6) 0.35 (13 10) 14.14 satisfied.

Tolerances are specified as j6/J7. If it wasn't satisfied, one would change tolerancesfor surfaces: j6/J6, j5/J6... calculates R and change tolerances until R < 30 issatisfied.

8/6/2019 ME308_Tutorial5_2011

7/9


PROBLEM 4

The pinion shaft, transmitting a torque T is shown in the figure schematically. Toprovide the overhung pinion with maximum rigidity, two tapered roller bearings arearranged in O- arrangement. The radial, axial and tangential forces acting on thepinion are:

Wr= 2540 N , Wa = 4500 N, and Wt = 12000 N.

The bearing K is 323 08 whereas the bearing L is selected as 302 08. The pitch radiusof the pinion is rp=36mm. The shaft rotates at 1000 rpm. The bearings are lubricatedby SAE 40 oil and the operating temperature is 60

oC.Calculate the rating life of these

tapered rolling bearings at reliability of %.99.

Solution

Lets first determine the the position of the bearing reactions .Examining table 2.2/2

in page 37 of STEYR catalogue for e,X,Y values and page 236 of STEYR cataloguefor CISO, d , D,a and T values:

Designation e X Y CISO d(mm) D(mm) a(mm) T(mm)

323 08(K) 0.34 0.4 1.75 101 40 90 23 35.25302 08(L) 0.38 0.4 1.6 53.6 40 80 16 19.75

dK=22.75 mm+(T-a) mm=22.75+(35.25-23) mm=35 mm for bearing KdL=22.75 mm+86-(T-a) mm=108.75 mm-(19.75-16)=105 mm for bearing LL1= dK=35 mm; L2= dL- dK=70 mm

Let us first make a static analysis to determine the radial forces acting on the bearingK & L.

The free body diagram of the shaft in x-y plane is;

_ _0

y r K y L yF W R R Eq. 1

_ 1 _ 20

K z r L yM M W L R L Eq.2

From Eq.2 1_

2

1044.3

rL yM W L

R NL

and from Eq.1;

_ _1495.7

K y r L y R W R N

The free body diagram of the shaft in x-z plane is;

_ _0 z t K t L t F W R R Eq. 3_ 1 _ 2

0 K y T L Z M W L R L Eq.4

From the Equation 4

1_

2

6000tL zW L

R NL

and from the Equation 3,

_ _18000

K z t L z R W R N

The resultant radial forces acting on the bearings K&L are

2 2

_ _18062 K K y K z R R R N

2 2

_ _ 6090 L L y L z R R R N These equivalent forces must be modified by load factors suggested on page 38.

Average values like fz = 1.2 can be used for gearing applications.

y

x

22.75 mm 86 mm

Wa

WtWr

T

K L

r

O

Pinion gear

z

x

RK_zWt RL_z

L2L1

y

x

RK_yWr RL_y

L2L1M

8/6/2019 ME308_Tutorial5_2011

8/9


7308 L z L R f R N and from the Equation 3,

21674 K z K R f R N

5400 a z aF f F N

As the radial forces are known, now let us designate the bearings.

From the STEYR catalogue, page 42, bearing A is defined as that bearing which hasto absorb the external force. If one looks at the figure it is easily seen that the bearing

located at K should absorb the external axial force, Wa. Therefore bearing K isdesignated with A, and L is designated with B.

Rewrite the axial and radial forces with the new designation.

Note that each external radial force acting on the tapered roller bearing gives rise to aninduced axial force, due to taper angle, which again turns into an external force for the

opposing bearing. The equivalent axial loads acting on the bearings K&L can befound from the formulas in Figure 2.2/8 of STEYR catalogue (Page 42).

Induced axial force due to rAF

216746193

2 2 1.75

rK

K

K

FN

Y

Induced axial force due torBF

73082284

2 2 1.6

rL

L

L

FN

Y

It is seen that K L

Also 5400 3909 a K L

F N

This means that for the selected bearings for K &L, the net axial shaft force is actingon bearing K.

Then the resultant axial forces acting on the bearings K&L are,

0aLF

7684 aK L aF F N

The estimated dynamic equivalent loads at bearings K&L are

22.12

7684( 0.355 0.34)

21674

7.31

K a ra a Ka

Ka

Kr

L

P X F Y F kN

Fe

F

P kN

For bearing K:

p=10/3 for tapered roller bearings.

Life adjustment factor a1 for 99% reliability is 0.21, from Figure 1 of STEYRcatalogue.

Life adjustment factor a23 depends on the viscosity of the oil.

Dynamic viscosity of SAE40 oil at 60oC; 36 . mPa s (from Figure

12.13 of Notes to be used in Examination)

The density of the SAE40 oil:3

860 /kg m The kinematic viscosity is

3 6

22

2

3 9

3

36 10 10 .

41.86 ( )

860 10 10

Ns

mmmmcSt

sN s

mm

mm

From Figure 2 of STEYR catalogue, find the required minimum

viscosity,1

, for n=1000 rpm and mean diameter for the selected bearing

323 08

2

1

40 9065 , 1000 17 /

2 2

m

d Dd mm n rpm v mm s .

The viscosity ratio1

41.86

2.4617 v

8/6/2019 ME308_Tutorial5_2011

9/9


From Figure 4 of STEYR catalogue, for normal operating conditions whichwe are not sure about, choose the lower side of the yellow region, then life

adjustment factor is; a23=1.8

Life equation

6

1 23

10

60

p

ISOh

C L a a

n P

10/3610 1010.21 1.8 995

60 1000 22.12

h

kN L hour

kN

For bearing L:

p=10/3 for tapered roller bearings.

Life adjustment factor a1 for 99% reliability is 0.21, from Figure 1 of STEYRcatalogue.

Life adjustment factor a23 depend on the viscosity of the oil.

Dynamic viscosity of SAE40 oil at 60oC; 36 . mPa s (from Figure12.13 of Notes to be used in Examination)

The density of the SAE40 oil:3860 /kg m

The kinematic viscosity is

3 6

22

2

3 9

3

36 10 10 .

41.86 ( )

860 10 10

Ns

mmmmcSt

sN s

mm

mm

From Figure 2 of STEYR catalogue, find the required minimum

viscosity,1

, for n=600 rpm and mean diameter for the selected bearing 302

08

2

1

40 80

60 , 1000 18 /2 2

md D

d mm n rpm v mm s .

The viscosity ratio

1

41.862.33

18

v

From Figure 4 of STEYR catalogue, for normal operating conditions whichwe are not sure about chose the lower side of the yellow region, then lifeadjustment factor is; a23=1.7

Life equation

6

1 23

10

60

p

ISOh

C L a a

n P

10/3610 53.6

0.21 1.7 456160 1000 7.31

h

kN

L hour kN

Documents

ME308_Tutorial5_2011