me307-11 tutorial 9

METU – ME 307 Machine Elements I – Fall 2011

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Prepared by: Hüseyin Enes Salman Room: G-156 Phone: 5246 E-mail: [email protected] Date: 02/12/2011

ME 307 – MACHINE ELEMENTS I TUTORIAL 9

“DESIGN OF SHAFTS” Problem 1: The shaft in the figure is subjected to the loadings as shown. The flywheel is mounted on the shaft by means of the key and the snap ring (retaining ring) which is set on the groove to secure the flywheel on the + x direction. The material of the shaft is annealed AISI 1050 HR. The surface of the shaft is machined. Determine the diameter of the shaft for:

a.) a factor of safety of 2.5, if the loads are static. b.) a factor of safety of 2, if the loads are completely reversed with the

given magnitudes. Take 0.85bk and the reliability is 95 percent.

(Take: r/d=0.10, D/d=1.02, t/D=0.125, b/D=0.25)

100 , 200 , 400F N M N m T N m

Solution: a.) The normal stress due to axial force is uniform over the area:

2

4axial

F F

A d

The maximum stresses due to the bending moment and the twisting torque are:

bending

Mc

I ,

twisting

Tc

J

The Maximum torsional shear due to the twisting torque occurs at the outer surface of the cross-section and the maximum normal stress due to bending occurs on the stress element at the upper point of the shaft for static loading. So this point is the critical point. The stresses at this point due to these loadings can be expressed in terms of the diameter as:

3

32bending

Mc M

I d

,

3

16twisting

Tc T

J d

The total normal stress and the shear stresses are:

2 3

4 32x

F M

d d

3

16xz

T

d

Using the distortion-energy theory,

2 2

2 2

2 3 3

4 32 16' 3 3x xz

F M T

d d d

The material properties from Table A-18 (Shigley 8

th Edition)

620utS MPa , 340yS MPa

The diameter of the shaft can be found using the safety factor value.

'

ySn

M

T T

M

x

y

z F F

b

t

r

Snap ring

Flywheel

d

фD


2

2 2

2 3 3

2.54 32 16

3

yS

F M T

d d d

2 23 3

2 3 3

2.5

4 100 32(200 10 ) 16 400 103

yS

d d d

which gives 31.1d mm , by iteration.

Note that, taking the axial force as 0F , 31.0d mm can be found. This value

can be used as an initial value to the iterations. Then, by using the given diameter ratio D/d=1.02, shaft diameter, D is found as 31.7mm. Select an appropriate value for diameter from the table A-17 (Preferred Sizes and Renard Numbers Table). Preferred size for shaft diameter is D = 32mm. Also, note that, stress calculations on the section where the keyway is located are performed by using the whole shaft area. Since, the area at the keyway is very small compared to the shaft cross-section as shown below, the area of the shaft is used in calculations.

2 2

0.25 0.1250.04

4 4

key

shaft

A b t D D

D DA

b.) For the fatigue case the points at the keyway and at the groove of the snap-ring should be investigated to determine the critical section. The stress formulations used above will be the alternating stress for this case. The mean components of the stresses will be zero due to completely reversed loading. The stress calculation at the keyway: Since, both the axial loading and bending moment cause normal stress on the cross-

section, it is assumed that 1fK can be taken as the stress concentration factor for

axial loading as well.

1 2 3

4 32a f

F MK

D D

2 3

16a f

TK

D

Alternating component of the von Mises stress is:

2 2

2 2

, 1 22 3 3

4 32 16' 3 3a keyway a a f f

F M TK K

D D D

where

1fK (stress concentration factor for bending for profile keyways): 1.6

2fK (stress concentration factor for torsion for profile keyways): 1.3

(From Stress Concentration Factors for Threads and Keyways Table under the Useful Information link of the course web page)

2 23 3

2 2

, 2 3 3

4 100 32 200 10 16 400 10' 3 1.6 3 1.3a keyway a a

D D D

Since the normal stress from axial loading is so small when compared with stress due to bending moment, the axial stress term is neglected. So,

12

, 6

31.6 10'a keyway

D

The stress calculation at the groove:

3 2 3

4 32a f

F MK

d d

3 3

16a f

TK

d

Alternating component of the von Mises stress is:

22

4 52 2

, 3 2 3 3

4 32 16' 3 3

f f

a groove a a fb b

K KF M TK

kd d k d

From Figure 7-14 in the Formula Sheet, q is taken as 0.95 assuming that notch radius is 3 mm.

1.60tK (from Figure A-13-13 in the Formula Sheet, grooved round bar in tension)

1.55tK (from Figure A-13-14 in the Formula Sheet, grooved round bar in bending)

1.30tsK (from Figure A-13-15 in the Formula Sheet, grooved round bar in torsion)


3

So,

3 1 ( 1) 1 0.95(1.60 1) 1.57f tK q K

4 1 ( 1) 1 0.95(1.55 1) 1.52f tK q K

3 1 ( 1) 1 0.95(1.30 1) 1.28f tsK q K

2 23 3

, 2 3 3

4 100 1.52 32 200 10 1.28 16 400 10' 1.57 3

0.85 0.85a groove

d d d

Since the normal stress from axial loading is so small when compared with stress due to bending moment, the axial stress term is neglected.

12 12

, 6 6

41.5 10 42.3 10'a groove

d D

Since, the stresses at the groove are more critical, those will be taken in consideration for the determination of shaft diameter. The Endurance limit of the shaft is to be determined. The endurance limit of the test specimen is:

' 0.5 0.5 620 310e utS S MPa

The endurance limit modifying factors are:

0.75ak for 620utS MPa and machined surface.

0.85bk given in the problem.

0.868ck for 95% reliability.

1dk nothing mentioned.

1ek since stress concentrations are considered in stress

calculations.

1fk nothing mentioned.

0.75 0.85 0.868 1 1 1 310 172eS MPa MPa

The diameter of the shaft can be found using the safety factor value. Since the mean stress is zero:

'

e

a

Sn

12

6

1722

42.3 10

MPa

N

D

which gives D=42.28 mm Select an appropriate value for diameter from the table A-17 (Preferred Sizes and Renard Numbers Table). Preferred size for shaft diameter is D = 45mm. Note that, if both the alternative and mean stress components exist, the critical section can be found by using one of the failure theorems. Note that, if the neglected term of axial stress is included in the equations, a factor of safety of 1.9 is determined.

2 23 3

, 2 3 3

4 100 1.52 32 200 10 1.28 16 400 10' 1.57 3

(42.28 /1.02) 0.85 (42.28 /1.02) 0.85 (42.28 /1.02)a groove

,' 90.52

a grooveMPa

1721.9 2

' 90.52

e

a

Sn


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Problem 2 : In the figure; the shaft transmits 7.5 kW from the pulley P to the gear G at 900 rev/min under steady load conditions. The shaft is machined. The ultimate tensile strength is 590 MPa and the yield point in tension is 380 MPa. The pulley diameter is 250 mm and the pitch diameter of the gear is also 250 mm. The mass of the pulley P and the gear G are both 12.2 kg. Neglect the weight of the shaft. The ratio of the belt tensions is T1/T2 = 2.5. The gear pressure angle is 20

o. Take reliability

as 0.99 and working temperature is 100 Co. Neglect the effect of keyway. Factor of

Safety is 1.5. Determine the shaft size, d.

Solution: ka=0.75 (Fig 7-8, machined) kb=0.85 Same for bending and torsion, assuming 8<d<50 for all sections on shaft kc=0.814 (Table 7-7, 99% Reliability) kd=0.9 (T=100 C

o)

0.5e a b c d utS k k k k S

(0.75)(0.85)(0.814)(0.9)(590)(0.5) 138eS MPa

The torque Mt on the shaft is found from:

2, 7500 , 900 / min

60

tNMP P W N rev

So, 79.6 .tM N m

The belt forces are found from:

1 2( )( ) tT T R M or 1 2( )(0.125) 79.6T T and 1 2/ 2.5T T

From which:

1 1060T N , 2 424T N

The weights of the gear and the weight of the pulley are each 12.2 x 9.81 = 119.6 N; say 120 N.

The tangential gear force / 79.6 / 0.125 636.8t tF M R N

The vertical force is tan 636.8tan 20 231.8o

r tF F N

The loading and bending moment diagrams are shown with the figure; the belt forces and gear forces are assumed concentrated as shown. The weight of the pulley and gear are included.


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Consider the section at the gear first. The stress concentration factor in bending due to the fillet where the shaft portion of diameter d joins with the shaft diameter 2d depends upon the radius of the fillet. Even if the ratio of the radius of the fillet to the shaft d is specified to obtain the theoretical stress concentration factor Kt, the actual stress concentration factor Kf depends upon the notch sensitivity, which is a function of the radius of the fillet. Thus it is necessary to use a trial and error solution, by approximating the diameter d and the radius of the fillet and checking with the calculated value of d. The values used in the following calculations are those which agree with the final values: r/d= 0.1, for which r=2.4 mm For r/d = 0.1, D/d = 2, and r=2.4 mm: Kts=1.47 Kt=1.71 q1=0.8 q2=0.94

11 ( 1) 1 0.8(1.71 1) 1.568f tK q K

21 ( 1) 1 0.94(1.47 1) 1.442fs tsK q K

The bending is constant, 65.6 N.m but a particle on the surface is subjected to a complete reversal of stress. The mean bending stress is zero. The variable stress is:

3 3 3

32 (32)(65.6) 1048(1.568)a f

MK

d d d

The shear stress due to torsion is constant, since the torque is constant. The variable shear stress is zero. The mean shear stress is:

3 3 3

16 (16)(79.6) 584.6(1.442)m fs

TK

d d d

Kfs is used in mean shear stress for additional safety although material is not a

particularly high strength material. Based on the distortion-energy theory the equivalent stresses are:

'a a 2

3

1012.6' 3m m

d

Using modified Goodman relation:

' '1 a m

e utn S S

3 3

6 6

1 1048/ 1012.6 /

1.5 138 10 590 10

d d

d=24.1 mm; r=2.41 mm are found. The shaft at the right bearing will be analyzed next. Specifically, the section with diameter d at the fillet will be considered, with the bending moment at the centerline of the bearing taken as acting at the section with the fillet. The stress concentration factor Kf cannot be determined directly here. The same procedure as used at the section with the gear will be used, trial and error, with final values given

r/d= 0.1, for which r=3.5 mm For r/d = 0.1, D/d = 2, and r=3.5 mm: Kts=1.47 Kt=1.71 q1=0.83 q2=0.96

11 ( 1) 1 0.83(1.71 1) 1.589f tK q K

21 ( 1) 1 0.96(1.47 1) 1.451fs tsK q K

The mean bending stress is zero.

3 3 3

32 (32)(223) 3609(1.589)a f

MK

d d d

The variable shear due to torque stress is zero. The mean shear stress is:

3 3 3

16 (16)(79.6) 588.2(1.451)m fs

TK

d d d

Based on the distortion-energy theory the equivalent stresses are:


6

'a a 2

3

1018.8' 3m m

d

Using modified Goodman relation:

' '1 a m

e utn S S

3 3

6 6

1 3609 / 1018.8 /

1.5 138 10 590 10

d d

d=34.731 mm; r=3.47 mm are found. Use The size of trhe shaft is then determined by the stresses acting in the shaft at the right bearing.

Problem 3: The figure shows a shaft with two gears and a drum mounted on it. Gears 1 and 2 are identical with weight 70N. The weights of the shaft and the drum are 140N and 100N, respectively. Find the first critical speed of the shaft using Rayleigh’s equation.

Take the flexural rigidity of the shaft as 11 24 10EI N mm .

Solution: The Rayleigh’s equation gives the first critical speed of a shaft as:

√ ∑

∑ where is the weight of the i

th element on the shaft and is the

deflection at the ith

body location. The shaft can be analyzed by dividing it into 3 elements as in Figure 2. The deflection of the gears, drum and each element on the shaft are to be determined.

The positions 1 and 5 are the positions of the gears. The positions 2, 3 and 4 are the centers of the 3 elements of the shaft. Position 3 is also the position of the drum. Note that the lowest critical speed of the shaft is related with the first bending mode of the shaft. The shape of the first mode is given in Figure 3.

The deflections at the specified positions will be calculated considering the weights are acting in the directions shown in Figure 4, that will result in the deflection curve of the first mode shape given in Figure 3.

20 100

140

280

G1 G2

Drum

Figure 1

Figure 2

50 50 40 40 50 50

1 2 3 4 5

A B

Figure 3

A B

Figure 4 F1 F2 F3 F4 F5

A B

35d mm


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The shaft in the problem can be considered as an overhanging simply supported beam. To find the deflections at position 3, the forces outside the supports will be replaced with equivalent force and moment system at the supports. The deflections will be determined by using the formula from the deflection table for simply supported and cantilever beams.

Using the equivalent system in Figure 5, the deflection at position 3 can be calculated. Superposition of the three loadings; two end moments and the center load will give the total deflection. Also for the positions outside the supports, the slopes at the supports due to these loads are needed.

Figure 6 shows the superposition of the loads over the shaft. Note that equivalent force due to overhanging forces will not make any deflection or slope, since they are acting on the supports. The deflection at 3 and the slopes at A and B due to end moment at B:

( ), (from beam deflection tables)

,, ,

Note that the sign of the end moment is taken as negative due to the sign convention used in beam deflection formulas.

(from beam deflection tables)


The deflections at 3 and the slopes at A and B due to end moment at A: Using the symmetry,

The deflections at 3 and the slopes at A and B due to the center load:

(

), (from beam deflection tables),



Figure 5 F1 F2 F3 F4 F5

A B A B

FR FR FR FR Feq

F3

Meq Meq

A B

Feq Feq

F3

Meq Meq

A B Meq

A B

Meq

F3

A B A B

Feq Feq

Feq

Figure 6


8

To find the deflections outside the supports, the slope at supports A and B of the simply supported beam given in Figure 6 is obtained and the deflection of the overhanging locations due to these forces are obtained by simply multiplying the distance with the slope. In addition to this, the overhanging forces are considered as they are acting on a beam cantilevered at the support locations. Therefore, the total defections at the overhanging parts are obtained by simply adding the deflections due to the slope of simply supported beam (including moments) and the cantilever beam. It will be sufficient to find the deflections at positions 4 and 5, because of symmetry. These deflections will be the same as that of 2 and 1, respectively.

The deflections at 4 and 5 due to the forces inside the supports: ,

,

The deflections at 4 and 5 due to the concentrated force at 4:

(

), ,

( )

The deflections at 4 and 5 due to the concentrated force at 5:

(

), ,

(

), ,

The total deflections at 4 and 5:

The total deflections at 1 and 2:

The weight at each position is also needed.

The first critical speed of the shaft can be calculated using the Rayleigh’s equation. Note that during the calculation of all the deflections, y axis has been selected as pointing upwards. However, since Rayleigh’s equation is obtained from the equality of potential energy and kinetic energy all the forces and deflections used in Rayleigh’s equation should be taken as positive. Note that energy is a positive quantity and it cannot be negative.

Figure 7 B

F5 F4


9

∑ (

)

∑ ∑ ( ) ( ) ( )

( ) ( )

√

The analysis can be simplified if the shaft masses for the overhanging locations are lumped onto the positions of the gears. This way, deflections of only 3 locations are to be found, i.e. the locations of the gears and the drum.

me307-11 tutorial 9

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