SHAFT ANALYSIS

SHAFT 1

i. Free body diagram l

= FORCE

= FIXED

ii. Moment and shear diagram for X-Z plane

For x- z plane

-28.75 (28.5) + RBY (150) = 0

RBY = 5.4625 N

+↑ ΣF y=0

RAY -28.75 + 5.4625

RAY = 23.2875 N

Data Pinion 1

FT 78.99 N

FR 28.75 N

iii. Moment and shear diagram for X-Y plane

For x-y plane

-78.99 (28.5) + RBZ (150) = 0

RBZ = 15.0081 N

+ ΣF z = 0

RAZ – 78.99 + 15.0081 = 0

RAZ = 63.9819 N

M = √0.6637 ²+1.8235 ²

=1. 9405 Nm

Tm = Ft (Dp2

)

= 78.99 (0.016

2)

= 0.6319 Nm

Mm = Ta = 0

Shaft Material Selection

We Choose Steel AISI 1080 HR

SUT =770 Mpa SY = 420Mpa

Estimate,

Kt=1.7 and Kts = 1.5 . To ease and quick, conversation pass, assume Kt= Kf, Kts= Kfs.

Endurance Limit Modifying Factor

Se = KaKbKcKdKeKfSe’

Se’ = 0.5Sut

Se’ = 0.5(770Mpa)

Se’ = 385 Mpa

1. Surface Factor, Ka

Ka = aSutb

= 4.51 (770)-0.265

= 0.775

1. Size Factor, Kb

Assume Kb = 0.9

2. Loading Factor, Kc

The combination of bending and torsion must take the higher value that is 1. Kc = 1

3. Temperature Factor, Kd

We assume the temperature as 20° as a room temperature.Kd=1

4. Reliability Factor, Kc

We assume Ke =1 for easy calculation.

5. Miscellaneous-Effects Factor, Kf

Kf = 1

Se = KaKbKcKdKeKfSe’

= (0.775) (0.9) (1) (1) (1) (1) (385)

= 268.5096 Mpa

Minimum diameter required at each gear

Use DE- Goodman Criterion

Assume the safety factor is 2.5

Tm =0.6319Nm

M = 1.9405 Nm

d = {16nπ [ 1

Se ((4 {K f M a }2 )1/2+ 1Sut

(3 {K fsT m }2 )1 /2)]}1/3

d = {16 (2.5)π [ 1

268.5 ((4 {(1.7)(0.6319)}2)1 /2+ 1

770(3 {(1.5)(1.9405)}2)1/2)]}

1 /3

d = 6.97 mm ≈ 7 mm

Fatigue and yield factors of safety (shaft)

Small diameter at shoulder is 6.97 mm. Typical D/d ratio for support at a shoulder is (D/d = 1.2)

1. Typical ratio :Dd

=1.2

D = 1.2(6.97) = 8.36 mm

Assume fillet radius, r=1mm

17=¿ =0.1429

2. Calculation for get real Endurance limit Modifying Factors

Se = KaKbKcKdKeKfSe’

Se’ = 0.5Sut

Se’ = 0.5(770Mpa)

Se’ = 385Mpa

I. Surface Factor, Ka

Ka = aSutb

= 4.51 (770)-0.265

= 0.775

II. Size Factor , kb

Since d= 6.97 mm

7 < 51 we use as shown below:

kb¿1.24d−0.107

= 1.24 (6.97)−0.107

= 1.007

IV. Loading Factor, Kc

The combination of bending and torsion must take the higher value that is 1. Kc = 1

V. Temperature Factor, Kd

We assume the temperature as 20° as a room temperature.Kd=1

VI. Reliability Factor, Ke

Since Reliability = 0.99

Ke = 0.814

VII. Miscellaneous-Effects Factor, Kf

Kf = 1

Se = KaKbKcKdKeKfSe’

= (0.775) (1.0074) (1) (1) (0.814) (1) (385)

= 244.6745 Mpa

3. Stress Concentration and Notch Sensitivity

kf = 1+q(kt-1) and kfs = 1+q(kts-1)

Based on info D/d =1.2

r/d = 17

= 0.1429

Based on graph above Kt = 1.48

Based on graph above Kts = 1.28

q = 0.78

qs = 0.81

Kf = 1 + q (Kt -1)

= 1 + (0.78) (1.48-1)

= 1.3744

Kfs = 1 + qs (Kts -1)

= 1 + (0.81) (1.28-1)

= 1.2268

Ma = √547.2563 ²+1503.5 ² = 1600Nmm

Ma = 1.6 Nm

σ ' a=32K f M aπd ³

σ ' a=32 (1.3744 )(1.6)π (0.0 07) ³

= 65.3 MPa

Tm = 0.6319Nm

σ 'm=[3 (16K fsT mπd ³ )

2]12

σ 'm=[3 (16 (1.2268 )(0.6319)π (0.0 07) ³ )

2]12

= 19.9 MPa

Using Goodman Criterion

1nf

=σ' aSe

+ σ'mSut

1nf

= 65.3244.67

+ 19.9770

nf = 3.41

ny = Sy

σ ' max>

Sy

σ ' a+σ ' m>

42065.3+19.9

= 4.93