1

ME 452: Machine Design II Name of Student:___________________________

Problem 1. The effective masses of the pistons of the two-cylinder engine shown in Figure 1 are

1 2m m m 20 kg.= = = The lengths of the connecting rods are 1 2L L L 300 mm= = = and the lengths of the throws of each crank are 1 2R R R 150 mm.= = = The two cranks are oriented at

o240 to each other and the crankshaft is rotating counterclockwise with a constant angular speed ω 250 rad/s.= Both pistons are in the same XY plane.

Determine the X and Y components of the primary shaking force acting on the ground bearing; and the magnitudes and angular locations of the correcting masses which balance the primary shaking force. The radial distances of the correcting masses are C1 C2 CR R R 50 mm.= = =

On the figure to the right, which corresponds to the crank 1 posture ,oθ 30= show the magnitude and direction of the primary shaking force; and the locations of the correcting masses.

Figure 1. A two-cylinder engine. (Not drawn to scale).

θ = 30o

Reference Line -ω

Y

1

1 X

ω

2

ME 452: Machine Design II Name of Student: __________________________ Problem 2.

The AISI 1030 cold drawn steel bar shown in Figure 2 is simply supported at each end. The circular bar is not rotating and the diameters are D = 2.4 in and d = 1.6 in and the radius of the groove is 0.4 in. The bar is subjected to the repeated load 0 ≤ F ≤ 1300 lb at section A which is in-phase with the repeated load 0 ≤ F ≤ 500 lb at section B. The Marin modification factors for load, temperature, reliability, and miscellaneous-effects are k = k = k = k = 1. (i) Neglecting the effects of stress concentration, determine the factor of safety guarding against yielding for the critical element of the bar using the Langer line. (ii) Including the effects of stress concentration and the notch sensitivity q = 0.88, determine the fatigue factor of safety for the critical element using the Gerber parabola criterion of failure.

Figure 2. A simply supported steel bar. (Not drawn to scale).

O d = 1.6 in B A C

7 in 5 in 2 in

F = 1300 lb F = 500 lb

X

Y

D = 2.4 in

r = 0.4 in

3

ME 452: Machine Design II Name of Student: __________________________ Problem 3.

A rotating steel shaft is simply supported by two rolling element bearings at A and B as shown in Figure 3. The length of the shaft is 1.45 m and two flywheels with weight 1 2W W 300 N= = are rigidly attached to the shaft at the locations shown in the figure. One flywheel is 0.35 m to the right of the left bearing at A and the other flywheel is 0.35 m to the left of the right bearing at B. The weight of the shaft can be neglected. The known influence coefficients are

511 12.600 x 10 cm/Nδ −= and 5

21 9.250 x 10 cm/N.δ −= (i) Determine the first and second critical speeds of the shaft using the exact solution. (ii) Determine the first critical speed using the Dunkerley approximation. (iii) Determine the first critical speed using the Rayleigh-Ritz equation.

Figure 3. A Steel Shaft Simply Supported by Two Rolling Element Bearings.

4

ME 452: Machine Design II Name of Student:___________________________ Problem 4.

The face widths of the pinion and the gear of a pair of mating steel spur gears are 0.75 inches and the gearset transmits a load of 40 lbs. To estimate the contact stresses between the pinion and the gear assume that the teeth can be modelled as cylindrical profiles.

At the point of contact between the pinion and the gear assume that the radius of the pinion is 0.47 inches and the radius of the gear is 0.62 inches. (i) Determine the maximum principal normal stress and the minimum principal normal stress. (ii) Estimate the maximum contact pressure and the maximum contact shear stress experienced by the pinion or the gear.

5

ME 452: Machine Design II Name of Student: __________________________ Problem 5.

A full journal bearing is lubricated with SAE 30 grade oil and operates at o40 C. The diameter of the journal is 55 mm and the diameter of the bearing is 55.05 mm. The rotational speed of the journal is 50 rev / s. Part I. If the radial load acting on the bearing is 2.5 kN and the length of the bearing is 55 mm then determine: (i) The Sommerfeld number. (ii) The eccentricity. (iii) The coefficient of friction. Part II. If the radial load is increased to 5 kN and the length of the bearing is increased to 110 mm then using the Raimondi and Boyd interpolation determine the position of the minimum film thickness.

6

Problem 6. The gear of a commercial spur gearset is the input and is rotating at 500 rpm and is transmitting

6 horsepower to the output pinion. The pitch diameter of the pinion is 5 inches, the diametral pitch of the gear is 1,5 inches− and the gear ratio is 2. The gearset has full-depth teeth with a pressure angle of 20o and the face widths of the gear and pinion are 0.8 inches. The gear is subjected to a light shock load and the pinion a moderate shock load where the loads are applied at the tips of the teeth. The gear and pinion operate at room temperature. The gear is nitrided through-hardened Grade 2 steel with a Brinell hardness BH 300.= The known AGMA factors for the gear are: dynamic factor vK 1.25,= size factor sK 1.0,= and load distribution factor mK 1.0.= The backup ratio for the gear is Bm 1.5.= (i) Determine the load transmitted by the gearset. (ii) Determine the AGMA bending stress on the teeth of the gear. (iii) Determine the AGMA safety factor guarding against bending fatigue failure for the gear for

810 load cycles and 95 percent reliability.

7

Solution to Problem 1. The X and Y components of the resultant of the primary shaking force for the two-cylinder

reciprocating engine can be written in the form

(1a) and

(1b) where the design parameters can be written as

(2a)

(2b)

(2c)

and

(2d)

The two correcting forces can be written as

(3a)

and

(3b) Also, the location angles of the correcting forces can be written as

(4a) and

2tan C BD A

γ +=−

(4b)

Note that the numerator and the denominator of Eq. (4a) and Eq. (4b) give the correct quadrants for the two angles. Knowing the quadrants it is common practice to write Eq. (4a) as

1tan C BD A

γ −=+

(4c)

However, Eq. (4c) does not give the correct quadrant for the angle.

A cos B sin= +PXS θ θ

C cos D sin= +PYS θ θ

2

1A cos ( ) cosi i i i

iP ψ φ ψ

=

= −

2

1B sin ( ) cosi i i i

iP ψ φ ψ

=

= −

2

1C cos ( )sini i i i

iP ψ φ ψ

=

= −

2

1D sin ( )sini i i i

iP ψ φ ψ

=

= −

( ) ( )2 21

12

F A D B C= + + −

( ) ( )2 22

12

F A D B C= − + +

1tan( )

−=− +

B CA D

γ

8

The X and Y component of the first harmonic force for cylinder 1 can be written as

(5a) and the Y component of the first harmonic force for cylinder 1 can be written as

(5b) The X and Y components of the first harmonic force for cylinder 2 can be written as

(6a) and

(6b)

The X-component of the resultant of the first harmonic forces can be written as

(7a) and the Y-component of the resultant of the first harmonic forces can be written as

(7b)

The primary shaking force due to each of the two pistons can be written as

21 1 1P m r Pω= = and 2

2 2 2P m r Pω= = (8)

To determine the coefficients A, B, C, and D. The angles are

and (9) Substituting Equations (9) into Equations (5a) and (6a), and the results into Equation (7a), the X-component of the resultant of the first harmonic forces can be written as

cos ( 135 )cos135 cos ( 180 240 )cos180o o o o oPx

S P Pθ θ= − + − + (10a)

that is 0 (P cos ) (0.866 0.5) P sin 0 (P cos ) 0.366 (P sin )Px

S θ θ θ θ= + − = + (10b)

Substituting Equations (9) into Equations (5b) and (6b) and the results into Equation (7b), the Y-component of the resultant of the first harmonic forces can be written as

cos ( 135 )sin135 cos ( 180 240 )sin180o o o o oPy

S P Pθ θ= − + − + (11a)

that is 0.5 (P cos ) 0.5 (P sin )Py

S θ θ= − + (11b)

Comparing Equations (10b) and Equation (11b) with Equations (1a) and (1b), the coefficients are

0, 0.366 , 0.5 , and 0.5A B P C P D P= = + = − = + (12)

1 1 1 1cos( )cosXS P θ ψ ψ= −

1 1 1 1cos( )sinYS P θ ψ ψ= −

2 2 2 2 2cos( )cosXS P θ ψ φ ψ= − +

2 2 2 2 2cos( )sinYS P θ ψ φ ψ= − +

1 2= +PX X XS S S

1 2= +PY Y YS S S

1 135 ,= °ψ 2 180 ,ψ = ° 2 240= °φ

9

The magnitude of the resultant of the first harmonic forces is

(13) Substituting Equations (10b) and (11b) into Equation (13), the magnitude of the resultant of the first harmonic forces is

2 21 1(0.366 P sin ) ( P sin P cos )2 2PS θ θ θ= + − (14a)

that is 2 2P 0.384 sin 0.25 cos 0.5 sin cosPS θ θ θ θ= + − (14b)

The direction of the resultant of the first harmonic forces is

1tan PY

PX

SS

τ − =

(15a)

Substituting Equations (10b) and (11b) into Equation (15a), the direction of the resultant of the first harmonic forces as a function of the crank position is

1 10.5 P sin 0.5 P cos 0.5 sin 0.5 costan tan0.366 Psin 0.366 sin

θ θ θ θτθ θ

− − − −= =

(15b)

Substituting the coefficients in Equation (12) into Equation (3a), the first correcting force is

( ) ( )2 21

1 0 0.5 0.366 0.52

F P P P P= + + + (16a)

which can be written as 1 0.5PF = (16b)

Substituting the coefficients in Equation (12) into Equation (3b), the second correcting force is

( ) ( )2 22

1 0 0.5 0.366 0.52

F P P P P= − + − (17a)

which can be written as

21 0.2680 0.2588P2

F P= = (17b)

Substituting the given data into Equation (8), the primary shaking force is

220 0.15 250 187500 N 187.5 kNP = × × = = (18a) Substituting Equation (18) into Equations (12), the coefficients are

0, 68.625 kN, 93.75 kN, and 93.75 kNA B C D= = + = − = + (18b) Then substituting Equation (18b) into Equations (16b) and (17b), the correcting forces are

2 2= +P PX PYS S S

10

1 0.5 187500= 93750 NF = × (19a) and

2 0.2588 187500= 48525 NF = × (19b) Equations (19) indicate that two correcting masses are required to balance the primary shaking force. The correcting masses can be obtained from the relation

(20a)

Substituting the given data and Equations (19) into Equation (20a), the two correcting masses are

11 2 2

1

93750 30 kg0.05 250C

C

FmR ω

= = =×

(20b)

and 2

2 2 22

48525 15.528 kg0.05 250C

C

FmR ω

= = =×

(20c)

Substituting Equations (12) into Equation (4a), the location angle for the first correcting mass is

10.366 0.5 0.866tan 1.732

(0 0.5 ) 0.5P P

P Pγ + += = = − − + −

(21a)

Since the numerator (or the sine of the angle) is positive and the denominator (or the cosine of the angle) is negative then the angle is in the second quadrant, that is

1 120γ = ° (21b) Substituting Equations (12) into Equation (4b), the angle for the second correcting mass is

20.366 0.5 0.134tan 0.268

0.5 0 0.5P PP P

γ − −= = = − + − + (22a)

Since the numerator (or the sine of the angle) is negative and the denominator (or the cosine of the angle) is positive then the angle is in the fourth quadrant, that is

2 15 345γ = − ° = + ° (22b)

For the arbitrary crank position the primary shaking force (or the first harmonic force) and the location of the correcting masses are shown in Figure 1.

Note that the magnitude of the resultant of the first harmonic forces PS and the direction of the resultant of the first harmonic forces τ are given by Equation (14b) and Equation (15b), respectively.

2C

CC

FmR ω

=

,θ

11

Figure 1. Magnitude and location of the correcting masses for an arbitrary crank position θ.

For the crank position o30θ = . Substituting this value into Equation (14b), the magnitude of the resultant of the first harmonic forces is

0.096 0.1875 0.2165PS P= + − that is

0.067 187500 x 0.2588 48525 NPS P= = = (23)

Substituting the crank position o30θ = into Equation (15b), the direction of the resultant of the first harmonic forces can be written as

1 10.183tan tan ( 1)0.183

τ − − −= = − + (24a)

Since the numerator (or the sine of the angle) is negative and the denominator (or the cosine of the angle) is positive then the direction of the resultant of the first harmonic forces is in the fourth quadrant, that is

45 315τ = − ° = + ° (24b) The primary shaking force and the location of the correcting masses are shown in Figure 2.

- (θ +345o)

θ + 120o

Reference Line Crank Angle θ

-ω

Y

1

1 X

ω

F1 = 93.750 kN

F2 = 48.525 kN

SP

θ

τ

mC1 = 30 kg mC2 = 15.528 kg

12

Figure 2. Magnitude and location of the correcting masses for the crank position o30θ = .

150°

Reference Line θ = 30° -ω

Y

1

1 X

ω

- 15°

F1 = 93.750 kN

F2 = 48.525 kN

SP = 48.525 kN

- 45°

mC1 = 30 kg

mC2 = 15.528 kg

13

Solution to Problem 2. (i) The sum of the moments about the right-hand bearing C can be written as ΣM = 0 (1a) that is 14R − 7F − 2F = 0 (1b)

Substituting the maximum forces at A and B, that is, F = 1300 lb and F = 500 lb into Eq. (1b), the maximum reaction force at bearing O is R = 721.43 lb (2)

The sum of the forces in the Y-direction can be written as ΣF = 0 (3a) that is R + R − F + F = 0 (3b)

Substituting the maximum force at A and B and Eq. (2) into Eq. (3b), the maximum reaction force at bearing C is R = 1078.57 lb (4)

The shear force diagram showing the maximum forces on the bar is shown in Figure 1a.

Figure 1a. The shear force diagram of the bar.

The maximum bending moment at section A is M = 7 R = 7 × 721.43 = 5050.01 lb ∙ in and the minimum bending moment at section A is M = 0 (5a)

14

The maximum bending moment at section B is M = 12 R − 5 F = 2157.16 lb ∙ in and the minimum bending moment at section B is M = 0 (5b)

The bending moment diagram of the bar is shown in Figure 1b.

Figure 1b. The bending moment diagram of the bar.

Note that the bending moment at section A is the largest and the diameter at section A is the smallest. (Note that stress concentration effects are only at section A.) Therefore, the critical section of the bar is section A. The critical element is at the outer edge of the critical section.

The factor of safety guarding against yielding using the Langer line can be written from Eq. (6-43), see page 330, as 𝑛 = (6) The yield strength of AISI 1030 cold drawn steel, see Table A-20, page 1056, is S = 64 kpsi (7)

From Eq. (5a), the alternating component of the bending moment at section A is M = |MAmax MAmin| = 2525lb. in (8a)

and the mean component of the bending moment at section A is M = (MAmax MAmin) = 2525lb. in (8b)

The alternating and mean components of the normal stress at section A are σ = = and σ = = (9)

15

Substituting Eqs. (8a) and (8b) into Eq. (9), the alternating and mean components of the normal stress are σ = ×. = 6.28 kpsi and σ = ×. = 6.28 kpsi (10)

Note that the effects of stress concentration are neglected here. Substituting Eqs. (7) and (10) into Eq. (6), the factor of safety guarding against yielding is n = . . = 5.1 (11)

(ii) The Gerber criterion of fatigue failure can be written from Eq. (6.48), see page 334, as

1/22 221 ( )( ) 1 1 ( )

2a ut m e

fe m a ut

S SnS Sσ σ

σ σ

= − + +

(12)

The fully corrected endurance limit of the bar can be written from Eq. (6-17), see page 309, as

S = k k k k k k S (13)

The uncorrected endurance limit of the bar from Eq. (6-10), see page 305, is S = 0.5 S (14)

The ultimate tensile strength of AISI 1030 cold drawn steel, from Table A-20, see page1056, is S = 76 kpsi (15)

Substituting Eq. (15) into Eq. (14), the uncorrected endurance limit of the bar is S = 0.5 × 76 = 38 kpsi (16)

For cold-drawn steel, the surface modification factor from Eq. (6-18), see page 311, is k = aS = 2 × 76 . = 0.781 (17)

The size modification factor. Since the round bar is not rotating then the equivalent diameter can be written from Eq. (6-23), see page 313, as d = 0.370 d = 0.370 × 1.6 = 0.592 in (18) Then the size modification factor can be written from Eq. (6-19), see page 312, as k = 0.879d . = 0.879 × 0.592 . = 0.930 (19)

Substituting Eqs. (16), (17), and (19) into Eq. (13), the fully corrected endurance strength is S = 0.781 × 0.930 × 38 = 27.6 kpsi (20)

The fatigue stress concentration factor for the critical element from Eq. (6-32), see page 321, is K = 1 + q(K − 1) (21a)

16

Given D = 2.4 in, d = 1.6 in, and r = 0.4 in, the theoretical stress concentration factor for the critical element from Figure A-15-14, see page 1046, is K = 1.5 (21b)

Substituting the notch sensitivity q = 0.88 and Eq. (21b) into Eq. (21a), the fatigue stress concentration factor is K = 1 + 0.88 × (1.5 − 1) = 1.44 (22)

The alternating and mean components of the normal stress on the critical element, including the fatigue stress concentration factor at the groove, can be written as σ = K σ , and σ = K σ , (23) Substituting Eqs. (10) and (22) into Eq. (23), the alternating and mean components of the normal stress on the critical element are σ = 1.44 × 6.28 = 9.04 kpsi and σ = 1.44 × 6.28 = 9.04 kpsi (24) Substituting Eqs. (15), (20), and (24) into Eq. (12), the fatigue factor of safety as predicted by the Gerber failure criterion for infinite life can be written as 𝑛 = . . . −1 + 1 + ( × . × .. × ) /

(25a) Therefore, the fatigue factor of safety as predicted by the Gerber failure criterion is 𝑛 = 2.7 (25b)

17

Solution to Problem 3. The exact solutions for the first and second critical speeds of the shaft can be written as

2

11 1 22 2 11 1 22 2 11 22 12 21 1 22 21 2

( a m a m ) ( a m a m ) 4 ( a a a a ) m m1 1,ω ω 2

+ ± + − −=

(1) From the symmetry of the loading, the influence coefficients

511 22a a 12.600 x 10 cm/N−= = (2)

From Maxwell's reciprocity theorem, the influence coefficients

521 12a a 9.250 x 10 cm/N−= = (3)

The mass of the flywheels are

21 2

300m m m 0.30612 N.s /cm980

= = = = (4) Substituting Equations (2), (3), and (4) into Equation (1), the exact solutions can be written as

11 212 21 2

1 1, a m a mω ω

= ± (5)

Equation (5) can be written as

2 21 2

11 21

1ω ,ωa m a m

=± (6)

Substituting the numerical values into Equation (6), the exact solutions can be written as

52 2 21 2

10ω ,ω (rad / sec)0.3061 (12.600 9.250)

=× ±

(7)

Using the positive sign in the denominator of Eq. (7), the first critical speed of the shaft can be obtained from the relation

5 52 4 21

10 10ω 1.495 10 (rad / sec)0.3061 21.850 6.689

= = = ×× (8)

Therefore, the first critical speed of the shaft is

1ω 122.3 rad / sec= (9) Similarly, using the negative sign in the denominator of Eq. (7), the second critical speed of the shaft can be obtained from the relation

5 52 4 22

10 10ω 9.751 10 (rad / sec)0.3061 3.350 1.026

= = = ××

(10)

18

Therefore, the second critical speed of the shaft is

2ω 312.3 rad / sec= (11) Note that the second critical speed is about three times the first critical speed. (ii) Determine the first critical speed of the shaft using the Dunkerley approximation.

The Dunkerley approximation to the first critical speed of the shaft can be written as

11 22 1121

1 a m a m 2 m aω

= + = (12)

Substituting the numerical values into Equation (12), the Dunkerley approximation to the first critical speed of the shaft is

5 5 221

1 2 0.3061 12.6 10 7.7137 10 secω

− −= × × × = × (13)

Therefore, the Dunkerley approximation to the first critical speed of the shaft is

1ω 113.9 rad / sec= (14) Note that the the Dunkerley approximation to the first critical speed of the shaft is less than the exact answer, see Eq. (9); i.e., the Dunkerley approximation always gives a lower bound. (iii) Determine the first critical speed of the shaft using the Rayleigh-Ritz equation.

The Rayleigh-Ritz equation can be written as

2 1 1 2 21 2 2

1 1 2 2

W x W xω gW x W x += +

(15)

where the deflections are

5 51 11 1 12 2x a W a W 300 ( 12.600 9.250) 10 6555 x 10 cm− −= + = + × = (16a)

and 5 5

2 21 1 22 2x a W a W 300 ( 12.600 9.250) 10 6555x 10 cm− −= + = + × = (16b) Substituting these values into Equation (15), the Rayleigh-Ritz equation can be written as

2 41 2 5

2 W x 1 980ω g g 1.495 102 W x x 6555 10−

= = = = × × (17)

Therefore, the Rayleigh-Ritz equation to the first critical speed of the shaft is

1ω 122.3 rad / sec= (18) Note that the Rayleigh-Ritz equation to the first critical speed of the shaft is the same as the exact answer, see Eq. (9). In general, the Rayleigh-Ritz equation will give a slightly greater value than the exact answer; i.e., the Rayleigh-Ritz equation will give an upper bound.

19

Solution to Problem 4. From Table A-5, see page 1023, Poisson’s ratio and the modulus of elasticity of carbon steel

gears, respectively, are

v = 0.292 and E = 30 Mpsi (1a) The length (or face width) and the diameters of the two cylindrical profiles are

l = 0.75 in, d1 = 2 x 0.47 = 0.94 in, and d

2 = 2 x 0.62 = 1.24 in (1b)

The maximum pressure can be written from Eq. (3-74), see page 148, as

max2 Fp

b lπ =

(2)

The half-width of the contact patch for two cylinders can be written from Eq. (3-73), see page 148, as

2 21

1

1

2

2

2

(1 )

1

(1 )

12 Eb

d

ElF

dπ

− ν − ν+ = +

(3a)

Substituting Eqs. (1a) and (1b) and the transmitted load 40 lbF = into Eq. (3a), the half-width of the contact patch is

1/22 632 x 40 2(1 0.292 / (30 x10 ) 1.052 10 in

x 0.75 1/ 0.94 1/1.24)b

π− −= = × +

(3b)

Then substituting 40 lbF = and Eqs. (1b) and (3b) into Eq. (2), the maximum pressure is

max 32 40 32275 psi 32.275 kpsi

1.052 10 0.75p

π −

×= = =× × ×

(4)

The maximum shear stress can be written from Eq. (3-16), see page 109, repeated as Eq. (3-

72), see page 147, as 1 3

max 2σ στ −= (5a)

Assume here that the maximum principal normal stress 1 yσ σ= , then from page 148 the ratio

0.436zb

≥ (5b) Therefore, the maximum principal normal stress can be written from Eq. (3-76), see page 148, as

20

2

1 max 2

1 22

1y

zzbpbz

b

σ σ

+ = = − − +

(6a)

The minimum principal normal stress can be written from Eq. (3-77), see page 148, as

m3 2

ax

1Z

zb

pσ σ= = − +

(6b)

The maximum and minimum principal normal stresses, see Eqs. (6a) and (6b), depend on the maximum pressure and width of the contact patch which depend on Poisson’s ratio, see Eq. (3a). Check: The maximum shear stress from Figure 3-40, see page 149, can be approximated as

max max0.30 pτ = (7a) Substituting Eq. (4) into Eq. (7a), the maximum shear stress is

max 0.3 32275 9682.5 psi 9.6825 kpsi (cw)τ = × = = (7b) Note that the maximum shear stress according to Figure 3-40, see page 149, occurs at the depth

0.786z b= (8) Substituting Eqs. (4) and (8) into Eq. (5a), the maximum principal normal stress can be written as

( )( )

2

1 2

1 2 0.78632275 2(0.786) psi

1 0.786σ

+ = − − +

(9a)

Therefore, the maximum principal normal stress is

12.23559232275 1.572 5991.6 psi1.271926

σ = − − = − (9b)

Substituting Eqs. (4) and (8) into Eq. (5b), the minimum principal normal stress is

3 2

32275 32275 25374.9 psi1.2719261 (0.786)

σ = − = − = −+

(10)

Substituting Eqs. (9b) and (10) into Eq. (5a), the maximum shear stress is

max5991.6 ( 25374.9) 9691.65 psi 9.69 kpsi

2τ − − −= = = (11)

which is in good agreement with Eq. (7b).

21

Solution to Problem 5. Part I. (i) The radial clearance, see Example 12-5, page 652, is

d db j 55.05 55 0.025 mm22c −= =

−= (1)

The viscosity of SAE 30 lubricant at 40 oC , from Figure 12-3, see page 629, is

377.5 x10 Pa.sμ −= (2)

The nominal bearing pressure, (that is, the load per unit of projected bearing area), see page 630, and Example 12-1, page 645, can be written as

2W

Pr l

= (3a)

Substituting W 2.5 kN,= 27.5 mm,r = and 55 mml = into Eq. (3a), and rearranging, the nominal bearing pressure is

25000.82645 MPa3 32 27.5 10 55 10

P = =− −× × × × (3b)

The Sommerfeld number can be written from Eq. (12-10), see page 632, as

2r N

Sc P

μ=

(4a)

Substituting 27.5 mm,r = N 50 rev/s,= and Eqs. (1) and (3b) into Eq. (4a), and rearranging, the Sommerfeld number is

2 327.5 77.5 10 505.760.025 0.82645 10

S−× ×

= × =×

(4b)

(ii) For l/d = 1, the minimum film thickness variable can be obtained from Figure 12-15, see page 641. From the right-hand side of the y-axis of this figure, the eccentricity ratio is

0.018ec

ε = = (5a)

Therefore, the eccentricity is 30.018x0.025 0.45x10 mme −= = (5b)

Check: The eccentricity can be written from page 633 as

oe c h= − (6a) The left-hand side of the y-axis of Figure 12-15, see page 641, gives

22

0.982ohc

= (6b)

Substituting Eq. (1) into Eq. (6a), and rearranging, the minimum film thickness is

0.982 0.025 0.02455 mmoh = × = (7) Therefore, the eccentricity is

30.025 0.02455 0.00045 mm 0.45x10 mmoe c h −= − = − = = (8) Note that this answer agrees with Eq. (5b). (iii) The coefficient of friction variable can be written from Figure 12-17, see page 642, as

110r

fc

= (9)

Also, the coefficient of friction variable can be written from Eq. (12-11), see page 632, as

22r

f Sc

π= (10a)

Substituting Eq. (4b) into Eq. (10), and rearranging, the coefficient of friction variable is

2 22 2 (5.7) 112.51r

f Sc

π π= = = (10b)

Note that this answer is based on the Petroff equation and is not as accurate as Eq. (9). From the given journal diameter and Eq. (1), the radial clearance ratio is

27.5 11000.025

rc

= = (11)

Substituting Eq. (11) into Eq. (9), and rearranging, the coefficient of friction is

1100.1

1100f = = (12)

Part II. Substituting W 5000 N= and 110 mml = into Eq. (3a), the nominal bearing pressure is

50000.82645 MPa3 32 27.5 10 110 10

P = =− −× × × × (13)

Since the nominal bearing pressure is the same as Eq. (3b) then the Sommerfeld number is as given by Eq. (4b), that is

5.7S =

23

To determine the position of the minimum film thickness 𝜙, we use the design chart, Figure 12-16, see page 642. However, the specified value of / 2l d = does not appear on this chart, therefore, we must use the Raimondi and Boyd interpolation formula.

The desired value, in the interval / 1/ 4l d∞ > > , can be written from Eq. (12-20), see page 649, as

1 1 132 4

1 1 1 1 1[ (1 )(1 2 )(1 4 ) (1 2 )(1 4 ) (1 )(1 4 ) (1 )(1 2 ) ]8 3 4 24l

d

l l l l l l l l ly y y y yd d d d d d d d dl

d

∞= − − − − + − − − − − + − −

(14) The position of minimum film thickness 𝜙, from Figure 12-16, see page 642, for the Sommerfeld number S = 5.7 is 𝑙/𝑑 ∞ 1 0.5 0.25 𝜙 70° 84° 83° 70° Substituting these values and 𝑙/𝑑 = 2 into Eq. (14), the position of the minimum film thickness can be written as 𝜙 = 1(2) −18 (1 − 2)(1 − 4)(1 − 8)70° + 13 (1 − 4)(1 − 8)84° − 14 (1 − 2)(1 − 8)83°+ 124 (1 − 2)(1 − 4)70° (15a) or as 𝜙 = × 21 × 70° + × 21 × 84° − × 7 × 83° − × 21 × 70° = 183.75° + 588° −145.25° − 8.75° (15b) or as 𝜙 = . ° (15c) Therefore, the position of the minimum film thickness is 𝜙 = 79.41° (16)

24

Solution to Problem 6. (i) The transmitted (or tangential) load can be written from Eq. (13-35), see page 712, as

33000tt

HWV

= (1)

The pitch line velocity can be written from Eq. (13-34), see page 711, as

ft/min12 12

G G p pt

d nd nVππ= = (2)

The gear ratio (or speed ratio), see page 699 and Eq. (14-22), see page 759, can be written as

G GG

P P

N dmN d

= = (3)

Substituting the given gear ratio 2Gm = and pitch diameter of the pinion 5 inPd = into Eq. (3), and rearranging, the pitch diameter of the gear is

2x5 10 inchesG G Pd m d= = = (4)

Substituting Eq. (4) and the speed of the gear 500 rpmGn = into Eq. (2), the pitch line velocity is

10 500 1309 ft/ min12tV π × ×= = (5)

Substituting Eq. (5) and the given horsepower 6 hpH = into Eq. (1), the transmitted load is

33000 6 151.26 lb1309tW ×= = (6)

Check: The horsepower input can be written from Eq. (13-33), see page 711, as

63000GinH

nT= (7)

Rearranging Eq. (7), the input torque can be written as

63000G

inHT

n= (8)

Substituting the given horsepower input 6 hpH = and the speed of the gear 500 rpmGn = into Eq. (8), the input torque is

63000 x 6 756 lbs-in500inT = = (9)

25

Check: The input torque can be written from Eq. (b), see page 711, and Figure 13-29, see page 711, as

Gin tT r W= (10a)

Substituting Eq. (9) and the radius of the pitch circle of the gear 5 in,Gr = see Eq. (4), and into Eq. (10a), and rearranging, the transmitted load acting on the pinion is

756 151.2 lbs5tW = = (10b)

Note that this answer is in good agreement with Eq. (6). (ii) The AGMA bending stress in the gear can be written in US customary units from Eq. (14-15), see page 751, as

( )to v s m B

W P K K K K KF J

σ = (11)

The diametral pitch of the gear and the pinion can be written from Eq. (13-1), see page 684, as

G P

G P

N NPd d

= = (12)

Substituting Eq. (4) and the given diametral pitch 15 inchesP −= into Eq. (12), and rearranging, the number of teeth on the gear is

5 10 50G GN P d= = × = (13a) and the number of teeth on the pinion is

5 5 25p pN P d= = × = (13b)

Since the loads are applied at the tips of the teeth and the number of teeth on the gear is 50,GN = see Eq. (13a), then the bending strength geometry factor for the gear teeth from Figure 14-6, see page 759, is

0.28J = (14)

The problem states that the input gear is subjected to a light shock load and the output pinion is subjected to a moderate shock load, therefore, the overload factor for the gearset from the table at the bottom of page 772, see Figure 14-17, is

1.5oK = (15a) The dynamic factor for the gear, see Section 14.7, page 763, is given in the problem statement as

1.25vK = (15b)

The size factor for the gear, see Eq. (a), page 765, is given in the problem statement as

1.0sK = (15c)

26

The load distribution factor for the gear, see Eq. (14-30), page 765, is given in the problem statement as

mK = 1.0 (15d)

The backup ratio for the gear is given as Bm 1.5,= therefore, the rim-thickness factor for the gear from Eq. (14-40), see page 770, is

BK = 1.0 (15e)

Substituting the given face width of the gear F 0.8 inches,= diametral pitch 15 inchesP −= and Eqs. (6), (14), and (15) into Eq. (11), the AGMA bending stress in the gear can be written as

151.26 5 (1.5 1.25 1.0 1.0 1.0) kpsi0.8 0.28

σ ×= × × × × ××

(16a)

Therefore, the AGMA bending stress in the gear is

26.33 kpsi = 6331 lbs/inσ = (16b) (iii) The AGMA safety factor guarding against bending fatigue failure for the gear can be written from Eq. (14-41), see page 771, as

/ ( )t N T RF

S Y K KSσ

= (17) First, determine the bending strength geometry factor for the gear. For nitrided through-hardened Grade 2 steel the repeatedly applied bending strength (for 810 load cycles and 95 percent reliability) from Figure 14-3, see page 753, is

2B108.6H 15890 lbs/instS = + (18a)

Substituting the specified Brinell hardness BH 300= into Eq. (18a), the repeatedly applied bending strength for the gear is

248470 lbs/ins 48.47 kpsitS = = (18b) Second, determine the modification factors for the gear. For commercial gears (nitrided through-hardened Grade 2 steel), the repeatedly applied bending strength stress-cycle factor (for load cycles equal to, or greater than, 810 ) from Figure 14-14, see page 769, is

0.0178nY 1.3558 N−= (19a)

Substituting the number of load cycles specified in the problem statement, that is, 810 cycles,N = into Eq. (19a), the bending strength stress-cycle factor for the gear is

8 0.0178nY 1.3558 (1x10 ) 0.9768−= = (19b)

Note the procedure of obtaining the bending strength stress-cycle factor from Figure 14.14, see page 769, is not acceptable since a scale is not provided.

27

Since the gear is operating at room temperature then the temperature factor is

KT = 1 (20)

The reliability factor for the specified 95 percent reliability can be written from Eq. (14-38), see page 769, is

0.658 0.0759ln (1 )RK R= − − (21a) Substituting the reliability specified in the problem statement, that is 0.95,R = into Eq. (21a), the reliability factor is

0.658 0.0759ln (1 0.95) 0.8854RK = − − = (21b) Finally, determine the AGMA safety factor guarding against bending fatigue failure for the gear. Substituting Eqs. (16b), (18b), (19b), (20), and (21b) into Eq. (17), the AGMA safety factor guarding against bending fatigue failure for the gear can be written as

1 48.47 0.97686.33 1.0 0.8854FS

×= × (22)

Therefore, the AGMA safety factor guarding against bending fatigue failure for the gear is

8.5FS = (23)