3. Table 1. Technical Design Requirements of the Gearbox Assembly
Length (in) ≤ 5.0 5.0 By Design 0 %
Height (in) ≤ 10.0 9.4 By Design 6 %
Input Length (in) 2.0 2.0 By Design 0 %
Output Length (in) 4.0 4.0 By Design 0 %
Input/Output Length (in) 4.0 4.0 By Design 0 %
Input and Output Coupling Diameter (in) ≤ 1.5 1.255 By Analysis 16.3 %
Four ¼28 UNF Mounting Bolts Comply Comply By Design N/A
Speed (rpm) 650 Comply By Design N/A
Operating Torque (inlb) 1500 Comply By Design N/A
Limit Torque Static Yield (inlb) 3500 Comply By Design N/A
Ultimate Torque yield ok, but no fracture (inlb) 5250 Comply By Design N/A
1 Life Cycle (106
revs) 40 Infinite By Analysis N/A
Fatigue Life Requirement 6 N/A N/A *
Unit Must be Sealed Comply Comply By Design N/A
Breakout Torque (inlb) < 10.0 8.727 By Design 12.73 %
Temperature Comply Comply By Design N/A
Surface Wear Comply Comply By Design N/A
Vibration Comply No By Design **
Corrosion Resistant Comply Comply By Design N/A
*This value was not necessary to calculate due to the assumption of infinite life. Infinite life is
discussed in Section C of shaft analysis.
3
6. 2 Output
Housing
(Blue)
Protect gears from damage
and keeps lubricant inside,
contains bore hole for output
shaft
Has a length of 5
in, height of 9.4
in, and width of
1.6 in.
Aligned to left
housing using
dowel pins, and
fastened with six
screws
Fracturing,
deformation,
loose
connection from
screws or dowel
pins
3 Input Shaft
(Yellow)
Supplies initial drive to
system from hydraulic
power control unit
Has a small
diameter of
1.245in, a large
diameter of
1.866in, and a
length of 5in.
Shaft inputted
into the input
housing and held
in place with the
bearings and
retaining rings.
Fracturing and
bending.
4 Output Shaft
(Gray)
Accepts drive from the input
shaft and transmits power to
the rotary actuators of the
wing.
Has a small
diameter of
1.245in, a large
diameter of 1.866,
and a length of
3.85in.
Aligned parallel
to the input shaft,
and is held in
place in the right
housing using
bearings and
retaining rings.
Fracturing and
bending.
5 Gear (x2)
(Green)
Transfers direction of power
from the input shaft to the
output shaft
Has 16 teeth, a
pressure angle of
20o
, pitch
diameter of 4in
and width
0.995in, and tooth
height of
0.5625in.
Held onto Input
or Output Shaft
by Retaining
Rings and a key.
Improper
alignment, tooth
bending, and
surface wear.
6 SKF CRL 10A
Bearings (x4)
[2]
(Dark Blue)
Keeps shaft stationary, and
prohibits movement between
the housing and the shafts
(both input and output shaft)
Inner diameter of
1.25in, outer
diameter of
2.75in, and bore
of 0.6875in.
In the housing,
contains either
input or output
shaft.
Loose fit, tight
fit,
misalignment,
lubricant failure,
fracturing
corrosion
7 Gasket
(Black)
Used as a sealant to
pressurize the gearbox, keep
lubrication inside the
assembly, and keep debris
out of the assembly.
Made to fit the
walls of both the
input and output
housings.
Press fitted
between the
input and output
housing.
Improper sizing
or fit in between
the two
housings.
8 Key (x2)
(Orange)
Holds the gear to the shaft
(one for input shaft and one
for output shaft).
Height and width
of 0.345in and a
length of 1 in
Held in place by
the shaft keyway
and gear keyway.
Improper fit to
keyway of shaft
or gear.
9 ORing (x2) Seals the input and output
shaft. Keeps lubrication
inside the gearbox and keeps
debris from entering the
Thickness of
0.01in, and
diameter of
1.247in.
Press fitted
between the
coupling hole of
either housing
Improper sizing
or fit between
the shaft and
6
13. from a larger diameter to a smaller diameter, where stress will concentrate. Meanwhile, the
keyway also has a change in geometry, plus a higher moment than the shoulder, thus creating a
critical stress point. To find the factors of safety for yielding, stress must first be calculated using
the equation below [2].
( ) + ) ]σ ′ = [ πd3
k 32(M +M )t a m 2
( πd3
k 16(T +T )fs a m 2
1/2
(1)
is the stress, kf is the fatigue concentration factor, kfs is the fatigue concentration factor forσa
shear, Ma is the alternating moment, Mm is the midrange moment, Ta is the alternating torque, Tm
is the midrange torque and d is the diameter of the shaft. It is important to note that the diameters
for the shoulder and keyway had been estimated in a previous iteration, and for simplicity, were
used for this calculation. Because these diameters are related through a ratio of 1.5, the diameter
value used for this calculation will not affect the determination of the lower factor of safety.
Thus
Kf and Kfs are found using the equation below [2].
(k )Kf = 1 + q t − 1 (k )Kfs = 1 + qsheer ts − 1 (2)
where, q and qshear are the notch sensitivities which are obtained from Figure 620 and Figure
621[2]
, and kt and kts are the stress concentration factors which are obtained from Figure A159
and Figure A158. Factor of safety is then found by using the equation below [2].
n = σ′
Sy
(3)
where n is the factor of safety, Sy is the yield strength of 4340 normalized steel (103 kpsi), and σ′
is the stress.
The factor of safety found for the keyway was found to be 2.81, which was much less
than the factor of safety for the shoulder, 2.96 as observed below in Table 6. Therefore the
keyway was considered as the critical stress point of the shaft.This can be compared to the FEA
analysis seen below in Figure 6, which demonstrates analysis on the input shaft. The results of
this FEA yield similar results to the data calculated analytically above.
13
15. (psi)σ ′ 36595.63 34698.89 24397 32132.6
Sy (psi) 102976.79 102976.79 102976.79 102976.79
n 2.81 2.96 4.22 4.45
Although the yielding factor of safety is very similar between the shoulder and the keyway the
factor of safety for the keyway was always slightly smaller. Since the factor of safety was
always slightly smaller for the keyway the keyway was used as the critical stress point for further
calculations.
B. Determining the Minimum Diameter of the Shaft:
The minimum diameter of the shaft was determined using the ultimate torque and the
forces of the bearings A and B at the ultimate torque. The material selected for the shaft was
4340 normalized steel which has an ultimate strength of 161 kpsi, and a yielding strength of 103
kpsi. In order to determine the minimum diameter of the shaft the roadmap in Chapter 7 of
Shigley’s Mechanical Engineering Design was used [2].
Equation 1 was used to determine the test specimen endurance limit:
e 0.5SS′ = ut (4)
Where S’e is the test specimen endurance limit and Sut is the ultimate strength.
After S’e is determined, various factors must also be found in order to solve for the endurance
limit at the critical location of the shaft. The equations used to determine the factors ka, kb, kc, kd,
ke, and kf can be seen below.
Ska = a b
ut (5)
ka is the surface modification factor, a and b are factors used for machined equipment, for this
situation, a= 2.7, b= 0.265.
.879dkb = 0 −0.107
(6)
kb is the size modification factor, and d is the smaller diameter of the shaft. Initially the smaller
diameter is assumed in order to complete the process.
15
16. The loading factor (kc) is dependent on the load applied to the shaft. In this case the shaft was
subjected to a completely reversing bending stress, for which kc=1.
The temperature factor (kd) accounts for the effects of temperature variation on the shaft. In this
case the shaft must be able to survive being exposed to both extremely low and high
temperatures. The temperature requirement for the shaft was survival from 101.2 o
F to 230 o
F.
The equation to determine kd can be seen below.
(7).975 .432(10 )T .115(10 )T .104(10 )T .595(10 )Tkd = 0 + 0 −3
F − 0 −5 2
F + 0 −8 3
F − 0 −12 4
F
TF is the highest temperature (in this case 230 o
F). The hottest temperature was used because
yield strength drops as temperature increases. It is also known that materials are also more
susceptible to creep at higher operating temperatures.
The reliability of the shaft was assumed to be 95% resulting in a reliability factor (ke) of 0.868.
There were no miscellaneous effects to be considered so kf (the miscellaneouseffects
modification factor) can be considered to equal 1.
After all the factors were determined, the endurance limit at the critical location (the keyway)
could be calculated. The equation used to determine the endurance limit (Se) can be seen below.
k k k k k SSe = ka b c d e f ′e (8)
After the endurance limit is calculated, stress concentration factors must be determined before
the minimum diameter can be calculated. Since the critical stress point was found to be at the
keyway, the critical stress factors for torsion (kts) and bending (kt) were found using the geometry
of the keyway. For the first iteration of minimum diameter calculation, a kt value of 2.14 and kts
value of 3 were taken from Table 71.[2]
Using the notch radius of the keyway and the ultimate
strength of the shaft material, the notch sensitivity factors (q and qts) can be determined using
Figure 620 and 621 [2]. The values kt, kts, q, and qts can be used to solve for the fatigue stress
concentration factors (Kf and Kfs) using the Equation 2. Assuming a factor of safety (n) of 1.5,
16
17. the first trial can be done to determine the minimum shaft diameter, d. The equation used to
determine the minimum shaft diameter can be seen below:
[4( ) ( ) ( ) ( ) ] }d = { π
16n
Se
K Mf a 2
+ 3 Se
K Tfs a 2
+ 4 Sy
K Mf m 2
+ 3 Sy
K Tfs m 2 1/2 1/3
(9)
However, Ta and Mm are 0 reducing Equation 9 to:
[4( ) ( ) ] }d = { π
16n
Se
K Mf a 2
+ 3 Sy
K Tfs m 2 1/2 1/3
The minimum diameter found is then used to repeat the process in order to increase the accuracy
of the result. Each time the process was repeated the ratio of the large diameter and the small
diameter (D/d) was always kept at 1.5. After each trial, the minimum diameter that was found
was used in the next trial. Three trials were performed in order to determine an accurate value
for the minimum diameter of the shaft, and these results can be observed below in Table 7, with
a final value of 1.2303 inches for the diameter.
Table 7. Endurance Limit Variable Values for Each Trial
Trial 1 Trial 2 Trial 3
Sut(psi) 160991.89 160991.89 160991.89
Se’ (psi) 80495.95 80495.95 80495.95
ka 0.702 0.702 0.702
kb 0.879 0.860 0.860
kc 1 1 1
kd 1.02 1.02 1.02
ke 0.868 0.868 0.868
Se (psi) 443193.77 43224.26 43221.75
kt 2.14 2.14 2.14
kts 3.00 3.00 3.00
q 0.82 0.82 0.82
qs 0.83 0.83 0.83
kf 1.93 1.93 1.93
17
18. kfs 2.66 2.66 2.66
Assumed FOS 1.5 1.5 1.5
d (in) 1.23036 1.23103 1.2303
After this process was completed a shaft diameter of 1.245in was used in order to decrease the
shafts chance of failing. 1.245in was also chosen in order to find bearings which would closely
fit around the shaft.
C. Shaft Fatigue Factor of Safety
The number of lives that the shaft can withstand can be found by determining the factor
of safety for fatigue. If the value is greater than 1, than infinite life can be assumed. However if
the value is less than 1, finite life is assumed and further analysis can be done to find the number
of cycles to failure. The forces and moments associated with the operating torque were used for
this equation, since the shaft is most likely to experience operating torque a majority of the time.
This factor of safety is found using the equation below.
nf = Se
σa′
(10)
is the factor of safety for fatigue, Se is the endurance limit which is found from Equations 4,5,nf
6 ,7, and 8, and ’ is stress which is determined from Eq. 1 and 2. The factor of safety foundσa
from this equation is larger than 1, which concludes that the shaft has an infinite life at operating
torque.
Table 8. A Table of Parameters for the Factor of Safety for Yielding of the Shaft.
Se’(psi) 80495.95
ka 0.702
kb 0.859
kc 1
kd 1.02
ke 0.868
Se (psi) 40316.5
18
19. Ma(lbin) 213.465
kf 1.935
d (in) 1.235
’ (psi)σa 2234.51
nf 19.25
D. Shaft Deflection Analysis
In order to analyze the deflection and slope of the shaft from the applied loads, Equation
11 was used. The part of the shaft between the bearings was analyzed and assumed to be
symmetrical. The length between each bearing was 2.02 inches. The elastic modulus for AISI
4340 normalized steel is 29,732,736.22 psi. The mass moment of inertia was calculated for a cylinder
of diameter 1.25 inches.
(4x l) y = Fx2
48EI − 3 (11)
The variable x is the distance from one bearing to where the maximum deflection would occur.
This is at the middle of the section of the shaft being examined. The maximum deflection was
calculated to be 0.0000115 inches which is within the requirements as a spur gear can have a
maximum deflection of 0.010 in when the pitch diameter is 4 inches. By taking the derivative of
Equation 11 with respect to x, the slope of the spur gear can be calculated. Equation 12 is the
formula used and the maximum slope should occur in between the bearing and the point of
maximum deflection (x divided by 2). The resulting slope was 0.000034 radians.
L − ( )θ = − dx
dy
= L 4EI
Fx2
− 8EI
3FxL
(12)
The calculated slope is acceptable as the slope for an uncrowned spur gear must be less than
0.0005 radians as defined by Table 72 [2].
Table 9. A Table of Parameters for the Deflection of Analysis of the Shaft.
L (in) distance between the bearings 2.018
F (lbs) 955
x at max deflection (in) 1.009
19
20. x at max slope (in) 0.504
E (psi) 29732736.22
I 0.1198
y (in) 0.0000115
(rad)θ 0.000034
E. Key Factor of Safety
The factor of safety for yielding of the key was found through the following relation.
n
Ssy
= F
tL/2 (13)
Ssy is the shear yield strength, which is 57.7% of the yield strength of 4340 normalized steel. The
variable n is the factor of safety, F is the force from the applied torque, t is the thickness of the
key, and L/2 is the length of the key divided by two to account for the possibility of crushing.
The values used for this equation were from the limit torque and the operating torque.
Table 10. A Table of Parameters for the Factor of Safety for Yielding of the Shaft.
Ssy (ksi) 59417
F (at ultimate torque, lbf) 5668
F(at limiting torque, lbf) 3242
t (in) 0.810
L(in) 0.346
n (at ultimate torque) 1.47
n (at yielding torque) 2.57
Table 11. A Table of Parameters for the Shaft
Parameter Requirement Estimated
Capability
Basis of Estimate Margin of
Difference (%)
Factor of Safety
for Yielding at
Ultimate Torque
1> 2.81 By Analysis 181%
20
21. Factor of Safety
for Yielding at
Limiting Torque
1> 4.22 By Analysis 322%
Diameter Size (in) <1.5 1.25 By Analysis 16.7%
Fatigue Life
Factor of Safety
(lives)
6 Infinite By Analysis N/A
Life cycle 40 N/A By Analysis *
Deflection (in) <0.003 0.0000115 By Analysis 99.62%
Deflection
(radians)
<0.0005 0.00034 By Analysis 32%
Key Factor of
Safety at Ultimate
>1 1.47 By Analysis 47%
Key Factor of
Safety at Limiting
.1 2.57 By Analysis 157%
Gear (Trent)
Applied Loads
Using the ultimate torque (T=5250 inlb), the gear speed ( 650 rpm) and Equation 14,ω =
the power required of the shaft was found to be about 54 hp which must be transmitted through
the gear.
H = Tω
63025 (14)
Equation 14 was then used to find the tangential force applied to the gear which was 2625 lbf.
The pitchline velocity, V, was calculated using Equation 15 where n is the gear speed and d is
the diametral pitch of the gear.
V = 12
πdn
(15)
Wt
= V
33000H
(16)
A 20° degree pressure angle was assumed as it is a common pressure angle for commercial spur
gears. Thus, the radial force was found to be about 955 lbf and the resultant force was about
21
23. The factor of safety for spur gear bending was found using Equation 18. The material
chosen for the gear was a grade 3 carburized and hardened steel.
SF = σ
S Y /K Kt N t R
(18)
The bending stress was calculated first using several variables and assumptions. The geometry
factor, J, was found to be 0.27 from Figure 146 with 16 teeth in the gear design.[2]
The overload
factor, , was assumed to be 1 as the first iteration for the gear design was calculated at theKo
ultimate torque so an overload adjustment was not necessary. The dynamic factor, , wasKv
found using Equation 19 where Equation 20 was used to solve for variables A and B with an
assumption of 8 for the quality factor, .Qv
Kv = ( )A
A+√V
B
(19)
0 6(1 )A = 5 + 5 − B
.25(12 )B = 0 − Qv
2/3
(20)
The size factor, , was found using Equation 21 where the Lewis form factor, Y, was 0.27 forKs
16 teeth, F was the gear width of 1 in and P was the pitch diameter of 4 teeth per inch.
.192( )Ks = 1 P
F√Y
0.0535
(21)
The rimthickness factor, , was assumed to be 1 as the gear design did not have a rim. TheKB
load distribution factor, , was found using Equation 22.Km
(C C C )Km = 1 + Cmc pf pm + Cma e (22)
The gear has uncrowned teeth so was 1. was calculated to be 0.00525 using the gearCmc Cpf
width of 0.79 in and the diameter pitch of 4 in with Equation 23. was assumed to be 1.1 asCpm
the value is half the gear width making the total distance between the two bearings on eachS1
shaft about 2.83 inches; however, the shaft length will be less than 2 inches, thus, making the
value 1.1. Equation 24 and Table 149 was used to calculate . The condition assumedCpm Cma
for Table 149 was that it was a commercial, enclosed unit. was assumed to be 1 “for all otherCe
conditions” [2].
.025Cpf = F
10dp
− 0 (23)
F FCma = A + B + C 2
(24)
23
24. The bending stress that the gear experiences under the ultimate torque is about 100 ksi, 67 ksi at
the limit torque and 29 ksi for operating torque. Equation 25 was used to do this.
K K Kσ = Wt
o v s F
Pd
J
K Km B
(25)
Figure 1414 was used to find the stress cycle factor, , where N was 240 millionY N
revolutions. The temperature factor, , was designated as 1 due to the design parameterKT
survival temperatures between 74°C to 110°C, which is below 120°C. The reliability factor, KR
was calculated using Equation 26, assuming a 99% reliability in order to use Table 143 for steel
gears [2].
.5 .109ln(1 )KR = 0 − 0 − R (26)
The bending stress number, , was found to be 75,000 psi for a steel grade 3 carburized andSt
hardened material from Table 143. The bending factor of safety for the gear under the ultimate
torque was calculated to yield at a value of 1.16. At the limit torque, the factor of safety was
calculated to be 1.74. At operating torque the factor of safety for bending was calculated to be
4.06. The gear will survive the operating torque and can withstand the ultimate torque without
yielding.
C. Surface Analysis:
For the spur gear wear analysis, the gear contact stress was calculated using Equation 27.
σc = Cp
√W K K Kt
o v s
Km
d Fp I
Cf
(27)
The elastic coefficient, , was found to be 2300 from Table 148 where the gear andCp √psi
pinion were made of steel.[2]
The overload factor , the dynamic factor , the size factorK )( o K )( v
and the load distribution factor were the same values as the bending analysis. wasK )( s K )( m Cf
assumed to be 1 as given by the roadmap on page 759 [2]. Equation 26 for an external gear was
used to calculate the surfacestrength geometry factor, I, where the pressure angle was 20° and
the speed ratio, , and the loadsharing ratio, , were 1.mG mN
I = 2mN
cosϕsinϕ mG
m +1G
(28)
24
25. The gear contact stress under the ultimate torque, limit torque and operating torque was
calculated to be 251 ksi, 205 ksi and 134 ksi respectively.
For the other variables in the spur gear wear factor of safety equation, Table 146 was
used to find the repeated applied contact strength, , at 275,000 psi for a steel grade 3Sc
carburized and hardened material. The stress cycle factor for wear, , was determined usingZn
Figure 1415 with 240 million revolutions.[2]
The hardnessratio factor, was assumed to be 1,CH
as the same material was used for both of the gears. The reliability factor, , was determinedKR
using Equation 26 with an assumed 99% reliability. The temperature factor, , was the same asKT
the spur gear bending value. The spur gear wear factor of safety at the ultimate torque, limit
torque and operating torque were 1.01, 1.51 and 3.52 respectively. The factors of safety for
bending are higher than those of the surface wear, so when the gearbox undergoes the ultimate
torque, the surface will have pits and wear before the gear yields. The material for the gear, steel
grade 3 carburized and hardened, was chosen for its high bending and surface stress number. The
gears will be sealed within the gearbox containing a lubricant. As long as the gearbox is correctly
sealed, the gears will not be susceptible to corrosion. Further comparisons of the factors of safety
can be found in Table 3 and indepth calculations can be found in the appendix.
Table 13. A table of parameters for the gear.
Parameter Requirement Estimated
Capability
Basis of
Estimate
Margin of
Difference (%)
Factor of Safety
for bending at
ultimate torque
(5250 inlb)
>1.00 1.16 By analysis 15.9 %
Factor of safety
for bending at
limit torque
(3500 inlb)
>1.00 1.75 By analysis 73.8 %
Factor of safety
for bending at
operating torque
(1500 inlb)
>1.00 4.07 By analysis 306 %
25
26. Factor of safety
for wear at
ultimate torque
(5250 inlb)
>1.00 1.01 By analysis 0.90 %
Factor of safety
for wear at limit
torque
(3500 inlb)
>1.00 1.52 By analysis 51.4 %
Factor of safety
for wear at
operating torque
(1500 inlb)
>1.00 3.53 By analysis 252 %
Sealants, Lubricants, Bearings (Charlton)
A. Bearings:
The selected bearing for the gearbox was an SKF CRL 10A cylindrical roller bearing.
Based on Equation 29 below, the desired life hours for the bearing was calculated to be roughly
6154 hours.
(N)( ) LD = 1
ω
1 hr
60 min (29)
A bore diameter for the bearing was specified based on the diameter of shaft, which was 1.25
inches. The desired rating was then calculated using Equation 30. These factors as well asC10
the operating speed of 650 rpms was taken into account when selecting a bearing. Dimensions of
the bearing can be found in Table 14.
( )C10 = FD L n 60R R
L n 60D D
1/a
30)(
Table 14. Key features of the SKF CRL 10A roller bearing
Face width (in) 0.6785
Bore diameter (in) 1.25
Outer diameter (in) 2.75
Operating speed, (rpm)Nr 19,000
rating (lbf)C10 10,000
The reliability of the bearing was then calculated using Equation 31, which was found to be 0.99.
26
27.
xp(− ) )R = e ( θ−x0
x ( ) −xD C10
a Ff D a
0 b
(31)
B. Sealants:
Rubber Gasket
The housing was sealed using a hightemperature silicone rubber gasket material found
on McMasterCarr’s website. The gasket was shaped on Solidworks to fit the housing, while also
being careful to include holes for the screws and dowels were properly aligned. This material has
an operating temperature range from 80 degrees Fahrenheit to 450 degrees Fahrenheit and can
also withstand pressures up to 1000 psi. The breakout torque for the seal was 8.727 lb in and∙
was calculated using the following equation:
65 diameter) rpm) T = . * ( 2
* ( 1/3
(32)
Figure 7: HighTemperature Silicone Rubber Gasket
ORing
For the shaft, an oring seal was used because of its many advantages. They are very easy
to maintain because there are no bolts to tighten or adjust. They also are advantageous as they do
not require any additional adhesives or alternative application methods. In the housing, the
diameter for the hole where both the input and output shafts are located is 1.25 inches while both
the shafts themselves have 1.24 inch diameters at that same point. This meant that the oring had
to have an outer diameter slightly larger than 1.25 inches so that it could be pressfit into the
housing. For this report and design, a McMasterCarr’s Extreme Chemical
27
35. 625 lbf an(20 ) 55 lbfWr
= 2 * t ° = 9
Yielding Factor of Safety for Shaft Analysis:
At the Keyway of the Shaft:
Eq. 2: = =1.93(k )Kf = 1 + q t − 1 .82(2.14 )1 + 0 − 1
= =2.66(k )Kfs = 1 + qsheer ts − 1 .83(3 )1 + 0 − 1
Eq. 1: =( ) + ) ]σa′ = [ πd3
k 32(M +M )t a m 2
( πd3
k 16(T +T )fs a m 2
1/2
( ) + ) ][ π 1.235in*
3
1.93 32(747in−lbs)* 2
( π 1.235in*
3
2.66 16(5250in−lbs)* 2 1/2
=36.5kpsi
Eq. 3: n = σ′
Sy
= = 2.81103kpsi
36.5kpsi
At the Shoulder of the Shaft:
Eq. 2: = =1.57(k )Kf = 1 + q t − 1 .82(1.69 )1 + 0 − 1
= =1.40(k )Kfs = 1 + qsheer ts − 1 .83(1.48 )1 + 0 − 1
Eq. 1: =( ) + ) ]σa′ = [ πd3
k 32(M +M )t a m 2
( πd3
k 16(T +T )fs a m 2
1/2
( ) + ) ][ π 1.235in*
3
1.57 32(747in−lbs)* 2
( π 1.235in*
3
1.40 16(5250in−lbs)* 2 1/2
=34.7kpsi
Eq. 3: n = σ′
Sy
= = 2.96103kpsi
34.7kpsi
Determining the Minimum Diameter of the Shaft:
These sample calculations show the process using the numbers from the first trial where d was
assumed to be 1in and D was assumed to be 1.5in.
Eq. 4: =0.5*161kpsi = 80.5kpsie 0.5SS′ = ut
Eq. 5: =2.7* (0.265)
= 0.702Ska = a b
ut 161kpsi)(
Eq. 6: = =0.879.879dkb = 0 −0.107
.879 in0 * 1 −0.107
kc=1 (for bending)
Eq. 7: .975 .432(10 )T .115(10 )T .104(10 )T .595(10 )Tkd = 0 + 0 −3
F − 0 −5 2
F + 0 −8 3
F − 0 −12 4
F
= =1.02.975 .432(10 )230 .115(10 )230 .104(10 )230 .595(10 )2300 + 0 −3
− 0 −5
2
+ 0 −8
3
− 0 −12
4
ke=0.868 (for a reliability of 95%)
kf=1 (there were no miscellaneous factors to be considered)
Eq. 8: = = 44.2kpsik k k k k SSe = ka b c d e f ′e .702 .879 .02 .868 0.5kpsi0 * 0 * 1 * 1 * 0 * 1 * 8
35
36. Kf = 1.93 (found previously in yielding factor of safety section)
Kfs = 2.66 (found previously in yielding factor of safety section)
n = 1.5 (was used as the minimum factor of safety)
Eq. 9: =[4( ) ( ) ] }d = { π
16n
Se
K Mf a 2
+ 3 Sy
K Tfs m 2 1/2 1/3
= 1.23in[4( ) ( ) ] }{ π
16 1.5*
44.2kpsi
1.93 747in−lbs* 2
+ 3 103kpsi
2.66 5250in−lbs* 2 1/2 1/3
Shaft Fatigue Factor of Safety:
The values of the different factors and the endurance limit from the final trial of the minimum
shaft diameter calculations can also be used to determine the fatigue factor of safety.
Eq. 10: = = 26.7 (only bending stress was accounted for as specified in thenf = Se
σa′
43kpsi
1.6kpsi
requirements)
Shaft Deflection Analysis:
Eq. 11: = )= 1.15E(5)(4x l) y = Fx2
48EI − 3 (4 .0088in .0175in955lbf 1.0088in*
2
48 29.7Msi 0.119* * * 1 − 3 * 2
Eq. 12: − L )L .0175in( )θ = dx
dy
= − (4EI
Fx2
− 8EI
3Fxl = − 2 4 29.7Msi 0.119* *
955lbf 1.0088in*
2
− 8 29.7Msi 0.119* *
3 955lbf 1.0088in 2.0175in* * *
= 2.44E(4)rad
Key Factor of Safety Analysis:
Eq. 13: = = 1.8n = 2 F*
S t Lsy* *
2 5.6kip*
59.4kpsi 0.346in 1in* *
Spur Gear Bending Analysis
Figure 146 .37J = 0
Assume , , Ko = 1 KB = 1 T FKT = 1 < 250°
Eq. 1427 .21874Kv = ( )A
A+√V
B
= 1
Eq. 1428 0 6(1 ) 0 6(1 .629961) 0.7222A = 5 + 5 − B = 5 + 5 − 0 = 7
.25(12 ) .629961B = 0 − Qv
2/3
= 0
AssumingQv = 8
.192( ) .192( ) .0713Ks = 1 P
F√Y
0.0535
= 1 4 teeth/in
1 in√0.27
0.0535
= 1
36
37. where the Lewis Form Factor, Y=0.27
(C C C ) .15598Km = 1 + Cmc pf pm + Cma e = 1
where uncrowned teethCmc = 1
.025 Cpf = F
10dp
− 0 = 0
F F .127 .0158(1) .930(10 )(1) .142707Cma = A + B + C 2
= 0 + 0 − 0 −4 2
= 0
all other conditionsCe = 1
K K K 8016.88 psi 8 ksiσ = Wt
o v s F
Pd
J
K Km B
= 5 = 5
Assuming Steel carburized and hardened grade 3
40 0 )6 .4 0 revolutionsN = ( * 1 6
= 2 * 1 8
.6831(N) .902462Y N = 1 −0.0323
= 0
Assuming a reliability factor of 0.99
.5 .109ln(1 ) .002 KR = 0 − 0 − R = 1
S 5000 psit = 7
.1579 .16SF = σ
S Y /K Kt N t R
= 1 = 1
Spur gear wear analysis
Force at Operating Torque
35714.5 psi (14 6) σc = Cp
√W K K Kt
o v s
Km
d Fp I
Cf
= 1 − 1
where 3, 00 3, 00 50 lbfWt
= 3 0 V
H
= 3 0 15.4701 hp
680.678 ft/min = 7
300 Cp = 2 √psi
Cf = 1
.080348 (14 3) external gearsI = 2mN
cosϕsinϕ mG
m +1G
= 2(1)
cos20°sin20° 1
1+1 = 0 − 2
in F in dp = 4 = 1
.53 (14 2)SH = σc
S Z C /K Kc n H T R
= 3 − 4
where for carburized and hardened grade 3 steel75, 00 psiSc = 2 0
Figure 1415.4488(2.4 ) .92953Zn = 1 * 108 −0.023
= 0
as .2CH = 1 HBG
HBP
= 1 < 1
as T 50°FKT = 1 < 2
37
38. for 99% reliability.5 .109ln(1 ) .002 KR = 0 − 0 − R = 1
Bearing Reliability
(N)( ) (2.4 )( ) 153.85 hoursLD = 1
ω
1 hr
60 min = 1
650 * 108 1 hr
60 min = 6
( ) 55.42( ) 946.13 lbfC10 = FD L n 60R R
L n 60D D
1/a
= 9 106
6153.85 650 60* * 3/10
= 4
40xD = LR
L N 60D D
= 10 rev6
6153.85 hours 650 rpm 60* *
= 2
xp(− ) ) .99R = e ( θ−x0
x ( ) −xD C10
a Ff D a
0 b
= 0
where , , , , , .2af = 1 a = 3
10
0, 00 lbfC10 = 1 0 .459θ = 4 .02x0 = 0 .483b = 1
Sealants
65 diameter) rpm) 65 1.245 in) 650 rpm) .727 lb nT = . * ( 2
* ( 1/3
= . * ( 2
* ( 1/3
= 8 ∙ i
38