Design of an Intermediate Gearbox for the Trailing Edge Flaps in a High Lift System for the Airbus A350

“Gearing up for Senior Capstone”

Team #4 Mech 310 May 1, 2015

Trent Hosokawa (Gear Analysis, Bearing Selection, Deflection Analysis) Joey Miyamoto (Shaft Analysis, 2D Drawings, Shear/Moment Diagrams)

Sonja Feck (Shaft Analysis, 2D Drawings, Shaft CAD) Nina Lepp (Housing Design, CAD Assembly, FEA, Housing Analysis) Charlton Azuoma (Lubrication/Sealant Selection, Gasket Design, O­Ring

Assembly)

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ABSTRACT

This report studies a gearbox used for the Airbus A350, specifically for the trailing edge

flaps of a high lift system. The gearbox assembly is comprised of a sealed housing, an input and

an output shaft, two spur gears, and four roller bearings. The gears within the assembly have a

1:1 ratio and are used to simply change the direction of rotation. Analysis was conducted to

determine forces applied on the shaft from the spur gears as well as the reaction forces from the

bearings. These forces were found using an operating speed of 650 rpm, an operating torque of

1500 in­lb, a limit torque of 3500 in­lb and an ultimate torque of 5250 in­lb. The location of the

critical stress point was identified to be at the shoulder of the shaft and analysis was conducted at

this location, specifically in the keyway. Using Solidworks, these parts were designed and

assembled as seen below in Figure 1. The gearbox design met all specified parameters, including

requirements added throughout the design and analysis process.

DESIGN REQUIREMENTS

Through computational and theoretical analysis, the required parameters were determined

and displayed below in Table 1. In this table, color­coding was used to specify the status of a

requirement. Green shading is used for positive margins, yellow for margins near zero, and red

for negative margins. All values in Table 1 are green, indicating that the values all have a

positive margin which means all of the requirements were met. Some requirements had to be met

exactly with no variation, such as the input and output length. It can be seen in Table 1 that the

gearbox assembly complied with these requirements and they have a positive margin. The

overall length of the assembly is 5 in, and the height is 9.40 in. The input and output coupling

diameter is 1.25 in, which results in a margin of 0. The basis of estimate “By Design” specifies

that the value was found through computational and/or theoretical analysis, with the necessary

equations which can be seen in the Analysis section of the report. In similar fashion, values

found “By Analysis” were analyzed using the SolidWorks 3D design and program. All the

values below in Table 1 meet the minimum requirements, with the smallest margin being at 0%

for numerous requirements, and the largest margin of 16.3%, for the input and output coupling

diameter, which allows for more than enough space between the design and the requirement.

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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.

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**The Vibration Testing could not be completed, because the version of Solidworks used for the

simulation could not load properly to produce frequency results.

COMPONENT CONFIGURATION

Table 2. Gearbox Assembly Overview

Assembly Overview

General Description This assembly is a gearbox assembly for an A350 high lift actuation system. The gearbox is used to change the direction of the rotation of the shaft. The gearbox is sealed and pressurized in order to ensure survival at a range of pressures and temperatures. Torque transmitted to the assembly can potentially vary so it is important that the assembly is able to withstand high torque values.

Basic Operation The central hydraulic power control unit (PCU) of the A350 high lift actuation lift system supplies the necessary power to operate flap panels on each wing. Mechanical power is transmitted through the mechanical transmission shaft to the rotary actuators, which move the flaps on the tracks of the wing. The gearboxes in the system are necessary for larger direction changes for power transmission.[1]

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Figure 1. Exploded View of Gearbox Assembly with Primary Component Callouts

Table 3. Description of the Primary Gearbox Components

ID #

Primary Components

Primary Function Geometry Positioning Failure Modes

1 Input Housing (Red)

Protect gears from damage and keeps lubricant inside, contains bore hole for input shaft

Has a length of 5 in, height of 9.4 in, and width of 1.6 in.

Held as a fixed part in the Assembly.

Fracturing, deformation, loose connection from screws or dowel pins

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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

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assembly through the shaft coupling hole.

and the input and output shaft.

coupling hole of the housing.

ANALYSIS:

Housing/Assembly (Nina)

The housing was designed based on the gear housing from a similar mechanism. This

shape was chosen due to its ease of manufacturing, light weight, and strength to maintain the

shafts and gears under the ultimate torque of 5250 in­lb. The left housing and right housing have

nearly identical designs; however the left housing, which contains the input shaft, has a 4­ ¼ 28

UNF mounting bolt pattern, while the right housing has no holes in the side of the piece. With a

sealant, the two housings are connected using six screws with a size of ¼” diameter. In order to

keep the required pressure of 15 psi, the two housing parts are connected with a rubber gasket

sealant, and there is a 1.25” hole in the housing surrounding both the input and output shafts,

sealed by an o­ring. This can be seen above in Figure 1.

Many of the given requirements above left no room for variation and thus have a margin

of difference of zero. The gearbox housing has a width of exactly 5.00 inches and a height of

9.40 inches, both of which are less than or equal to the requirements of 5 and 10 inches

respectively. These dimensions allow for efficient use of materials, while also keeping true to the

requirements assigned by the customer, Airbus. The values for input length, output length, and

the length between the input and output shafts were all specific dimensions, and these values

were taken into consideration when conducting analysis, especially on the shaft design.

Therefore, these values can be observed in the SolidWorks design. The input and output coupling

diameter, assigned to be the diameter of the input and output shafts was found to be 1.25”, to

meet the requirements of design. On the left side of the housing that contains the input shaft,

there is a four bolt pattern, of ¼ inch diameter, with 28 threads/inch with a unified fine threading.

This pattern can be observed by design in Part #1 in the Bill of Materials above.

The Unit is sealed, both with a gasket between the left and right housing and also o­rings

between the shafts and housings in order to both keep the lubricant in the housing and also keep

foreign debris and other materials from entering the housing. The specified temperature

requirement is met by the materials selected and also the lubricants, seals, and gaskets, to ensure

that all materials will maintain their properties in the extended temperature range. The surface

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wear requirement of selecting a lubricant compatible with the seal was taken into account when

selecting the gasket, o­ring, and a lubricant.

Corrosion Resistance/ Material Selection

Category T requires that the housing should be resistant to corrosion that may occur when

the aircraft is parked or hovering near the sea.[2] Therefore, cast iron was selected as the material

for its resistance to salt corrosion. It also acts as a damper for vibrations and can save weight.

Analysis was conducted using the Solidworks simulation software to better understand

the effects of the force on the housing. It can be observed that the Factor of Safety is lowest

closest to the area where the force is applied.

Figure 2. FEA Analysis of Housing Factor of Safety

Table 4 below, corresponds to the image of the Assembly in Figure 1, to show the overall

assembly of the gearbox and also how each part is necessary for the system. It can be seen how

the torque on the input shaft is transferred to the output shaft through the gears, maintaining the

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revolutions per minute, and meeting all the specified requirements for both the housing part and

the entire assembly.

Table 4. Overall Housing Assembly Requirements

Parameter Requirement Estimated

Capability

Basis of

Estimate

Margin of

Difference (%)

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

Unit Must be Sealed 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 (See

Above)

By Test? **

Corrosion Resistant Comply Comply By Design N/A

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Shaft (Sonja/Joey)

Applied Loads

The input and output shafts were designed to be very similar, excluding the length of the

shafts. Due to this and the gear ratio of 1, the forces and moments found in the input shaft are the

same as those found in the output shaft in their respective locations, and consequently, analysis

of only the input shaft is presented below. Important loads required to perform analysis on the

shafts of the assembly were ultimate torque (5250 in­lb), limiting torque (3500 in­lb), and

operating torque (1500 in­lb). Important forces for shaft analysis were the reaction forces of both

bearings B and D, and the force applied to the gear W, which were found through analysis and

can be seen in Table 5. The ultimate torque was used to determine the minimum shaft diameter

to be used in the gearbox assembly. The ultimate torque was used to ensure that the shaft would

survive at all torques which it may be subjected to. Infinite life of the shaft was found using the

operating torque because the shaft will be subjected to this torque for the majority of its life. The

yielding factor of safety was found at the ultimate, limiting, and operating torque to ensure that

the shaft did not yield at any of these values.

Table 5. Applied Forces to the Shaft

Component Ultimate Torque (5250 in­lb)

Limiting Torque (3500 in­lb)

Operating Torque (1500 in­lb)

Bearing B (lbf) 1396.66 931.01 399.07

Bearing D (lbf) 1396.66 931.01 399.07

Gear (lbf) 2793.38 1862.32 798.14

Theoretical Approach and Results:

The results found through analysis of the shaft can be seen in Table 11 at the end of this section.

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A. Critical Stress Points and Factors of Safety for Yielding

Forces B, D, and W were broken into components, where B and D are in both the y and z

components and W is in both the radial and tangential components. These forces are represented

in Figure 3. The shear and moment diagrams were then found and can be seen below in Figures 4

and 5. Ultimate torque was considered for the shear and moment diagrams because this torque

will determine value of factors of safety for yielding and the location of the factors of safety

along the shaft regardless of the torque type. Limited torque is also calculated without shear and

moment diagrams.

Figure 3: Free Body Diagram of Shaft and Gear (Charlton)

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Figure 4. Shear Diagram of the Shaft at Ultimate Torque

Figure 5. Moment Diagram of the Shaft at Ultimate Torque

As found in the moment diagram, two potential critical stress points were discovered, one

at the shoulder and the other at the keyway. The shoulder has a change in geometry that goes

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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 ′ = [ πd3k 32(M +M )t a m 2

( πd3k 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 )K f = 1 + q t − 1 (k )K fs = 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.

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Figure 6. Factor of Safety FEA using Solidworks

Table 6. A Table of Parameters for the Factor of Safety for Yielding of the Shaft

Ultimate Torque (Keyway)

Ultimate Torque (Shoulder)

Yield Torque (Shoulder)

Yield Torque (Keyway)

kt 1.69 2.14 1.69 2.14

kts 1.48 3.00 1.48 3.00

q 0.82 0.50 0.82 0.50

qshear 0.83 0.20 0.83 0.20

kf 1.93 1.57 1.93 1.57

kf,s 2.66 1.40 2.66 1.40

Ma (lb­in) 747.19 537.69 875 358.435

Tm (in­lb) 5250 5250 3500 3500

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(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 but (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.

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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 oF to 230 oF.

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 −3F − 0

−5 2F + 0

−8 3F − 0

−12 4F

TF is the highest temperature (in this case 230 oF). 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,

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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 = π16n

SeKMf a 2 + 3 Se

K Tfs a 2 + 4 SyKMf 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 = π16n

SeKMf 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

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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 =Seσa′ (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

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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 = Fx248EI − 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 = − dxdy = L 4EI

Fx2 − 8EI3FxL (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

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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.

nSsy = 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%

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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 = Tω63025 (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 = 12πdn (15)

W t = V33000H (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

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2793 lbf. The same process was used to find the resultant force at the limit torque of 3500 in­lb

and the operating torque of 1500 in­lb. The operating torque forces were found to be 750 lbf for

the tangential force, 273 lbf for the radial force and 1838 lbf for the resultant force. The forces

were assumed to be applied at the center of the gear teeth due to their point of contact.

Theoretical Approach and Result

A. Gear Design:

In order to determine the minimum amount of teeth for the gear to avoid interference,

Equation 17 was used where full­depth teeth, a one­to­one gear ratio change, and a 20° pressure

angle were assumed.

(1 ) Np = 2k3 sin ϕ2 +√1 sin ϕ+ 3 2 (17)

The number of teeth was designated to be 16 as it was a common value found in the tables. The

input to output distance between the two shafts was required to be 4 in. so the diametral pitch of

the gear was designated to be 4 in. Table 12 shows the key features of the first iteration of the

gear that will be used later on.

Table 12. Key Features of the Gear Used in the Gearbox

Material Steel Grade 3 carburized and hardened

Pitch diameter, (teeth/in)P d 4

Pressure Angle, (degrees)ϕ 20

Diametral pitch, (in)dp 4

Number of teeth 16

Gear Speed , (rpm)ω 650

Pitch­line velocity, V (m/s) 680.678

Face Width (in) 0.995

B. Bending Analysis:

22

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 = σS Y /KKt 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 = ( )AA+√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 PF√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 = F10dp − 0 (23)

F FCma = A + B +C 2 (24)

23

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 = W to v s F

PdJ

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 Kto v s

Kmd 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 C f

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 = 2mNcosϕsinϕ mG

m +1G (28)

24

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

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 = 1ω

1 hr60 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 D1/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

xp(− ) )R = e ( θ−x0

x ( ) −xD C10

a Ff D a0 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

polytetrafluoroethylene (PTFE) #24 o­ring was used. These o­rings have exceptional temperature

resistance as it can withstand a range of ­100 degrees Fahrenheit to 500 degrees Fahrenheit. The

outer diameter of the o­ring is 1.254 inches and has a thickness of 0.070 inches. It can observed

in the assembly in Figure 1 that there is no specific o­ring part installed in the housing. This is

because of the properties of Solidworks and the tight tolerances that the o­ring contains. In the

true assembly, the o­ring would be press­fit into this space, allowing for a tight pressure to be

maintained, but also allowing for movement of the shafts and thus the gears.

Lubricant

Parker Super O­Lube is an all­purpose lubricant that operates at the required temperature

range of ­65.2 degrees Fahrenheit to 399.2 degrees Fahrenheit. Although it is an­purpose

lubricant, this material has the capacity to function as a high viscosity silicone oil, which makes

it a good choice as a seal lubricant. This lubricant is also compatible with the PTFE as well as

silicone, which is the material the o­ring and gasket sealant are made of, respectively. Below,

Table 15 shows the properties of these materials for the gasket, o­ring, and lubricant, and to

show how these properties meet the temperature range requirement.

Table 15. Properties of Materials to be used in Assembly

Item Brand Temperature Range

Rubber Gasket Sealant High­Temperature Rubber Gasket Sealant (McMaster­Carr)

­80 deg F to 450 deg F

O­Ring Extreme Chemical O­Ring #024 (McMaster­Carr)

­100 deg F to 500 deg F

Lubricant Parker Super O­Lube ­65.2 deg F to 399.2 deg F

2D DRAWINGS (Sonja/Joey)

28

Figure 8: Input Shaft Solidworks Drawing

29

Figure 9: Output Shaft Solidworks Drawing

30

Figure 10: Input Gearbox Solidworks Drawing

31

Figure 11: Output Gearbox Solidworks Drawing

32

Figure 12: Gear Solidworks Drawing

33

REFERENCES [1] Recksiek, M. (2009). Advanced high lift system architecture with distributed electrical flap

actuation. Proceedings of the 2nd International Workshop on Aircraft System Technologies, 49­59.

[2] Experimental Conditions and Test Procedures for Airborne Equipment. Thesis. RTCA D0­160F,

2007. N.p.: n.p., n.d. Print. [3] Budynas, Richard G., J. Keith. Nisbett, and Joseph Edward. Shigley. Shigley's Mechanical

Engineering Design. 10th ed. N.p.: n.p., 2014. Print. [4] ISSF. "The Salt Spray Test and Its Use in Ranking Stainless Steel." (2008): n. pag. Web. APPENDIX Sample Calculations: Power & Torque:

250 in bsτ = 5 ∙ l

50 rpmω = 6 = n

57, 56 lb n/sH = 3 3 ∙ i

e = product of driven tooth numbersproduct of driving tooth mumbers = 1

32Np =

teeth/inP d = 8

/P 2/8 indp = Np d = 3 = 4

/N /32 125m = d p = 4 = .

(m) 393pn = π = .

80.678 ft/minV = 12πdn = 12

π(4)(650) = 6

Force Analysis at Ultimate Torque:

(10)4.14 horsepowerH = Tω63025 = 63025

5250in∙lbs 650 rpm* = 5

80.678 feet/minV = 12πdn = 12

π(4in)(650rpm) = 6

624.77 lbf 625 lbfW t = V33000H = 680.678 f t/min

33000 H (54.14 horsepower) = 2 = 2

34

625 lbf an(20 ) 55 lbfW r = 2 * t ° = 9

Yielding Factor of Safety for Shaft Analysis:

At the Keyway of the Shaft:

Eq. 2: = =1.93(k )K f = 1 + q t − 1 .82(2.14 )1 + 0 − 1

= =2.66(k )K fs = 1 + qsheer ts − 1 .83(3 )1 + 0 − 1

Eq. 1: =( ) + ) ]σa′ = [ πd3k 32(M +M )t a m 2

( πd3k 16(T +T )fs a m 2 1/2 ( ) + ) ][ π 1.235in* 3

1.93 32(747in−lbs)* 2( π 1.235in* 32.66 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 )K f = 1 + q t − 1 .82(1.69 )1 + 0 − 1

= =1.40(k )K fs = 1 + qsheer ts − 1 .83(1.48 )1 + 0 − 1

Eq. 1: =( ) + ) ]σa′ = [ πd3k 32(M +M )t a m 2

( πd3k 16(T +T )fs a m 2 1/2 ( ) + ) ][ π 1.235in* 3

1.57 32(747in−lbs)* 2( π 1.235in* 31.40 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 but 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 −3F − 0

−5 2F + 0

−8 3F − 0

−12 4F

= =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

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 = π16n

SeKMf a 2 + 3 Sy

K Tfs m 2 1/2 1/3

= 1.23in[4( ) ( ) ] π16 1.5*

44.2kpsi1.93 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 =Seσa′

43kpsi1.6kpsi

requirements)

Shaft Deflection Analysis:

Eq. 11: = )= ­1.15E(­5)(4x l) y = Fx248EI − 3 (4 .0088in .0175in955lbf 1.0088in* 2

48 29.7Msi 0.119* * * 1 − 3 * 2

Eq. 12: − L )L .0175in( )θ = dxdy = − (4EI

Fx2 − 8EI3Fxl = − 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 = ( )AA+√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 PF√Y 0.0535

= 1 4 teeth/in1 in√0.27 0.0535

= 1

36

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 = F10dp − 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 = W to v s F

PdJ

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 = σS Y /KKt N t R = 1 = 1

Spur gear wear analysis

Force at Operating Torque

35714.5 psi (14 6) σc = Cp√W K K Kto v s

Kmd Fp I

Cf = 1 − 1

where 3, 00 3, 00 50 lbfW t = 3 0 VH = 3 0 15.4701 hp

680.678 f t/min = 7

300 Cp = 2 √psi C f = 1

.080348 (14 3) external gearsI = 2mNcosϕsinϕ mG

m +1G= 2(1)cos20°sin20° 1

1+1 = 0 − 2

in F in dp = 4 = 1

.53 (14 2)SH = σcS 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 HBGHBP = 1 < 1

as T 50°FKT = 1 < 2

37

for 99% reliability.5 .109ln(1 ) .002 KR = 0 − 0 − R = 1

Bearing Reliability

(N)( ) (2.4 )( ) 153.85 hoursLD = 1ω

1 hr60 min = 1

650 * 108 1 hr60 min = 6

( ) 55.42( ) 946.13 lbfC10 = FD L n 60R R

L n 60D D1/a

= 9 1066153.85 650 60* * 3/10 = 4

40xD = LRL N 60D D = 10 rev6

6153.85 hours 650 rpm 60* * = 2

xp(− ) ) .99R = e ( θ−x0

x ( ) −xD C10

a Ff D a0 b = 0

where , , , , , .2af = 1 a = 310 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