Transmission Design Report

4-‐Speed Transmission

10/20/15

Group 44

Ryan Rampolla

Ryland Ballingham

Dr. Griffis

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Table of Contents

1.0 Preliminary Need ............................................................................................................. 3

2.0 Preliminary Specifications ............................................................................................ 3 2.1 Constraints Imposed by Boss ................................................................................................. 3

3.0 Design ................................................................................................................................... 4 3.1 Countershaft Assembly ............................................................................................................ 5 3.2 Countershaft Layout .................................................................................................................. 5

4.0 Theoretical Overview ...................................................................................................... 6 4.1 Theoretical Loads ...................................................................................................................... 6 4.2 Estimated Performance ........................................................................................................ 17 4.3 Countershaft Bearings ........................................................................................................... 20 4.4 Countershaft Critical Locations .......................................................................................... 21 4.5 Theoretical Critical Locations ............................................................................................. 21 4.6 Theoretical Keys ...................................................................................................................... 25

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1.0 Preliminary Need

The preliminary need provided was, “For a conventional rear-‐wheel drive sporty vehicle, there is a need for a 4-‐speed manual transmission that can operate in any gear over an infinite life.” While designing this transmission it will be crucial to take the strength and torque requirements of all the components into consideration.

2.0 Preliminary Specifications

The specifications that were provided are as follows:

1. Design Point (continuous): 40 HP at 2000 RPM (input shaft) 2. Design factor (countershaft): nd = 1.75 3. Design factor (gears): ng = 1.25 4. Design factor (keys): nk = 1.1 5. Size (Max):

a. Length: 2 ft (target 1 ft on main body, 1 ft on tail piece) b. Width: 14 in c. Height: 14 in

6. Target Ratios: a. 1st gear: 3.2 to 1 (+/-‐ 5%) b. 2nd gear: 2.2 to 1 (+/-‐ 5%) c. 3rd gear: 1.6 to 1 (+/-‐ 2%) d. 4th gear: direct drive

2.1 Constraints Imposed by Boss The constraints given by the boss were:

1. Assume constant torque at the design point (even though it isn’t). 2. Input-‐output shafts collinear (so 4th can be direct connect). 3. Helical, parallel axis gearing, all gears same diametral pitch. 4. Simplification: no reverse gear here. 5. Gears Keyed to Countershaft (idler shaft). 6. Leave 3 in spacing between 1st & 2nd gears and also between 3rd & 4th for

synchronizers. 7. Gear teeth (counts) should be chosen for even wear. 8. Passenger side view of countershaft, maintain ordering of gears: starting on left: 1st,

2nd, 3rd gear with countershaft input driver gear on right. 9. Objective is to fully design the countershaft (idler shaft) of the transmission. The

case, the input, output shafts, the synchronizers are not being designed in this project. For those parts, include enough mainshaft detail (input, output shafts) and case detail to allow for simulation, provide space for synchronizers, use solid works mates to simulate synchronizers.

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3.0 Design

The above figure is the overall design of the 4-‐speed transmission. The view is oriented from the passenger side of the vehicle showing the input shaft on the top left and the output on the top right. The shaft assembly on the bottom is the countershaft with a total of four gears. The gear on the far right is the idler gear, which is driven by the driving gear directly above it at 2000 RPM. To the left of the idler are the 1st, 2nd, and 3rd gears (left to right), which are all attached to their corresponding output gears. This design incorporates multiple snap rings, key ways, bearings, and spacers to keep the gears in place.

The above illustration is used to declare important values of the gears and overall transmission design. The image provides the face width values for the output and driving gears along with the tooth numbers for all gears. Gear teeth numbers were chosen in order to achieve the desired gear ratios. In addition the pitch for all gears was established to be 5 T/in and a diametral pitch of 7.011. Also, a centerline distance from the mainshaft to countershaft was chosen to be 5.9 in which allowed gears enough clearance to work properly while achieving the desired torque values.

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Since location A only experiences radial loading and spins at an RPM greater than 100, we chose to use a 1in roller bearing at this location. At location B, axial and radial loading occurs and thus we need to a 1.25in tapered-‐roller bearing to account for this extra force component. As for the keys, we chose to use the same key/keyway for all the gears. We chose to use a woodruff key with a length of 0.75in and a radius of 0.303 in. The keyways were sized accordingly to this size with a depth of 0.06in and length of 0.44 in. The keys are made out of 316 stainless steel. In order to fix bearing A to the countershaft, two 1in snap rings are used on both sides of the bearing. Bearing B is held in place by the 1.5 in shoulder and a 1.25in snap ring. There is nothing to hold the gears in place axially, which is a design flaw in our design.

3.1 Countershaft Assembly The countershaft can easily be assembled due to its design by following a series of simple steps. The first step in the countershaft assembly is to insert all the keys into their mating key ways, since all the keys are the same it doesn’t matter which key goes to which key way. Once all the keys are installed, gear 3 would be slid over the countershaft followed by a 0.125in spacer and gear 2 would be slid over the countershaft and against the spacer. Following this, a 3in spacer and gear 1 would be slid over the counter shaft. Now the idler gear along with a spacer can be slid onto the end by gear 3 followed on the end by a 1.25in snap ring. Finally the 1in, 1.25in roller bearings would be held in place onto the ends of the counter shaft with the appropriately sized snap rings.

3.2 Countershaft Layout

The image above is a passenger view of the countershaft alone. On both the far left and far right of the countershaft are flat ends which is where the specific rolling bearing will be implicated in order to allow the shaft to spin freely within the case. Starting from the left

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side of the shaft we have the two 1in snap rings (to hold the roller-‐bearing in place) followed by the 1st gear, 2nd gear, 3rd gear keys ways followed by the 1.25in shoulder and idler gear key way followed by the 1.50 in shoulder and finally the 1.25 in snap ring to hold the tapered roller bearing in place.

4.0 Theoretical Overview

4.1 Theoretical Loads The following figures are the fbd’s of the countershaft gears. All dimensions are detailed along with locations of all the forces. All images are a passenger view of the countershaft and its gears.

The figure above is the fbd of the 1st gear.

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The figure above is the fbd of the 2nd gear.

The figure above is the fbd of the 3rd gear.

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The following table shows the calculations used to find the various tooth and reaction forces along with torques from the three countershaft fbd’s above.

Table I Diameter of drive gear 𝑑! = 7𝑖𝑛 Pressure angle ϕ! = 14.5° Helix angle ψ = 45° Length L=15.62in B coordinate 𝑥! = 14.4575𝑖𝑛 Input-‐shaft torque 𝑇! =

40 𝐻𝑃2000 𝑟𝑝𝑚

5252 𝑙𝑏𝑓 𝑓𝑡 12 𝑖𝑛 𝑟𝑝𝑚𝑓𝑡 𝐻𝑃

= 1260 𝑙𝑏𝑓 𝑖𝑛

Counter-‐shaft torque 𝑇! =

𝑁!𝑁!

𝑇! = 3524

1260 𝑙𝑏𝑓 𝑖𝑛 = 1838 𝑙𝑏𝑓 𝑖𝑛

Tangential tooth force at drive gear 𝐹!" = 2

𝑇!𝑑!

= 21838 𝑙𝑏𝑓 𝑖𝑛

7 𝑖𝑛= 525 𝑙𝑏𝑓

Normal tooth force at drive gear 𝐹! =𝐹!"

cos ϕ! cos (ψ)=

525 𝑙𝑏𝑓cos 14.5 cos (45)

= 767 𝑙𝑏𝑓

Radial tooth force at drive gear 𝐹!" = 𝐹! sin ϕ! = 767 𝑙𝑏𝑓 sin 14.5 = 192 𝑙𝑏𝑓 Axial tooth force at drive gear 𝐹!" = 𝐹! cos ϕ! sin ψ = 767 𝑙𝑏𝑓 cos 14.5 sin 45 = 525 Drive gear coordinate 𝑥! = 𝑥! − 𝑥! = 11.975 𝑖𝑛 Drive gear pitch radius 𝑦! =

7𝑖𝑛2= 3.5𝑖𝑛

Diameter of 1st gear 𝑑! = 3.6𝑖𝑛 Tangential tooth force at 1st gear

𝐹!! =𝑑!𝑑!

𝐹!" =7 𝑖𝑛3.6 𝑖𝑛

525 𝑙𝑏𝑓 = 1021 𝑙𝑏𝑓

Normal tooth force at 1st gear 𝐹! =𝐹!!

cos ϕ! cos (ψ)=

1021 𝑙𝑏𝑓cos 14.5 cos (45)

= 1491 𝑙𝑏𝑓

Radial tooth force at 1st gear 𝐹!! = 𝐹! sin ϕ! = 1491 𝑙𝑏𝑓 sin 14.5 = 373 𝑙𝑏𝑓 Axial tooth force at 1st gear 𝐹!! = 𝐹! cos ϕ! sin ψ = 1491 𝑙𝑏𝑓 cos 14.5 sin 45

= 1021 𝑙𝑏𝑓 1st gear torque

𝑇! = 𝐹!!𝑑!2

= 525 𝑙𝑏𝑓3.6𝑖𝑛2

= 1838 𝑙𝑏𝑓𝑖𝑛

1st gear output torque 𝑇!! = 𝐹!!

𝑑!!2

= 525 𝑙𝑏𝑓8.2𝑖𝑛2


1st gear coordinate 𝑥! = 2.02125𝑖𝑛 1st gear pitch radius

𝑦! =𝑑!2

=3.6𝑖𝑛2

= 1.8𝑖𝑛

Right-‐side bearing reaction (axial) 𝑅!" = 𝐹!! − 𝐹!" = 496 𝑙𝑏𝑓 Right-‐side bearing reaction (vertical) 𝑅!" =

1𝑥!

𝑥!𝐹!! + 𝑦!𝐹!! + 𝑥!𝐹!" − 𝑦!𝐹!" = 211 𝑙𝑏𝑓

Right-‐side bearing reaction (horizontal) 𝑅!" =1𝑥!

𝑥!𝐹!" − 𝑥!𝐹!! = 292 𝑙𝑏𝑓

Left-‐side bearing reaction (vertical) 𝑅!" = 𝐹!! + 𝐹!! − 𝑅!" = 354 𝑙𝑏𝑓 Left-‐side bearing reaction (horizontal) 𝑅!" = 𝐹!" − 𝐹!! − 𝑅!" = −788 𝑙𝑏𝑓 Diameter of 2nd gear 𝑑! = 4.6𝑖𝑛 Tangential tooth force at 2nd gear

𝐹!! =𝑑!𝑑!

𝐹!" =7 𝑖𝑛4.6 𝑖𝑛

525 𝑙𝑏𝑓 = 799 𝑙𝑏𝑓

Normal tooth force at 2nd gear 𝐹! =𝐹!!

cos ϕ! cos (ψ)=

799 𝑙𝑏𝑓cos 14.5 cos (45)

= 1167 𝑙𝑏𝑓

Radial tooth force at 2nd gear 𝐹!! = 𝐹! sin ϕ! = 1167 𝑙𝑏𝑓 sin 14.5 = 292 𝑙𝑏𝑓

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Axial tooth force at 2nd gear 𝐹!! = 𝐹! cos ϕ! sin ψ = 1167 𝑙𝑏𝑓 cos 14.5 sin 45= 799 𝑙𝑏𝑓

2nd gear torque 𝑇! = 𝐹!!

𝑑!2

= 799 𝑙𝑏𝑓4.6𝑖𝑛2


2nd gear output torque 𝑇!! = 𝐹!!

𝑑!!2

= 799 𝑙𝑏𝑓7.2𝑖𝑛2


2nd gear coordinate 𝑥! = 6.33625𝑖𝑛 2nd gear pitch radius

𝑦! =𝑑!2

=4.6𝑖𝑛2

= 2.3𝑖𝑛

Right-‐side bearing reaction (axial) 𝑅!" = 𝐹!! − 𝐹!" = 274 𝑙𝑏𝑓 Right-‐side bearing reaction (vertical) 𝑅!" =

1𝑥!

𝑥!𝐹!! + 𝑦!𝐹!! + 𝑥!𝐹!" − 𝑦!𝐹!" = 287 𝑙𝑏𝑓


𝑥!𝐹!" − 𝑥!𝐹!! = 85 𝑙𝑏𝑓

Left-‐side bearing reaction (vertical) 𝑅!" = 𝐹!! + 𝐹!" − 𝑅!" = 197 𝑙𝑏𝑓 Left-‐side bearing reaction (horizontal) 𝑅!" = 𝐹!" − 𝐹!! − 𝑅!" = −359 𝑙𝑏𝑓 Diameter of 3rd gear 𝑑! = 5.6𝑖𝑛 Tangential tooth force at 3rd gear 𝐹!! =

𝑑!𝑑!

𝐹!" =7 𝑖𝑛5.6 𝑖𝑛

525 𝑙𝑏𝑓 = 656 𝑙𝑏𝑓

Normal tooth force at 3rd gear 𝐹! =𝐹!!

cos ϕ! cos (ψ)=

656 𝑙𝑏𝑓cos 14.5 cos (45)

= 959 𝑙𝑏𝑓

Radial tooth force at 3rd gear 𝐹!! = 𝐹! sin ϕ! = 959 𝑙𝑏𝑓 sin 14.5 = 240 𝑙𝑏𝑓 Axial tooth force at 3rd gear 𝐹!! = 𝐹! cos ϕ! sin ψ = 959 𝑙𝑏𝑓 cos 14.5 sin 45 = 656 𝑙𝑏𝑓 3rd gear torque

𝑇! = 𝐹!!𝑑!2

= 656 𝑙𝑏𝑓5.6𝑖𝑛2


3rd gear output torque 𝑇!! = 𝐹!!

𝑑!!2

= 656 𝑙𝑏𝑓6.2𝑖𝑛2


3rd gear coordinate 𝑥! = 7.03125𝑖𝑛 3rd gear pitch radius

𝑦! =𝑑!2

=5.6𝑖𝑛2

= 2.8𝑖𝑛

Right-‐side bearing reaction (axial) 𝑅!" = 𝐹!! − 𝐹!" = 131𝑙𝑏𝑓 Right-‐side bearing reaction (vertical) 𝑅!" =

1𝑥!

𝑥!𝐹!! + 𝑦!𝐹!! + 𝑥!𝐹!" − 𝑦!𝐹!" = 276𝑙𝑏𝑓


𝑥!𝐹!" − 𝑥!𝐹!! = 116𝑙𝑏𝑓

Left-‐side bearing reaction (vertical) 𝑅!" = 𝐹!! + 𝐹!" − 𝑅!" = 156𝑙𝑏𝑓 Left-‐side bearing reaction (horizontal) 𝑅!" = 𝐹!" − 𝐹!! − 𝑅!" = −247𝑙𝑏𝑓

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The following images show the shear and bending moments of all the gears in order.

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4.2 Estimated Performance The following images are the results for all the gears in the transmission.

1st gear results.

1st gear output results

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2nd gear results.

2nd gear output results.

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3rd gear results.

3rd gear output results.

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Drive gear results.

Input gear results. When comparing the torque calculations in 4.1 to the rush gear results above for each gear, it is clear that the gear will be strong enough to handle the transmitted torque.

4.3 Countershaft Bearings Since our countershaft at the design point spins at 1371 RPMs, we must select bearings that are rated for speeds greater than 1371RPMs. Since the radial loads at location A equals: 𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 788 lbf+359 lbf +247 lbf= 1394 lbf

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𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 354 lbf+197 lbf +156 lbf= 707 lbf

Since 1394 lbf >707 lbf, we need to select a bearing that can handle that is rated for more than 1394 lbf of dynamic loading. Since we have only radial loads at point A and 𝜔 > 100 RPMS, we chose to use a double sealed ball bearing. The one chosen is rated up to 2500 RPMs, has a maximum dynamic load capacity of 1480 lbf and its designed for a 1in diameter shaft. Now we analyze the bearing at point B. Since point B has axial and radial loading, we chose to use a tapered-roller bearing. First we will analyze the radial loading. 𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! = 292 lbf+85 lbf +116 lbf= 493 lbf

𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! =211 lbf+287 lbf +276 lbf= 774 lbf

Since 774 lbf >493 lbf, we need to choose a bearing that can handle more than 774 lbf radially. Now we will analyze the axial loading 𝑅!" = 𝑅!"! + 𝑅!"! + 𝑅!"! =496 lbf+274 lbf +131 lbf= 901 lbf So we need a bearing that can handle more than 901 lbf axially and 774 lbf radially. Because of this we chose a tapered steel roller bearing with a maximum radial dynamic load rating of 2,130 lbf, a maximum axially axial dynamic load rating of 1,500 lbf and is designed for a shaft with a diameter of 1.25 in.

4.4 Countershaft Critical Locations Critical locations occur at locations, which undergo high bending moments, high stress concentrations, small cross sectional area, and/or heavy torque. Throughout the countershaft design there are several critical locations that can be recognized. These locations include key ways, snap ring grooves, and shoulders. The keyways and keys themselves have a very small cross sectional area and the gears are torqueing the keys. The snap rings are critical locations because they involve keeping the gears from walking around on the countershaft.

4.5 Theoretical Critical Locations The Mcomp graph image in section 4.1 will be used to find the correct moment values used for the analysis of the critical locations. This section is used primarily to determine what material and dimensions of the countershaft should be used in order for the shaft to be strong enough for the applied forces. Tables and equations were all found in the McGraw-‐Hill textbook. First we will analyze the critical point at the shoulder to the right of gear 3. The material used for the countershaft was AISI 1045 steel CD with a diameter, d= 1in. From the material properties in Solid Works, Sut= 91kpsi, and Sy= 77kpsi. In order to find Se we must find ka &

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kb first. From table 6-‐2 for cold drawn finish we get, a= 2.7 & b= -‐0.265. From Eq. 6-‐19,20 we get:

𝑘! = 𝑎𝑆!"! 𝑘! = 2.7(91𝑘𝑝𝑠𝑖)!!.!"#

𝑘! = 0.817 𝑘! = 0.879𝑑!!.!"#

𝑘! = 0.879(1𝑖𝑛)!!.!"# 𝑘! = 0.879

𝑆! = 𝑘!𝑘!𝑆!"2

𝑆! = (0.817)(0.879)91𝑘𝑝𝑠𝑖2

𝑆! = 32.68 𝑘𝑝𝑠𝑖

For the shoulder at gear 3, its estimated that Ma= 2000lbf in. Using

!!= 1.5𝑖𝑛 & !

!=

0.1𝑖𝑛, gives kt= 1.65, and kts= 1.425 when using table A15-‐7,8. Next from figure 6-‐20,21 q= 0.825, and qs= 0.84. All these values will now be used in Eq. 6-‐32:

𝑘! = 1 + 𝑞 𝑘! − 1 𝑘! = 1 + 0.825 1.65 − 1

𝑘! = 1.53625

𝑘!" = 1 + 𝑞 𝑘!" − 1 𝑘!" = 1 + 0.84 1.425 − 1

𝑘!" = 1.357 Next the Alternating and Midrange Von Mises Stresses will be solved for with all the given values from above.

σ!! =32𝑘!𝑀!

π𝑑!

σ!! =32(1.53625)(2000 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 31.296 𝑘𝑝𝑠𝑖

σ!! = 316𝑘!"𝑇!π𝑑!

σ!! = 316(1.357)(1838 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 22.002 𝑘𝑝𝑠𝑖 Finally the fatigue factor of safety and Langer will be calculated:

1𝑛!=σ!!

𝑆!+σ!!

𝑆!"

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1𝑛!=31.296 𝑘𝑝𝑠𝑖32.68 𝑘𝑝𝑠𝑖

+22.002 𝑘𝑝𝑠𝑖91 𝑘𝑝𝑠𝑖

𝑛! = 0.834

1𝑛!

=σ!!

𝑆!+σ!!

𝑆!

1𝑛!

=31.296 𝑘𝑝𝑠𝑖77 𝑘𝑝𝑠𝑖


𝑛! = 1.44 These are not acceptable factors of safety for the shoulder to the right of gear 3. The target factor of safety was at least 1.75. In order to obtain this factor of safety the material of the countershaft could be changed to a stronger material but more importantly the diameter of 1in should be increased. Next we will find the Alternating/Midrange Von Mises Stresses and fatigue/Langer factors of safety of the shoulder to the right of the drive gear. At this point it is estimated that Ma= 690 lbf in and the midrange stress is equal to the previous shoulders.

σ!! =32𝑘!𝑀!

π𝑑!

σ!! =32(1.53625)(690 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 10.8 𝑘𝑝𝑠𝑖

1𝑛!=σ!!

𝑆!+σ!!

𝑆!"

1𝑛!=

10.8 𝑘𝑝𝑠𝑖32.68 𝑘𝑝𝑠𝑖


𝑛! = 1.75

1𝑛!

=σ!!

𝑆!+σ!!

𝑆!

1𝑛!



𝑛! = 2.35 These are acceptable factors of safety for the shoulder to the right of the drive gear. The target factor of safety was at least 1.75, which was just obtained. Next we will find the Alternating Von Mises Stresses and fatigue/Langer factor of safety of the snap ring groove closest to gear 1, directly to the right of the bearing. For this snap ring groove, its estimated that Ma= 200lbf in. Using

!!= 0.2𝑖𝑛 & !

!= !.!

!.!= 1.2, gives kt= 4.35 when using table A15-‐16.

Next from figure 6-‐20,21 q= 0.58, when using r= (0.2)(0.5)= 0.1. All these values will now be used in Eq. 6-‐32:

𝑘! = 1 + 𝑞 𝑘! − 1 𝑘! = 1 + 0.58 4.35 − 1

𝑘! = 2.943

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Next the Alternating Von Mises Stresses will be solved for with all the given values from above.

σ!! =32𝑘!𝑀!

π𝑑!

σ!! =32(2.943)(200 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 6 𝑘𝑝𝑠𝑖 Finally the fatigue/Langer factor of safety will be calculated, since there is no torque acting at this location the midrange stress is equal to zero:

1𝑛!=σ!!

𝑆!+σ!!

𝑆!"

1𝑛!=σ!!

𝑆!

1𝑛!=

6 𝑘𝑝𝑠𝑖32.68 𝑘𝑝𝑠𝑖

𝑛! = 5.45

1𝑛!

=σ!!

𝑆!+σ!!

𝑆!

1𝑛!

=σ!!

𝑆!

1𝑛!

=6 𝑘𝑝𝑠𝑖77 𝑘𝑝𝑠𝑖

𝑛! = 12.8 These exceed the acceptable factor of safety for the snap ring groove to the right of the bearing closest to gear 1. The target factor of safety was at least 1.75. Next we will find the Alternating Von Mises Stresses and fatigue/Langer factor of safety of the snap ring groove to the right of the bearing closest to the drive gear. At this point it is estimated that Ma= 600 lbf in.

σ!! =32𝑘!𝑀!

π𝑑!

σ!! =32(2.943)(600 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 18 𝑘𝑝𝑠𝑖 Finally the fatigue/Langer factor of safety will be calculated, since there is no torque acting at this location the midrange stress is equal to zero:

1𝑛!=σ!!

𝑆!+σ!!

𝑆!"

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1𝑛!=σ!!

𝑆!

1𝑛!=

18 𝑘𝑝𝑠𝑖32.68 𝑘𝑝𝑠𝑖

𝑛! = 1.82

1𝑛!

=σ!!

𝑆!+σ!!

𝑆!

1𝑛!

=σ!!

𝑆!

1𝑛!

=18 𝑘𝑝𝑠𝑖77 𝑘𝑝𝑠𝑖

𝑛! = 4.28 These are acceptable factors of safety for the snap ring groove to the right of the bearing closest to the drive gear. The target factor of safety was at least 1.75.

4.6 Theoretical Keys Next we will analyze the keyway critical location at gear 2, all keyways are critical locations but the keyway here has the highest bending moment. At this point its estimated that Ma= 2500lbf in. Using !

!= 0.02𝑖𝑛, gives kt= 2.14, and kts= 3.0 when using table 7-‐1. Next from

figure 6-‐20,21 q= 0.65, and qs= 0.7. All these values will now be used in Eq. 6-‐32:

𝑘! = 1 + 𝑞 𝑘! − 1 𝑘! = 1 + 0.65 2.14 − 1

𝑘! = 1.741

𝑘!" = 1 + 𝑞 𝑘!" − 1 𝑘!" = 1 + 0.7 3.0 − 1

𝑘!" = 2.4 Next the Alternating and Midrange Von Mises Stresses will be solved for with all the given values from above.

σ!! =32𝑘!𝑀!

π𝑑!

σ!! =32(1.741)(2500 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

σ!! = 44.33 𝑘𝑝𝑠𝑖

σ!! = 316𝑘!"𝑇!π𝑑!

σ!! = 316(2.4)(1838 𝑙𝑏𝑓 𝑖𝑛)

π(1𝑖𝑛)!

26

σ!! = 38.91 𝑘𝑝𝑠𝑖 Finally the fatigue factor of safety and Langer will be calculated:

1𝑛!=σ!!

𝑆!+σ!!

𝑆!"

1𝑛!=44.33 𝑘𝑝𝑠𝑖32.68 𝑘𝑝𝑠𝑖


𝑛! = 0.56

1𝑛!

=σ!!

𝑆!+σ!!

𝑆!

1𝑛!



𝑛! = 0.93 These are not acceptable factors of safety for the keyway at gear 2. The target factor of safety was at least 1.1. In order to obtain this factor of safety the material of the countershaft could be changed to a stronger material but more importantly the diameter of 1in should be increased. Finally we will calculate the static failure factor of safety for the keys. The keys used had a length and width of 0.065 and 0.25 inches. The key is made out of 316 stainless steel which has a σy= 30 kpsi.

𝑉 =2𝑇!𝑑

=2(1838 𝑙𝑏𝑓 𝑖𝑛)

1𝑖𝑛= 3.7 𝑘𝑝𝑠𝑖

τ =𝑉ℎ𝑙=

3.7 𝑘𝑝𝑠𝑖(0.065𝑖𝑛)(0.25𝑖𝑛)

= 226 𝑘𝑝𝑠𝑖

𝑛!"" =σ!2τ

=30 𝑘𝑝𝑠𝑖

2(226 𝑘𝑝𝑠𝑖)= 15.1

This factor of safety exceeds the target factor of safety at 1.1.

Transmission Design Report

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