GP03 Trash Compactor G07

Group Assignment 03Design a Big Trash Compactor

Date prepared 12/01/2016Date due 12/07/2016

Group 07: King, MichaelLewis, Parker

Stribrny, CaseyTang, Willie

MEE 342 – Principles of Mechanical DesignFall 2016

Arizona State UniversitySchool for Engineering of Matter, Transport, and Energy

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SummaryThe goal of this assignment was to design a trash compactor like the one

outside the ERC building. The total cost of the new trash compactor is $14,800,000. The inner dimensions of the compacting cylindrical volume is: Diameter = 1.00 m and length = 2.20 m and the outside dimensions of the whole trash compactor are 2.43m x 2.43 m x 6.71m. This trash compactor can compact a load by 75.0% of its original size in the time of 8.80 seconds, which achieves a compaction rate performance design factor of 1.5 (bonus) while the performance of the crushing time has a design factor of 1.15. The maximum speed of the compaction plate is 0.386 m/s during normal operation. The motor is a WEG motor with 450 hp (336 kW) of power, 2130 N*m of rated torque at the shaft, and 1485 rpm of rated rotational speed. The performance design factor of the motor is 1.18, which is greater than 1.15 (bonus). This compactor has a mechanical advantage due to its simple two stepped gear reduction. The gear reduction of the system is 5.6:1 which produces an output torque of 10,010 Nm under operating conditions. The maximum deflection in shaft 1 was 0.078 mm (refer to appendix for this SkyCiv result) with a shear stress nd of 3.88. For shaft 2 the maximum deflection was 0.553 mm with a shear stress nd of 2.11, the smallest component failure design factor found in the trash compactor. The largest movable load for our system is 144000 kg of trash during normal operation performance. All of the analysis done on this trash compactor factors in the mass moments of inertia for all of the shafts and gears (bonus).

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Nomenclature and Symbols

Table 1. Symbols used in this report. Notice that there is a general symbol followed by any specific instances. Alphabetize symbols (Latin first, then greek)

Symbol Meaning/Usage

nd Design factor

ndshaft Design factor for shafts

ndGear Design factor for the gears

ndpushplate Design factor for the push plateσ b Stress due to bending

Sy Yeild Strength

W Weightg Gravity constant of accelerationT TorquePcr Critical force for buckling

P Load force for bucklingA AreaB Base lengthH Heightt ThicknessI Second area moment of inertiaF t Tangential force of the gearF r Radial force of the gearRt Reaction force tangentialRr Reaction force radial

π pit min Minimum thickness

E Young’s modulus of elasticityJ Second polar moment of inertiaV Shear force

M Moment

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σ Stressσ l Longitudinal stressσ t Hoop stressPi Inside pressurePo Outside pressureD Diametert ThicknessF Force

F pr Force of the pushrodF load Force of the loadw Rotational speed (angular velocity)d Pitch diameter

m ModuleN Number of teethρ Density of the materiala Bearing type

C10 Bearing forceLd Design life of bearingsl Hours of bearings runningn Revolutions per minuteR ReliabilityI d Inside diameterOd Outside diameterR0 Outer radiusRi Inner radiusC Value for buckling based of off conditions

F edge Force from the edge of the platePcompaction Pressure from compaction

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Table of Contents1. Project Description...........................................................................................................8

1.1 Problem Statement........................................................................................................81.2 Overall design solution.................................................................................................8

1.3 Summary of design approach and philosophy............................................................102. Approach.........................................................................................................................11

2.1 Problem Statements....................................................................................................112.1.1 General................................................................................................................11

2.1.2 Usage..................................................................................................................112.1.3 Load (Bonus)......................................................................................................11

2.1.4 Trash Pressure vessel..........................................................................................112.1.5 Push Rod.............................................................................................................11

2.1.6 Push Plate............................................................................................................112.1.7 Motor..................................................................................................................12

2.1.8 Gears...................................................................................................................122.1.9 Shafts..................................................................................................................12

2.1.10 Shafts Mass Moment of Inertia (bonus).........................................................122.1.11 Bearings..........................................................................................................12

2.1.12 Housing...........................................................................................................122.1.13 Door hinge......................................................................................................12

2.2 Find.............................................................................................................................122.2.1 Usage..................................................................................................................12

2.2.2 Load (Bonus)......................................................................................................122.2.3 Trash vessel.........................................................................................................12

2.2.4 Push Rod.............................................................................................................122.2.5 Push Plate............................................................................................................13

2.2.6 Motor..................................................................................................................132.2.7 Gears...................................................................................................................13

2.2.8 Shafts..................................................................................................................132.2.9 Shafts Mass Moment of Inertia (bonus).............................................................13

2.2.10 Bearings..........................................................................................................132.2.11 Housing...........................................................................................................13

2.2.12 Door hinge......................................................................................................132.3 Free Body Diagrams of Problem................................................................................13

2.3.1 Usage..................................................................................................................132.3.2 Load (Bonus)......................................................................................................13

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2.3.3 Trash Pressure Vessel.........................................................................................142.3.4 Push Rod.............................................................................................................14

2.3.5 Push Plate............................................................................................................162.3.6 Motor..................................................................................................................16

2.3.7 Gears...................................................................................................................162.3.8 Shafts..................................................................................................................17

2.3.9 Shafts Mass Moment of Inertia (bonus).............................................................182.3.10 Bearings..........................................................................................................18


2.4 Givens.........................................................................................................................202.4.1 Usage..................................................................................................................20

2.4.2 Load (Bonus)......................................................................................................202.4.3 Trash Pressure Vessel.........................................................................................20

2.4.4 Push Rod.............................................................................................................202.4.5 Push Plate............................................................................................................20

2.4.6 Motor..................................................................................................................202.4.7 Gears...................................................................................................................21

2.4.8 Shafts..................................................................................................................212.4.9 Shafts Mass Moment of Inertia (bonus).............................................................21

2.4.10 Bearings..........................................................................................................212.4.11 Housing...........................................................................................................21

2.4.12 Door hinge......................................................................................................222.5 Assumptions................................................................................................................22

2.5.1 Usage..................................................................................................................222.5.2 Load (Bonus)......................................................................................................22

2.5.3 Trash Pressure Vessel.........................................................................................222.5.4 Push Rod.............................................................................................................23

2.5.5 Push Plate............................................................................................................232.5.6 Motor..................................................................................................................23

2.5.7 Gears...................................................................................................................242.5.8 Shafts..................................................................................................................24

2.5.9 Shafts Mass Moment of Inertia (bonus).............................................................242.5.10 Bearings..........................................................................................................24


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2.6 Equations....................................................................................................................252.6.1 General................................................................................................................25

2.6.2 Usage..................................................................................................................262.6.3 Load (Bonus)......................................................................................................26

2.6.4 Trash pressure vessel..........................................................................................262.6.5 Push Plate............................................................................................................26

2.6.6 Push Rod.............................................................................................................272.6.7 Motors.................................................................................................................27

2.6.8 Gears...................................................................................................................282.6.9 Shafts..................................................................................................................28

2.6.10 Shafts Mass Moment of Inertia (bonus).........................................................282.6.11 Bearings..........................................................................................................28


3. Proposed Design.............................................................................................................293.1 Summary of usage model...........................................................................................30

3.2 Summary of proposed design.....................................................................................304. Solutions/Calculations....................................................................................................31

4.1 Analysis of usage........................................................................................................314.2 Analysis of loading for push assembly.......................................................................34

4.3 Analysis of Load (Bonus)...........................................................................................344.4 Analysis of push plate.................................................................................................35

4.5 Analysis of push rod...................................................................................................384.6 Analysis of hoop and tangential stresses in pressure vessel.......................................43

4.7 Analysis of Motor.......................................................................................................454.8 Analysis of Gears........................................................................................................50

4.9 Analysis of Shafts.......................................................................................................524.10 Shafts Mass Moment of Inertia (bonus).................................................................56

4.11 Analysis of Bearings...............................................................................................574.12 Analysis of Housing................................................................................................59

4.13 Door Hinge/Door Latch/Motor Bolts.....................................................................625. Overall Solution..............................................................................................................65

5.1 Bill of Materials..........................................................................................................666. Appendix.........................................................................................................................67

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1. Project Description1.1 Problem Statement

Our trash compactor consists of a 450hp motor that is connected to a gear train that has a gear ratio of 5.6 . It needs to compact trash to 50% compaction in less than 10 seconds with a performance design factor of 1.15. the design factor for component failure must be 2 for all components. The trash compactor must fit within the dimensions that the trash compactor outside the ERC sits in. The final gear in the gear train pushes a pushrod, which pushes a plate to compact the trash. The output torque at the last gear is 1.00*10^4 Nm and the force that is exerted is 78500 N during normal operation. The plate can handle a load of 207000 N and this load results in a minimum nd of 2.11 in the entire system (occurs at shaft 2). A schematic of the overall assembly is shown in Fig. 1.

1.2 Overall design solution

Figure 1. Isometric view of entire Trash compactor

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Figure 2. Top view with outer box

Table 1. Dimensions of major components and corresponding design factors

Part Vendor Length Radius Thickness nd Operating nd Stall

[-] [-] [m] [mm] [mm] [-] [-]

Push Rod McMaster-CARR 2.00 55.0 n/a 6.72 2.55

Push Plate McMaster-CARR n/a 500 28.3 7.44 2.50

Trash Compartment McMaster-CARR 2.20 500 20.0 862 131

Hinge / Latch McMaster-CARR 0.20 12.5 n/a 82.4 12.6

Table 2. Summary of Motor Specifications

Part # Vendor Power Speed Trated Tmax

[-] [-] [hp] [kW] [rpm] [ft-lbf] [N-m] [ft-lbf] [N-m]50018EP3GKD580Z

-W22 WEG 450 336 1490 1570 2130 3450 4680

Table 3. Summary of Gear Specifications. All gears have a 20˚ pressure angle and are made from titanium with assumed E = 119 GPa and Sy = 889 MPa. Gears were purchased from McMaster Carr. All loading and stress analysis reported at design load of 78500 N; please see analysis section for maximum loading and stresses when motor is a stall or maximum torque.

Gear # Mounted On Part # m N b h t T Ft Fr σb nd

[-] [-] [-] [mm] [-] [mm] [mm] [mm] [N-m] [N] [N] [MPa] [-]1 Motor KSUS4-15 17 15 170 34 26.7 1800 14100 5130 23.7 37.12 Shaft 1 KSUS4-30J22 17 30 170 34 26.7 3580 14100 5130 23.7 37.1

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3 Shaft 1 KSUS4-15 17 15 170 34 26.7 3580 28100 10200 47.3 18.64 Shaft 2 KSSG4-42 17 42 170 34 26.7 10000 28100 10200 47.3 18.65 Shaft 2 KSUS4-15 17 15 170 34 26.7 10000 78500 28600 132 6.66

Table 4. Summary of Shaft Specifications. Shafts made from titanium with assumed E = 119 GPa and Sy = 880 MPa. Shaft stock purchased from McMaster-CARR and machined to size. All loading and stress analysis reported at design load of 78,500 N; please see analysis section for maximum loading and stresses when motor is a stall or maximum torque.

Shaft # Diameter LengthLocations RLeft

Bearing

RRight

Bearingσmax nd

Gear A Gear B Left Bearing

Right Bearing

[-] [mm] [mm] [mm] [mm] [mm] [mm] [N] [N] [MPa] [-]1 120 550 150 315 30 520 17,600 25,500 50.1 10.12 120 550 150 315 30 520 68,900 40,500 132 3.85

Table 5. Summary of Bearing Specifications. Bearings were purchased from Grainger and nd based upon rated load, not accounting for wear and usage. ID = inner diameter, OD = outer diameter. All loading and stress analysis reported at design load of 78,500 N; please see analysis section for maximum loading and stresses when motor is a stall or maximum torque.

Bearing #

Mounted On Part # ID OD Rated

LoadLoad from

Shaft C10 nd

[-] [-] [-] [mm] [mm] [kN] [N] [N] [-]

1 Shaft 1 NU2322-E-TVP2 110 240 750 17,600 18,000 41.8

2 Shaft 1 NU2322-E-TVP2 110 240 750 25,500 26,000 28.8

3 Shaft 2 NU2322-E-TVP2 110 240 750 68,900 70,300 10.7

4 Shaft 2 NU2322-E-TVP2 110 240 750 40,500 41,300 18.2

1.3 Summary of design approach and philosophyWe were required to achieve a compaction of 50% in 10 seconds for this trash compactor, so we

started with these requirements and worked backwards to design our system. We applied our performance design factors and found the forces that achieving this would put on the push plate. We then found how this force transferred to the push rod and through the gear train (factoring in the mass moment of inertia of the gears and shafts) and determined the operational torque for a motor that would be required. We then picked a motor based on the required torque and speed and tweaked the gear ratios to suit our requirements.

Once we had the motor, we worked forward to make sure that every component of our trash compactor could handle the maximum amount of torque that our motor could output (the stall torque). We worked forward through the motor bolts, gear train, push rod, plate, and door hinges/latch, finding the design factors at motor stall for all of these. In the end, the smallest design factor in the trash compactor is

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2.11 and the smallest performance design factor is 1.15, showing our team’s commitment to making a functioning product.

2. Approach2.1 Problem Statements

2.1.1 General

Design a Trash compactor with similar dimensions as the trash compactor outside the ERC building on ASU campus. This trash compactor must include a gear train with no less than 2 gear reductions. It must have a circular compacting volume with a radius of 1 meter and a length of 1/3 the overall length of the entire trash compactor. This compactor must compact at least 50% of the compacting volume within 10 seconds. The performance design factor must be at least 1.15 and the compaction performance must be at least 1.5. all other components in the trash compactor must have a design factor of 2.

2.1.2 Usage

Find the appropriate Acceleration to compact at least 50% of the trash in 10 seconds with the nd

time of 1.15 and nd compaction of 1.5. Length of pressure trash vessel is 1/3 the overall length of the trash compactor.

2.1.3 Load (Bonus)

The trash compactor has to compact a load of trash that consists of just cardboard in a pressure vessel with a diameter of 1m. The compactor has to compact the trash to 50% of its original volume in at least 10 seconds and we wanted a design factor of 1.15 for the time and 1.5 for the compaction.

2.1.4 Trash Pressure vessel

The hoop stress that pressure vessel experiences comes from the compaction of the trash and from the sealing of the push plate edge. The vessel is a thin walled pressure vessel that is made from titanium.

2.1.5 Push Rod

A titanium pushrod will be used to accelerate the titanium push plate that will be compacting the trash. The radius of the pushrod was assumed in the previous section in the analysis of push plate and was assumed to be 0.055 m. The pushrod must withstand the forces from the push plate and the gear teeth so that buckling does not occur. As for the support of the shaft, the shaft will be supported by a frictionless channel.

2.1.6 Push Plate

Since we know the force required to move the trash within our time frame with an nd= 1.15, then need to find the force required to move the push plate with an nd = 1.5 for compaction and an acceleration with an nd = 1.15.

2.1.7 Motor

2.1.8 Gears

Calculate the design factor of the gears during normal operation and when the motor is at its stall torque.

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2.1.9 Shafts

The shaft must withstand the calculated reactions from where the bearings are reacting and also the gears. In order to calculate all of the stresses on the shaft we used Skyciv. Then the design factor, nd, can be determined.

2.1.10 Shafts Mass Moment of Inertia (bonus)

Find the mass moment of inertia of 2 shafts and the 5 gears in the gear train.

2.1.11 Bearings

The bearings must withstand forces that come from the shaft and the gears meshing with each other. The bearings can be modeled as either roller bearings or ball bearings depending on the design.

2.1.12 Housing

To solve for the nd for the bearing housing we were able to run a simple FEA analysis to solve for the Max stress. This max stress was calculated under the stall torque under max loading. The reaction forces on the bearings were solved for in the shaft analysis

2.1.13 Door hinge

The trash compactor has a door at the end of the compactor vessel. This door must withstand the forces of compaction at 75% compaction when our pushrod assembly stops. This includes hinges and latch. Of the door assembly.

2.2 Find

2.2.1 Usage

Acceleration and deceleration; a x at vmax t at vmax vmax compaction %

2.2.2 Load (Bonus)

Acceleration needed to have a nd = 1.15 Vmax with nd =1.15 FT, the force required to move the Trash at desired speed

2.2.3 Trash vessel

Outer radius; ro

Thickness; t

2.2.4 Push Rod

Ft , The tangential force from the pinon gear to the rack. Pcr

nd at operational torque

2.2.5 Push Plate

Pcompaction

Fedge

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Fpr, the force from the push rod required to move the push plate nd

2.2.6 Motor

2.2.7 Gears

Design factor during operation and at motor stall.

2.2.8 Shafts

Maximum Von Mises Stress, σmax Design Factor, nd


I; the mass moment of inertia of each shaft and gear.

2.2.10 Bearings

For each bearing: C10 Design factor of each bearing; nd

2.2.11 Housing

Design factor of housing; nd

2.2.12 Door hinge

nd hinge

2.3 Free Body Diagrams of Problem

2.3.1 Usage

N/A

2.3.2 Load (Bonus)

(a) (b)

Figure 3. (a) Schematic of the trash in the pressure vessel where r is the radius of the pressure vessel and LT is the length of the trash. (b) FBD of the Trash where Fcompaction = Fwall.

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2.3.3 Trash Pressure Vessel

Figure 4. (a) This is an FBD of the pressure vessel and it shows the forces acting on the outside. (b) FBD of the inside of the pressure vessel and it shows how the forces are pushing against the inner wall.

2.3.4 Push Rod

(a) (b)

Figure 5. Push rod b) FBD and a) schematic

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Figure 6. To solve for the push-rod it is necessary to analyze the gear train and solve for Ft, above is the FBD of the gears being analized.

Figure 7. (a) Schematic of the pushrod where D is the diameter and L is the length of the rod. (b) FBD of the pushrod where -Fpr + Ft

=ma.

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2.3.5 Push Plate

(a) (b)

Figure 8. (a) Schematic of the push plate with thickness, t, and a radius, r. (b) FBD of the push plate where Ff is the force from the friction edge of the plate, Fcompaction is the resultant force due to compaction, FT, is the force from the trash, and Fpr is the force from the push rod.

2.3.6 Motor

2.3.7 Gears

Figure 9. Schematic of how we found the gear tooth dimensions based off our CAD

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Figure 10. We assumed the gear teeth to be cantilevered beams with the dimensions gathered.

2.3.8 Shafts

Figure 11. FBD of shaft 1. Note: bearings 1 and 2 go left to right, respectively for each shaft.

Figure 12. This is the schematic of shaft 1 produced by Skyciv with dimensions and forces.

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Schematics: Refer to figure 12 above.

(FBD not applicable in this section, we are only finding the moment of inertia)

2.3.10 Bearings

Figure 13. Above are the reaction forces on the bearings generated from Skyciv.

Figure 14. FBD of Bearing 1

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2.3.11 Housing

Figure 15. Schematic showing the applied forces

2.3.12 Door hinge

(a) (b) (c)

Figure 16. (a) Schematic of the pin and door assembly looking from the side. (b) FBD of the door with the distributed load due to compaction. (c) FBD of the bolt with shear forces.

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2.4 Givens

2.4.1 Usage

Loverall = 6.7056 [m] nd compaction = 1.5 nd time = 1.15 Lvessel = 2.2352 [m] C = 50% = 0.50 t = 10 [s]

2.4.2 Load (Bonus)

Dimensions of Trash Compactor bcompactor = 8 ft = 2.44 m hcompactor = 8 ft = 2.44 m Lcompactor = 22 ft = 6.71 m


Inner radius; ri = 0.50 m

2.4.4 Push Rod

Fpr = 78538.53 N a = -0.08868 m/s^2 for operation (Maximum stresses in the plate will occur at

maximum compaction, when the system is decelerating) a = 2*-0.08868 m/s^2 for motor stall calculations (Worst case scenario: Increased

deceleration causes increased stress on the push rod and plate, so we will assume a more rapid deceleration than will be experienced under normal operation to be conservative.)

2.4.5 Push Plate

r = .5 m t = 0.0283 m, thickness determined from motor stall a = -0.0887 m/s^2 A negative acceleration is used because the maximum compaction occurs when the

plate is decelerating. FT=8.56 N q = distributed load (Pcompaction) [Pa] w =distributed load (FEdge) [N/m] a = part radius [m] b = support radius [m] t = plate thickness [m] ro = radius of push rod [m] (inner radius of the distributed load) v = poison’s ratio

2.4.6 Motor

Table 6. Summary of assumed motor performance from vendor specifications (note, add additional information as needed)


[-] [-] [hp] [kW] [rpm] [ft-lbf] [N-m] [ft-lbf] [N-m]

50018EP3GKD580Z WEG 450 336 1490 1570 2130 3450 4680

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-W22

2.4.7 Gears

Table 7. Summary of Gear Specifications. All gears have a 20 ˚ pressure angle and are made from Titanium alloy (Ti-6Al-4V) with assumed E = 119 GPa and Sy = 880 MPa. Gears are custom machined but were based of of gears purchased from SD-SI.

Gear # Mounted On Part # m N b h t

[-] [-] [-] [mm] [-] [mm] [mm] [mm]1 Motor KSUS4-15 17.0 15 170 34 26.72 Shaft 1 KSUS4-30J22 17.0 30 170 34 26.73 Shaft 1 KSUS4-15 17.0 15 170 34 26.74 Shaft 2 KSSG4-42 17.0 42 170 34 26.75 Shaft 2 KSUS4-15 17.0 15 170 34 26.7

2.4.8 Shafts

Stepped shafts must be used nd ≥ 2


Density of Titanium alloy: 4430 kg/m^3 Pitch radius of gear 1: r = 0.1275 m Thickness of gear 1: t = 0.17 m

2.4.10 Bearings

The trash compactor will last 100 years C10 R = C10 rated C10 D = C10 designed Fr = 17600 N

2.4.11 Housing

Below are the Y-Z components of the reaction forces from the shafts that were input into ANSYS FEA at each bearing.

F1y = 3500 N F1z = -45800 N F2y = 17000 N F2z = -64600 N F3y = 1900 N F3z = 66100 N F4y = -726 N F4z = 40500 N

2.4.12 Door hinge

r = .5 m Bolts are made out of Titanium(Ti-6Al-4V) E = 119 GPa = 1.19E+11 Pa Sy = 880 MPa = 8.80E+08 Pa

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Pcompaction=40000 Pa, refer to analysis of push plate Pcompaction=262957 Pa (AT PEAK COMPACTION), refer to hoop stress analysis

2.5 AssumptionsWe assumed that all of the components except the motor are made from titanium(Ti-4Al-4V). The

properties of the material are the following: E= 1.19E+11, Sy=8.80E+08, Shear strength=5.50E+08, and the ultimate strength= 9.50E+08. The material is perfectly uniform and there are no irregularities that effect the trash compactor.

The final analysis of the motor base and shaft locations were analyzed by using ANSYS finite element analysis software. We assumed that this analysis perfectly calculated the stresses in the base and where the bearings are located. Another piece of software that was used to analyze the shafts for this project was Skyciv. Our team assumed that Skyciv calculated the stresses in the shafts and at the bearing locations with the correct forces being applied.

2.5.1 Usage

Acceleration /Deceleration is constant during usage Material can be compacted

2.5.2 Load (Bonus)

The trash only consists of cardboard

Density of Cardboard: ρ=55 kgm3

Dimensions of Trash: r = .5 m LT = 1/3 * Lcompactor = 2.2352 m a = constant Vmax @ t/2 nd = 1.15 for time


The forces that are acting on the pressure vessel are perfectly uniform and act orthogonal to the surface of the inside and outside of the vessel.

Pressure outer is 0 Pa. Pressure inner is from the pushrod force over the area of the plate. Also, assume an outer radius since an outer radius was not provided. The material for the vessel is titanium and it has the following characteristics:

o Young’s modulus: 119 GPao Yield strength: 880 MPao Outer radius= 0.52 mo Isotropic homogeneous material

2.5.4 Push Rod

The tangential force from the pinon gear, Ft, acts at a length of 2 m for the buckling analysis, more than it will ever travel during operation. This will be the worst case scenario.

Homogeneous isotropic material Titanium (Ti-6Al-4V)

Young’s modulus= 119 GPa = 1.19E+11 Pa Sy = 880 MPa = 8.80E+08 Pa

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ρ=4430 kgm3

Frictionless surface for support Radial force from the gear does not contribute significantly to buckling of the rod D = .110 m r = 0.0550 m Lpr = 2.00 m

2.5.5 Push Plate

Homogeneous isotropic material Static Equilibrium Plate is made out of Titanium (Ti-6Al-4V) E = 119 GPa = 1.19E+11 Pa Sy = 880 MPa = 8.80E+08 Pa Ν = 0.342 Ρ =4430 kg/m3 C = .75 with nd = 1.5 a = 0.5 m q = 4.00E+04 Pa w = N/m b = ro = 0.0425 m The pressure due to compaction, Pcompaction, is uniform on the plate and the force of friction due

to the edge, Fedge, of the plate is uniform around the edge of the plate. For calculations at stall:

o Assume that the force from the push rod is only resisted by the Fedge. This is the worst-case scenario for loading of the plate.

o Assume that the system is decelerating at twice the normal operating deceleration (a = 2*-0.08868 m/s2). This puts more stress on the plate, making it a conservative assumption.

2.5.6 Motor

Assume that motor performance matches vendor specifications.Table 8. Summary of assumed motor performance from vendor

specifications (note, add additional information as needed)



-W22 WEG 450 336 1490 1570 2130 3450 4680

2.5.7 Gears

Homogeneous isotropic material Static Equilibrium Titanium (Ti-6Al-4V)

E = 119 GPa = 1.19 E+09 Pa Sy = 880 MPa = 8.80 E+06 Pa

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Assume that the radial force acting on the gear tooth is negligible for calculating the stress in the tooth.

2.5.8 Shafts

Homogeneous isotropic material Static Equilibrium All shafts are titanium

E = 119 GPa = 1.19 E+09 Pa Sy = 880 MPa = 8.80 E+06 Pa

Diameters: 100mm, 110mm, 120mm (steps on the same shaft) Total length of the shaft: 550mm

i. see figure 12 for exact shaft dimensions Each step increases the diameter by 5mm Stress concentrations was included in Skyciv analysis


Uniform density of material Gears and shafts spin perfectly about their centers Assume gears are perfect cylinders with no hole in the center (This will yield a higher

moment of inertia than reality and is thus a conservative assumption)

2.5.10 Bearings

a = 10/3 b = 1.483 R = 0.99999875 L = 8.82 hours nd = rev/min af = 1.02 X0 = 0.02 θ = 4.459 degrees L10= 106 rev

2.5.11 Housing

Base is a fixed support to the base of the floor housing

2.5.12 Door hinge

Isotropic homogeneous material Bolts and hinge are made out of titanium Steady state Static equilibrium For analysis at motor stall assume that the force from the plate is only resisted by the

pressure in the vessel (at peak compaction). This will produce the maximum pressure on the door for the worst case scenario.

rBolt = 0.025 m rDoor = 0.50 m

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2.6 Equations

2.6.1 General

nd=S y

σmax

I= 112

w t 3

∑ F y=( m) (a )

∑ M z=Iα=0

∑ F x=ma

σ bending=McI

=lFyI

=¿

Torque T=r × F

Mass M=Wg

Volume V=b ×h × w WeightW ¿−g× M x=∫ vdt

v=∫ adt=¿

x=∫ vdt

σ bending=McI

=lFyI

=¿

τ t=Tcj

I=b h3

12

I=π d 4

64

J= π d4

32

x=∫ vdt → dxdt

=v

v=∫ adt → dvdt

=a

Torque T=r × F

Mass m=WgVolume V=b ×h × wWeight W ¿−g× m

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τ max3 D =

σ1−σ3

2

σ 1,2=σ x+σ y

2±√( σ x−σ y

2 )2

+( τ xy )2

nd=Sy

2 τmax3D

τ max=16

π d3 √( Mr )2+(T )2

∑ Fx=ma V=π r2 L m=Vρ

2.6.2 Usage

x=∫ vdt → dxdt

=v

v=∫ adt → dvdt

=a

a=constant v=at

x=a t 2

2

2.6.3 Load (Bonus)

F compaction=F wall∑ F y=ma=0∑ Fx=maV=π r2 L

m=Vρ

2.6.4 Trash pressure vessel t=ro−ri Pi=10,000+2,000∗v

1−c Po=0

σ l=( Pi∗D i

4∗t ) σ Tmax=

Pi ( Di+t )2∗t σ Tavg=

r i∗P i

t

26

σ l=σ x σ Tmax=σ y T xy=0

σ 1,2=( σ x+σ y

2 )±√( σ x−σ y

2 )2

+T xy2

T 3dmax=( σ1−σ3

2 ) nd=( Sy

2∗T 3 dmax )2.6.5 Push Plate

Aplate=π r2 Co=2 πr PCompaction=

10000+2000 v1−C

FCompaction=PCompaction∗Aplate F edge=

5000∗C1−C

F f =Fedge∗Co Bending stress of plate due to compaction pressure and friction edgeσ bending=

6 Mt 2

M dist=−q a2

C8[ C9

2ab (a2−ro2 )−L17 ], Moment due to pressure

C8=12 [1+v+ (1−v )( b

a )2]

C9=12 [ 1+v

2⋅ ln a

b⋅ 1−v

4 [1−( ba )

2]] L17=

14 {1−1−v

4 [1−( ro

a )4]−( r o

a )2

[1+ (1+v ) ⋅ ln ar o ]}

M edge=−w a2

b[1−v2 ] C9

C8, Moment due to friction edge

σ max=σdist+σedge=6t2 (M dist+M edge), superposition

nd=S y

σmax

∑ F y=ma=0 ∑ Fx=ma V=π r2 L m=Vρ

27

2.6.6 Push Rod

∑ F y=ma=0 ∑ Fx=ma V=π r2 L m=Vρ Pcr

A=C∗π2∗E

( lk )

2

k=√ IA I=π∗D2

64 A=π∗r2

c= 14

( lk )

1=√ 2C π2 E

S y nd=Pcr

P

2.6.7 Motors

nd=T rated

T op

( Nout

N ¿)

min=

F t r5

T rated

( N out

N ¿)

max=

ωrated

vmax

r5

2.6.8 Gears

I=bh3

12; σ b=

McI M max=Ft h

σ b=McI

=F t h( t

2 )1

12b t 2

=6 Ft hb t 2

nd=S y

σ b→σb=

S y

nd

28

S y

nd=

6 F t hb t2 → S y b t 2

6 F t h=nd

2.6.9 Shafts

nd=( 0.577∗Syσb

)


m=ρV V=π r2 L

I=12

mr 2

2.6.11 Bearings Ld=( L∗60∗nd )

X d=( Ld

L10)

C10=(af ∗Fdesign )∗( Xd

(X0+(θ−X 0 )∗(1−R )1b) )

1a

Nd=( C10 R

C10 D)

2.6.12 Housing

nd=( 0.577(Sy)σ b

)

2.6.13 Door hinge

∑ Fx=ma

A=π r2 F=P∗A

τ max=4 V3 A

nd=Sy

2 τmax

29

3. Proposed Design

Figure 17. CAD model of major trash compactor components.

Figure 18. Exposed gear train

30

3.1 Summary of usage modelOur trash compactor is designed to operate under normal operating conditions for 100 years. It is

expected to be used once every day during this time period. This means that over this 100 year period it will operate for a total of 8.82 hours. Normal operating conditions are for compacting trash less than 144000 kg in 8.70 seconds to a compaction of 75%. It will be located in an urban environment near offices, so it is not expected that incredibly high loads will be applied regularly.

3.2 Summary of proposed design

The main details of our design are listed below. For more technical information on specific components, please refer to the tables throughout our report.

Overall: speed: 8.70 secondscosts: $14,800,000maximum load capable of being pushed: 144000 kg

Motor: power: 450 HP (336 kW)rated torque at shaft: 1570 ft*lb (2130 N*m)RPM: 1485 (156 rad/s)Voltage: 380 Vno-load current: 163 Amax current: 604 A

Gears: material: Titanium alloy (Ti-6Al-4V)pitch diameter: 0.225, 0.51, 0.225, 0.714, 0.225 (units in m)number of teeth: 15, 30, 15, 42, 15hub style: hublesspressure angle: 20 degreesbore size: 110 mmface width: 170 mm

Shafts: material: Titanium alloy (Ti-6Al-4V)diameter(s): 100mm, 110mm, 120mm (steps on a single shaft)length: 550mm

Bearings: bearing type: roller bearingsmax radial load: 750,000 NSmallest nd of all components: 2.11 (shaft 2)Smallest performance nd: 1.15 (compaction time)

31

4. Solutions/Calculations4.1 Analysis of usage

1) Problem statement:

Find the appropriate Acceleration to compact at least 50% of the trash in 10 seconds with the nd

time of 1.15 and nd compaction of 1.5. Length of pressure trash vessel is 1/3 the overall length of the trash compactor.

2) Find: Acceleration and deceleration; a X at vmax

t at vmax

vmax

compaction %3) FBD:

N/A

4) Givens: Loverall = 6.7056 [m] nd compaction = 1.5 nd time = 1.15 Lvessel = 2.2352 [m] C = 50% = 0.50 t = 10 [s]

5) Assumption: Acceleration /Deceleration is constant during usage Material can be compacted

6) Equations:

x=∫ vdt → dxdt

=v

v=∫ adt → dvdt

=a

a=constant v=at

x=a t 2

27) Solve:

t= 101.15

=8.6956=8.70[s ]

C=.5(1.5)=0.75=75.0 %

t v max=t2=8.6956

2=4.3782

32

xVmax=Lvessel C

2=

(2.2352)(.75)2

=0.838

x=a t 2

2→a=2x

t2 =2(x¿¿Vmax )

t v max2 =

(2 ) (0.8382 )(4.3782 )2

=0.08868 ¿

a=± 0.09[ ms2 ]

Figure 19. Acceleration / Deceleration of our push plate assembly is +/- 0.09 [m/s2]

v=at vmax=a ( t vmax )= (0.08868 ) ( 4.3782 )=0.38825

vmax=0.39[ms ]

33

Figure 20. Max Velocity is 0.39 [m/s]

x=∫ vdt

x=∫0

t

vdt=∫0

t vmax

v dt+∫t vmax

t

v dt

x=a t 2

2 =2a (t v max )2

2 =a (t v max )2=(0.08868 ) (4.3782 )2=1.69

x total=1.70[m ]

Figure 21. Max position is 1.70 m at 8.80 seconds

C=x total

Lvessel= 1.69987

2.2352=0.75

34

C=75%

Figure 22. Max % Compaction is 75.0% at 8.70 [s]

Table 9. % Compaction at various points of interest.

Compactionnd

compaction

Original Length Length Acceleration Velocity Time nd time

[%] [-] [m] [m] [m/s2] [m/s] [s] [-]

75.0 1.50 2.24 1.70 0.09 0.00 8.80 1.15

38.4 1.50. 2.24 0.86 -0.09 0.39 4.40 1.15

4.2 Analysis of loading for push assembly

This analysis is covered throughout the other analysis sections in order to keep component analyses together.

4.3 Analysis of Load (Bonus)1) Problem statement:

The trash compactor has to compact a load of trash that consists of just cardboard in a pressure vessel with a diameter of 1m. The compactor has to compact the trash to 50% of its original volume in at least 10 seconds and we wanted a design factor of 1.15 for the time and 1.5 for the compaction.

2) Find:

Acceleration needed to have a nd=1.15Vmax with nd=1.15FT, the force required to move the Trash at desired speed

3) FBD:

35

(b) (b)

Figure 23. (a) Schematic of the trash in the pressure vessel where r is the radius of the pressure vessel and LT is the length of the trash. (b) FBD of the Trash where Fcompaction = Fwall.

4) Givens:

Dimensions of Trash Compactor bcompactor = 8 ft = 2.4384 m hcompactor = 8 ft = 2.4384 m Lcompactor = 22 ft = 6.7056 m

5) Assume: The trash only consists of cardboard

Density of Cardboard: ρ=55 kgm3

Dimensions of Trash: r = .5 m LT = 1/3 * Lcompactor = 2.2352 m a = constant Vmax @ t/2 nd = 1.15 for time

6) Equations:F compaction=F wall∑ F y=ma=0∑ Fx=maV=π r2 Lm=Vρ

7) Solve:Since Fcompaction = Fwall, taking the sum of forces in the x-direction results in the following: V=π r2 L=π (.5 )2 (2.24 )=1.75 m3

V=1.76 m3

m=Vρ= (1.75 ) (55 )=96.5 kg m=96.6 kg

∑ Fx=ma=FT

36

FT=ma=(96.5 ) (1.75 )=8.56 N FT=8.56 NTherefore, the force required to move the trash 50% of its original volume in the time frame

with an nd=1.15 isFT=8.56 N .

Table 10. Summary of Compaction analysis with pressure and normal load on push plate

Design Load Design Load w/nd

[Pa] [N] [Pa] [N]

10.9 8.56 12.5 9.84

Using the same methods as shown above and in the analysis for the push plate, find Fcompaction and Fedge

at the worst case for a massive load: when the system is at maximum velocity and the highest compaction during acceleration (a = 0.08868 m/s^2, C = 0.75/2 = 0.375, vmax = 0.3856 m/s^2). From these values we can calculate Pcompaction and Medge, and once we have those, can calculate Fcompaction and Fedge . Now, performing a similar analysis as above but in reverse starting with the calculated force that push rod provides during operation (Fpr = 78540 N), we can work to find mmax, the largest possible load that this trash compactor can push and maintain its operational specs (operational torque input from motor, compaction time, compaction ratio). After doing all the algebra, we end up with this equation:

mload ,max=F pr−F edge−Fcompaction−mplate a

a=

78540 N−9425 N−13536 N−(98.5 kg )(0.3856 ms2 )

0.3856 ms2

mload=144000 kg

4.4 Analysis of push plate

1) Problem statement:Since we know the force required to move the trash within our desired time frame with an

nd= 1.15, we now need to find the force required to move the push plate that has a design factor, nd = 1.5 for compaction and an acceleration nd of 1.15. We also need to find the maximum load that our trash compactor can compress during normal operation.

2) Find Pcompaction

FEdge

t, thickness of the plate Fpr, the force from the push rod required to move the push plate

3) FBD:

37

(b) (b)

Figure 24. (a) Schematic of the push plate with thickness, t, and a radius, r. (b) FBD of the push plate where Ff is the force from the friction edge of the plate, Fcompaction is the resultant force due to compaction, FT, is the force from the trash, and Fpr is the force from the push rod.

4) Givens: r = 0.50 m a = -0.08868 m/s^2 FT=8.56 N q = distributed load (Pcompaction) [Pa] w =distributed load (FEdge) [N/m] a = part radius [m] b = support radius [m] t = plate thickness [m] ro = radius of push rod [m] (inner radius of the distributed load) v = poison’s ratio

5) Assume Plate is made out of Titanium(Ti-6Al-4V) E = 119 GPa = 1.19E+11 Pa Sy = 880 MPa = 8.80E+08 Pa ν=0.342, Poisson’s ratio

ρ=4430 kgm3

C = 0.75 with nd = 1.50, Compaction Ratio a = 0.50 m q = 4.00E+04 Pa w = 15.0E+03 N/m b = ro = 0.055 m We used a negative acceleration because the maximum compaction occurs when the

plate is decelerating. The pressure due to compaction, PCompaction, is uniform on the plate and the force of

friction due to the edge, F edge, of the plate is uniform around the edge of the plate. Forces are at max at max compaction of the system

6) Equations:Aplate=π r2 Co=2 πr

38

PCompaction=10000+2000 v

1−C

FCompaction=PCompaction∗Aplate

F edge=5000∗C

1−C

F f =F edge∗Co

Bending stress of plate due to compaction pressure and friction edge

σ bending=6 Mt 2

Moment due to pressure

M dist=−q a2

C8[ C9

2 ab (a2−ro2 )−L17]

C8=12 [1+v+ (1−v )( b

a )2]

C9=12 [ 1+v

2⋅ ln a

b⋅ 1−v

4 [1−( ba )

2]] L17=

14 {1−1−v

4 [1−( ro

a )4]−( r o

a )2

[1+ (1+v ) ⋅ ln ar o ]}

Moment due to friction edge

M edge=−w a2

b[1−v2 ] C9

C8

Superposition

σ max=σdist+σedge=6t2 (M dist+M edge)

nd=S y

σmax

∑ F y=ma=0 ∑ Fx=ma V=π r2 L m=Vρ

7) Solve:V=π r2 t=π (0.5 )2 (0.0283 )=0.0222m3 m=Vρ= (0.0222 ) ( 4430 )=98.5 kg ∑ Fx=ma=F pr−FT−FCompaction−F f F pr=FT+FCompaction+F f +ma=8.56+31415+47123+(98.5 ) (−0.0887 ) F pr=78539 N=785 kN Aplate=π r2=π (0.5 )2=0.785 m2

Co=2 πr=2 π (0.5 )=3.14 m

39

PCompaction=10000+2000 v

1−C=10000+2000(0)

1−.75=40000Pa

FCompaction=PCompaction∗Aplate=( 40000 ) (0.785 )=31415 N=31.4 kN

F edge=5000∗C

1−C =(5000 )(0.75)

1−0.75 =15000 Nm

F f =F edge∗Co=(15000 ) (3.14 )=47123 N=47.1 kN

Calculating the stress on the titanium plate

C8=12 [1+v+ (1−v )( b

a )2]=1

2 [1+(0.342)+ (1−0.342 )( 0.0550.5 )

2]=0.675

C9=12 [ 1+v

2⋅ ln a

b⋅ 1−v

4 [1−( ba )

2]]=12 [ 1+0.342

2⋅ ln 0.5

0.055⋅1−0.342

4 [1−( 0.0550.5 )

2]]=0.120

L17=14 {1−1−v

4 [1−( ro

a )4]−( r o

a )2

[1+ (1+v ) ⋅ ln ar o ]}=1

4 {1−1−0.3424 [1−( 0.055

0.5 )4]−( 0.055

0.5 )2[1+(1+0.342 )⋅ ln 0.5

0.055 ]}=0.197

M dist=−q a2

C8[ C9

2 ab (a2−ro2 )−L17]=−( 40000 ) 0.52

0.675 [ 0.1202 (0.5 ) (0.55 )

(0.52−0.0552 )−0.197 ] M dist=5090 Nm

M edge=−w a2

b[1−v2 ] C9

C8=

− (15000 )0.52

0.055[1−0.3422 ] 0. .120

0.675=10700Nm

σ max=σdist+σedge=6t2 ( M dist +M edge )= 6

0.02832 (5090+10700 )

σ max=118 MPaNow that we know the maximum bending stress we can obtain an nd for the titanium push

plate by doing the following:

nd=S y

σmax=880 MPa

118 MPa=7.44

The nd calculated above was is for the push plate at operating load. The nd for the push plate at motor stall was calculated using the same method. We assumed that the only force acting on the plate to resist the force applied from the rod was from FEdge while the plate was decelerating at 2*-0.08868 m/s^2. Having all the load at the edge of the plate is the worst case scenario, making this a conservative assumption.

∑ F x=F pr−Fedge=ma

F edge=−ma+F pr=−(98.5 kg )(−2∗0.08868 ms2 )+206509N

(The value for Fpr is calculated in the analysis of the pushrod) F edge=206526 N

M edge=Fedge

2 πr= 206526

2 π∗0.5 m=65739 N

m

We used this value of Medge calculated at motor stall and plugged it into the equations the same way as was done with Medge at operating loads. The resulting design factor at motor stall is shown below.

40

nd=S y

σmax=880 MPa

352 MPa=2.50

Figure 25. Max operating compaction of our system occurs at 0.75 compaction. Above shows functions of load force vs. %compaction

The table below displays the diameter of the plate, the thickness, the max bending stress of the plate at design load, and the max bending stress at motor stall with nd’s for both the max stress.

Table 11. Summary of Push Plate Specifications. Made from titanium with E = 119 GPa, Sy = 880 MPa, and input loads of 8.56 N from the trash, 31.4 kN from the compaction load, and 47.1 kN from the force of friction due to the edge of the plate.

Diameter Thickness

Design load Stresses

nd @ design load

Motor Stall Stresses

nd @ Motor Stall

[m] [mm] [MPa] [MPa] [-]

1 28.3 118 7.44 352 2.5

4.5 Analysis of push rod1) Problem statement:

A titanium pushrod will be used to accelerate the titanium push plate that will be compacting the trash. The radius of the pushrod was assumed in the previous section in the analysis of push plate and was assumed to be 0.055 m. The pushrod must withstand the forces from the push plate and the gear teeth so that buckling does not occur. As for the support of the shaft, the shaft will be supported by a frictionless channel.

41

2) Find: Ft , The tangential force from the pinon gear to the rack. Pcr

nd at operational torque3) FBD:

(a) (b)

Figure 26. a): dimension of Push rod b) FBD of pushrod

42

Figure 27. (a) Schematic of the pushrod where D is the diameter and L is the length of the rod. (b) FBD of the pushrod where -Fpr + Ft

=ma.

4) Givens: Fpr = 78538.53 N a = -0.08868 m/s^2 for operation (Maximum stresses in the plate will occur at

maximum compaction, when the system is decelerating) a = 2*-0.08868 m/s^2 for motor stall calculations (Worst case scenario: Increased

deceleration causes increased stress on the push rod and plate, so we will assume a more rapid deceleration than will be experienced under normal operation to be conservative.)

5) Assumptions: The tangential force from the pinon gear, Ft, acts at a length of 2 m for the buckling

analysis, more than it will ever travel during operation. This will be the worst case scenario.

Homogeneous isotropic material Titanium (Ti-6Al-4V)

o Young’s modulus= 119 GPa = 1.19E+11 Pao Sy = 880 MPa = 8.80E+08 Pa

o ρ=4430 kgm3

Frictionless surface for support Radial force from the gear does not contribute significantly to buckling of the rod D = .110 m r = 0.0550 m Lpr = 2.00 m

6) Equations∑ F y=ma=0 ∑ Fx=ma V=π r2 L m=Vρ

PcrA

=C∗π2∗E

( lk )

2

k=√ IA I=π∗D2

64 A=π∗r2

c= 14

( lk )

1=√ 2 C π2 E

S y nd=Pcr

P7) Solve:

43

Calculations for Operating Load:Taking the sum of forces in the x direction then solving for Ft results in the following: V=π r2 L=π (0.055 )2 (2 )=0.0190 m3

m=Vρ= (0.0190 ) ( 4430 )=84.2kg F t=ma+F pr=F t=(84.2 ) (−0.08868 )+(78538.53 )=78531.07 N

F t=78.5 kN

Now that we calculated the forces acting on the pushrod, we are able to determine if the rod buckles due to the forces when the motor is running at the operational torque. The following calculations for the buckling analysis are shown below:

I= π∗D2

64=

π (0.11 )2

64=7.19∗10−6 m4

A=π∗r2=π (0.055 )2=9.50∗10−3 m2

k=√ IA

=√ 7.19∗10−6

9.50∗10−3 =0.0275 m

lk= 2

0.0275=72.73

( lk )

1=√ 2 C π2 E

S y=√ 2(.25)π2(1.19E+11)

8.80E+08=25.8

Since lk>( l

k )1, it is appropriate to use Euler’s equation for buckling to find the critical load, Pcr.

Pcr=C∗π 2∗E

( lk )

2 A=(0.25 ) π 2 (1.19E+11 )

( 20.0275 )

2 ( 9.50∗10−3 )=527554.54 kN

Pcr=528 kN

Now that we have obtained the critical load, Pcr, we are able to determine the design factor of the pushrod using the following equation:

nd=PcrP

= PcrFt

=527554.5478531.07

=6.72

Therefore the design factor for the titanium pushrod when the motor is running at operating torque is 6.72.

Calculations at Motor Stall (Accounting for Mass Moment of Inertia of Gears (bonus))

The first thing to find is the force of the tooth on gear 5 acting on the push rod when the motor is outputting its stall torque.

Gear 1:

44

α gear 1=

N 2

N 1∗N 4

N3∗a

r5=

30 teeth15 teeth

∗42teeth

15 teeth∗2∗−0.08868 m

s2

0.1275 m =−7.790 rad / s2

∑ M =T stall−Ft ,12r 1=I gear 1 α gear 1

F t ,12 r1=T stall−I gear 1 α gear 1=( 4683 Nm )−(0.3126 kg∗m2)¿)

F t ,12=4685 Nm0.1275 m

=36745 N

Gear 2:

α gear 1=

N4

N3∗a

r 5=

42 teeth15 teeth

∗2∗−0.08868 ms2

0.1275 m=−3.895 rad /s2

∑ M =−T2+F t , 12r2=( I¿¿ gear 2+ I shaft)(α gear 2)¿

T 2=Ft ,12 r2+( I ¿¿gear 2+ I shaft) (−α gear 2 )¿T 2=(36745 N ) ( 0.255m )+(5.002+0.03472kg∗m2)¿3.895 rad/s^2) T 2=9389.7 Nm

Gear 3:

∑ M =T2−F t ,34 r3=(I ¿¿ gear 3)(−α gear 3)¿

F t ,34 r3=T 2+( I ¿¿ gear 3) (−αgear 3 )=9389.7 Nm+( 0.3126 kg m2 )(3.895 rads2 )=9390.9 Nm¿

F t ,34=9390.9 Nm0.1275 m

=73653 N

Gears 4 and 5:

To find the force of gear 5 on the rack attached to the push rod, repeat the same process used above for gears 2 and 3. This yields the result below:

F t=206494 N

If we had not accounted for the mass moment of inertia of the gears and shafts, we would end up with this smaller value:

45

F t , noinertia r5=

N2

N1∗N 4

N3∗T stall=

30 teeth15 teeth

∗42teeth

15 teeth∗4683 Nm=26225 Nm

F t , noinertia=26225 Nm0.1275 m

=205686 N

The calculations above show that accounting for the inertia in the gears during deceleration at twice the rate experience during operation increases the load on the push rod slightly, thus making it a conservative assumption.

The tangential force applied to the rack by gear 5 is known, so now we can calculate the maximum buckling force on the push rod.

∑ F x=F t−F pr=mrod a

F pr=F t−mpr a=(206494 N )− (84.20 kg )(2∗−0.08868 ms2 )=206509 N

F pr=206509 N

The max force compressing the push rod during motor stall is 206509 N, so use this value as the value of P.

nd=PcrP

= PcrFpr

=527554206509

=2.55

Table 12. Summary of Push Rod Specifications. Made from Titanium alloy (Ti-6Al-4V) with E = 119 GPa, Sy = 880 MPa, and input loads of 78500 N during operation and 207000 N at stall from the compaction load.

Radius Length ( lk )

1( l

k )sys

Pcr

Design/Operating Max. Loading

Load nd Load nd

[mm] [m] [-] [-] [N] [N] [N] [-]

55.0 2.00 25.8 72.7 528000 78500 6.72 207000 2.55

4.6 Analysis of hoop and tangential stresses in pressure vessel

1) Problem Statement:The hoop stress that pressure vessel experiences comes from the compaction of the trash

and from the sealing of the push plate edge. The vessel was modeled as a thin walled pressure vessel that is made from titanium. We need to find the design factor at operating pressures and the pressures called at motor stall.

46

2) Find: The design factor for the chosen thickness of our cylindrical vessel.

3) FBD:

Figure 28. (a) This is an FBD of the pressure vessel and it shows the forces acting on the outside. (b) FBD of the inside of the pressure vessel and it shows how the forces are pushing against the inner wall.

4) Givens:Inner radius= 0.5mLength=2.20m

5) Assume: The forces that are acting on the pressure vessel are perfectly uniform and act

orthogonal to the surface of the inside and outside of the vessel. Pressure outer is 0 Pa. Pressure inner is from the pushrod force over the area of the plate. Also, assume an outer radius since an outer radius was not provided. The material for the vessel is titanium and it has the following characteristics: Young’s modulus: 119 GPa Yield strength: 880 MPa Outer radius= 0.52 m

6) Equations: t=ro−r i

Pi=10,000+2,000∗v

1−c Po=0

σ l=( Pi∗D i

4∗t )

47

σ Tmax=Pi ( Di+t )

2∗t

σ Tavg=ri∗P i

t σ l=σ x

σ Tmax=σ y

T xy=0

σ 1,2=( σ x+σ y

2 )±√( σ x−σ y

2 )2

+T xy2

T 3 dmax=( σ1−σ3

2 ) nd=( Sy

2∗T 3 dmax )7) Solve:

Pi=10,000+2,000∗v

1−c=10,000+2,000∗0

1−0.75=40,000 Pa

Po=0

ro=0.52 m

t=ro−r i= 0.52-0.5 = 0.02 m

σ l=( Pi∗D i

4∗t )= ( 40,000 Pa )∗(1m)4∗(0.52 m−0.5 m )

=500,000 Pa

σ Tmax=Pi ( Di+t )

2∗t=

40,000 Pa (1m+0.02 m )2∗0.02 m

=1,020,000 Pa

σ 1,2=( σ x+σ y

2 )±√( σ x−σ y

2 )2

+T xy2

σ 1,2=( 0.5 MPa+1.02 MPa2 )±√( 0.5 MPa−1.02 MPa

2 )2

+02=760,000 ± 260,000

σ 1,2=1,020,000 Pa∧500,000 Pa

T 3 dmax=( σ1−σ3

2 )=1,020,000−02

=510,000 Pa

nd=( Sy2∗T 3 dmax )=( 880,000,000 Pa

2∗(510,000 Pa ) )=863

The value above was calculated for the hoop stress under operating conditions and the same steps were followed for the hoop stress at stall. The following calculations show the hoop stresses at stall.

48

F rod=206,509 N(Frod came from the push-plate analysis)

a=−2∗(0.08868( ms2 ))=−0.17736 m

s2

m plate=98.96 kg

F compaction=206,509 N+98.96∗(2∗0.08868 ms2 )=207,000 N

F compaction

A plate=206,526 N

π∗(0.5 m )2=262,957 Pa

Pi=262,000 Pa Po=0 ro=0.52 mt=ro−r i= 0.52-0.5 = 0.02 m

σ l=( Pi∗D i

4∗t )= (262,957 Pa )∗(1 m)4∗( 0.52m−0.5m )

=3,284,000 Pa=3.28 MPa

σ Tmax=Pi ( Di+t )

2∗t=

26,957 Pa (1m+0.02 m )2∗0.02 m

=6,700,000 Pa=6.70 MPa

σ 1,2=( σ x+σ y

2 )±√( σ x−σ y

2 )2

+T xy2

σ 1,2=( 3.28 MPa+6.70 MPa2 )±√( 3.28 MPa−6.70 MPa

2 )2

+02=4.99 MPa ±1.71 MPa

σ 1,2=6.70 MPa∧3.28 MPa

T 3dmax=( σ1−σ3

2 )=6.70 MPa−02

=3.35 MPa

nd=( Sy2∗T 3 dmax )=( 880 MPa

2∗(3.35 MPa ) )=131.3

Table 13. Summary of Push Plate Specifications. Made from titanium with E = 119 GPa, Sy = 880 MPa, and input loads of 31,400 N and 206,500 N from the compaction load. (note, add additional information as needed)

ro ri tDesign/Operating Max. Loading

σl σt nd σl σt nd

[m] [m] [m] [MPa] [MPa] [-] [MPa] [MPa] [-]

0.52 0.50 0.02 0.50 1.02 863 3.28 6.70 131

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4.7 Analysis of Motor

1) Problem Statement:

Calculate the design factor for the motor based upon its rated torque and the torque required to operate the trash compactor. Ensure that the chosen gear ratio is within the allowable range based upon the specifications of the motor.

2) Find: The design factor of the motor during intended operation and the minimum and

maximum gear ratios needed for our system to operate. Also find the gear ratio of our actual gear train.

3) Schematics and FBD:

Figure 29. The torque applied from gear 1 onto the shaft of the motor during operation.

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Figure 30. Free body diagram of gears and shafts including the mass moments of inertia.

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4) Given:

Figure 31. Motor torque curve.

Table 14. Summary of Motor Specifications (note, add additional information as needed)



-W22 WEG 450 336 1490 1570 2130 3450 4680

No-load current: 163 AMaximum current: 603 AVoltage: 380 V (3-phase motor)

Tangential force on teeth of gear 5 during operation: F t=¿78530 N

5) Assumptions: The motor specifications provided by the vendor are accurate. The rated torque of the motor is the maximum torque that the motor can output

during normal operation (the torque at which the motor fails). The mass moment of inertia in the gears and shafts are significant. The acceleration of the push rod in the system is the acceleration during operation

o a = 0.08868 m/s2

Max velocity of the push rod: vmax = 0.386 m/s

6) Equations:

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nd=Trated

T op

( Nout

N ¿)

min=

F t r5

Trated

( N out

N ¿)

max=

ωrated

vmax

r5

7) Solution:The torque applied on the motor shaft, Top , was calculated factoring in the mass

moments of inertia for all of the gears and the shafts. As such, the calculations take up a lot of space. For the process of calculating the torques and tangential forces from gear to gear, see the push rod analysis at motor stall for an example of the way in which Top

below was calculated.T op=1797 Nm

We verified the calculations of Top above by calculating the operational torque neglecting the mass moment of inertia in the shafts and gears.

T op ,no inertia=

N 1

N 2∗N3

N4∗F t r5=

15 teeth30teeth

∗15teeth

42teeth∗(78530 N ) (0.1275 m )

T op ,no inertia=179 Nm

As seen above, including the mass moments of inertia for the gears and shafts increases the torque required during normal operation by a small margin. Now calculate the design factor using the Top calculated factoring in inertia.

nd=T rated

T op= 2129 Nm

1797 Nm=1.18>1.15(BONUS POINTS )

To ensure that the chosen motor is capable of making our trash compactor operate as designed, the minimum and maximum values for the gear ratio were calculated based upon the rated motor torque and the rated motor angular velocity, respectively.

( N out

N¿)

min=

F t r5

Trated=

(78530 N ) (0.1275 m )2129 Nm

=4.70

( N out

N ¿)

max=

ωrated

vmax

r5

= 156 rad /s0.386 m/ s0.1275m

=51.5

The gear ratio of the actual gear train is shown below:

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N out

N ¿=

N2

N1∗N 4

N 3=

30 teeth15teeth

∗42 teeth

15teeth=5.60

The gear ratio of the gear train is within the required range, so the trash compactor will be able to operate successfully with this motor.

4.8 Analysis of Gears

1) Problem statement: Calculate the design factor of the gears during normal operation and when the motor is at its stall torque.

2) Find: Design factor during operation and at motor stall.

3) Schematics and FBDs:

Figure 32. How we found the gear tooth dimensions

Figure 33. We assumed the gear teeth to be cantilevered beams with the dimensions gathered.

4) Given:

Table 15. Summary of Gear Specifications. All gears have a 20 ˚ pressure angle and are made from Titanium alloy (Ti-6Al-4V) with assumed E = 119

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GPa and Sy = 880 MPa. Gears are custom machined but were based of of gears purchased from SD-SI.

Gear # Mounted On Part # m N b h t

[-] [-] [-] [mm] [-] [mm] [mm] [mm]1 Motor KSUS4-15 17.0 15 170 34 26.72 Shaft 1 KSUS4-30J22 17.0 30 170 34 26.73 Shaft 1 KSUS4-15 17.0 15 170 34 26.74 Shaft 2 KSSG4-42 17.0 42 170 34 26.75 Shaft 2 KSUS4-15 17.0 15 170 34 26.7

5) Assumptions: Homogeneous isotropic material Static Equilibrium Titanium (Ti-6Al-4V)

Young’s modulus= 119 GPa = 1.19E+11 Pa Sy = 880 MPa = 8.80E+08 Pa

ρ=4430 kgm3

Assume that the radial force acting on the gear tooth is negligible for calculating the stress in the tooth.

6) Equations:

I=b t3

12; σ b=

McI

M max=Ft L

7) Solution: Analysis of gear 5 at operation loads:

σ b=McI

=F t L( t

2 )1

12b t2

=6 F t Lbt 2 =

6(78539 N )(0.034 m)(0.170m)(0.0267 m)2 =1.32∗108 Pa

nd=S y

σ b=

8.80∗108 Pa1.32∗108 Pa

nd=6.66

This report accounts for the inertia of the gears and shafts, so finding how the torques and forces transfer through the gears is complex takes a lot of space. Please refer to the stall calculations done for the push rod for an example of how the force on the teeth was calculated from gear to gear. Once the tangential force on a gear tooth has been calculated, use the same process as above to calculate the design factor.

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Table 16. Summary of Gear loading and stress analysis. Worst case scenario assumed at motor stall. NOTE: design load is the load necessary to compact the trash under operating conditions.

Gear # At Design Loads At motor stall (worst case scenario)

T Ft Fr σb nd T Ft Fr σb nd

[-] [N-m] [N] [N] [MPa] [-] [N-m] [N] [N] [MPa] [-]

1 1800 14100 5130 23.7 37.1 4680 36700 13400 61.8 14.2

2 3580 14100 5130 23.7 37.1 9390 36700 13400 61.8 14.2

3 3580 28100 10200 47.3 18.6 9390 73700 26800 124 7.09

4 10010 28100 10200 47.3 18.6 26300 73700 26800 124 7.09

5 10010 78500 28600 132 6.66 26300 206,50

0 75200 348 2.53

4.9 Analysis of Shafts

1) Problem statement:The shaft must withstand the calculated reactions from where the bearings are reacting

and the gears. In order to calculate all of the stresses on the shaft we used Skyciv. Then the design factor, nd, can be determined.

2) Find: Maximum Von Mises Stress, σmax Design Factor, nd

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3) FBDs and Schematics:

Figure 34. FBD of shaft 1. Note: bearings 1 and 2 go left to right, respectively for each shaft.

Figure 35. This is the schematic of shaft 1 produced by Skyciv with dimensions and forces.

4) Givens: Stepped shafts must be used nd ≥ 2

5) Assume:

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All shafts are titanium E = 119 GPa = 1.19 E+09 Pa Sy = 880 MPa = 8.80 E+06 Pa Diameters: see figure 35 Length of each section: see figure 35 Each step increases the diameter by 5mm Stress concentrations was included in Skyciv analysis

6) Equations:

nd=(0.577(Sy)

σ b)

7) Solve:

Figure 36. This shows the reaction forces that are acting on the shaft.

Figure 37. This is the results table from the Skyciv analyzing the shaft.

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Figure 38. Von Mises Stress and stress concentrations.

nd=( Syσb )=( 0.577∗880 MPa

50.133 MPa )=10.1

This is the nd under operating conditions and the nd at motor stall is calculated the same way using the Von Mises Stress for stall torque. The value for the maximum Von Mises Stress was pulled from the table that Skyciv produced and it is attached in the appendix.

nd=( Syσb )=( 0.577∗880 MPa

131.149 MPa )=3.87

Table 17. Summary of Shaft Specifications. Shafts made from titanium with assumed E = 119 GPa and Sy = 880 MPa. Shaft stock purchased from McMaster-CARR and machined to size.

Shaft # Diameter LengthLocations

Gear A Gear B Left Bearing

Right Bearing

[-] [mm] [mm] [mm] [mm] [mm] [mm]1 120 550 150 315 30 5202 120 550 150 315 30 520

Table 18. Summary of Shaft loading and stress analysis. Worst-case scenario assumed at motor stall.

Shaft #At Design Loads At motor stall (worst case scenario)

RLeft

Bearing

RRight

Bearingσmax nd RLeft Bearing RRight Bearing σmax nd

[-] [N] [N] [MPa] [-] [N] [N] [MPa] [-]1 17,600 25,500 50.1 10.1 45,900 66,700 131 3.882 68,900 40,500 132 3.85 106,000 181,000 240 2.11

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4.10 Shafts Mass Moment of Inertia (bonus)1) Problem Statement:

Find the mass moment of inertia of 2 shafts and the 5 gears in the gear train.

2) Find: The mass moment of inertia of each shaft and gear.

3) Schematic: see figure 35(FBD not applicable in this section, we are only finding the moment of inertia)

4) Assumptions: Uniform density of material Gears and shafts spin perfectly about their centers Assume gears are perfect cylinders with no hole in the center (This will yield a higher

moment of inertia than reality and is thus a conservative assumption)5) Given:

Density of Titanium alloy: 4430 kg/m^3 Pitch radius of gear 1: r = 0.1275 m Thickness of gear 1: t = 0.17 m

6) Equations:

m=ρV V=π r2 L

I=12

mr 2

7) Solve:

Find the moment of inertia of each part of the shaft that has a different radius and then sum those values to find the total moment of inertia of each shaft.

msmall=ρ π r2 L=(4430 kgm3 )∗π∗(0.05 m )2∗(0.06 m∗2 )=4.175 kg

mmedium=ρ π r2 L=(4430 kgm3 )∗π∗(0.055m )2∗(0.175 m∗2 )=14.735 kg

mlarge= ρ π r2 L=(4430 kgm3 )∗π∗(0.06 m )2∗(0.08 m )=4.008 kg

I small=12

m r2=0.5∗4.175 kg∗(0.05m )2=0.005219 kg∗m2

I medium=12

mr2=0.5∗14.74 kg∗(0.055m )2=0.02229 kg∗m2

I large=12

m r2=0.5∗4.008 kg∗(0.06 m )2=0.007214 kg∗m2

I shaft=I small+ I medium+ Ilarge=0.005219+0.02229+0.007214 kg∗m2

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I shaft=0.0347 kg∗m2

mgear 1=ρ π r2 t=(4430 kgm3 )∗π∗(0.1275 m )2∗(0.17 m )=38.46 kg

I gear 1=12

m r2=0.5∗38.46 kg∗(0.1275 m )2=0.3126 kg∗m2

I gear 1=0.313 kg∗m2

4.11 Analysis of Bearings1) The bearings must withstand forces that come from the shaft and the gears meshing with each

other. The bearings can be modeled as either roller bearings or ball bearings depending on the design.

2) Find C10 Nd

3) FBD:

Figure 39. Forces acting on Bearings From Skyciv shaft analysis

Figure 40. FBD of Bearing 1

4) Givens: The trash compactor must last 100 years C10 R = C10 rated

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C10 D = C10 designed Fr = 17600 N

5) Assume: Bearing one analysis a = 10/3 b = 1.483 R = 0.99999875 L = 8.82 hours nd = rev/min af = 1.02 X0 = 0.02 θ = 4.459 degrees L10= 106 rev

6) Equations: Ld=( Lsquiggly∗60∗nd squiggly )

X d=( Ld

L10)


(X0+(Theta−X0 )∗(1−R )1B ) )

1A

N d=( C10 R

C10 D)

7) Solve: F r=17,600 N

Ld=( L∗60∗nd )=8.82hrs∗60∗742.5( revmin )=393,000 rev

X d=( Ld

L10)=( 393,000

106 )=0.393


(X0+(θ−X 0 )∗(1−R )1b) )

1a

C10=(1.02∗17,600 )∗( 0.393

0.02+(4.459−0.02 )∗(1−0.99999875 )1

1.483 )1103

C10=17,962.91 N

Nd=( C10 R

C10 D)=( 750,000 N

17,900 N )=41.753

N d=41.8

This is the nd under the operating conditions of the system. Calculating the nd for the stall conditions used the same equations and approach but it utilized a different number for the force radial which came from Skyciv and can be seen in the appendix for shaft 1 at stall.

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F r=45,916.62 N

Ld=( Lsquiggly∗60∗nd squiggly )=8.82 hrs∗60∗742.5( revmin )=392,931rev

X d=( Ld

L10)=( 392,931

106 )=0.393


(X0+(Theta−X0 )∗(1−R )1B ) )

1A

C10=(1.02∗45,916.6 )∗( 0.393

0.02+ (4.459−0.02 )∗(1−0.99999875 )1

1.483 )1

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C10=125,413.5 N

N d=( C10 R

C10 D)=( 750,000 N

125,413.5 N )=5.98

Nd=5.98Table 19. Summary of Bearing Specifications. Bearings were purchased

from Grainger. ID = inner diameter, OD = outer diameter.

Bearing #

Mounted On Part # ID OD Rated

Load

[-] [-] [-] [mm] [mm] [kN]1 Shaft 1 NU2322-E-TVP2 110 240 7502 Shaft 1 NU2322-E-TVP2 110 240 7503 Shaft 2 NU2322-E-TVP2 110 240 7504 Shaft 2 NU2322-E-TVP2 110 240 750

Table 20. Summary of Bearing load analysis. nd based upon rated load, not accounting for wear and usage. ID = inner diameter, OD = outer diameter.

Bearing #

Rated Load

At Design Loads At motor stallLoad from

Shaft ndLoad from

Shaft nd

[-] [kN] [N] [-] [N] [-]1 750 17,600 41.8 45,900 5.982 750 25,500 28.8 66,800 4.113 750 68,900 10.7 106,200 3.654 750 40,500 18.2 181,000 2.14

4.12 Analysis of Housing

1) Problem statementTo solve for the nd for the bearing housing we were able to run a simple FEA analysis to

solve for the Max stress. This max stress was calculated under the stall torque under max loading. The reaction forces on the bearings were solved for in the shaft analysis

2) Find

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Design factor of housing; nd

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3) Free body diagrams and schematics:

Figure 41. Schematic showing the applied forces. NOTE: We accidentally entered the forces in lbf instead of Newtons like we intended. However, since 1 lbf is 4.45 N, this actually means that we stressed our bearing holders further than they will actually experience and they still passed with a high design factor, making our mistake actually a conservative assumption.

4) Givens:Below are the Y-Z components of the reaction forces from the shafts that were input into

ANSYS FEA at each bearing. F1y = 3500 NF1z = -45800 NF2y = 17000 NF2z = -64600 NF3y = 1900 NF3z = 66100 NF4y = -726 NF4z = 40500 N

5) Assume:Base is a fixed support to the base of the floor housing

6) Equations:

nd=(0.577(Sy)

σ b)

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7) Solve:

Figure 42. Mesh of nodes created in FEA ANSYS

Figure 43. Above is a FEA plot showing Von-Mises stresses at stall torque

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nd=( 0.577 (Sy )σb )

nd=( 0.577 (880,000,000 )(6,510,000 ) )

nd=¿78.0

Figure 44. Above shows an additional analysis of max deformation. The max deformation occurs at the top of the bracket and it is 0.10 mm. this shows that the defection is negligible.

4.13 Door Hinge/Door Latch/Motor Bolts1) Problem statement:

The trash compactor has a door at the end of the compactor vessel. This door must withstand the forces of compaction at 75% compaction when our pushrod assembly stops. The door is held in place by two pins of equal material and dimensions, but one can be removed via a handle, making this pin serve as a latch. Find the design factor for the identical pins that serve as the hinge pin and the door latch. Also find the design factor for the motor bolts holding the motor in place.

2) Find: nd hinge nd bolts

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3) FBDs and Schematics:

(a) (b) (c)

Figure 45. (a) Schematic of the pin and door assembly looking from the side. (b) FBD of the door with the distributed load due to compaction. (c) FBD of the bolt with shear forces.

Figure 46. CAD image of door hinge design

4) Givens:

r = 0.50 m Hinge pins are made out of Titanium(Ti-6Al-4V)

o E = 119 GPa = 1.19E+11 Pao Sy = 880 MPa = 8.80E+08 Pa

Motor bolts are made out of black oxide alloy steelo Sy = 1.17 GPa

Pcompaction=40000 Pa, refer to analysis of push plate Pcompaction=262957 Pa (AT PEAK COMPACTION), refer to hoop stress analysis

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5) Assume: Isotropic homogeneous material Hinge pins are made out of titanium with material properties listed above Motor bolts are made out of black oxide alloy steel with properties listed above Steady state Static equilibrium For analysis at motor stall assume that the force from the plate is only resisted by the

pressure in the vessel (at peak compaction). This will produce the maximum pressure on the door for the worst case scenario.

rBolt = 0.025 m rDoor = 0.50 m

6) Equations:∑ Fx=ma

A=π r2 F=P∗A

τ max=4 V3 A

nd=Sy

2 τmax

7) Solve:

ADoor=π r2=π (0.5 )2=0.785 m2 FDoor=(P¿¿Compaction∗A)=(40000)(0.785)=31.4 kN ¿

Solving for FBolt in Fig. 45b by summing the forces results in the following: ∑ Fx=ma=F Door−2 FBolt

FBolt=F Door

2=31.4 kN

2=15.7 kN

Now that we know the force on the bolts from the door, can determine the shear forces on both edges on the bolt in Fig45c.

∑ Fx=ma=−FBolt+2V

V=FBolt

2=15.7 kN

2=7.85 kN

Since we have the shear force on the bolt we can now determine nd

ABolt=π r2=π (0.025 )2=0.00196 m2

τ max=4 V3 A

= ( 4 ) (7.85 kN )(3 ) (0.00196 )

=5.34 MPa

nd=Sy

2 τmax= 880 MPa

(2 ) (5.34 MPa )=82.4

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The design factor at operating conditions is 82.4. For calculating the design factor at motor stall the same procedure was used as above.

FDoor=(P¿¿Compaction @ peak∗A)=(262957)(0.785)=206 kN ¿

FBolt=F Door

2=206 kN

2=103 kN

V=FBolt

2=103 kN

2=51.5 kN

τ max=4 V3 A

=( 4 ) (51.5 kN )(3 ) (0.00196 )

=35.0 MPa

nd=Sy

2 τmax= 880 MPa

(2 ) (35.0 MPa )=12.6

Therefore the design factor at motor stall is nd = 12.6.

The bolts holding the motor were calculated in the same manner as above. The only difference is that instead of 2 shear forces resisting the applied load, there is just a single shear force. The applied load for on these bolts is ¼ of the tangential force acting on gear.

During operation: F t ,12=T op

r1=1797 Nm

0.1275 m=14090 N

At motor stall: F t ,12=T op

r1= 4683 Nm

0.1275 m=36730 N

Applying these values in the same manner as above for the bolt dimensions, we end up with a design factor of 226 during operation and 86.6 at motor stall.

5. Overall SolutionOverall:

speed: 8.70 secondscosts: $14,800,000maximum load capable of being pushed: 144000 kg

Motor: power: 450 HP (336 kW)rated torque at shaft: 1570 ft*lb (2130 N*m)RPM: 1485 (156 rad/s)Voltage: 380 Vno-load current: 163 Amax current: 604 A

Gears: material: Titanium alloy (Ti-6Al-4V)pitch diameter: 0.225, 0.51, 0.225, 0.714, 0.225 (units in m)

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number of teeth: 15, 30, 15, 42, 15hub style: hublesspressure angle: 20 degreesbore size: 110 mmface width: 170 mm

Shafts: material: Titanium alloy (Ti-6Al-4V)diameter(s): 100mm, 110mm, 120mm (steps on a single shaft)length: 550mm

bearings: bearing type: roller bearingsmax radial load: 750,000 NSmallest nd of all components: 2.11 (shaft 2)Smallest performance nd: 1.15 (compaction time)

5.1 Bill of MaterialsTable 21. Bill of Materials

Part Name Vendor Part # Units $/unit Qty. # of machining

operations$/unit +

machining costs Extended Costs

[-] [-] [-] [-] [$] [-] [-] [$] [$]Gear 1 SDP-Si KSUS4-15 1 $17,900 1 1 $17,900 $17,900Gear 2 SDP-Si KSUS4-30J22 1 $50,000 1 1 $50,000 $50,000Gear 3 SDP-Si KSUS4-15 1 $17,900 1 1 $17,900 $17,900Gear 4 SDP-Si KSSG4-42 1 $39,700 1 1 $39,700 $39,700Gear 5 SDP-Si KSUS4-15 1 $17,900 1 1 $17,900 $17,900

Sidewall McMaster-CARR Plate-01 1 $274,000 4 4 $274,020 $1,097,000Front/Back plate

McMaster-CARR Plate-02 1 $99,700 2 2 $99,710 $199,000

Bearings Grainger NU2322-E-TVP2 1 $2,390 4 1 $2,400 $9,580Pushing

plate McMaster-CARR Plate-03 1 $143,000 1 1 $143,000 $143,000

Vessel McMaster-CARR Plate-04 1 $908,000 1 1 $908,005 $908,000Bearing

bolts McMaster-CARR 91290A156 1 $56.50 8 0 $56.5 $451

Shaft 1 McMaster-CARR Round stock-01 1 $2,200 1 6 $2,200 $2,200Shaft 2 McMaster-CARR Round stock-02 1 $2,200 1 6 $2,200 $2,200

Motor WEG 50018EP3GKD580Z-W22

1 $26,400 1 0 $26,400 $26,400

Base McMaster-CARR Block-01 1 $12,200,000 1 6 $12,200,000 $12,200,000Hinge McMaster-CARR Block-02 1 $5,140 1 4 $5,160.00 $5,160.00

*It is rounded because of 3 sig figs

Total Cost = $14,800,000

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6. Appendix Solving for the cost of our parts: Our group went about solving for the cost of gears and bearings by

creating a few plots that show cost versus gear size and cost versus load. These graphs are very non-linear and when using tools in matlab the cost equation that is generated is not precise. The way we went about creating an accurate cost difference for titanium versus steel components is we first chose equivalent pieces of steel and titanium based on volume. The volume we chose to go with for bar stock was 6.15 E-06 m^3 and the cost for the steel was $1.37, while the titanium it was $39.58. This produced a cost of 29 times more for titanium than steel. When it came to picking gears, bearings, shafts, and the mount we designed all of it in 3D modeling software and got the volume of each of the components and multiplied it by the cost of titanium. This produced an accurate cost of each of component. Also, each machining procedure came with a cost of $5 and each of the procedures were accounted for.

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Documents

GP03 Trash Compactor G07