Final_Report_Example_group_1

Final_Report_Example_group_1 - Group 1: Bicycle HandGroup...

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Unformatted text preview: Group 1: Bicycle HandGroup warming Device Design Considerations Design Problems: Cold fingers sting and chap Cold hands and gloved hands are Cold unresponsive unresponsive Goal: Provide effective hand-warming solution solution Primary Consideration: Warming of hands Secondary Consideration: Durable Current Solutions Current Cumbersome to wear Excess Luggage Water/Wetness Issue Makes Hands Sweat Rain/Snow problems TRIZ Analysis TRIZ temperature vs. volume of object Universality. Make a part or object perform Universality. multiple functions; eliminate the need for other parts. other the device might have enough volume so as the to deflect chap-inducing cold winds. The housing of the device might be curved to direct the wind away from the hands aiding in the primary function. the TRIZ Analysis TRIZ improve temperature without adversely improve affecting the adaptability affecting An inexpensive short-life object instead of An an expensive durable one. Use a battery powered device so that battery Use can be changed as opposed to replacing whole system. whole “Since when can you buy just one sock or just Since one glove? one TRIZ Analysis TRIZ maximize volume with a constraint of maximize shape shape Nesting. Place one object inside another; Nesting. place each object, in turn, inside the other; make one part pass through a cavity in the other. other. Place heating device within the cavity of outer Place shell. [Or] Have hollow tube and fill with components. components. Design Blueprint Design Focus of Performance Index Focus Bracket Hollow Cylinder Construction Experiences both a force and a torque Requires Von Mises failure criteria Electrical Wire Cylinder rod conductor Dissipates heat P = V*I Bracket Performance Index Derivation Derivation Objective: Minimize Mass m = 2 ⋅π ⋅ L ⋅ t ⋅ ρ Constraint: Avoid Failure Von Mises Failure Criteria 2 2 σ ' = (σ x + 3τ xz )1 / 2 τ xz T ⋅r 16 ⋅ T ⋅ d = = J π ⋅ (d o4 − d i4 ) σx = M ⋅C 32 ⋅ L ⋅ F ⋅ d = I π ⋅ (d o 4 − d 4 ) L ⋅ d [4 ⋅ F 2 ⋅ L2 + 3 ⋅ T 2 ]1 / 2 σ= 4 ⋅π ⋅ r 3 ⋅ t ' t= L ⋅ d [4 ⋅ F ⋅ L + 3 ⋅ T ] 4 ⋅π ⋅ r 3 ⋅σ ' 2 2 2 1/ 2 L ⋅ d [4 ⋅ F 2 ⋅ L2 + 3 ⋅ T 2 ]1 / 2 m= 2⋅r2 ρ ' σ σ' M= ρ Bracket Material Choices Bracket Material Steel Alloys Steel alloy A36, hot rolled Steel alloy 1020, hot rolled Steel alloy 1020, cold drawn Stainless Steels Stainless alloy 304 Stainless alloy 316 Stainless alloy 405 Stainless alloy 440A Aluminum Alloys, Heat treated Alloy 1100 Alloy 2024 Alloy 6061 Alloy 7075 Alloy 356.0 Magnesium Alloys Alloy AZ31B, Extruded Alloy AZ91D, as cast Titanium Alloys Alloy Ti-5Al-2.5Sn Alloy Ti6Al-4V Polymers LDPE HDPE UHMWPE PMMA PVC Density, ρ (g/cm^3) Yield Strength, σ (Mpa) performance index, M=σ/ρ 7.85 7.85 7.85 220 210 350 28.02547771 26.75159236 44.58598726 8 8 7.8 7.8 205 205 170 415 25.625 25.625 21.79487179 53.20512821 2.71 2.77 2.7 2.8 2.69 117 345 276 505 164 43.17343173 124.5487365 102.2222222 180.3571429 60.96654275 1.77 1.81 200 100 112.9943503 55.24861878 4.48 4.43 760 830 169.6428571 187.3589165 0.925 0.959 0.94 1.19 1.4 14 33 28 73 44 15.13513514 34.41084463 29.78723404 61.34453782 31.42857143 Callister, W.D. (2000), Materials Science and Engineering: An Introduction, 5th ed., John Wiley (New York). Bracket Material Choice Final Selection Selection Material Steel alloy 1020, cold drawn Stainless alloy 440A Alloy 6061 Alloy AZ31B, Extruded Alloy Ti6Al-4V PMMA Density, ρ (g/cm^3) 7.85 7.8 2.7 1.77 4.43 1.19 Yield Strength strength, σ (Mpa) performance index, M=σ/ρ Cost ($US/kg) 350 44.58598726 415 53.20512821 276 102.2222222 200 112.9943503 830 187.3589165 73 61.34453782 Aluminum Alloy 6061 is the best material choice because of its low density, high strength, reasonable cost and availability. Choose Aluminum Alloy 6061 ¾” OD tubing at $1.12/foot. http://www.windevor.com/tubing.htm 0.85 5 6.2 8.8 60 4.2 Design Specifications Design Bracket Tube Aluminum Alloy 6061 Schedule 10 Tubing • ¾” O.D. • 0.083” Thickness • Estimated Cost $1.12/foot Wire Wire Objective: Minimize Objective: Mass Mass Constraint: Power Constraint: Dissipation Dissipation Free Variable: cross Free sectional area sectional P =V ⋅I m = A ⋅ L ⋅ ρd P= R=ρ⋅ L A m= P⋅L ⋅ L ⋅ ρd V 2 ⋅σ σ M= ρd => I= V R V2 R ρ= V 2 ⋅σ ⋅ A P= L V = I ⋅R 1 σ => L σ⋅A P⋅L A= 2 V ⋅σ R= P m= 2 V 2 ρd ⋅ L ⋅ σ () Note: It makes sense that a good conductor will dissipate heat rapidly. If you take a wire and short a battery, the wire will immediately become unbearably hot. We don’t want such a large conductor so as to destroy a battery, but enough so as to achieve a pleasant temperature. Wire Design Specifications Wire Specifications: Supplied Voltage, V = 9 V Length of Resistor, L = 0.10 m Power Dissipation, P = 10 W Assumptions: Voltage difference between terminals of the battery is the voltage difference on Voltage the terminals of the wire the Battery is capable of maintaining power output Wire Materials Selection Wire Material Perf. Index, M=(10^5)*σ/ρ Low Alloy Steels Alloy 1020 7.961783439 7.961783439 Alloy A36 Aluminum Alloys, Annealed Alloy 6061 100.1001001 Alloy 1100 127.2426517 Alloy 2024 106.179656 Copper Alloys C11000 65.39880189 C36000 17.82531194 18.90859585 C26000 C93200 7.776533532 21.26528442 C17200 C71500 2.98284862 Magnesium Alloys 32.49918752 Alloy AZ91D Alloy AZ31B 61.40997298 Precious Metals Silver 64.8495814 Gold 22.02546143 4.398117606 Platinum Refractory Metals 18.81679964 Molybdenum Tungsten 9.776126699 Tantalum 4.462293619 Other Nonferrous Alloys Tin 12.56486612 Nickel 200 11.84062519 3.869265265 Zirconium, reactor grade 702 Kovar 2.441167855 Ceramics Graphite 0.835421888 Tungsten Carbide 1.207160878 Fiber Materials Carbon (PAN precursor), High Mod 0.581564408 M/(cost/kg) 15.923567 12.637751 22.750023 17.550711 12.06587 16.3497 5.57041 5.402456 1.7281186 0.8506114 0.3509234 8.5524178 6.978406 0.3814681 0.0023185 0.0003858 0.2213741 0.1261436 0.0114418 1.834287 0.6231908 0.0879378 0.0793876 0.1670844 0.0146075 0.0033232 Callister, W.D. (2000), Materials Science and Engineering: An Introduction, 5th ed., John Wiley (New York). Wire Materials Selection Check Wire Material Low Alloy Steels Alloy A36 Alloy 1020 Aluminum Alloys, Annealed Alloy 1100 Alloy 2024 Alloy 6061 Copper Alloys C11000 C17200 C26000 C36000 C71500 C93200 Magnesium Alloys Alloy AZ31B Alloy AZ91D Precious Metals Gold Platinum Silver Refractory Metals Molybdenum Tantalum Tungsten Miscellaneous Nonferrous Alloys Nickel 200 Kovar Tin Zirconium, reactor grade 702 Ceramics Graphite Tungsten Carbide Fiber Materials Carbon (PAN precursor), High Mod Feasibility Check (gm) 1.551E-03 1.551E-03 9.702E-05 1.163E-04 1.233E-04 1.888E-04 5.806E-04 6.529E-04 6.926E-04 4.139E-03 1.588E-03 P 2 ρd m = 2 ⋅ L ⋅ V σ () 2.010E-04 3.799E-04 P = 10 W 5.605E-04 2.807E-03 1.904E-04 V=9V 6.561E-04 2.767E-03 1.263E-03 1.043E-03 5.057E-03 9.826E-04 3.191E-03 1.478E-02 1.023E-02 2.123E-02 L = 0.10 m Wire: Feasibility cont. Wire: 4 A 0.1 m rod of aluminum alloy 6061 with mass 1.233x10--4 gm would have a cross sectional area of 457x10-12 m2, or a radius of 12 μm. have or If we assumed something more imaginable such as P =5-11 W, V = If 9 V, L = 0.1 m, and r = 0.1 in ≈ 0.25 cm wires, the conductivity result would be 304.6 - 670 (ohm m)-1, or a resistivity of (1.49-3.28)x10-3 would ohm m. ohm Materials having a resistivity yielding a power rating of include: Polyphthalamide, Stainless Fiber, EMI [ 2.0 – 160.] [1.54gm/cc] ComAlloy Comshield® 229 ABS, Electrically Conductive - Stainless ComAlloy Steel Fiber Filled [1.0 – 10.] [1.16gm/cc] Steel Emerson & Cuming C429-2 Eccobond® One-Component Silver Filled Emerson Electrically Conductive Epoxy Adhesive [2.0] [4.2g/cc] Electrically http://www.matweb.com/ Statistics Statistics Assume power dissipating off a wire has an Assume average of 10 W and standard deviation of 0.5 watts. Determine what minimum length we should make the wire such that there is at least a 95% chance the power remains under 15 W. Assume exact voltage supplied is 9 V, r = 2.54x10-3 m, and resistivity is 2.0x10-3 ohm m. 2.54x10 π ⋅V 2 ⋅ r 2 P= L ⋅ ρ resis X −P = 1.645 X = 10.8225 W sp L ≥ 75.8x10 -3 m Economic Analysis Economic Using the geometric present worth factor and the following projections we calculated that we will need to increase our sales by 43.7% for the next 5 years in order to break even on the initial investment. The 43.7% is what marketing has to increase sales by each year in order for the company to break even on the initial investment in five years. As the product becomes more accepted and reviewed, the number of new users per year is expected to increase. •Initial Investment = $250000 •Interest Rate I=10% •Manufacture Cost/Unit = $15 •Selling Price/Unit = $40 •Projected production in the first year = 200 units •Projected sales volume in the first year = 150 units •First year returns = $3000 Conclusion Conclusion Bracket material: Aluminum Alloy 6061 Wire-material: Aluminum Alloy 6061 … or not. Feasibility analysis suggests that we ought to use Emerson & Cuming’s C429-2 Eccobond® One-Component Silver Filled Electrically Conductive Epoxy Adhesive Electrically 3-year economic break-even time period ...
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This note was uploaded on 06/10/2011 for the course EMA 7414 taught by Professor Mecholsky during the Spring '11 term at University of Florida.

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