Motorized Skateboard Project Based On Mechanical Knowledge - Report Example

Motorized Skateboard Project (2nd Order) Based On Mechanical Knowledge Safety factor analysis for the deck According to (Hughes & Drury, 2013), the essence of the safety factor is highly vital in design of the deck. The deck is made up of composite material consisting of 7 layers of carbon fiber, a layer of balsa wood , 7layers of carbon fiber and 2 layers of fiber glass from top to bottom. And also considering that the orientation for the carbon fiber layers is arranged in a typical 0o, 450, 45o (the opposite angle), 0, 90, 45, 45(the opposite angle)) and for the fibreglass is (0, 45). The safety factor and design measures on this product aim to avail a maximum design stress which falls below the ultimate tensile stress (UTS). It will also present a nice design margin over the expected theoretical design capacity to permit uncertainty in the deck design process. with regards to the selection of the appropriate safety factor, extreme regards are placed on uncontrollable uncertainties on the deck which mainly entails the uncertainty concerning exact properties of material entailed. Uncertainties may also arise from the varied methods of machining the deck and the yield strength is best specified under certain notable ranges. Undue uncertainties due unexpected assembly process falling within a range including joining the various parts of the skate model. Normally, there is extreme involvement or effect of time on the reduction of the strength of the material as there is gradual deterioration created by the environment on the material forming the design part of the skate. They shall also be expected to be applied under prevailing average conditions of setting, load and stress by users. Every component material has differing contribution on the material strength as in accordance with the provided layering of the structures. The various components or material layering avails varying and inherent qualities to the strength of the deck (Hughes & Drury, 2013). Fig a : layer of materials Assumptions 1. The composiste body acts like a single material body 2. Fiber carbon has ultimate strength of (xt=100 Mpa) 3. Balsa wood has ultimate strength of (xt=40 Mpa) 4. Fiber glass has ultimate strength of (xt=90 Mpa) 5. Consider a maximum design load of (human weight + skate weight) =934N 6. Given dimensions of the board: 39.1inch in length, 9.2inch in width, and 0.5inch in height and wheel base of 14inch. Hence Average ultimate strength of the composite = (100+40+100+90)/4 =82.5mpa=82.5 Mn/m2 Cross-sectional area of skateboard=9.2inch in width x 0.5inch Convert to metrics (2.54 x9.2) x (0.5x2.54) =29.67736cm2=2.968x10-3 m2 Failure force or load of the skateboard deck= ultimate strength of the composite xcross-sectional area =82.5Mn/m2x2.968x10-3 m2 =244860N [S.F=material strength / design load} Normal design load = 934N Assumption: In case of great impact on the skateboard for instance during jumps, load on the skate increases to very high amounts from 934N to as high as 102025N. Maximum design load 102025N, hence selected to ensure of highest safety standards. [S.F=maximum material strength load/ maximum design load weight} =244860N/102025N =2.4 In selection of the outlined safety factor, there is critical observation and compromise of the entailed additional cost and the weight gained. For the skate weight refers to better stability and increased output in terms of productivity. The strength is highly relevant in determination of the resilience of the skate, long lasting and reliability process. Analyzing of free body diagrams The free body diagrams reveal turning, braking and acceleration of the skateboard. While turning there is nearly equal application of force both front and back pedals but turning force is normally applied on the y-axis direction. Hence forces on the front and back pedal are typically half of the total weight of the person and skateboard. While braking, acceleration assumes a negative figure until it reaches zero velocity. The force applied at the front is typically high in comparison to force applied at the back. This is 2/3 fraction of weight at front but 1/3 fraction of weight at the back. Lastly, while accelerating, the force applied at the back is contrastingly larger than the force applied in the front. While accelerating, it is worth noting that acceleration is greater than zero. Fig a:free body diagram for turning deck 1. While turning person’s weight remain constant weight=(mg+ma), 2. a=0, zero acceleration 3. hence = weight=(mg- ma)=mg Fig c:free body diagram for accelerating deck 1. While accelerating person’s weight is less, weight=(mg-ma), 2. Acceleration (a) is greater than zero, 3. Resultant weight weight=(mg-ma), 4. Result Weight mg Gear analysis Given two gears, motor power 400w, wheel diameter 5inch, Velocity of 15mph=6.7m/s, weight of 210 lb=934.12 N Power required = (F drag + F friction) velocity (1/2 x Cd Apv2+umg) velocity (1/2x 1.1 x 0.451612x6.72 x0.02x934.12) x6.7 Torque needed =power needed / angular velocity=200/ (6.7/0.0635) Torque needed amounts to 1.89552 nm However, motor power equals= 400w Power needed is half power released by the motor Gear ratio =number of teeth driven gear/ number of teeth driver gear =30/15 Hence gear ratio =2 Gear ratio & chain length Fig e: gear representation For a gear ratio of 2 Assume that Let r2=1/3”, d2=2/3” R1=2/3”, d1= 4/3” Distance x=6” Length of the polychain Length of the polychain=3.14/2x(2/3+4/3)+2x6+(4/3-2/3)2/(4x6) =3.14/2x (6/3)+12+4/216 =pi+12+0.18519 =15.1601 The ratio of the polychain and gears in general emanates from the inherent dimensions of the gears which give rise to the eventual dimension of the chain. Confirmable to (Hughes & Drury, 2013), gears are highly important to this project as it they enhanced reduction in slippage the detrimental effects of slippage is majorly a transmission lagging behind process. The gears also created conducive essence of mechanical advantage on the device while on motion thus facilitating the motion of the motor and the driving force essence. The gears would also appropriate in this case in that they would allow constant source of energy to be variable in input is terms of speed and power or torque. The provided power transmission process in this case mainly associates with vivid and elaborate gear ratio or mechanical advantage. The appropriate inclusion of the gear ratio systematically increases the output torque or entailed output pace of the entailed mechanism. The trade off is typically due to the essence of the law of energy conservation and the point is conventionally chief or key concept of entailed mechanical advantage (Hughes & Drury, 2013). With proper balancing and alteration of the gear ratio, it is possible to have the high velocity or considerably high torque output or even both of the two variable output. The most essential contributor factor is the law of energy conservation that normally states the no one can ever acquire more energy in the output movement or motion than availed by the energy source. The apposite application of the belt would astoundingly reduce the chances of slippage. The motor stability is equally relevant in determination of the success of the project as great impact of weight or force on the device would eventually trigger malfunction of the gadget. Power efficiency =ratio of power output/power input Assuming power output is 340kw Power Motor input Efficiency Assumptions =Efficiency in % Hp –name plate rated horsepower=1.4 Load-output power in %=360 Pi=three phase power in terms of KW=400 Efficiency in % = (0.7457x1.40 x360)/400 =0.9395=93.95% Assumptions There is negligible power loss in or on the: 1. Gears 2. Belts 3. Any delivery process 4. Wheels are well lubricated hence no friction experienced Ultimate Power loss= input –output =400kw-360kw=40kw General efficiency= output power/input power =340/400 =0.85=85% References Hughes, A., & Drury, B. (2013), Electric motors and drives: Fundamentals, types and applications. Oxford: Newnes. Read More