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Theoretical derivation for Mechanical Booster

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Mechanical Booster is the machine which takes energy from humans and converts it into useful form of mechanical energy.The following is the theoritical derivation...

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Theoretical derivation for Mechanical Booster

  1. 1. Theoretical Derivation:Let W= Force Applied at outer end A of spring B= Width of strip T= Thickness of strip L= Length of strip forming spring Y= Distance of Center of gravity of spring from A I= Moment of inertia of spring section= bt3/12 Z=Section modulus of spring section= bt2/6When end A of spring is pulled by force W, then bending moment on spring at distance Y from line ofaction of W is given by M= W*YThe greatest Bending moment occurs in spring at B ,which is at maximum distance from applicationof W - Bending Moment at B, MB = MMax = M*2Y= 2WY=2M - Maximum Bending stress induced in spring material, --b = Mmax /Z= 12WY/bt2= 12M/bt2Assume both of springs are clamped, the angular deflection of spring is given by O=ML/EI= 12ML/Ebt3& Deflection , d= O*Y= MLy/EI=12WY2L/Ebt3 = --bYL/EtThe strain energy stored in spring= ½ MO= M2L/2EI= 6W2Y2L/Ebt3= --b2/24E *btL = (--b2/24E)* btL= (--b2/24E)* Volume of springLet us take spiral spring of flat strip of 6mm wide & 0.25mm thickness and length of 2.5 m. And alsoMaximum stress withstand of spring is 800 Mpa and Elasticity of 200 KN/mm2 - Bending moment in spring,--b= 12M/bt2  M= 800/32= 25 N-mm - Number of turns to windup spring, O= 12ML/Ebt3= 40 rad= 7 turns - Strain energy stored in spring , = ½ MO=1/2 *25*40= 500 N-mm
  2. 2. If I am using about 100 springs in one energy packet, strain energy stored= 500*100=50 KN-mm& since I am also using about 20 elastic threads in one energy packet, Energy stored= 20*1200 N-mm= 24000=24KN-mmTotal energy stored in one energy packet= 24+40=64 KN-mmWe know that energy stored in the springs or elastic material is in the form of potential energyFor appllication purpose , Potential energy must convert to work In terms of acceleration , workdone= potential energy  M*a*S= 74000  9as=74000  Distance moved S= ( 8222/a ) metersIf a= 10 m/sec2 (very high)  s=822.2 m  By means of one energy packet, 9 Kg Mass can move 822 metres with acceleration of 10m/sec2 (Without friction...)  90 Kg mass can move 82.2 metres with acceleration of 10 m/sec2 (Without friction...)  90 Kg mass can move 75 metres with acceleration of 10 m/sec2 (With friction...)  65 kg mass can move about 100 metres with acceleration of 10 m/sec2(With friction...)Since there are about 20 such energy packets in this machine, Mass can move about 20*100= 2Kmi.e, 65 Kg person can move on this machine for 2 Km with constant acceleration of 10 m/sec2And Practically, 65 Kg person can move on this machine for 4 Km with constant acceleration of 5 m/sec2And also by considering Typical Indian Roads and Low speeds,65 Kg person can travel on this vehicle for 6km with constant acceleration of 3 m/sec2Efficiency= Output power/ input power(Supply by exercise) =mas/ PE= 70/100 KN-mm= 70%
  3. 3. Further detailsBharadwaj Ravipatiannangi.bharat@gmail.com

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