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52070-467995SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY, COIMBATORE-10(Approved by AICTE, New Delhi – Affiliated to Anna University, Chennai)Department of Mechanical Engineeringcenter-629920Class: III YearSemester: VSubject: Dynamics of Machinery Max Marks: PROBLEM 2Link 1 = Fixed LinkLink 2 = CrankLink 3 = Connecting RodLink 4 = SliderLink 4 acted upon by 3 Forces F, F14, F34 (Scale 1mm = 50N)Free Body Diagram of Link 4Force Polygon F=1500 N;F34 = 1564 NLink 3 acted upon by 2 Forces F23,F43 Link 2 : h=35mmFrom the above relationship-F34 = F43 = -F23 = F32 = 1564 NMoment:= Force * Horizontal Distance= 564 * 35= 54740 N mm= 54.740 N.mTherefore the required input Torque is 55 N.m (Anti Clockwise)A horizontal steam engine running at 240 r.p.m has a bore of 200mm and stroke of 360mm. The piston rod is 20mm in diameter and connecting rod length is 900mm. the mass of the reciprocating parts is 7Kg and the frictional resistance is equivalent to a force of 500N. Determine the following when the crank is at 120? from the IDC, the mean pressure being 5000N/m2 on the cover side and 100N/m2 on the crank side.Thrust on the connecting rod,Thrust on the cylinder wallsLoads on the bearingsTurning moment on the crankshaftGiven Data:Speed (N)= 240 rpmBore diameter (d)= 200 mm=0.2 mStroke Length (L)= 360 mm=0.36 mPiston rod diameter (d)= 20 mm=0.02 mLength of the connecting rod (l) = 900 mm= 0.9 mMass of the reciprocating Parts = 7kgFrictional Resistance= 500 NAngle (θ)= 120 °Pressure on cover side (P1)= 5000 N/m2Pressure on crank side (P2)= 100 N/m2 To Find:Thrust on the connecting rod (Fc),Thrust on the cylinder walls (Fn)Loads on the bearings (Fr)Turning moment on the crankshaft (T)Solution:ω=2πN60= 25.13 rad/s r = L/2= .18mn=l/r= .9/.18 = 5sinβ= sinθn =0.173β = 9.96°Force acting on the Piston:F = Fp-Fb-FfForce due to Gas Pressure (Fp) = P1A1-P2A2= (5000*(π/4)*(0.22))-(100*(π/4)*(0.22-0.022))= 157.08 – 3.11= 153.97 NInertia Force (Fb)= mrω2cosθ+cos2θn= 7*0.18*25.132(cos 120° + (cos 120°/5))= 795.71 * (-0.6)= -477.43 NFrictional Resistance= 500 NF= Fp-Fb-Ff= 153.97+477.43-500= 131.4 NThrust on the connecting rod (Fc),Fc = F/cos β= 131.4/cos 9.96= 131.4/0.98= 134.08 NThrust on the cylinder walls (Fn)Fc = F tan β = 131.4 * tan 9.96°=131.4/0 .18= 730 NLoads on the bearings (Fr)Fr = (F/cosβ ) (cos (θ+β)) = 134.4 * (-0.64)= -84.1 NTurning moment on the crankshaft (T)Fr = (F/cosβ) (sin (θ+β)) r = (134.4) (0.77) (0.18) = 18.63 NmA machine punching 38 mm holes in 32 mm thick plate requires 7 N-m of energy per sq. mm of sheared area, and punches one hole in every 10 seconds. Calculate the power of the motor required. The mean speed of the flywheel is 25 metres per second. The punch has a stroke of 100 mm. find the mass of the flywheel required, if the total fluctuation of speed is not to exceed 3% of the mean speed. Assume that the motor supplies energy to the machine at uniform rate.Given: d = 38 mm; t = 32 mm; E1 = 7 N-m/mm2 of sheared area; v = 25 m/s;s = 100 mm; v1 ? v2 = 3% v = 0.03 vPower of the motor requiredWe know that sheared area,A = πd .t = π × 38 × 32 = 3820 mm2Since the energy required to punch a hole is 7 N-m/mm2 of sheared area, therefore total energy required per hole,E1 = 7 × 3820 = 26 740 N-mAlso the time required to punch a hole is 10 second, therefore energy required for punching work per second= 26 740/10 = 2674 N-m/sThere fore Power of the motor required= 2674 W = 2.674 kW Ans.Mass of the flywheel requiredLet m = Mass of the flywheel in kg.Since the stroke of the punch is 100 mm and it punches one hole in every 10 seconds, therefore the time required to punch a hole in a 32 mm thick plate= (10/2*100)*32= 1.6 sEnergy supplied by the motor in 1.6 seconds,E2 = 2674 × 1.6 = 4278 N-mEnergy to be supplied by the flywheel during punching or the maximum fluctuation of energy,ΔE = E1 ? E2 = 26 740 ? 4278 = 22 462 N-mCoefficient of fluctuation of speed,Cs= v1-v2v= 0.03We know that maximum fluctuation of energy (ΔE) ,22 462= m.v2 .CS = m × (25)2 × 0.03 = 18.75 mThere fore, m = 22 462 / 18.75 = 1198 kg Ans.A shaft carries four masses of magnitude 200kg, 300kg, 400kg and 200kg respectively and revolving at radii 80mm, 70mm, 60mm and 80mm in planes measured from 1 at 300mm, 400mm and 700mm. The angles between the cranks measured anticlockwise are 1 to 2 is 45°, 2 to 3 is 70° and 3 to 4 is 120°. The balancing masses are to be placed in planes X and Y. The distance between the planes A and X is 100mm between X and Y is 400mm and Between Y and D is 200mm. If the balancing masses revolve at a radius of 100mm fine their magnitudes and angular positions.Given Data:m1 = 200 kgr1 = 80 mm θ1 = 0°m2 = 300 kgr2 = 70 mmθ2 = 45°m3 = 400 kgr3 = 60 mmθ3 = 115°m4 = 200 kgr4 = 80 mmθ4 = 235°To Find:Magnitude of X = mx Magnitude of Y = my Angular Position of X = θx Angular Position of Y = θySolution:Plane Diagram & Space Diagram: PlaneMass(kg)Radius(m)Force (mr) kg.mDistance From RP (m)Couple(kg.m2)12000.08016-0.100-1.6X(RP)mx0.1000.100 mx0023000.070210.2004.234000.060240.3007.2Ymy0.1000.100 my0.4000.04my42000.080160.6009.6Couple Polygon:0.04my=73.62/10my=7.362/0.04my= 184.05 kgθy=167.20+180°θy=347.20°Force Polygon:0.1mx=35.29my=35.29/0.1my= 352.9 kgθy=33.38+180°θy=213.38°Analytical Solution:Considering Couple:ΣFH = 0:m1r1l1cosθ1+ mxrxlxcosθx+ m2r2l2cosθ2+ m3r3l3cosθ3+ myrylycosθy+ m4r4l4cosθ4=0-1.6cos0+0+4.2cos45+7.2cos115+0.04mycosθy+9.6cos235 = 0ΣFv = 0:m1r1l1sinθ1+ mxrxlxsinθx+ m2r2l2sinθ2+ m3r3l3sinθ3+ myrylysinθy+ m4r4l4sinθ4=0-1.6sin0+0+4.2sin45+7.2sin115+0.04mysinθy+9.6sin235 = 0Considering Force:ΣFH = 0:m1r1cosθ1+ mxrxcosθx+ m2r2cosθ2+ m3r3cosθ3+ myrycosθy+ m4r4cosθ4=016 cos0+0.1 mxcosθx +21 cos45+24 cos115+mycosθy+16 cos235 = 0ΣFv = 0:m1r1sinθ1+ mxrxsinθx+ m2r2sinθ2+ m3r3sinθ3+ myrysinθy+ m4r4sinθ4=016 sin0+0.1 sinθx +21 sin45+24 sin115+mysinθy+16 sin235 = 0The following data apply to an outside cylinder uncoupled locomotive : Mass of a rotating parts per cylinder = 300 kg;Mass of reciprocating parts per cylinder = 450 kg ; Angle between cranks = 90°; Length of Each crank = 350mmDistance between Wheels = 1.6mCylinder centers = 1.9 m;Diameter of Driving Wheel = 2m Radius of balance masses = 0.8 m; If whole of the rotating and two-thirds of reciprocating parts are to be balanced in planes of the driving wheels, find": Magnitude and angular positions of balance masses. speed in kilometers per hour at which the wheel will lift of the rails when the load on each driving wheel is 35 kN Swaying couple at speed arrived at in (ii) above. Total Mass (m2)= ma = md = (m1+cm)= (300+ (2/3)450)= 600 kgPlaneMass (m)KgRadius (r)mCentrifugal Force /ω2 (mr) kg.mDistance from RP (m)Couple / ω2(mrl) kg.m2A6000.35210-0.15-31.5B(RP)mb0.80.8 mb00Cmc0.80.8 mc1.61.28 mcD6000.352101.75367.5Using Couple Polygon:Σmrl=0Horizontal Components:m1r1l1cosθ1+m2r2l2cosθ2+m3r3l3cosθ3+m4r4l4cosθ4=-31.5 cos 0 + 0+ 1.28 mc cos θ3+ 367.5cos 90 = 0-31.5+0+1.28 mc cos θ3+0 =01.28 mc cos θ3=31.5Vertical Components:m1r1l1sinθ1+m2r2l2sinθ2+m3r3l3sinθ3+m4r4l4sinθ4=-31.5 sin 0 + 0+ 1.28 mc sinθ3+ 367.5 sin 90 = 00+0+1.28 mc sin θ3+367.5 =01.28 mc sin θ3=-367.5mc2 = (31.52+367.52)/1.282mc= 285 kgθ3=275Using Force Polygon:Σmr=0Horizontal Components:m1r1cosθ1+m2r2cosθ2+m3r3cosθ3+m4r4cosθ4=210 cos 0 + 0.8m2 cos θ2+ 0.8m3 cos θ3+ 210 cos 90 = 0Vertical Components:m1r1sinθ1+m2r2sinθ2+m3r3sinθ3+m4r4sinθ4=210 sin 0 + 0.8m2 sin θ2+ 0.8m3 sin θ3+ 210 sin 90 = 0mb= 285 kgθb= 175Speed at which the wheel will lift off the rails:Given:P = 35kND = 2mR = 1mω= PB.bB= c*Receiprocating MassTotal Mass to be Balanced*Balancing MassB= 23*450600*285=142.5 kgω= 35*1000142.5*0.8=17.52v=Rω= 1*17.52=17.52 m/s=63 km/hrSwaying Couple:Swaying Couple = l21-cmrω2= 1.414*(1-2/3) (450*17.522*0.35)= 21.65 kN.mThe stroke of each piston of a six cylinder two stroke in line engine is 320 mm and the connecting rod is 800mm long. The cylinder centre lines are spaced at 500 mm. The cranks are at 60° apart and the firing orders are 1-4-5-2-3-6. The reciprocating mass per cylinder is 100kg and the rotating parts are 50kg per crank. Determine the out of balance forces and couples about the mid plane if the engine rotates at 200rpm.PlaneMass (m)Radius (r)Force (mr)Distance from RP (m)Couple AnglekgmKg mmKg m2θ2θ11500.1624-1.25-300021500.1624-0.75-1818036031500.1624-0.25624048041500.16240.2566012051500.16240.751812024061500.16241.2530300600Primary Force Polygon: Primary Couple Polygon:Secondary Force Polygon: Secondary Couple Polygon:Primary Force, Couple and Secondary Force Polygons are closedFrom the secondary Couple Polygon mrl = 55.43There fore,The unbalanced couple = 55.43*(ω?/n)= 55.43(1/5)((2π*200)/60)?= 4863 NmThe lengths of the upper and lower arms of a Porter governor are 200 mm and 250 mm respectively. Both the arms are pivoted on the axis of rotation. The central load is 150 N, the weight of each ball is 20 N and the friction on the sleeve together with the resistance of the operating gear is equivalent to a force of 30 N at the sleeve. If the limiting inclinations of the upper arms to the vertical are 30° and 40°, determine the range of speed of the governor. (Nov/Dec 2006)Given Data:Arm Length= 200mmArm Length= 250 mmBall Weight (m)= 20NSleeve Weight (M)= 150NFriction Force (F)= 30NAngle(α1)= 30?Angle (α2)= 40?To Find:Equilibrium Speeds N1 and N2Speed RangeSolution:Minimum Speed:Minimum radius of rotation r1= BD = AB sin α1 = 0.2 sin 30?= 0.1 mHeight of Governor h1= AD = AB cos α1=0.2 cos 30?= 0.173 mCD= BC2-BD2= 0.252-(0.1)?= 0.229 mtan β1= BD/CD= 0.1/0.229= 0.436tan α1= tan 30?= 0.577q1 = tan β1/ tan α1= 0.436/0.577= 0.756The minimum Speed is given by,N1? = mg+Mg-F2(1+q1)mg*895h1=20+150-302(1+0.756)20*8950.173= 32389N1 =179.97rpmMaximum Speed:Maximum radius of rotation r2= BD = AB sin α2 = 0.2 sin 40?= 0.129 mHeight of Governor h2= AD = AB cos α2=0.2 cos 40?= 0.153 mCD= BC2-BD2= 0.252-(0.129)?= 0.214 mtan β2= BD/CD= 0.129/0.214= 0.603tan α2= tan 40?= 0.839q2 = tan β2/ tan α2= 0.603/0.839= 0.718The maximum Speed is given by,N2? = mg+Mg+F2(1+q1)mg*895h2=20+150+302(1+0.718)20*8950.153= 51.73N2 =226rpmRange of Speed:=Maximum Speed –Minimum Speed= 226-179.97=46.03rpmIn a Porter governor, the mass of the central load is 18 kg and the mass of each ball is 2 kg. The top arms are 250 mm while the bottom arms are each 300 mm long. The friction of the sleeve is 14 N. If the top arms make 45° with the axis of rotation in the equilibrium position, find the range of speed of the governor in that position. (April/May 2008)Given Data:Mass of sleeve (M)= 18kgMass of Ball (m)= 2kgTop Arm Length = 250mm = 0.25mBottom Arm Length= 300mm= 0.3mFriction (F)= 14NAngle (α)= 45?To Find:Range of speedSolution:Minimum Speed:Minimum radius of rotation r1= BD = AB sin α1 = 0.25 sin 45?3676650609250= 0.177 mHeight of Governor h1= AD = AB cos α1=0.25 cos 45?= 0.177 mCD= BC2-BD2= 0.32-(0.177)?= 0.242 mtan β1= BD/CD= 0.177/0.242= 0.731tan α1= tan 45?= 1q1 = tan β1/ tan α1= 0.731/1= 0.731The minimum Speed is given by,N1? = mg+Mg-F2(1+q1)mg*895h1=(2*9.81)+(18*9.81)-142(1+0.731)20*8950.177N1 =203.27rpmThe maximum Speed is given by,N2? = mg+Mg+F2(1+q1)mg*895h2=(2*9.81)+(18*9.81)+142(1+0.731)20*8950.177N2 =218.10rpmRange of Speed:Range of speed = Maximum speed – Minimum Speed= 218.10-203.27Range= 14.822rpmThe intercepted areas between the output torque curve and the mean resistance line of a turning moment diagram for a multi cylinder engine, taken in order from one end are as follows:-0.35,4.10,-2.85,3.25,-3.35,2.60,-3.65,2.85,-2.6 sq cm. The diagram drawn into a scale of 1cm=700Nm and 1cm = 45°. The engine speed is 900rpm and the fluctuation of speed is not to exceed 2% of the mean speed. Find the suitable diameter and cross section of the flywheel rim if the safe centrifugal stress is limited to 7MPa. The density of the material is 7200kg/m3. The rim is rectangular with the width 2 times the thickness. Neglect the effect of arms.Solution:Let Flywheel KE at a = Eat a= Eat b= E-0.35E-0.35 (Min Energy)at c = E-0.35+4.10E+3.75at d= E+3.75-2.85E+0.9at e= E+0.9+3.25E+4.15 (Max Energy)at f= E+4.15-3.35E+0.8at g= E+0.8+2.60E+3.4at h= E+3.4-3.65E-0.25at i= E+2.85-0.25E+2.6at j= E+2.6-2.6EMax Energy: E+4.15Min Energy: E-0.35Maximum Fluctuation of energy:ΔE = Max Energy- Min Energy= E+4.15-(E-0.35)ΔE= 4.5 cm2Scale:1cm= 700 Nm;1cm = 45°1mm2 in turning moment diagram = (45*π/180) *700= 549.78 Nm4.5cm2= 2474.01Nmσ= ρυ?7*10^6=7200* υ?υ= 31.18m/sυ=πDN/6031.18= (π*D*900)/60D= 0.66mω= 2 πN/60=(2*π*900)/60= 94.25 rad/sΔE= Iω2Cs= mk2ω2Cs2474= (m*0.332*94.252*18)m= 0.14kgm = πDAρ= π*D*b*t*ρ0.14= π*0.66*2t2*7200t= 0.218mm,b= 0.436mm ................
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