Advanced Kinemtics and Dynamics of Mechanisms

1、Advanced Kinemtics and Dynamics of Mechanisms1Inertia and shaking force2Inertia and shaking torque3Total engine torque4FlywheelCONTENTS5Pin force in the single-clylinder engine6Balancing the single-cylinder engine7Design trads-offs and ratiosINERTIA AND SHAKING FORCEinertia force and torques: result

2、ing from the high speed motionshaking force and torques on the grounddAlember theory:conbinating constrains , driving force and inertia force13-12(a)Simplified and Lumped Mass Dynamic model The acceleration of point A can be gained by derivation:The total inertia force is the sum of the inertia forc

3、e at point A and point B:breaking it into x and y components:The expression shows that the x directed inertia forces have primary components at crank frequency and secondary force at twice crank frequency. From the free-body diagram for link1 in Figure 13-12, the shaking force can be expressed as:j

4、tri traj tri trRAAsincossincos22BBAAiamamFtrAmiyFtlrtrBmtrAmixFsin22coscos2cos24141222sin2coscoscosiiAsyBAsxFFtrmFtlrtrmtrmFiFFrom the left-figures,we can found the side force of the piston on the cylider was which was cancelled.Therefore, the shaking force is equal to the inertia force.13-12(b) fre

5、e-body diagramsisFF41iFsFThe gas force dont contribute to the shaking force for the gas force as a internal force was concelled by itself dynamic motion, while only inertia and external forces are felt as shaking forces.13-13 shaking force in an unbalanced slider-crank linkage1 Cylinder4 stroke Cycl

6、eBore=3 inStroke=3.54B/S=0.85L/R=3.5RPM=3400Due to the high speed of piston only on single component,the force on x is larger than that on y.And the forces are seen to be quite large despite this being little engine running at the moderate speed as in figure 13-13,which will be eliminated later.INER

7、TIA AND SHAKING TORQUESThe inertia torques results from the action of the inertia forces at a moment arm. the inertia force at point A has two components:radius and tangential. the radius have no arm,but the tangential has a moment arm r. if the crank angular velocity is zero, the mass of point A ca

8、nnt result in the inertia torque. At the same time, the inertia force at B only have nonezero component vertial the motion of piston. the inertia torque is:when substituting the former gained parameters, it can be expressed as:kxFkxFTiii414121ktlrtrlrlxmTBi2cos4cos4tan221 then,simplifying it by trig

9、onometry,the expression shows as follows:the expression shows that the inertia torque has a third hamonic term as well as a first and second. the second harmonics is the dominant term as it has the largest coefficient.The shaking torque is equal to the inertia torque:ttttttt2sincossin2sin3sin2cossin

10、2ktlrttlrrmTBi3sin232sinsin221222121isTT1 Cylinder4 stroke CycleBore=3 inStroke=3.54B/S=0.85L/R=3.5RPM=340013-14 Inertia torque in the slider-crank linkageThe figure shows a plot of the inertia torque for this built-in example engine, noting the dominance of the second harmonic.Entirely,its average

11、value is always zero,so it contributes nothing to net driving torque.It merely creates large positive and negative vibration in the total torqueTOTAL ENGINE TORQUEThe total engine torque is the sum of the gas and inertial torque: the gas torque is less sensitive to engine speed than is the inertia t

12、orque,which is a function about . So the relative contbutions of these two components to the total torque will vary with the velocity of engine.igtotalTTT2800rpm3400rpm6000rpm13-15 the total torque functions shape and magnitude vary with crankshaft speedFLYWHEELSThe single-cylinder engine is a prime

13、 candidate for the use of flywheel, which was infered to reduce the vibration in the torque-time function.The intermittent nature of its power strokes makes one mandatory as it will store the kinetic energy needed to carry the piston through the bottom of the four-strokes. Usually the two-stroke eng

14、ine needs a flywheel to drive the piston up to suppress the gas.13-16 the total torque functions shape and magnitude vary with crankshaft speed800rpm3400rpm6000rpmPIN FORCE IN THE SINGLE-CYLINDER ENGINEIn addition to calculationg the overall effects on the ground plane of the dynamic forces present

15、in the engine. We also need to know the forces at the pin joints. Those forces will dictate the design of the pins and the bearings at the joints. Though we were able to lump the mass due to both conrod and piston，or conrod and crank at points A and B for an overall analysis of the linkages effects

16、on the ground plane,we cannot do so for the pin force calculations . This is because the pins feel the effect of the conrod pulling on one side and the piston pulling on the other side of the pin as shown in figure 13-17. Thus we must seperate the effects of the masses of the links joined by the pin

17、s.We will calculate the effect of each component due to the various masses, and the gas force and then superpose them to obtain the complete pin force at each joint. We need to keep tack of all these components with simplified markers,that is : MARKERS1 the gas force , 2 the inertia force due to the

18、 piston mass,3 the inertia force due to the mass of the conrod at the wrist pin,4 the inertia force due to the mass of the conrod at the crank pin,5 the inertia force due to the mass of the crank at the crank pin,13-17 Forces on a pivot pingFipFiwFicFirFThe figure shows the free-body diagrams for th

19、e forces due only to the acceleraion of the mass of the piston, . Those components are:13-18 Free-body diagrams for forces due to the piston mass322134324434441tantanipipipipBBipBipFFFFjamiamFjamF4mThe figure shows the free-body diagrams for the forces due only to the acceleraion of the mass of the

20、conrod locared at the wrist pin, .Those components are:13-19 Free-body diagrams for forces due to the conrod mass concentrated at the pin322141343332341tantaniwiwiwiwBbBbiwBbiwFFFFjamiamFjamFbm3The figure shows the free-body diagrams for the forces due only to the acceleraion of the mass of the conr

21、od and crank located at crankpin, .Those components are:am3j ti trmFj ti trmFamFairaicAaicsincossincos2221233233213-20 Free-body diagram for forces due to masses at the crankpinWe can now sum the components of the forces at each pin joint. for the sidewell force of the piston against the cylinder wa

22、ll:The total force on the wrist pin is:jFammjamjamjFFFFFgBbBbBgiwipgtantantantan343441414141jammFiamFjamjamiamjFiFFFFFBbgBgBbBBggiwipgtantantantan3443443434343434F41F jFammtrmiFammtrmj ti trmjamiamjamiamjFiFFFFFFgBbagBbaaBbBbBBggiciwipgtansincossincostantantan432343232333443232323232The total force

23、on the crankpin is：32FThe total force on the main pin is:Note that,unlike the inertia force as before13.14, which was unaffected by the gas force, these pin forces are a function of the gas force as well as of the -ma force. Engine with larger piston diameters will experience greater pin forces as a

24、 result of the explosion pressure acting on their larger piston area.213221irFFF21Frpm3400rpm6000rpm800rpm340013-21 Forces on the wrist pin of the single-cylinder engine at various speeds 13-22 Pin force variationThe first and third figures show that the variation in the inertia force component on t

25、he crankpin and wrist pin, respectively over one full revolution of the crank as the engine speed increased from zero to the max-imum. And the second and fourth figures show the variation in the total force on the same respective pins with both the inertia and gas force components included. In conclusion, the two plots show only the first of crank revolution where the gas force in a four-stroke cylinder occurs90BLANCING THE SINGLE-CYLINDER ENGINE According to the former sections, gas forces and inertia and shaking forces will result in the significant forc