Presentation on theme: "Acceleration Polygon for a Four-bar Mechanism"— Presentation transcript:
1Acceleration Polygon for a Four-bar Mechanism IntroductionAcceleration Polygon for a Four-barAcceleration Polygon for a Four-bar MechanismThis presentation shows how to construct the acceleration polygon for a four-bar mechanism. It is assumed that the dimensions for the links are known and the analysis is being performed at a given (known) configuration. Since the four-bar has one degree-of-freedom, the angular velocity and acceleration of one of the links must be given as well.As an example, for the four-bar shown, we first construct the velocity polygon. Then we will learn:How to construct the acceleration polygon shown on the rightHow to extract acceleration information from the polygonOAABAABAAABAO4BO2ω2α2
2ExampleAcceleration Polygon for a Four-barAO4BO2ω2α2ExampleThis example shows us how to construct the acceleration polygon for a typical four-bar, such as the one shown on this slide.It is assumed that: all the lengths are known and the four-bar is being analyzed at the configuration shown; the angular velocity and acceleration of the crank are given as well.
3Vector loop, differentiation Acceleration Polygon for a Four-barBRBAVector loopWe follow the same process as we did for the velocity polygon.We first define four position vectors to obtain a vector loop equation:RAO2 + RBA = RO4O2 + RBO4The first time derivative provides the velocity loop equation:VtA + VtBA = VtBThe second time derivative provides the acceleration loop equation:AA + ABA = ABWe split each acceleration vector into a normal and a tangential component:ARBO4RAO2►O4RO4O2O2AtA + AnA + AtBA + AnBA = AtB + AnBWe need the velocities to calculate some of the accelerations. Therefore we perform a velocity analysis first.
4The direction is found by rotating RAO2 90° in the direction of ω2 Velocity analysisAcceleration Polygon for a Four-barVtABRBAVelocity polygonWe calculate VtA:VtA = ω2 ∙ RAO2The direction is found by rotating RAO2 90° in the direction of ω2The direction of VtBA is perpendicular to RBAThe direction of VtB is perpendicular to RBO4Now we construct the velocity polygonNext we determine the angular velocities.We will use these results to calculate the normal components of acceleration vectors.ARBO4ω2RAO2O4►RO4O2O2►A►VtAOV►VtBAVtBBVtA + VtBA = VtB
5The absolute value of the angular velocities are computed as: Acceleration Polygon for a Four-barVtABω3RBAAngular velocitiesThe absolute value of the angular velocities are computed as:ω3 = VtBA ∕ RBAω4 = VtB ∕ RBO4RBA has to be rotated 90° cw to head in the direction of VtBA. Therefore ω3 is cwRBO4 has to be rotated 90° ccw to head in the direction of VtB. Therefore ω4 is ccwWe will use these results to calculate accelerations.ARBO4ω2RAO2ω4O4RO4O2O2►AVtAOVVtBA►VtBB
6AtA + AnA + AtBA + AnBA = AtB + AnB Normal componentsAcceleration Polygon for a Four-barAtA + AnA + AtBA + AnBA = AtB + AnBBω3RBANormal componentsWe first calculate the magnitude of all the normal components:AnA = ω22 ∙ RAO2AnBA = ω32 ∙ RBAAnB = ω42 ∙ RBO4The direction of each normal component of acceleration is opposite to the corresponding position vectorARBO4ω2RAO2ω4O4RO4O2O2BAnBA►AAnBAnAO4O2
7Tangential components Acceleration Polygon for a Four-barAtA + AnA + AtBA + AnBA = AtB + AnBBRBATangential componentsWe first calculate the magnitude of AtA:AtA = α2 ∙ RAO2The direction is found by rotating RAO2 90° in the direction of α2We also know that AtBA is on an axis perpendicular to RBASimilarly we know that AtB is on an axis perpendicular to RBO4ARBO4RAO2α2O4RO4O2O2►►AtABAnBAA►AnBAnAO4O2
8AtA + AnA + AtBA + AnBA = AtB + AnB Acceleration polygonAcceleration Polygon for a Four-barAtA + AnA + AtBA + AnBA = AtB + AnBAcceleration polygonNow we are ready to draw the acceleration polygon.First we select the origin and add AtA and AnA to obtain AAAnBA is a added at AWe also know the axis of AtBA which would be added to AnBAAnB is added to the originWe also know the axis of AtB which would be added to AnBThe two lines intersect at B. We complete the acceleration polygon by drawing the missing accelerationsNext we determine the angular accelerations.AtABAnBAAAnB►AnA►O4O2AtA►►AnAOAAAAnB►AnBAAABABA►AtBAAtBB
9Angular accelerations Acceleration Polygon for a Four-barBα3RBAAngular accelerationsThe absolute values of the angular accelerations are computed as:α3 = AtBA ∕ RBAα4 = AtB ∕ RBO4RBA has to be rotated 90° cw to head in the direction of AtBA. Therefore α3 is cwRBO4 has to be rotated 90° ccw to head in the direction of AtB. Therefore α4 is ccwThis completes the acceleration analysis of this four-bar.ARBO4RAO2α4α2O4RO4O2O2AtAAnAOA►AAAnBAnBA►ABABAAtBAAtB