Presentation on theme: "Velocity Analysis with Instant Centers for a Four-bar Mechanism"— Presentation transcript:
1 Velocity Analysis with Instant Centers for a Four-bar Mechanism IntroductionInstant Centers for a Four-barVelocity Analysis with Instant Centers for a Four-bar MechanismThis presentation shows how to perform velocity analysis for a four-bar mechanism with the method of instant centers. 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 of one of the links (or one other velocity information) must be given as well.For a given four-bar mechanism the velocity analysis consists of two steps:Finding the instant centersFinding velocitiesNote that the instant center method is a graphical method.
2 Four-bar mechanismInstant Centers for a Four-barFour-bar mechanismAssume that for this four-bar mechanism all the link lengths are known and the angular velocity of the crank is given as ω2 ccw.In the configuration shown we can perform a velocity analysis with the instant center method.BAω2O2O4
3 Number of instant centers Instant Centers for a Four-barNumber of instant centersThe first task is to determine how many instant centers exist for a four-bar.The number of links in a four-bar isn = 4Between n links, there aren (n − 1) ∕ 2instant centers. That means in a fourbar there are 4 (4 − 1) ∕ 2 = 6 instant centers.A small circle will help us keep track of locating each center. On the circumference of the circle we put as many marks as the number of links.Each time we find a center between two links, we draw a line between the corresponding marks on the circle.BAω2O212O4►43
4 Finding the instant centers Instant Centers for a Four-barFinding the instant centersFour of the centers are already known: They are the four pin joints.We don’t know I1,3 but we know that it lies on the same line as I4,1 and I3,4.I1,3 also lies on the same line as I2,1 and I3,2.The point of intersection is I1,3.I2,4 is also unknown but it lies on the same line as I3,4 and I2,3.I2,4 also lies on the same line as I4,1 and I1,2.The point of intersection is I2,4.Now we have found all 6 centersB = I3,4A = I2,3I2,4►►ω2O2 = I1,2►►12►O4 = I4,1►►43I1,3
5 A (or I2,3) is a point on link 2, therefore: Finding ω3, knowing ω2Instant Centers for a Four-barFinding velocitiesA (or I2,3) is a point on link 2, therefore:VA = ω2 ∙ RAI1,2Its direction is obtained by rotating RAI1,2 90° in the direction of ω2.A (or I2,3) is also a point on link 3, which rotates around I1,3, this means:VA = ω3 ∙ RAI1,3Since we already know VA, we can solve for ω3:ω3 = VA ∕ RAI1,3ω3VAB = I3,4A = I2,3I2,4RAI1,2ω2►O2 = I1,2RAI1,3O4= I4,1►I1,3
6 Since we know ω3, we can find VB. B is a point on link 3: Finding ω4, knowing ω3Instant Centers for a Four-barFinding velocitiesSince we know ω3, we can find VB. B is a point on link 3:VB = ω3 ∙ RBI1,3Its direction is obtained by rotating RBI1,3 90° in the direction of ω3.B is also a point on link 4 which yields:VB = ω4 ∙ RBI1,4We already know VB so we can solve for ω4:ω4 = VB ∕ RBI1,4VBω3VAB = I3,4A = I2,3I2,4RBI4,1ω2►O2 = I1,2ω4O4= I4,1RBI1,3►I1,3
7 We could have determined ω4 without knowing ω3: Finding ω4, knowing ω2Instant Centers for a Four-barAnother approachWe could have determined ω4 without knowing ω3:I2,4 is a point on link 2, therefore:VI2,4 = ω2 ∙ RI2,4 I1,2Its direction is obtained by rotating RI2,4 I1,2 90° in the direction of ω2.I2,4 is also a point on link 4, which rotates around I4,1. This means:VI2,4 = ω4 ∙ RI2,4 I4,1We already know VI2,4 so we can solve for ω4:ω4 = VI2,4 ∕ RI2,4 I4,1B = I3,4A = I2,3I2,4VI2,4RI2,4 I1,2ω2O2 = I1,2ω4►RI2,4 I4,1O4= I4,1I1,3►
8 Velocity of a coupler point Velocity of coupler pointInstant Centers for a Four-barVPVelocity of a coupler pointThe instant center method makes it easy to find the velocities of additional points.For example suppose link 3 is a triangular plate and we want to determine the velocity of P. Since P is a point on link 3, we have:VP = ω3 ∙ RPI1,3The direction is found by rotating RPI1,3 90° in the direction of ω3.PB = I3,4A = I2,3I2,4RPI4,1ω2O2 = I1,2O4= I4,1►I1,3
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