Mechanics of Machines Dr. Mohammad Kilani Class 3 Position Analysis

TYPES OF MECHANISM ANALYSESE

Position Analysis  Position Analysis  Given θ 12 find θ 23, θ 34 and θ 14  Find maximum and minimum values of θ 14.  Find the location of each point on the mechanism for a given value of θ 12 and the curve traced by the a point during motion. θ 12 θ 41 θ 23 θ 34

Velocity Analysis  Position Analysis  Given θ 12 and ω 12 find ω 23, ω 34 and ω 14  Find the speed of a point in the mechanism θ 12, ω 12 θ 23, ω 23 θ 34, ω 34 θ 41, ω 41

Acceleration Analysis  Position Analysis  Given θ 12, ω 12 and α 12 find α 23, α 34 and α 14  Find the acceleration of a point in the mechanism θ 12, ω 12, α 12 θ 23, ω 23, α 23 θ 34, ω 34, α 34 θ 41, ω 41, α 41

Question  Why is acceleration analysis important? θ 12, ω 12, α 12 θ 23, ω 23, α 23 θ 34, ω 34, α 34 θ 41, ω 41, α 41

GRAPHICAL POSITION ANALYSIS

Graphical Position Analysis  Use graphical methods, (pen, ruler, compass and protractors to solve position analysis problems.  CAD drafting methods may also be used as a convenient alternative. θ 12 θ 41 θ 23 θ 34

VECTOR POSITION ANALYSIS

Vector Position Analysis  Use vector loop closure equation to solve position analysis problems

Loop Closure Equation  Use vector loop closure equation to solve position analysis problems

Planar Four Bar Loop Closure Equation  How many scalar equations can be written from the vector loop-closure equation?  How many unknowns can be solved for

Planar Four Bar Loop Closure Equation  How many scalar equations can be written from the vector loop- closure equation?: Two scalar equations could be written from a vector equation  How many unknowns can be solved for: Two unknowns: In a standard problem, θ 1 and θ 2 are given. θ 3 and θ 4 are to be solved for.  We will focus on planar mechanisms in this course

Planar Four Bar Loop Closure Equation  The closure condition expresses the condition that a loop of a linkage closes on itself.  For the four-bar linkage shown the closure equation is

Planar Four Bar Loop Closure Equation  The closure condition expresses the condition that a loop of a linkage closes on itself.  For the four-bar linkage shown the closure equation is  The following two scalar equations are produced

Analytical Solution when θ 2 is known  The above two equations could be solved for two unknowns. One approach is to eliminate one of the unknown angles by isolating the trigonometric function involving the angle on the left-hand side of the equation.  Squaring both sides of both equations and using the identity we obtain

Analytical Solution when θ 2 is known  To obtain an explicit expression for θ 4 in terms of θ 2 and the constant angle θ 1, we combine the coefficients of cos θ 4 and sin θ 4 in the equation above as follows:  where

 The above equation can be solved for θ 4 by using the following half angle identities  After substitution and simplification, we get where Analytical Solution when θ 2 is known

 Solving for t we obtain Analytical Solution when θ 2 is known

The ± sign identifies the two possible assembly modes of the linkage Given the values of θ 1 and θ 2 for a four bar mechanism of known r 1, r 2, r 3 and r 4 the output angle θ 4 is calculated as: Analytical Solution when θ 2 is known

The ± sign identifies the two possible assembly modes of the linkage Given the values of θ 1 and θ 2 for a four bar mechanism of known r 1, r 2, r 3 and r 4 the output angle θ 4 is calculated as: Analytical Solution when θ 2 is known

 Note that –π/2 ≤ tan -1 (t) ≤ –π/2. Therefore, θ 4 will have the range –π ≤ θ 4 ≤ –π.  Unless the linkage is a Grashof type II linkage in one of the extreme positions of its motion range, there are two valid solutions for θ 4. These correspond to two assembly modes or branches for the linkage. Analytical Solution when θ 2 is known

 Because of the square root in the expression for t, it can be complex if (A 2 + B 2 ) < C 2. In this case, the mechanism is a Grashof type II linkage in one of the extreme positions of its motion range and it cannot be assembled in specified values of θ 1 and θ 2.  The assembly configurations would then appear as shown below Analytical Solution when θ 2 is known

 After θ 4 is known, an expression for θ 3 can be obtained by solving the loop closure equation to obtain:  Note that it is essential that the sign of the numerator and denominator be maintained to determine the quadrant in which the angle θ 3 lies. This can be done by using the ATAN2 function. The form of this function is: ATAN2 (sin θ 3, cos θ 3 ) Analytical Solution when θ 2 is known

 Once all of the angular quantities are known, it is relatively straightforward to compute the coordinates of any point on the mechanism.  In particular, the coordinates of Q,P, and R are given by Analytical Solution when θ 2 is known

 If the coupler angle θ 3 is given, and θ 2 and θ 3 are to be determined, graphical solution will require an iterative trial and error solution.  The analytical procedure, in contrast, follows exactly the same procedure as when θ 2 is given. It starts by writing the loop closure equations with θ 2 replacing θ 3 as the variable to be eliminated.  The equations set are of exactly the same form except that the indices 2 and 3 are interchanged. Therefore, we can use directly the position solution derived for the case of known θ 2 while interchanging the indices 2 and 3. Analytical Solution when θ 3 is known

 When the coupler angle θ 3 is given, there is an assembly-mode ambiguity similar to that occurine when θ 2 is given.  It is necessary to know the appropriate mode of the linkage before the analysis is begun;. The mode is determined by the + or – sign used for the square root term when calculating t.  Once the assembly mode is determined, it remains the same for any position of the input link unless the linkage is a class III linkage, and passes through a singular (indeterminate) position. Analytical Solution when θ 3 is known

 A four bar linkage with r 1 = 1, r 2 = 2, r 3 = 3.5, r 4 = 4, and θ 1 = 0, find θ 3 and θ 4 for each of the solution branches when the driving crank is in the positions θ 2 = 0, π/2, π,and - -π/2. Example: Position Analysis of a Four Bar Linkage

 A four bar linkage with r 1 = 1, r 2 = 2, r 3 = 3.5, r 4 = 4, and θ 1 = 0, find θ 3 and θ 4 for each of the solution branches when the driving crank is in the positions θ 2 = 0, π/2, π,and - -π/2. Example: Position Analysis of a Four Bar Linkage

POSITION ANALYSIS FOR A RIGID BODY WHEN TWO POINTS ARE KNOWN

 Given the kinematic properties of one point on a rigid body and the angular position, angular velocity, and angular acceleration of the body, we can compute the position, velocity, and acceleration of any defined point on the rigid body.  For the rigid body shown. Assume that A and B are two points attached to an arbitrary link, say link 5, and a third point is defined relative to the line between points A and B by the angle β and the distance r C/A, which is represented as r 6. Then the position of point C can be computed directly if r A and θ 5 are known. Position Analysis for a Rigid Body When Two Points are Knows

 The position of point C is given as:  If θ 5 If is known, the equation above can be used to calculate the location of point C directly.  We often know the position vectors of two points A and B on the rigid body. The value of θ 5 can be calculated from the from the x and y components of the position vectors for A and B using Position Analysis for a Rigid Body When Two Points are Knows

POSITION ANALYSIS FOR A SLIDER-CRANK MECHANISM

HW#2 (Prob. 4-6 and 4-7, with data in row (a) Table P4-1). (4-10, 4-12, 4-18(f))

 Next to the four- bar linkage, the slider-crank is probably the most commonly used mechanism.  It appears in all internal combustion engines and in numerous industrial and household devices. Position Analysis for a Slider-Crank Mechanism

 To develop the closure equations, locate vectors r 2 and r 3 as was done in the regular four-bar linkage.  One of the other two vectors is taken in the direction of the slider velocity and the other is taken perpendicular to the velocity direction. The loop closure equation is Position Analysis for a Slider-Crank Mechanism Offset r3r3 r2r2 r1r1 r4r4 rprp

Position Analysis for a Slider-Crank Mechanism  Writing the loop closure equation in terms of the vector angles, we obtain  Two scalar equations are produced. The equations can be solved for two unknowns. r3r3 r2r2 r1r1 r4r4 rprp

Position Analysis for a Slider-Crank Mechanism  Unlike the four-bar linkage loop closure equations where all link lengths are known, the piston displacement r 1 is an unknown in the slider-crank equation. The constraint resulting from a known r 1 is replaced by the constraint θ 4 = θ 1 + π/2.  The following problem statements are possible  Crank angle θ 2 given, find θ 3 and r 1  Piston displacement r 1 given, find θ 2 and θ 3  Coupler angle θ 3 given, find θ 2 and r 1. r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when θ 2 is known  The analytical solution procedure follows the same major steps as in the four-bar linkage case. To eliminate θ 3, first isolate it in the loop closure equations as follows:  Squaring both sides of both equations and using the identity we obtain r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when θ 2 is known  The expression gives r 1 in a quadratic expression involving θ 2 and the other known variables. To obtain a solution, collect together the coefficients of the different powers of r 1 as follows r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when θ 2 is known r3r3 r2r2 r1r1 r4r4 rprp  The expression for A could be simplified by noting that

Analytical Solution when θ 2 is known  Solving for r 1 gives  The ± sign indicates two possible assembly modes for the same θ 2.

Analytical Solution when θ 2 is known  Because of the square root in the expression for r 1, it becomes complex when A 2 < 4B.  If this happens, the mechanism cannot be assembled in the position specified.

Analytical Solution when θ 2 is known  Once a value for r 1 is determined, the closure equations can be solved for θ 3 to give  As in the case of the four-bar linkage, it is essential that the signs of the numerator and denominator in the above expression be maintained to determine the quadrant in which the angle θ 3 lies. r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when θ 2 is known  Once all of the angular quantities are known, it is relatively straightforward to compute the coordinates of any point on the vector loops used in the closure equations.  In particular, the coordinates of Q and P are given by r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when r 1 is known  The analytical solution procedure follows the same major steps as in the previous case.  After eliminating θ 3 from the loop closure equation, we simplify the resulting equation as follows  where r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when r 1 is known  The trigonometric half-angle identities can be used to solve the equation above for θ 2. Using these identities and simplifying gives  Solving for t gives r3r3 r2r2 r1r1 r4r4 rprp

Analytical Solution when r 1 is known  The ± sign indicates two possible assembly mode. Typically, there are two solutions for θ 2 and they are both valid.  Because tan -1 has a valid range of π/2 ≤ tan -1 (t) ≤ –π/2, θ 2 will have the range –π ≤ θ 2 ≤ –π.

Analytical Solution when r 1 is known  Because of the square root in the equation for the variable t, it becomes complex when (A 2 + B 2 ) < C 2  If this happens, the mechanism cannot be assembled for the specified value of r 1.

Analytical Solution when r 1 is known  Knowing θ 2, the closure equations can be solved for θ 3. As in the previous cases, it is essential that the signs of the numerator and denominator be maintained to determine the quadrant in which the angle θ 3 lies.

Analytical Solution when θ 3 is known  When the coupler angle θ 3 is the input link, the analytical procedure for solving the position equations follows the same major steps as when θ 2 is the input.  we can assume that θ 1, θ 3, θ 3 are known and that θ 2 and r 1 are to be found. The link lengths r 2 and r 2 and the angles θ 1, and θ 4 are constants.  For the position analysis, again begin with the loop closure equations and isolate the terms with θ 2 The resulting equations of the same form obtained when θ 2 is the input except that the indices 2 and 3 are interchanged. Therefore, we can use directly the position solution derived for the case when θ 2 is the input and interchange the indices 2 and 3.