13216-14-9P AID: 1825 | 23/09/2014

Show the free body diagram of the given four-bar linkage as shown in Figure (1).

Let , , and be the unit vectors.

Calculate the position difference vectors , , , , and on the four-bar linkage in the vector form from Figure (1).

Here, the position difference vector between points , and is , the position difference vector between points , and is , the position difference vector between points , and is , the position difference vector between points , and is , the position difference vector between points , and is , and the position difference vector between points , and is .

Use the following steps to draw the velocity polygon for the given four-bar linkage.

Use the scale factor 1:1 to draw the velocity polygon.(3.6 cm for 3.6 m/s) Mark a point on a blank sheet as a reference point.

Construct the line perpendicular to the link 2 from the known value of .

From the terminus of point construct a line in a direction perpendicular to link .

From the point draw a line in a direction perpendicular to link to intersect the line at .

Measure the lines , and , and calculate the angular velocities of the links 3, and 4 .

Draw two arcs with centers at points , and , and radius as known velocities of the links , and . These two arcs intersect at the point .

Draw two arcs with centers at points , and , and radius as known velocities of the links and . These two arcs intersect at the point .

Construct lines , and from the points and respectively in a direction perpendicular to the position vectors and . These two lines intersect at point .

Construct lines , and from the points , and respectively in a direction perpendicular to the position vectors , and . These two lines intersect at point

Show the velocity polygon for the given four-bar linkage as shown in Figure (2).

Use the following steps to draw the acceleration polygon for the given four-bar linkage.

Use the scale factor 1:1 to draw the acceleration polygon. Mark a point on a blank sheet as a reference point. From the known angular velocity of the link 2, draw the normal acceleration of point .

From the point draw a line , in a direction parallel to the position vector.

Calculate the normal acceleration of the link 4 using the angular velocity of the link 4. Draw a line from the point in a direction parallel to the link 4, and length equal to the normal acceleration of link 4.

From the terminus of the line draw a line in a direction perpendicular to the link 4.

From the terminus of the line draw a line in a direction parallel to the link , this denotes the normal acceleration of the link .

Measure the angles ,,,,,,, and from Figure (2).Use these angles to construct the lines , , , , , , , and respectively.

From the construction of the above lines, the points C, D, , and are obtained. Measure the lengths from the acceleration polygon to get the total acceleration for the required links.

Show the acceleration polygon for the given four-bar linkage as shown in Figure (3).

Perform the kinematic analysis of the given four bar linkage.

Express the velocity difference equation relating the points A, and .

Here, the velocity of the point A is , the velocity of the point is , and the velocity difference vector of points A, and is .

Substitute 0 for , and for .

Here, the angular velocity of the link 2 is , and the position difference vector between the points A, and is .

Substitute 12 for , and for .

Hence, the velocity of the point A is .

Express the velocity difference equation relating the points A, and B.

(1)

Here, the velocity of the point B is, and the velocity difference vector of points A, and B is .

Express the velocity difference equation relating the points B, and .

(2)

Here, the velocity of the point is , and the velocity difference vector of points B, and is .

From Equations (1), and (2).

Substitute for , for , 0 for , and for .

Substitute for , for , for , and for .

Consider unit vector

(3)

Consider unit vector

(4)

Solve Equations (3), and (4) for , and . Multiply Equation (3) with 1.304, and Equation (4) with 0.740.

(5)

(6)

Add Equations (5), and (6) to find .

Substitute 3.607 for in Equation (3) to find .

Hence, the angular velocity of link 3 is , and the angular velocity of the link 4 is .

Substitute 0 for, for , 3.607 for , and for in Equation (2) to find .

Hence, the velocity of the point B is .

Express the acceleration difference equation relating the points A, and .

(7)

Here, the acceleration of the point A is , the acceleration of the point is , and the acceleration difference vector of points A, and is .

Express as the sum of normal, and tangential components of acceleration, and substitute 0 for in Equation (7).

Here, the normal acceleration difference vector of the points A, and is , the tangential acceleration difference vector of the points A, and is , and the angular acceleration of the link 2 is .

Substitute 12 for , for , and 0 for .

Hence, the acceleration of the point A is .

Express the acceleration difference equation relating the points A, and B.

Here, the acceleration of the point B is , and the acceleration difference vector of points A, and B is

Express the total acceleration of as the sum of normal, and tangential accelerations.

(8)

Here, the normal acceleration difference vector of the points A, and B is , the tangential acceleration difference vector of the points A, and B is , and the angular acceleration of the link 3 is .

Express the acceleration difference equation relating the points B, and O4.

Here, the acceleration of the point is , and the acceleration difference vector of the points B, and is .

Express the total acceleration of as the sum of normal, and tangential accelerations, and substitute 0 for .

(9)

Here, the normal acceleration difference vector of the points B, and is , the tangential acceleration difference vector of the points B, and is , and the angular acceleration of the link 4 is .

From equations (8), and (9)

Substitute for , for , for , for , for , for , and for .

Consider unit vector

(10)

Consider unit vector

(11)

Solve equations (10), and (11) for , and . Multiply Equation (10) with 1.304, and Equation (11) with 0.740.

(12)

(13)

Add Equations (12), and (13) to find .

Substitute for in Equation (10) to find .

Hence, the angular acceleration of link 3 is , and the angular acceleration of link 4 is .

Express the acceleration difference equation relating the points G3, and A.

Here, the acceleration of the point is , and the acceleration difference vector of points , and A is

Express the total acceleration of as the sum of normal, and tangential accelerations.

Here, the normal acceleration difference vector of the points , and A is , the tangential acceleration difference vector of the points , and A is

Substitute for , for ,for , and for .

Hence, the acceleration of the point is .

Express the acceleration difference equation relating the points , and .

Here, the acceleration of the point is , the acceleration of the point is , and the acceleration difference vector of points , and is

Express the total acceleration of as the sum of normal, and tangential accelerations, and substitute 0 for .

Here, the normal acceleration difference vector of the points , and is , the tangential acceleration difference vector of the points , and is

Substitute for , for , and for .

Hence, the acceleration of the point G4 is .

Show the free body diagram of link 2 as in Figure (4).

Show the free body diagram of link 3 as in Figure (5).

Show the free body diagram of link 4 as in Figure (6).

Find the DAlembert inertia forces, torques, and offsets for link 2, link 3, and link4 of the four-bar linkage.

Express the DAlembert inertia force.

(14)

Here, the inertia force for the link is , the mass of the link is , and the acceleration about the mass center of the link is .

Express the DAlembert inertia torque.

(15)

Here, the inertia torque for the link is , the mass moment of inertia for the link is , and the angular acceleration of the link is .

Express the DAlembert inertia offset.

(16)

Here, the offset distance from the center of mass at the which the inertia force acts on the link is .

Substitute 2 for i in Equation (14) to find

Substitute 0 for , and 5.2 kg for .

Hence, the inertia force acting on link 2 is .

Substitute 2 for i in Equation (15) to find

Substitute 0 for , and 4.2 for

Hence, the inertia torque acting on link 2 is 0.

Substitute 2 for i in Equation (16) to find

Substitute 0 for

Hence, the offset of the inertia from the mass center of link 2 is 0 m.

Substitute 3 for i in Equation (14) to find

Substitute for , and 65.8 kg for

Hence, the inertia force acting on link 3 is .

Substitute 3 for i in Equation (15) to find

Substitute -0.406 for , and 4.2 for .

Hence, the inertia torque acting on link 3 is.

Substitute 3 for i in Equation (16) to find

Substitute for , and 2657.057 N for.

Hence, the offset of the inertia force from the center of mass of link 3 is 0.001 m.

Substitute 4 for i in Equation (14) to find

Substitute for , and 21.8 kg for .

Hence, the inertia force acting on link 4 is

Substitute 4 for i in Equation (15) to find

Substitute 2.51 for , and -40.820 for .

Hence, the inertia torque acting on link 4 is .

Substitute 4 for i in Equation (16) to find

Substitute 102.458for , and 420.098 N for .

Hence, the offset of the inertia force from the center of mass for link 4 is 0.244 m.

From Figure (6), take moment about point .

Here, the external force acting on the point D is , and the transverse component of force acting on link 4 due to link 3 is .

Substitute for, for, for, for,for ,for , and only direction for to find the magnitude of .

Consider unit vector .

Hence, the transverse component of force acting on link 4 due to link 3 is .

From Figure (5) the transverse component of force acting on link 3 due to link 4 is . From Figures (5), and (6) forces , and are equal in magnitude but opposite in direction.

Hence, the transverse component force acting on link 3 due to link 4 is .

From Figure (4), take moment about point A.

Here, the radial component of force acting on link 3 due to link 4 is.

Substitute for , for , for , for , direction vector for , and for .

Consider unit vector.

Hence, the radial component of force acting on link 3 due to link 4 is.

From Figures (5), and (6), forces , and are equal in magnitude, and opposite in direction.

Hence, the force acting on link 4 due to link 3 is , and the force acting on link 3 due to link 4 is.

From Figure (6), take the summation of all the forces.

Here, the force acting on link 4 due to link 1 is .

Substitute for , for, and for .

Hence, the force acting on link 4 due to link 1 is .

From Figure (5), take the summation of all the forces.

Here, the force acting on link 3 due to link 2 is .

Substitute for , and for .

Hence, the force acting on link 3 due to link 2 is .

From Figure (4) the force acting on link 2 due to link 3 is. From Figures (4), and (5) forces , and are equal in magnitude but opposite in direction.

Hence, the force acting on link 2 due to link 3 is .

From Figure (4), take the summation of all the forces.

Here, the force acting on link 2 due to link 1 is.

Substitute for.

Hence, the force acting on link 2 due to link 1 is.

From Figure (4), take moment about point .

Here, the external torque required to drive link 2 is.

Substitute for , and for .

Hence, the external torque required to drive link 2 at the given velocity is , and it acts in the counter clockwise direction.