College

Problem 2 (8 points)

A crank-rocker is to be designed so that the output rocker \(O_4B\) makes a 45-degree swing angle. It is required that the rocker takes 0.8 seconds moving from extreme right to extreme left and 0.6 seconds to move back.

a) Assuming that the rocker length \(l_4\) is 50 mm, use a graphical design technique to find suitable link lengths \(l_1\), \(l_2\), and \(l_3\). Show your work. (6 points)

b) At what speed should crank \(O_2A\) rotate to ensure the specified timing? (2 points)

Answer :

The lengths of links 11, 12, and 13 can be measured from the drawing.

Crank O₂A should rotate at a speed of approximately 4.487 rad/s to ensure the specified timing.

a) To find suitable link lengths 11, 12, and 13 using graphical design technique, we can follow these steps:

Step 1: Draw a rough sketch of the crank-rocker mechanism with the given dimensions. Place the fixed pivot point O at one end and the desired swing angle position for rocker O&B.

Step 2: Choose a suitable length for link 14, which is given as 50 mm.

Step 3: Determine the fixed pivot point for the rocker, denoted as P. Place point P on the desired swing angle position for rocker O&B. Draw a line connecting points O and P.

Step 4: Select a length for link 11. Draw a line parallel to the line OP, and locate point A on this line.

Step 5: Select a length for link 12. Draw a circle with center at point O and a radius equal to the length of link 12. The circle should intersect the line OP at point B.

Step 6: Select a length for link 13. Draw a line connecting points A and B.

b) To determine the speed at which crank O₂A should rotate to ensure the specified timing, we need to consider the angular velocity of the crank. The angular velocity, ω, can be calculated using the formula:

ω = (2π) / T

where T is the time taken for one complete revolution of the crank, which is equal to the sum of the time taken to move from extreme right to extreme left and the time taken to move back.

Given that it takes 0.8 s to move from extreme right to extreme left and 0.6 s to move back, the total time for one revolution is:

T = 0.8 + 0.6 = 1.4 s

Substituting this value into the formula, we get:

ω = (2π) / 1.4 ≈ 4.487 rad/s

Know more about graphical design here:

https://brainly.com/question/11299456

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