# ch07 - Solutions to Chapter 7 Exercise Problems Problem 7.1...

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- 277 - Solutions to Chapter 7 Exercise Problems Problem 7.1 A coupler curve has the approximate straight-line section shown in the figure below. Design a four-bar linkage that will generate the portion of the curve shown. Describe the linkage in sufficient detail that it can be manufactured. X Y y A x A A B y B x B x A = 6 y A = 3 x B = 18 y B = 12 Solution: To solve the problem, we need to find a coupler curve with an approximate straight-line section. We can use any of the mechanisms identified in section 7.1.2. We will use the Watt straight-line mechanism shown in Fig. 7.1. Assuming the crank is the driver, the coupler curve is shown in Fig. 7.1.1 for the coupler length equal to 5/3 times the crank length. This drawing is based on the results from the program fourbar_analysis.m included on the disk with this book. The linkage is a Grashof, Type 2 double rocker. The straight line part of the coupler curve is given by segment SU, and this distance is approximately equal to the length of the cranks. The line SU is also bisected by the line of centers PQ, and vice versa. Also, line SU is inclined at an angle of approximately -50.2 ˚ to the line PQ. The length of the line of centers can be computed using the Pythagorean theorem when the coupler and cranks are at right angles to each other as shown in Fig. 7.1.2. Using a crank length of 3 and a coupler length of 5, the center distance is found to be 2(3.9051) = 7.8102. 50.2 ˚ P Q R S T U V Fig. 7.1.1: Coupler curve for Watt's mechanism when the coupler length is 5/3 times the length of the cranks.

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- 278 - 3 2.5 2.5 3 3.9051 50.2 ˚ P Q R T V Fig. 7.1.2: Determination of the distance between centers. To use this linkage for the given problem, it is necessary to map point S to point A and point U to point B. The length of the line AB is given by L = ( x B ± x A ) 2 + ( y B ± y A ) 2 = (18 ± 6) 2 + (12 ± 3) 2 = 15 and the angle of inclination of the line AB is given by ± = tan ² 1 y B ² y A x B ² x A [ ] = tan ² 1 18 ² 6 12 ² 3 [ ] = tan ² 1 12 9 [ ] = 53.13 ˚ Therefore, to use the Watt's linkage to generate the curve, we need only scale the linkage so that the length of SU is 15 and rotate the linkage so that the line SU is at and angle of 53.13 ˚ to the horizontal. The rotation required is 53.13 ˚ - 50.20 ˚ or 2.93 ˚ . The length of line SU in Fig. 7.1.1 is 3. Therefore, the scaling factor for the linkage is (15/3) or 7. The final link lengths then are: r 1 = 5 ± 7.8102 = 39.051 r 2 = 5 ± 3 = 15.000 r 3 = 5 ± 5 = 25.000 r 4 = 5 ± 3 = 15.000 The coupler point T which generates the coupler curve is located midway between the two ends of the coupler. In the position corresponding to Fig. 7.1.2, the point T has the following x and y coordinates: x T = ( x B + x A ) / 2 = (18 + 6) / 2 = 12 y T = ( y B + y A ) /2 = (12 + 3) / 2 = 7.5 The coordinates of point P are given by: x P = x T ± PT cos ² = 12 ± (39.051/ 2)cos(2.93) = 12 ± 19.5255cos(2.93) = ± 7.500 y P = y T ± PT sin ² = 7.5 ± (39.051 /2)sin(2.93) = 7.5 ± 19.5255sin(2.93) = 6.502 And the coordinates of point Q are given by:
- 279 - x Q = x T + TQ cos ± = 12 + (39.051 / 2)cos(2.93) = 12 + 19.5255cos(2.93) = 31.500 y Q = y T + TQ sin ± = 7.5 + (39.051 /2)sin(2.93) = 7.5 + 19.5255sin(2.93) = 8.498 The final solution linkage is shown in Fig. 7.1.3.

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