Acceleration in slider-crank mechanism

ACCELERATION IN SLIDER-CRANK MECHANISM

A slider-crank mechanism is shown in Fig.2.14(a). Crank OA rotates clockwise with uniform angular velocity ω rad/s. It is required to draw the acceleration diagram of the configuration.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale, as shown in Fig.2.14(a).

Step 2: Velocity of input link: When length of input link OA and its angular velocity ω_OA are known, then the velocity of input link (i.e., crank OA) is given by

v_OA = ω_OA × OA (clockwise about O)

Step 3: Velocity diagram: Now draw the velocity diagram, as shown in Fig.2.14(b), to some suitable scale, using the procedure given in Section 2.7.

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

Velocity of the connecting rod, v_BA = vector ab

and Velocity of the slider, v_BO = vector ob

The angular velocity of the connecting rod AB (ω_BA) can be determined using the relation

Step 5: Construction of acceleration diagram:

Using the velocity of various links that are obtained with the help of velocity diagram, the values of radial and tangential components of acceleration of various link can be calculated "as shown in Table 2.3 below.

Table 2.3 Radial and tangential components of acceleration of various links

Now using the known values of magnitude and direction of acceleration components, the acceleration diagram can be constructed, as shown in Fig.2.14(c), to some suitable scale, using the procedure given below.

1. Since the link AG is fixed, therefore take o' and g' as one point.

2. From point o', draw vector o'a' such that  in the direction parallel to OA to represent the radial component of acceleration of link OA (i.e., a^r_AO). Since a^t_AO = 0, therefore a_AO = a^r_AO.

3. From point a', draw vector a'x such that a'x = a^r_BA = v²_BA/AB in the direction parallel to BA to represent the radial component of acceleration of link AB (i.e., a^r_BA). Now from point x, draw vector xb' perpendicular to AB to represent the tangential component of acceleration of link AB (i.e., a^t_BA) whose magnitude is unknown.

4. We know that the acceleration of slider B acts in the direction parallel to the line of motion of slider B. So from point o', draw a vector o'b' parallel to BO, intersecting the vector xb' at b'.

5. Join points a' and b'.

Step 6: Acceleration of various links:

Now by measurement from the acceleration diagram, the various components of acceleration of links can be found.

Acceleration of slider, a_BO = vector b'o'

Radial component of acceleration of the connecting rod, a^r_BA = vector dx

Tangential component of acceleration of the connecting rod, a^t_BA = vector b'x

Total acceleration of the connecting rod, a_BA = vector b'a'

Also the angular acceleration of the connecting rod can be determined as

Example 2.6

ABCD is a four-bar chain with link AB fixed. The lengths of the links are AB 62.5 mm; BC 175 mm; CD = 112.5 mm; and AD = 200 mm. The crank AB rotates at 10 rad/s clockwise. Draw the velocity and acceleration diagram when angle BAD 60° and B and C lie on the same side of AD. Find the angular velocity and angular acceleration of links BC and CD.

[A.U., Nov/Dec 2011]

Given data:

AB = 62.5 mm; BC = 175 mm; CD = 112.5 mm; AD = 200 mm; ω_BA = 10 rad/s (CW);  BAD = 60°.

Solution: Relative velocity method.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale Gay, 1 cm = 40 mm), as shown in Fig.2.15(a).

Step 2: Velocity of input link AB:

Velocity of input link AB, v_BA = ω_BA × AB

= 10 × 0.0625 = 0.625 m/s

Step 3: Velocity diagram: Now draw the velocity diagram, to some suitable scale (say, 1 cm = 0.4 m/s), as shown in Fig.2.15(b), using the procedure given in Example 2.1.

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

v_CB = vector bc = 0.35 m/s

and v_CD = vector dc = 0.44 m/s

The angular velocity of links BC and CD are given by

Step 5: Acceleration diagram: The values of radial and tangential components of acceleration of various links are calculated as shown in Table 2.4.

Table 2.4. Radial and tangential components of acceleration of various links

Now using the known values of magnitude and direction of acceleration components, the acceleration diagram can be constructed, to some suitable. scale (say, 1 cm = 1.5 m/s²), as shown in Fig.2.15(c), using the procedure given in Section 2.11.

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

a^t_CB = vector xc' = 4.2 m/s²

a^t_CD = vector yc' = 5.5 m/s²

The angular acceleration of links BC and CD are given by

Example 2.7

In a four-bar mechanism ABCD, the link lengths are as follows: Input link AB = 25 mm; Coupler link BC = 85 mm; Output link CD = 50 mm; Frame AD = 60 mm.

The angle between the frame and the input link is 100° measured amiclockwise. The velocity of point B is 1.25 m/s in the clockwise direction.

(i) Sketch the mechanism.

(ii) Find the angular velocity and angular acceleration of links BC and CD.

(iii) Determine the velocity and acceleration of the mid-point of the link BC.

Given data:

AB = 25 mm; BC = 85 mm; CD = 50 mm; AD = 60 mm; BAD = 100°; v_B = 1.25 m/s

Solution: Relative velocity method.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale (say, 1 cm = 15 mm), as shown in Fig.2.16(a).

Step 2: Velocity of input link AB:

Velocity of input link AB, v_BA = 1.25 m/s (clockwise) (given)

Step 3: Velocity diagram: Now draw the velocity diagram, to some suitable scale (say, 1 cm = 1 m/s), as shown in Fig.2.16(b), using the procedure given in Example 2.1.

Note

In order to find the velocity of mid-point E on link BC, divide the vector bc in the same ratio as E divides BC. Since E is the mid-point of BC, therefore e is also mid-point of vector bc.

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

Velocity of link BC, v_CB  = vector cb = 0.8 m/s

Velocity of link CD, v_CD = vector cd = 1.55 m/s

Velocity of mid-point, E = v_E = vector ed = 1.3 m/s Ans. 

Angular velocity of links BC and CD can be determined as

Step 5: Acceleration diagram: The values of radial and tangential components of acceleration of various links are calculated as shown in Table 2.5.

Table 2.5. Radial and tangential components of acceleration of various links

Now using the known values of magnitude and direction of acceleration components, the acceleration diagram can be constructed, to some suitable scale (say, 1 cm = 10 m/s²), as shown in Fig.2.16(c) using the procedure given in Section 2.11.

Note

 In order to find the acceleration of the mid-point E on link BC, divide the vector b'c' in the same ratio as E divides BC. Since E is the mid-point of BC, therefore e' is also mid-point of vector b'c'.

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

a^t_CB = vector xc' = 12 m/s²

a^t_CD = vector yc' = 18 m/s²

Acceleration at mid-point E, a_E = vector a'e' = 58 m/s² Ans.

Now the angular acceleration of links BC and CD are given by

Example 2.8

In a small steam engine running at 600 rad/min clockwise, length of crank is 80 mm and•ratio of connecting rod length to crank radius is 3. For the position when crank makes 45° to horizontal, determine:

(i) the velocity and acceleration of the piston;

(ii) the angular velocity and angular acceleration of the connecting rod; and

(iii) the linear velocity and acceleration of a point X on connecting rod 80 mm from crank pin.

Given data:

ω_OA = 600 rad/min = 600/60 = 100 rad/s (CW); OA = 80 mm; n = l/r = 3 or l = 3 r = 3 (80) = 240 mm; BOA = 45°

Solution: Relative velocity method.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale (say, 1 cm = 40 mm), as shown in Fig.2.17(a).

Step 2: Velocity of input link:

Velocity of crank OA is given by

v_OA = ω_OA × OA = 100 × 0.080 = 8 m/s

Step 3: Velocity diagram: Now draw the velocity diagram, as shown in Fig.2.17(b), to some suitable scale (say, 1cm = 2 m/s), using the procedure given in Section 2.7.

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

Velocity of piston, v_B = vector ab = 7 m/s  Ans.

Velocity of connecting rod, v_BA = vector ab = 6 m/s

Velocity of point X on AB, v_X = vector x = 7 m/s Ans. 

Note

In the velocity diagram, the position of point x on the connecting rod can be obtained as below.

Now the angular velocity of the connecting rod AB is given by

Step 5: Acceleration diagram: The values of radial and tangential components of acceleration of various links are calculated as shown in Table 2.6.

Table 2.5. Radial and tangential components of acceleration of various links

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

Note

 In the acceleration diagram, the position of point x' on the connecting rod a'b' can be obtained as below.

Now the angular acceleration of the connection rod AB is given by

Example 2.9

In the mechanism, as shown in Fig.2.18, the crank OA rotates at 20 rpm anticlockwise and gives motion to the sliding blocks B and D. The dimensions of the various links are OA = 300 mm; AB = 1200 mm; BC≈ 450 mm and CD = 450 mm.

For the given configuration, determine: (i) velocities of sliding at B and D; (ii) angular velocity of CD; (iii) linear acceleration of D; and (iv) angular acceleration of CD.

Given data:

N_OA = 20 rpm (CCW); OA = 300 mm; AB = 1200 mm; BC = 450 mm; CD = 450 mm

Solution: Relative velocity method.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale (say, 1 cm = 200 mm), as shown in Fig.2.19(a).

Step 2: Velocity of input link:

Velocity of crank OA, v_OA = ω_AO × ΑΟ

= 2.09 × 0.3 = 0.627 m/s

Step 3: Velocity diagram: Now draw the velocity diagram, to some suitable scale (say, 1 cm = 0.2 m/s), as shown in Fig.2.19(b).

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

Step 5: Acceleration diagram: The values of radial and tangential components of acceleration of various links are calculated as shown in Table 2.7.

Table 2.7. Radial and tangential components of acceleration

Now using the known values of magnitude and direction of acceleration components, the acceleration diagram can be constructed, to some suitable scale (say, 1 cm = 0.3 m/s²), as shown in Fig.2.19(c).

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

Example 2.10

In the toggle mechanism shown in Fig.2.20, the slider D is constrained to move on a horizontal path. The crank OA is rotating in the counter clockwise direction at a speed of 180 rpm increasing at the rate of 50 rad/s. The dimensions of the various links are as follows:

OA = 180 mm; CB = 240 mm; AB = 360 mm; and BD = 540 mm.

For the given configuration, determine:

(i) the velocity of the slider D;

(ii) the angular velocity and angular acceleration of links AB, BC and BD; and

(iii) the linear acceleration of the slider D.

Given data:

N_OA = 180 rpm (CCW); α_OA = 50 rad/s²; OA = 180 mm; CB = 240 mm; AB = 360 mm; BD = 540 mm

Solution: Relative velocity method.

Procedure:

Step 1: Configuration diagram: First of all, draw the configuration diagram, to some suitable scale (say, 1 cm = 20 mm), as shown in Fig.2.21(a).

Step 2: Velocity of input link:

Step 3: Velocity diagram: Now draw the velocity diagram, to some suitable scale (say, 1 cm = 0.75 m/s), as shown in Fig.2.21(b).

Step 4: Velocity of various links:

By measurement from the velocity diagram, we get

v_D = vector cd = 2.12 m/s Ans.

v_BA = vector ab = 1 m/s

v_BC = vector cb = 2.92 m/s

v_DB = vector bd = 2.48 m/s

The angular velocity of links AB, BC and BD are given by

Step 5: Acceleration diagram: The value of radial and tangential components of acceleration of various links are calculated as shown in Table 2,8.

Table 2.8. Radial and tangential components of acceleration

Now using the known values of magnitude and direction of acceleration components, the acceleration diagram can be constructed, to some suitable scale (say, 1 çm 10 m/s²), as shown in Fig.2.21(c).

Step 6: Acceleration of various links:

By measurement from the acceleration diagram, we get

The angular acceleration of links AB, BC and CD are given by