Answer to Question #227545 in Mechanical Engineering for Trilok

Question

The following data relate to a four-link mechanism:

Link

AB

BC

CD

AD

Length

60mm

200mm

100mm

140mm

Mass

MOI about an axis through centre of mass

80 kg. mm 1600 kg. mm

kg. mm

AD is the fixed link. The centres of mass for the links BC and CD lie at their mid-point whereas the centre of mass for link AB lies at A. Find the drive torque on the link AB at the instant when it rotates at an angular velocity of 47.5 rad s counter clockwise and angle DAI is 135 degree. Neglect gravity effects.

Answer 1

The velocity of AB $V_{AB} = w_{AB} *AB = 47.5*0.06 = 2.85 m/s^2$

The radial acceleration of AB is given by $a^r_{AB} = \frac{V_{AB}^2 }{ AB} = \frac{2.852 }{0.06} = 135.38 m/s^2$

Since A is fixed, the tangential component is zero. Also, the radial acceleration is parallel to the link AB.

Velocity Diagram:

1. Draw velocity vector of Ab parallel to link AB on a suitable scale.

2. draw the line at an angle perpendicular to BC from B on the velocity diagram.

3. Similarly, draw an extension line from D on the velocity diagram at an angle perpendicular to link CD. Both extension lines intersect at C to give the velocity diagram.

Measure the velocities we get

$V_{AB} =2.85 m/s\\ V_{CD} =2.71 m/s\\ V_{BC} =2.35 m/s\\$

Acceleration Diagram :

We Know that

a^r_AB = V_AB² / AB = 2.85² /0.06 = 135.38 m/s²

Similarly

a^r_BC = V_BC² /BC = 2.35² /0.2 = 27.61 m/s²

a^r_CD = V_CD² /CD = 2.71² /0.1 = 73.44 m/s²

Steps:

1. Since AB have only the radial acceleration draw the radial component parallel to link AB

2.Draw the radial component of BC from B and draw tangential component perpendicular to radial component

3Draw the radial component of CD from D and draw tangential component perpendicular to radial component

From acceleration diagram we get the acceleration of eac link

a_B = 135.38 m/s²

a_BC = 37.78 m/s²

a_C = 100.37 m/s²

To find the acceleration due to masses we use similarity ratios

since the mass m1 is at the orgin the acceleration due to the mass is zero

$\frac{BC}{B_2} =\frac{a_BC}{a_m^2}$

B2- distance between Band mass m₂

$\frac{200}{100} = \frac{37.78}{a_{m_2}}\\ a_{BM_2} = \frac{a_{BC} }{ 2} = 18.89 m/s^2$

(Since the center of mass is at center)

$a_{m_3} =\frac{a_C}{ 2} = 50.19 m/s^2$

Plotting the point in the acceleration diagram and measure it to get true acceleration due to the mass. The acceleration due to the mass is drawn in green color.

From the drawing we get

$a_{m_2} = 117.64 m/s^2$

Then the inertia force is given by

F = -ma

Therefore;

$F_1 = - 0.2 * 0 =0\\ F_2 = -0.4 * 117.64 = - 47.06 N\\ F_3 = - 0.6*50.19 = -30.11 N$

The negative sign indicates the forces are acting in opposition to the direction of acceleration.

Also, the forces are displaced from the center of mass. Let the distance be h.

Then we get

F*h = I

I - Moment of inertia

h - Perpendicular distance between the force and center of gravity.

 - Angular acceleration.

 = a_t / Length of link

₂ = 25.78/0.1 = 257.8 rad/s

₃ = 90.65 /0.2 = 453.25 rad/s

Therefore

$h_2 = 1600 *10-6 *\frac{257.8 }{ 47.06} = 0.0088m = 8.8 mm\\ h_3= 400*10-6 *\frac{453.25}{ 30.11} = 0.006 m = 6 mm$