Answer to Question #196810 in Mechanical Engineering for muhammed mudhasir

Question

In a four bar chain ABCD, link AD is fixed and the crank AB rotates at 10 rad/sec clockwise. Length of the links are AB=60

mm; BC=CD=70 mm; DA= 120 mm. When the angle DAB = 60o and both B and C lie on the same side of AD, find 1.angular

velocities (magnitude and direction) of BC and CD; and 2.angular acceleration of BC and CD.

Answer 1

By measurement , we find that-

VCD$=$ Vector dc $=$ $4.6 \times scale=4.6\times0.1=0.46m/sec$

VCB$=$ VBC$=$ Vector C$=$ 4.4$\times scale=4.4\times 0.1=0.44m/sec$

Angular velocity if link BC-

$VBC= Vector bc =0.44m/sec$

wBC$= \frac{VBC}{BC}= \frac{0.44}{0.07}=6.28rad/sec$

wBC $=6.28rad/sec$

Angular velocity of link CD

$VCD=vector \space dc=0.46m/sec$

wCD$=\frac{VCD}{CD}=\frac{0.46}{0.07}=6.57rad/sec$

wCD$=6.57rad/sec$

Angular Acceleration of links BC &CD

-since the angular acceleration of the crank AB is not given, therefore there will be no tangential component of the acceleration of B with respect to A.

-we know that the radial component of the acceleration of B with respect to A ( on the acceleration of B)

$a^r_{BA}=a_{BA}=a_B=\frac{v^2_{BA}}{AB}=\frac{(0.6)^2}{0.06}=6m/s^2$

-Radial component of the acceleration of C with respect to $B_1$

$a^r_{CB}=\frac{V^2CB}{BC}=\frac{(0.44)^2}{0.07}=2.76m/s^2$

and,

-radial component of the acceleration of C with respect to D (or the acceleration of C)

$a^r_{CD}=a_{CD}=a_c=\frac{v^2CD}{DC}=\frac{(0.46)^2}{0.07}=3.02m/s^2$

(i) since A and D are fixed points ,therefore these points lie at one place in the acceleration diagram. Draw vector a'b' parallel to AB ,to some suitable scale , to represent the radial component of acceleration of B with it $a^r_{BA}$ or $a_B$ such that,

vector a'b' $=a ^r_{BA}=a_B=6M/S^2$

(ii) from point b' , draw vector b' parallel to BC to represent the radial component of acceleration of C with respect to B i.e $a^r_{CB}$ such that

$vector b'x=a^r_{CB}= 2.76m/s$

(iii) from point x , draw vector xc' perpendicular to BC to respect the tangential component of acceleration of C with respect to B i.e $a_{CB}^t$ while magnitude is not yet known.

(iv) now from point d' draw vector d'y parallel to DC represent the radial component of the acceleration of C with respect to D

$Vector d'y=a^r_{CD}=3.02m/s^2$

(v) from point y, draw vector yc' perpendicular ti DC to represent the tangetial acceleration of C with respect to D i.e

$a^t_{CD}$

(vi) the vector xc' and yc' intersect at join a'c' and b'c' . By measurement we find that-

$a_{CB}^r= vector xc'= 0.7\times 1=0.7m/s^2$

$A_{CD}^t= vector yc' =7.2m/s^2$

now ,angular acceleration of link BC,

$\alpha BC =\frac{a_{CB}^t}{BC}=\frac {0.7}{0.07}=10 rad/s^2$

$\alpha CD = \frac{a_{CD}^t}{DC}=\frac {7.2}{0.07}=102.85 rad/s^2$