Answer to Question #200279 in Mechanical Engineering for Muhammad Idrees

Question

The engine mechanism shown in Fig. 8.38 has crank OB = 50 mm and length of connecting rod A B = 225 mm. The centre of gravity of the rod is at G which is 75 mm from B. The engine speed is 200r.p.m. For the position shown, in which OB is turned 45° from O A, Find 1. the velocity of G and the angular velocity of A B, and 2. the acceleration of G and angular acceleration of A B.

Answer 1

BG=75 mm

$N_B=200 rpm \implies \omega_{BO}=\frac{2 \pi N_D}{60}=\frac{2 \pi*200}{60}=20.94 rad/s$

$V_{BO}=\omega _{BO}*BO=20.94*0.05=1.0472 m/s$

The velocity of link BO (crank) on velocity diagram will be $BO=v_{BO}*50=1.0472*50=52.36 mm$

$\frac{bg}{ba}=\frac{BG}{BA} \implies bg=\frac{75}{225}*37.5=12.5 mm$

This can be drawn on the diagram considering the following values.

$oa =42.92; ba=37.5; V_{oa}=\frac{oa}{50}=\frac{42.92}{50}=0.858 m/s$

$v_g=\frac{46.14}{50}=0.923m/s; V_{BA}=\frac{37.5}{50}=0.75 m/s$

Radial accelertion of the components

$OB =a^rob=\frac{1.0472^2}{0.05}=21.93 m/s^2 ; AB =a^rab=\frac{0.75^2}{0.22}=2.5 m/s^2$

$\frac{b'g'}{b'a'}=\frac{BG}{BA} \implies b'g'= \frac{75}{225}*77.35=25.784 mm$

For acceleration diagram

$o'g' =93.31 mm$ then $ag=\frac{93.31}{5}=18.662 m/s^2$

Angular accleartion of AB

$a^t_{AB}=\frac{76.33}{5}=15.266 m/s ^2$

$\alpha_{AB}=\frac{15.266}{0.255}=67.85 rad/s$