Answer to Question #206791 in Mechanical Engineering for pspsps3333

Question

The crank and connecting rod of a theoretical steam engine are 0.5 m and 2 m long respectively. The crank makes 180 r.p.m. in the clockwise direction. When it has turned 45° from the inner dead centre position, determine : 1. Velocity and acceleration of the piston 2. Angular velocity and angular acceleration of the connecting rod 3. Velocity and acceleration of point E on the connecting rod 1.5 m from the gudgeon pin 4. Position and linear velocity of any point G on the connecting rod which has the least velocity relative to crank shaft 

Answer 1

$N _{ BO} =180 r.p.m. \implies ω_{ BO} ​ =2π× \frac{180}{60}=18.852rad/s$

$v _{BO} ​ =v{ B} ​ =ω _{ BO} ​ ×OB=18.852×0.5=9.426m/s$

By measurement, we find that velocity of piston P, v_P ​= vector op=8.15m/s

v_PB = vector bp=6.8m/s

Since the length of connecting rod PB is 2 m, therefore angular velocity of the connecting rod,

$ω _{ PB} ​ = \frac{v_{PB}}{PB} = = \frac{6.8}{2} ​ =3.4rad/s( Anticlockwise )$

By measurement, we find that velocity of point E, $v _ E ​ = vector oe=8.5m/s$

$\frac{BE}{BP} ​ = \frac{ bp}{ be} \space ​ or \space be= \frac{BE×bp}{BP} ​$

We know that velocity of rubbing at the pin of crank-shaft

$= \frac{d_ O}{2} ​ ​ ×ω_{ BO} ​ = 2 0.05 ​ ×18.85=0.47m/s$

Velocity of rubbing at the pin of crank

$= \frac{d _ B}{2} ​ ​ (ω _{ BO} ​ +ω_{ PB} ​ )= \frac{ 0.06}{2} ​ (18.85+3.4)=0.6675m/s$

∵ω_BO ​ is clockwise and ω_PB ​ is anticlockwise and velocity of rubbing at the pin of cross-head

$= \frac{ d_ C}{2} ​ ​ ×ω _{ PB} ​ = \frac{ 0.03}{2} ​ ×3.4=0.051m/s$

∴At the cross-head, the slider does not rotate and only the connecting rod has angular motion.

By measurement, we find that vector bg = 5 m/s

The position of point G on the connecting rod is obtained as follows

$\frac{b_g}{b_p} ​ = \frac{BG}{BP} ​ or BG= \frac{bg}{bp} ​ ×BP= \frac{5}{6.8} ​ ×2=1.47m$

By measurement, we find that the linear velocity of point G, v_G ​= vector og=8m/s