written 6.7 years ago by | • modified 6.6 years ago |
Subject: Kinematics of Machinery
Topic: Kinetics of Rigid Bodies
Difficulty: Medium
written 6.7 years ago by | • modified 6.6 years ago |
Subject: Kinematics of Machinery
Topic: Kinetics of Rigid Bodies
Difficulty: Medium
written 6.6 years ago by |
This mechanism is mostly used in shaping and slotting machines. In this mechanism the link CD (link 2) forming the turning pair is fixed ,as shown in fig. .The link 2 corresponds to a crank in a reciprocating steam engine. the driving crank CA (link 3) rotates at a uniform angular speed. The slider (link 4) attached to the crank pin at A slides along the slotted bar PA (link 1) which oscillates at a pivoted point point D. The connecting rod PR carries the ram at R to which a cutting tool is fixed. The motion of the tool is constrained along the line RD produced, i.e. along a line passing through D and perpendicular to CD.
When the driving crank CA moves from the position $CA_{1}$ to $CA_{2}$ (or the link DP from the position $DP_{1}$ to $DP_{1}$) through an angle α in the clockwise direction, the tool moves from the left hand end of its stroke to the right hand end through a distance 2 PD.
Now when the driving crank moves from the position $CA_{1}$ to $CA_{2}$ (or the link DP from $DP_{1}$ to $DP_{2}$) through an angle $\beta$ in the clockwise direction, the tool moves back from right hand end of its stroke to the left hand end. A little consideration will show that the time taken during the left to right movement of the ram (i.e. during forward or cutting stroke) will be equal to time taken by the driving crank to move from $CA_{1}$ to $CA_{2}$. Similarly, the time taken during the right to left movement of the ram (or during the idle or return stroke) will be equal to the time taken by the driving crank to move from $CA_{2}$ to $CA_{1}$.
Since the crank link CA rotates at uniform angular velocity therefore time taken during the cutting stroke (or forward stroke) is more than the time taken during return stroke.in other words ,the mean speed of the ram during cutting stroke is less than the mean speed during the return stroke, The ratio between the time taken during the cutting and return strokes is given by
$\quad\quad\quad\quad\quad\quad\quad$ $\frac{Time\space of\space cutting\space stroke}{Time\space of\space return\space stroke}=\frac{\alpha}{\beta}=\frac{\alpha}{360 ^{\circ}-\alpha}\space or\space\frac{360 ^{\circ}-\beta}{\beta}$