Answer to Question #277510 in Mechanical Engineering for Suraj

Question

In a crank and slotted lever quick return mechanism, as shown in Fig. 5.37, the driving crank length is 75 mm. The distance between the fixed centres is 200 mm and the length of the slotted lever is 500 mm. Find the ratio of the times taken on the cutting and idle strokes. Determine the effective stroke also. [Ans. 1.67 ; 380 mm] 3.

Answer 1

Solution;

Draw the extreme positions of the crank as shown below;

Let the angle covered during the cutting stroke be $\alpha$ and that of idle stroke be $\beta$

The trigonometric expression of angle $\beta$ ;

$cos \frac{\beta}{2}=\frac{BC}{AC}$

Which $BC=75mm$ and $AC=200mm$

Therefore;

$cos \frac{\beta}{2}=\frac{75}{200}$

$\beta=2×cos^{-1}\frac{75}{200}=135.95°$

From the figure,angle $\alpha$ is;

$\alpha+\beta=360°$

$\alpha=360-135.95=224.05°$

The ratio (T) of the time take on the two strokes is;

$T=\frac{\alpha}{\beta}=\frac{224.05}{135.95}=1.648$

The stroke Length (R) is;

$R=\frac{2×s×c}{C}$

Where ;

s is the length of slotted lever =500mm

c is the crank length =75mm

C is the length of connecting rod=200mm

$R=\frac{2×500×75}{200}=375mm$