Answer to Question #248581 in Mechanical Engineering for Kay

Question

Question 1 [20]

A simple gear train of 30 and 50 teeth for pinion and gear wheel respectively, is used to drive an electric wheelchair at a velocity that is double the maximum velocity of slip. The pinion has an equal addendum as the gear wheel, a module of 3 mm, and a pressure angle of 14.5o. The wheelchair has a maximum velocity of 5 m/s when its pinion rotates at 800 r.p.m. Calculate:

1.1. The

1.2. The

1.3. The

1.4. The

1.5. The

1.6. The

magnitude of the velocity of slip of gears in mesh. (2) angular speed of the gear wheel in rad/s. (4) maximum length of a path of the approach in mm. (2) length of a path of a recess in mm. (7) addendum of the gears in mm. (7) contact ratio. (8)

Answer 1

Solution;

The given data;

No. of teeth on pinion(t)=30

No. of teeth on gear wheel (T)=50

Module(m)=3mm

Pressure angle($\phi$ )=14.5°c

Velocity of wheel chair($V_w$)=5m/s

Rotation of pinion($N_p$)=800r.p.m

(1.1) Magnitude of velocity of slip;

$V_s=2×V_w$

$V_s=2×5m/s$

$V_s=10m/s$

(1.2)Angular speed of the gear wheel ;

$\frac{N_p}{N_g}=\frac{T_g}{T_p}$

$N_g=\frac{N_p×T_p}{T_g}$

$N_g=\frac{800×30}{50}$

$N_g=480 r.p.m$

Convert into angular speed;

$w_g=\frac{2πN_g}{60}$ =$\frac{2×π×480}{60}$

$w_g=50.26rad/s$

(1.3) Maximum path of approach;

$p_a=rsin(\phi)$

r is the radius of pinion pitch circle.

$r=\frac{mT_p}{2}$ =$\frac{3×30}{2}=45mm$

$p_a=45sin(14.5°)$

$p_a=11.27mm$

(1.4)length of path of recess;

Pitch radius of gear wheel,R;

$R=\frac{mT_g}{2}=\frac{3×50}{2}$ =75mm

Addendum radius of pinion is given by;

$r_a=r\sqrt{1+\frac Rr(\frac Rr+2)sin^2\phi}$

$r_a=45\sqrt{1+\frac{75}{45}(\frac{75}{45}+2)sin^2(14.5)}$

$r_a=52.92mm$

Path of recess,$p_r$ ;

$p_r=\sqrt{r_a^2-r^2cos^2\phi}-rsin\phi$

$p_r=\sqrt{52.92^2-45^2cos^2(14.5)}-45sin(14.5)$

$p_a=18.77mm$

(1.5) Addendum of the gears ;

$A_g=R[\sqrt{1+\frac{r}{R}(\frac{r}{R}+2)sin^2\phi}-1$

$A_g=75[\sqrt{1+\frac{45}{75}(\frac{45}{75}+2)sin^214.5}-1$

$A_g=3.585mm$

(1.6) contact ratio;

Path of contact;

$p_c=p_a+p_r$

$p_a=\sqrt{R_a-R^2cos^2\phi}-Rsin\phi$

$p_a=\sqrt{78.585^2-75^2cos^214.5}-75sin(14.5)$

$p_a=11.2754mm$

Hence;

$p_c=11.2754+118.77=30.05mm$

The pitch circumference;

$C_p=3π$

Hence the contact ratio;

$\frac{p_c}{C_p×cos\phi}$ =$\frac{30.05mm}{3×π×cos(14.5)}$

$=3.293$