Answer to Question #248575 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

Question 2

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;

Given;

No. of teeth of pinion,T_p=30

No. of teeth on gear,T_g=50

Module,m=3mm

Pressure angle,$\phi$ =14.5°

Velocity of wheel chair ,v_w=5m/s

Rotation of pinion,N_p=800r.p.m

Pitch radius of pinion,r=$\frac{mT_p}2$ =45mm

Pitch radius of gear,R=$\frac{mT_g}{2}=75mm$

(1.1) Magnitude of velocity of slip;

$v_s=2×v_w$

$v_s=2×5m/s$ =10$m/s$

(1.2) Angular speed of gear wheel;

From velocity ratio;

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

$N_g=\frac{N_p×T_p}{T_g}$ =$\frac{800×30}{50}$ =480rpm

Convert to angular velocity;

$w_g=\frac{2πN_g}{60}$ =$\frac{2×π×480}{60}=50.27rad/s$

(1.3) maximum length of approach;

$P_a(max)=rsin\phi$

$P_a(max)=45sin(14.5°)$

$P_a(max)=11.27mm$

(1.4)path of recess in mm;

Addendum radius of pinion;

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

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

$r_a=52.92mm$

Hence,path of recess is ;

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

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

$p_r=18.7741mm$

(1.5) Addendum of the gears;

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

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

$A_g=3.585mm$

(1.6) contact ration,c.r;

$c.r=\frac{p_c}{C_p×cos\phi}$

p_c is the path of contact

C_p is the circular pitch

Path of contact=path of recess +path of approach

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

$p_a=\sqrt{78.585^2-75^2cos^2(14.5)}-75sin(14.5)=11.2754mm$

$c.r=\frac{11.2754+18.7741}{π×3×cos(14.5°)}$ =3.293