A cone clutch is used to transmit 20kW at 1200 rpm. The coefficient of friction of 0.1, while the permissible intensity of pressure is $0.1 \mathrm{N} / \mathrm{mm}^{2}$.The semi cone angle is 20 degree, the larger diameter of the friction surface is 500mm. Overload factor is 1.25. Design the clutch (shaft diameter, spline design, friction plate dimensions, spring design).

Given Data

Overload factor $K =1.25$

Power $(\mathrm{P})=20 \mathrm{k} \mathrm{W}$

Speed $(\mathrm{n})=1200 \mathrm{rpm}$

Coefficient of friction $(\mu)=0.1$

Pressure $P_{max}=0.1 \mathrm{N} / \mathrm{mm}^{2}$

Semi cone angle $\alpha=20^{\circ}$

Diameter $\left(D_{\max }\right)=500 \mathrm{mm} \ and \ \left(r_{\max }=250 \mathrm{mm}\right)$

Solution:

Part A: Design of shaft and splines

1) Transmitted Torque

$\quad M_{t}=\frac{9.73 \times 10^{6} \mathrm{kW}}{\mathrm{n}}=\frac{9.73 \times 10^{6} \times 20}{1200}=162166.66 \mathrm{Nm}$

2) Design torque $\left[M_{t}\right]=K . M_{t} \quad$ [PSG 7.96]

$\left[M_{t}\right]=1.25 \times 162166.66=243250 \mathrm{Nmm}$

3) Shear strength of shaft $M_{T}=\frac{\pi}{16} \tau d^{3}$ ,

4) Assuming $\mathrm{C} 45$ material for shaft, the diameter of shaft is [PSG 7.96]

$ d=120\left[\frac{(k W) K}{n}\right]^{\frac{1}{3}}=120\left[\frac{(20) 1.5}{1200}\right]^{\frac{1}{3}}=35.08 \cong 35 \mathrm{mm} $

5) Hub Diameter, $\mathrm{D}=1.5 \times \mathrm{d}=1.5 \mathrm{x} 35=52.5 \mathrm{mm}$

6) Select standard Splined Hub (No of Splines $\times d \times D ) \quad$ [PSG5.30]

$(8 \times 36 \times 40)$

Part B: Clutch dimensions

7) Torque $=\left[M_{T}\right]=A \times \mu \times p \times r_{m}$ where $A=2 \pi r_{m} b$

$[243250]=2 \pi r_{m} b \times \mu \times p \times r_{m}=2 \pi r_{m}^{2} b \times 0.1 \times 0.1$---------Eq(1)

8) Substituting $\mathrm{b}=0.4 r_{m}$ in equation $1,$ we get $r_{m}=213.11$ and $b=85.22 \mathrm{mm}$

9) Width $b=\left(r_{\max }-r_{\min }\right) / \sin \alpha \therefore 85.22=\left(r_{\max }-r_{\min }\right) / \sin 20 $----------Eq(2)

10) Mean Radius $-r_{m}=\frac{r_{\max }+r_{\text {min}}}{2} \therefore 213.11=\frac{r_{\text {max}}+r_{\text {min}}}{2}$----------Eq(3)

Solving Eq 2 and 3 Clutch dimensions $r_{\max }=227.68$ and $r_{\min }= 198.53 \mathrm{mm}, b=85.22 \mathrm{mm}$ (Size restriction is satisfied)

Part C: Spring Design

11) Force on spring during engagement $=F_{e}=\frac{M_{t}}{\mu r_{m}}(\sin \alpha+\mu \cos \alpha)$

$ F_{e}=\frac{243250}{0.1 \times 213.11}(\sin 20+0.1 \cos 20)=4976.51 \mathrm{N} $

12) During disengagement $=F_{d}=\frac{M_{t}}{\mu r_{m}}(\sin \alpha-\mu \cos \alpha)$

$ F_{d}=\frac{243250}{0.1 \times 213.11}(\sin 20-0.1 \cos 20)=2831.32 \mathrm{N} $

13) Assume static design and Spring Index as $5,$ Spring material Grade 4 wire $\sigma_{u}=1880 M P a$ and $\tau_{\max }=0.7 \sigma_{u}=1316 \mathrm{MPa}$ from page. [PSG 7.105]

find diameter of spring wire and coil diameter. [PSG7.100]

$\tau=k_{S}\left[\frac{8 F_{e} C}{\pi d^{2}}\right] \therefore 1316=1.3\left[\frac{8 \times 4976.51 \times 5}{\pi d^{2}}\right]=4.05 \mathrm{mm}$

Diameter of wire $d =4 \mathrm{mm}$ and Coil diameter $=\mathrm{D}=\mathrm{d} \times\mathrm{C}=4 \times 5=20 \mathrm{mm}$

14) Assume deflection of spring $y=25 \mathrm{mm}$

$y=\left[\frac{8 P C^{3} n}{G d}\right] \therefore 25=\left[\frac{8 \times 4976.51 \times 5^{3} n}{80000 \times 4}\right]$

Number of coils $=1.82 \cong 2$

For square and ground ends, $\mathrm{n}=2+2=4$

15) Find the number of coils. [PSG 7.100]