0
3.4kviews
Determine .

A cone clutch with a semi – cone angle of 150 transmits 10 kw at 600 rpm. The normal pressure intensity between the surfaces in contact is not to exceed $100kN/m^2$. The width of the friction surfaces is half to mean diameter. Assume μ=0.25. Determine

  1. the outer and the inner diameter of the plate.
  2. width of the cone face.
  3. the axial force to engage the clutch.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: Dec 2015

1 Answer
0
180views

$α = 15˚ \\ P=10kw \\ N = 600 rpm \\ Pn = 100 KN/m^2 \\ Width l = (Mean Radius)/2 \\ = \dfrac{\dfrac{(R_1+R_2)}{2}}{2} \\ = \dfrac{R_1+R_2}{4} \\ µ = 0.25 \\ P=\dfrac{2πNT}{60}=10 \times 10^3=\dfrac{2 \times π \times 600 \times T}{60} \\ \therefore,T=159.15 N-m \\ =159.15 \times 10^3 N-mm \\ T = \dfrac12 \times \dfrac{u.w(R1+R2)}{sin⁡∝} -(I) Since,\dfrac{(R_1+R_2)}{2}=Rm \dfrac{u.w.Rm}{\sin⁡∝} \\ P=\dfrac{w}{π[R_1^2-R_2^2]}=\dfrac{w}{π[R_1+R_2][R_1-R_2]} \\ =\dfrac{W}{π \dfrac{[R_1+R_2]}{2}×2×[R_1-R_2]} \\ =\dfrac{W}{πRm×2×[R_1-R_2]} \\ =\dfrac{w}{π[l.\sin α]×Rm×2} \ \ \ \ \ \ \ Since,R_1-R_2=l \sin α \\ w= ρ×π×l \sinα×Km×2 \\ \text{Substituting the value of w in equation (I)} \\ T=\dfrac{1×μ×P×π×l \sinα×Rm×2Rm}{ \sinα} \\ =2×μ×ρ×πl×Rm^2 \\ 159.15 = 2 × 0.25 ×π× 100 × 10^3×Rm^2×0.5×2×R_m \\ \therefore, Rm^3 = 1.0131 × 10^{-3} \\ Rm = 0.1004m \\ Rm = 100.4mm \\ Dm = 2 × Rm \\ Dm = 200.87mm \\ l = 0.5 Dm \\ = 0.5 × 200.87 \\ = 100.43 mm \\ R1 + R2 = 2Rm \\ R1 + R2 = 2 × 100.43 \\ R1 + R2 = 200.87 \\ R1 – R2 = l \sin α \\ = 100.43 × sin 15˚ \\ R1 – R2 = 25.993 mm$

Solving $R_1 \& R_2$ we get

$2R_1 = 226.86 \\ R_1 = 113.43mm \\ And R_2 = 87.4385 mm$

$P_{\max} = c/R_2 \\ \therefore, c = P_{\max} \times R_2 \\ =100 \times 10^3 \times 87.4385/1000 \\ \therefore, c = 8743 N/m \\ \text{Axial load m clutch} \\ w = 2 \times π c (R_1 – R_2) \\ = 2 \times π \times 8743 \times 25.983 \\ W = 1427 N$

Please log in to add an answer.