Answer to Question #202465 in Mechanical Engineering for Brian

Question

The power transmitted between two shafts 3.5 metres apart by a cross belt drive round the two pulleys 600 mm and 300 mm in diameters, is 6 kW. The speed of the larger pulley (driver) is 220 r.p.m. The permissible load on the belt is 25 N/mm width of the belt which is 5 mm thick. The coefficient of friction between the smaller pulley surface and the belt is 0.35. Determine : 1. necessary length of the belt ; 2. width of the belt, and 3. necessary initial tension in the belt.

Answer 1

$L = \fracπ2 (d_1 − d_2) + 2𝑥 + \frac{(d1−d2)^2}{4𝑥}=\fracπ 2(0.6 − 0.3) +(2 × 3.5) + \frac{(0.6−0.3)^2}{4×3.5}= 8.472m$

$v =\frac{ πd_1N_1}{ 60 }= \frac{π × 0.6 × 220}{ 60 }= 6.911m/s$

$α = sin^{−1} (\frac{02.6×−3}{2*5.3}) = 7.387°$

$θ = (180 − 2α) × \frac{π}{ 180} = (180 − (2 × 7.387)) × \frac{π}{ 180 }= 3.399 rad$

$\frac{T1}{ T2} = e^{μθ} = e^{0.3×3.399} = 3.286 … … . a$

$P = (T_1 − T_2) × v → 6000 = (T_1 − T_2) × 6.911$

$→ T_1 − T_2 =\frac{ 6000}{ 6.911 }= 868.181N → T_1 = 868.181 + T_2 … … . b$

sub a in b

$\frac{868.181 + T_2}{ T_2} = 3.286 → T_2 = 379.781N$

$T_1 = 868.181 + 379.781 = 1247.926N$

$b = \frac{ T1}{ T′ }= \frac{ 1247.926}{ 25 }= 49.9mm$

$Tₒ = \frac{ T1 + T2}{ 2} =\frac{1247.926 + 379.78}{ 2 }= 889.08$