Question

6. The following band brake system as shown in figure 6, has the braking capacity of 180 Nm when the drum is rotating in clockwise direction. The coefficient of friction between the drum and the band is 0.2. Determine (a) The value and the position of the maximum pressure in the band (b) The value of acting force F (c) Calculate the braking capacity of this system if the drum is rotating in counter clockwise direction and the maximum band pressure is limited to the value found in part (a). All dimensions are in mm. width of the band is 60 mm 80 + F 50 100 -100

Answer 1

Solo 6- Given Data :- 80 mm Breaking Cefacity Tb= 180 Nm M = 0.2 , b=60mm Radies of drum het Ty be the tension in tight side and To be the tension in slack gide. Then, Ta ya 0.2% 31 2 Tai Ta = 2.5663 T (1) Th= (Ta-Tz) x x 180 1.5663 Tz x 0.08 T2= 1436. SAN

T2= 2.5 663 X 1436.51 2 Ta= 3686 52 N Ta Maximum pressure in band, P= Width * Radies of Drum p = Ta br p= 3.686.52 768025 Pa 0.06 X 0.08 P= 768.025 kPa 162 the FBD Drowning the beam , we get - Tz f $ e с. 100 100 Shifting Ty and Tz at point B, we get is that var → 0.02 Ty EMAZO (f X0.2) + (0.02 TL) - (TAX 0.1) - Co.os 1]=0

0.28= 0.1 T2+ O.OST2 3 0.02 TL 0.24 = 0.08 Tat 0.os Tz 0.27 - (0.08 x 3686 · 52 ) + (0.05 x 2436. 52) - F= 1833.74 N counter-clockwise rotating in as the pressure is limited to same as in Case cal If the drum is direction, the bands of tight side and sleck side will interchange. But, the value of Te and Iz will remain unchanged. the value of tension is unchanged. As fight side tension and stack side Braking capacity will so also be same : Braking capacity - 280 Nm