Question

Qib) The two ends of the simple brake bands are attached on the brake drum such a way that one end is fastened to a fixed pin at fulcrum and the other end connected to the brake arm which is 200 mm from fulcrum. The drum diameter 1.5 m makes an angle contact of 230" while the band passing over it. The brake is to absorb a power of 120 kW at 900 rpm, the coefficient of friction between the blocks and the drum is 0.45. Draw simple band brake diagram to represent the given data. Consider the drum rotates in the clockwise direction. Calculate the operating force required at the end of the lever 3 m long from fulcrum of a simple and brake. Also Calculate width and thickness of steel band if thickness is 0.25 times of the width required for this brake band. The maximum tensile stress is not to exceed 70 MPa? (05 marks)

Answer 1

Q 1b) Aff F = Operating founce + TI AT2 10 8 .0.2m, Fulcrum 3m Given Data : Drum Diametee D=1.5m Coefficient of friction u= 0.45 N = 900 rpm powel, P = 120kW = 120x103 W max z masimum Tensile stress = 7OMPa t = 0.25b Ti = Tension on Tight Side, T2 = Tension om slack side Ponies, p= = 2nNT 69 T = P x 60 = 120X10 3 60 =1273.2 NAM ATTN 2 ñ x 900 Braking Torque BT = (T. - Tz) * radius of drum T= (T.-T2) * 0.75 1273.2=1T.-T2) *0.75 |(T. -T2) = 1697.6

0.45 230 K T = e no se 180 = 6.086 T2 Ti = 6.086 Te Substitute Ti Volue in equation 2 6.086 T2 - T2 = 1697.6 Tz = 33.7N TI - 2031.4 N moment about Fulcrum 66 Hinge Taking (EM) Hinge F*3= Te *0.am F = 22.4 N Ttight Imaxe (bxt) 2031.4 = 70 *b* 0.25b b = 116.08 mm = 0.116om t = 29.02mm = 0.029 m