Machine Design Two helical springs of the same material and of equal circular cross-section, length and number of turns, but having radii 80 mm and 40 mm, kept concentrically (smaller radius spring within the larger radius spring), are compressed between two parallel planes with a load W. The inner spring will carry a load equal to W/9 2W/3 8W/9 W/2 W/9 2W/3 8W/9 W/2 ANSWER DOWNLOAD EXAMIANS APP
Machine Design When the belt is transmitting maximum power, the belt speed should be(Where, m = mass of belt per meter (kg/m) and Pmax = maximum permissible tension in belt (N)) √(Pmax / m) √(Pmax / 2m) √(3m /Pmax) √(Pmax / 3m) √(Pmax / m) √(Pmax / 2m) √(3m /Pmax) √(Pmax / 3m) ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ratio of belt tensions (p1/p2) considering centrifugal force in flat belt is given byWhere m = mass of belt per meter (kg/m)v = belt velocity (m/s)f = coefficient of frictiona = angle of wrap (radians) (P1 - mv²)/ (P2 - mv²) = e–ᶠα P1 / P2 = e–ᶠα P1 / P2 = eᶠα (P1 - mv²)/ (P2 - mv²) = eᶠα (P1 - mv²)/ (P2 - mv²) = e–ᶠα P1 / P2 = e–ᶠα P1 / P2 = eᶠα (P1 - mv²)/ (P2 - mv²) = eᶠα ANSWER DOWNLOAD EXAMIANS APP
Machine Design The condition for maximum power transmission is that the maximum tension in the flat belt should be equal to(Where, Pc = tension in belt due to centrifugal force) Pc/3 3 Pc 1 Pc 2 Pc Pc/3 3 Pc 1 Pc 2 Pc ANSWER DOWNLOAD EXAMIANS APP
Machine Design A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is τ. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be τ/2 4 τ 2 τ τ τ/2 4 τ 2 τ τ ANSWER DOWNLOAD EXAMIANS APP
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