Machine Design The dedendum circle diameter is equal to (where φ = Pressure angle) Addendum circle diameter × cosφ Clearance circle diameter × cosφ Pitch circle diameter × sinφ Pitch circle diameter × cosφ Addendum circle diameter × cosφ Clearance circle diameter × cosφ Pitch circle diameter × sinφ Pitch circle diameter × cosφ ANSWER DOWNLOAD EXAMIANS APP
Machine Design According to Indian standard specifications, a Grey cast iron designated by 'FG 200' means that the Maximum shear strength is 200 N/mm² Maximum compressive strength is 200 N/mm² Minimum tensile strength is 200 N/mm² Carbon content is 2% Maximum shear strength is 200 N/mm² Maximum compressive strength is 200 N/mm² Minimum tensile strength is 200 N/mm² Carbon content is 2% ANSWER DOWNLOAD EXAMIANS APP
Machine Design The taper on cotter varies from 1 in 24 to 1 in 20 1 in 15 to 1 in 10 1 in 32 to 1 in 24 1 in 48 to 1 in 24 1 in 24 to 1 in 20 1 in 15 to 1 in 10 1 in 32 to 1 in 24 1 in 48 to 1 in 24 ANSWER DOWNLOAD EXAMIANS APP
Machine Design A hollow saddle key is A taper key which fits in a key way of the hub and the bottom of the key is shaped to fit the curved surface of the shaft A taper key which fits half in the key way of hub and half in the key way of shaft Provided in pairs at right angles and each key is to withstand torsion in one direction only A taper key which fits in a key way of the hub and is flat on the shaft A taper key which fits in a key way of the hub and the bottom of the key is shaped to fit the curved surface of the shaft A taper key which fits half in the key way of hub and half in the key way of shaft Provided in pairs at right angles and each key is to withstand torsion in one direction only A taper key which fits in a key way of the hub and is flat on the shaft ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ratio of circumferential stress to longitudinal stress in a thin cylinder subjected to an internal pressure is 1 4 43832 2 1 4 43832 2 ANSWER DOWNLOAD EXAMIANS APP
Machine Design A shaft is subjected to fluctuating loads for which the normal torque (T) and bending moment (M) are 1000 N-m and 500 N-m respectively. If the combined shock and fatigue factor for bending is 1.5 and combined shock and fatigue factor for torsion is 2, then the equivalent twisting moment for the shaft is 2136 N-m 2050 N-m 2100 N-m 2000 N-m 2136 N-m 2050 N-m 2100 N-m 2000 N-m ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS