Machine Design An elastic bar is fixed at the upper end and loaded at the lower end by a falling weight. The shock load produced can be reduced by Decreasing the modulus of elasticity of the bar Decreasing the cross-sectional area of the bar All of these Increasing the length of bar Decreasing the modulus of elasticity of the bar Decreasing the cross-sectional area of the bar All of these Increasing the length of bar ANSWER DOWNLOAD EXAMIANS APP
Machine Design The plasticity is the property of a material which enables it to Retain deformation produced under load permanently Regain its original shape after deformation when the external forces are removed Draw into wires by the application of a tensile force Resist fracture due to high impact loads Retain deformation produced under load permanently Regain its original shape after deformation when the external forces are removed Draw into wires by the application of a tensile force Resist fracture due to high impact loads ANSWER DOWNLOAD EXAMIANS APP
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 2W/3 W/9 8W/9 W/2 2W/3 W/9 8W/9 W/2 ANSWER DOWNLOAD EXAMIANS APP
Machine Design A tapered key which fits in a key way in the hub and is flat on the shaft, is known as Gib-head key Wood ruff key Flat saddle key Feather key Gib-head key Wood ruff key Flat saddle key Feather key ANSWER DOWNLOAD EXAMIANS APP
Machine Design The efficiency of a square threaded screw is maximum if the helix angle is equal to(Where φ = Angle of friction) 45° + φ/2 45° - φ 90° - φ 45° - φ/2 45° + φ/2 45° - φ 90° - φ 45° - φ/2 ANSWER DOWNLOAD EXAMIANS APP
Machine Design If P1 and P2 are the tight and slack side tensions in the belt, then the initial tension Pi (according to Barth) will be equal to (Where, Pc is centrifugal tension) [(√P1 + √P2)/2]² ⅟₂× (P1 + P2) P1 + P2 [⅟₂ × (P1 + P2)] + Pc [(√P1 + √P2)/2]² ⅟₂× (P1 + P2) P1 + P2 [⅟₂ × (P1 + P2)] + Pc ANSWER DOWNLOAD EXAMIANS APP
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