Machine Design A type of brake commonly used in motor cars is a Internal expanding brake Band and block brake Shoe brake Band brake Internal expanding brake Band and block brake Shoe brake Band brake ANSWER DOWNLOAD EXAMIANS APP
Machine Design If T₁ and T₂ are the tensions on the tight and slack sides of the belt respectively, and Tc is the centrifugal tension, then initial tension in the belt is equal to T₁ + T₂ + Tc (T₁ - T₂ + Tc)/2 T₁ - T₂ + Tc (T₁ + T₂ + Tc)/2 T₁ + T₂ + Tc (T₁ - T₂ + Tc)/2 T₁ - T₂ + Tc (T₁ + T₂ + Tc)/2 ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ratio of the ultimate stress to the design stress is known as Elastic limit Strain Factor of safety Bulk modulus Elastic limit Strain Factor of safety Bulk modulus ANSWER DOWNLOAD EXAMIANS APP
Machine Design In a band and block brake, the ratio of tensions on the tight and slack sides of band is given by (where μ = Coefficient of friction between the blocks and the drum, θ = Semi-angle of each block subtending at the centre of drum, and n = Number of blocks) T₁/T₂ = (μθ)n T₁/T₂ = [(1 - μ tanθ)/ (1 + μ tanθ)]n T₁/T₂ = μθ × n T₁/T₂ = [(1 + μ tanθ)/ (1 - μ tanθ)]n T₁/T₂ = (μθ)n T₁/T₂ = [(1 - μ tanθ)/ (1 + μ tanθ)]n T₁/T₂ = μθ × n T₁/T₂ = [(1 + μ tanθ)/ (1 - μ tanθ)]n ANSWER DOWNLOAD EXAMIANS APP
Machine Design A fast and loose pulley drive is used when Shafts are arranged parallel and rotate in the same directions Driven shaft is to be started or stopped whenever desired without interfering with the driving shaft Shafts are arranged at right angles and rotate in one definite direction Shafts are arranged parallel and rotate in the opposite directions Shafts are arranged parallel and rotate in the same directions Driven shaft is to be started or stopped whenever desired without interfering with the driving shaft Shafts are arranged at right angles and rotate in one definite direction Shafts are arranged parallel and rotate in the opposite directions ANSWER DOWNLOAD EXAMIANS APP
Machine Design When a circular beam of diameter d is subjected to a shear force F, the maximum shear stress induced will be 6F/ πd² 16F/ 3πd² 8F/ πd² 4F/ πd² 6F/ πd² 16F/ 3πd² 8F/ πd² 4F/ πd² ANSWER DOWNLOAD EXAMIANS APP
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