Machine Design If a bearing is designated by the number 305, it means that the bearing is of Medium series whose bore is 5 mm Light series whose bore is 5 mm Light series whose bore is 25 mm Medium series whose bore is 25 mm Medium series whose bore is 5 mm Light series whose bore is 5 mm Light series whose bore is 25 mm Medium series whose bore is 25 mm ANSWER DOWNLOAD EXAMIANS APP
Machine Design A locking device extensively used in marine type connecting rod ends is a Jam nut Ring nut Castle nut Sawn nut Jam nut Ring nut Castle nut Sawn nut ANSWER DOWNLOAD EXAMIANS APP
Machine Design In hydrostatic bearings Grease is used for lubrication The oil film is maintained by supplying oil under pressure The Oil film pressure is generated only by the rotation of the journal Do not require external supply of lubricant Grease is used for lubrication The oil film is maintained by supplying oil under pressure The Oil film pressure is generated only by the rotation of the journal Do not require external supply of lubricant ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ends of the leaves of a semi-elliptical leaf spring are made triangular in order to Permit each leaf to act as an overhanging beam Make M/I constant throughout the length of the leaf Obtain variable moment of inertia (I) in each leaf Have variable bending moment (M) in each leaf Permit each leaf to act as an overhanging beam Make M/I constant throughout the length of the leaf Obtain variable moment of inertia (I) in each leaf Have variable bending moment (M) in each leaf ANSWER DOWNLOAD EXAMIANS APP
Machine Design The ratio of belt tensions (p1/p2) considering centrifugal force in flat belt is given by Where m = mass of belt per meter (kg/m) v = belt velocity (m/s) f = coefficient of friction a = angle of wrap (radians) (P1 - mv²)/ (P2 - mv²) = eᶠα P1 / P2 = e–ᶠα (P1 - mv²)/ (P2 - mv²) = e–ᶠα P1 / P2 = eᶠα (P1 - mv²)/ (P2 - mv²) = eᶠα P1 / P2 = e–ᶠα (P1 - mv²)/ (P2 - mv²) = e–ᶠα P1 / P2 = eᶠα 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 τ τ/2 4 τ τ 2 τ τ/2 4 τ ANSWER DOWNLOAD EXAMIANS APP
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