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Determine the percentage savings in the material:
A hollow shaft containing an inside diameter 0.5 times the outer diameter is to replace the solid shaft of the similar material transmitting the similar power at the similar speed. Determine the percentage savings in the material.
Solution
di = 0.5 do
PH = PS
NH = NS
∴ TH = TS
(τmax ) H = (τmax )S
(Z P ) H = (Z P )S
R = do /2
(Z P )Solid = (π /16 )d3
D = 0.62 do
% saving in material
= % reduction in area
= (AH - AS / AH )× 100
= (0.75 - 0.38 /0.75 )× 100
= 49.3%
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