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From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi
The cross-sectional area of the aluminum tube is given by
Set this expression equal to the minimum area and solve for the maximum inside diameter d
The outside diameter D, the inside diameter d, and the wall thickness t are related by
Therefore, the minimum wall thickness required for the aluminum tube is
The cross-sectional area of the stainless steel tube is
The normal stress in the tube can be expressed as
The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.
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