Question

The tube A made from an aluminum alloy (E = 73 GPa has an outer diameter of 75 mm. It is used to support the rod B which is made from steel (E = 200 GPa) and has a 25 mm diameter. Calculate the minimum thickness required for the tube if the maximum deflection of the loaded end of the rod must be limited to .4 mm.

Answer 1

Maximum load = $73× 10⁹ ×π(\dfrac{75×10^{-3}}4)²= 8.06×10⁷N$

Considering allowable stress = $200×10⁹\ Pa$

$\sigma = \dfrac PA ,\ A =\dfrac P{\sigma}$

$A =\dfrac{8.06×10^7}{200×10^9} = 4.03×10^{-4}\ m²$

Considering allowable extension = $4× 10^{-4}\ m$

$\delta =\dfrac{PL}{AE}, \ A = \dfrac{PL}{\delta E}$

$A= \dfrac{8.06× 10⁷ ×1}{4×10^{-4}×73×10^9}= 2.76\ m²$

Larger area governs,

$A =π(\dfrac{d²}4)$

$d =\sqrt{\dfrac{4A}π} =\sqrt{\dfrac{4× 2.76}{π}} =1.87\ m$