Answer to Question #236027 in Mechanical Engineering for Deven

Question #236027
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.
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Expert's answer
2021-09-12T07:24:41-0400

Maximum load = 73×10×π(75×1034)²=8.06×10N73× 10⁹ ×π(\dfrac{75×10^{-3}}4)²= 8.06×10⁷N


Considering allowable stress = 200×10⁹ Pa200×10⁹\ Pa


σ=PA, A=Pσ\sigma = \dfrac PA ,\ A =\dfrac P{\sigma}


A=8.06×107200×109=4.03×104 m²A =\dfrac{8.06×10^7}{200×10^9} = 4.03×10^{-4}\ m²


Considering allowable extension = 4×104 m4× 10^{-4}\ m


δ=PLAE, A=PLδE\delta =\dfrac{PL}{AE}, \ A = \dfrac{PL}{\delta E}


A=8.06×10×14×104×73×109=2.76 m²A= \dfrac{8.06× 10⁷ ×1}{4×10^{-4}×73×10^9}= 2.76\ m²


Larger area governs,

A=π(d²4)A =π(\dfrac{d²}4)


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

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