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Question: Four identical hollow cylindrical columns, support a big structure of mass M. The inner and another ...

Four identical hollow cylindrical columns, support a big structure of mass M. The inner and another radii of a column are R1R_{1}and R2R_{2}respectively. Assuming the load distribution to be uniform, the compressional strain of each column is

(where Y is Young’s modulus of the column)

A

Mgπ(R22R12)Y\frac{Mg}{\pi(R_{2}^{2} - R_{1}^{2})Y}

B

Mg4π(R22R12)Y\frac{Mg}{4\pi(R_{2}^{2} - R_{1}^{2})Y}

C

Mgπ(R12R22)Y\frac{Mg}{\pi(R_{1}^{2} - R_{2}^{2})Y}

D

Mg4π(R12R22)Y\frac{Mg}{4\pi(R_{1}^{2} - R_{2}^{2})Y}

Answer

Mg4π(R22R12)Y\frac{Mg}{4\pi(R_{2}^{2} - R_{1}^{2})Y}

Explanation

Solution

: Area of cross-section of each column,

A=π(R22R12)A = \pi(R_{2}^{2} - R_{1}^{2})

Since each column supports one-quarter of the load

F=Mg4\therefore F = \frac{Mg}{4}

Compressional strain of each column

=FAY=Mg4π(R22R12)Y= \frac{F}{AY} = \frac{Mg}{4\pi(R_{2}^{2} - R_{1}^{2})Y}