Mechanical Properties of Solids
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Question : 7
Total: 21
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. The Young’s modulus of steel is 2.0 × 10 11
Solution:
Here, Y = 2.0 × 10 11 P a
Inner radius of each column,r 1 = 30 cm = 0.3 m ;
Outer radius of each column,r 2 = 60 cm = 0.6 m
Therefore, area of cross-section of the each column,
A = π ( r 2 2 − r 1 2 )
= π ( 0.6 2 − 0.3 2 ) = 0.27 π m 2
The mass supported on the four columns,
M = 50,000 kg
The whole weight of the structure will be shared by the four columns.
Therefore, compressional force on one column,
F =
=
N , Y =
Therefore, compression strain,
=
=
= 7.22 × 10 − 7
Inner radius of each column,
Outer radius of each column,
Therefore, area of cross-section of the each column,
The mass supported on the four columns,
M = 50,000 kg
The whole weight of the structure will be shared by the four columns.
Therefore, compressional force on one column,
Therefore, compression strain,
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