Question

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 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer 1

Given, the total mass carrying all four cylinders is 50000 kg; the inner and outer radius of each column is 30 cm and 60 cm respectively. The Young’s modulus of the mild steel is 2× 10 11  Pa.

The total load on each column is,

F= mg 4

Substituting the values in the above expression, we get:

F= 50000 kg×9.8 m/ s 2 4 =122500 N

The Young’s modulus of each column is given by the equation,

Y= F Aε ε= F AY

The compressive strain of all four columns is given by the equation,

ε T = 4F AY = 4F π( r 2 2 − r 1 2 )Y

Substituting the values in the above expression, we get:

ε T = 4( 122500 N ) π[ ( 0.6 m ) 2 − ( 0.3 m ) 2 ]( 2× 10 11  Pa ) =2.8× 10 −6

Thus, the compressional strain of each column is 2.8× 10 −6 .