(I) A marble column of cross-sectional area \({\bf{1}}{\bf{.4}}\;{{\bf{m}}^{\bf{2}}}\) supports a mass of 25,000 kg. (a) What is the stress within the column? (b) What is the strain?

When a force (F) is applied to an object made of a certain material, the length of the object gets changed by an amount that is proportional to the original length (\({l_0}\)) and inversely proportional to the cross-sectional area (A), i.e., \(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\).

Here, E is the constant of proportionality and is termed the elastic modulus or Young’s modulus; the ratio of force per unit area is termed stress; the ratio of change in length to the original length is termed as strain.

Thus, the stress is directly proportional to the strain.

In this problem, the stress within the marble column is directly proportional to the strain in it. The value of the constant of proportionality or the value of Young’s modulus of the marble is\(E{\bf{ = 50 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

The cross-sectional area of the marble column is\(A = 1.4\;{{\rm{m}}^2}\).

Mass supported by the marble column is\(m = 25,000\;{\rm{kg}}\).

Force of gravity on mass m is calculated as follows:

\(\begin{array}{c}F = mg\\ = \left( {25,000\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}}^2}} \right)\\ = 245,000\;{\rm{kg}} \cdot {{\rm{m}}^2}\end{array}\)

The stress within the marble column is calculated as follows:

\(\begin{array}{c}Stress = \frac{F}{A}\\ = \frac{{245,000\;{\rm{kg}} \cdot {{\rm{m}}^2}}}{{1.4\;{{\rm{m}}^2}}}\\ = 175,000\;{\rm{kg}}\\ = 1.8 \times {10^5}{\rm{ kg}}\end{array}\)

Thus, the stress within the marble column is \(1.8 \times {10^5}{\rm{ kg/}}{{\rm{m}}^{\rm{2}}}\).

The strain in the marble column is calculated as follows:

\(\begin{array}{c}Stress = E\left( {Strain} \right)\\Strain = \frac{{Stress}}{E}\\ = \frac{{175,000{\rm{ kg}}}}{{50 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}\\ = 3.5 \times {10^{ - 6}}\end{array}\)

Thus, the strain in the marble column is \(3.5 \times {10^{ - 6}}\).