91Ó°ÊÓ

When a force (F) is applied to an object, the length of the object gets changed by an amount 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, 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 called strain. Thus, the stress is directly proportional to the strain.

In this problem, the ratio of the force of gravity per unit cross-sectional area of steel girder is stress, and the ratio of change in length of the steel girder to the original length of the steel girder is strain.

The mass of the sign hanging on the bottom of the vertical steel girder is m = 1700 kg.

The cross-sectional area of the bottom of the steel girder is\(A = 0.012\;{{\rm{m}}^{\rm{2}}}\).

The original length of the steel girder is\({l_0} = 9.50\;{\rm{m}}\).

The force of gravity acting on the steel girder is calculated as follows:

\(\begin{array}{c}F = mg\\ = \left( {1700\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}}^2}} \right)\\ = 16,660\;{\rm{N}}\end{array}\)

The value of Young’s modulus of the steel is \(E = 200 \times {10^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let\(\Delta l\)be the change in length of the steel girder.

The stress within the girder is calculated as follows:

\(\begin{array}{c}Stress = \frac{F}{A}\\ = \frac{{16,660\;{\rm{N}}}}{{0.012\;{{\rm{m}}^2}}}\\ = 1.388 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\\ = 1.4 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\end{array}\)

Thus, the stress within the marble column is \(1.4 \times {10^{ - 6}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

The stress within the steel girder is directly proportional to the strain on the girder, i.e.,

\(\begin{array}{c}Stress = E\left( {Strain} \right)\\Strain = \frac{{Stress}}{E}\\ = \frac{{1.388 \times {{10}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}{{200 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}\\ = 6.94 \times {10^{ - 6}}\\ = 6.9 \times {10^{ - 6}}\end{array}\)

Thus, the strain on the girder is \(6.9 \times {10^{ - 6}}\).

The expression for the strain on the girder is written as follows:

\(\begin{array}{c}Strain = \frac{{\Delta l}}{{{l_0}}}\\\Delta l = Strain \times {l_0}\\ = \left( {6.94 \times {{10}^{ - 6}}} \right) \times \left( {9.50\;{\rm{m}}} \right)\\ = 6.6 \times {10^{ - 5}}\;{\rm{m}}\end{array}\)

Thus, the length of the steel girder is increased by \(6.6 \times {10^{ - 5}}\;{\rm{m}}\).