Solution

Statement

$3.6.8.$ The column of St. Isaac's Cathedral in St. Petersburg has a height of $30$ m. By how much is it compressed under its own weight? The density of granite is $2.7 \cdot 10^3$ kg/m$^3$, and its Young's modulus is $10^{11}$ Pa.

Solution

The main feature of this problem is that the compression force of the column is non-uniform: the upper part is not loaded with anything, while the very base bears the entire huge weight of the column. Let us consider two methods for finding the total deformation.

Method 1. Using average force

Since the column is uniform and has a constant cross-section, the pressure force (weight) increases linearly as we move from the top to the base. At the very top, the force is zero ($F_{\text{top}} = 0$), and at the base, it is maximal and equal to the total weight of the column ($F_{\text{bottom}} = Mg = \rho S h g$).
Due to this strictly linear dependence, we can use the average force acting along the entire length of the column:
$$F_{\text{avg}} = \frac{F_{\text{top}} + F_{\text{bottom}}}{2} = \frac{0 + \rho S h g}{2} = \frac{\rho S h g}{2}$$
Let us write Hooke's law for the entire column as a whole, substituting the average force into it:
$$\Delta h = \frac{F_{\text{avg}} h}{E S} = \frac{\frac{\rho S h g}{2} \cdot h}{E S} = \frac{\rho g h^2}{2E}$$

Method 2. Using integration

Let us divide the column into infinitely small horizontal layers of thickness $dy$. Let the coordinate $y$ be measured from the top of the column downwards (from $0$ to $h$).
An isolated micro-layer located at depth $y$ is pressed by the weight of the granite located strictly above it. The mass of this part is $m(y) = \rho S y$, and the pressure force is:
$$F(y) = m(y)g = \rho S g y$$
According to Hooke's law, the absolute compression $d(\Delta y)$ of this infinitely small layer is equal to:
$$d(\Delta y) = \frac{F(y) dy}{E S} = \frac{\rho S g y dy}{E S} = \frac{\rho g y}{E} dy$$
To find the total compression of the entire column, let us integrate (sum) the deformations of all such layers from the top ($y = 0$) to the base ($y = h$):
$$\Delta h = \int_{0}^{h} \frac{\rho g y}{E} dy = \frac{\rho g}{E} \int_{0}^{h} y dy = \frac{\rho g}{E} \left[ \frac{y^2}{2} \right]_{0}^{h} = \frac{\rho g h^2}{2E}$$
As we can see, both methods naturally lead to exactly the same result.

Calculation:
Let us substitute the given values, assuming the acceleration of gravity is $g \approx 9.8$ m/s$^2$:
$$\Delta h = \frac{2700 \cdot 9.8 \cdot 30^2}{2 \cdot 10^{11}} = \frac{2700 \cdot 9.8 \cdot 900}{2 \cdot 10^{11}} \approx 1.19 \cdot 10^{-4} \text{ m}$$
*(When using the classical value for this problem book, $g = 10$ m/s$^2$, the calculated value will be $\approx 1.215 \cdot 10^{-4}$ m, which, when rounded to two significant figures, gives the final answer).*