1163. What shape should a homogeneous vertical column with a circular cross-section have so that the pressure from the load $P$ it supports and its own weight, per unit area of horizontal cross-section, is the same everywhere? (Column of equal pressure.) The specific weight of the column material is $\delta$, and the radius of its upper base is $r$.

Find then the radii of the upper and lower bases of a bridge pier so that the pressure in any of its horizontal sections
is $3000 \kappa \Gamma / \partial m^{2}$, if the specific weight of the pier material is 2.5, its height is $12 m$, and the load it supports is $90000 \kappa \Gamma$.
Solution. Let the section of the column by a vertical plane passing through its axis of symmetry have the shape shown in Fig. 210.

Choosing a rectangular coordinate system $x O y$, we intersect the column with a horizontal plane passing through an arbitrary point $M(x, y)$ of the desired curve $A A_{1}$ and determine the pressure of the load $P$ and the weight of the upper cut-off part of the column per unit area of the resulting horizontal section $M N$.

The volume of the upper cut-off part of the column, as the volume of a body formed by rotating the curvilinear trapezoid $O A M B$ adjacent to the axis $O x$ around the axis $O x$ (Chapter $V, \S 5$),

$$
v=\pi \int_{0}^{x} y^{2} d x, \text { and its weight } Q=\delta v
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-412.jpg?height=341&width=290&top_left_y=409&top_left_x=901)

Fig. 210

Taking the ratio of $P+Q$ to the area $S=\pi y^{2}$ of the section $M N$, we obtain the pressure per unit area of this section, which, according to the problem statement, should be equal to the pressure per unit area of any other horizontal section.

The pressure per unit area of the upper base of the column is $\frac{P}{\pi r^{2}}, r=O A$, which follows from the problem statement. Therefore,

$$
\begin{gathered}
\frac{P+Q}{S}=\frac{P}{\pi r^{2}} \text { or } P+Q=\frac{P S}{\pi r^{2}} \\
P+\pi \delta \int_{0}^{x} y^{2} d x=\frac{P}{r^{2}} y^{2}
\end{gathered}
$$

Differentiating both sides of this equation, we obtain the differential equation of the curve $A A_{1}$

$$
\pi \delta y^{2} d x=\frac{2 P}{r^{2}} y d y
$$

Solving it as a separable variable equation, we find

$$
d x=\frac{2 P}{\pi \delta r^{2}} \frac{d y}{y} ; \quad x+c=\frac{2 P}{\pi \delta r^{2}} \ln y
$$

From the condition $y=r$ when $x=0$, we find that the constant $c=\frac{2 P \ln r}{\pi r^{2}}$. Therefore, the equation of the curve $A A_{1}$ is

$$
x=\frac{2 P}{\pi \delta r^{2}} \ln \frac{y}{r}
$$

and the desired shape of the column of equal pressure is the surface formed by rotating this curve around the axis $O x$:

$$
x=\frac{2 P}{\pi \delta r^{2}} \ln \frac{\sqrt{y^{2}+z^{2}}}{r}
$$

With this shape of the column, the pressure at all its points will be the same.

For the specified bridge pier, the radius of the upper base is determined from the equality

$$
\frac{90000}{\pi r^{2}}=3000 ; \quad r \approx 3.09 \partial \mu
$$

and the radius of the lower base by substituting the known values into the equality (1)

$$
120 \approx \frac{2 \cdot 90000}{\pi \cdot 2.5 \cdot 3.09^{2}} \ln \frac{r_{1}}{3.09} ; \quad r_{1} \approx 3.24 \mu
$$

Similarly, the shape of long rods or cables, which under the action of their own weight and some load have the same tension in all cross-sections, is determined.