By means of a turnbuckle A a tensile force P is produced in one of the radial bars of the hinged regular octagon shown in fig. Determine the force produced in the other bars of the system

To determine the forces in the other radial bars of a hinged regular octagon when a tensile force \( P \) is applied through a turnbuckle, we need to analyze the geometry and the load distribution within the structure. The octagon has symmetrical properties, which simplifies our calculations significantly.

Understanding the Structure

A regular octagon consists of eight equal sides and angles. When a tensile force is applied to one of the radial bars, it affects the entire structure due to the interconnected nature of the bars. Each radial bar connects the center of the octagon to its vertices, and the symmetry of the octagon means that the forces will distribute evenly among the bars.

Force Distribution in Radial Bars

Let’s denote the radial bars as \( R_1, R_2, R_3, R_4, R_5, R_6, R_7, \) and \( R_8 \). If we apply a tensile force \( P \) to one of the bars, say \( R_1 \), we can analyze how this force is transmitted through the structure.

• The force \( P \) in \( R_1 \) will create tension in the adjacent bars \( R_2 \) and \( R_8 \).
• Due to symmetry, the forces in the other bars \( R_3, R_4, R_5, R_6, R_7 \) will be influenced by the tension in \( R_1 \) as well.

Calculating Forces in Each Bar

To find the force in each of the radial bars, we can use the concept of equilibrium and symmetry. Since the octagon is regular, the angles between the bars are equal, specifically \( 45^\circ \) between adjacent bars. The force in each bar can be determined using trigonometric relationships.

Assuming the tensile force \( P \) is applied to \( R_1 \), the forces in the adjacent bars can be calculated as follows:

• The force in \( R_2 \) and \( R_8 \) will be \( \frac{P}{\sqrt{2}} \) each, due to the angle of \( 45^\circ \) between \( R_1 \) and these bars.
• The forces in the remaining bars \( R_3, R_4, R_5, R_6, R_7 \) will be zero, as they are not directly affected by the tensile force applied to \( R_1 \).

Summary of Forces

In summary, when a tensile force \( P \) is applied to one radial bar of a regular octagon:

• Force in \( R_1 \): \( P \)
• Force in \( R_2 \): \( \frac{P}{\sqrt{2}} \)
• Force in \( R_8 \): \( \frac{P}{\sqrt{2}} \)
• Forces in \( R_3, R_4, R_5, R_6, R_7 \): \( 0 \)

This analysis shows how the tensile force propagates through the structure and how the symmetry of the octagon plays a crucial role in determining the forces in the radial bars. Understanding these principles is essential for analyzing similar structures in engineering and physics.