## Stability of Compressed Rods

## Stability of rods working within the elastic range

Consider the ball equilibrium condition (Figure 2.44, a - * c). If the ball is given a small deviation in the horizontal direction, in the first case it will return to its initial position (steady state), in the second - will not return (unstable position), and in the third case will roll back to the side and stop (the position of indifferent equilibrium). Similar phenomena are observed in elastic bodies.*

* Fig. 2.44 *

Let the rod be loaded with the longitudinal force f (Figure 2.45). Apply to the rod a small side force and remove it. As in the cases with the ball (see Figure 2.44), after removing the load, the rod can: 1) return to its original position; 2) do not return to the starting position; 3) remain twisted, i. E. The elastic rod can have three states: a stable, unstable and indifferent state of equilibrium. These three positions, however, here essentially depend on the force * F* * and the geometric dimensions of the rod. Similar phenomena are observed in many thin-walled elastic systems, shells, plates, etc. At a certain value of the force* * F. * called * critical * , the rod ns straightens, but retains a curved position, i.e. an indifferent state of equilibrium. With a slight excess of the critical force, the rod will begin to strongly curl and collapse. The permissible value of the longitudinal force

(2.96)

where is the margin of stability, which is often set approximately equal to the margin of strength relative to the yield point:

* Fig. 2.45 *

The critical stress of the compressed rod may be less than the elastic limit, and more.

For the first time the problem of the stability of a compressed rod was solved by L. Euler in 1744. He derived the formula for the critical force for long rods operating within the elastic range from considering the curvilinear shape of the rod equilibrium at a constant load. The Euler formula has the form

(2.97)

where * E * is the modulus of elasticity of the material, - the minimum moment of inertia of the rod section, - length the rod;

- the coefficient of reduction of the length of the rod, depending on the number of half-waves * n * of the curved rod, which in turn depends on the type of pin fixation (Figure 2.46).

* Fig. 2.46 *

We calculate by the Euler formula the critical stresses:

(2.98)

It is customary to designate where is the radius of inertia of the section.

Then the formula (2.98) takes the form

(2.99)

where - the flexibility of the rod.

If we denote the value of the rod's flexibility at which the critical stress reaches the elastic limit, then from formula (2.99) we obtain

(2.100)

Euler's formula is applied within the elasticity of the material of the rod , i.e. for flexibilities, large critical .

## Stability of rods beyond elasticity

For each material, you can plot the critical stresses using the mechanical characteristics of the material.

The graph of critical stresses (Figure 2.47) has three areas:

I - area of plastic deformations, in which short struts (rods) work on compression. Here the loss of stability does not occur, and ;

II - area of elastic-plastic deformations. Critical stresses in this region are determined by the Yasinskii formula (the equation of a straight line):

(2.101)

where * a * are the coefficients of the straight line equation.

Since the line * A B * connects the points * A * and * B, * then, substituting the coordinates of the points A and * B * into equation (2.101), we obtain the coefficients * a * and * b *;

III - the region of elastic deformations - the Euler region. Here, is determined by the Euler formula expressed in voltages (2.99).

Consider the problem of determining critical stresses.

Problem 2.1. Given a rack in shape and size. Known material. It is required to find a critical load.

The problem can be solved using the curve of critical stresses. In any case, we define , then . Further, we find the flexibility of the stance from the previously calculated minimum radius of inertia . The values and * A * are taken from the tables or calculated.

From the found flexibility, determine in which area the rack works, and, depending on this, determine by co -

* Fig. 2.47 *

the corresponding formula for σκρ or according to the graph. Having found , define . If the stability margin is given, the allowable load is found by formula (2.96).

Problem 2.2. Limit permissible force and stability margin are known , i.e. is known . Given the design of the rack, all material. It is required to determine the size of the section.

This task is undefined, because you can not find , where , because and * A * are unknown.

In this case, they are given by section sizes and solve a direct problem (type one). Determine the value of and compare it with the given. If the resulting value is greater than the specified value, take a smaller section and again determine ; if the resulting value is less than the specified value - increase the cross section. Thus, with the help of several samples, the required cross-sectional dimensions are determined.

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