Simplified Calculation of the Inertia Moment of the Cross Section of the Console under Loading

Introduction. Numerous building structures contain elements in the form of rods, which undergo elastic deformations during manufacture or operation [1]. The bending strength of a rod or beam determines the bearing capacity of the structure [2]. The ability of a beam to elastic deformation is characterized by stiffness, defined as the ratio of load P to the elastic deflection of the beam λ_e : m_x = P / λ_e [3]. As a rule, under laboratory conditions, stiffness is checked on a console beam. One end of it is embedded in a rigid base, and an external force, directed perpendicular to the axis of the beam, acts on the other one [4]. Modeling and calculations of deformation and fracture characteristics of rods, as a rule, are associated with solving the problem of deflection of a console beam, or console, under the action of an external applied force [5]. At the same time, there are no publications on simple and valid methods for determining the inertia moments of complex sections relative to the action of an external applied force. The solution to this problem is described in the presented article.

This work is aimed at the creation of a valid, rapid method for calculating the inertia moments of complex console sections under the action of an external applied force. The new approach provides analytical determination of the shear stress distribution in the section corresponding to the action of an external applied force. It should be noted that earlier, tangential stresses were not taken into account in such calculations. In addition, for the first time, examples of calculating the moment of inertia for complex figured sections are given.

Materials and Methods. In a number of works on the resistance of materials, e.g., in [6], a universal formula is given for calculating the elastic deflection of console λ_e. According to this formula, the rigidity of the console is:

(1)

where E — Young's modulus; L — length of the console; I_x — inertia moment of the cross-section of the beam relative to x-axis, passing through the center of gravity of the section perpendicular to the applied force P.

It follows from equation (1) that a fundamentally important characteristic of the console is the inertia moment of section I_x, whose value depends on the geometry of the cross-section of the beam and the direction of x-axis [7]. It should be emphasized that in equation (1), inertia moment I_x refers to x-axis, which is perpendicular to the direction of the external applied force P. Specifically, the inertia moment of rectangular section a×b relative to the axis of symmetry x is equal to [8]:

(2)

Here, a — thickness of the console, b — its width. Force P is directed parallel to side b of the rectangle.

Substituting (2) into (1), we obtain the well-known equation for the deflection of a rectangular console [9]:

The cross sections of console beams, or consoles, are different. Figure 1 shows a simple example of a square section console under the action of external force P, directed along the diagonal of a square.

The inertia moment of the section relative to x-axis is called the sum, or integral, of the products of elemental areas ds = dxdy by the squares of distances y of the areas to x-axis: [10]. The integrand function is virtually the inertia moment of the elemental area dxdy relative to x-axis.

The console can be represented as a pack of extremely thin plates with thickness dx and length L, oriented parallel to the force vector P. Under the action of P, all plates bend by the same amount λ_e. With a given orientation of the console section, a separate plate is not affected by deformation of the rest of the volume. Then, the inertia moment of the section of the console as a whole will be determined by the integral sum of the inertia moments of the sections of all the plates in the pack.

The half-length of the strip in Figure 1 is equal to h = x. From equation (3), we find:

It can be seen that turning x-axis by 45° does not change the inertia moment of the square section. Consequently, when the external applied force is rotated by 45°, the rigidity of the console does not change either (1).

Substituting x in place of h (4), we obtain a shear stress distribution in the section of the square console corresponding to deflection λ_e:

Triangle. Consider the section in the form of an isosceles triangle (Fig. 4). x-axis is directed along the altitude of the triangle (Fig. 4 a). At distance x from the base of the triangle, the half-length of the strip is equal to h = b (a – x)/2a.

Integrating expression (4) over x from 0 to a determines the following value of the inertia moment:

(7)

Consider the case when x-axis passes through the center of gravity and is parallel to the base of the triangle (Fig. 4 b).

The strip at distance x from the left corner of the triangle has half-length (Fig. 4 b). The center of gravity of the strip is located at distance h from the base of the triangle (Fig. 4 b). Therefore, according to the Huygens–Steiner theorem for the cross section, the inertia moment of the strip relative to the base of the triangle is:

Integrating the resulting expression over variable x from –b/2 to b/2 determines the inertia moment of the triangle relative to the base:

(8)

The center of gravity of the triangle is located at distance a/3 from the base. According to the Huygens–Steiner theorem [13], the inertia moment of a triangle relative to its own center of gravity is less than (8) by:

where S = ab/2 — area of triangle.

In this manner, the inertia moment of the triangle relative to its own center of gravity is equal to:

(9)

The result obtained corresponds exactly to the tabular value of the inertia moment of the section relative to the axis through the center of gravity parallel to the base of the triangle.

Figure 5 shows the shear stress distributions in the triangular section at λ_e = 2 mm for cases where the deflection force is directed along the base (a) and along the altitude of the triangle (b).

When the triangle is oriented as in Figure 4 a, voltage τ(x) decreases gradually from τ_max at the base to 0 at the apex (Fig. 5 a). With the orientation of the triangle as in Figure 4 b, distribution τ(x) (Fig. 5 b) is similar to the distribution for a square section when the force is oriented along the diagonal (Fig. 2 b).

Regular hexagon. Figure 6 shows two orientation of the hexagon relative to x-axis: parallel (a) and perpendicular (b) to its diagonal.

The section of the hexagon in Figure 6 a consists of the following:

• rectangle with width a and height a√3,
• two isosceles triangles with apex a/2 and base a√3.

Examples of Simplified Calculation of the Inertia Moment of Complex Sections

Six-pointed star. Using the proposed method, we calculate the inertia moment of a non-standard section in the form of a regular 6-pointed star with side a. To determine the inertia moment relative to the minor diagonal of the star, we use the diagram in Figure 8 a. Let us highlight three zones in the diagram: zone I with width a/2, zone II for the rest of the half-figure, and the adjacent auxiliary zone III in the form of a triangle.

Half-length of the strip in zone I at distance x from the diagonal is h_I = √3(a – x). Using expression (3), for two zones I in Figure 6 a, we obtain the inertia moment:

The inertia moment of zone II is equal to the inertia moment of a rectangle with height a√3 and width a without the inertia moment of triangle III. The half-length of the strip in the rectangle is equal to h_II = a √3/2. From equation (3), we find the inertia moment of the rectangle:

The half-length of the strip in the triangle is equal to h_III = x√3. According to (4), the inertia moment of the triangle is:

Hence, the inertia moment of zone II is equal to:

The doubled sum of the inertia moments of zones I and II determine the inertia moment of the 6-pointed star:

(12)

When the star rotates by 30°, voltage distribution τ changes significantly, acquiring the outlines of a dovetail (Fig. 9 b). Two peaks are observed with value τ_max = 28.12 MPa. The distribution pattern is symmetrical. However, with distance from the axis of symmetry, the voltage first increases from 12.5 to 28.12 MPa, and then drops to 3.125 MPa. In this regard, two peaks are observed on dependence τ(x). Further on, the voltage decreases rapidly to zero.

Figured cross. Consider a solution for the inertia moment of a non-standard section in the form of a figured cross (Fig. 10), each side of which is the fourth part of the circle of radius R. The distance between the apices of the figure is R√2. We first find the inertia moment of the cross relative to the major diagonal (Fig. 10 a).

Half-length of strip h in Figure 10 a is equal to:

Taking into account the symmetry and using equation (3), we obtain the following value of the inertia moment of the section of the figured cross:

(13)

We determine the inertia moment of the cross when it is rotated by 45° relative to x-axis (Fig. 10 b).

Figured cross fits into the circle of radius R. Figure 10 b shows that there are four figures in the form of an oval with sharp corners around the cross. To determine the inertia moment of the cross, it is sufficient to subtract the inertia moments of the four ovals from the inertia moment of the circle.

It follows from (6) that the inertia moment of the circle is equal to:

We determine the inertia moments of the ovals relative to the center of gravity of the cross.

The inertia moment of the oval on the right is equal to the inertia moment of the oval on the left. Half-height of the strip at the oval on the left is:

We take into account equation (3), as well as the symmetry of these parts and their location. By integrating from 0 to R – R/√2, the following value of the inertia moment of these two parts is obtained:

(14)

By (4), we calculate the moment of inertia of this pair:

Value h in (3), according to Figure 10 b, (oval at the bottom of the circle) is equal to Therefore, for a pair of ovals at the top and bottom of the circle, the inertia moment relative to their own centers of gravity can be recorded as:

(15)

We determine the resulting inertia moment of the figured cross. To do this, we subtract the inertia moments of the four ovals from inertia moment (13) of the circle:

Comparing the result obtained to result (13), we can see that the inertia moment of this cruciform section of the console does not change when x-axis is rotated by 45°.

Figure 11 a shows the distribution of shear stress in the cross section of the figured cross, corresponding to λ = 2 mm, and the action of an external applied force along the axis of the cross (Fig. 10 a). A sharp voltage peak is observed on the vertical axis of the cross, where τ_max = 80.0 MPa.

When the cross is rotated relative to the applied force by 45°, voltage distribution τ changes significantly. As the axis of symmetry deviates, voltage first gradually increases from τ = 13.73 MPa to τ _max = 20.32 MPa. Then, voltage τ drops sharply to zero. Therefore, two symmetrical peaks (Fig. 11 b) are observed in distribution τ at a distance of 2 mm from the axis of symmetry.

Discussion and Conclusion. The proposed simplified calculation method provides for the rapid determination of the inertia moments of complex cross sections of the console. In this case, the shear stress field in the sample section corresponding to the action of an external applied force is uniquely determined. In addition, it is shown that the stress distribution in the section qualitatively and quantitatively depends on the orientation of the section relative to the direction of the external applied force.

To validate the method, the inertia moments were calculated not only for known sections of simple geometry (which showed the absolute identity of the calculated and published results in the literature), but also for two new complex sections in the form of a regular six-pointed star and a figured cross. It is shown that the rigidity of the console does not change when an external force applied perpendicular to the axis of symmetry is rotated by 30° for a section in the form of a 6-pointed star, and by 45° — for a square and a figured cross. The method and the solutions obtained can be used by engineers and mechanics in modeling and calculating strength and stiffness of rod structural elements.