Structural mechanics

The indicated method of calculating the Mohr integrals can lead to significant difficulties, since a very complex formula either can be obtained, or cannot be obtained at all for an indefinite integral F(x).

analytically or using numerical integration.

But for the case when the bar has constant rigidity in the integration area, that is EJ f3(x) const, and one of the function f 1 (x) or

f 2 (x) is linear, the method proposed by A. K. Vereshchagin is usually

used. This method is one of the most effective methods of calculating definite integrals. Let us explain its essence.

We plot the graphs of functions f1(x) and f2 (x) , that is, diagrams of bending moments Mi (x) and M F (x) on the integration area (Figure 7.18).

Figure 7.18

Suppose, for example, diagram Mi is rectilinear (Figure 7.18, b). The reference point is the intersection point of the bar axis with the diagram inclined line. Then Mi (x) x tg , and the Mohr integral is converted to

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ea of the diagram M F (Figure 7.18, a) relative to the axis y . The static

moment is equal to the product of the area of this diagram by the distance from its center of gravity to the axis, that is:

b

x M F dx x0.

a

Given the ratio x0 y 0 /tg , we get:

Thus, the Mohr integral is calculated by multiplying the area of the curvilinear diagram with the ordinate of the rectilinear diagram, taken under the center of gravity of the curvilinear one.

The process of calculating the integrals by Vereshchagin's method is sometimes called the “multiplication” of diagrams. The positive sign of

the product у0 is taken when the diagram M , whose area is denoted by, and the ordinate y have the same signs, i.e., when they are located

on one side of the bar. In practice, one can be guided by a simpler rule: if both diagrams of efforts for certain section of the bar are located on one side of its axis, the result of their “multiplying” is accepted as positive, if diagrams are located on opposite sides of the bar, the result of their “multiplying” is accepted as negative.

When using the Vereshchagin's rule, complex diagrams of the internal forces should be represented as a sum of simple ones, for each of which formulas for area calculation and gravity center position are known. Examples of the simple diagrams are bending moment diagrams

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for cantilever or single-span beams loaded with concentrated force or uniformly distributed load (Figure 7.19).

Figure 7.19

To obtain simple diagrams, the principle of independence of the action of forces should sometimes be used.

E x a m p l e. Determine the vertical displacements of points A and B (Figure 7.20) of the beam with constant rigidity.

Diagrams of bending moments for a beam from a given load and unit forces are shown in Figure 7.20.

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To determine the vertical displacement of a point D we load the beam with force F1 1 (Figure 7.21, c) and construct the corresponding diagram of bending moments (Figure 7.21, d).

Using the principle of superposition, we represent the diagram M F

in the form of two simple ones (Figure 7.22) and determine the displacement according to the Vereshchagin's rule:

Figure 7.22

The auxiliary state for determining the angle of rotation of crosssection C is shown in Figure 7.21,e, and the corresponding diagram of bending moments is shown in Figure 7.21,f.

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E x a m p l e. Find the horizontal displacement of point A of the frame shown in Figure 7.23 a.

The auxiliary state (state 1) is shown in Figure 7.22, b. The bending moment diagrams corresponding to the frame states are shown in Figure 7.23, c, d.

Figure 7.23

In this example, the required displacement is calculated as the sum of the integrals over three members. In each of them, the functions M1(x) and M F (x)

have well-defined analytical expressions. If through the length of one element the diagrams of moments are described by different functional dependencies, the element must be divided into the corresponding sections, the integrals must be calculated separately for each section, and the calculation results should be summarized.

Once again, we note that the Vereshchagin's method cannot be applied in the case when both diagrams are non-linear. So, for example, it cannot be applied to calculating the area of the diagram of the deflections of a beam loaded with a uniformly distributed load.

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The same rule for calculating integrals can be applied to the other two terms in the Mohr's formula for determining displacements.

The value of a definite integral, as it is known, can be calculated using formulas of numerical integration, that are based on replacing the integral with a finite sum:

where xk are the points of the segment a, b : ck are the numerical coefficients.

Given equality, generally approximate, is called the quadrature formula, points xk are the nodes of the quadrature formula, and numbers

ck are called coefficients of the quadrature formula. The error of the quadrature formula

depends both on the location of the nodes and on the choice of coefficients. Most often, a uniform grid of nodes is used in practical applications to the problems of structural mechanics; in this case, the initial integral is represented as the sum of the integrals over partial segments, on each of which a quadrature formula is applied.

The simplest quadrature formulas for one interval are the rectangle formula

and the trapezoid formula

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Naturally, even in the case of functions close to linear, the use of these formulas will lead to an error in the calculations of displacements.

If concentrated forces or uniformly distributed load act on a system composed of rectilinear elements, the diagram of bending moments on separate sections of the element is limited to a straight line or parabola. If it is necessary for this system to determine the linear or angular displacement of some point, in the auxiliary state, the contour of the diagram “ M ” due to the load F1 1 will be determined by linear relation-

degree. Then, on the segments of elements with constant rigidity, the Mohr integral can be calculated exactly using T. Simpson's formula (parabola formula):

where y1, y2 , y3 are the values of the function at the end points of the segment and in the middle of it (Figure 7.24).

Figure 7.24

Simpson's formula is exact for any polynomial not higher than the third degree.

Using the Simpson's formula, we determine the vertical displacement of the cross-section D and the angle of rotation of the cross-section C for the beam shown in Figure 7.21:

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The obtained values of the displacements coincide with those ones found according to the Vereshchagin's rule.

E x a m p l e. Determine the angle of mutual rotation of the ends of the beams, adjacent to the hinge C (Figure 7.25). The bending rigidity of the beams is constant.

Diagrams of bending moments for a beam from a given load and unit force are shown in Figure 7.25.

Figure 7.25

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6EJ4 (4 1.5 60 2 160) 12803EJ

We offer the reader to show how the same value of the displacement can be calculated easier.

E x a m p l e. Determine the horizontal displacement of the end of the cantilever broken beam (Figure 7.26, a).

The diagram of bending moments caused by a given load is shown in Figure 7.26, b, from unit force F1 1 is shown in Figure 7.26, c.

Figure 7.26

“Multiplication” of diagrams on a vertical element is made according

to the Vereshchagin's rule, on an inclined one (its length is 10 m) – according to Simpson's formula:

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