Structural mechanics

If a function in a certain section of the element is a more complex than a polynomial of the third degree, which is possible for elements of curvilinear shape, or the rigidity changes along the axis of the element, or the load is non-uniformly distributed on it, the result of the calculation using the Simpson's formula will be approximate.

On a partial section, the error is estimated as follows:

h5 M , 2880

where

x a, b

that is, on this section the Simpson's formula has accuracy O h5 , on the whole section accuracy is O h4 , while the trapezoid formula, like the formula of rectangles, has a second order of accuracy.

E x a m p l e . Using the Simpson's formula, determine the area of the deflection's diagram of the cantilever beam with a constant cross-section, loaded with a uniformly distributed load.

Diagrams M F and M1 are shown in Figure 7.27.

Figure 7.27

Lat. supremus is the highest.

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Here:

f (x) M F (x) M1(x) qx44 .

For the variant with one section of length l we get:

The exact solution has been obtained earlier by direct integration.

gration interval, the error is estimated as follows:

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In some cases, quadrature formulas are applied with a large number of equally spaced nodes. In particular, such a formula, built on four nodes, is the following one:

This formula is sometimes convenient to use to multiply linear diagrams of the internal forces. The result of this calculation is accurate. For example, if the multiplied diagrams have the form shown in Figure 7.28, the Mohr's integral in this section will be equal to:

Figure 7.28

In general, formulas with a large number of equally spaced nodes are applied relatively rarely.

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7.8. Determining Displacements Caused by the Thermal Effects

Suppose that for a system in state а (Section 7.6) the external influence is thermal one: the temperature of systems elements has changed with respect to some initial state. For an infinitely small element (Figure 7.29) of this system, we take the temperature of the lower fiber equal

to t1, the upper one equal to t2. And the temperature distribution along the cross-section height is accords to the linear law.

Figure 7.29

The temperature on the axis passing through the center of gravity of

t t1 t2 . 2

Under the influence of temperature, the element moves to a new position (it is indicated by a dashed line). In this new position, all the fibers are extended by an amount d t dx t dx and each lateral face is

rotated by an angle d t relative to the axis passing through the center 2

of gravity.

The elongation of the lower fiber is equal to t1 dx, and the upper one is equal to t2 dx, ( is the coefficient of linear expansion). Then, due to small deformations, we obtain:

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where t t1 t2 is the temperature difference.

Since temperature deformations do not cause a cross-sectional shear, substituting d t and d t in the general formula (7.6) for determining

displacements and replacing the index a in the designation ia by t (indicates the reason that caused the displacement), we obtain:

Note that each of the integrals in this expression determines the work of the internal forces of the auxiliary state of the system on displacements caused by a change in temperature. Therefore, the values of the integrals are accepted positive on the integration interval in the case when the corresponding directions of the element deformations, caused by the forces of the i-th (auxiliary) state and by thermal action, coincide.

If the values , t, t and h remain unchanged in some parts of the elements, the expression (7.11) is converted to the form:

are the areas of the diagrams of longitudinal forces and bending moments on the segments of the members with the specified features.

E x a m p l e. Determine the horizontal displacement of the frame support B (Figure 7.30, a) from thermal action indicated on the figure. Unchanged cross-sections through the length of each element are as-

sumed to be symmetrical. The height of the vertical element is h1, the height of the horizontal one is h2.

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The temperature along the axis of members is:

the temperature differences are:

Figure 7.31

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The auxiliary state of the frame and the diagrams of internal forces M1 and N1 are shown in Figure 7.31, b, c, d.

We calculate the required displacement:

7.9. Determination of Displacements Caused by the Settlement of Supports

Suppose that the support connections of a given statically determinate system (Figure 7.32, a) under the influence of some actions moves to the positions shown in Figure 7.32, a: rigid support turned clockwise by an angle c1, and the hinged-movable support moved upward by c2. We

denote this state of the system as state с. To determine the displacement of a point, for example, the horizontal displacement of the node D , we apply a force Fi 1 in the auxiliary state in the direction of the required

displacement (Figure 7.32, b).

Figure 7.32

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We define the work of the forces of the i-th state of the system at its displacements in the state c. There are no internal forces in a state c:

displacements of the supports of a statically determinate system do not cause forces in its elements. Therefore, only external forces, which include support reactions, will do the work on the displacements of the state c . In accordance with the principle of virtual displacements, we obtain:

The sign of the product Rki ck is assumed to be positive if the directions of Rki and ck coincide.

For this example, we get:

E x a m p l e. Determine the horizontal displacement of the frame cross-section K (Figure 7.33, a) caused by the settlement of supports indicated on the figure.

According to (7.13), the expression for the requied displacement is:

horizKC Rki ck .

The auxiliary state of the frame for determining support reactions caused by a unit concentrated force applied to the cross-section K in the horizontal direction is shown in Figure 7.33, b.

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