Structural mechanics

Thus, the calculation of displacements in arches, as well as in other curvilinear bars of variable section, is much more time-consuming than the calculation of displacements in rectilinear bars of constant section. The calculation of definite integrals according to the rule of multiplying diagrams (according to the Vereshchagin rule) is not feasible here, since under the signs of the definite integrals there is a product of several nonlinear functions. The Simpson formula should be considered as one of the options for the approximate (numerical) calculation of the definite integrals.

Taking into account the above assumptions about neglecting longitudinal and shear deformations of the arch, it is permissible to use a simpler method of numerical integration to calculate displacements in the arch. This is the rectangle method. For this purpose, the span of the arch is divided into sufficiently small sections, preferably of the same length. They are numbered in a certain sequence, and in the middle point of each section, the values of all integrand functions are calculated. As a result, the procedure for taking a definite integral is replaced by calculating the final sum of the products of the integrands values in the middle of the sections:

where k is the section number, n is the amount of sections. Typically, all calculations are carried out in tables (Table. 13.1).

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The sum of the elements of the penultimate column gives the part of the unit displacement due to the bending deformations of the arch. The complete unit displacement is found as the sum

The tie internal force is found from the solution of the canonical equation

Ntie X1 1F ,

11

Internal forces in any section of a two-hinged arch can be found using the same formulas as in a three-hinged arch

M x M x0 X1 yx ;

Qx Qx0 cos x X1 sin x ;

Nx (Qx0 sin x X1 cos x ) .

The index x in these formulas denotes an arbitrary cross section of the arch.

13.3. Influence of the Tie Longitudinal Rigidity on the Tie Internal Force

In statically indeterminate systems, the distribution of internal forces between elements depends on the ratio of their rigidity Therefore, tie

deformability (the magnitude l / (EAtie ) ) will affect the value of the tie

internal force. The formula obtained above for calculating the tie force can be rewritten as follows:

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A graphical representation of the dependence of the tie rigidity on the tie force is given in Figure 13.4.

Figure 13.4

If you gradually reduce the tie longitudinal rigidity, the value EAtie ,

then the tie force will decrease. The weaker the tie, the smaller force it perceives. In the limit, with a tie of zero rigidity, i.e. in the absence of it, the arch with a tie turns into a simple curved beam; tie force is zero.

On the other hand, if the tie longitudinal rigidity is gradually increased, the tie force is also increased, but to a much lesser extent. In the limit, when the tie rigidity tends to infinity the tie force will asymptotically tend to

H 1F ,

11(M )

where H is a quantity numerically equal to the thrust of a two-hinged arch without a tie on hinged immovable supports (Figure 13.1, a).

That is, in this limiting case, when the tie is absolutely inextensible, the two-hinged arch with the tie turns into a two-hinged arch on hinged immovable supports, as it were, into an ordinary two-hinged arch. Thus, the relatively weak tie does not allow using all the advantages of the arch with the tie as a thrust system.

In contrast, an overly rigid tie is practically useless. The magnitude of the force in a rigid tie cannot exceed the magnitude of the thrust in the ordinary two-hinged arch, calculated by the last formula.

When calculating double-hinged arches without a tie, the thrust is also taken as the primary unknown (H X1). The primary system of the

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force method is obtained by discarding the horizontal support link of one of the supports. The diagram of bending moments due to the primary unknown equal to one and the diagram of load bending moments in the primary system of a two-hinged arch without a tie are the same as for an arch with a tie (Figure 13.3, c, d). The thrust of a two-hinged arch without a tie is calculated by the formula

13.4. Hingeless Arches. Features of Calculation

A hingeless arch is three times statically indeterminate. To determine the three primary unknowns of the force method, it is necessary to compose and solve three canonical equations. By the appropriate choice of a rational primary system of the method of forces, one can even achieve complete separation of the system of canonical equations into three separate equations, each with one unknown only. This takes place with an arbitrary outline of the axis of the arch and an arbitrary load. Variants of the primary systems shown in Figure 13.5, a, b, c, allow you to reset to zero the secondary coefficients of the canonical equations and bring the equations to the form:

11X1 1F 0 ;

22X2 2F 0 ;

33 X3 3F 0 .

Such a result can be obtained at the cost of additional calculations to determine the length of the hard consoles (Figure 13.5, a, b). So, for ex-

ample, in a symmetric arch (Figure 13.5, b), the primary unknown X1 is

skew-symmetric and is separated from two other direct-symmetric unknowns X2 and X3 . The direct-symmetric primary unknowns can be

separated by selecting the length of absolutely rigid consoles so that the

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displacements 23 32 are zero. The point at which the ends of abso-

lutely rigid consoles are located is called the elastic center of the arch. A complete separation of the primary unknowns can also be achieved by placing the end of a single console in the elastic center (Figure 13.5, a).

The same result can be obtained with the primary system in the form of a three-hinged arch (Figure 13.5, c), grouping the primary unknowns

X1 and X3 , and determining the position of the extreme hinges from the

conditions so that the secondary coefficients of the canonical equations of the force method vanish.

However, in the age of electronic calculators and computers, solving systems of linear algebraic equations of the second-third order does not present any difficulties. Therefore, you can abandon the choice of the singular primary systems and additional calculations.

Figure 13.5

So the primary system, obtained by cross-cut of the hingeless arch by the axis of symmetry (Figure 13.5, d), allows you to divide immediately three joint canonical equations into one independent equation relative to

the skew-symmetric primary unknown X1 and into a system of two joint equations relative to two symmetric primary unknowns X2 and X3:

11X1 1F 0 ;

22 X 2 23 X 3 2F 0 ;

32 X 2 33 X 3 3F 0 .

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The primary system in the form of a curved beam leads to the same results (Figure 13.6, b). Let us consider this option in more detail, since the techniques for constructing a series of diagrams of internal forces in such a primary system have already been reflected in the calculation of two-hinged and three-hinged arches.

Figure 13.6

We group the unknown support moments, decomposing them into a skew-symmetric group unknown X1 and a symmetric group unknown

X2 . The primary unknown X1 1 will cause a linear skew-symmetric

diagram of bending moments in the curved beam (Figure 13.6, c). The ordinates of this unit diagram can be calculated in the usual coordinate system with the origin on the left support according to the equation

M1(x) 1 2x / L.

The unknown X2 1 will cause constant positive bending moments

in the curved beam (Figure 13.6, d). In the primary system, the unknown X3 1 will cause a symmetric diagram of bending moments with nega-

tive ordinates, which coincides with the outline of the axis of the arch (Figure 13.6, e). The load diagram of bending moments in the primary system coincides with the beam diagram of bending moments (Figure 13.6, f).

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The curvilinear integrals along the length of the arch, determining the coefficients and free terms of the canonical equations, are calculated according to the rule of rectangles, dividing the arch span into n sections. In view of the notation introduced above, we accordingly obtain:

In the above formulas, the index k denotes the section number when calculating the Mohr integrals according to the rectangle rule. The values of the integrands are usually calculated in the middle of the sections.

After determining the primary unknowns from the solution of the canonical equations, one can determine the internal forces in any cross section of the hingeless arch just like in three-hinged and two-hinged arches.

The primary unknown X1 causes vertical support reactions in the primary system

In the primary system the primary unknown X2 does not cause support reactions. The primary unknown X3 causes only the horizontal reaction in the primary system

H A3 X3 .

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Therefore, to calculate the internal forces in an arbitrary cross section x of a hingeless arch, we will have the following formulas:

M x M x0 X1(1 2x / L) X 2 yx X3 ;

Qx (Qx0 2X1 / L)cos x X 3 sin x ;

Nx (Qx0 2X1 / L) sin x X3 cos x .

13.5.Calculating Combined and Suspension Statically

Indeterminate Systems

Remember that design schemes of structures are called combined systems in which some of the bars work on bending, and the rest only on compression-tension. Bars that work on bending usually have a more powerful cross-section and are called rigid members. Rods that accept only compressive or, especially, tensile forces are lighter. They are called flexible elements. Some statically determinate combined systems were considered above in the sixth topic. The types of statically indeterminate combined systems are practically immense.

All types of arches with ties can be attributed to combined systems. An arch itself is a rigid element. Tie elements are flexible members.

Examples of some other statically indeterminate combined systems are shown in Figure 13.8.

Figure 13.8

Suspension systems are those systems whose main load supporting elements work in tension. Suspension systems include hanging arches

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(Figure 13.9, a), various cable-stayed and pure cable systems of bridges, roofs and others constructions (Figure 13.9, b, c, d). Many combined systems can also be attributed to suspension systems (Figure 13.8, b, c).

The calculation of a hanging (stretched) two-hinged arch (Figure 13.9, a) differs from the calculation of an ordinary (compressed) twohinged arch only in the fact that the thrust of the hanging arch is directed outward from the span. If the hanging arch is made of flexible elements (cables, ropes), then it turns into a flexible thread (Figure 13.9, c). The calculation of flexible threads, as well as other hanging, cable-stayed and combined systems of large spans, is carried out in a nonlinear formulation according to the deformed design scheme. In this section, we consider the features of calculation of some combined systems (including suspension) in the classical linear formulation according to an undeformed design scheme.

The beams with two-post hinged chain (Figure 13.8, a) and with mul- ti-post hinged chain (Figure 13.8, c) are examples of combined systems with one redundant link. A suspension bridge in the form of a chain with a continuous stiffening beam (Figure 13.8, b) is three times statically indeterminate. A truss with a continuous upper chord (Figure 13.8, d) contains four redundant links.

Figure 13.9

The cable-stayed combined system (Figure 13.9, b) with a continuous beam and two cables is statically indeterminate once, provided the cables

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do not turn off from work. The cable truss (Figure 13.9, d) is statically indeterminate three times. Such trusses must be pre-tensioned so that all their elements are not switched off. Under this condition, the calculation of cable trusses does not differ from the calculation of the usual statically indeterminate trusses considered above.

Calculation of combined systems of small and medium spans with a small number of primary unknowns can be performed by the method of forces according to an undeformed design scheme. It is recommended to choose the primary system of the force method for combined systems so that internal forces from a given load arise only in its rigid elements. This is possible if the entire load is applied to the rigid elements. Figure 13.10 shows options of the primary systems of the force method for two combined systems. The given load and the unit values of primary unknowns cause the bending moments only in their rigid elements. The primary unknowns also cause longitudinal efforts in flexible elements. But there will be no internal forces in the flexible elements of such primary systems from external loads.

Figure 13.10

The coefficients at the unknowns and the free terms of the canonical equations of the force method in combined systems are calculated by the two-term Maxwell-Mohr formulas:

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