Structural mechanics

E x a m p l e 4. The displacements of the supports of the continuous beam are shown in Figure 11.12, a. It is necessary to construct the bending moment diagram, taking c1 0.01 rad, с2 с3 0.06 m.

Figure 11.12

Calculation of the beam subjected by the displacements of the supports is carried out by the methods considered earlier. We show the solution by the force method.

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Let the primary system be the same as shown in Figure 11.12, b. The free terms of the canonical equations are determined by the formula (7.13). Using the distribution of reactions in the supports (Figures 11.12, c–d), we find:

1с Rк1ск 1 с1 с1 0.01;

The coefficients at unknowns have the same values as in example 1. We write the canonical equations corresponding to the given dis-

placements of supports:

X3 6.0 10 3 EJ kN·m.

The diagram M is shown in Figure 11.12, f.

11.3. Constructing Influence Lines for Internal Forces

To construct influence lines for internal forces by the static method (see Sections 8.11, 9.11), it is necessary, in the general case, to calculate a continuous beam on the action of force F 1 , applied at a number of characteristic points of each span and compose an influence matrix LS

of internal forces. Based on the values of the i-th row elements, you can construct an influence line Si .

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We calculate the beam on action of the unit force in the sections indicated in Figure 11.13, a and, based on the calculation results, construct the influence lines for internal forces.

The shape of the influence lines for each span, as a rule, is determined by the values of its ordinates in three intermediate points. For example, to build influence line Mc (Figure 11.13, b) or influence line M5

(Figure 11.13, c) it is enough to find the corresponding bending moments Mc or M5 when the force F 1 locates in the cross-sections

dividing the span into four parts.

The shape of influence line for bending moments can have some features when it is constructed for sections located near the supports. So, when constructing the influence line for bending moment for the section

K2 (Figure 11.13, d), it turns out that the force F 1 located to the right-

hand of the second span does not cause a bending moment in this crosssection (such point K2 is called the left focus of the second span).

If a certain cross-section K1 is located between the points B and K2 , then the influence lines for the bending moment in this cross-

section of this span will be have double-sign (Figure 11.13, e). Therefore, in order to avoid errors, in the part of the span to which the effort under investigation belongs, the number of trial installations of force F 1 should be taken as increased.

It is known that the shape of the influence line, in accordance with the kinematic method (see Section 8.11), is similar to the diagram of beam deflections caused by the displacement of the corresponding link in its direction by value equal to one. For example, to get the outline of inf.line Q5 (Figure 11.13, f) it is necessary to move apart the ends of the beam

adjacent to the fifth cross-section vertically by a length equal to one so that this ends remain parallel to each other.

The numerical values of the ordinates of the influence line are conveniently calculated by the static method.

To establish the form of Influence Line VB (Figure 11.13, g) it is nec-

essary to remove the support rod at the point B in the beam and give in its direction the displacement equal to unit. The outline of the curved axis of the beam will correspond to the outline of the required influence line. The ordinates of the influence line are determined by the static method.

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11.4. Enveloping Diagrams of the Internal Forces

Continuous beams, like most other structures, are loaded with both constant and temporary loads, the nature of the action of which, in the general case, turns out to be rather arbitrary: it can be in all spans of the beam or only in some of them. Extremal efforts in beam cross-sections are dependent on unfavorable loadings the location of which is determined by using influence lines (see chapter 3).

However, this method of finding extremal efforts is quite complicated and, moreover, does not provide a clear idea of the distribution of the maximum and minimum efforts along the length of the beam.

The problem of determining extremal efforts is solved more simply using enveloping diagrams of internal forces. Consider the problem of constructing enveloping diagrams of bending moments in a continuous beam loaded with a constant (Figure 11.14, a) and temporary loads (Figure 11.14, b). The moment diagram due to the constant load is shown in Figure 11.14, c. The moment diagrams due to the serial loading of each span with a temporary load are shown in Figures 11.14, d–g.

The maximum and minimum bending moments in the beam crosssections are determined by the expressions:

where Mconst is the bending moment in the given cross-section due to the constant load;

Mtemp is a positive bending moment in the given cross-section due to temporary loads in a corresponding span;

Mtemp is a negative bending moment in the given cross-section due to temporary loads in a corresponding span.

For example, max M7 34.03 33.63 0.90 68.56 kN m; max M10 44.03 2.26 2.64 39.13 kN m; min M1 21.04 24.39 1.02 46.45 kN m; min M11 5.99 10.9 4.28 9.19 kN m.

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Connecting by the smooth curve the points corresponding to the values of max M , we obtain the enveloping diagram of the maximum mo-

ments (Figure 11.14, h). The enveloping diagram of the minimum moments corresponds to the values of min M.

From the constructed graphs it follows that in some parts of the beam the stretched fibers of the beam are located only below (or only at the top), and in other parts, the stretched fibers can be located both below

and above. In the cross-section 11 max M11 15.24 kN m (not shown in Figure 11.14), and min M11 9.19 kN m.

Information about the distribution of calculated efforts is used in the design of beams.

A similar approach to the construction of enveloping diagrams of bending moments, shear and longitudinal forces can be applied in the calculation of other structures.

11.5. Calculating Continuous Beams on Elastic Supports

Examples of elastic supports are long columns, on which a continuous beam rests (Figure 11.15, a), transverse beams of the carriageway of a metal bridge, on which longitudinal continuous beams rest, as well as pontoons, which serve as supports of the floating bridge.

Figure 11.15

In the design scheme of the beam, such supports are depicted in the form of springs (Figure 11.15, b). If the elastic supports are linearly deformable, then the displacements of the support points of the beam are proportional to the reactions of the supports:

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ym cm Rm ,

where cm – is the pliability coefficient of the m-th support, m/kN.

Calculation of continuous beams on elastic supports is conveniently performed by the force method. The primary system of the force method is accepted the same as in the calculation of beams on non-deformable (absolutely rigid) support rods. A fragment of the primary system of a multi-span continuous beam is shown in Figure 11.16, a. Emphasizing the physical meaning of the primary unknowns of the force method, in practical calculations the notation Xi is replaced with Mi .

Figure 11.16

Displacement in the direction of the unknown M n (the angle of mu-

tual rotation of the cross-sections of the beams adjacent to the n-th support) will be caused only by the support moments M n 2 , M n 1 , M n ,

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M n 1 , M n 2 and the load located in the spans ln 1 , ln , ln 1 , ln 2 , therefore, the corresponding canonical equation of the force method has the form:

n,n 2Mn 2 n,n 1Mn 1 nn Mn

n,n 1Mn 1 n,n 2Mn 2 nF 0.

It is called an equation of five moments.

The deformed state of the primary system caused by Mn = 1 is shown in Figure 11.16, b. In Figures 11.16, c the moments diagram and the values of the support reactions due to Mn = 1 are presented. The moments diagram and the reactions due to Mn–2 = 1 are given in Figures 11.16, d.

The coefficients and free terms of the equations are determined by the Mohr-Maxwell formula, taking into account the influence of bending moments in the beam and reactions in elastic supports:

ik Mi Mk dx cmRmi Rmk ,

EJ

iF Mi M F dx cmRmi RmF ,

EJ

where Mi , Mk are the moments in the beam, respectively, due to Mi = 1

and Mk = 1;

M F are the moments in the beam due to given load;

Rmi , Rmk are the reactions in the support m, respectively, due to

Mi = 1 and Mk = 1;

RmF is the reaction in the support m due to given load; cm – is the pliability coefficient of the m -th support.

Influence lines for internal forces in beams on elastic supports, as in beams on absolutely rigid supports, are constructed by static and kinematic methods.

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