Structural mechanics

The explanations for the matrix formation: the values of the ordinates of the diagram M9 (Figure 3.20, d) are recorded in the ninth column of

the matrix, the values of the ordinates of the influence line for M2

(Figure 3.20, d) are recorded in the second row.

Performing the load transformation, we get the vector of concentrated forces in the form:

Having preliminary information that the bending moments are equal to zero in sections 1, 5, and 9, we can delete the corresponding rows of the matrix LM . Since the concentrated forces above the supports do not

affect the outline of the diagram of moments, columns 1, 4, and 8 can be deleted in the matrix. As a result, we obtain a matrix LM of size (6 6):

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Figure 3.21, a shows the diagrams of bending moments in the beam with a given load. Figure 3.21, b shows one in the beam with a converted load. The ordinates in the considering cross-sections are the same.

Figure 3.21

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THEME 4. CALCULATING OF THREE-HINGED ARCHES

AND FRAMES

4.1. General Information and Principles of Creation

A system consisting of two disks interconnected by a hinge and joined with the ground using immovable hinged supports is called a three-hinged system (Figure 4.1).

Three-hinged systems where discs are represented by polygonal bars are called three-hinged frames (Figure 4.2).

Three-hinged systems where the disks are represented by curved bars are called three-hinged arches (Figure 4.3). According to their shape, arches are divided into circular, parabolic, sinusoidal, etc. arches.

Three-hinged systems are formed by the triangles method. Therefore, they are geometrically unchangeable and statically determinate. All threehinged systems belong to the class of thrusting systems (Figures 1.24, 4.1–4.3).

To eliminate the effect of the horizontal pressure due to the thrust on the underlying structures, the supporting hinges of the three-hinge systems can be connected by horizontal hinged rods or ties. In such cases, one of the supports should be hinged movable. For example, a threehinged arch with a tie (or a tightrope) at the level of the supports is shown at Figure 4.4.

Three-hinged arches with a tie are externally non-thrusting systems. A vertical loads cause only vertical reactions in supports of such arches.

Arches with an elevated (Figure 4.5) or polygonal complex tie (Figure 4.6) are applied in order to rationally use the space under the arches.

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Therefore, vertical reactions of the arch are often referred to as beam reactions. And this is true with arbitrary vertical load.

Three independent equilibrium equations have already been used to determine the support reactions of the arch. The equilibrium equation in the form of the sum of all external forces projections on the vertical axis is usually used to verify the correctness of the vertical reactions calculation.

Y VA VB F 0.

There is just a need to find the value H of the arch thrust. To determine the arch thrust, we will use the distinguishing property of the arch compared to the equivalent beam. In the intermediate hinge C of the arch (Figure 4.7, a) there is no bending moment. There is no hinge in the corresponding cross-section of the equivalent beam, and the bending moment in this cross-section of the beam (Figure 4.7, b), in the general case, is not equal to zero.

Therefore, defining the bending moment in the hinge C of the arch as the sum of the moments relative to this cross-section of all external forces, for example, located to the left of it, we must equate the resulting expression to zero.

MC ΣMCleft VA 2l F(2l a) Hf 0.

Taking into account, that

VA 2l F(2l a) MC0 ,

where MC0 is the bending moment in the cross-section C of the equivalent beam, we can eventually find the thrust H.

M 0

H fC .

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Thus, the arch thrust is directly proportional to the beam bending moment in cross-section C of the equivalent beam and inversely proportional to the rise of the arch in the intermediate hinge.

To check the calculated thrust value, the beam bending moment in the cross-section C is usually calculated once again through the sum of the moments of external forces applied to the beam to the right of this section. For our example, it is possible to write

MC0 MCright VB 2l .

After calculating the support reactions, the determination of the internal forces in the cross-sections of three-hinged arches is usually carried out by the section method, as in any other bars systems.

Consider the features of applying the section method to a threehinged arch with supports at the same level (Figure 4.7, a). To do this, we cut the arch at some cross-section x-x and consider the equilibrium of the left-hand part (Figure 4.8). The action of the discarded right-hand part is replaced by three internal forces: bending moment M x , transver-

sal force Qx , and longitudinal (normal) force N x .

The bending moment M x in the cross-section x-x of the arch is cal-

culated as the sum of the moments of only external forces acting on the left part of the arch relative to the center of gravity of the cross-section x-x of the arch

M x M xleft VA x F x aFi Hy.

Taking in to account, that

VAx F(x a) M x0,

where M x0 is the bending moment in the cross-section x-x of the equiv-

alent beam (Figure 4.7, b), the bending moment in the cross-section x-x of the arch may be finally found using a formula:

M x M x0 Hy.

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Thus, the transversal force in the cross-sections of the arch is expressed through the projection of the beam transversal force Qx0 in the

corresponding cross-section of the equivalent beam and the projection of the thrust H on the normal to the arch axis in the considered crosssection of the arch.

Similarly, from the sum of the projections of all the forces on the axis tangent to the axis of the arch in section x-x, we find the longitudinal force in this section of the arch

Nx VA sin x F sin x H cos x

(VA F )sin x H cos x ,

or

Nx Qx0 sin x H cos x .

The longitudinal force in the cross-section of the arch is also expressed through the projection of the beam transversal force Qx0 in the cor-

responding cross-section of the equivalent beam and the projection of the thrust H on the tangent to the arch axis in this cross-section of the arch.

Compared with simple beams in three-hinged arches, the transversal forces, as well as bending moments, are much smaller. But unlike the beams, longitudinal compressive forces occur in the cross-sections of the arches. While no longitudinal forces are present in simple horizontal beams with vertical loads.

The final diagrams of the internal forces in the arch along its entire length would be curvilinear. Curvilinear diagrams, like any graphs, can be built by calculating the values of the corresponding internal forces in a number of predetermined (characteristic) cross-sections of the arch (the more sections are presented the more accurate the diagram).

Let us illustrate the definition of reactions and internal forces using the example of a circular three-hinged arch with a span of l = 36 m with a rise of f = 8 m (Figure 4.9). The arch is loaded with a concentrated force F = 24 kN and a uniformly distributed load q = 2 kN/m.

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