Structural mechanics

The sum of the projections of all external forces on the vertical axis confirms the result:

Y 25 35 24 2 18 60 60 0.

In the cross-section C, the bending moment of the beam is calculated and checked:

MC0 MCleft 2518 24 6 306 kN m,

MC0 MCright 3518 2 18 9 306 kN m. Then the arch thrust is calculated:

To plot the diagrams of the internal forces it is necessary to assign characteristic arch cross-sections. Firstly, these are the supports A and B and the intermediate hinge C. Secondly, these are the point of application of concentrated force and the beginning and the end of the arch segment where the distributed load acts. Thirdly, these are additional intermediate cross-sections necessary for constructing curvilinear segments of the diagrams with sufficient accuracy. In this example, there are at least seven of these characteristic points.

They are located along the arch span in increments of 6 m. To plot the diagrams of the transversal and longitudinal forces at the point of application of the concentrated external force, it is necessary to consider two infinitely close points: one to the left of the application point of the external force, the second to the right of this point. At this cross-section, there will be a jump on the indicated diagrams of the internal forces, and a fracture on the diagram of bending moments. When constructing diagrams of internal forces and moments, it is necessary to monitor their correspondence with each other and the load. The differential depen-dencies between bending moments, transversal forces, and the load must be fulfilled.

To determine the geometric characteristics of the arch, calculate the value of the arch axis radius

R 4 82 362 24.25 m. 8 8

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All further calculations are summarized in the following tables. Calculations in tables can be performed on a calculator, plotting manually, using patterns and other drawing tools. But it is possible to use computers: universal mathematical and engineering software, programming languages, tabular and graphic editors and other modern software tools that automate the process of computing and plotting graphic objects.

Table 4.1 Calculation of bending moments in a three-hinged arch

So, to build the below diagrams of internal forces in the arch, modern software was used that automates the process of performing calculations and graphing. The diagram of bending moments (Figure 4.10), the diagram of transversal forces (Figure 4.11) and the diagram of longitudinal forces (Figure 4.12), are built on the horizontal projection of the arch axis using the graphic software. Of course, the number of characteristic cross-sections along the span has to be significantly increased.

Figure 4.10. Diagram M

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At the points where the transversal forces diagram passes through zero, there are the extremums on the diagram of bending moments. At the point where the segment of the distributed load begins, there is a fracture on the diagram of the transversal forces. In areas where the transversal forces diagram is ascending, the bending moments diagram is convex up. In areas where the transversal forces diagram is downward, the bending moments diagram is convex down.

Such conclusions follow from the differential dependences known from the resistance of materials, according to which the transversal force in the cross-sections of the arch is the first derivative along the length of the arch arc from the function of bending moments. And the load is the first derivative from the transversal force function.

Table 4.2

Arch parameters for calculation of the transversal and longitudinal forces in the arch

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4.3. Calculating a Three-Hinged Arch with a Tie

Three-hinged tied arches are externally non-thrusting systems. Vertical loads cause only vertical reactions in supports of such arches. These vertical reactions are determined as in simple beams. The horizontal reaction of their immovable hinged support is equal to zero under vertical loads.

But internally such arches are thrusting systems. Their thrust is an internal longitudinal force in ties.

To determine the tightening force, it is necessary to cut an arch by a section through the key hinge of this arch. For example, in a three-hinged arch with a complex tie it is a cross-cut 1-1 passing through the intermediate hinge C (Figure 4.13). The equilibrium equation of the left part of the arch in the form of the sum of the moments of all forces relative to the key hinge C gives a possibility to determine the arch thrust H.

MCleft 0; RA 2l H f f0 0.

Figure 4.13

The first term in the resulting equation is the bending moment in section C of the equivalent beam:

RA 2l MC0 .

Therefore, to determine the tightening force (thrust), the following expression can be obtained:

M 0 H f Cf0 .

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The internal forces in the other members of the complex tie and in the cross-sections of the arch may be calculated by the usual method of sections.

4.4. Influence Lines in Three-Hinged Arches

Consider an arch loaded with a single vertical force, the position of it is determined by the abscissa xF (Figure 4.14, a). To determine the vertical support reactions, we compose the equilibrium equations in the form of sums of the moments of all the forces acting on the arch relative to the left and right supports:

M A 0; 1xF RBl 0;M B 0; 1(l xF ) RAl 0.

From these equations we find the functions of changing the vertical support reactions depending on the position of the unit force

The obtained dependences of the change in the values of the support reactions completely coincide with the corresponding dependences for the support reactions of a simple two-support beam. Therefore, the influence lines for the vertical reactions (Figure 4.14, c, d) in the arch coincide with the influence lines (Inf. Lin.) for the reactions in the corresponding equivalent beam (Figure 4.14, b).

The thrust H of the arch under the action of vertical loads is determined by the expression:

M 0

H fC .

Hence

Inf . Lin. Н Inf . Lin. МС0 f .

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The influence lines of internal forces in the cross-sections of the arches will be built using the previously obtained dependencies expressing the internal forces in the arches through the corresponding internal beam forces and the arch thrust.

So the bending moment in the section K of the arch (Figure 4.14, a) is determined by the expression

MK MK0 HyK .

Since the ordinate yK of the cross-section K of the arch is constant, for the influence line for MK we get

Inf . Lin. M K Inf . Lin. M K0 Inf . Lin. H yK .

In accordance with this expression, we separately construct the influence line for the bending moment Inf .Lin.M K0 in the section K of the

equivalent beam (Figure 4.14, g) and the influence line for the thrust H Inf .Lin.H , multiplied by a factor yK (Figure 4.14, h). Subtracting the

ordinates of the second influence line from the ordinates of the first, we get the influence line for the bending moment in the section K of the arch

Inf .Lin.M K (Figure 4.14, i).

The transversal force in the cross-section K of the arch is determined by the dependence

QK QK0 cos K H sin K .

Therefore, the influence line for the transversal force in this arch section can be represented as follows:

Inf .Lin. QK Inf .Lin. QK0 cos K Inf .Lin. H sin K .

We build the influence line for the thrust H (Figure 4.15, b), and the influence line for the transversal force in the section K of the equivalent beam (Figure 4.15, c). Then we build intermediate influence lines, multi-

plying all the ordinates of Inf .Lin.QK0 on cos K (Figure 4.15, d), and the ordinates Inf .Lin.H on sin K (Figure 4.15, e). Subtracting the ordinates of the second from the ordinates of the first influence line, we get

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the desired influence line for transversal force in the section K of the arch (Figure 4.15, f).

Figure 4.15

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