Structural mechanics

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

NK QK0 sin K H cos K .

Accordingly, the expression

Inf .Lin. NK Inf .Lin. QK0 sin K Inf .Lin. H cos K ,

is used to construct the influence line for this longitudinal force.

By building intermediate lines of influence (Inf .Lin. QK0 )sin K (Figure 4.15, g) and Inf .Lin. H cos K (Figure 4.15, h), we sum up

them. Changing the sign of the result to the opposite, we obtain the desired influence line for the longitudinal force in the section K of the arch (Figure 4.15, i).

4.5. The Rational Axis of the Arch

Rational is called the axis of the arch, if bending moments in the cross-sections of the arch are zeros or close to zeros.

The condition

M x M x0 Hy(x) 0

means that bending moments are absent in all cross sections of the arch. This condition allows you to find the equation of the rational arch axis:

M 0 y(x) Hx .

Whence it follows that under the action of vertical loads the ordinates of the rational arch axis are proportional to the bending moments in the equivalent beam having the same span and the same load as the arch. The reciprocal of the thrust H is in this case a proportionality coefficient.

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For an example, we define the rational axis of a three-hinged arch when a vertical, evenly distributed load acts on the arch (Figure 4.16, a).

q

The bending moment in an arbitrary cross-section x of the equivalent beam is defined as the sum of the moments of external forces applied to the beam to the left of section x:

M x0 M xleft ql2 x (qx) 2x qx2 (l x).

Dividing the resulting expression by the thrust, we obtain the equation of the rational axis of the three-hinged arch with a uniform load over the span:

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The resulting equation is a quadratic parabola equation. A parabolic arch with a load evenly distributed over the span does not have bending moments. Only longitudinal forces occur in the arch cross-sections.

In the key (in the middle of the span) of the arch, the longitudinal force is

NC H ql2 . 8 f

In heels (supports), the longitudinal forces are equal

If the arch is outlined in a circle arc, then from the equilibrium conditions of an infinitesimal arch element of length ds, it can be proved that the arch circular axis will be rational when the arch is loaded with a uniformly distributed radial load (Figure 4.16, b). With a uniform radial load in a circular arch, there are no bending moments, and the longitudinal forces will be constant along the length of the arch and equal

N qr.

We invite the reader to carry out the corresponding evidence independently.

4.6. Three-Hinged Arches with a Superstructure

Arches that serve as supporting structures for bridges usually have over-the-top or under-arch superstructures. The moving load on the main structure of such arches is not transmitted directly, but through the auxiliary vertical members (links) at certain points – at nodes.

Three-hinged arches with a superstructure are generally regarded as statically determinate, and are complex systems in which an auxiliary part (overor under-arch superstructure) rests on the main part (threehinged arch).

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The analysis of systems for a moving load is carried out in the same way as for beams with nodal transfer of the load. Initially, the movement of a unit force F = 1 directly along the axis of the main arch is considered, and the influence lines for the factors under study are constructed. Then ordinates are fixed on these influence lines under the nodes. It can be proved that during load nodal transfer, sections of the influence lines between nodal points will be rectilinear. Therefore, if the ordinates fixed under the nodes are connected by straight lines, then the influence lines adjusted in this way will correspond to the influence lines for arches with a superstructure.

E x a m p l e. For a three-hinged arch with a under-arch superstructure (Figure 4.17, a), draw an influence line of the bending moment in section K.

Solution:

1. First, we construct the influence line for the bending moment M K*

(Inf. Line MK* ) as if the unit force F = 1 moved directly along the axis of

the three-hinged arch (Figure 4.17, b). To construct this influence line, we will use the arguments presented in Section 4.4:

Inf.Line M K* Inf.Line M K0 Inf.Line H yK .

The influence line for the bending moment MK0 arising in section K

of the equivalence beam is shown in Figure 4.17, c, and the influence line for the thrust H multiplied by yK is shown in Figure 4.17, d.

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this influence line under the nodal points 2, 3, 4 and 5. The ordinates under the nodal points 1 and 6 have zero values (Figure 4.17, f).

3. We connect the calculated ordinates with straight lines. The resulting graph is the influence line for the bending moment in the section K (Figure 4.17, f) under the condition that the load on the arch is transmitted through the over-arch superstructure.

4.7. Determining Support Reactions and Internal Forces

in Three-Hinged Frames

Consider the process of determining support reactions in a threehinged frame with supports at different levels (Figure 4.18).

The frame is loaded with a horizontal uniformly distributed load of intensity q = 4 kN/m and a vertical concentrated force F = 12 kN. The expected directions of support reactions are shown in Figure 4.18.

Figure 4.18

As usual, we compose the sum of the moments of all external forces relative to the support B:

M B 4 6 1 12 2 VA 8 H 4 0.

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Since the equation contains two unknown quantities VA and HA, we compose the second equation in the form of the sum of the moments of the left forces only, relative to the joint С:

MCleft 4 6 5 VA 4 H A 8 0.

The resulting equation includes the same two unknown quantities VA and HA. Solving the system of two joint equations, we find the values of the support reactions of the right support A:

Accordingly, the reactions of support B will be found from the sum of the moments of all external forces relative to the support A and the sum of the moments of only the right forces relative to the intermediate joint C:

M A 4 6 3 12 10 VB 8 HB 4 0,

MCright 12 6 VB 4 HB 4 0.

Solving the resulting system of two equations, we find

The calculated values of all supporting reactions are positive. Therefore, their directions shown in Figure 4.18 are valid.

We will check the results. We compose the sum of all forces projections on the X and Y axes, as well as the sum of all external forces moments relative to, let's say, the point D in the middle of the left rack (the moment from the distributed load and the moment from the reaction VA at this point are zero, which reduces the amount of calculations):

X 4 5 20 4 24 24 0,Y 10 22 12 22 22 0,

M D 20 3 12 10 22 8 4 1 180 180 0.

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All three checking equilibrium equations are satisfied identically. Summing up, we can recommend the following rules for calculating

support reactions in arbitrary three-hinged arches and other three-hinged systems with arbitrary external loads.

Usually, four equations are composed to calculate the four support reactions, and three more equations are used to verify the results.

The support reactions of the left support (VA, and HA) are calculated from two equations.

The first is the sum of the moments of all external forces relative to the right support B:

M B 0.

The second is the sum of the moments relative to the intermediate joint C of only external forces located to the left of the joint C:

MCleft 0.

The support reactions of the right support (VВ and HВ) are calculated from two more equations.

The third is the sum of all external forces moments relative to the right support A:

M A 0.

Fourth is the sum of the moments relative to the intermediate joint C of only external forces located to the right of the joint C:

MCright 0.

To verify the results, the sums of all external forces projections on the coordinate axes and the sum of all external forces moments relative to any point not previously used as a moment point are written.

After determining the support reactions, the diagrams of internal forces in the bars of the three-hinged frame are constructed, as in any other bars systems. The calculated support reactions are considered as

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known external forces. Internal forces are calculated according to general rules in given characteristic cross-sections. For the considered frame from cross-sections with nonzero bending moments, six characteristic sections have been selected (Figure 4.19): this is the beginning and end of each bar, the middle of the distributed load application segment.

Figure 4.19

We calculate the bending moments in the indicated sections:

left

M1 M1bottom 20 3 4 31.5 42 kN m.

left

M2 M2bottom 20 6 4 6 3 48 kN m .

M3 M3left 20 6 4 6 3 48 kN m .

M4 M4left 20 10 4 6 7 10 8 48 kN m .

M5 M5right (12 2) 24 kN m .

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The diagram of bending moments is plotted in Figure 4.20.

Figure 4.20

We begin the calculation of transversal and longitudinal forces from the support section A:

On a length A-2 of the bar, where a uniformly distributed load is applied, the transversal forces linearly decrease, and the longitudinal forces are constant. Therefore, we calculate:

Q2 20 4 6 4 kN , N2 N A 10 kN.

On an inclined bar section 3-4, the transversal and longitudinal forces are constant. The tangent of the angle of bar inclination to the horizon on this length is equal tg 4 / 8 0.5. Therefore sin 0.4472, and

cos 0.8944. Next we calculate:

Q3 Q4 10 0.8944 (20 4 6) 0.4472 10.73 kN ,

N3 N4 10 0.4472 (20 4 6) 0.8944 0.8944 kN.

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