Structural mechanics
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Figure 6.2
The obtained dependence coincides with the corresponding dependence of a simple beam of a span of l. Therefore, the influence line of the
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support reaction RA is constructed as in a simple beam (Figure 6.2, b). Similarly, we obtain the dependence for the vertical component of the
reaction of the support B:
RB xlF .
The influence line for the reaction RB is shown in Figure 6.2, c.
The horizontal component of the support reactions, i.e. thrust, may be defined, as in a three-hinged arch, according to the formula:
M 0
H fC .
Therefore, the influence line for the thrust is the influence line for the beam bending moment in the beam cross section located under the intermediate hinge of the arch truss, taken with a coefficient 1/ f (Figu-
re 6.2, d).
The internal force N1 may be calculated using section I–I and mo-
ment point 1 (Figure 6.2, a).
If the unit force is located to the left of the section, then, considering the equilibrium of the right part of the truss, we get:
M1right 0; RB 8d H 4a N1a 0; N1 4H 8ad RB.
It means that
Inf .Line N1 (Inf .Line H ) 4 (Inf .Line RB ) 8ad .
When the unit force moves to the right of the dissected panel, from the equation of equilibrium of the left forces we find:
M1left 0; |
RA 2d H 4a N1a 0; |
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It means that
Inf .Line N1 (Inf .Line H ) 4 (Inf .Line RA) 2ad .
In the length of the dissected panel, we draw a transition line. The influence line for the internal force N1 is shown in Figure 6.2, d.
6.2. Calculation of Combined Systems
Structural systems, some of the elements of which work on bending, shear and tension-compression, and the other part only on tensioncompression, are called combined systems. Such systems, for example, include: a beam with a hinged arch (Figure 6.3, a), three-hinged systems (arches¸ frames) with ties of various kinds (Figures 6.3, b, c, d), a beam with a hinged chain (Figure 6.4, a), a suspension hinged chain with a stiffening beam (Figure 6.5) and many others.
Figure 6.3
Features of the combined systems calculation will be discussed below on the examples of the calculation of a beam with a hinged chain (Figure 6.4, a) and a suspension system. (Figure 6.5, a)
6.3. Calculation of a beam with a hinged chain
A geometrically unchangeable and statically determinate beam with a hinged chain (Figure 6.4, a) is a structure, where the horizontal bars AC
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and CB connected by an intermediate hinge are strengthened by a polygonal hinged chain with vertical struts.
Figure 6.4
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The horizontal reaction of support A is zero under any vertical load. Vertical support reactions caused by a given load, we find from the
equilibrium equations of the entire system:
M A 0; |
RB 5d q 2d d 0; |
RB 0.4 q d; |
M B 0; |
RA5d q 2d 4d 0; |
RA 1.6 q d. |
We begin the calculation of internal forces by determining the force H in the rod 4-6 of the hinged chain. To do this, we draw section I–I through the named rod and hinge C. Considering the equilibrium of the right part, we obtain
MCright 0; |
RB 2.5d Hh 0; |
H RB |
2.5d |
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Then the internal forces in the rods of the chain and in the struts can be determined as in the rods of any truss (Figure 6.4, b).
After determining the forces in the elements of the chain and in the struts, the horizontal bars are calculated on the action of a given load and the forces transmitted by the members of the strengthening system (Figure 6.4, d), like a simple beam. We recommend that the reader perform the corresponding calculations on their own.
Diagrams of internal forces are shown in Figure 6.4, e, f, g.
6.4. Calculation of a suspension system
The features of the influence lines construction for internal forces in the elements of combined systems can be considered using an example of a suspension system such as a hinged chain with a stiffening beam (Figure 6.5, a).
The procedure for determining the forces in the elements of this system is as follows.
To find the support reactions from the action of the load applied to the stiffening beam, the hinged chain must be cut at the points A and B located vertically above the supports A and B (Figure 6.5, a). The longitudinal forces in the cut rods can be decomposed into horizontal and
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librium equations of the lower part of the system in the form of sums of
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nents R |
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1x R l 0; R |
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From the equations of equilibrium of the hinged chain nodes at the junctions of the vertical suspensions (Figure 6.6, b, c, d) or of a fragment (Figure 6.6, a) it follows that the horizontal component of the longitudinal forces in the chain elements is constant and equal to the thrust of system H.
To find the thrust H, we draw section II–II, passing through the hinge C and the horizontal chain rod (Figure 6.5, a). Having compiled the sum of the moments of forces relative to the hinge C for one of the parts of the system, for example, for the left, we get:
MCleft 0;
Or, considering that
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Hh H (h f ) 0. |
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RA 2l 1 2l x MC0 ,
get the formula for determining the horizontal component H
M 0
H fC .
(6.3)
(6.4)
From the conditions for the expansion of the longitudinal force in the chain element at the point A (Figure 6.6, a), we find the vertical com-
ponent V :
A
V H tg .
A 3
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Figure 6.5
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Figure 6.6
Similarly, the component V is determined. |
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After that we find the support reactions VA and VB: |
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V |
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R |
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With the known horizontal component H, the total forces in the chain elements will be equal
Ni |
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Suspension forces are determined from the equilibrium equations of the nodes (Figures 6.6, b, c, d).
To determine the internal forces in the section K of the beam, we draw a strictly vertical section through K and consider the equilibrium of the left part (Figure 6.7).
We decompose the longitudinal force in the cut chain element into the horizontal and vertical components H and V1. The bending moment and the transverse force in the cross section K will be equal to:
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x) H (h f ) H (h f y |
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x) HyK M K0 |
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RAxK 1(xK |
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V ) F H tg |
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where M K0 and QK0 are the bending moment and the transverse force in
the corresponding section of a simple two-support beam having the same span and the same load as the system under consideration.
Figure 6.7
Based on the obtained dependencies for determining the support reactions and efforts, it is possible to construct the necessary lines of influ-
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So, |
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formulas (6.1) and (6.2), we build the influence lines |
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(Figure 6.5, b) and R |
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(Figure 6.5, c). As for a |
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simple beam, the influence line for the beam bending moment MC0 is
constructed (Figure 6.5, d). According to the formula (6.4), the influence line for the component H is built (Figure 6.5, e), and on the basis of (6.5) the influence line for the reaction VA (Figure 6.5, e) is constructed.
According to formulas (6.6) and (6.7), the influence lines of the bending moment M K (Figure 6.5, g) and the transverse force QK (Figu-
re 6.5, h) are plotted.
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THEME 7. BASIC THEOREMS OF STRUCTURAL MECHANICS
AND DETERMINATION OF DISPLACEMENTS
7.1. Bars Systems Displacements. General Information
When the load is applied to a structure (we will denote this factor by F), when the temperature changes ( t ) or the supports are displaced
(c), linear deflections of its points and the angles of rotation of its cross-
sections appear.
In Figure 7.1 the solid line shows the initial state (before the external load applied) of the frame elements, the dashed line shows the state after loading (deformed state). The cross-section K has moved to the position K1. The angle describes the rotation of the cross-section, the section
KK1 (not shown in the diagram) describes the linear displacements of the cross-section K.
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Figure 7.1 |
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The linear displacement of the cross-section K in a direction that does not coincide with the true one can be determined by finding the projection of the segment KK1 on this direction. In engineering calcula-
tions, the displacements of the cross-section in the vertical and horizontal directions are often determined.
The displacements are determined by checking the rigidity of structures, by calculating them for stability and vibrations, and also by calculating statically indeterminate systems.
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