WO2010081973A2 - Novel approach for non-accessible large-span roofing - Google Patents

Abstract

The invention relates to a novel approach for making non-accessible large-span roofing. The problem to be solved by the invention is to develop a classical static frame (three-hinge gantry operating under flexure stress) with mixed operation in which axial forces (compression forces) are also significant. This implies that the two members of the gantry are arc-shaped and also includes adaptation measures, which are impossible as the arc will quickly have to operate as any other beam under flexion. Another drawback relates to the realisation of arcs in general, which remain money and time consuming due to their geometric complexity and do not respond to modularity and prefabrication requirements, unlike encompassed by the present invention. Considering these two problems, the solution to the first one is to conjugate the two aspects of the arc (its potential to resist axial forces and the potential of its section to resist flexion). The arc will be deployed before applying external loads so as to balance them, thereby resulting in a prestressed arc aspect or so-called hunter's arc. Considering the second problem, the realisation of the arc structures can be simplified by the implementation of an innovative geometry as well as by the modulation, prefabrication and even the access to areas not accessible before, i.e. pre-engineered (ready-to-use structures).

Description

 New approach to non-accessible roofs of large spans

Description Technical Area

Belonging to the process technology in the arc-shaped structure, this invention claims to extract from a new static frame (on one side) and adapted manipulation of its constituents (on the other) unexploited potentials so far in terms of economy and reliability In its other part, this invention addresses a new approach to the execution of arc-shaped structures, an approach that fits in the logic modulation and prefabrication, thus breaking with the disadvantage that makes this type of structures - by their geometrical complexities - more difficult in terms of money and time of execution.

PRIOR ART Introduction It should be remembered that the operation of an arc (characterized by the transformation of moments into axial forces) relies on restricting the displacement of each of its two ends, totally (against the foundations), or partially (by cables) whose elongation gives rise to secondary moments. Being of secondary order, interest in such moments remained, so far, as well; few quantitative studies or qualitative assessments of the role that these moments are likely to play have been undertaken (at least not in the sense of this invention)

Presentation of the invention A - The conceptual aspect (A - 1 The static frame A - 2 The interactivity) A - 1 The static frame (the three - jointed arc):

This term "three joints" automatically refers to its source "three-hinged gantries" where the transport of loads is through the moments of sag; yet the presentation will demonstrate that in our we remain well in the logic of arc to convey the charges through the axial forces.

This frame is the union of two segments of arcs, comparable to the bow of the hunter (in the sense that they are exposed before the application of the external charges to the initial tension of their strings), and thus generating a reservoir of compensatory moments, comparable to those counterweights of construction cranes give rise to their vertical elements (towers); but unlike the latter (which are fixed in intensity) these moments are reactive and adapt, positively, in intensity as in sense of application, to the different states of the charges, throughout the life of the structure. It's two segments are two joints each; they are united through their common articulation at the top of the structure (the third that we will call it "floating"); the remaining two rely on the foundations (see static aspects, figure 1)

It must be emphasized that to stay in the logic of the bow you will have to have your strings in perpetual state of tension; no matter how intense, however, with a certain margin of safety.

It should be emphasized that this state of tension is the indicator that tells us that the arc is still functioning as such, and in our case this has been ensured by two measures: The first is introduced through the initial tension to the strings to create the compensatory moment (mentioned above) but also so that the strings gain in statue of tension whose release (which puts the stability of the structure in question) will be in no case tolerable.

The second is to play with the interruption of the continuity between the two segments (at their point of meeting) so that the application of external loads pushes the floating joint upwards thus accentuating the initial tension of the strings; by moving them further away from their point of relaxation.

These two measures combined together, they play on the quantitative side, and this leads us to the interactivity (elaborated below) which addresses the behavioral side of our structure A - 2 Interactivity

A - 2 - 1 the spring effect analysis

It is a question of assimilating the body of the arc segment on one side, and its strings on the other, with two interactive springs, which offer, through a manipulation adapted to their level of their constant, the optimal condition in the meaning of minimizing the demand of these segments in terms of compensatory moments; and since the intensity of these is directly related to the floating of the articulation (under different states of charge) this condition is translated - as the next paragraph explains - by the equidistance of the floating above and below the hypothetical neutral position (to be defined as the place where this articulation would occupy if the strings did not exist)

Theoretical elaboration (see diagram of the floating figure 3)

Legend Kc The spring constant (strings).

Ka The constant of the spring (arc).

Pt Total Charges = Permanent Pp

+ Ps service

Nc (T) t The projection, on the axis of the strings T, of the axial forces Nc under the total loads = Permanent Nc (Op

+ service Nc (T) s)

TO The initial tension at the strings, chosen arbitrarily equal to Nc (T) p.

RO reaction of the bow to the initial powering of the ropes; RO = TO Tt Tension in the strings under the total loads Pt = TO + Tp + Ts.

Rp The reaction of the arc due to the intervention of Pp

Rt The reaction of the arc due to the intervention of Pt w the elongation of the ropes under the initial tension; (distance not referenced) x The difference between the floating joint and its neutral position under TO y The difference between the floating joint and its neutral position

95 under Pp z The gap of the floating joint on the other side of its neutral position under

Pt α Coefficient of charge distribution α = Ps / Pp = Nc (T) s / Nc (T) p

Equilibrium equations

100 The phase of the initial tension (before the application of external loads) The distance of the floating joint from its neutral position = x; TO = RO; it results w = Tp / Kc = Rp / Kc x = Tp / Kα = Rp / Kα (relative to the neutral position)

The phase of the application of permanent loads Pp

105 The gap of the floating joint from its neutral position is reduced from x to y, and the moments at the segment are also reduced, but their effects remain in alliance with the charges Pp against the strings.

Tp = Nc (T) p + Rp, Kc {w + (x-y)} = Nc (T) p + Ka * y; and since w = Fp / Kc

110 y = Kc * x / (Kc + Ka) (relative to the neutral position)

The phase of the application of total charges Pt

The floating articulation passes on the other side of its neutral position, at a distance from that ci = z, the moments change of sign and alliance and associate with the ropes against the total charges Pt; The result is 115 Nc (T) t = Rt + Tt

Nc (T) p + Nc (T) S = Ka * z + Kc (w + x + z) = (Kα + Kc) z + Kc.w + Kc.x)

= (Kα + Kc) z + Nc (t) p + Kc * x 120 Nc (T) S = (Kα + Kc) z + Kc * xz = (Nc (T) S - Kc * x) / (Kα + kc)

For y = z (equidistant floating condition around the neutral position)

Kc * x = (Nc (T) s - Kc * x)

Nc (T) S = 2 Kc * x = 2 Kc * Nc (T) p / Kα;

125 Nc (T) S / Nc (T) p = Ps / Pp = 2 Kc / Kα

Ps / Pp = α = 2 Kc / Kα

It should be emphasized that in the analysis of the spring effect only the forces liable to generate moments of deflection are taken into account, therefore the compressibility of the arc under the axial forces was excluded from our

130 analysis.

This compressibility has only the effect of a general slip of the coordinates of these flutations, without incident on the values of these

A - 2 - 2 The virtual approach to calculates

In my analyzes I started from a symmetrical case of loads; this is almost always true when with permanent loads in this type of structure, but far from being the case as regards the service charges (a combination wind and snow for example, or under loads of dynamic type, earthquake, etc.) where the asymmetry of the charges, on each of its two blanks, is essential, and it is to this critical aspect (arising from the non-equilibrium of the horizontal forces at the level of the floating articulation 140) that it is necessary to face

This answer I propose under a case of virtual symmetry which produces the same elongation of the strings as the one subjected to the blankest loaded of the two, under the case of real asymmetry; with all that this entails in terms of other load distribution coefficient, and other constant springs. (see example 145 explanatory, further) B - The execution aspect

Far from any theoretical aspect and even independently of our framework, it is a question here of rationalizing the execution of the arcs in general; in our case the segment is the main repetitive module; it is the expression of a continuous function which reflects the state of the external charges (for simplicity our example has adopted a second-degree function); it is of equilateral triangular section variable (maximum in the center and minimum in the level of its articulations); it is in the form of a lattice of which the three supporting components (the longitudinal elements located at each of its three angles) are of closed section (tubular)

155 B - 1 Curvature of linear constituents B -1 - 1 The theoretical deduction of functions

Starting from the chosen parameters of our arc segment, and from the degree of its continuous function (which is the expression of the geometric locus of its neutral axis), and from the choice of section (a triangular lattice), this gives us all that it is necessary to deduce 160 the equations of the two types of functions concerned

yl which inclines the vertical plane of an angle μ and which defines the two upper load cells. y2 which lies in a vertical plane and defines the lower bearing element so as to cover with the other two the requirement of an equilateral section 165 (see geometrical aspects;

B -1 - 2 The geometry of the three straight bearing elements

It is known that the integral double of the diagram of moments, which are exerted on an element, gives the state of its deformation under the loads which occasioned these moments, so it suffices in our case to expose the rectilinear element ( the tube) to

170 two moments at each of their two ends to produce - by M = cte - a function of deformation of second degree, j This artificial setting in geometry must not pose a problem nor damage the tubes, since the moments concerned are sufficiently low, considering the coefficient L / Φ of each of them is (in our case equal to 400) and the low 175 value of the arrow concerned (in our case is 2.5 m). Once the tubes are in place and in geometry it will be time to proceed to the triangulation which will be of two types, rigid simple and flexible double (crossed).

This approach is, in a way, comparable to the Lamellecollé process, where it is relatively easy to bend the non-jointed wooden planks, before the grip of the neck (in this case before triangulation) to obtain a much more resistant section after

B - 2 The pre-engineered aspect

Frames made of pre-enginnered components (ready-to-wear) obviously do not make a novelty, yet one can imagine that those in lattice, and of this type 185 of geometry and variable sections, and surplus made from components rectilinear, can make one; especially when one adds on all this the logic of modulation and prefabrication.

B - 2 - 1 Modulation

Starting from the largest to the smallest our modulation begins with segment 190 (as one of two identical components of the three-jointed arc) which in turn is divided into three sub-modules of two different types; central, 12 m in length (the industry standard in the manufacture of tubes); and peripheral on each side of a complementary length (according to the recommended production range, which can go up to 12 m each, thus totaling a length 195 maximum of 36 m for each segment, and which can cover (with the conjunction of two) a span of about 65 m.

This being so, it is not impossible to exclude, of course, custom-made in particular cases of solicitation, or when the maximum span exceeds the mentioned limit of 65 m.

200 B - 2 - 2 Prefabrication

Under the pre-engineered range prefabrication covers

• Support tubes of two types (central and peripheral) severed (when less than 12 m in length), perforated, and prepared to receive the nodes, and to join together (in order to meet the requirement of the total length of the segment. • The parts of the rigid triangulation (the tubes which delimit the transverse equilateral triangles) and the flexible triangulation (the strings which restrict the nodes in the longitudinal direction, see detail node figure 4)

• Accessories and these are the components of the nodes (designed to receive rigid and flexible rigid and rigid support tubes and triangulation); the components of the peripheral elements, (designed to ensure the structural behavior of these articulations and to receive the strings of the arcs (see geometrical aspects figure 2, details node figure 4, peripheral elements figure 5)

215 Encrypted and simplified example

Geometric determinants

δ = 2.50 m; AC - 30.0 m; S (overbasing) - δ / AC = 1/12; tg φ = 4 * S = 1/3 h = 13.33 m; λ = 26.66 m; tg θ = 2h / = = 1/2 d = 80 cm;

220 h = d * sin 60 ° = 69 cm; h / 3 = 23 cm; 2h / 3 = 46 cm cotg μ = 6.825

Sin φ = 0.44; sin φ = 0.32; sin (θ + φ) = 0.71; sin (θ - φ) = 0.13 cos θ = 0.89; cos φ = 0.95; cos (θ + φ) = 0.71; cos (θ - φ) = 0.98

ylH: The geometric equation of the horizontal projection of yl 225 ylV: The geometric equation of the vertical projection of yl

ylH = - 0.00195 x ² + d / 2 ylV = - ylH cotg μ y2 = ylV - / 3 ylH (The condition of the equilateral triangle) 230 y2 = - 0.0121 X ² + [(2.50 + 0.23) -

(/ 3 d / 3 = 69)]

yl = - 0.0123 x ² + 2.76 y 2 = - 0.0090 x ² + 2.04

The static calculation 235 External loads

Permanent Pp = 0.4 t / m ¹ On duty Ps = 0.6 t / m 'Total Pt = 1.0 t / m ¹ ; α (real) = (Pt - Pp) / Pp = Ps / Pp = 1.5

Symmetrical charges (the general case); in tons

240 Table 1

Asymmetrical loads

Table 2

L'articulation « C » est exposée à une poussée = Hcl-Hc2 ; en tonnes ; les données critiques sont Na = 45,06 t ;

The "C" joint is exposed to a thrust = Hcl-Hc2; in tons; critical data is Na = 45.06 t;

245 Nc (T) I + (HcI - Hc2) / 2 cos θ = 9.55 + 2.19

= 11.74 t (voltage)

Nc (T) 2 - (HcI - Hc2) / 2 cos θ = 6.37 - 2.19 = 04.18 t (voltage) The critical calculation will be based on a virtual symmetric load distribution that produces a Nc load ( T) t = 11.47 t

Therefore α (virtual) = (Pt - Pρ) / Pρ = Ps / Pp = Nc (T) virtual S / Nc (T) p α (virtual) = 11.74 - 3,18 / 3,18 = 2, 69

Calculates sections

In this calculating step it is realized that on the value of Na (of 45.06 1) one must add the charge arising from the tension of the cords against both ends of the arc segment; on the other side we are also aware that the value of Nc (T) t we must subtract the contribution of the arc segment in terms of compensatory moments, so our pre-dimensioning must take into account.

From the steel quality, for the tubes σ (tubes) = 1.8 t / cm ² ; and that of the strings σ (strings) = 15 t / cm ² 260 and for the critical loads listed above we arrive at

The tubes: 3 Φ 75 mm; A - 3 X 9.42 cm ² ; t = 4 mm; A total = 28.26 cm ² The strings: 4 Φ 5 mm; A - 4 X 0.20 cm ²

A total = 0.80 cm ²

Determine the constants of the springs

265 The constant of the spring strings Kc = 1 / d €; where d € under a unit load = IXt / EA;

Kc = 0.58 From the equation α (virtual) = 2Kc / Kα we deduce the value of the other constant

Ka = 0.43

The question is now reduced to determining the inertia of an equilateral triangular arc (whose section of its three tubular carrying elements is predetermined), a conventional calculation of displacement of one end relative to the other under a unitary force. applied on the axis of the strings and balanced by the resistant moment of the arc (comparable to a spring whose constant Ka = 0.43) 275 1 / Ka = (UE) JS M dx / I; (I being variable is part of the integral);

I mαx = 28,200 cm4; h / 3 = 23 cm; h = 69 cm; the side of the triangle d = 80 cm

Asymmetric loads Critical blank Case 0 (external loads = 0); Nc (T) O = O t; TO = 3.18 t

280 w = Tp / Kc; 3.18 / 0.58 = -5.48 cm; x = Tp / Kα; 3.18 / 0.43 = - 7.40 cm NaO = TO + 0 = 3.18 t

MO = (x) Kα * δ = - 7.4 * 0.43 * 2.5 = -

7.95 t.m

The case Pp (external loads = Pp); Nc (T) p = 3.18 t; Tp = 5.00 t 285 Y = Kc.x / (Kα + Kc) = 0.58 (7.40) / l, 01 =

- 4.23 centimeters

Nα, p = Tp + Nα, p = 5.00 + 15.02

20.02 t

Mp = (y) Kα * δ = - 4.23 * 0.43 * 2.50 =

290 - 4.55 t.m

The case Pt (the external charges = Pt); Nc (T) t = 11.74 t; Tt = 9.92 t z = [Nc (T) s-Kc.x] / (Kα + Kc) = - y = + 4.23 cm (above the neutral position)

Na, t = Tt + Na, t = 9.92 + 45.06 t = 55.0 t

Mt = (z) Ka * δ = 4.23 X 0.43 X 2.5 = + 4.55 tm (M changes sign) 295 Table 3

The position of ¹¹ C "Arc Strings w = 5.48 cm Na Moments

Case 0 x = - 7,40 cm TO = 3,18 t ^r NaP = 3,18 t ¹ - 7,95 tm

Case Pp y = - 4.23 cm Tp =: 5.00 t Na, p = 20, O2t i - 4.55 t.m

Case Pt Z = - y = + 4.23 cm Tt =: 9.92 t i Na, t = 55.0 t! + 4.55 t.m

As the table of the solicitation (hereafter) and the state of the compensatory moments (see diagram figure 6) the critical demand in term of compression σ = (1,95 + l, 49) / 2 + 0,37 = 2, 09 t / cm ²

Table 4

300 The non-critical blank

Case 0 (external loads = 0); Nc (T) O = O t; TO = 3.18 t w = Tp / Kc; 3.18 / 0.58 = -5.48 cm; x = Tp / Kα; 3.18 / 0.43 = - 7.40 cm NaO = TO + 0 = 3.18 t

MO = (x) Ka * δ 7.4 * = 0.43 * 2.5 = 7.95 t.m

305 The case Pp (external loads = Pp); Nc (T) p = 3.18 t; Tp = 5.00 t Y = Kc * x / (Kα + Kc) = 0.58 (7.40) / 1.01 = - 4.23 cm Nα, p = Tp + Nα, p = 5.00 + 15.02 = 20.02 t

Mp = (y) Ka * δ = 4.23 * 0.43 * 2.50 = - 4.55 t.m

The case Pt (the external charges = Pt); Nc (T) t = 4.18 t; Tt = 6.22 t 310 z = Nc (T) S - Kc * x / (Kα + Kc) z = - 2.16 cm Nα, t = Tt + Nα, t = 6.22 + 30.22 = 36 , 44 t Mt = (z) Kα * δ = 2.16 * 0.43 * 2.50 = - 2.32 tm

Table 5

The symmetrical case

Case 0 (external loads = 0); Nc (T) O = O t; TO = 3.18 t w = Tp / Kc; 3.18 / 0.58 = -5.48 cm; x = Tp / Kα; 3.18 / 0.43 = - 7.40 cm NaO = TO + 0 = 3.18 t

MO = (x) Kα X δ = 7.4 * 0.43 * 2.5 = 7.95 t.m

The case Pp (external loads = Pp); Nc (T) p = 3.18 t; Tp = 5.00 t 320 Y = Kc.x / (Kα + Kc) = 0.58 (7.40) / 1.01 = - 4.23 cm

Nα, p = Tp + Nα, p = 5.00 + 15.02 = 20.02 t

Mp = (y) Kα X δ = - 4.23 * 0.43 * 2.50 = - 4.55 t.m

The case Pt (the external charges = Pt); Nc (T) t = 7.96 t; Tt = 7.75 t z = Nc (T) S-Kc.x / (Kα + Kc) = 0.48 cm 325 Nα, t = Tt + Nα, t = 7.75 + 37.55 = 20.02 t

Mt = (z) Kα X δ = 0.48 * 0.43 * 2.50 = + 0.52 t.m

Table 6

! The position of "C" Arc Strings

, W = 5.48 cm Na Moments

Case 0 X = - 7,40 cm TO = 3,18 t Na = 03,18 t - ^~ 7 £ 5 ^~ t7m_j

Case Pp i y = - 4.23 cm Tp = 5.00 t Na = 20.02 t - 4.55 t.m

Pt case! z = - * 0.48 cm Tt = 7.75 t Na = 45.30 t + 0.52 tm Brief description of the drawings

These are three types of drawings; static, geometric and theoretical diagrams (Figures 1, 2 and 3); details of execution (figures 4 and 5)

(Figure 1) Static aspects defining the static frame of the floating joint and the general equations of its equilibrium

(Figure 2) Geometric aspects

Defining the geometrical frame of the arc segment and the equations of the coordinates of the nodes

(Figure 3) Floating diagram

In relation to the interactivity between the body of the arc segment and its strings (both assimilated to two springs), defining the floating equations in general and thus the condition of the equidistant floating below and above the neutral position.

(Figure 4) Node detail

One way to execute it.

(Figure 5) Peripheral elements

Details 1 (gyroscopic seal). Detail 2 ((foundation level joint).

One way to execute them.

(Figure 6) The data of the example

The stress diagram under the critical asymmetric case

Claims

 claims 1) Arc structure characterized by its new static frame which is a development of a traditional static frame (portal with three joints) in order to make it possible to operate its two sides (in this case two segments of arcs) under forces axial instead of bending moments; this static frame is the union of two segments of exposed arcs, before the application of the external loads (as is the case of the bow of the hunter), to the initial tension adapted of their strings (this tension will vary according to the variation of the external loads but it must always be positive (thus in perpetual state of tension as it is the case with a prestressed element); each segment presenting two articulations, and the two segments are connected together by their joint joint at the top of the structure, called floating joint (C), the two other joints (A, B) remaining supported on the foundations, (Figure 1) 2) Arc structure according to claim 1 characterized in that the two arcs segments are lattice-shaped and of equilateral triangular section, of varying sides, maximum in the center and minimum in the level of its joints, (Figure 2) 3) Arc structure according to claim 2, characterized by the transposition of the geometry setting, used in the field of glued laminated structures, that of metal structures, where non-solidary wood blades keep their final geometric disposition to through the glue while (in our case) the metallic profiles do the same through the triangulation 4) Arched structure according to claim 2 characterized by the particularity of the triangulation of our lattice (rigid single in the transverse direction and flexible crossed in the longitudinal direction, Figure 2 and 4) 5) Arc structure according to claim 2 characterized in that the nodes of this lattice are in this way to meet its geometric specificity (Figure 4) 6) Arc structure according to claim 2 characterized in that the peripheral elements of this lattice are in this way to meet its geometric specificity (Figure 5) 7) Arc structure according to claim 2 characterized in that the main module (each of the two segments) consists of three sub-modules of two different types, a central 12 m long, and two peripherals on each side and of a complementary length (according to the span to be covered and which can go up to 12m of each side), totaling thus a rectilinear length of 36 m; whose combination of two can cover (using standard components of 12 m) a span on the order of 65 m.