KR101381115B1 - Steel pipe strut - Google Patents

Abstract

The present invention relates to a square steel tube brace to improve the performance as a brace by changing the radius of curvature of the corner rounding. The square steel tube brace using the curvature radius adjustment effect of the present invention has a square cross-section and rounded corners, and the radius of curvature R of the corner rounding part is 8 to 20t. According to the present invention, the curvature of the square steel tube using the curvature radius adjustment effect is to change the corner radius of curvature R of the square steel pipe based on the same weight as that of the standardized square steel pipe, thereby compressing the rectangular steel pipe against total buckling and local buckling. It can improve performance. By improving the compressive resistance performance of the square steel pipe, it is possible to reduce the working weight of the steel and to obtain economic effects.

Description

Square pipe support using curvature radius adjustment effect {STEEL PIPE STRUT}

The present invention relates to a square steel tube brace, and to a square steel tube brace to improve the performance as a brace by changing the radius of curvature of the corner rounding.

The strut is one of the supporting balls for supporting the retaining wall. The strut is a structure supporting the retaining wall by transferring the load acting on the wall to the opposite wall. Even if a variety of earth walls are used, when the support is used as the support, the belt is installed on the wall and the support is installed on the wall to connect in the same way as the opposite wall.

The middle pile is installed in the middle of the brace, and it has the function of adjusting the buckling length of the brace and supporting the mold beam to support the load of road, office, and equipment moving passage on the top of the retaining wall.

There are three types of jibo ball: strut, raker, tie-rod, earth anchor, and soil nail.

The brace is not limited to the depth of excavation compared to the raker, and it is easy to check the stress and deformation of materials by installing in the excavation surface, so it belongs to the easy method of safety management. It is the most widely used method because it is relatively easy to construct and does not depend on the ground conditions. However, when the excavation width is larger than 50m, it is unsuitable to use and has various disadvantages of the ground excavation.

Due to the characteristics of the H-beam, it has a weak axis and a weak axis. When it is used as a compression member such as a brace, it is disadvantageous to the buckling in the weak axis direction. For this purpose, a reinforcement such as a bracing should be designed in parallel. This reinforcement also has the advantage of interconnecting all the braces, but has the disadvantage that the dangerous work of the worker is installed directly up on the braces and the impact to other braces due to sudden weak axial destruction. In addition, the reinforcement acted as an obstacle to the work during the ground excavation and the construction of this structure, and caused various construction costs to increase the construction cost.

On the other hand, steel pipe brace has no structural division between steel shaft and weak shaft, so it has a structural cross section that is advantageous for buckling and torsion. It does not need horizontal and vertical bracing, so construction cost, construction period, and constructability (support / disassembly, ground excavation, construction of main structure) Due to its relative advantage over H-beams, it is used in most bracing forms overseas.

The radius of curvature of the corner rounding part 11 of the square steel pipe brace 10 of the steel pipe braces that are conventionally used as a brace in the earthenware temporary installation is 2 to 3 tons (t is the thickness of one side of the square steel brace) ). The radius of curvature of the rounded part does not take into account the optimum compressive resistance performance, and it is necessary to optimize the radius of curvature of the rounded part of the square steel brace in order to reduce the amount of steel used and to obtain economic effects by exhibiting the optimal compressive resistance performance. .

The present invention has been made to solve the conventional problems as described above, to provide a square steel tube brace using the curvature radius adjustment effect of optimizing the radius of curvature of the corner rounded portion to exhibit the optimal compression resistance performance as a brace.

The square steel tube brace using the present invention's curvature radius adjustment effect is a rectangular steel pipe with a rounded corner and rounded corners, and the radius of curvature R of the corner rounding part is 8 to 20t. Where t is the thickness of one side (mm) and R is the radius of curvature (mm)

The rounding radius of curvature R of the square steel brace is preferably 8 to 15 t.

According to the present invention, according to the square steel pipe support using the curvature radius adjustment effect, the compression resistance performance of the square steel pipe against total buckling and local buckling by changing the radius of curvature (R) of the rounded portion of the square steel pipe based on the same weight as the standardized square steel pipe Can improve. By improving the compressive resistance performance of the square steel pipe, it is possible to reduce the working weight of the steel and to obtain economic effects.

1 is a cross-sectional view of a square steel tube brace according to the prior art.
Figure 2 is a cross-sectional view of the square steel tube brace using the present invention curvature radius adjustment effect.
Figure 3 is a graph showing the axial force according to the radius of curvature when the length of the brace is 5m.
Figure 4 is a graph showing the axial force according to the radius of curvature when the length of the brace is 6m.
5 is a graph showing the axial force according to the radius of curvature when the length of the brace is 7m.
6 is a graph showing the axial force according to the radius of curvature when the length of the brace is 8m.
7 is a graph showing the axial force according to the radius of curvature when the length of the brace is 9m.
8 is a graph showing the axial force according to the radius of curvature when the length of the brace is 10m.
9 is a graph showing the axial force according to the radius of curvature when the length of the brace is 11m.

Hereinafter will be described in detail with reference to the accompanying drawings, a rectangular steel tube brace using the present invention curvature radius adjustment effect.

The square steel tube support 100 using the curvature radius adjustment effect of the present invention is formed as a square steel tube having a square cross section. At this time, the four corners of the square steel pipe is rounded, the radius of curvature (R) of each rounding portion 110 can improve the compressive resistance performance of the square steel pipe for full buckling and local buckling when 8 ~ 20t. Here, t is the thickness (mm) of one side of the square steel pipe.

The design process of the square steel tube support 100 of the present invention goes through the process of checking the ground survey data, obstacles, etc. in the excavation area, and is assumed to apply the temporary program to calculate the deformation and axial force and moment of the wall according to the excavation step by step Examine the stability of one section.

The brace is to be considered as a flexural compressive member having axial force and self-weight transmitted from the band and, in some cases, moments due to loading.

The process of examining the cross section of the square steel brace is as follows: ① section force calculation, ② working stress calculation, ③ allowable stress calculation, and ④ stress review.

① Section force calculation

The design compressive force (P _max ) is the sum of the axial force (R _max ) and the temperature load (T) due to earth pressure.

That is, P _max = R _max + T... ... ... ... ... ... Equation 1

The design bending moment (M _max ) is calculated from the sum of the work loads (W) and the length of the support beams (L).

That ^{_{is, M max = W × L 2}} /8 ... ... ... ... ... ... Equation 2

② Calculation of working stress

Stress (f _b ) is the design bending moment (M _max ) divided by the section modulus (Z _x ).

I.e. f _b = M _max / Z _x . ... ... ... ... ... Equation 3

Compressive stress (f _c ) is the design compression force (P _max ) divided by the cross-sectional area (A).

That is, f _C = P _max / A. ... ... ... ... ... Equation 4

③ Calculation of allowable stress

Allowable stress calculation is applied when the plate thickness of one side of the square steel brace is less than 40mm. The allowable stress increase factor of the temporary structure is 1.5 (50% premium is applied to the allowable stress specified in the general specifications for temporary cladding structures), and the allowable stress correction factor for steel is 0.9 (allowed stress considering steel reuse and corrosion). Reduction factor).

Permissible axial compressive stress (f _ca ) is multiplied by permissible axial compressive stress (f _cag ) not considering local buckling and permissible stress (f _cal ) for local buckling. The upper limit of stress (f _cao ) divided by

That is, f _ca = f _cag f _{cal /} f _cao … ... ... ... ... ... Equation 5

It allows bending compression stress (f _ba) is derived from the relationship between the allowable stress (f _cal) and the upper limit value of the allowable bending compression stress to the local buckling (f _bao). If transverse buckling is difficult to occur, such as a box-shaped cross section, the allowable bending compressive stress cannot be greater than the allowable stress value for local buckling.

That is, f _ba = min. (F _cal , f _bao ). ... ... ... ... ... Equation 6

Euler's buckling stress (f _ea ) acts on long braces, such as jangju.

f _ea = 1.5 × 0.9 × 1,2000,000 ÷ (L / R) ² . ... ... ... ... ... Formula 7

Where L is the effective buckling length (cm) and R is the total cross-sectional secondary radius (cm) of the member.

④ Stress Review

If the allowable axial compressive stress (f _ca ) is greater than the axial compressive stress, the design criterion is considered. If the allowable flexural compressive stress (f _ba ) is greater than the flexural compressive stress, the design criterion is satisfied.

If the combined stress of the brace that receives both the axial direction and the bending moment is examined, it is suitable for the design criteria.

That is, combined stress = f _ba / f _ba + f _ba / (f _ba x (1- (f _ba / f _ba ))) <1.0 ... ... ... ... ... Equation 8

[Example]

Calculate the design axial force while changing the curvature radius (R) of the corner rounding part and the length of the brace of the square steel tube STKT590 with the same load and cross-sectional area and a square cross-section of 9 mm on each side. It was. The design axial force was calculated using Equations 1 to 8 above when setting the combined stress to 1.0. At this time, the spacing of the braces was 5.5 m.

R
W
A
B × H × t
Compression Resistance Performance (STKT590) 5m 6m 7m 8m 9m 10m 11m 2t 94.7 120.67 350 × 350 × 9 214.9 192.1 168.8 145.3 122.3 98.0 78.1 5t 94.7 120.67 361.59 × 361.59 × 9 297.5 268.6 238.6 208.0 177.8 145.5 118.9 8t 94.7 120.67 373.18 × 373.18 × 9 337.0 305.3 272.2 238.3 204.4 168.7 138.5 9t 94.7 120.67 377.04 × 377.04 × 9 337.8 306.3 273.5 239.7 205.9 170.3 140.0 10t 94.7 120.67 380.90 × 380.90 × 9 337.7 306.2 273.3 239.5 205.6 170.0 139.7 11t 94.7 120.67 384.77 × 384.77 × 9 337.7 306.2 273.3 239.5 205.6 170.0 139.7 12t 94.7 120.67 388.63 × 388.63 × 9 337.5 305.8 272.8 238.9 204.9 169.3 139.0 15 t 94.7 120.67 400.22 × 400.22 × 9 337.0 305.2 271.9 237.8 203.6 167.9 137.6 20 t 94.7 120.67 419.53 × 419.53 × 9 332.8 299.9 265.4 230.1 195.0 158.8 129.0

R: Angular Curvature Radius (mm)

W: weight (kg / m)

A: cross-sectional area (cm 2)

B × H × t: Square steel pipe specification (mm)

Referring to [Table 1] showing the results, the design axial force value according to the radius of curvature for each length of the support beam is shown in the form of a graph in FIGS.

The X axis of each graph represents the radius of curvature R (mm) of the brace and the Y axis represents the axial force (ton) of the brace. Here, t is 9 mm in thickness of each side of the brace.

As shown in each graph, when the difference between the maximum design force value and the minimum design force value is 1, the percentage of the design force of the brace has more than 85% when the radius of curvature R is in the range of 8 to 20t. When it is 8 ~ 15t, the percentage of design force of props is over 95%.

When the radius of curvature R is 5t or less, the percentage of design axial force for each length of the brace is within 67%, 66.9%, 66.6%, 66.4%, 66.3%, 65.6%, and 65.9%, respectively. Therefore, the design axial force difference is more than 18% compared with the curvature radius R of 8 to 20t.

This difference leads to a large difference in the stability of the structure. Therefore, in order to exhibit the function as the optimum brace, it is preferable to use a square steel pipe brace having a radius of curvature R of 8 to 20t, and more preferably to use a square steel brace that has a radius of curvature R of 8 to 15t. Do.

This technical feature is prominent when using high strength steel as a square steel tube brace.

While the present invention has been particularly shown and described with reference to exemplary embodiments thereof, it is to be understood that the invention is not limited to the disclosed embodiments, but, on the contrary, is intended to cover various modifications and equivalent arrangements included within the spirit and scope of the appended claims.

10, 110: square steel pipe brace 11, 110: corner rounding

Claims (2)

Square steel pipe with rounded corners and rounded corners, and the radius of curvature of the rounded corners using the curvature radius adjustment effect, characterized in that the curvature radius (R) of 8 ~ 20t.
Where t is the thickness of one side (mm) and R is the radius of curvature (mm) delete

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