RU2486972C2 - Method of rolling double-tee from low-alloy steel - Google Patents

Abstract

FIELD: process engineering.

SUBSTANCE: invention relates to metallurgy. Double-tee section is made from low-alloy steel and has flanges with parallel faces and wall. Optimum distribution of steel over beam cross-section is ensured by section flexibility equal to wall height-to-wall thickness ratio does not exceed limit flexibility λ_{wall lim} that ensures wall stability without intermediate stiffness ribs and equals 65 while wall thickness, height and limit flexibility are specified by mathematical relations. Note here that section cross-section is distributed as follows: 50% - section wall, 25% - each flange. Besides, wall width-to-wall thickness does not exceed limit ratio that ensures local flange stability to add to strength That is, moment of resistance W_x cm³ and stiffness are increase.

EFFECT: decreased metal input

1 dwg, 2 tbl

Description

The present invention relates to the improvement of hot-rolled I-sections and to the metal structures of industrial and civil buildings with beam ceilings and frames, as well as to bridge structures from such profiles.

Known I-beam rolling profile GOST 26020-83 [1]. An assortment of hot-rolled I-beams with parallel flange faces. We will take this profile as an analogue. In an analogue, the section proportions do not depend on which steel the I-beam is rolled from: low carbon or low alloy.

, где h_ст - высота стенки; t_ст - толщина стенки) переменная и колеблется для разных номеров двутавров. Кроме того, материалоемкость стенки отличается от оптимальной материалоемкости 50%, напримерThe disadvantage of the analogue is that the flexibility of the profile wall ( $${\lambda }_{\mathrm{from}t}=\frac{h_{\mathrm{from}t}}{t_{\mathrm{from}t}}$$

where h _article - wall height; t _article - wall thickness) is variable and varies for different numbers of I-beams. In addition, the material consumption of the wall differs from the optimal material consumption of 50%, for example

I-Beam I The flexibility of the wall λ _st The material consumption of the wall K W _x cm ³ J _x cm ⁴ I 100 B1 59.25 0.52 9011 446,000 I 100 B2 55.76 0.49 10350 516400 I 100 B3 52.67 0.519 11680 597700 I 100 B4 48.6 0.562 12940 655400

I 100B4-h = 101.3 cm, t _p = 3.25 cm, t _article = 1.95 cm, h _article = 94.8 cm, $${\lambda }_{\mathrm{from}t}=\frac{9\mathrm{four},8}{\mathrm{one},95}=\mathrm{four}8,6$$

Known I-beam rolling profile for rolling beams (from mild steel), proposed by K. K. Nezhdanov and developed with graduate students [2, RU No. 2383401]. We will take this profile as a prototype.

In the prototype, a range of new rolling profiles from low-carbon steel C255 was developed (B St3 Sp5, Gost 27772-88).

However, I-shaped rolled sections of low alloy steel must have different parameters. The range of I-sections from low alloy steel has not been developed.

Currently, rolled profiles of mild and low alloy steel have the same cross-sectional shape, which is erroneous, since the wall flexibility for alloy steel profiles is less than the ultimate wall flexibility, namely, λ _{st. prev} = 65.

, где значение 2,5 - при наличии местных напряжений в профиле.According to current standards [2, p.27] the stability of the walls of the beams does not need to be checked if the conditional flexibility of the walls $$\overline{{\lambda }_{\mathrm{from}t}}=\frac{h_{\mathrm{from}t}}{t_{\mathrm{from}t}}\sqrt{\frac{R_{\mathrm{at}}}{E}}<2,5$$

, where the value is 2.5 - in the presence of local stresses in the profile.

, отсюда предельная гибкость, когда не требуется постановка промежуточных ребер жесткости, равна λ_{ст пред}=2,5×26,204=65,51.For low alloy steels, for example 12G2Sgr1, TU 14-1-43 23-88, 09G2S, 14G2, 15KHSND according to GOST 19282-73 * [2, p.64], with a design resistance of R _y = 300 MPa with a thickness of t = 20 ... 40 mm and elastic modulus E = 206000 MPa reduced flexibility $$\overline{{\lambda }_{\mathrm{from}t}}={\lambda }_{\mathrm{from}t}\sqrt{\frac{300}{206000}}<2,5$$

, hence the ultimate flexibility, when the setting of intermediate stiffeners is not required, is equal to λ _{st before} = 2.5 × 26.204 = 65.51.

With a certain margin, we assign the ultimate flexibility of the wall of the I-beam rolling profile to λ _{st pre} = 65.

Optimum distribution of material over the beam

A significant reduction in the material consumption of the I-beam rolling profile can be achieved by the optimal distribution of metal over the cross section of the beams. We set the task to roll the beam of greatest strength, that is, with a maximum resistance moment max W _x from the workpiece with a cross-sectional area of A cm ² .

Obviously, a decrease in wall thickness t _article leads to an increase in the height of the beam h and an increase in the moment of resistance W _x . The cross-sectional material should be optimally distributed between the wall and the girdle of the beam. We introduce the coefficient K, which determines the material consumption of the wall.

Then $$A_{\mathrm{from}t}=K\cdot A\text{(one)}$$

We introduce a constant coefficient of wall flexibility:

$${\lambda }_{\mathrm{from}t}=\frac{h}{t_{\mathrm{from}t}}=\frac{h\cdot t_{\mathrm{from}t}}{t_{\mathrm{from}t}^2}=\frac{K\cdot A}{t_{\mathrm{from}t}^2}=const\text{(2)}$$

From here

$$t_{\mathrm{from}t}=\sqrt{\frac{A}{{\lambda }_{\mathrm{from}t}}}\cdot K^{0.5}\text{(3)}$$

Then the distance between the centers of gravity of the belts h _cc will depend only on the coefficient of material _consumption of the wall K:

$$h_{cc}=t_{\mathrm{from}t}\cdot {\lambda }_{\mathrm{from}t}=\sqrt{A\cdot {\lambda }_{\mathrm{from}t}}\cdot K^{0.5}\text{(four)}$$

Cross sectional area of two belts

$$2A_P=A-A_{\mathrm{from}t}=A\cdot (\mathrm{one}-K)\text{(5)}$$

Neglecting our own moments of inertia of the beam belts, we write down the moment of inertia as the main one (see Fig. 1)

$$J_x=2A_P{\left(\frac{h_{cc}}{2}\right)}^2+A_{\mathrm{from}t}\frac{h_{cc}^2}{12}\text{(6)}$$

Dividing (6) by 0.5 · _hcc , we find the moment of resistance W _x at the height of the centers of gravity of the belts:

$$W_x=\frac{h}{2}\left(2A_P+\frac{A_{\mathrm{from}t}}{3}\right)=\frac{\mathrm{one}}{2}A\sqrt{A\cdot {\lambda }_{\mathrm{from}t}}\cdot K^{0.5}(\mathrm{one}-\frac{2}{3}K)\text{(7)}$$

So, the moment of resistance depends only on the material intensity K of the wall. We define the extremum W _x by taking the derivative with respect to K.

$$\frac{d{W}_X}{dK}=\frac{\mathrm{one}}{2}A\sqrt{A\cdot {\lambda }_{\mathrm{from}t}}\left[\frac{\mathrm{one}}{2}{K}^{0.5}(\mathrm{one}-\frac{2}{3}K)-\frac{2}{3}{K}^{0.5}\right]=\text{0 (8)}$$

Hence K = 0.5.

Therefore, at K = 0.5, the wall material consumption is 50% of the material consumption of the entire beam section.

Then the optimal section height is

$$h_{\mathrm{about}Pt}=\frac{0.5A}{t_{\mathrm{from}t}}\text{(9)}$$

This height corresponds to the greatest moment of resistance equal to

$$W_x^2=\frac{A^3{\lambda }_{\mathrm{from}t}}{\mathrm{eighteen}},\text{(eleven)}$$

and corresponding wall flexibility

$${\lambda }_{\mathrm{from}t}=\frac{0.5A}{t_{\mathrm{from}t}^2}.\text{(12)}$$

Substituting (12) in (11), we obtain W _x depending on A and t _article

$$W_X=\frac{A^2}{6\cdot t_{\mathrm{from}t}}$$

and corresponding minimum beam cross-sectional area

$$A=\sqrt{6\cdot W_X\cdot t_{\mathrm{from}t}},\text{(13)}$$

or depending on the flexibility of the wall

$$A=\sqrt[3]{\frac{\mathrm{eighteen}{W}_X^2}{{\lambda }_{\mathrm{from}t}}}\text{(fourteen)}$$

From (11) and (12) we obtain

$$t_{\mathrm{from}t}=\sqrt[3]{\frac{\mathrm{1,5}{W}_X}{{\lambda }_{\mathrm{from}t}^2}},\text{(fifteen)}$$

it is easy to verify that the radius of the core of the cross section r in this case is equal to

r = h / 3.

Let us analyze the material consumption and flexibility of the analog wall. For example, for I-beams with a section height of 100 cm according to GOST 26020-83.

Table 1 Dimensions in cm, area in cm ² h b t _st t _p h _st A A _st λ _st TO 100 B1 99 32 1,6 2.1 94.8 293.82 151.68 59.25 0.52 100 B2 99.8 32 1.7 2.5 94.8 328.9 161.16 55.76 0.49 100 B3 100.6 32 1.8 2.9 94.8 364 170.64 52.67 0.519 100 B4 101.3 32 1.95 3.25 94.8 400.6 184.86 48.62 0.562

The analysis shows that the material consumption of the wall differs from the optimal L = 0.5, and the flexibility of the wall for low alloy steel is less than the limiting value λ _{st before} = 65.

Thus, there are reserves to reduce the material consumption of the profile.

The technological task of the invention is the maximum reduction in the material consumption of the I-beam rolling profile, when it is rolled, by the optimal distribution of material between the belts and the wall of the profile, ensuring that the moment of resistance of the section of the profile relative to the main axis X reaches its maximum value.

The technological task of implementing the method of distribution of steel over the cross-section of the I-beam profile, containing shelves with parallel faces and connecting their wall, is solved as follows.

The difference is that when rolling on a rolling mill of an I-profile, the flexibility of the beam wall is assigned as the limit for this low-alloy steel grade, which ensures the stability of the wall without stiffeners. The bloom is heated to a temperature of 600 ... 650 ° C and squeezed in a cage by rolls on a rolling mill on four sides, the section is deformed into an I-beam profile containing shelves with parallel faces and a wall that integrally joins the shelves, with the formation of an I-beam, when rolled from low alloy steel ( for example, 12G2Sgr1, TU14-1-4323-88, with design resistance R _y = 300 MPa and a thickness of 20… 40 mm)

The section cross-sectional area during rental is distributed over the cross-section in the following proportion: 50% on the profile wall and 25% on each of the belts.

The wall thickness of the profile is determined from the equation

$$t_{\mathrm{from}t}=\sqrt{\frac{0,5\mathrm{BUT}}{{\lambda }_{\mathrm{from}t}}}$$

where A is the cross-sectional area of the profile;

λ _st - the ultimate flexibility of the profile wall, ensuring its stability without stiffening ribs.

The wall height is found from the formula h _article = λ _article · t _article , where the width of the shelf and its thickness are assigned such that the ratio of the width of the shelf to its thickness does not exceed the limit ratio that ensures local stability of the flat belt.

The main moment of inertia of the profile is determined by the formula

$$J_x=\frac{b\cdot h^3-(b-t_{\mathrm{from}t})\cdot h_{\mathrm{from}t}^3}{12}$$

where b is the width of the profile shelf;

h is the height of the cross section of the profile.

The moment of resistance is calculated by the formula $$W_x=\frac{2J_x}{h}$$

Figure 1 shows a cross section of an I-beam rolling profile. Shelves have a width b. The thickness of the shelf t _p The profile section height is h. The wall has a height of h _Art . The wall thickness is equal to t _Art .

The cross-sectional area of each of the shelves is equal to A _p = b · t _p .

The cross-sectional area of the wall is equal to A _st = h _st · t _article

The total cross-sectional area is A = 2A _p + A _st .

Concrete implementation example

We will increase the strength of the I-beam I100B4 in accordance with GOST 26020-83 [1]. The main dimensions of the profile and its characteristics are given in table 1.

The calculations are performed in the following sequence.

1. By the formula 12 we find the required wall thickness of the new I-beam at a given cross-sectional area A = 400.6 cm ² and the ultimate flexibility of the wall $${\lambda }_{\mathrm{from}t.PRed}=65\Rightarrow t_{\mathrm{from}t}=\sqrt{\frac{0,5\mathrm{BUT}}{{\lambda }_{\mathrm{from}tPRed}}}$$

2. We find the cross-sectional area of the wall A _article = 0.5A.

, округляем ее.3. Then the wall height $$h_{\mathrm{from}t}=\frac{{\mathrm{BUT}}_{\mathrm{from}t}}{t_{\mathrm{from}t}}$$

, round it.

4. The actual cross-sectional area of the wall A _{article f} = h _article · t _article

5. We find the cross-sectional area of the shelf And _p = 0.25 · A.

, при заданной ширине полки b_п, и округляем ее.6. Determine the thickness of the shelf $$t_P=\frac{{\mathrm{BUT}}_P}{b_P}$$

, for a given shelf width b _p , and round it.

7. Then the actual cross-sectional area of the shelf A _pf = b _p · t _p .

8. Determine the height of the beam section h = h _st + 2t _p .

9. The actual cross-sectional area A = 2A _pf + A _Art .

10. Determine the main moment of inertia $$J_x=\frac{b{h}^3-\left(b-t_{\mathrm{from}t}\right)h_{\mathrm{from}t}^3}{12}$$

11. The moment of resistance of the cross section $$W_x=\frac{2J_x}{h}$$

For example, according to the assortment for the I100B4 double tee, the cross-sectional area A = 400.6 cm ² . Assign the flexibility of the wall λ _{st pre} = 65.

1. Find the required wall thickness of the new I-beam at a given cross-sectional area and wall flexibility

Принимаем t_ст=1,76 см. $${\lambda }_{\mathrm{from}t}=\frac{0.5\mathrm{BUT}}{t_{\mathrm{from}t}^2}\Rightarrow t_{\mathrm{from}t}=\sqrt{\frac{0.5\mathrm{BUT}}{{\lambda }_{\mathrm{from}t\text{before}}}}=\sqrt{\frac{0.5\cdot 400.6}{65}}=\mathrm{1,7554.}$$

Take t _article = 1.76 cm.

2. We find the cross-sectional area of the wall A _article = 0.5A = 200.3 cm ² .

3. Then the wall height $$h_{\mathrm{from}t}=\frac{{\mathrm{BUT}}_{\mathrm{from}t}}{t_{\mathrm{from}t}}=\frac{200.3}{1.76}=\mathrm{113,807}\Rightarrow 200.3{\text{cm}}^{\text{2}}.$$

4. The actual cross-sectional area of the wall A _stf = h _st · t _st = 113.8 · 1.76 = 200.288 cm ² .

5. We find the cross-sectional area of the shelf And _p = 0.25 · A = 0.25 · 400.6 = 100.15 cm ² .

Leave the width of the shelf b _p = 32, then

6. $$t_P=\frac{100.15}{32}=\mathrm{3,1297}\Rightarrow 3.13\text{cm}\text{.}$$

7. Then the actual cross-sectional area of the shelf is A _pf = b _p · t _p = 32 · 3.13 = 100.16 cm ² .

8. Determine the height of the beam

9. h = h _st + 2t _p = 113.8 + 2 · 3.13 = 120.06 cm.

10. The actual cross-sectional area A = 2A _pf + A _st = 2 · 100.16 + 113.8 · 1.76 = 400.608 cm ² .

11. Determine the main moment of inertia $$J_x=\frac{b{h}^3-(b-t_{\mathrm{from}t})h_{\mathrm{from}t}^3}{12}=\frac{32\cdot {120.06}^3-(32-1.76)\cdot {113.8}^3}{12}=901040{\text{cm}}^{\text{four}}\text{(137}\text{,5\%)}$$

12. It was J _x = 655400 cm ⁴ (100%).

13. The moment of resistance $$W_x=\frac{2J_x}{h}=15009.8{\text{cm}}^{\text{3}}\left(\text{116\%}\right)$$

14. It was W _x = 12940 cm ³ (100%).

Bibliography

1. Hot-rolled steel I-beams with parallel flange faces. Assortment. GOST 26020-83. Reissue. October 1998, 9 p.

2. SNiP 11-23-81. Steel structures, Gosstroy of the USSR, Moscow, 1999, 96 p.

3. Vasilchenko V.T. et al. Handbook of the designer of metal structures, Kiev, Budivelnik, 1980, 288 pp.

4. Pisarenko G.S. et al. Handbook of resistance to materials, Kiev, “Naukova Dumka”, 1975, 704 pp.

5. Nezhdanov K.K., Nezhdanov A.K., Eidlin A.M. I-beam rolling profile. Russian patent No. 2383401, B21B 1/08 (2006.01). Application No. 2007 136405/02. Publication of the application 10.04.2009. Published on March 10th, 2010. Bull. Number 7.

Claims (1)

A method of rolling a profile of an I-section from low alloy steel with a design resistance of R _y = 300, including heating the bloom 20-40 mm thick to a temperature of 600-650 ° C and crimping it in a stand with rolls on a rolling mill on four sides, deformation into an I-profile having shelves with parallel faces as flat belts and a wall integrally connecting them, characterized in that the wall of the rolled profile with maximum flexibility, equal to the ratio of height to wall thickness it does not exceed the flexibility limit value λ _{article before,} with stavlyayuschey 65 and ensures stability of the walls without intermediate ribs, the profile transformed during hot rolling by crimping its rollers and forced area distribution over the cross section in a proportion of 50% in the profile wall and by 25% for each of the flat belt, wherein a thickness profile wall t _Art determined from the equation
$$t_{\mathrm{from}t}=\sqrt{0.5A/{\lambda }_{\mathrm{from}t\text{before}}}$$

,
where A is the cross-sectional area of the profile;
λ _st limit - the ultimate flexibility of the profile wall, ensuring its stability without intermediate stiffeners;
moreover, the wall height h _article is determined from the equation
h _article = λ _{article before} · t _article
where the shelf is rolled so that the ratio of the width of the shelf to its thickness does not exceed the limit ratio, which provides local stability of the flat belt, and the main moment of inertia of the profile J _x is determined by the formula
$$J_x=\left[b\cdot h^3-(b-t_{\mathrm{from}t})\cdot h_{\mathrm{from}t}^3\right]/12$$

,
where b is the width of the flange of the I-beam profile;
h is the height of the cross section of the I-beam profile,
and the maximum moment of resistance is determined by the formula:
W _x = 2J _x / h.

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