Eng.-Mech. P. A. DUNAEV.

CALIBRATION OF I-BEAM No. 25

The calibration of shaped sections presents great difficulties in preserving the same deformation of the metal in the individual parts of the pass.

In the calibration of shaped sections (I-beams, rails, channels, etc.), in practice these unequal deformations in the individual parts of the pass are permitted, which also corresponds to different elongations in the parts of the pass; the latter leads to overstressing of the metal in individual parts of the section of the rolled strip.

Often these overstresses are so great that the metal in the more overstressed places of the section cannot withstand them—it is destroyed, and hence the amount of defective product during rolling increases, as a consequence of incorrect calibration.

Consequently, when calibrating a shaped section, special attention must be paid to the choice of the elongation coefficient both in the individual pass and in its separate parts. The elongation coefficient must be selected in accordance with the decreasing temperature, i.e., at the beginning—larger, and at the end of rolling—smaller.

To determine (select) the elongation coefficient in the individual passes, it is necessary in advance to outline the law of its variation and, according to this outlined law of variation of the elongation coefficient, to calibrate the specified section.

To clarify the law of variation of the elongation coefficient, we have carried out three examples of calibration of I-beam No. 25. Example 1 is a calibration according to the Kirchberg principle, which “outlines” the law of variation of the elongation coefficient according to the law of variation of the tangent—the inclination of the inner face of the flange.

Example No. 2 is a calibration according to Tafel’s principle, i.e., without lateral pressure in the closed part of the pass, and the law of variation of the elongation coefficient is outlined by us according to the law of a straight line.

Example No. 3 is our calibration; the change in the elongation coefficient by passes is taken according to the law of a geometric progression.

Example No. 1—(Kirchberg principle).

The finished section of I-beam No. 25 in the cold state has the dimensions (Fig. 1):

\[ \begin{aligned} B &— 250\ \text{mm}\\ H &— 110\ \text{”}\\ T &— 9.0\ \text{”}\\ S &— 13.6\ \text{”} \end{aligned} \]

\(\tg \alpha = 0.14\) (slope of the inner face of the flange—14%).

If we take: the temperature at the end of rolling \(T = 1000^\circ \mathrm{C}\), the coefficient of expansion of the metal at this temperature \(\lambda = 1.015\), then the section in the hot state will have the dimensions:

\[ \begin{aligned} B_1 &= \lambda B = 253.8\ \mathrm{mm}\\ H_1 &= \lambda H = 111.6\ \text{''}\\ s_1 &= \lambda s = 9.2\ \text{''}\\ t_1 &= \lambda t = 13.8\ \text{''} \end{aligned} \]

As the cross-sectional area of the section \((Q)\) we shall take the area bounded by straight lines (Fig. 2).

For convenience and simplicity of calibration, we divide the area of the section into the following parts: a) the area of the web, which we calibrate as flat iron; b) four halves of the flange, which we calibrate as closed and open parts of the pass (Fig. 2).

a) Area of the web \((q)\):

\[ q_1 = B_1 s_1 = 253.8 \cdot 9.2 = 2335\ \mathrm{mm}^2 \]

b) the four halves of the flange, the area of each of them will be:

\[ R_1 = \frac{a_1 + b_1}{2} h_1, \]

where: \(a_1\)—the thickness of the flange at the edge

\[ b_1\text{— } \quad \text{''} \quad \text{''} \quad \text{at the point of connection with the web} \]

\(h_1\)—the height of the trapezoid (Fig. 2).

The trapezoid bounded by \(EFCD\) will henceforth be called the flange, and if this flange is rolled in the closed part of the pass, we shall call it the “closed flange,” and correspondingly in the open part—the “open flange.”

\(h_1\)—the height of the flange (closed and open)

\[ h_1 = \frac{H_1 - S_2}{2} = \frac{111.6 - 9.2}{2} = 51.2\ \mathrm{mm} \]

\[ a_1 = DC = OC \tg \alpha_1 \qquad \ldots\ldots\ldots (1) \]

\[ b_1 = EF = OE \tg \alpha_1 \qquad \ldots\ldots\ldots (2) \]

where:

\[ OC = OB - \frac{H_1}{4} = \frac{t}{\tg \alpha_1} - \frac{H}{4} = \frac{13.8}{0.14} - \frac{116.6}{4} = 77.7\ \mathrm{mm} \]

\[ OE = X = OC + h = 70.7 + 51.2 = 122\ \mathrm{mm} \]

\(X\)—the pole distance, and point \(0\)—the pole (or construction point).

Then:

\[ a_1=(x-h_1)\,\tg \alpha_1=(129-51.2)0.14=9.9\ \mathrm{mm} \]

\[ b_1=x\cdot \tg \alpha_1=122\cdot 0.14=17.0\ \mathrm{mm}. \]

Area of the flange (trapezoid)

\[ R_1=\frac{9.9+17.0}{2}\cdot 51.2=689\ \mathrm{mm}^2. \]

The total area of the hot section, or the area of the first pass \(Q_1\), will be:

\[ Q_1=2335+4\cdot 689=5091\ \mathrm{mm}^2 \]

Kirchberg takes the elongation \((f)\) in the first pass as equal to \(\tg \alpha\) (the slope of the flange), or, in percent, \(14\%\); consequently, the coefficient of elongation in the first pass will be

\[ \mu_1=1+\frac{f_1}{100}=1+\frac{14}{100}=1.14. \]

In the subsequent passes Kirchberg finds the elongation by the following formula:

\[ f_2=f_1\left(1+\frac{f_1}{100}\right)\ldots\ldots\ldots (3) \]

where: \(f_2\)—elongation of the subsequent pass
\(f_1\)—“ ” previous “

The radii of rounding of the blunt corners are taken in the following ratios:

\[ r=t;\qquad r'=t' \]

\[ \rho=0.5t;\qquad \rho'=0.5t' \]

\[ V=0.25t;\qquad V'=0.25t', \]

where \(t'_1\) refer to the closed part of the pass, and \(t\)—to the open part.

Pass No. 2.

Open flange:

\[ h_2=h_1+5=51.2+5=56\ \mathrm{mm} \]

\[ R_2=R_1\mu_1=689\cdot 1.14=786\ \mathrm{mm}^2 \]

\[ t_2=\frac{R_2}{h_2}=\frac{786}{56}=14\ \mathrm{mm} \]

Axial distance \(x=\mathrm{const}\)

\[ \tg \]

of the angle of inclination of the inner face of the flange will be

\[ tg\,\alpha_2=\frac{t_2}{X-0.5h_2}=\frac{14}{122-0.5\cdot56}=0.15 \]

\[ a_2=(x-h_2)tg\,\alpha_2=(122-56)0.15=9.9\ mm \]

\[ b_2=x\cdot tg\,\alpha_2=122\cdot0.15=18.3\ mm. \]

Closed flange:

\[ h'_2=h_1-1=51.2-1=50\ mm \]

\[ R'_2=R_2=786\ mm^2\ ^*) \]

\[ t'_2=\frac{R'_2}{h'_2}=\frac{786}{50}=15.7\ mm \]

\[ tg\,\alpha'_2=\frac{t'_2}{x-0.5h'_2}=\frac{15.7}{122-0.5\cdot50}=0.18 \]

\[ d_2=(x-h'_2)tg\,\alpha'_2=(122-50)0.18=12.9\ mm \]

\[ c_2=x\cdot tg\,\alpha'_2=122\cdot0.18=22.0\ mm. \]

Web:

Area of the web

\[ q_2=q_1\cdot\mu_1=2335\cdot1.14=2662\ mm^2 \]

\[ B_2=B_1-\beta=253.8-1.8=252\ mm, \]

where \(\beta\) is the spread

\[ S_2=\frac{q_2}{B_2}=\frac{2662}{252}=10.6\ mm. \]

Total area of the pass

\[ Q_2=q_2+2(R_2+R'_2)=2662+2(786+796)=5805\ mm^2 \]

Pass No. 3.

Elongation in % according to Kirchberg

\[ f_2=f_1\left(1+\frac{f_1}{100}\right)=14\left(1+\frac{14}{100}\right)=16\%. \]

Elongation coefficient:

\[ \mu=1+\frac{f_2}{100}=1+\frac{16}{100}=1.16. \]

\[ \overline{\phantom{xxxxxxxxxxxxxxxxxxxxxxxx}} \]

*) The elongation coefficient is the same in all parts of the pass.

Open flange:

\[ h_3=h'_2+5=50+=55\ \mathrm{mm} \]

\[ R_3=R'_2\mu_2=786\cdot1{,}16=912\ \mathrm{mm}^2 \]

\[ t_3=\frac{R_3}{h_3}=\frac{912}{55}=16{,}6\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_3=\frac{t_3}{x-0{,}5h_3} =\frac{16{,}70}{122-0{,}5\cdot55}=0{,}174 \]

\[ a_3=(x-h_3)\operatorname{tg}\alpha_3=(122-55)0{,}174=16{,}2\ \mathrm{mm} \]

\[ b_3=x\cdot\operatorname{tg}\alpha_3=122\cdot0{,}174=21{,}2. \]

Closed flange:

\[ h'_3=h_2-1=56-1=55\ \mathrm{mm} \]

\[ R'_3=R_3=912\quad " \]

\[ t'_3=t_3=16{,}6\quad " \]

\[ \operatorname{tg}\alpha'_3=\operatorname{tg}\alpha_3=0{,}174\quad " \]

\[ d_3=a_3=16{,}2\quad " \]

\[ c_3=b_3=21{,}2\quad " \]

Web:

\[ q_3=q_5\mu_2=2662\cdot1{,}16=3088\ \mathrm{mm}^2, \]

\[ B_3=B_2-3=252-3=249\ \mathrm{mm} \]

\[ S_3=\frac{q_3}{B_2}=\frac{3088}{249}=12{,}5\ \mathrm{mm}. \]

Total area of the pass

\[ Q_3=q_3+4R_3=3088+4\cdot912=6736\ \mathrm{mm}^2 \]

Pass No. 4.

\[ f_3=f_2\left(1+\frac{f_2}{100}\right) =16\left(1+\frac{16}{100}\right)=19\% \]

\[ \mu_3=1+\frac{f_3}{100}=1+\frac{19}{100}=1{,}19 \]

Open flange:

\[ h_4=h'_3+5=55+5=60\ \mathrm{mm} \]

\[ R_4=R'_3\mu_3=912\cdot1{,}19=1085\ \mathrm{mm}^2 \]

\[ t_4=\frac{R_4}{h_4}=\frac{1085}{60}=18.1\ \mathrm{mm} \]

\[ \tg \alpha_4=\frac{t_4}{x-0.5h_4}=\frac{18.1}{122-0.5\cdot 60}=0.20 \]

\[ a_4=(122-60)0.20=12.0\ \mathrm{mm} \]

\[ b_4=122\cdot 0.20=24.4\ \mathrm{mm} \]

Closed flange:

\[ h'_4=h_3-1=55-1=54\ \mathrm{mm} \]

\[ R'_4=R_4=1085\ \mathrm{mm}^2 \]

\[ t'_4=\frac{R'_4}{h_4}=\frac{1085}{54}=20\ \mathrm{mm} \]

\[ \tg \alpha'_4=\frac{t_4}{x-0.5h'_4}=\frac{20}{122-0.5\cdot 54}=0.21 \]

\[ d_4=(122-54)0.21=14.3\ \mathrm{mm} \]

\[ c_4=122\cdot 0.21=25.6\ \mathrm{mm} \]

Web:

\[ q_4=q_3\mu_3=3088\cdot 1.19=3675\ \mathrm{mm}^2 \]

\[ B_4=B_3-4=249-4=245\ \mathrm{mm} \]

\[ S_4=\frac{q_4}{B_4}=\frac{3675}{245}=15\ \mathrm{mm} \]

Full pass area

\[ Q_4=q_4+4R_4=3678+4\cdot 1085=8015\ \mathrm{mm}^2. \]

Pass No. 5.

Elongation

\[ f_4=f_3\left(1+\frac{f_3}{100}\right)=19\left(1+\frac{19}{100}\right)=23\%. \]

Elongation coeff. \(\mu_4=1+\frac{23}{100}=1.23.\)

Open flange:

\[ h_5=h'_4+5=54+5=59\ \mathrm{mm} \]

\[ R_5=R_4\mu_4=1085\cdot 1.23=1334\ \mathrm{mm}^2 \]

\[ t_5=\frac{R_5}{h_5}=\frac{1335}{59}=22,6\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_5=\frac{22,6}{122-0,5\cdot 59}=0,244 \]

\[ a_5=(122-59)\,0,244=15,2\ \mathrm{mm} \]

\[ b_5=122\cdot 0,244=30,0\ \mathrm{mm} \]

Closed flange:

\[ h'_5=h_4-1=60-1=59\ \mathrm{mm} \]

\[ R'_5=R_5=1334\ \mathrm{mm}^2 \]

\[ t'_5=t_5=28,6\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha'_5=\operatorname{tg}\alpha_5=0,244 \]

\[ d_5=a_5=15,2\ \mathrm{mm} \]

\[ c_5=b_5=30,0 \]

Neck:

\[ q_5=q_4\mu_4=3675\cdot 1,23=4490\ \mathrm{mm}^2 \]

\[ B_5=B_4-5=245-5=240\ \mathrm{mm} \]

\[ S_5=\frac{q_5}{B_5}=\frac{7490}{240}=18\ \mathrm{mm} \]

Full area of the pass

\[ Q_5=4499+4\cdot 1334=9826\ \mathrm{mm}^2. \]

Pass No. 6.

\[ f_5=f'_5\left(1+\frac{f_4}{109}\right)=23\left(1+\frac{23}{100}\right)=28\% \]

\[ \mu_5=1+\frac{f_5}{100}=1+\frac{28}{100}=1,28 \]

Open flange:

\[ h_6=h'_5+5=59+5=64\ \mathrm{mm} \]

\[ R_6=R'_5\mu_5=1335\cdot 1,28=1707\ \mathrm{mm}^2 \]

\[ t_6=\frac{R_6}{h_6}=\frac{1707}{64}=26,5\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_6=\frac{26,5}{123-0,5\cdot 64}=0,294 \]

\[ a_6=(122-64)0.294=17.0\ \text{mm} \]

\[ b_6=122\cdot 0.294=36.0\ \text{mm}. \]

Closed flange:

\[ h'_6=h_5-1=59-1=58\ \text{mm} \]

\[ R'_6=R_6=1707\ \text{mm}^2 \]

\[ t'_6=\frac{R_6}{h'_6}=\frac{1707}{58}=30.0\ \text{mm} \]

\[ \tg \alpha'_6=\frac{30}{122-0.5\cdot 58}=0.322 \]

\[ d_6=(122-58)0.322=20.7\ \text{mm} \]

\[ c_6=122\cdot 0.322=39.2\ \text{mm} \]

Web:

\[ q_6=q_5\mu_5=4490\cdot 1.28=5747\ \text{mm}^2 \]

\[ B_6=B_5-6=240-6=234\ \text{mm} \]

\[ S_6=\frac{q_6}{B_6}=\frac{5747}{234}=24.5\ \text{mm} \]

Total area of the pass

\[ Q_6=5747+4\cdot 1707=12575\ \text{mm}^2. \]

Pass No. 7.

Elongation in %%

\[ f_6=f_5\left(1+\frac{f_5}{100}\right)=28\left(1+\frac{28}{100}\right)=36\% \]

\[ \mu_6=1+\frac{36}{100}=1.36 \]

The open flange and the closed flange are equal:

\[ h_7=h'_6+5=58+5=63\ \text{mm} \]

\[ R_7=R\mu_6=1707\cdot 1.36=2322\ \text{mm} \]

\[ t_7=\frac{R_7}{h_7}=\frac{2322}{63}=36.8\ \text{mm} \]

\[ \tg \alpha_7=\frac{36.8}{122-0.5\cdot 63}=0.40 \]

\[ b_7=122\cdot 0.4=49.0\ \text{mm} \]

\[ a_7=(122-63)0.4=24.6\ \text{mm}. \]

Cheek:

\[ q_7 = q_6 \psi_6 = 5747 \cdot 1.36 = 7846\ \mathrm{mm}^2 \]

\[ B_7 = B_6 - 8 = 234 - 8 = 226\ \mathrm{mm} \]

\[ S_7 = \frac{q_7}{B_7} = \frac{7846}{226} = 34.6\ \mathrm{mm} \]

Gauge area

\[ Q_7 = 7846 + 4 \cdot 2322 = 17134\ \mathrm{mm}. \]

Gauge No. 8.

To determine the elongation \((f)\), Kirchberg’s formula is unsuitable; it is applicable only up to the 7th gauge. To determine the next gauge we shall use Kirchberg’s transition formulas; we shall take the closed part as equal to the open part of the gauge

\[ a_8 = a_7 + \frac{a_7^2}{200} = 24.6 + \frac{(24.6)^2}{200} = 27.6\ \mathrm{mm} \]

We reduce the axial distance by 10%, and then we shall have:

\[ h_8 = h'_7 + 5 = 63 + 5 = 68\ \mathrm{mm} \]

\[ x' = 0.9\,x = 122 \cdot 0.9 = 110\ \text{”} \]

\[ tg\,\alpha_8 = \frac{a_8}{x' - h_8} = \frac{27.6}{110 - 68} = 0.64. \]

\[ b_8 = x'\,tg\,\alpha_8 = 110 \cdot 0.64 = 70.4\ \mathrm{mm} \]

\[ R_8 = \frac{(a_8 + b_8)h_8}{2} = \frac{27.7 + 70.4}{2} \cdot 68 = 3242\ \mathrm{mm} \]

\[ \psi_7 = \frac{R_8}{R_7} = \frac{3242}{2322} = 1.40. \]

Cheek:

\[ q_8 = q_7 \mu_7 = 7846 \cdot 1.40 = 10984\ \mathrm{mm}^2 \]

\[ B_8 = B_7 - 10 = 266 - 10 = 216\ \mathrm{mm} \]

\[ S_8 = \frac{q_8}{B_8} = \frac{10984}{216} = 50.8\ \mathrm{mm} \]

Total gauge area

\[ Q_8 = 10984 + 4 \cdot 3242 = 23952\ \mathrm{mm}^2. \]

Pass No. 9

We reduce the pass-to-pass distance by another 10%:

\[ X'' = X' 0.9 = 110 \cdot 0.9 = 99\ \mathrm{mm} \]

\[ a_9 = a_8 + \frac{a_8^2}{200} = 27.6 + \frac{(27.6)^2}{200} = 31.0\ \mathrm{mm} \]

\[ h_9 = h_8 = 68\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_9 = \frac{a_9}{X'' - h_9} = \frac{31.0}{99 - 68} = 1.00 \]

\[ b_9 = X'' \operatorname{tg}\alpha_9 = 99 \cdot 1.0 = 99\ \mathrm{mm} \]

\[ R_9 = \frac{(a_9 + b_9)h_9}{2} = \frac{31 + 99}{2}\cdot 68 = 4420\ \mathrm{mm}^2 \]

\[ \mu_8 = \frac{R_9}{R_8} = \frac{3420}{3342} = 1.36. \]

Neck:

\[ q_9 = q_8 \mu_8 = 10984 \cdot 1.36 = 14938\ \mathrm{mm}^2 \]

\[ B_9 = B_8 - 12 = 216 - 12 = 204\ \mathrm{mm} \]

\[ S_9 = \frac{q_9}{B_9} = \frac{14938}{204} = 72\ \mathrm{mm}. \]

Total area of the pass:

\[ Q_9 = 14938 + 4 \cdot 4420 = 32620\ \mathrm{mm}^2 \]

The billet will have the following dimensions:

Height

\[ H_{\mathrm{bil.}} = 2h_9 + S_9 + 20 = 2 \cdot 68 + 72 + 20 = 208\ \mathrm{mm}. \]

Width

\[ B_{\mathrm{bil.}} = B_9 - 14 = 204 - 14 = 190\ \mathrm{mm}. \]

Billet area

\[ Q_{\mathrm{bil.}} = H_{\mathrm{bil.}} \cdot B_{\mathrm{bil.}} = 208 \cdot 190 = 4370\ \mathrm{mm}^2 \]

The elongation coefficient in the 9th pass (edging) will be equal to:

\[ \mu_9 = \frac{Q_{\mathrm{bil.}}}{Q_9} = \frac{43709}{32620} = 1.34. \]

Overall elongation coefficient from the billet to the finished section

\[ \mu_{\mathrm{tot.}} = \mu_1 \cdot \mu_2 \cdot \mu_3 \cdot \mu_4 \cdot \mu_5 \cdot \mu_6 \cdot \mu_7 \mu_8 \cdot \mu_9 = 8.65. \]

All calculation results have been compiled in Table No. 1 (p. 13). From this table a graph has been constructed (Fig. 3) of the change in elongation coefficient \(\mu\), the change in section area by passes (passes) \((Q)\), the change in spread \((B)\), and the change in lateral reduction has also been plotted—

Table No. 1. Calibration of I-beam No. 25 (after Kirchberg).

Pass No.	Upper flange, $\mu$	Upper flange, $R$	Upper flange, $h$	Upper flange, $t$	Upper flange, $\Delta t$	Lower flange, $\mu$	Lower flange, $R$	Lower flange, $h$	Lower flange, $t$	Lower flange, $\Delta t$	Web, $\mu$	Web, $q$	Web, $\beta$	Web, $B$	Web, $S$	Overall, $\mu$	Overall, $Q$
1	1.14	680	51.2	13.6	2.1	1.14	689	51.2	13.6	0.4	1.14	23.35	1.8	253.8	9.2	1.14	5090
2	1.16	786	50	15.7	1.1	1.16	786	56	14.0	2.6	1.16	26.62	3	252	10.6	1.16	5806
3	1.19	912	55	16.6	3.4	1.19	912	55	16.6	1.5	1.19	3088	4	249	12.5	1.19	6736
4	1.23	1085	54	20.0	2.6	1.23	1085	60	18.1	4.5	1.23	3675	5	245	15.0	1.23	8015
5	1.28	1334	59	22.6	7.4	1.28	1334	59	22.6	3.9	1.28	4490	6	240	18.0	1.28	9826
6	1.36	17.07	58	30.0	6.8	1.36	1706	64	26.5	10.3	1.36	5747	8	234	24.5	1.36	12575
7	1.40	2322	63	36.8	1202	1.40	2322	63	36.8	12.2	1.40	7846	10	226	34.5	1.04	17134
8	1.36	3242	62	49.0	24	1.36	3242	68	49.0	240	1.36	10984	12	216	50.8	1.36	23952
9	1.34	4420	68	650	—	1.34	4420	68	65.0	—	1.34	14938	14	204	72	1.34	32620

Billet $230 \times 190$.

the flange, both open and closed \((\Delta t)\). Along the abscissa axis the passes are plotted, and along the ordinate axis the corresponding quantities from those listed above, i.e. \((\mu_1, Q_1, \beta\) and \(\Delta t)\).

Example No. 2 (Tafel principle).

The Tafel principle consists in not allowing lateral pressure in the closed part of the pass; hence it naturally follows that the elongation coefficient \((\mu)\) in the closed part of the pass will be less than in the open part. Consequently, according to Tafel, it is necessary to design the pass with different elongation coefficients in the separate parts of the pass. Pass No. 1 will obviously have the same dimensions as in example No. 1; we shall not calculate its dimensions, but shall proceed to the determination of pass No. 2. We shall assign the elongation in pass No. 1 as 12% and equal in all parts of the pass, in order to avoid unequal internal stresses in the finished section. The elongation coefficient common to the whole pass will be:

\[ \mu_{\text{total}}=\mu_{\text{sh}1}=\mu'_1=\mu_1=1.12, \]

where: \(\mu_{\text{sh}1}\) = elongation coefficient of the neck
\(\mu_1\) — “ ” of the open flange
\(\mu'_1\) — “ ” of the closed “

Pass No. 2.

Open flange.

\[ R_2=R_1\mu_1=689\cdot1.12=772\ \mathrm{mm}^2 \]

The open flange of pass No. 2 goes into the closed part of pass No. 1, and since there must be no lateral pressure in the closed part, we have:

\[ t_2=t'_1=13.8\ \mathrm{mm}, \]

flange height

\[ h_2=\frac{R_2}{t_2}=\frac{772}{13.8}=57\ \mathrm{mm}, \]

\[ \operatorname{tg}\alpha_2=\frac{13.8}{122-0.5\cdot56}=0.147 \]

\[ a_2=(122-56)0.147=9.7\ \mathrm{mm} \]

\[ b_2=122\cdot0.147=17.9\ \text{“} \]

Closed flange:

\[ h'_2=h_1-0.2=51.2-0.2=51\ \mathrm{mm}. \]

\[ R'_2=R_2=772\ \mathrm{mm}^2 \]

\[ t'_2=\frac{R'_2}{h'_2}=\frac{772}{51}=15.2\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_2=\frac{15.2}{122-0.5\cdot51}=0.16 \]

\[ d_2=(122-51)0.16=11.5\ \mathrm{mm};\quad C_2=122\cdot 0.16=19.5\ \mathrm{mm}. \]

Neck:

\[ q_2=q_1\mu_1=2335\cdot 1.12=2615\ \mathrm{mm}^2 \]

\[ B_2=B_1-1.8=253.8-1.8=252\ \mathrm{mm} \]

\[ S_2=\frac{q_2}{B_2}=\frac{2615}{252}=10.4\ \mathrm{mm}. \]

Total area of the pass

\[ Q_2=2615+4\cdot 772=5703\ \mathrm{mm}^2 \]

Pass No. 3.

Open flange:

We obtain the open flange of pass No. 3 from the previous pass (pass No. 2) by closing its closed part.

The increase in flange height, \(n\), is determined on the basis of the difference of tangential velocities*).

Beam No. 25 is rolled in rolls with mean diameter

\[ D=(550+650)\ \mathrm{mm}. \]

The allowance for reduction \((n)\)—(increase in height)—according to the formulas of L. Gez (Fig. 4):

\[ r_0=\sqrt{R_0^2-2XR^0}\ . . . . . . . . . . . . . . . (1) \]

\[ n=R_0-(r_0+x)\ . . . . . . . . . . . . . . . . . . . . (2) \]

where: \(R_0\) — mean radius of the roll for the neck
\(\phantom{where: }r^0\) — ″ ″ ″ flange
\(\phantom{where: }x\) — height of the trapezoid = height of the flange of the preceding pass \((h'_2)\)
\(\phantom{where: }n\) — increase in flange height.

\[ n=R_0-\left(\sqrt{R_0^2-2h'_2R_0+h'_2}\right) =275-\left(\sqrt{285^2-2\cdot 51\cdot 275+51}\right)=6\ \mathrm{mm} \]

The flange thickness is equal to the thickness of the closed flange of the preceding pass, i.e. \(t_3=t'_2=15.2\ \mathrm{mm}\) (since there is no lateral pressure in the closed flange)

\[ h_3=h'_2+n=51+6=57\ \mathrm{mm} \]

\[ \operatorname{tg}\alpha_3=\frac{15.2}{122-0.5\cdot 57}=0.163 \]

\[ b_3=122\cdot 0.163=19.8\ \mathrm{mm} \]

\[ a_3=(122-57)0.163=10.6\ \mathrm{mm}. \]

) L. Gez, Roll Pass Design*, publ. 1929, p. 31.

The elongation coefficient in the closed flange of pass No. 2 will be:

\[ \mu'_2=\frac{R_3}{R'_2}=\frac{866.4}{772}=1.12. \]

The elongation coefficients in the passes, as well as in the individual parts of the pass, are also found graphically (see Fig. 5).

The elongation coefficient in the closed flange of pass No. 2 was obtained as \(\mu'_2=1.12\), and the maximum in the 8th pass is taken as \(\mu'_8=1.25\) (closed flange). We specify the change in the elongation coefficient according to the law of a straight line; drawing the straight line, we obtain a series of values of the elongation coefficient in the individual passes.

Open flanges:

In the second pass we take \(\mu_2=1.20\), and the maximum in the 8th is \(\mu_8=1.55\). Drawing a straight line (Fig. 5), we obtain a series of coefficient values in the open parts of the passes. For the web we take the arithmetic mean of the open and closed flanges, which will correspond to the general elongation coefficient for the entire pass; although this is not entirely exact, for practice the accuracy is sufficient:

\[ \mu_{ш2}=\frac{\mu_2+\mu'_2}{2}=\frac{1.20+1.12}{2}=1.16. \]

Thus, from the graph of Fig. 5 we obtain Table No. 2.

Table No. 2.

Nos.	Open flange	Closed flange	Web (arith. mean)
1	\(\mu_1 = 1.12\)	\(\mu'_1 = 1.12\)	\(\mu_{ш1}=\dfrac{1.12+1.12}{2}=1.12\)
2	\(\mu_2 = 1.20\)	\(\mu'_2 = 1.12\)	\(\mu_{ш2}=\dfrac{1.20+1.12}{2}=1.16\)
3	\(\mu_3 = 1.26\)	\(\mu'_3 = 1.14\)	\(\mu_{ш3}=\dfrac{1.26+1.14}{2}=1.20\)
4	\(\mu_4 = 1.32\)	\(\mu'_4 = 1.16\)	\(\mu_{ш4}=\dfrac{1.32+1.16}{2}=1.24\)
5	\(\mu_5 = 1.38\)	\(\mu'_5 = 1.18\)	\(\mu_{ш5}=\dfrac{1.38+1.18}{2}=1.28\)
6	\(\mu_6 = 1.43\)	\(\mu'_6 = 1.21\)	\(\mu_{ш6}=\dfrac{1.43+1.25}{2}=1.32\)
7	\(\mu_7 = 1.49\)	\(\mu'_7 = 1.23\)	\(\mu_{ш7}=\dfrac{1.48+1.21}{2}=1.36\)
8	\(\mu_8 = 1.55\)	\(\mu'_8 = 1.25\)	\(\mu_{ш8}=\dfrac{1.56+1.25}{2}=1.40\)

Closed flange:

\[ h'_3 = h_2 - 1 = 56 - 1 = 55\ mm \]

\[ R'_3 := R_2 = 772 \cdot 1{,}20 = 926\ mm^2 \]

\[ t'_3 = \frac{R_3}{h'_3} = \frac{926}{55} = 16{,}9\ mm \]

\[ tg\,\alpha'_3 = \frac{16{,}9}{122 - 0{,}5 \cdot 55} = 0{,}18 \]

\[ d_3 = (122 - 54)0{,}18 = 12\ mm \]

\[ c_3 = 122 \cdot 0{,}18 = 22\ mm \]

Neck.

\[ q_3 = q_2 \mu_{ш2} = 2615 \cdot 1{,}16 = 3033\ mm \]

\[ B_3 = B_2 - 3 = 252 - 3 = 249\ mm \]

\[ S_3 = \frac{q_3}{B_3} = \frac{3033}{249} = 12{,}2\ mm \]

The total area of the pass

\[ Q_3 = 3033 + 2{,}926 + 2 \cdot 866{,}4 = 6620\ mm^2 \]

Passes Nos. 4, 5, 6, 7, and 8 are calculated analogously to example No. 1, using the calculation of elongation coefficients listed in Table No. 2; we shall not repeat them here, but will present the results of the calculation in Table No. 3, “Calibration of I-beam No. 25 according to Tafel.”

Pass No. 9.

Pass No. 9 is a split pass, in which the open and closed flanges are equal to each other in height, but their areas are different, owing to the different elongation coefficients in pass No. 8.

\[ h_8 = h'_9 = h'_8 = 68\ mm \]

\[ R_9 = R_8 \mu'_8 = 2150 \cdot 1{,}25 = 3940\ mm^2 \]

\[ t_9 = \frac{3940}{68} = 58{,}0\ mm. \]

To determine the small base of the flange \((a_9)\), we use Kirchberg’s transition formula:

\[ a_9 = a_8 + \frac{a_8^2}{200} = 28{,}6 + \frac{(28{,}6)^2}{200} = 32{,}6\ mm. \]

Let us determine the pole distance \((X')\). From the similarity of triangles COD and AOB (Fig. 6), we have:

\[ X'=\frac{h_9 b_9}{q_9-a_9} =\frac{h_9(2t_9-a_9)}{2(t_9-a_9)} =\frac{68(2\cdot 58-32.6)}{2(58-32.6)}=113, \]

where \(b_9=2t_9-a_9=2\cdot 58-32.6=83.4\ \mathrm{mm}\).

This is the flange going into the closed part of pass No. 8 (in Table No. 3 it is called the “upper” one). The flange going into the open part of the pass will have the dimensions:

\[ R'_9=R_8\mu_8=2900\cdot 1.55=4495\ \mathrm{mm^2}. \]

\[ t'_9=\frac{4495}{68}=66\ \mathrm{mm} \]

\[ \tg \alpha'_9=\frac{66}{113-0.5\cdot 68}=0.839 \]

\[ d_9=(113-58)\cdot 0.839=37.8\ \mathrm{mm} \]

\[ c_9=113\cdot 0.839=94.8\ \mathrm{mm}. \]

Web:

\[ q_9=q_8\cdot \mu_{\mathrm{w}8}=10380\cdot 1.40=14540\ \mathrm{mm^2} \]

\[ B_9=B_8-13=216-12=204\ \mathrm{mm} \]

\[ S_9=\frac{q_9}{B_9}=\frac{14540}{204}=70\ \mathrm{mm} \]

The total area of pass No. 9:

\[ Q_9=14540+2(3940+4495)=31410\ \mathrm{mm^2}. \]

The billet for this beam No. 25 will have the following dimensions:

Height

\[ H_{\mathrm{bil.}}=2h_9+S_9+15=2\cdot 68+70+15=220\ \mathrm{mm}. \]

The billet width, if we take the broadening equal to \(14\ \mathrm{mm}\),

\[ B_{\mathrm{bil.}}=B_9-\beta=204-14=190\ \mathrm{mm}. \]

Area of the billet:

\[ Q_{\mathrm{bil.}}=H_{\mathrm{bil.}}B_{\mathrm{bil.}}=220\cdot 188=41800\ \mathrm{mm^2}. \]

Coefficient of elongation in the edging pass:

\[ \mu_9=\frac{Q_{\mathrm{bil.}}}{Q_9}=\frac{41800}{31410}=1.33. \]

Using Table No. 3, we construct the graph in Fig. 5, where the changes by passes are plotted: \(Q,\mu,\mu',\mu_{\mathrm{w}}=\mu_{\mathrm{total}},\Delta t\) and \(\beta\).

Table No. 3. Calibration of I-Beam No. 25 (after Tafel).

No. of passes	Upper flange: $\mu$	Upper flange: $R$	Upper flange: $h$	Upper flange: $t$	Upper flange: $\Delta t$	Lower flange: $\mu$	Lower flange: $R$	Lower flange: $h$	Lower flange: $t$	Lower flange: $\Delta t$	Web: $\mu_m$	Web: $q$	Web: $B$	Web: $S$	Web: $\beta$	Total: $\mu$	Total: $Q$
1	1.12	689	51.2	13.8	0.4	1.12	689	51.2	13.8	0	1.12	2335	258.8	9.4	1.8	1.12	5090
2	1.12	772	51	15.2	0	1.20	772	56	13.8	3.1	1.16	2615	252	10.4	3	1.16	5703
3	1.26	866	57	15.2	4.3	1.14	926	55	16.9	0	1.20	3033	249	12.2	4	1.20	6620
4	1.16	1092	56	19.5	0.5	1.32	1056	61	16.9	6.6	1.24	3640	245	14.9	5	1.24	7934
5	1.38	1265	63	20.0	8.3	1.18	1365	60	23.5	1.0	1.28	4514	240	18.8	6	1.28	9840
6	1.21	1750	62	28.3	2.5	1.43	1650	67	25.5	11.3	1.32	5780	234	24.8	8	1.32	12580
7	1.49	2115	69	30.8	15.6	1.23	2360	66	35.8	3.2	1.36	7630	226	33.8	10	1.36	16580
8	1.25	3150	68	46.4	11.6	1.55	2900	74	39.0	27	1.40	10380	216	48	12	1.40	22480
9	1.33	3940	68	58	—	1.33	4450	68	66	—	1.33	14540	204	70	14	1.33	31410

Preparation $220 \times 190$.

Example No. 3

In the preceding examples there were the following shortcomings in the choice of elongation coefficient: in example No. 1 we determined the elongation coefficients for the passes by Kirchberg’s formulas, which give suitable results only in six grooves, while in the remaining grooves they must be determined by other means; moreover, this method of determining elongations by grooves is not entirely convenient, since the number obtained has to be rounded off, as a result of which in the last grooves (passes) the elongation increases greatly. In example No. 2 the elongation coefficients are different in the individual parts of the groove, which leads to nonuniform stressing of the metal over the section of the profile during rolling.

We propose determining the elongation coefficients for the passes according to the law of a geometric progression. Here it is necessary to know only the elongation coefficient in the 1st groove and in the groove up to which we wish to increase the elongation coefficients according to the law of a geometric progression (up to the 8th inclusive), and also the total coefficient up to the given groove (the 8th groove). This is especially important when we are required to obtain a billet of definite dimensions for the given profile.

In the present example we shall perform the pass design by elongation coefficients varying in the individual grooves according to the law of a geometric progression.

In the preceding examples we obtained a billet with section

\(Q_{bil.} = 4370\ mm^2\)

and a total elongation coefficient:

\[ \mu_{tot.}=\frac{Q_{bil.}}{Q_1}=\frac{43700}{5000}=8,6. \]

Here we shall adhere to these quantities in order to have a complete analogy with the preceding examples; for this same purpose the same profile number is considered in all three examples.

In the edging groove (groove No. 9) it is recommended to take small elongation coefficients in order to ensure good filling of the groove. Tentatively we take:

\[ \mu_9=1,20 \]

\[ \mu_1=1,14. \]

Then the total up to the 8th groove (inclusive) will be:

\[ \mu_{tot.8}=\frac{\mu_{tot.}}{\mu_9}=\frac{8,60}{1,20}=7,10. \]

Using the formula of the geometric progression, we determine the elongation coefficients in all grooves (passes):

\[ \mu_{tot.n}=\left(\mu_1\cdot q_0^{\frac{n-1}{2}}\right)^n \qquad \ldots\ldots\ldots\ldots (1) \]

where: \(\mu_{tot.n}\) — total elongation coefficient up to the \(n\)-th groove (inclusive)

\(\mu_1\) — elongation coefficient in the first groove

$n$ — number of passes (grooves)

$q_0$ — common ratio of the geometric progression.

From formula (1) we determine the common ratio of the geometric progression $(q_0)$:

\[ lg\, q_0=\frac{2(lg\,\mu_{\mathrm{total},n}-n\,lg\,\mu_1)}{n(n-1)} \ldots \ldots \ldots \ldots \ldots (2) \]

The $n$-th term of the geometric progression:

\[ \mu_n=\mu_1\cdot q_0^{\,n-1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots (3) \]

By formula (2) we determine the common ratio $(q_0)$ for the given example:

\[ lg\, q_0=\frac{2(lg\,7{,}10-8\,lg\,1{,}14)}{8(8-1)}=0{,}0144. \]

By formula (3) we determine the draft coefficients by passes (grooves) according to formula (3):

\[ lg\,\mu_n=lg\,\mu_1+(n-1)lg\,q_0. \]

According to this expression, the coefficients calculated by passes will be:

\[ \begin{aligned} &\text{1st pass} && \mu_1=1{,}14\\ &\text{2nd } && \mu_2=1{,}18\\ &\text{3rd } && \mu_3=1{,}22\\ &\text{4th } && \mu_4=1{,}26\\ &\text{5th } && \mu_5=1{,}30 \qquad \mu_{\mathrm{total}}=8{,}60\\ &\text{9th } && \mu_6=1{,}34\\ &\text{7th } && \mu_7=1{,}39\\ &\text{8th } && \mu_8=1{,}43\\ &\text{9th } && \mu_9=1{,}20 \end{aligned} \]

From these coefficients the grooves are calculated. The calculation of the groove dimensions is performed in the same way as in example No. 1. The results of the calculations are brought together in Table No. 4; the calculation itself has been omitted by us in order to avoid repeating what was done in example No. 1. Grooves Nos. 1 and 2 will be exactly the same as in example No. 1, but grooves Nos. 3, 4, 5, 6, 7, and 8 will differ from the grooves of example No. 1 owing to the different draft coefficients (see Table No. 4). Here we shall calculate the dimensions of groove No. 9 (the slitting groove), since it is needed in order to check its filling when the billet is slit.

Groove No. 9.

Here the open and closed flanges are equal to each other:

\[ h_9=h_8+5=62+5=67\ mm \]

\[ R_9=R'_8\cdot\mu_8=3440\cdot1{,}43=4910\ mm^2 \]

\[ t_9=\frac{R_9}{h_9}=\frac{4910}{67}=73\ mm. \]

We increase the pole distance by 10% in order to have a sharper pass:

\[ x' = x \cdot 1.10 = 122 \cdot 1.10 = 134\ \mathrm{mm}. \]

\[ \tg \alpha_9 = \frac{73}{1.34 - 0.5 \cdot 67} = 0.72 \]

\[ b_9 = 1.34 \cdot 0.75 = 965\ \mathrm{mm} \]

\[ a_9 = (134 - 67)0.72 = 48.0\ \mathrm{mm} \]

Web:

\[ q_9 = q_8 \cdot \mu_8 = 11410 \cdot 1.47 = 16320\ \mathrm{mm}^2 \]

\[ B_9 = B_8 - 12 = 216 - 12 = 204\ \mathrm{mm} \]

\[ S_9 = \frac{q_9}{h_9} = \frac{16320}{204} = 80\ \mathrm{mm}. \]

Total area of the pass

\[ Q_9 = 16320 + 4 \cdot 4919 = 35960\ \mathrm{mm}^2. \]

The billet for this beam will have the following dimensions:

Billet width

\[ B_{\mathrm{bil.}} = B_9 - \beta = 204 - 14 = 190\ \mathrm{mm}. \]

Height of pass No. 9

\[ H_9 = 2h_9 + S_9 = 2.67 + 80 = 214\ \mathrm{mm}. \]

The billet height will be determined from the conditions of filling the roughing pass by Ekelund’s method*):

\[ H_{\mathrm{bil.}} = 4F_4\left(\frac{1}{h_4} - \frac{B_{\mathrm{bil.}}}{Q_9}\right) + h_4 \ldots \]

where \(H_{\mathrm{bil.}}\) is the height of the billet that must fill the pass,

\(B_{\mathrm{bil.}}\) is the billet width,

\(Q_9\) is the sectional area of the pass,

\(F_4\) is the area of the pass most subject to deformation
(see Fig. 7, the hatched area),

\(h_4\) is the mean height of the hatched area in Fig. 7.

If the area of the pass is divided by its width, we obtain the height \(h_a\), and this mean height will divide the pass into zones, of which the two outer zones will be less subject to deformation, while the middle zone will be subject to maximum deformation; its area is denoted by \(F_4\).

We calculate the area \(F_4\) graphically and obtain \(F_4 = 9300\ \mathrm{mm}^2\), width \(b_4 = 80\ \mathrm{mm}\), and maximum height \(h_a = 176\ \mathrm{mm}\).

Consequently, the mean height of the area (hatched in Fig. 7) will be:

\[ h_4 = \frac{F_4}{b_4} = \frac{9600}{80} = 120\ \mathrm{mm}. \]

Substituting all the quantities found into Ekelund’s formula, we find

\[ H_{\mathrm{bil.}} = 4 \cdot 9600\left(\frac{1}{120} - \frac{190}{35960}\right) + 129 \simeq 235\ \mathrm{mm}. \]

) See the journal Domes, 1930, No. 4, abstract of the article by W. Tafel and Wegener, “Das Füllen von Kalibern.” St. u. Eis.*, 1930, No. 5.

Table No. 4. Calibration of beams No. 25 (example No. 3).

Pass No.	Lower flange, $\mu$	Lower flange, $R$	Lower flange, $h$	Lower flange, $t$	Lower flange, $\Delta t$	Upper flange, $\mu$	Upper flange, $R$	Upper flange, $h$	Upper flange, $t$	Upper flange, $\Delta t$	Web, $\mu$	Web, $q$	Web, $B$	Web, $S$	Web, $\beta$	General, $\mu$	General, $Q$
1	1.14	689	51.2	13.8	1.9	1.14	189	51.2	13.8	0.2	1.14	2335	253.8	9.2	1.8	1.14	5091
2	1.18	786	50	15.7	1.1	1.18	786	56	14.0	2.8	1.18	2662	252	10.6	3	1.18	5806
3	1.22	928	55	16.8	4.0	1.22	828	55	16.8	2.0	1.22	3143	249	12.6	4	1.22	6855
4	1.26	1130	54	20.8	3.4	1.26	1130	60	18.8	5.4	1.26	3840	245	15.7	5	1.26	8360
5	1.30	1425	59	24.2	7.7	1.30	1425	59	24.2	4.8	1.30	4740	240	20	6	1.30	10440
6	1.34	1852	58	31.9	7.5	1.34	1852	65	29.4	10.4	1.34	6150	234	26	8	1.34	13560
7	1.38	2485	63	39.4	16.1	1.385	2485	63	39.4	11.1	1.385	8250	226	36	10	1.380	18190
8	1.43	3440	62	55.5	16.5	1.43	3440	68	50.5	21.5	1.43	11410	216	52.8	12	1.43	25170
9	1.24	4860	67	73	—	1.24	4860	67	73	—	1.24	16720	204	80	15	1.24	35960

Billet 235 × 190.

We obtain the area of the billet:

\[ Q_{\text{bil.}} = H_{\text{bil.}} \cdot B_{\text{bil.}} = 235 \cdot 190 = 44650\ \text{mm}. \]

The elongation coefficient in the slitting pass will be:

\[ \mu_9 = \frac{Q_{\text{bil.}}}{Q_9} = \frac{44650}{35940} = 1.24. \]

\[ \mu_{\text{tot.-0}} = \frac{Q_{\text{bil.}}}{Q_1} = \frac{44650}{5090} = 8.77. \]

We have obtained a total coefficient of elongation somewhat greater than was specified, for it is very difficult to arrive at the billet dimensions from the slitting groove while preserving the principle of filling the pass when the bloom is slit.

If, during the slitting of the bloom, the slitting pass were not filled, then this non-filling could persist chronically from pass to pass and affect the finished section; that is, the finished section might come out with unfilled edges at the flanges.

The results of the calculations are brought together in Table No. 4, and from this table curves (Fig. 8) have been constructed showing the changes, by passes, of \(Q, \beta, \Delta t\), and \(\mu\).

SUMMARY.

This work represents a theoretical study on the calibration of H-girders and especially on the calculation of coefficients of draught. The advantage of the method of Kirchberg for gauging H-girders consists in even draughts in all parts of the roll-groove and in the consequent absence of unequally strained in the section of the H-gider. But this method has a defect with regard to the calculation of coefficients of draught, the formulae of Kirchberg permitting to determine the coefficient of draught only as far as the 7 draught but failing farther. The good side of the method of gauging by the principle of Tafel consists in the absence of the lateral stress in the closed part of the roll-groove, thus increasing the durability of great rolls, but this method has also a great defect, the coefficients of draughts being various in differents parts of the roll-groove. This leads to different strains of the metal along the section of the rolled bar. The calculation of coefficients of draught in roll-grooves must certainly be submitted to some law with regard to the fall of temperature of the rolled bar. The practice shows that coefficients of draught are selected arbitrarily, their alteration by passes ranging in broad limits. This fact leads to an unequal wear of roll-groove and to the consequent rapid wear of the roll. Our gauging, example No. 3, gives the law of changes of coefficients of draught by passes according a geometric progression, the difference of which must be calculated depending on the fall of temperature in the rolled bar and depending on the number of passes. The coefficients of draught in roll-grooves will consequently diminish with the fall of temperature and this will lead to a rational exploitation of rolls and to an proportional gauging of H-girders.

Fig. 1.

Fig. 2.

Fig. 4.

Fig. 3.

Visible graph labels: \(Q\), \(\mu\), \(\Delta t\), \(\lambda\), \(C\), \(\beta\); \(\Delta t\)—lower flange; \(\Delta t\)—upper flange.

Visible scale markings: \(Q = 0, 60, 120, 180, 240, 300\); \(\mu = 0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0\); horizontal markings \(9, 8, 7, 6, 5, 4, 3, 2\).

Fig. 5.

Visible labels in the figure: \(Q\), \(\mu\), \(\Delta t\); \(\mu\)—lower flange; \(\mu\)—upper flange; \(\mu\)—web; \(\Delta t\)—lower flange; \(\Delta t\)—upper flange; \(Q\); \(\beta\).

Fig. 6.

Fig. 7.

Fig. 8.