Flexure Members

A plate girder is a built-up I-beam (web plate with flange plates welded or bolted to it) used when rolled sections cannot carry the moment or shear. Its size is fixed by an economical (minimum weight) proportioning followed by checks.

Process (step by step)

1. Loads and forces. Find the factored maximum bending moment $M$ and shear force $V$ from the span and loads (include an assumed self-weight, about 1.5 to 2.5 kN/m per 10 m span, to be revised).
2. Economical depth of web. Take $d \approx L/10$ to $L/12$ for a start, then use the minimum-weight condition. If the flanges carry all the moment, flange area $A_f = M/(f\,d)$ and web area $A_w = d\,t_w = d^2/k$ where $k = d/t_w$. Total area $A = 2M/(f d) + d^2/k$. For minimum $A$, $dA/dd = 0$:

$$d = \left(\frac{M\,k}{f}\right)^{1/3}, \qquad f = \frac{f_y}{\gamma_{m0}}$$

Here $k$ is chosen about 100 to 200 for a web with stiffeners (a deeper girder with a thinner web is lighter but needs more stiffeners). 3. Web thickness. $t_w = d/k$, then check limits of IS 800:2007 cl. 8.6.1: $d/t_w$ not more than about $200\varepsilon$ to $270\varepsilon$ depending on the stiffener spacing (serviceability), and $d/t_w \le 345\varepsilon_f^2$ to prevent the compression flange buckling into the web. Also the shear check $V \le V_d = d\,t_w f_{yw}/(\sqrt3\gamma_{m0})$ (with post-buckling strength if stiffened). The thickness is rounded up (not less than 8 mm; 6 mm in sheltered work). 4. Flange area. Flanges are assumed to take the whole moment (the web is conservatively ignored, or a part $A_w/6$ is added):

$$A_f = \frac{M\gamma_{m0}}{f_y\,(d + t_f)} - \frac{A_w}{6} \ (\text{approx.})$$

5. Flange plate size. Choose the flange width $b_f \approx d/3$ to $d/5$ (or $L/40$ to $L/45$ for lateral stability), then $t_f = A_f/b_f$ (round up to standard plate thickness). Check the outstand $b/t_f \le 8.4\varepsilon$ (plastic) or $9.4\varepsilon$ (compact), and up to $13.6\varepsilon$ for semi-compact welded flanges.
6. Check the trial section. Compute $I_{zz}$, $Z_{zz}$, the moment capacity (with lateral buckling if the flange is unsupported), the shear capacity, the deflection, and then design stiffeners, flange-web welds and splices.

Sketch of the girder section

   |<-- b_f -->|
   +-----------+  ---  t_f
        | |
        | |  d  (web, t_w)
        | |
   +-----------+  ---  t_f

The flanges give the moment resistance and the web gives the shear resistance. The depth, not the thickness, is the most effective way to gain strength, so the web is made as deep and as thin as the code limits allow.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
• Effective span $L = 6.30$ m, bearing length of support $= 300$ mm.

Loads and design forces

The hall is 6 m wide, so the beams span 6 m and are spaced 3 m c/c. With 300 mm wall bearing, effective span $= 6.0 + 0.3 = 6.3$ m. Unit weight of RCC $= 25$ kN/m³ (assumed); no floor finish is given.

• Slab: $0.12\times25 = 3.00$ kN/m², per metre of beam $= 3.00\times3 = 9.00$ kN/m
• Imposed load: $3\times3 = 9.00$ kN/m

Factored load $w_u = 1.5(9.00 + 9.00) = 27.00$ kN/m (plus self-weight).

Moment without self-weight $M = 134.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 134.0\times10^6\times1.1/250 = 589$ cm³ (self-weight adds a little).

Trial section

Try ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $R_1 = 14$ mm; $I_{zz} = 8990$ cm⁴; $Z_e = 599$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670$ cm³; mass $46.0$ kg/m.

Self-weight $= 46.0\times9.81/1000 = 0.451$ kN/m, factored $= 1.5\times0.451 = 0.677$ kN/m. Total factored UDL $w = 27.00 + 0.68 = 27.68$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{27.68\times6.30^2}{8} = 137.3\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 87.2\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 31.9 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 245.8$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The compression flange is supported throughout its length by the slab, so lateral buckling is prevented (cl. 8.2.2); this is the check for lateral stability.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 152.4$ kN·m.

$$M_d = 152.4\ \text{kN·m} > M_u = 137.3\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned}$$

$V_u = 87.2$ kN $< V_d$; also $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 31.9 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times9.00\times(6300)^4}{384\times2\times10^5\times8990\times10^4} = 10.27\ \text{mm}$$

$\delta = 10.27$ mm $< L/300 = 21.00$ mm. OK.

Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length $b_1 = 300$ mm. Clear web depth $d = 245.8$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 122.9$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$; curve c gives $f_{cd} = 140.4$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 122.9)\times7.7\times140.4/10^3 = 457.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 67.8)\times7.7\times250}{1.1\times10^3} = 643.6\ \text{kN} \end{aligned}$$

Support reaction $R = 87.2$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 300 (46.0 kg/m): $M_d = 152.4$ kN·m $> M_u = 137.3$ kN·m, $V_d = 303$ kN $> V_u = 87.2$ kN, $\delta = 10.3$ mm $< 21.0$ mm.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
• Effective span $L = 6.00$ m.

Loads and design forces

Span $L = 6$ m. The "30 kN" imposed load is read as 30 kN/m (a UDL, like the dead load of 10 kN/m); dead load is taken to exclude the beam's own weight. Factored load $w_u = 1.5(10 + 30) = 60$ kN/m, plus factored self-weight. The beam is "sufficiently stiff against bearing", so the web buckling and bearing checks are not required.

Moment without self-weight $M = 270.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 270.0\times10^6\times1.1/250 = 1188$ cm³ (self-weight adds a little).

Trial section

Try ISMB 450 (IS 808): $D = 450$ mm, $b_f = 150$ mm, $t_w = 9.4$ mm, $t_f = 17.4$ mm, $R_1 = 15$ mm; $I_{zz} = 30400$ cm⁴; $Z_e = 1350$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1534$ cm³; mass $72.4$ kg/m.

Self-weight $= 72.4\times9.81/1000 = 0.710$ kN/m, factored $= 1.5\times0.710 = 1.065$ kN/m. Total factored UDL $w = 60.00 + 1.07 = 61.07$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{61.07\times6.00^2}{8} = 274.8\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 183.2\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (150/2)/17.4 = 4.31 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 41.0 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 385.2$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The compression flange is restrained by the RCC slab along its full length, so lateral-torsional buckling is prevented.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 348.7$ kN·m.

$$M_d = 348.7\ \text{kN·m} > M_u = 274.8\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned}$$

$V_u = 183.2$ kN $< V_d$; also $V_u < 0.6V_d = 333.0$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 41.0 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times30.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 8.33\ \text{mm}$$

$\delta = 8.33$ mm $< L/300 = 20.00$ mm. OK.

Web bearing/buckling at supports is not checked because the question states the beam is sufficiently stiff against bearing.

Final design

Provide ISMB 450 (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 274.8$ kN·m, $V_d = 555$ kN $> V_u = 183.2$ kN, $\delta = 8.3$ mm $< 20.0$ mm.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
• Effective span $L = 6.00$ m.

Loads and design forces

Span $L = 6$ m. Dead load $25$ kN/m (taken to exclude the beam's own weight) and imposed load $20$ kN/m. Factored $w_u = 1.5(25 + 20) = 67.5$ kN/m plus factored self-weight. The beam is "sufficiently stiff against bearing", so no web bearing check is needed.

Moment without self-weight $M = 303.8$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 303.8\times10^6\times1.1/250 = 1336$ cm³ (self-weight adds a little).

Trial section

Try ISMB 450 (IS 808): $D = 450$ mm, $b_f = 150$ mm, $t_w = 9.4$ mm, $t_f = 17.4$ mm, $R_1 = 15$ mm; $I_{zz} = 30400$ cm⁴; $Z_e = 1350$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1534$ cm³; mass $72.4$ kg/m.

Self-weight $= 72.4\times9.81/1000 = 0.710$ kN/m, factored $= 1.5\times0.710 = 1.065$ kN/m. Total factored UDL $w = 67.50 + 1.07 = 68.57$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{68.57\times6.00^2}{8} = 308.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 205.7\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (150/2)/17.4 = 4.31 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 41.0 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 385.2$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The compression flange is restrained by the RCC slab along its full length, so lateral-torsional buckling is prevented and $M_d$ is the plastic moment capacity.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 348.7$ kN·m.

$$M_d = 348.7\ \text{kN·m} > M_u = 308.5\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned}$$

$V_u = 205.7$ kN $< V_d$; also $V_u < 0.6V_d = 333.0$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 41.0 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 5.55\ \text{mm}$$

$\delta = 5.55$ mm $< L/300 = 20.00$ mm. OK.

Web buckling and bearing are not checked because the beam is stated to be sufficiently stiff against bearing.

Final design

Provide ISMB 450 (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 308.5$ kN·m, $V_d = 555$ kN $> V_u = 205.7$ kN, $\delta = 5.6$ mm $< 20.0$ mm.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
• Effective span $L = 6.25$ m, bearing length of support $= 250$ mm.

Loads and design forces

The hall is 6 m wide, so the beams span 6 m at 3 m c/c. Wall bearing 250 mm: effective span $L = 6.0 + 0.25 = 6.25$ m. RCC unit weight 25 kN/m³ (assumed).

• Slab: $0.11\times25 = 2.75$ kN/m², on 3 m width $= 8.25$ kN/m
• Imposed: $4\times3 = 12.00$ kN/m

$w_u = 1.5(8.25 + 12.00) = 30.375$ kN/m plus self-weight. Stiff bearing $b_1 = 250$ mm.

Moment without self-weight $M = 148.3$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 148.3\times10^6\times1.1/250 = 653$ cm³ (self-weight adds a little).

Trial section

Try ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $R_1 = 14$ mm; $I_{zz} = 8990$ cm⁴; $Z_e = 599$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670$ cm³; mass $46.0$ kg/m.

Self-weight $= 46.0\times9.81/1000 = 0.451$ kN/m, factored $= 1.5\times0.451 = 0.677$ kN/m. Total factored UDL $w = 30.38 + 0.68 = 31.05$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{31.05\times6.25^2}{8} = 151.6\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 97.0\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 31.9 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 245.8$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The slab, cast over the beam, supports the compression flange throughout its length, so lateral buckling cannot occur; this satisfies the check for lateral stability (cl. 8.2.2).

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 152.4$ kN·m.

$$M_d = 152.4\ \text{kN·m} > M_u = 151.6\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned}$$

$V_u = 97.0$ kN $< V_d$; also $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 31.9 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times12.00\times(6250)^4}{384\times2\times10^5\times8990\times10^4} = 13.26\ \text{mm}$$

$\delta = 13.26$ mm $< L/300 = 20.83$ mm. OK.

Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length $b_1 = 250$ mm. Clear web depth $d = 245.8$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 122.9$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$; curve c gives $f_{cd} = 140.4$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (250 + 122.9)\times7.7\times140.4/10^3 = 403.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(250 + 67.8)\times7.7\times250}{1.1\times10^3} = 556.1\ \text{kN} \end{aligned}$$

Support reaction $R = 97.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 300 (46.0 kg/m): $M_d = 152.4$ kN·m $> M_u = 151.6$ kN·m, $V_d = 303$ kN $> V_u = 97.0$ kN, $\delta = 13.3$ mm $< 20.8$ mm.

Data

• Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
• Effective span $L = 6.30$ m, bearing length of support $= 300$ mm.

Loads and design forces

Effective span of joist (spans the 6 m width): clear span $+$ bearing $= 6.0 + 0.3 = 6.3$ m. Joists at 3 m c/c, so each joist carries a 3 m width of floor.

Dead load $= 10.08$ kN/m (excluding joist self-weight), live load $= 12.00$ kN/m. Factored load (IS 800 Table 4, $\gamma_f = 1.5$ for DL and IL): $w_u = 1.5(10.08 + 12.00) = 33.12$ kN/m.

Moment without self-weight $M = 164.3$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 164.3\times10^6\times1.1/250 = 723$ cm³ (self-weight adds a little).

Trial section

Try ISMB 350 (IS 808): $D = 350$ mm, $b_f = 140$ mm, $t_w = 8.1$ mm, $t_f = 14.2$ mm, $R_1 = 14$ mm; $I_{zz} = 13600$ cm⁴; $Z_e = 779$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 877$ cm³; mass $52.4$ kg/m.

Self-weight $= 52.4\times9.81/1000 = 0.514$ kN/m, factored $= 1.5\times0.514 = 0.771$ kN/m. Total factored UDL $w = 33.12 + 0.77 = 33.89$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{33.89\times6.30^2}{8} = 168.1\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 106.8\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/14.2 = 4.93 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 36.2 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 293.6$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The RCC slab is cast over the joist and holds the compression flange in position along its length, so the beam is laterally supported and lateral-torsional buckling does not govern. The check for lateral stability is therefore satisfied by the full restraint of the compression flange (cl. 8.2.2: no reduction when the flange is continuously restrained).

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{877\times10^3\times250}{1.1\times10^6} = 199.3\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times779\times10^3\times250/(1.1\times10^6) = 212.5\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 199.3$ kN·m.

$$M_d = 199.3\ \text{kN·m} > M_u = 168.1\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 350\times8.1 = 2835\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2835\times250}{\sqrt3\times1.1\times10^3} = 372.0\ \text{kN} \end{aligned}$$

$V_u = 106.8$ kN $< V_d$; also $V_u < 0.6V_d = 223.2$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 36.2 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times12.00\times(6300)^4}{384\times2\times10^5\times13600\times10^4} = 9.05\ \text{mm}$$

$\delta = 9.05$ mm $< L/300 = 21.00$ mm. OK.

Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length $b_1 = 300$ mm. Clear web depth $d = 293.6$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 146.8$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 87.9$; curve c gives $f_{cd} = 124.1$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 146.8)\times8.1\times124.1/10^3 = 449.3\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 70.5\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 70.5)\times8.1\times250}{1.1\times10^3} = 682.1\ \text{kN} \end{aligned}$$

Support reaction $R = 106.8$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 350 (52.4 kg/m): $M_d = 199.3$ kN·m $> M_u = 168.1$ kN·m, $V_d = 372$ kN $> V_u = 106.8$ kN, $\delta = 9.0$ mm $< 21.0$ mm.

Approach. IS 800:2007 cl. 8.2: design bending strength $M_d = \beta_b Z_p f_y/\gamma_{m0}$ for a laterally supported beam, and $M_d = \beta_b Z_p f_{bd}$ with $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$ for an unrestrained beam. $f_y = 250$ N/mm² (Fe 410), $\gamma_{m0} = 1.10$. Safe load found from $M_d = wL^2/8$.

Section data: ISLB 600 @ 99.5 kg/m (IS 808)

$D = 600$ mm, $b_f = 210$ mm, $t_w = 10.5$ mm, $t_f = 15.5$ mm, $R_1 = 20$ mm, $I_{zz} = 72800$ cm⁴, $I_{yy} = 1820$ cm⁴, $Z_e = 2430$ cm³.

Plastic modulus (from dimensions): $Z_p = b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 2752$ cm³.

Classification: flange $b/t_f = 105/15.5 = 6.77 < 9.4$; web $d/t_w = 50.4 < 84$. The section is plastic ($\beta_b = 1$).

(a) Compression flange restrained against lateral buckling

$$\begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{2752\times10^3\times250}{1.1\times10^6} = 625.6\ \text{kN·m} \\ \text{Limit } 1.2Z_ef_y/\gamma_{m0} &= 662.7\ \text{kN·m} \\ M_d &= 625.6\ \text{kN·m} \end{aligned}$$

Safe factored UDL (including self-weight): $w = 8M_d/L^2 = 8\times625.6/7^2 = 102.1$ kN/m.

Shear check: $V = wL/2 = 357.5$ kN $< V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 827$ kN, and below $0.6V_d$, so bending is not reduced.

Service load $= 102.1/1.5 = 68.1$ kN/m. Deducting the self-weight ($99.5\times9.81/1000 = 0.98$ kN/m) the superimposed service load is $\approx 67.1$ kN/m.

Answer (a): $M_d = 626$ kN·m; safe factored load $\approx 102.1$ kN/m ($\approx 68.1$ kN/m unfactored, including self-weight).

(b) Not restrained against lateral buckling (torsion restrained, flanges free to warp)

For simply supported ends with torsion restrained and warping free, $L_{LT} = 1.0L = 7000$ mm (Table 15).

$$\begin{aligned} I_t &= \frac{2b_ft_f^3 + (D-2t_f)t_w^3}{3} = 74.1\ \text{cm}^4 \\ h_f &= D - t_f = 584.5\ \text{mm},\quad I_w = \frac{I_yh_f^2}{4} = 155.4\times10^{10}\ \text{mm}^6 \\ M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 296.1\ \text{kN·m} \end{aligned}$$

(with $G = 0.769\times10^5$ N/mm²). Rolled section, so $\alpha_{LT} = 0.21$ (Table 13):

$$\begin{aligned} \lambda_{LT} &= \sqrt{Z_pf_y/M_{cr}} = 1.524 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 1.801 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.362 \\ f_{bd} &= \chi_{LT}f_y/\gamma_{m0} = 82.3\ \text{N/mm}^2 \\ M_d &= Z_pf_{bd} = 226.7\ \text{kN·m} \end{aligned}$$

Safe factored UDL: $w = 8M_d/L^2 = 8\times226.7/49 = 37.0$ kN/m (service $\approx 24.7$ kN/m).

Answer (b): $M_d = 227$ kN·m; safe factored load $\approx 37.0$ kN/m ($\approx 24.7$ kN/m unfactored, including self-weight).

Lack of lateral restraint reduces the strength to about 36% of the restrained value.

Approach. Plastic design strength of the built-up section (IS 800:2007 cl. 8.2.1), then reduction of moment capacity under high shear (cl. 9.2), and the load found from the governing check. $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. The beam is laterally restrained, so lateral-torsional buckling is ignored. Plates are assumed fully connected to the flanges (continuous welds or adequate bolts), one 200×12 plate on each flange. Effective span $L = 3.5$ m.

Section properties

ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $I_{zz} = 8990$ cm⁴, mass 46.0 kg/m. Plastic modulus of ISMB 300 from its dimensions: $Z_{p,I} = 670$ cm³.

$$\begin{aligned} \text{Overall depth} &= 300 + 2\times12 = 324\ \text{mm} \\ y_c &= 150 + 6 = 156\ \text{mm (plate centroid from N.A.)} \\ I_{zz} &= 8990\times10^4 + 2\left[\frac{200\times12^3}{12} + 200\times12\times156^2\right] = 20677\ \text{cm}^4 \\ Z_e &= \frac{I_{zz}}{162} = 1276\ \text{cm}^3 \\ Z_p &= Z_{p,I} + 2(200\times12)(156) = 670 + 749 = 1419\ \text{cm}^3 \end{aligned}$$

Classification: plate outstand from the web face $= (200 - 7.7)/2 = 96.2$ mm, $b/t = 96.2/12 = 8.0 < 9.4\varepsilon = 9.4$ (plastic); ISMB web $d/t_w = 31.9 < 84$. So $\beta_b = 1$.

Moment capacity

$$\begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{1419\times10^3\times250}{1.1\times10^6} = 322.6\ \text{kN·m} \\ 1.2Z_ef_y/\gamma_{m0} &= 348.1\ \text{kN·m} \end{aligned}$$

$M_d = 322.6$ kN·m. Load limited by moment: $w = 8M_d/L^2 = 8\times322.6/3.5^2 = 210.7$ kN/m.

Shear capacity

Shear is resisted by the web of the rolled section: $A_v = D t_w = 300\times7.7 = 2310$ mm² (plates ignored, conservative).

$$V_d = \frac{A_vf_y}{\sqrt3\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN}$$

Load limited by shear: $w = 2V_d/L = 2\times303.1/3.5 = 173.2$ kN/m. Shear governs (173.2 < 210.7).

Moment capacity under high shear (cl. 9.2.2)

At $w = 173.2$ kN/m: $V = V_d = 303.1$ kN $> 0.6V_d$, so the moment capacity is reduced. $M_{fd}$ is the plastic strength of the section without the shear area: $Z_{fd} = Z_p - t_wD^2/4 = 1419 - 173 = 1246$ cm³, $M_{fd} = 283.2$ kN·m.

$$\begin{aligned} \beta &= \left(\frac{2V}{V_d} - 1\right)^2 = 1.00 \\ M_{dv} &= M_d - \beta(M_d - M_{fd}) = 322.6 - 1.00(322.6 - 283.2) = 283.2\ \text{kN·m} \end{aligned}$$

Moment at this load $= wL^2/8 = 173.2\times3.5^2/8 = 265.2$ kN·m $< M_{dv} = 283.2$ kN·m. OK.

Result

Self-weight $= (46.0 + 2\times0.2\times0.012\times7850)\times9.81/1000 = 0.82$ kN/m.

Answer: the safe factored uniformly distributed load is $\approx 173$ kN/m (including self-weight; governed by shear), i.e. a service load of $\approx 115$ kN/m, or about 172 kN/m factored superimposed load after deducting the factored self-weight of 1.23 kN/m. Total factored reaction $= 303$ kN.

Lateral-torsional buckling

When a beam bends about its major axis, the compression flange behaves like a column but is restrained by the tension flange and web. If the compression flange is not held laterally, at a certain load the beam suddenly deflects sideways and twists at the same time. This failure by combined lateral bending and twisting is lateral-torsional buckling (LTB). It reduces the bending strength below the plastic moment of the section.

• The strength is governed by the elastic critical moment $M_{cr}$, which depends on the unbraced length $L_{LT}$, $EI_y$, the torsional constant $GI_t$ and the warping constant $EI_w$ (IS 800:2007 cl. 8.2.2, Annex E).
• Slenderness $\lambda_{LT} = \sqrt{\beta_bZ_pf_y/M_{cr}}$, then $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$ and $M_d = \beta_bZ_pf_{bd}$.
• It is avoided or reduced by closer lateral bracing, a slab connected to the compression flange, wider flanges (larger $I_y$), or closed sections (box, which have a high $I_t$).
• Short, stocky or fully restrained beams (for $\lambda_{LT} \le 0.4$) are not affected.

Curtailment of flanges in a plate girder

The bending moment in a simply supported girder is largest at mid-span and zero at the supports. A flange plate sized for the maximum moment is more than is needed near the ends, so the flange thickness or width is curtailed (reduced, or extra cover plates are stopped short) where the bending moment is smaller. This saves steel.

   +---------------------------+  <- full flange plate
 +-------------------------------+
 |        theoretical cut-off    |
 |       /                       \
 |  extra length for development   |

• The theoretical cut-off point is found where the moment of resistance of the reduced section $M_{d,red}$ equals the applied moment. For a UDL $w$ on span $L$: $w\,x(L-x)/2 = M_{d,red}$, giving $x = \dfrac{L}{2} - \sqrt{\dfrac{L^2}{4} - \dfrac{2M_{d,red}}{w}}$ from the support.
• The cut plate must be extended beyond the theoretical point (by enough length and enough bolts or welds to develop the force in the plate) so that it can take its share of the force, and the end is welded all round to avoid fatigue cracks. The welding or bolting between flange and plate over the extension must carry the plate force.
• Curtailing is done in stages to keep the section changes small, and the web-to-flange connection is checked at each change because the shear flow changes with the section.