Bridge compression member, ISLC 400, toe to toe

Approach. Two channels toe to toe, laced; IS 800:2007 cl. 7.1.2 (compressive strength) and cl. 7.6 (laced columns). $f_y = 250$ N/mm² (E250), $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$. Section properties of ISLC 400 @ 45.8 kg/m (IS 808; check against your steel table): $A = 58.3$ cm², $I_{xx} = 14000$ cm⁴, $I_{yy} = 462$ cm⁴, $C_{yy} = 2.37$ cm, $b = 100$ mm, $t_w = 8.0$ mm, $t_f = 14.0$ mm.

(i) Safe load of the member

The width over the backs is 400 mm, so the distance between the centroids of the channels is $h = 400 - 2C_{yy} = 400 - 47.4 = 352.6$ mm and the gap between the toes is $400 - 2\times100 = 200$ mm.

$$\begin{aligned} I_{zz} &= 2\times14000 = 28000\ \text{cm}^4,\quad r_{zz} = 15.50\ \text{cm} \\ I_{yy} &= 2[462 + 58.3\,(17.63)^2] = 37165\ \text{cm}^4,\quad r_{yy} = 17.85\ \text{cm} \\ \lambda_{zz} &= 8000/155.0 = 51.6,\quad \lambda_{yy} = 8000/178.5 = 44.8 \end{aligned}$$

Maximum slenderness $= 51.6$ (zz-axis). Laced column: $\lambda_{eff} = 1.05\times51.6 = 54.2$ (cl. 7.6.1.5). Buckling curve c ($\alpha = 0.49$, Table 10):

$$\begin{aligned} f_{cc} &= \frac{\pi^2E}{\lambda^2} = 672\ \text{N/mm}^2,\ \lambda_n = 0.610,\ \phi = 0.787,\ \chi = 0.779 \\ f_{cd} &= \chi f_y/\gamma_{m0} = 177.1\ \text{N/mm}^2 \\ P_d &= 2A f_{cd} = 2\times5830\times177.1/10^3 = 2065\ \text{kN} \end{aligned}$$

Answer (i): safe (design) load $\approx 2065$ kN (factored). Working load $= 2065/1.5 \approx 1377$ kN.

(ii) Design of lacing (M16, grade 4.6 bolts, single lacing)

Transverse shear $V_t = 2.5\%\,P = 0.025\times2065 = 51.6$ kN; per lacing plane $= 25.8$ kN (cl. 7.6.6.1).

Bolt lines are at mid-flange, so $s = 200 + 100 = 300$ mm. Angle of lacing $\theta = 45^\circ$ (between 40° and 70°).

$$\begin{aligned} l &= s/\sin\theta = 424\ \text{mm} \\ F &= \frac{V_t/2}{\sin\theta} = 36.5\ \text{kN} \\ a_1 &= 2s\cot\theta = 600\ \text{mm} \end{aligned}$$

Check of the main member between lacing points: $a_1/r_{y} = 600/28.2 = 21.3$, below the smaller of 50 and $0.7\lambda_{max} = 36.1$. OK (cl. 7.6.5.1).

Flat: width 60 mm ($\ge 3d = 48$ mm); thickness $\ge l/40 = 10.6$ mm and $\ge l\sqrt{12}/145 = 10.1$ mm, so use ISF 60 × 12 mm. $l/r = 122.5 \le 145$; $f_{cd} = 81.2$ N/mm², $P_{d,bar} = 720\times81.2/10^3 = 58.5$ kN $> 36.5$ kN. OK.

Bolts M16, grade 4.6: $V_{dsb} = \frac{400}{\sqrt3}\times\frac{0.78\times\pi\times16^2/4}{1.25\times10^3} = 29.0$ kN; $V_{dpb} = 2.5k_bdtf_u/\gamma_{mb} = 87.5$ kN ($k_b = 0.556$, $t = 12$ mm). Bolt value $= 29.0$ kN, so the number of bolts $= 36.5/29.0 = 1.26$; provide 2 bolts M16 at each end of each bar.

Net section of the flat in tension: $A_n = (60 - 18)\times12 = 504$ mm², $T_{dn} = 0.9f_uA_n/\gamma_{m1} = 148.8$ kN $> 36.5$ kN. OK.

End tie plates (cl. 7.6.8, 7.7.2): size $400\times360\times8$ mm (depth $\ge$ distance between centroids $= 353$ mm; thickness $\ge s/50 = 6$ mm).