--- title: "Bridge compression member, 1200 kN, tie bar lacing" subject: "Design of Steel and Timber Structure" chapter: 6 key: "semester_6/design_of_steel_and_timber_structure:bridge-compression-1200kn-tie-bar" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bridge-compression-1200kn-tie-bar site: BCT Engineering --- # Bridge compression member, 1200 kN, tie bar lacing Design of Steel and Timber Structure · Chapter 6: Compression Members - Asked in: 2071 Magh, 5+7 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/bridge-compression-1200kn-tie-bar - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-6 ## Question Design a bridge compression member using two channels placed back to back to carry a factorial load of 1200 kN, if effective length of column is 8.5 m. Also design the single lacing system using tie bar. ## Short answer Use two ISMC 300 channels back to back with 190 mm gap, overall width 370 mm. Factored capacity is 1324 kN > 1200 kN. Single lacing with ISF 60×10 mm flat tie bars, two M16 bolts per end. ## Answer **Approach.** Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$. #### Data - Factored axial load $P = 1200$ kN; effective length $KL = 8.5$ m (given, same for both axes). - Built-up member: buckling curve **c**, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of a laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5). #### Step 1: Trial section Try 2 × **ISMC 300** (IS 808): $A = 46.3$ cm² each, $I_{xx} = 6420$ cm⁴, $I_{yy} = 313$ cm⁴, $C_{yy} = 2.35$ cm, $B = 90$ mm, $t_w = 7.8$ mm, $t_f = 13.6$ mm. Total area $A_e = 2 \times 4630 = 9260$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 11.78$ cm. $$ \begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{8500}{117.8} = 72.2 \\ \lambda_{eff} &= 1.05 \times 72.2 = 75.8 \end{aligned} $$ For $\lambda_{eff} = 75.8$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 344$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.853$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 1.024$, $\chi = 0.629$. $$ \begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.629 \times 250/1.1 = 142.9\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 9260 \times 142.9/10^3 = 1324\ \text{kN} > 1200\ \text{kN} \end{aligned} $$ #### Step 2: Spacing of channels To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels: $$ \begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(6420 - 313) \times 10^4}{4630}} = 230\ \text{mm} \end{aligned} $$ Gap between webs $g = h - 2C_{yy} = 230 - 47.0 = 183$ mm. Provide **g = 190 mm** (centroid distance $h = 237$ mm). $$ \begin{aligned} I_{yy} &= 2[313 + 46.3(11.85)^2] = 13629\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 12.13\ \text{cm}, \quad \lambda_{yy} = 8500/121.3 = 70.1 \end{aligned} $$ $\lambda_{yy} = 70.1 \le \lambda_{zz} = 72.2$, so the zz-axis governs and the capacity $P_d = 1324$ kN $\ge 1200$ kN. **Safe.** Overall width of column $= 2B + g = 370$ mm. ```text +-------+ gap g +-------+ | ISMC |<--------->| ISMC | +-------+ +-------+ webs face each other, flanges point outward lacing bolted to the flanges (top and bottom) ``` #### Step 3: Single lacing system using flat tie bars (two planes) Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1200 = 30.0$ kN, shared by two lacing planes: $15.00$ kN each. Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 190 + 90 = 280$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4). $$ \begin{aligned} \text{Length of bar } l &= s/\sin\theta = 280/\sin 45^\circ = 396\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{15.00}{\sin 45^\circ} = 21.21\ \text{kN} \end{aligned} $$ Spacing of connections along one channel $a_1 = 2s\cot\theta = 560$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 560/26.0 = 21.5$, which is less than the smaller of 50 and $0.7\lambda_{max} = 50.5$. OK. **Tie bar (flat) size.** Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 9.9$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 9.5$ mm. Provide **ISF 60 × 10 mm**. $r = t/\sqrt{12} = 2.89$ mm, $l/r = 137.2 < 145$ OK. $f_{cd}$ (curve c) $= 68.4$ N/mm², $P_{d,lace} = 600 \times 68.4/10^3 = 41.0$ kN $> F = 21.21$ kN. OK. #### Step 4: Connection of lacing (M16, grade 4.6 bolts) $$ \begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 10 \times 410}{1.25 \times 10^3} = 72.9\ \text{kN} \end{aligned} $$ ($k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$.) Bolt value $= 29.0$ kN. Bolts needed $= 21.21/29.0 = 0.73$, so provide **1 bolt M16** at each end of every bar. Net section of flat in tension: $A_n = (60 - 18) \times 10 = 420$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 124.0$ kN $> 21.21$ kN. OK. #### Step 5: End tie plates Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth (distance between outermost bolts) $\ge h = 237$ mm; thickness $\ge s/50 = 5.6$ mm. Use 4 M16 bolts on each channel at 80 mm pitch (outermost bolts 240 mm apart) with 30 mm end distance, so the plate is **370 mm wide × 300 mm deep × 6 mm thick**. With $C = 560$ mm, $N=2$: $V_b = V_tC/(NS) = 30.0$ kN and $M = V_tC/2N = 4200$ kN·mm. Bolt distances from the group centre are ±40 and ±120 mm, $\Sigma r^2 = 32000$ mm². Bolt forces: from $M$ $= 4200 \times 120/32000 = 15.8$ kN, from shear $= 30.0/4 = 7.5$ kN, resultant $= 17.4$ kN $\le 29.0$ kN. OK. #### Final design - Column: 2 × ISMC 300 back to back, gap 190 mm, $P_d = 1324$ kN $> 1200$ kN. - Lacing: single, flats 60 × 10 mm at 45°, 1 M16 bolt at each end. - End tie plates 370 × 300 × 6 mm, 4 M16 bolts per channel. ```text +--------------------+ <- end tie plate | \ /\ /\ /| | \ / \ / \ / | single lacing | \/ \/ \/ | (same direction | /\ /\ /\ | on both faces) +--------------------+ <- end tie plate ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the laced column design, how is the effective slenderness ratio obtained from the actual maximum slenderness? A. Multiplied by 1.05 (cl. 7.6.1.5) B. Multiplied by 0.7 C. Left unchanged D. Divided by 1.05 Answer: A. For laced columns IS 800 requires the effective slenderness to be taken as 1.05 times the actual maximum slenderness ratio, which allows for shear deformation in the lacing. (Key point: Effective slenderness of laced column) 2. Why is the gap between the two channels chosen so that I_yy >= I_zz? A. To make the column equally strong about both axes B. To reduce the weight of the lacing C. To make the tie plates thinner D. To raise the yield strength of the steel Answer: A. Making the built-up I about the open axis at least equal to that about the web axis means neither axis is weaker, so the zz-axis slenderness governs and capacity is not reduced. (Key point: Spacing of channels) 3. What transverse shear is used to design the lacing for P = 1200 kN, and how is it shared? A. 2.5% of P = 30 kN, shared by the two lacing planes (15 kN each) B. 2.5% of P = 30 kN, taken entirely by one plane C. 5% of P = 60 kN, shared by two planes D. 10% of P = 120 kN, shared by two planes Answer: A. Lacing is designed for a transverse shear of 2.5% of the axial load. With two lacing planes, each takes half, 15 kN. (Key point: Transverse shear for lacing) 4. With lacing at 45° to the column axis and 15 kN shear per plane, what is the force in each bar and why? A. 21.21 kN, because F = (V/2)/sin 45° B. 15 kN, because the bar carries the shear directly C. 10.6 kN, because F = (V/2) x sin 45° D. 30 kN, because both planes load the same bar Answer: A. The bar is inclined, so its axial force is the plane shear divided by sin θ: 15/0.7071 = 21.21 kN. 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