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I-section purlin for industrial building

Short answer

Provide an ISMB 125 purlin with a sag rod at mid-span. The section satisfies biaxial bending, shear, and deflection; use cleat connections with two M12 bolts at each end.

Design an I-section purlin for an industrial building to support a galvanized corrugated iron sheet roof from the following data: Spacing of the trusses = 5 m Spacing of purlins = 1.5 m Inclination of main rafter to horizontal = 30° Weight of galvanized sheets taking into account laps and connecting bolts = 130 N/m² Wind load = 1 kN/m²

Approach. The purlin is a simply supported beam bent about both axes, because the loads are inclined to its principal axes. Design to IS 800:2007 (cl. 8.2 and 9.3), with loads from IS 875.

Data

  • Steel Fe 410 (E250): fy=250f_y = 250 N/mm², γm0=1.10\gamma_{m0} = 1.10. Purlin span L=5.0L = 5.0 m (truss spacing); inclination of rafter α=30.00∘\alpha = 30.00^\circ (cos⁡α=0.866\cos\alpha = 0.866, sin⁡α=0.500\sin\alpha = 0.500).
  • Sag rods are provided at mid-span to restrain the purlin in the plane of the roof. The 1 kN/m² wind load is taken as the net design pressure normal to the roof (acting either towards or away from the roof). Spacing of purlins along the slope = 1.5 m; trusses at 5 m.

Loads per metre run of purlin

  • Sheeting: 0.130×1.5=0.1950.130\times1.5 = 0.195 kN/m (dead, vertical), plus the self-weight of the purlin (trial section below).
  • Live load (IS 875 Part 2, roof slope above 10°): 0.75−0.02(α−10)=0.350.75 - 0.02(\alpha - 10) = 0.35 kN/m², but not less than 0.40 kN/m² on plan, so use 0.400.40 kN/m². Per metre of purlin: 0.40×1.5cos⁡α=0.5200.40\times1.5\cos\alpha = 0.520 kN/m (vertical).
  • Wind: given wind pressure 1.01.0 kN/m² ×1.5\times 1.5 m =1.500= 1.500 kN/m, acting normal to the roof.

Trial section and capacities

Try ISMB 125: D=125D = 125, bf=70b_f = 70, tw=5.0t_w = 5.0, tf=8.0t_f = 8.0 mm, Izz=445I_{zz} = 445 cm⁴, Iyy=38.5I_{yy} = 38.5 cm⁴, Zez=71.2Z_{ez} = 71.2 cm³, mass 13.3 kg/m (self-weight 0.1300.130 kN/m).

Zpz=bftf(D−tf)+tw(D−2tf)2/4=80.4 cm3,Mdz=Zpzfyγm0=18.27 kN⋅mZpy=2⋅tfbf24+(D−2tf)tw24=20.3 cm3,Mdy=min⁡(Zpy,1.2Zey)fyγm0=3.00 kN⋅m\begin{aligned} Z_{pz} &= b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 80.4\ \text{cm}^3,\quad M_{dz} = \frac{Z_{pz}f_y}{\gamma_{m0}} = 18.27\ \text{kN·m} \\ Z_{py} &= 2\cdot\frac{t_fb_f^2}{4} + \frac{(D - 2t_f)t_w^2}{4} = 20.3\ \text{cm}^3,\quad M_{dy} = \min(Z_{py}, 1.2Z_{ey})\frac{f_y}{\gamma_{m0}} = 3.00\ \text{kN·m} \end{aligned}

Section class: b/tfb/t_f and d/twd/t_w are within the plastic limits (9.49.4 and 8484).

Load combinations and moments (IS 800 Table 4)

Normal component wnw_n and tangential component wtw_t (kN/m), with Mz=wnL2/8M_z = w_nL^2/8 and My=wt(L′)2/8M_y = w_t(L')^2/8, where L′=L/2L' = L/2 (one sag rod at mid-span):

Combinationwnw_nwtw_tMzM_z (kN·m)MyM_y (kN·m)MdzM_{dz} usedMz/Mdz+My/MdyM_z/M_{dz} + M_y/M_{dy}
1.5(DL+LL)+1.100.633.430.5018.270.35
1.2(DL+LL+WL)+2.680.518.370.4018.270.59
1.5(DL+WL)+2.670.248.350.1918.270.52
0.9DL+1.5WL (uplift)-2.000.156.240.1111.360.59

For the uplift case the bottom flange is in compression and unrestrained between sag rods (LLT=L/2L_{LT} = L/2), so the lateral-torsional buckling capacity is used for MdzM_{dz}. The greatest ratio is 0.59 ≤1.0\le 1.0. Safe in biaxial bending.

Shear: Vmax=6.70V_{max} = 6.70 kN ≪Vd=Dtwfy/(3γm0)=82.0\ll V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 82.0 kN. OK.

Deflection (unfactored wind load of 1.5 kN/m normal to the roof, limit span/150 for elastic sheeting, Table 6): δ=5wL4/(384EI)=13.72\delta = 5wL^4/(384EI) = 13.72 mm <L/150=33.3< L/150 = 33.3 mm. OK.

Result

Provide an ISMB 125 purlin, simply supported on the rafters, with a sag rod (12 mm) at mid-span and cleat connections with two M12 bolts at each end.

✓ Answer checked by Claude Opus 5.5 (high effort)

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