--- title: "Wind load on industrial truss, 40 m span" subject: "Design of Steel and Timber Structure" chapter: 8 key: "semester_6/design_of_steel_and_timber_structure:wind-industrial-40m-span" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-industrial-40m-span site: BCT Engineering --- # Wind load on industrial truss, 40 m span Design of Steel and Timber Structure · Chapter 8: Design of Roof Trusses - Asked in: 2077 Chaitra, 10 marks - Video: [Calculation of Wind load | Design of steel structures and timber | IOE III/II PU MU |](https://www.youtube.com/watch?v=32ze2SwlHXk) — Parash Joshi - Civil Construction and Tutor - 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/wind-industrial-40m-span - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8 ## Question Determine wind load on a roof truss for an industrial building with 40m span and 100m length. The roofing is galvanized iron sheeting. The basic wind speed is 47m/s and terrain is open industrial area and building is class A. The clear height of building at the eaves level is 10m. ## Short answer Using IS 875, Vz = 47 x 1.0 x 1.036 x 1.0 = 48.7 m/s and pz = 0.6Vz^2 = 1.42 kN/m^2. With assumed 20 degree roof slope and Cpi = +/-0.2, the worst net pressure is (Cpe - Cpi) = -0.90, giving 1.28 kN/m^2 suction. Force per truss (5 m spacing) is 272.7 kN; whole roof 5454 kN. ## Answer **Assumptions (data not fully given).** $V_b = 47$ m/s (given). Terrain is an open industrial area: **Category 2** ("open terrain with scattered obstructions"); **Class A** (as stated). Life 50 years: $k_1 = 1.0$; flat site: $k_3 = 1$. The roof slope is not given, so $\alpha = 20^\circ$ is assumed (rise $= 20\tan20^\circ = 7.3$ m). Eaves height 10 m, so $h/w = 10/40 = 0.25$; mean roof height $= 10 + 7.3/2 = 13.6$ m. Normal permeability ($C_{pi} = \pm0.2$); GI sheeting. Truss spacing is assumed to be 5 m. #### Step 1: Design wind speed (IS 875 Part 3:1987, cl. 5.3) $$ V_z = V_b\,k_1k_2k_3 $$ - Basic wind speed $V_b = 47$ m/s. - Risk coefficient $k_1 = 1.00$ (50-year life; Table 1). - Terrain, height and size factor $k_2$: Category 2, Class A, at height 13.6 m: interpolating between 10 m (1.00) and 15 m (1.05): $k_2 = 1.036$ (Table 2). - Topography factor $k_3 = 1.000$ (flat site). $$ \begin{aligned} V_z &= 47\times1.00\times1.036\times1.000 = 48.71\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times48.71^2 = 1424\ \text{N/m}^2 = 1.424\ \text{kN/m}^2 \end{aligned} $$ #### Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building) Height ratio $h/w = 0.25$, roof angle $\alpha = 20.0^\circ$. The values are read directly from the $\alpha = 20^\circ$ row, so no interpolation is needed. Internal pressure coefficient $C_{pi} = \pm0.2$ (normal permeability), cl. 6.2.3. | Surface | $C_{pe}$ | $C_{pe} - C_{pi}$ ($C_{pi}=+0.2$) | $C_{pe} - C_{pi}$ ($C_{pi}=-0.2$) | $p_d$ (N/m²), $C_{pi}=+$ | $p_d$ (N/m²), $C_{pi}=-$ | |---|---|---|---|---|---| | Windward slope EF (wind 0°) | -0.40 | -0.60 | -0.20 | -854 | -285 | | Leeward slope GH (wind 0°) | -0.40 | -0.60 | -0.20 | -854 | -285 | | Slope EG (wind 90°) | -0.70 | -0.90 | -0.50 | -1281 | -712 | | Slope FH (wind 90°) | -0.60 | -0.80 | -0.40 | -1139 | -569 | Design pressure normal to the roof: $p_d = (C_{pe} - C_{pi})\,p_z$ (positive = towards the surface, negative = suction). The most severe case is slope EG (wind 90°), with $C_{pe} - C_{pi} = -0.90$, so $p_d = -1281$ N/m² $= -1.281$ kN/m². #### Step 3: Wind force on the roof / truss $F = (C_{pe} - C_{pi})\,A\,p_z$, acting normal to the roof surface. The most severe net coefficient, $-0.90$, is used for the maximum force. - One truss (both slopes, slope length 21.28 m each, spacing 5 m): $A = 212.8$ m², $F = 0.90\times212.8\times1.424 = 272.7$ kN (suction, i.e. acting away from the roof). - Whole roof, 40 m × 100 m plan: $A = 4256.7$ m², $F = 0.90\times4256.7\times1.424 = 5454.0$ kN (suction, i.e. acting away from the roof). #### Answer - Design wind speed $V_z = 48.7$ m/s; design wind pressure $p_z = 1424$ N/m² (1.42 kN/m²). - Net design pressure on the roof: up to $1281$ N/m² (1.28 kN/m²), suction. - One truss (both slopes, slope length 21.28 m each, spacing 5 m): 272.7 kN. - Whole roof, 40 m × 100 m plan: 5454.0 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What terrain category and class apply here? A. Category 2 (open terrain with scattered obstructions) and Class A as stated B. Category 4 and Class A C. Category 1 and Class C D. Category 3 and Class B Answer: A. An open industrial area is Category 2, and the question states Class A. (Key point: Terrain and class) 2. At what height is k2 evaluated and what is its value? A. Eaves height 10 m, k2 = 1.00 B. Mean roof height 13.6 m, k2 = 1.036 by interpolation between 10 m and 15 m C. Ridge height 17.3 m, k2 = 1.08 D. Mean height with k2 = 0.94 Answer: B. Mean roof height = 10 + 7.3/2 = 13.6 m; interpolating 1.00 to 1.05 gives 1.036. (Key point: Height for k2) 3. What are Vz and pz? A. Vz = 47 m/s; pz = 1325 N/m^2 B. Vz = 48.7 m/s; pz = 29.2 N/m^2 C. Vz = 48.7 m/s; pz = 1424 N/m^2 D. Vz = 50.6 m/s; pz = 1536 N/m^2 Answer: C. Vz = 47 x 1.036 = 48.7 m/s and pz = 0.6 x 48.7^2 = 1424 N/m^2. (Key point: Design speed and pressure) 4. Which case governs the roof and what net pressure results? A. Windward slope EF at 0 degrees: 854 N/m^2 B. Leeward slope GH: 285 N/m^2 C. Slope FH with Cpi = -0.2: 569 N/m^2 pressure D. Slope EG for wind at 90 degrees: Cpe - Cpi = -0.90, so 1281 N/m^2 suction Answer: D. EG at 90 degrees has the largest net coefficient, -0.90 x 1.424 = -1.281 kN/m^2. 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