--- title: "Wind pressure on 36 m building, core city" subject: "Design of Steel and Timber Structure" chapter: 8 key: "semester_6/design_of_steel_and_timber_structure:wind-core-city-36m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/wind-core-city-36m site: BCT Engineering --- # Wind pressure on 36 m building, core city Design of Steel and Timber Structure · Chapter 8: Design of Roof Trusses - Asked in: 2072 Asoj, 10 marks; 2072 Magh, 8 marks; 2071 Magh, 10 marks (asked 3 times) - 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-core-city-36m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-8 ## Question The building is to be constructed in core city area for a 50 years life. The size of the building is over 30 m. The height of the building is 36 m and is classified as 1st category building. Determine the wind pressure at the site and force on the truss. ## Short answer Design wind speed Vz = Vb×k1×k2×k3 = 47×1×0.966×1 = 45.4 m/s. Design pressure pz = 0.6Vz² = 1.237 kN/m². Net coefficient (Cpe - Cpi) = -1.30 gives roof suction pressure 1.608 kN/m². Force on one truss slope (area 91.4 m²) is 1.30×91.4×1.237 = 147.0 kN. ## Answer **Assumptions (data not given in the question).** $V_b$ is not stated; take **47 m/s** (Kathmandu region, IS 875 Part 3 Fig. 1). The building is in a core city, so terrain **Category 4** (large city centre with obstructions). Height 36 m, so the greatest dimension is 36 m and the structure is **Class B** (20 m to 50 m). A design life of 50 years gives $k_1 = 1.0$. The roof truss is taken on a plan width $w = 30$ m with a low roof pitch of $\alpha = 10^\circ$, truss spacing 6 m and normal permeability. $k_3 = 1$ (flat site). #### 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 $k_1 = 1.00$ (design life 50 years, Table 1). - Terrain, height and size factor $k_2$: Category 4, Class B, at height 36 m, interpolating between 30 m (0.93) and 50 m (1.05): $k_2 = 0.966$ (Table 2). - Topography factor $k_3 = 1.000$ (flat site, slope less than 3°). $$ \begin{aligned} V_z &= 47\times1.00\times0.966\times1.000 = 45.40\ \text{m/s} \\ p_z &= 0.6V_z^2 = 0.6\times45.40^2 = 1237\ \text{N/m}^2 = 1.237\ \text{kN/m}^2 \end{aligned} $$ #### Step 2: Pressure coefficients (Table 5, pitched roof of rectangular clad building) Height ratio $h/w = 1.20$ (band $1/2 < h/w \le 3/2$), roof angle $\alpha = 10.0^\circ$. The values for $\alpha = 10^\circ$ are read directly from the table. 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°) | -1.10 | -1.30 | -0.90 | -1608 | -1113 | | Leeward slope GH (wind 0°) | -0.60 | -0.80 | -0.40 | -989 | -495 | | Slope EG (wind 90°) | -0.80 | -1.00 | -0.60 | -1237 | -742 | | Slope FH (wind 90°) | -0.60 | -0.80 | -0.40 | -989 | -495 | 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 the windward slope EF (wind 0°), with $C_{pe} - C_{pi} = -1.30$, so $p_d = -1608$ N/m² $= -1.608$ 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, $-1.30$, is used for the maximum force. - One slope of a truss (span 30 m, slope length $15/\cos10^\circ = 15.23$ m, spacing 6 m): $A = 91.4$ m², $F = 1.30\times91.4\times1.237 = 147.0$ kN (suction, i.e. acting away from the roof). #### Answer - Design wind speed $V_z = 45.4$ m/s; design wind pressure $p_z = 1237$ N/m² (1.24 kN/m²). - Net design pressure on the roof: up to $1608$ N/m² (1.61 kN/m²), suction. - Force on one slope of a truss (span 30 m, slope length 15.23 m, spacing 6 m): 147.0 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the design wind speed in IS 875 Part 3? A. Vz = Vb + k1 + k2 + k3 B. Vz = Vb k1 k2 k3 C. Vz = 0.6 Vb^2 D. Vz = Vb / (k1 k2 k3) Answer: B. Vz = Vb k1 k2 k3 (risk, terrain/height/size and topography factors); then pz = 0.6 Vz^2. (Key point: Design wind speed) 2. For a 36 m high building in a city centre (Category 4), what class applies and how is k2 obtained? A. Class A; k2 read directly at 36 m B. Class C; k2 = 1.05 C. Class B (20 to 50 m); k2 interpolated between 0.93 at 30 m and 1.05 at 50 m = 0.966 D. Class B; k2 = 0.93 Answer: C. Greatest dimension 36 m lies in 20-50 m so Class B; k2 = 0.93 + 0.12 x 6/20 = 0.966. (Key point: Class and k2) 3. With Vb = 47 m/s, k1 = 1, k2 = 0.966, k3 = 1, what are Vz and pz? A. Vz = 45.4 m/s, pz = 1237 N/m^2 B. Vz = 47 m/s, pz = 1325 N/m^2 C. Vz = 45.4 m/s, pz = 2061 N/m^2 D. Vz = 49.4 m/s, pz = 1461 N/m^2 Answer: A. Vz = 47 x 0.966 = 45.4 m/s and pz = 0.6 x 45.4^2 = 1237 N/m^2. (Key point: Wind pressure result) 4. How is the governing wind force on a roof slope found? A. F = pz x plan area B. F = (Cpe - Cpi) A pz using the most severe net coefficient (-1.30), acting normal to the roof as suction C. F = Cpe A Vz D. F = 0.6 Vz^2 only Answer: B. Net coefficient Cpe - Cpi = -1.30 (windward slope, Cpi = +0.2) times area 91.4 m^2 and pz gives about 147 kN suction. 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