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IS 875-3 Wind Load Calculation Example

A fully worked example of IS 875-3 wind load calculations

In this article, an example wind load pressure calculation for a building in Walwane, Maharashtra, India (18.945695° N, 74.564866° E) will be shown. This calculation will be in accordance with IS 875-3:2015 wind load calculations. SkyCiv free wind load calculator recently added the IS 875-3 wind load calculations, hence, we are going to demonstrate how to calculate the wind loads, by using an S3D Barnhouse model below:

Figure 1. Barnhouse model in SkyCiv S3D as an example.

For this case study, the structure data are as follows:

Figure 2. Site location (from Google Maps).
LocationWalwane, Maharashtra, India (18.945695° N, 74.564866° E)
OccupancyMiscellaneous – Farm structure
TerrainFlat open land
DimensionsB = 4 m × L = 14 m in plan
H =
Eave height of 2.4 m
Apex height at elev. 3.4 m
Roof slope 1:2 (26.565°)
No opening
CladdingPurlins spaced at 0.745m
Wall studs spaced at 0.8m
Table 1. Building data needed for our wind calculation.

Using the IS 875-3: 2015, the design wind speed for the location and the design wind pressure for the rectangular building with pitched roof can be solved using the equations below:

Design wind speed at height z (in m/s): Vz = Vbk1k2k3k4 (1)

Where:
Vb is theBasic wind speed, m/s
k1 is the Probability factor (risk coefficient) based on 6.3.1 of IS 875-3
k2 is the Terrain roughness and height factor based on 6.3.2 of IS 875-3
k3 is the Topography factor based on 6.3.3 of IS 875-3
k4 is the Importance factor for the cyclonic region based on 6.3.4 of IS 875-3

Design wind pressure (in Pa): pd = KdKaKcpz (2)

Where:
Kd is the Wind directionality factor based on 7.2.1 of IS 875-3. Equal to 1.0 when considering local pressure coefficients.
Ka is the Area averaging factor based on 7.2.2 of IS 875-3
Kc is the Combination factor based on 7.3.3.13 IS 875-3
pz is equal to 0.60Vz2 in Pa
Note that pd should not be taken less than 0.70pz

From the design pressure pd obtained, the pressure will be distributed to the members of the by using:

Wind force on surface or members (in N): F = (Cpe – Cpi)Apd (3)

Where:
A is the surface area of the structural element or cladding unit
Cpe is the external pressure coefficients
Cpi is the internal pressure coefficients

We will dive deep into the details of each parameter below.

Basic Wind Speed Vb

From Figure 1 of IS 875-3, the site location is situation of the map where the basic wind speed Vb is equal to 39 m/s.

Figure 3. Basic wind speed data based on Figure 1 of IS 875-3: 2015.

SkyCiv can automate the wind speed calculations just defining the site location in India. Try our SkyCiv Free Wind Tool.

Probability Factor (Risk Coefficient) k1

Table 1 of IS 875-3 presents the risk coefficients for different classes of structures in different wind speed zones. For this structure, since it is a barnhouse and will be used to shelter some livestock animals, the structure is classified under “Buildings and structures presenting a low degree of hazard to life and property in the event of failure, such as isolated towers in wooded areas, farm buildings other than residential buildings.” Hence, from Table 1 of IS 875-3, the corresponding probability factor (risk coefficient) k1 is equal to 0.92.

Figure 4. Risk coefficients table from IS 875-3:2015.

Terrain Roughness and Height Factor k2

For this structure, it is located at the center of a farm where there are no immediate obstructions. Hence, the terrain can be classified as Category 1. Using Table 2 of IS 875-3:2015, we can obtain k2 values (which varies depending on the height being considered):

Heightk2
Reference height, H = 2.4 m1.05

Topography Factor k3

In order to account for topographic effects, we need to get the elevation data of the location for the eight (8) cardinal directions – N, S, W, E, NW, NE, SW, and SE – using Google elevation API. Based on the data, we can generally assume that the terrain is “Flat” for all directions. Hence, based on 6.3.3 of IS 875-3:2015, we can set our k3 equal to 1.0.

Importance Factor k4

Since the site location is not located within the east coast of India and the structure will only be used for agricultural purposes, the value of k4 is equal to 1.0 based on 6.3.4 of IS 875-3:2015

Design Wind Speed Vz

From the factors above, we can already solve the design wind speed Vz using Equation (1):

LevelVb m/sk1k2k3k4Vz m/s
H = 2.4 m39.00.921.051.01.037.674

From the design wind speed, we can calculate the design wind pressure pd.

Wind Directionality Factor Kd

From 7.2.1 of IS 875-3:2015, the Wind Directionality Factor Kd is equal to 0.9 for frames and when considering local pressure coefficients, will be equal to 1.0. For this example, we will use Kd equal to 1.0 for purlins and wall studies and for Kd equal to 0.9 for the columns and trusses.

Area Averaging Factor Ka

The Area Averaging Factor Ka can be calculated using Table 4 of IS 875-3:2015:

Ka = 1.0 for area less than or equal to 10 sq.m.
Ka = 0.9 for area equal to 25 sq.m.
Ka = 0.8 for area greater than or equal to 100 sq.m.

Note that Ka can be linearly interpolated between values. For this structure, we need to get the tributary areas of columns for windward (Zone A), leeward (Zone B), sidewalls (Zone C and D), and truss for the roof. Moreover, we will also consider the tributary area of the wall studs and purlins.

ComponentArea, sq.m.Ka
Column2.4×3.5 m = 8.4 sq.m.1.0
Truss4×3.5 m (projection) = 14 sq.m.0.97
Wall studs0.8×3.5 m = 2.8 sq.m.1.0
Purlins0.745×3.5 m = 2.608 sq.m.1.0

Combination factor Kc

Since we will considered the simultaneous action of wall and roof pressures and internal pressures, the assumed Combination factor Kc is equal to 0.9 as referenced in 7.3.3.13 of IS 875-3:2015.

Design Wind Pressure, pd

Using Equation (2), we can calculate the design wind pressure, pd, Note that pz = 851.598 Pa and pd should not be less than 0.7pz or 596.119Pa.

ComponentKaKdKcpzpd
Column1.01.00.9851.598766.438
Truss0.971.00.9851.598743.445
Wall studs1.01.00.9851.598766.438
Purlins1.01.00.9851.598766.438

From these data, we need to calculate the pressure coefficients in order to distribute the design pressure to the components.

Internal Pressure Coefficients Cpi

The internal pressure coefficients Cpi can be determined from 7.3.2 of IS 875-3:2015. For this structure, it is assumed the total opening on the wall is less than 5 percent of the total wall area. Therefore, the Cpi values for this example are +0.2 and -0.2.

External Pressure Coefficients Cpe

The External Pressure Coefficients Cpe depend on certain parameters such as height, width, length, roof angle, and roof profile.

Wall External Pressure Coefficients

The external pressure coefficients for walls depend on h/w and l/w ratio, where h is the eave height, w is the least dimension of the building, and l is the larger dimension of the building. For this example, h = H, l = L, and w = B. Therefore, h/w = 0.6 and l/w = 3.5. From Table 5 of IS 875-3:2015, the corresponding Cpe values are as follows:

Figure 5. Wall zones for rectangular building based on IS 875-3:2015.

For wind angle = 0 degrees:

Zone/SurfaceCpe
Zone A – Windward wall+0.7
Zone B – Leeward wall-0.3
Zone C – Sidewall-0.7
Zone D – Sidewall-0.7
Local zone
(0.25w from edge)
-1.1

For wind angle = 90 degrees:

Zone/SurfaceCpe
Zone A – Windward wall-0.5
Zone B – Leeward wall-0.5
Zone C – Sidewall+0.7
Zone D – Sidewall-0.1
Local zone
(0.25w from edge)
-1.1

Note that w = 4 m.

Roof External Pressure Coefficients

For this structure, since the roof profile is gable or duopitch, the roof external pressure coefficients will be calculated based on Table 6 of IS 875-3:2015. For this example since h/w = 0.6, and the roof angle is 26.565°, the Cpe values will be interpolated using the following values:

Figure 6. Roof zones for pitched/gable roof based on IS 875-3:2015 – plan view.

Note: y = 0.15w = 0.6m

For wind angle = 0 degrees:

Roof angleZone EF – WindwardZone GH – Leeward
20°-0.7-0.5
26.565°-0.109-0.5
30°-0.2-0.5

For wind angle = 90 degrees:

Roof angleZone EG – CrosswindZone FH – Crosswind
20°-0.8-0.6
26.565°-0.8-0.6
30°-0.8-0.6

For local pressures:

Roof angleGable endsRidge zones
20°-1.5-1.0
26.565°-1.172-1.0
30°-1.0-1.0

The final roof pressure coefficients will be:

Zone/SurfaceWind direction – 0 degreesWind direction – 90 degrees
Zone EF – Windward-0.109
Zone GH – Leeward-0.5
Zone EG – Crosswind-0.8
Zone FH – Crosswind-0.6
Gable ends-1.172-1.172
Ridge zones-1.0-1.0

Combined internal and external pressures

From the values above, the wind force can be calculated using Equation (3). However, for simplicity, we will just be getting the design pressure (not multiplying the values to the area A) and also will be considering the wind direction angle 0 degrees for the main frame (column and truss). The frame spacing is equal to 3.5m. Note that pd = 766.438 Pa for both column and wall studs.

For columns and wall studs – 0 degrees:

Zone/SurfaceCpeCpiCpeCpip = pd(Cpe-Cpi) PaFor Column
px3.5m N/m
For wall studs
px0.8m N/m
Zone A – Windward wall0.7+0.2
-0.2
+0.5
+0.9
383.219
689.795
1341.267
2414.281
306.575
551.836
Zone B – Leeward wall-0.3+0.2
-0.2
-0.5
-0.1
-383.219
-76.644
-1341.267
-268.253
-306.575
-61.315
Local zone (1m from edge)-1.1+0.2
-0.2
-1.3
-0.9
-996.370
-689.795
-3487.295
-2414.281
-797.096
-551.836

The pressures on the columns will be multiplied to 3.5m in order to get a uniform load. Moreover, for the wall studs, it will be multiplied with 0.8m. Note that a positive pressure means it is acting towards the surface and negative is acting away from the surface (suction).

For truss and purlins – 0 degress:

Zone/SurfaceCpeCpiCpeCpip = pd(Cpe-Cpi) PaTruss
px3.5m N/m
Purlins
px0.745m N/m
Zone EF – Windward-0.109+0.2
-0.2
-0.309
+0.091
-229.725
67.654
-804.036
236.787
-171.145
50.402
Zone GH – Leeward-0.5+0.2
-0.2
-0.7
-0.3
-520.412
-223.034
-1821.441
-780.617
-387.707
-166.160
Gable ends-1.172+0.2
-0.2
-1.372
-0.972
-1051.553
-744.978
-3680.437
-2607.423
Ridge zones-1.0+0.2
-0.2
-1.2
-0.8
-919.726
-613.151
-3219.041
-2146.027

The pressures on the truss will be multiplied to 3.5m in order to get a uniform load. Moreover, for the wall studs, it will be multiplied with 0.745m. Note that pd = 766.438 Pa for the purlins and pd = 743.445 Pa for the truss.

Considering one critical frame – spacing is 3.5m:

For pd(Cpe – +Cpi):

Figure 7. Distributed load on a critical frame using pd(Cpe – +Cpi)x3.5m.

For pd(Cpe – -Cpi):

Figure 8. Distributed load on a critical frame using pd(Cpe – -Cpi)x3.5m.

For the design of wall studs and purlins, you just need to get the absolute maximum pressure acting on it and use it as the basis in calculating the design forces. For this case, the design wind load are: -797.096 N/m for wall stud and -783.407N/m for the purlins,

These calculations can be all be performed using SkyCiv’s Load Generator Software for IS 875-3 and other codes as well. Users can enter in a site location to get wind speeds and topography factors, enter in building parameters and generate the wind pressures. Try our SkyCiv Free Wind Tool for wind speed and wind pressure calculations on gable structures.

Patrick Aylsworth Garcia Structural Engineer, Product Development
Patrick Aylsworth Garcia
Structural Engineer, Product Development
MS Civil Engineering

References:

  • Design Loads (Other than Earthquake) for Buildings and Structures — Code of Practice (Part 3 Wind Loads ed.). (2015). Bureau of Indian Standards.
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