In this article we explain wind load calculation on roof truss as per revised is code 875 -2015. Explained all steps of wind load calculation with solved example. So read the article till the end and comment if you got anything wrong in this article.

Table of Contents

**Steps of roof truss Wind load calculation as per is 875-2015.**

**Step-1 : Angle of roof truss**

**Angle of roof truss = tan ^{-1}( Rise/(Span/2))**

**Step-2 : Determining Basic wind Speed (V**_{b})

_{b})

Finding basic wind speed from page no 6 or 51 of IS 875 part-3 -2015 as per location.

**Step-3: Wind pressure calculation**

**1. Design Wind Speed (V**_{z}) :

_{z}) :

For finding design wind speed, formula given on page no 5 of IS 875 part-3 2015.

**V _{z} = V_{b }× K_{1} ×K_{2} × K_{3} × K_{4}**

Where,

- V
_{b }= Basic wind speed - K
_{1}= Risk Coefficient - K
_{2}= Terrain roughness and height factor - K
_{3 }= Topography factor - K
_{4 }= Importance factor for cyclonic region

**Find K**_{1}

_{1}

- K
_{1}is obtained from**page no 7, table-1, IS 875 part-3 2015** - K1 depends on the class and life of the structure.

**Find K**_{2}

_{2}

- K
_{2 }depends on the**terrain category**and**height of the structure.** - Terrain category decided on the basis of
**terrain of location of structure.** - K
_{2}is obtain from**table-2, page no 8, IS 875 part-3 2015**

**Find K**_{3}

_{3}

K_{3 }is obtain from clause 6.3.3, **page no 8, IS 875 part-3 2015.**

**Find K**_{4}

_{4}

- K
_{4 is obtained}**from clause no 6.3.4, page no 8, IS 875 part-3 2015.** **For hospitals**, schools, communication towers, cyclone shelters,**K**_{4 }is 1.30.- For industrial structures, K
_{4 }is 1.15. - All other structures, K
_{4 }is 1.00.

**2. Design Wind Pressure:**

**P _{d} = K_{d} ×K_{a} × K_{c} × P_{z}**

Where,

P_{z} = wind pressure

**P _{z} = 0.6 × V_{z}^{2}**

- K
_{d }= wind directionality factor - K
_{a }= Area averaging factor. - K
_{c}= Combination factor.

**Find K**_{d}

_{d}

K_{d }is obtain from clause no 7.2.1, **page no 9, IS 875 part-3 2015.**

**Find K**_{a}

_{a}

- K
_{a is obtained}**from clause no 7.2.2, page no 10, IS 875 part-3 2015.** - Ka is dependent on the
**tributary area.** **Tributary area = spacing or pitch × rise**

**Find K**_{c}

_{c}

K_{c }is obtain from clause no 7.3.3.13**, page no16, IS 875 part-3 2015**.

**Step-4: Wind load on individual members**

Wind load on individual members is determined by formula which is given in IS 875 part 3, clause 7.3.1, page no 10.

**F = ( C _{pe }– C_{pi} ) A × P_{d}**

Where,

- C
_{pe }= external pressure coefficient - C
_{pi }= internal pressure coefficient - A = surface area of structural element or cladding unit
- P
_{d }= design wind pressure

**Find C**_{pi}

_{pi}

- C
_{pi is obtained}from clause 7.3.2,**page no 11, IS 875 part 3 2015.** - C
_{pi }depends on the opening area in a structure.

**Find C**_{pe}

_{pe}

- C
_{pe }is obtain from clause 7.3.3,**page no 11, IS 875 part-3 2015.** - C
_{pe }is also obtain from table 6, IS 875 part-3 2015.

**Solved example of Wind load calculation as per IS 875-2015**

Roofing system of an industrial shed consists of trusses spaced at 6 m apart. The span of roof truss is18 m and rise is 3 m. The level of eaves is 7 m above the ground. Assume suitable configuration of truss. The shade is situated on flat terrain with sparsely populated buildings. The shed has less than 20% permeability. Prepare structural layout of industrial steel shed with suitable configuration. Determine the wind forces on the truss. Location Chennai.

**Given Data:**

- Spacing : 6m
- Span = 18m
- Rise = 3m
- Height = 7m
- Terrain: flat terrain with sparsely populated buildings
- Shed has less than 20% permeability

Solution:

Assume **howe type truss** for 18m span.

**Step-1: Angle of roof truss**

Angle of roof truss = tan^{-1}( Rise/(Span/2))

= tan^{-1}(3/(18/2))

= 18.43

**Step-2: Determining Basic wind Speed (V _{b})**

For Chennai, the basic wind speed is 50m/s from page no51, IS 875 part-3 2015.

**Step-3: Wind pressure calculation**

Direct finding K_{1}, K_{2}, K_{3, } K_{4 }from the IS 875 part-3. For find this coefficient is explain above in steps.

- K
_{1}= 1 - K
_{2}= 1 - K
_{3}= 1 - K
_{4}= 1.15

**V _{z} = V_{b }× K_{1} ×K_{2} × K_{3} × K_{4}**

= 50 × 1 × 1 × 1 × 1.15

= 57.5m/s

**Design Wind Pressure:**

P_{d} = K_{d} ×K_{a} × K_{c} × P_{z}

Direct finding **K _{d}, K_{a}, K_{c }**from the IS 875 part-3. For find this coefficient is explain above in steps.

- K
_{d }= 0.9 - K
_{a}= 0.92 (getting this by interpolation between 10 and 25) - K
_{c}= 0.9

**P _{z} = 0.6 × V_{z}^{2}**

= 0.6 × 57.5^{2}

=1983.75 N/m2

**P _{d} = K_{d} ×K_{a} × K_{c} × P_{z}**

= 0.9 × 0.92 × 0.9 × 1983.75

= 1478.29 N/m2

= 1.478 KN/m2

Design wind pressure is less than **0.7 × P _{z}**

= 0.7 × 1983.75

= 1388.62 (N/m2) Hence OK

**Step-4: Wind load on individual members**

F = ( C_{pe }– C_{pi} ) A × P_{d}

**Find C _{pi}**

The shed has less than 20% permeability

Therefore, C_{pi }= +/-(0.5)

**Find C _{pe}**

For truss angle 18.43 and,

h/w = 7/18

= 0.38

the C_{pe }is given below.

**Finding area:**

A = Spacing × ( ((span/2)^{2} × (rise)^{2})^(0.5))/number of panels )

= 6 × (((9)^{2} × (3)^{2}) ^ (0.5)) / 8 )

= 7.115 m2

Therefore,

**A × P _{d}**

_{ }= 7.115 × 1.478

= 10.52 KN

**Wind load calculation table**

**Also read:**

Hi,

Was working through this example – now have following questions:

A = Spacing × ( ((span/2)2 × (rise)2)^(0.5))/number of panels )

should be

A = Spacing × ( ((span/2)2 + (rise)2)^(0.5))/number of panels )?

As we appear to be calculating roof slope area on one side only, the number of panels should be 4 rather than 8? ie. where the purlins would match the truss panels on one side of the roof?

A minor typo?

The shade is situated on flat terrain with sparsely populated buildings

should read The shed is situated…

Other than those points, a great example well explained and sequenced.

Regards