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.

**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

**Basic Wind Speed as per IS 875 Part 3**

Basic wind speed at 10m height for some important cities of India as per is 875 part 3 2015 is tabulated below.

City/Town | Basic Wind Speed | City/Town | Basic Wind Speed |
---|---|---|---|

Agra | 47m/s | Kanpur | 47m/s |

Ahemadabad | 39m/s | Kohima | 44m/s |

Ajmer | 47m/s | Kolkata | 50m/s |

Almora | 47m/s | Kozhikode | 39m/s |

Amritsar | 47m/s | Kurnool | 39m/s |

Asansol | 47m/s | Lakshadeep | 39m/s |

Aurangabad | 39m/s | Lucknow | 47m/s |

Bahraich | 47m/s | Ludhiyana | 47m/s |

Bengaluru | 33m/s | Madurai | 39m/s |

Barauni | 47m/s | Mandi | 39m/s |

Bareilly | 47m/s | Mangalore | 39m/s |

Bhatinda | 47m/s | Moradabad | 47m/s |

Bhilai | 39m/s | Mumbai | 44m/s |

Bhopal | 39m/s | Mysore | 33m/s |

Bhubaneshwar | 50m/s | Nagpur | 44m/s |

Bhuj | 50m/s | Nainital | 47m/s |

Bikaner | 47m/s | Nashik | 39m/s |

Bokaro | 47m/s | Nellore | 50m/s |

Chandigarh | 47m/s | Panjim | 39m/s |

Chennai | 50m/s | Patiala | 47m/s |

Coimbatore | 39m/s | Patna | 47m/s |

Cuttack | 50m/s | Punducherry | 50m/s |

Darbhanga | 55m/s | Port Blair | 44m/s |

Darjeeling | 47m/s | Pune | 39m/s |

Dheradun | 47m/s | Raipur | 39m/s |

Delhi | 47m/s | Rajkot | 39m/s |

Durgapur | 47m/s | Ranchi | 39m/s |

Gangtok | 47m/s | Roorkee | 39m/s |

Guwahati | 50m/s | Rourkela | 39m/s |

Gaya | 39m/s | Shimla | 39m/s |

Gorakhpur | 47m/s | Srinagar | 39m/s |

Hyderabad | 44m/s | Surat | 44m/s |

Imphal | 47m/s | Tiruchirappalli | 47m/s |

Jabalpur | 47m/s | Trivandrum | 39m/s |

Jaipur | 47m/s | Udaipur | 47m/s |

Jamshedpur | 47m/s | Vadodara | 44m/s |

Jhansi | 47m/s | Varanasi | 47m/s |

Jodhpur | 47m/s | Vijayawada | 50m/s |

Vishakapatnam | 50m/s |

**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

Respected Sir, madam

As per is 875/part 3 /2015

We want wind speed,

Vb=50m/s,

Length=16m,

Breadth=18m,

Height=18.

We want value of vz

If i Assuming the Terrain Category 2 and other factors are 1, Than the Vz is 53.50m/s.