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 (Vb)
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 (Vz) :
For finding design wind speed, formula given on page no 5 of IS 875 part-3 2015.
Vz = Vb × K1 ×K2 × K3 × K4
Where,
- Vb = Basic wind speed
- K1 = Risk Coefficient
- K2 = Terrain roughness and height factor
- K3 = Topography factor
- K4 = Importance factor for cyclonic region
Find K1
- K1 is obtained from page no 7, table-1, IS 875 part-3 2015
- K1 depends on the class and life of the structure.
Find K2
- K2 depends on the terrain category and height of the structure.
- Terrain category decided on the basis of terrain of location of structure.
- K2 is obtain from table-2, page no 8, IS 875 part-3 2015
Find K3
K3 is obtain from clause 6.3.3, page no 8, IS 875 part-3 2015.
Find K4
- K4 is obtained from clause no 6.3.4, page no 8, IS 875 part-3 2015.
- For hospitals, schools, communication towers, cyclone shelters, K4 is 1.30.
- For industrial structures, K4 is 1.15.
- All other structures, K4 is 1.00.
2. Design Wind Pressure:
Pd = Kd ×Ka × Kc × Pz
Where,
Pz = wind pressure
Pz = 0.6 × Vz2
- Kd = wind directionality factor
- Ka = Area averaging factor.
- Kc = Combination factor.
Find Kd
Kd is obtain from clause no 7.2.1, page no 9, IS 875 part-3 2015.
Find Ka
- Ka 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 Kc
Kc 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 = ( Cpe – Cpi ) A × Pd
Where,
- Cpe = external pressure coefficient
- Cpi = internal pressure coefficient
- A = surface area of structural element or cladding unit
- Pd = design wind pressure
Find Cpi
- Cpi is obtained from clause 7.3.2, page no 11, IS 875 part 3 2015.
- Cpi depends on the opening area in a structure.
Find Cpe
- Cpe is obtain from clause 7.3.3, page no 11, IS 875 part-3 2015.
- Cpe 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 (Vb)
For Chennai, the basic wind speed is 50m/s from page no51, IS 875 part-3 2015.
Step-3: Wind pressure calculation
Direct finding K1, K2, K3, K4 from the IS 875 part-3. For find this coefficient is explain above in steps.
- K1 = 1
- K2 = 1
- K3 = 1
- K4 = 1.15
Vz = Vb × K1 ×K2 × K3 × K4
= 50 × 1 × 1 × 1 × 1.15
= 57.5m/s
Design Wind Pressure:
Pd = Kd ×Ka × Kc × Pz
Direct finding Kd, Ka, Kc from the IS 875 part-3. For find this coefficient is explain above in steps.
- Kd = 0.9
- Ka = 0.92 (getting this by interpolation between 10 and 25)
- Kc = 0.9
Pz = 0.6 × Vz2
= 0.6 × 57.52
=1983.75 N/m2
Pd = Kd ×Ka × Kc × Pz
= 0.9 × 0.92 × 0.9 × 1983.75
= 1478.29 N/m2
= 1.478 KN/m2
Design wind pressure is less than 0.7 × Pz
= 0.7 × 1983.75
= 1388.62 (N/m2) Hence OK
Step-4: Wind load on individual members
F = ( Cpe – Cpi ) A × Pd
Find Cpi
The shed has less than 20% permeability
Therefore, Cpi = +/-(0.5)
Find Cpe
For truss angle 18.43 and,
h/w = 7/18
= 0.38
the Cpe 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 × Pd = 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