Week 5 Challenge

1. Generate manual wind loading in the design report based IS code as per the following input

• Basic wind speed = 39m/s
• Terrain category 2

AIM

To Generate the wind load to a model as per Indian standard code reference.

PROCEDURE

Data

• Basic wind speed, Vb = 39m/s.
• Terrain category = 2
• Class of structure = A
• One horizontal dimension of structure, L = 18 m
• Another horizontal dimension of structure, W = 18 m
• Eaves level height of structure, h = 12 m

SOLUTION

• Ratio of wall and Roof Co-efficient

Wall , L/W = 18/18 = 1

h/w = 12/18 = 0.66

ROOF, h/w = 13.5/18 = 0.75

• As per reference of IS 875, Part – 3, Table =5.
• To find external pressure co-efficient(CPe) for walls.

The condition is satisfied in,

• To find Design Wind pressure (Pz).

Pz = 0.6*Vz^2.

• To find Vz,

Vz = Vb * k1 * k2 * k3* k4

Therefore,

As we take value of k1, k2, k3, k4 from IS 875.

K1 = 1 ( under basic wind speed)

K2 = 1.0 ( height upto 10m)

K3 = 1

K4 = 1.15 ( for industrial building )

Vz = 39 *1*1*1*1.15

Vz = 44.85 m/s

Then to find,

Pz = 0.6 * (44.85) ^2

Pz = 1.207 KN/m^2.

• To find Wind load on wall.

• To find the wind load on Roof.

The height of the Top roof level = 13.5m

Ratio =H/W = 13.5/18 = 0.75

2. Based on the above calculation apply the loadings on the model

ANSWER

• After the calculation of wind load, then to add the load conditions on model.
• Go to Load > load cases > to add the all wind direction categories.

• To add the wall load calculation.
• select the frame 4 and to add the wing load at condition of Face A.

• select the frame 1 and to add the wing load at condition of Face B.

• select the frame I and to add the wing load at condition of Face C.

• select the frame E and to add the wing load at condition of Face D.

• To add the wind load on roof.
• To select the wind load on both the top sides as per conditions.

3. On a separate model generate wind loading using wind wizard

ANSWER

• To produce the wing loading using wind wizard on structure.