Wind Bracing

7.2.1.1. Longitudinal bracing:

The wind load acting on the endwall is transmitted to the wind columns by sheeting and girts. The columns then transmit the wind force equally to their bases and to the building roof. The force transmitted to building roof travels in the roof through purlins, instantaneously gets divided equally in the braced bays, and eventually gets transferred to the sides of the building and down into the sidewall foundation.

The following sketch illustrates the flow of wind loads in the bracing system:

In the above sketch, the rods shown in heavy lines are those acting (in tension) to resist the wind acting in the direction shown. When the wind reverses directions, the other rods, shown dashed, will act instead.

Note that for the above example, the rods between grids C & D are assumed to take no load for whichever direction the wind is acting in. However, for the case where there is a load applied at the ridgeline of the building, these rods will act in tension.

Note that all strut members and their connections, used in the end bays and braced bays, shall be designed for the axial loads shown on the above diagram.

The standard end connections are (2 Nos.) 12 mm φ A307 bolts in purlins and (2 Nos.) 12 mm φ A325N bolts on eave struts with allowable shear loads of 22 kN and 47 kN (an increase of 33% for wind included) respectively.

When the actual loads exceed these allowable loads, the connection shall be modified by:

Either changing the A307 bolts to A325N bolts, or by

Adding a clip, thus increasing the number of bolts resisting the axial load in shear, or by A similar method to increase the capacity of the connection.

The various options are shown in the purlin connection details in clause 5.2.2.3. of this manual.

Example 1:

An example showing the generalized procedure for the computation of forces in bracing due to wind in an un-symmetrical building is presented. This method is based on the truss analogy assuming each braced bay acts like a truss as shown in the following sketch.

Building Data:

Building Width: 24m Eave Height (left): 6m Eave Height (right): 5.7m Slope (left): 0.5 to 10 Slope (right): 1.0 to 10

Ridge Distance (from left): 14m No. of braced bays: 2

Wind Pressure: 0.7 Bay Spacing: 7.5m Ridge Height: 6.7m Step 1:

Calculate the tributary area associated for each column

Tributary area for corner column = 6.075x3 = 18.225m²

Tributary area for 1^st interior column = 6.3x6 = 37.8m² and so on.

Step 2:

Calculate the force transmitted to each braced bay at the location of column lines.

With n = Nos. of braced bays =2

Force at corner column = 0.7x18.225/(2n) = 3.189 kN

Force at 1^st interior column = 0.7x37.8/(2n) = 6.615 kN and so on.

Step 3:

Calculate reactions at both ends using the forces using moment and force equilibrium equations.

Reaction @ right = (3.071x24 + 6.615x18 + 6.917x12 + 6.615x6)/24 = 13.145 kN Reaction @ left = 3.189 + 6.615 + 6.917 + 3.071 – 13.14 = 13.262 kN

Step 4:

Draw shear force diagram using the forces and reactions

The shear force at each column location indicates the axial force in the strut purlin at that location. This force is distributed to the bracing cable/rod.

Step 5:

Calculate cable/rod force:

Length of rod in the 1^st bay = √(7.5²+6.007²) = 9.61m

Bracing force in 1^st bay = axial force x length of cable/bay spacing = 10.073x9.61/7.5 = 12.91kN Bracing force in 2^nd bay = 3.458x9.61/7.5 = 4.43 kN

Bracing force at ridge (left) = 3.458x7.76/7.5 = 3.58 kN Bracing force at ridge (right) = 3.458x8.51/7.5 = 3.92 kN Bracing force in last bay = 10.073x9.62/7.5 = 12.93 kN Step 6:

Calculate cable/rod force of left sidewall:

Cable/rod length = 9.6m

Bracing force = 13.262x9.6/7.5 = 16.975 kN

Similarly Calculate cable/rod force of right sidewall:

Cable/rod length = 9.42m

Bracing force = 13.145x9.42/7.5 = 16.51 kN Step 7:

Calculate the bracing force transmitted to foundations.

Horizontal Reaction H @ near side wall = P (force at eave) = 13.262kN

Vertical Reaction V @ near side wall = P x E / B = 13.262x6/7.5=10.61kN

Where, E = Eave Height and B = Bay spacing

Horizontal Reaction H @ far side wall = P (force at eave) = 13.145kN

Vertical Reaction V @ near side wall = P x E / B = 13.145x5.7/7.5=10.0kN

Simplified Bracing calculations for Symmetric Frames:

A much simplified procedure may be adopted for a symmetric frame and bracing pattern. The following example elaborates this procedure.

Example 2:

The computation of forces in bracing due to wind in a symmetrical building is presented. This method is based on the truss analogy assuming each braced bay acts like a truss as shown in the following sketch.

Building Data:

Building Width: 24m Eave Height: 6m Slope: 1.0 to 10 No. of braced bays: 2 Wind Pressure: 0.7 Bay Spacing: 7.5m Step 1:

Calculate the tributary area associated for each column

Tributary area for corner column = 6.15x3 = 18.45m²

Tributary area for 1^st interior column = 6.6x6

= 39.6m² and so on.

Step 2:

Calculate the force transmitted to each braced bay at the location of column lines.

With n = Nos. of braced bays =2

Force at mid column = 0.7x39.6/(2x2)=6.93 kN Force on either side of ridge P1 = 3.7kN

Force at 1^st interior column P2 = 0.7x39.6/(2x2) + P1 = 10.63 kN

Force at corner column P3 = 0.7x18.45/(2x2) + P2 = 13.86 kN

Step 3:

Calculate cable/rod force:

Length of cable in the 1^st bay = √(7.5²+6.03²) = 9.623m Bracing force F1 = 3.7x9.623/7.5 = 4.75 kN

Bracing force F2 = 10.63x9.623/7.5 = 13.64 kN

Step 4:

Calculate cable/rod force of sidewalls:

Cable/rod length = 9.6m

Bracing force F3 = 13.86x9.6/7.5 = 17.74 kN Step 5:

Calculate the bracing force transmitted to foundations.

Horizontal Reaction H = P (force at eave) = 13.86kN Vertical Reaction V = P x E / B = 13.86x6/7.5=11.09kN Where, E = Eave Height and B = Bay spacing

7.2.1.2. Transversal bracing in P&B end walls

Wind bracing is required in by-framed P&B endwalls in the absence of diaphragm action. Bracing is designed for the portion of wind force acting on the sidewall on half end bay width. This transverse lateral force is transmitted to the endwall rafter as an axial force and then back to an endwall post foundation through a brace rod placed in the plane of the endwall.

P = Transverse lateral load = = q x GC_p x h x B / 4n Where,

GC_p = 0.5+0.7 = 1.2 @ End Zone (Table 5.4a of MBMA 1996)

B = Side Wall End Bay h = peak height

S = endwall braced bay spacing n = number of braced bay

Horizontal Reaction at column: H = P;

Vertical Reaction V = P h / S Tension in the rod

Note that the other brace rod, shown in a dashed line above, acts whenever the wind reverses directions.

S S P h

T

2 2 +

=