Wind Load Calculation On Walls [A Beginner’s Guide]

Wind loads on walls must be taken into account in every structural design because they influence the stability of the structure. 💨🌬️

In this blog post we show, step by step, how to calculate the wind loads on walls according to the Eurocode. 🧮🔢

We like to explain structural design with examples.

This time, we’ll calculate the wind loads on facade elements for a precast concrete office building.

Let’s get started. 🚀🚀

Process Of Calculating The Wind Loads On Walls

In order to calculate the wind load or wind pressure on external surfaces, we are going to do the following steps:

Define geometry and parameters from wind velocity pressure

This is a quick summary of the values we calculated to get the wind velocity pressure.

Fundamental value of basic wind velocity	vb.0	24 m/s
Orography factor	c0	1.0
Turbulence factor	kl	1.0
Density of air	$\rho$	1.25 kg/m3
Reference height of Terrain cat. II	z0.II	0.05 m
Roughness Length (Terrain cat. III)	z0	0.3 m
Terrain factor	kr	0.215
Turbulance Intensity	Iv	0.247
Roughness factor	cr	0.871
Mean wind velocity	vm	20.9 m/s
Peak velocity pressure	qp	0.746 kN/m2

Wind pressure on surfaces

In general, distinguishes Eurocode between wind pressure on external and internal surfaces. This article is focusing on the wind pressure on external surfaces.

Wind pressure on external surfaces w_e

The formula to calculate the wind pressure on external surfaces is

$$w_{e} = q_{p} \cdot c_{pe}$$

Where

qp	is the peak velocity pressure and
cpe	is the external pressure coefficient

The coefficient c_pe has 2 different values depending on the wind loaded area.

There is a value for a surface area of 1 m² and 10 m². These to values can also be written as

cpe.1	for the external pressure coefficient for an area of 1 m2 and
cpe.10	for the external pressure coefficient for an area of 10 m2

You are now probably wondering: “When do we ever design a load bearing element where only 1 m² is loaded?🤔”  

Well, you are right, that is almost never the case for beams and slabs. But for smaller elements like cladding elements of the facade or fixings, c_pe.1 comes into play (EN 1991-1-4 7.2.1 Note 1).

Enough explanation, let’s have a look at the values of the coefficients.

EN 1991-1-4 Table 7.1 gives recommendations for c_pe.10 and c_pe.1. This means that you have to double-check with your National annex because those values might be defined differently there.

Table 7.1 gives values for 5 different areas A, B, C, D and E of our building.

Those areas depend on where the wind comes from and can be seen in EN 1991-1-4 Table 7.1. For our office building, we can define the areas as

Wind from Front

Width of building	b	49.75 m
Length of building	d	20.2 m
Height of building	h	17.1m

From those dimensions we can define e which determines the width of Areas A and B according to EN 1991-1-4 Figure 7.5.

$$e = min(b, 2h)$$

$$e = min(49.75m, 2 \cdot 17.1m=34.4m) = 34.4m$$

For the case of e > d which is true for us, the width of Area A is defined as:

$$e/5 = 6.9m$$

So let’s visualize all of those numbers. 🖼️🖼️

We did not include the walls on the rooftop in the following picture due to better visualization. We also need to take them into account.

Okay, now let’s go back to the formula for the wind pressure for external surfaces and derive the values. The peak velocity pressure was calculated in the last article as

$$q_{p} = 0.746 \frac{kN}{m^2}$$

The external pressure coefficients for rectangular buildings can be taken from EN 1991-1-4:2005 Table 7.1.

For a height to width ratio h/d = 17.1m/20.2m = 0.85 which is < 1 and > 0.25 we get the following coefficients.

Area	cpe.10	cpe.1
Area A	-1.2	-1.4
Area B	-0.8	-1.1
Area D	0.8	1.0
Area E	-0.5	-0.5

Based on our coefficients, we can now calculate the Wind pressure on external surfaces.

Area	we.10	we.1
A	$-1.2 \cdot 0.75 \frac{kN}{m^2} = -0.9 \frac{kN}{m^2} $	$-1.4 \cdot 0.75 \frac{kN}{m^2} = -1.05 \frac{kN}{m^2}$
B	$-0.8 \cdot 0.75 \frac{kN}{m^2} = -0.6 \frac{kN}{m^2} $	$-1.1 \cdot 0.75 \frac{kN}{m^2} = -0.825 \frac{kN}{m^2}$
D	$0.8 \cdot 0.75 \frac{kN}{m^2} = 0.6 \frac{kN}{m^2} $	$1.0 \cdot 0.75 \frac{kN}{m^2} = -0.75 \frac{kN}{m^2}$
E	$-0.5 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2} $	$-1.1 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2}$

When you calculate the wind loads the first time ever, it might be very confusing in which direction you have to apply the loads.

So let’s apply the wind loads (for 10 m²) on our building.

To better understand it, we have a look at a 3D picture.

Now we also have to do the same for the case that wind comes from the side.

Wind from Side

In the scenario that wind comes from the side, we have to define the Area widths again. We have to redefine the geometry parameters.

Width of building	b	20.2 m
Length of building	d	49.75 m
Height of building	h	17.1m

From those dimensions we can define $e$ according to EN 1991-1-4 Figure 7.5

$$e = min(b, 2h)$$

$$e = min(20.2m, 2 \cdot 17.1m=34.4m) = 20.2m$$

For the case of e < d which is true for us, the width of Area A is defined as

$$e/5 = 4.04m$$

The Area B is calculated as

$$e \cdot4/5 = 16.16m$$

The Area C is therefore the rest of d and calculated as

$$d – e = 49.75m – 20.2m = 29.55m$$

So let’s visualize all of those numbers. 🖼️🖼️

For a height to width ratio h/d = 17.1m/49.75m = 0.34 which is < 1 and > 0.25 we get the following coefficients we get the same coefficients for A, B, D and E as for Wind from Front, but additionally we get now a value for C (EN 1991-1-4:2005 Table 7.1).

Area	cpe.10	cpe.1
A	-1.2	-1.4
B	-0.8	-1.1
C	-0.5	-0.5
D	0.8	1.0
E	-0.5	-0.5

Based on our coefficients, we can now calculate the Wind pressure on external surfaces.

Area	we.10	we.1
A	$-1.2 \cdot 0.75 \frac{kN}{m^2} = -0.9 \frac{kN}{m^2} $	$-1.4 \cdot 0.75 \frac{kN}{m^2} = -1.05 \frac{kN}{m^2}$
B	$-0.8 \cdot 0.75 \frac{kN}{m^2} = -0.6 \frac{kN}{m^2} $	$-1.1 \cdot 0.75 \frac{kN}{m^2} = -0.825 \frac{kN}{m^2}$
C	$-0.5 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2} $	$-0.5 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2}$
D	$0.8 \cdot 0.75 \frac{kN}{m^2} = 0.6 \frac{kN}{m^2} $	$1.0 \cdot 0.75 \frac{kN}{m^2} = -0.75 \frac{kN}{m^2}$
E	$-0.5 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2} $	$-1.1 \cdot 0.75 \frac{kN}{m^2} = -0.375 \frac{kN}{m^2}$

Those wind area loads we can now visualize in a floor plan.

Reduction of Wind force on Areas D and E

According to EN 1991-1-4 7.2.2 (3) the wind loads can be reduced in the direction of the wind (Areas D and E) if they meet the following criteria:

$$h/d <1 $$

Wind from Front

$$h/d = 17.1m/20.2m = 0.85 < 1$$

The criteria are met, and the forces in Areas D and E can be reduced by the factor 0.85.

Area	we.10	we.1
D	$0.85 \cdot 0.6 \frac{kN}{m^2} = 0.51 \frac{kN}{m^2} $	$0.85 \cdot 0.75 \frac{kN}{m^2} = 0.64 \frac{kN}{m^2}$
E	$0.85 \cdot (-0.375 \frac{kN}{m^2}) = -0.32 \frac{kN}{m^2} $	$0.85 \cdot (-0.375 \frac{kN}{m^2}) = -0.32 \frac{kN}{m^2}$

Wind from Side

$$h/d = 17.1m/49.75m = 0.34 < 1$$

The criteria are met, and the forces in Areas D and E can be reduced by the factor 0.85.

Area	we.10	we.1
D	$0.85 \cdot 0.6 \frac{kN}{m^2} = 0.51 \frac{kN}{m^2} $	$0.85 \cdot 0.75 \frac{kN}{m^2} = 0.64 \frac{kN}{m^2}$
E	$0.85 \cdot (-0.375 \frac{kN}{m^2}) = -0.32 \frac{kN}{m^2} $	$0.85 \cdot (-0.375 \frac{kN}{m^2}) = -0.32 \frac{kN}{m^2}$

Conclusion

The wind loads on buildings must always be calculated for 2 directions. ⬇️➡️ The wind loads have different values for the different wall areas A, B, C, D and E, which are calculated with the pressure coefficients and the peak velocity pressure.

In addition, the wind load must also be calculated for the roof. We have explained this in detail in the following articles:

• Wind load on pitched roofs
• Wind load on flat roofs
• Wind load on arched structures

Wind Loads On Walls FAQ