Wind Loads

Structural Design · FE Reference Handbook section

Structural Design

5 formulas

10 exam-style examples

~55 min

All Structural Design lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

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Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Velocity pressure and windward wall pressure — Wind Loads

A building in a region with V = 115 mph has Kz = 0.79, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 115 mph$
$Kz = 0.79$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 115^{2} = 13,225$
Substituting
$qz = 0.00256(0.79)(1.0)(0.85)(13,225)$
Evaluate
$qz = 22.73 psf$
Wall pressure
$p = qz G Cp = 22.73(0.85)(0.8) = 15.46 psf$

Answer:

qz ≈ 22.7 psf, windward p ≈ 15.5 psf

Why the other options are there

0.20 psf (V not squared)
22.7 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 2

Velocity pressure and windward wall pressure — Wind Loads (2)

A building in a region with V = 115 mph has Kz = 0.82, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 115 mph$
$Kz = 0.82$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 115^{2} = 13,225$
Substituting
$qz = 0.00256(0.82)(1.0)(0.85)(13,225)$
Evaluate
$qz = 23.60 psf$
Wall pressure
$p = qz G Cp = 23.60(0.85)(0.8) = 16.05 psf$

Answer:

qz ≈ 23.6 psf, windward p ≈ 16.0 psf

Why the other options are there

0.21 psf (V not squared)
23.6 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 3

Velocity pressure and windward wall pressure — Wind Loads (3)

A building in a region with V = 115 mph has Kz = 0.90, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 115 mph$
$Kz = 0.90$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 115^{2} = 13,225$
Substituting
$qz = 0.00256(0.90)(1.0)(0.85)(13,225)$
Evaluate
$qz = 25.90 psf$
Wall pressure
$p = qz G Cp = 25.90(0.85)(0.8) = 17.61 psf$

Answer:

qz ≈ 25.9 psf, windward p ≈ 17.6 psf

Why the other options are there

0.23 psf (V not squared)
25.9 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 4

Velocity pressure and windward wall pressure — Wind Loads (4)

A building in a region with V = 130 mph has Kz = 1.03, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 130 mph$
$Kz = 1.03$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 130^{2} = 16,900$
Substituting
$qz = 0.00256(1.03)(1.0)(0.85)(16,900)$
Evaluate
$qz = 37.88 psf$
Wall pressure
$p = qz G Cp = 37.88(0.85)(0.8) = 25.76 psf$

Answer:

qz ≈ 37.9 psf, windward p ≈ 25.8 psf

Why the other options are there

0.29 psf (V not squared)
37.9 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 5

Velocity pressure and windward wall pressure — Wind Loads (5)

A building in a region with V = 90 mph has Kz = 0.89, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 90 mph$
$Kz = 0.89$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 90^{2} = 8,100$
Substituting
$qz = 0.00256(0.89)(1.0)(0.85)(8,100)$
Evaluate
$qz = 15.69 psf$
Wall pressure
$p = qz G Cp = 15.69(0.85)(0.8) = 10.67 psf$

Answer:

qz ≈ 15.7 psf, windward p ≈ 10.7 psf

Why the other options are there

0.17 psf (V not squared)
15.7 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 6

Velocity pressure and windward wall pressure — Wind Loads (6)

A building in a region with V = 115 mph has Kz = 0.72, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 115 mph$
$Kz = 0.72$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 115^{2} = 13,225$
Substituting
$qz = 0.00256(0.72)(1.0)(0.85)(13,225)$
Evaluate
$qz = 20.72 psf$
Wall pressure
$p = qz G Cp = 20.72(0.85)(0.8) = 14.09 psf$

Answer:

qz ≈ 20.7 psf, windward p ≈ 14.1 psf

Why the other options are there

0.18 psf (V not squared)
20.7 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 7

Velocity pressure and windward wall pressure — Wind Loads (7)

A building in a region with V = 130 mph has Kz = 0.74, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 130 mph$
$Kz = 0.74$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 130^{2} = 16,900$
Substituting
$qz = 0.00256(0.74)(1.0)(0.85)(16,900)$
Evaluate
$qz = 27.21 psf$
Wall pressure
$p = qz G Cp = 27.21(0.85)(0.8) = 18.50 psf$

Answer:

qz ≈ 27.2 psf, windward p ≈ 18.5 psf

Why the other options are there

0.21 psf (V not squared)
27.2 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 8

Velocity pressure and windward wall pressure — Wind Loads (8)

A building in a region with V = 90 mph has Kz = 0.84, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 90 mph$
$Kz = 0.84$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 90^{2} = 8,100$
Substituting
$qz = 0.00256(0.84)(1.0)(0.85)(8,100)$
Evaluate
$qz = 14.81 psf$
Wall pressure
$p = qz G Cp = 14.81(0.85)(0.8) = 10.07 psf$

Answer:

qz ≈ 14.8 psf, windward p ≈ 10.1 psf

Why the other options are there

0.16 psf (V not squared)
14.8 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 9

Velocity pressure and windward wall pressure — Wind Loads (9)

A building in a region with V = 115 mph has Kz = 0.73, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 115 mph$
$Kz = 0.73$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 115^{2} = 13,225$
Substituting
$qz = 0.00256(0.73)(1.0)(0.85)(13,225)$
Evaluate
$qz = 21.01 psf$
Wall pressure
$p = qz G Cp = 21.01(0.85)(0.8) = 14.29 psf$

Answer:

qz ≈ 21.0 psf, windward p ≈ 14.3 psf

Why the other options are there

0.18 psf (V not squared)
21.0 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Example 10

Velocity pressure and windward wall pressure — Wind Loads (10)

A building in a region with V = 90 mph has Kz = 1.00, Kzt = 1.0 and Kd = 0.85. Find the velocity pressure and the windward wall design pressure using G = 0.85 and Cp = 0.8.

Given

$V = 90 mph$
$Kz = 1.00$
$Kzt = 1.0$
$Kd = 0.85$
$G = 0.85$
$Cp = 0.8$

Find

qz and windward pressure p

Start with the thinking

Velocity pressure varies with the SQUARE of the wind speed.
0.00256 already converts mph² to psf — do not convert V to ft/s.

Step-by-step solution

Velocity pressure
$qz = 0.00256 Kz Kzt Kd V^{2}$
Speed squared
$V^{2} = 90^{2} = 8,100$
Substituting
$qz = 0.00256(1.00)(1.0)(0.85)(8,100)$
Evaluate
$qz = 17.63 psf$
Wall pressure
$p = qz G Cp = 17.63(0.85)(0.8) = 11.99 psf$

Answer:

qz ≈ 17.6 psf, windward p ≈ 12.0 psf

Why the other options are there

0.20 psf (V not squared)
17.6 psf reported as the wall pressure (G and Cp omitted)

Reference: FE Reference Handbook — Structural Design → Wind Loads

Wind Loads

Handbook notes for this section

Core formulas for this FE topic

qz=0.00256KzKztKdV2(lb/ft2)qz = 0.00256 Kz Kz t Kd V 2 (lb/ft_{2})qz=0.00256KzKztKdV2(lb/ft2​)

Kd=winddirectionalityfactor=0.85formoststructuresKd = wind directionality factor = 0.85 for most structuresKd=winddirectionalityfactor=0.85formoststructures

Kz=velocitypressureexposurecoefficientKz = velocity pressure exposure coefficientKz=velocitypressureexposurecoefficient

Kzt=topographicfactor=1.0forflatgroundKzt = topographic factor = 1.0 for flat groundKzt=topographicfactor=1.0forflatground

V=basicwindspeed(mph)V = basic wind speed (mph)V=basicwindspeed(mph)

Worked exam-style examples