wind webinar series #3: asce 71 0 wind loads for...

WIND WEBINAR SERIES #3:

ASCE 7�10 Wind Loads for Signs, Other Structures, Roof�Top Structures &

Equipment, and Other Special Conditions

Robert Paullus, P.E., S.E., SECB

Paullus Structural Consultants [email protected]

26 February 2013

Wind Webinar #326 February 2013

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Wind Loads for Solid Signs,

Other Structures, Roof-Top

Structures & Equipment, and

Other Special ConditionsBob Paullus, P.E., S.E.

Paullus Structural Consultants


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1. 1. Chapter 29– Other Structures (MWFRS

Directional Method)a. Conditions

b. Limitations

c. Solid Freestanding Walls or Signs

d. Solid Attached Signs

e. Design Wind Loads on Other Structures

f. Design Wind Loads on Rooftop Structures

and Equipment on Buildings

g. Parapets

h. Roof Overhangs

i. Minimum Design Wind Loadings

Outline


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1. 2. Chapter 30 – Part 6 - Components &

Cladding for Building Appurtenances

and Rooftop Structures and Equipment

(Directional Procedure)a. Parapets

b. Roof Overhangs

c. Rooftop Structures and Equipment for

Buildings with h ≤ 60 ft (18.3 m)

2. 3. Examples

Outline


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• 1. The structure is a regular-shaped

structure as defined in Section 26.2.2.

• Section 26.2.2 - BUILDING OR OTHER

STRUCTURE, REGULAR-SHAPED: A

building or other structure having no

unusual geometrical irregularity in spatial

form.

Section 29.1.2 - Conditions


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• 2. The structure does not have response

characteristics making it subject to across-

wind loading, vortex shedding, or instability

due to galloping or flutter; or it does not

have a site location for which channeling

effects or buffeting in the wake of upwind

obstructions warrant special consideration.

Section 29.1.2 - Conditions


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• 1. This chapter DOES consider: load

magnification effect caused by gusts in

resonance with along-wind vibrations of

flexible structures.

• 2. This chapter DOES NOT consider:

Unusual shapes or configurations that lead

to effects listed in Section 29.1.2 –

Conditions.

Section 29.1.3 – Limitations


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• 3. If your structure does not fall within the

listed limitations, it should probably be the

subject of Wind Tunnel Study.

Section 29.1.3 – Limitations


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• 1. No reductions allowed for apparent

shielding by buildings, other structures, or

terrain features.a. Individual hills

b. Individual trees or small groves of

trees

c. Individual levees and similar

features.

• 2. Reductions are afforded for Terrain

Features in determining Exposure

Categories in Chapter 26.

Section 29.1.4 – Shielding


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Steps


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• 1. Steps 1-4 are the same as in Chapters

26-30.a. Chapter 29, like Chapters 26-30, has its

own table, Table 29.3-1, for Kh and Kz

b. Step 5: Eq. 29.3-1

qz = 0.00256KzKztKdV2 (lb/ft2)

• 2. Be careful to use qz or qh, as directed

under each section.

Steps to Determine Wind Loads


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• 1. Hollow Signs and Walls are not covered.

a. Signs which have openings that can be

pressurized

1) Boxed signs

2) Signs made from sea containers

3) Signs with large internal areas for lights

with translucent panels

2. Research is being conducted at Texas

Tech University by Douglas Smith, PhD

Section 29.4.1 – Solid Freestanding Walls

and Solid Freestanding Signs


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• 3. Solid Signs can have openings up to

30% of the Gross Area.

a. Reduction factor can be applied to solid

signs with openings.

b. Reduction factor (1 - (1 - ε)1.5)

c. ε = ratio of solid area to gross area

4. If the area of openings exceeds 30% of

the gross area, it is an open sign.–Proceed to Section 29.5 - Design Wind

Loads—Other Structures




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• 5. Basic Equation: (Eq. 29.4-1)

F = qhGCfAs (lb)• a. qh = the velocity pressure evaluated at

height h (defined in Fig. 29.4-1) as

determined in accordance with

Section 29.3.2

» h = top of the wall or sign

» Note: qh is at the top of the sign or wall

and Kd in Eq 29.3-1 is the Kd of Solid

Freestanding Walls and Solid Free-

standing and Attached Signs in Table

26.6-1




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• b. G = gust-effect factor from Section 26.9

c. Cf = net force coefficient from Fig. 29.4-1

d. As = the gross area of the solid free-

standing wall or freestanding solid sign, in

ft2 (m2)




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Figure 29.4-1


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• Load Cases to Consider

• Case A – Load applied to the centroid of

the area

• Case B – Load applied with an

eccentricity of 0.2*B (width of the wall or

sign)

• Case C – Stepped application of reduced

wind pressures as the distance decreases

from the windward edge.




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• Case C

–Reduction in loads for walls or signs with

returns at the ends

» Up to 40 % reduction

– For Elevated Signs or walls: where s/h > 0.8,

force coefficients shall be multiplied by the

reduction factor (1.8 - s/h)

» Accounts for reduced wind pressures

when free air flow under the wall or sign

is reduced.




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a.1. Requirements to use method in Section

29.4.1

a. The plane of the sign is parallel to and in

contact with the supporting wall

b. Edges of the sign do not extend past the

supporting wall

c. Use Component & Cladding Wall pressures

calculated in Chapter 30

d. Set the Internal Pressure Coefficient (GCpi)

equal to 0

Section 29.4.1 2 - Solid Attached Signs


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a. 2. Procedure can also be used for

signs attached to but not in contact

with the supporting wall.

a. Sign must be parallel to the supporting wall.

b. Sign must not be more than three (3) feet

from the wall.

c. Edges of the sign must be at least (3) feet in

from the free edges of the supporting wall:

a. Top of the supporting wall.

b. Bottom of the supporting wall

c. Side Edges of the supporting wall

Section 29.4.1 2 - Solid Attached Signs


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Section 29.5: Design Wind Loads—Other

Structures


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Basic Equation: (Eq. 29.5-1)

F = qzGCfAf (lb) (N)

a. qz = velocity pressure evaluated at height z as

defined in Section 29.3, of the centroid of area Af

» Note qz is at the centroid of the area and Kd in

Eq 29.3-1 is the Kd of the structure type in

Table 26.6-1

b. G = gust-effect factor from Section 26.9 (these

structures may often be flexible)

c. Cf = force coefficients from Figs. 29.5-1


Structures


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d. Af = projected area normal to the wind except

where Cf is specified for the actual surface area,

in ft2 (m2)


Structures


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Figure 29.5-1


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Figure 29.5-2


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Figure 29.5-3


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1. No guidance is given for rooftop

structures on buildings > 60 feet.

2. Research in the ASCE 7 committee

suggests that it is probably acceptable to

use loads from this section for rooftop

structures on buildings > 60 feet, but this

has not been confirmed yet.

3. Equation 29.5-2 gives lateral pressure

on the rooftop structure.

Section 29.5-1 – Rooftop Structures and

Equipment For Buildings with h ≤ 60 feet


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• Lateral Wind force on Rooftop Structures

– Basic Equation: (Eq 29.5-2)

Fh = qh(GCr)Af (lb) (N)

» (GCr) = 1.9 for rooftop structures and

equipment with Af less than (0.1Bh). (GCr)

shall be permitted to be reduced linearly

from 1.9 to 1.0 as the value of Af is

increased from (0.1Bh) to (Bh).

» qh = velocity pressure evaluated at mean

roof height of the building




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• Lateral Wind force on Rooftop Structures

» Note, qh is at the mean roof height and Kd

in Eq 29.3-1 is the Kd of the building, in

Table 26.6-1, on which the rooftop

structure sits.

» Af = vertical projected area of the rooftop

structure or equipment on a plane normal

to the direction of wind, in ft2 (m2)




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• Vertical Wind force on Rooftop Structures

– Basic Equation: (Eq 29.5-3)

Fv = qh(GCr)Ar (lb) (N)

» (GCr) = 1.5 for rooftop structures and

equipment with Ar less than (0.1BL). (GCr)

shall be permitted to be reduced linearly

from 1.5 to 1.0 as the value of Ar is

increased from (0.1BL) to (BL).

» qh = velocity pressure evaluated at mean

roof height of the building




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• Vertical Wind force on Rooftop Structures

» Note qh is at the mean roof height and Kd in

Eq 29.3-1 is the Kd of the building, in Table

26.6-1, on which the rooftop structure sits.

» Ar = horizontal projected area of rooftop

structure or equipment, in ft2 (m2)




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• Lateral C & C pressure (psf) shall be equal

to the lateral force (Lbs) calculated with

equation (29.5-2) DIVIDED BY the

RESPECTIVE WALL Surface area of the

Rooftop Structure considered.

– Forces (psf) shall be considered to act inward

and outward

Section 30.11 – Component & Cladding

Loads for Rooftop Structures and

Equipment for Buildings with h ≤ 60 feet


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• Vertical C & C pressure (Lbs) shall be

equal to the vertical force (Lbs) calculated

with equation (29.5-3) DIVIDED BY the

Horizontal projected area of the roof of

the Rooftop Structure considered.

– The pressures are ONLY required to be

considered to act in the UPWARD direction.





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• Comment: If the Rooftop Structure is

large (10’x20’ or larger), consider looking

at the downward pressures from the

building C&C loading figures and make

some judgment about downward wind

loading rooftop structures that resemble

small buildings (penthouses for instance).

– Vertical Wind Load would act in addition to

Dead and Roof Live Loads or Snow Loads.

– Use appropriate load combinations.





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Other Resources


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Other Resources

• Prepared by: Task

Committee on Wind-

Induced Forces of the

Petrochemical Committee

of the Enginery Division of

ASCE

• Several of those on the

Task Committee are on

the ASCE 7 Wind

Subcommittee

• Based on ASCE 7-05


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• “Wind loads on parapets are specified in

Section 27.4.5 for buildings of all heights

designed using the Directional Procedure

and in Section 28.4.2 for low-rise

buildings designed using the Envelope

Procedure.”

• Method presented is the Directional

Procedure

Section 29.6 – Parapets (MWFRS)


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• Chapter 28 – The Envelope Method, is

exactly the same.

– Chapter 28 uses the velocity pressure

determined with the Envelope Method,

rather than the velocity pressure in Chapter

27 using the Directional Method.

Section 29.6 – Parapets (MWFRS)


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• MWFRS pressures due to parapets

– Rigid or Flexible Buildings

– Applies to Flat, Gable, or Hip Roofs

Section 27.4.5 - Parapets


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• Basic Equation: (Eq 27.4-4)

pp = qp(GCpn) (lb/ft2)

– pp = combined net pressure on the parapet

due to the combination of the net pressures

from the front and back parapet surfaces.

Plus (and minus) signs signify net pressure

acting toward (and away from) the front

(exterior) side of the parapet

Section 27.4.5 – Parapets (MWFRS)


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– qp = velocity pressure evaluated at the top of

the parapet

» (GCpn) = combined net pressure

coefficient

– = +1.5 for windward parapet

– = –1.0 for leeward parapet



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• FIGURE C29.7-1 Design Wind Pressures on Parapets


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• Applicable to All Building Types

• Applicable to All Building Heights

– Except Where the Provisions of Part 4

are used (Simplified Method for

Buildings with h ≤ 160 feet)

Section 30.9 – C & C Loading on Parapets


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• Basic Equation: (Eq. 30.9-1)

p = qp((GCp) – (GCpi))

– qp = velocity pressure evaluated at the top of

the parapet

– (GCp) = external pressure coefficient given in

» Fig. 30.4-1 for walls with h ≤ 60 ft (48.8 m)

» Figs. 30.4-2A to 30.4-2C for flat roofs,

gable roofs, and hip roofs

» Fig. 30.4-3 for stepped roofs



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– (GCp) = external pressure coefficient given in

» Fig. 30.4-4 for multispan gable roofs

» Figs. 30.4-5A and 30-5B for monoslope

roofs

» Fig. 30.4-6 for sawtooth roofs

» Fig. 30.4-7 for domed roofs of all heights

» Fig. 30.6-1 for walls and flat roofs with h >

60 ft (18.3 m)

» Fig. 27.4-3 footnote 4 for arched roofs



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– (GCpi) = internal pressure coefficient from

Table 26.11-1, based on the porosity of the

parapet envelope.

• Consider Two (2) Load Cases when

evaluating C & C pressures on parapets

– Case A – Pressures on the surfaces of the

Windward parapet

– Case B – Pressures on the surfaces of the

Leeward parapet



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• Specifics of Case A:

– Windward Parapet shall consist of applying the

applicable positive wall pressure from Fig. 30.4-1

(h ≤ 60 ft (18.3 m)) or Fig. 30.6-1 (h > 60 ft (18.3

m)) to the windward surface of the parapet while

applying the applicable negative edge or corner

zone roof pressure from Figs. 30.4-2 (A, B or C),

30.4-3, 30.4-4, 30.4-5 (A or B), 30.4-6, 30.4-7, Fig.

27.4-3 footnote 4, or Fig. 30.6-1 (h > 60 ft (18.3

m)) as applicable to the leeward surface of the

parapet.



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• Specifics of Case B:

– Leeward Parapet shall consist of applying the

applicable positive wall pressure from Fig. 30.4-1

(h ≤ 60 ft (18.3 m)) or Fig. 30.6-1 (h > 60 ft (18.3

m)) to the windward surface of the parapet, and

applying the applicable negative wall pressure

from Fig. 30.4-1 (h ≤ 60 ft (18.3 m)) or Fig. 30.6-1

(h > 60 ft (18.3 m)) as applicable to the leeward

surface. Edge and corner zones shall be arranged

as shown in the applicable figures. (GCp)



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• FIGURE C29.7-1 Design Wind Pressures on Parapets

• If internal pressure

is present, both

load cases should

be evaluated

under positive and

negative internal

pressure.


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• Step 1: Determine risk category of building,

see Table 1.5-1

• Step 2: Determine the basic wind speed, V,

for applicable risk category, see Figure 26.5-

1A, B or C

Table 30.9-1 – Steps to Determine C&C

Wind Loads on Parapets


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• Step 3: Determine wind load parameters:

–Wind directionality factor, Kd , see Section 26.6

and Table 26.6-1

» Use Kd for Buildings C&C (0.85)

– Exposure category B, C or D, see Section 26.7

–Topographic factor, Kzt, see Section 26.8 and Fig.

26.8-1

–Enclosure classification, see Section 26.10

– Internal pressure coefficient, (GCpi), see Section

26.11 and Table 26.11-1

» Open, Partially Enclosed, or Enclosed (parapet or bldg)




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• Step 4: Determine velocity pressure

exposure coefficient, Kh, at top of the

parapet see Table 30.3-1

• Step 5: Determine velocity pressure, qp, at

the top of the parapet using Eq. 30.3-1




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• Step 6: Determine external pressure

coefficient for wall and roof surfaces

adjacent to parapet, (GCp)

–Walls with h ≤ 60 ft., see Fig. 30.4-1

– Flat, gable and hip roofs, see Figs. 30.4-2A to

30.4-2C

– Stepped roofs, see Fig. 30.4-3

–Multispan gable roofs, see Fig. 30.4-4

–Monoslope roofs, see Figs. 30.4-5A and 30.4-5B

– Sawtooth roofs, see Fig. 30.4-6




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• Step 6: (Continued)

–Domed roofs of all heights, see Fig. 30.4-7

–Walls and flat roofs with h > 60 ft., see Fig. 30.6-1

–Arched roofs, see footnote 4 of Fig. 27.4-3

• Step 7: Calculate wind pressure, p, using Eq.

30.9-1 on windward and leeward face of

parapet, considering two load cases (Case A

and Case B) as shown in Fig. 30.9-1.




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Figure 30.6-1

• Note 7 defines parapets > 3 feet as tall

parapets• Reduced corner pressures on parapet

• Similar note on other figures


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• “Wind loads on roof overhangs are

specified in Section 27.4.4 for buildings of

all heights designed using the Directional

Procedure and in Section 28.4.3 for low-

rise buildings designed using the Envelope

Procedure.”

• Present Direction Method in Section 27.4.1

–Envelope Method in Section 28.3.1 is similar

–Uses different factor for Cp

Section 29.7 – Roof Overhangs (MWFRS)


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• The positive external pressure on the

bottom surface of windward roof

overhangs shall be determined using Cp =

0.8 and combined with the top surface

pressures determined using Fig. 27.4-1.



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• Must consider cases with positive internal

pressure and negative internal pressure

• From Figure 27.6-3


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Section 30.10 – C & C Loading on Roof

Overhangs

• Applicable to All Building Types

• Applicable to All Building Heights

– Except Where the Provisions of Part 4

are used (Simplified Method for

Buildings with h ≤ 160 feet)


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Overhangs

• Basic Equation: (Eq. 30.10-1)

p = qh((GCp) – (GCpi))

– qh = velocity pressure from Section 30.3.2

evaluated at mean roof height h using

exposure defined in Section 26.7.3

– (GCp) = external pressure coefficients for

overhangs given in Figs. 30.4-2A to 30.4-2C

(flat roofs, gable roofs, and hip roofs),

including contributions from top and bottom

surfaces of overhang.


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Overhangs» The external pressure coefficient for the

covering on the underside of the roof

overhang is the same as the external

pressure coefficient on the adjacent wall

surface, adjusted for effective wind area,

determined from Figure 30.4-1 or Figure

30.6-1 as applicable

– (GCpi) = internal pressure coefficient

given in Table 26.11-1


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Figure 30.4-2A


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Figure 30.4-1


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• Step 1: Determine risk category of building,

see Table 1.5-1

• Step 2: Determine the basic wind speed, V,

for applicable risk category, see Figure 26.5-

1A, B or C


Wind Loads on Roof Overhangs


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• Step 3: Determine wind load parameters:

–Wind directionality factor, Kd , see Section 26.6

and Table 26.6-1

» Use Kd for Buildings C&C (0.85)

– Exposure category B, C or D, see Section 26.7

–Topographic factor, Kzt, see Section 26.8 and Fig.

26.8-1

–Enclosure classification, see Section 26.10

– Internal pressure coefficient, (GCpi), see Section

26.11 and Table 26.11-1

» Open, Partially Enclosed, or Enclosed (overhang or bldg)




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• Step 4: Determine velocity pressure

exposure coefficient, Kh, see Table 30.3-1

• Step 5: Determine velocity pressure, qh, at

mean roof height h using Eq. 30.3-1




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• Step 6: Determine external pressure

coefficient, (GCp), using Figs. 30.4-2A

through C for flat, gabled and hip roofs.

• Step 7: Calculate wind pressure, p, using Eq.

30.10-1. Refer to Figure 30.10-1




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Figure 30.10-1


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• The design wind force for other structures

shall be not less than 16 lb/ft2 (0.77 kN/m2)

multiplied by the area Af.

–16 psf is 10 psf from ASCE 7-05 times 1.6 to bring

the load to a strength level load.

–Apply to the full projected area in each

orthogonal direction.

Section 29.8 – Minimum Design Wind

Loading (MWFRS)


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• Loads on many shapes in industrial plants

–Tanks, Silos, Pipe racks, Partially Clad

Frames, etc.

–See ASCE report: “Wind Loads for

Petrochemical and Other Industrial

Facilities”

• Wind Loads on roof mounted Solar

Photovoltaic Arrays

–See new SEOC Guide

Material in Chapters 29 Not Covered


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Other Ressources


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• Office Complex

• Location: Wichita, KS

• Freestanding Sign

• 3-Story Office

–5-foot tall parapet

–Roof Top Unit

• Well Pump & Maintenance Building

–Roof Overhang

• Chemical Storage Silo

Example


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• Site Parameters Common to All Examples

• Exposure (Terrain Roughness): C– Location: N 37.7500, W 97.1683

– Section 26.7.3 (Open Farmland)

• Risk Category II Structures – Table 1.5-1

• Wind Velocity: 115 mph– Fig. 26.5-1A

– http://www.atcouncil.org/windspeed/index.php

• Topographic Factor, Kzt: 1.00– Section 26.8

• All Loads are Calculated to LRFD Levels

Example


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Example


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• Solid Billboard Sign at Ground Level• Dimensions: 30’ Wide x 10’ High

• Reference Figure 29.4-1 (MWFRS)

– s=10’

– B=30’

– h=10’

• Kh = 0.85 (Table 29.3-1)

• Kd = 0.85 (Solid Freestanding Walls & Signs)(Table 26.6-1)

• qh = 0.00256 KzKztKdV2 (psf)

• qh = 0.00256(0.85)(1.00)(0.85)(115)2 = 24.46 psf

• B/s = 30’/10’ = 3.0

• s/h = 10’/10’ = 1.0

Example-Freestanding Sign


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• Enter Figure 29.4-1 for Cf

–Applies to Cases A & B– For B/s = 2: Cf = 1.40

– For B/s = 4: Cf = 1.35

– Interpolating for B/s 3.0, Cf = 1.375

• G = 0.85 (Section 26.9 – Rigid Structure)

• F = qhGCfAs (Lb) (Eq 29.4-1)

• F = (24.46 psf)(0.85)(1.375)As = 28.59 psf*As

– As = Af = 30’x10’ = 300 ft2

– 28.59 psf*As > 16 psf * Af (Section 29.8 Min. Load)

• F = 28.59 psf(300 ft2) = 8577 Lbs

– For CASE A, Load is applied at the plan C.L. and at

– (s/2)+(0.05h) = 5.5’ above base



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» See Cross-Section View, Figure 29.4-1

– For CASE B, Load is applied @ 5.5’ above base and at 0.2B

offset, either side of plan C.L.

» 0.2B = 0.2(30’) = 6.0’ either side of plan C.L.

• Check to see if CASE C must be considered

– Note 3, Figure 29.4-1

– If B/s ≥ 2.0, CASE C must be considered

– B/s = 30’/10’ = 3.0 > 2.0, therefore consider CASE C

• Enter Figure 29.4-1 for Cf, under CASE C

– 0-s (0’-10’): Cf = 2.60

– s-2s (10’-20’): Cf = 1.70

– 2s-3s (10’-30’): Cf = 1.15



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– For CASE C, where s/h > 0.8, Cf may be multiplied by the

reduction factor (1.8 – s/h)

– s/h = 1.0 > 0.8

– (1.8 – s/h) = (1.8 – 1.0) = 0.8

– F = qhGCfAs (Lb) (Eq 29.4-1)

– F1 = (24.46psf)(0.85)(2.60)(0.8)(10’x10’) = 4324 Lbs

– F2 = (24.46psf)(0.85)(1.70)(0.8)(10’x10’) = 2828 Lbs

– F3 = (24.46psf)(0.85)(1.15)(0.8)(10’x10’) = 1913 Lbs

» Apply F1, F2, and F3 at plan C.L. of each plan length, s,

from each end of sign. See Figure.

» Apply F1, F2 and F3 at 5.5’ above base of each plan

length, s



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• Parapet (MWFRS) (Section 29.6)

• Office Building– L = 200 ft., B = 100 ft.

–Roof Height: h = 40 ft.

–Parapet Height: hp = 45 ft.

–Roof Slope, Flat: 0.25:12

» Ridge parallel to 200’ side

–Exposure Category: C

• Section 29.6 references Section 27.4.5 for

directional procedure for MWFRS Parapet load

determination.

Example-Parapet (MWFRS)


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– pp = qp(GCpn) (psf)

– qp = 0.00256KhKztKdV2 (psf) (Eq 27.3-1)

– Kh @ hp = 45’, Kh = 1.065

– Kzt = 1.00 (for complex)

– Kd = 0.85 (Building MWFRS)(Table 26.6-1)

– V = 115 mph (for complex)

– qp = 0.00256(1.065)(1.00)(0.85)(115)2 = 30.65 psf

– GCpn = +1.5 for windward parapet (Section 27.4.5)

– GCpn = -1.0 for leeward parapet (Section 27.4.5)

• Windward Parapet

– pp = (30.65 psf)(1.5) = 45.98 psf acting toward building

• Leeward Parapet

– pp = (30.65 psf)(-1.0) = -30.65 psf acting away from

building

Example-Parapet (MWFRS)


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• Parapet (C & C Loads) (Section 30.9)

• Office Building – same as previous

• qp = 0.00256(1.065)(1.00)(0.85)(115)2 = 30.65 psf

– from parapet MWFRS, above

• p = qp((GCp) – (GCpi)) (Eq 30.9-1)

• Parapet can be pressurized along with building

– See Figure

• GCpi = ±±±± 0.18 (Enclosed Building – Table 26.11-1)

Example-Parapet (C & C)


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• Studs @ 16” o.c. (both faces)

• Determine Effective Wind Area of Studs

– Greater of Tributary Area or Effective Width

» Effective Wind Area Definition (Section 26.2)

– Greater of 16”/12” = 1.33’ or

– Length/3 = 5’/3 = 1.67’ (governs)

» Effective Wind Area: l2/3 = 52/3 = 8.33 ft2

» If Effective Wind Area > 700 ft2, use MFWRS loads

• Determine which figure to reference from Table 30.9-1,

Step 6

– Figure 30.4-1 for wall pressures, h ≤ 60 ft.



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– Figure 30.4-2A for roof loads, h ≤ 60 ft. and gable roofs θ

≤ 7°°°°

– Determine “a” distance (Figure 30.4-1 and 30.4-2A)

» Lesser of 10% of B = 0.10(100’) = 10’ and 0.4h =

0.4(40’) = 16’

– 10’ governs

» Not less than the greater of 4% of B = 0.04(100’) = 4’

or 3’

» a = 10’

– Entering Figure 30.4-1 for pressure coefficients on

exterior surfaces of the parapets:

» Zone 4 Positive Pressure: GCp = 1.0

» Zone 5 Positive Pressure: GCp = 1.0



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» Zone 4 Negative Pressure: GCp = -1.1


– Note 5 says that values of GCp may be reduced

by 10% when θ ≤ 10°°°°

» Zone 4 Positive Pressure: GCp = (0.9)1.0 = 0.9

» Zone 5 Positive Pressure: GCp = (0.9)1.0 = 0.9 Zone 4

Negative Pressure: GCp = (0.9)-1.1 = -1.0

» Zone 5 Negative Pressure: GCp = (0.9)-1.4 = -1.26

– Entering Figure 30.4-2A for pressure coefficients on

interior (roof side) surfaces of parapet:

» Effective Wind Area: A= 52/3 = 8.33 ft2

» a = 10 ft. as in Figure 30.4-1

» Zone 1, 2, and 3 Positive Pressure: GCp = 0.3



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– Note 5: “If a parapet equal to or higher than 3 ft

(0.9m) is provided around the perimeter of the

roof with θ ≤ 7°°°°, the negative values of GCp in

Zone 3 shall be equal to those for Zone 2 and

positive values of GCp in Zones 2 and 3 shall be

set equal to those for wall Zones 4 and 5

respectively in Figure 30.4-1.”

» Parapet height hp = 5’ > 3’

– Zone 3 Negative GCp is Not Applicable

– Zone 2 and Zone 3 Positive Pressure GCp are

those of Wall Zone 4 and Zone 5, respectively.



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• Step 7: Calculate wind pressure, p using Eq 30.9-1 on

windward and leeward faces of parapet, considering two

load cases (CASE A and CASE B) as shown in Figure 30.9-1

– Note: As wind direction changes, each parapet with shift

from a windward parapet to a leeward parapet.

• CASE A – Windward Parapet

– Exterior Face Wall Studs

» p = qp((GCp) – (GCpi)) (Eq 30.9-1)

» With Positive Internal Pressure

– Zone 4 = Zone 5

– P = (30.65 psf)((0.9)-(0.18)) = 22.06 psf

» With Negative Internal Pressure

– Zone 4 = Zone 5

– p = (30.65 psf)((0.9)-(-0.18)) = 33.10 psf



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» Apply to Tributary Area, not Effective Wind Area

– 33.10 psf(1.33’) = 44.12 plf

– Interior Face (roof side) Parapet Studs

» p = qp((GCp) – (GCpi)) (Eq 30.9-1)


– Zone 2 (Zone 3 also treated as Zone 2)

– p = (30.65 psf)((-1.8)-(0.18)) = -60.69 psf


– Zone 2 (Zone 3 also treated as Zone 2)

– p = (30.65 psf)((-1.8)-(-0.18)) = -40.65 psf


– -60.69 psf(1.33’) = -80.72 plf



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• CASE B – Leeward Parapet

– Interior Face Parapet Studs (load toward parapet)

» p = qp((GCp) – (GCpi)) (Eq 30.9-1)


– Substitute Zone 4 and Zone 5 pressures for roof

Zone 2 and Zone 3 pressures, respectively. Zone

4 = Zone 5

– P = (30.65 psf)((0.9)-(0.18)) = 22.06 psf


– Zone 4 = Zone 5

– p = (30.65 psf)((0.9)-(-0.18)) = 33.10 psf


– 33.10 psf(1.33’) = 44.12 plf



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– Exterior Face Wall Studs (load away from parapet)

» p = qp((GCp) – (GCpi)) (Eq 30.9-1)


– Zone 4 pressure

– p = (30.65 psf)((-1.0)-(0.18)) = -36.17 psf

– Zone 5 pressure

– p = (30.65 psf)((-1.26)-(0.18)) = -44.14 psf


– Zone 4 pressure

– p = (30.65 psf)((-1.0)-(-0.18)) = -25.13 psf

– Zone 5 pressure

– p = (30.65 psf)((-1.26)-(-0.18)) = -33.10 psf



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– Zone 4

– -36.17 psf(1.33’) = -48.11 plf

– Zone 5

– -44.14 psf(1.33’) = -58.71 plf

– Summary

» Exterior Wall Studs extended past roof into parapet:

– Zone 4: 44.12 plf (toward building)

– Zone 4: -48.11 plf (away from building)

– Zone 5 is anything within 10ft of the corner

– Zone 5: 44.12 plf (toward building)

– Zone 5: -58.71 plf (away from building)



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» Interior (roof side) Parapet Studs

– Zone 4: 44.13 plf (toward parapet)

– Zone 4: -80.72 plf (away from parapet)

– Zone 5 is anything within 10ft of the corner

– Zone 5: 44.13 plf (toward parapet)

– Zone 5: -80.92 plf (away from parapet)



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• Rooftop Equipment for Buildings (MWFRS)

(Section 29.5.1)

• Office Building – same as previous

• Plan Dimensions: 10’ wide x 20’ long

• RTU Height: 4’ over 1’ tall curb

• Projected Height: 4’+1’=5’

• Lateral Force: Fh = qh(GCr)Af (Lb) (Eq 29.5-2)

• Vertical Force: Fv = qh(GCr)Ar (Lb) (Eq 29.5-3)• qh calculated at mean roof height of building

• Kh @ h = 40’, Kh = 1.04 (Table 29.3-1)

• Use Kd for building NOT Kd for rectangular Other Structures

Example-Rooftop Equip. (MWFRS)


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• Kd = 0.85 (Table 26.6-1)

• Other parameters as previously defined for building

• qh = 0.00256(1.04)(1.00)(0.85)(115)2 = 29.93 psf

• HORIZONTAL WIND FORCE

– Check projected area of side compared with least

projected area of building

» B*h = 100’(40’) = 8000 ft2

» Af (max) = 20’(5’) = 100 ft2

» Af <0.1Bh: 100 ft2 < 800 ft2

– GCr = 1.9

– Fh = (29.93 psf)(1.9)(Af) = 56.87 psf(Af) < 16 psf(Af)

» Minimum Load from Section 29.8

– Fh = (29.93 psf)(1.9)(100 ft2) = 5687Lbs(Eq 29.5-2)



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» Perpendicular to long side

– Fh = (29.93 psf)(1.9)(50 ft2) = 2843 Lbs (Eq 29.5-2)

» Parallel to long side

– Horizontal wind forces applied to geometric center of

vertical projected plane of unit

• VERTICAL WIND FORCE

– Check projected area of roof compared with that of

building

» B*L = 100’(200’) = 20,000 ft2

» Ar = 20’(10’) = 200 ft2

» Ar <0.1BL: 200 ft2 < 2,000 ft2

– GCr = 1.5

– Fv = (29.93 psf)(1.5)(200 ft2) = 8979Lbs(Eq 29.5-2)



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» Vertical Up

– Vertical wind forces applied to geometric center of

horizontal projected plane of unit

• Note: The UPLIFT pressure on the top of the rooftop

equipment acts SIMULTANEOUSLY with either the Lateral

pressure parallel to or perpendicular to the long edge of the

rooftop equipment or structure.

• The same procedure is used for a roof-mounted penthouse.



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• Rooftop Equipment for Buildings (C&C)

(Section 30.11)• Loads for Designing the Equipment cabinet enclosure or the

wall components for a penthouse

• Lateral C & C pressures

• Fh = 5687 Lbs (from previous)

• C & C Lateral Loads: Fh/Af = 5687 Lbs/100 ft2 = 56.87 psf

– Load is applied toward or away from unit on all sides

• C & C Vertical Loads: Fv/Ar = 8979 Lbs/200 ft2 = 44.90 psf

– Load is applied only in the Upward direction, away from

the top of the unit

Example-Rooftop Equip. (C & C)


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• Personal Recommendation (Not in the Standard)

– If the unit is large (over 200 ft2), consider a minimum

downward wind load.

– C & C Loading from Figure 30.4-2A

» GCp = 0.2 (downward component)

» Fv = qhGCp = (29.93psf)(0.2) = 6 psf

» For higher wind loads and low snow loads,

particularly less than 10 psf, this may produce a

controlling load combination

Example-Rooftop Equip. (C & C)


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• Roof Overhang (MWFRS) (Section 29.7)

• Equipment Building– L = 60 ft., B = 30 ft.

–Eave Height: 10 ft.

–Overhang Width: 3 ft.

–Roof Slope: 4.375:12 (θ = 20°°°°)

» Ridge parallel to 60’ side

–Exposure Category: C

–Average Building Height: h = 13.28’ < 15’

• Section 29.7 references Section 27.4.4 for

directional procedure for MWFRS Roof Overhang

load determination.

Example-Roof Overhang (MWFRS)


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– Use Cp = 0.8 in Eq 27.4-1 for determining roof and wall

loads over and adjacent to roof overhang

– p = qGCp-qi(GCpi) (psf) (Eq 27.4-1)

– G = 0.85 (rigid structure) (Section 26.9)

– Kzt = 1.00 (for complex)

– Kd = 0.85 (Building MWFRS)(Table 26.6-1)

– V = 115 mph (for complex)

– Kz @ z = 10’ for soffit, Kz = 0.85 (Table 27.3-1)

– qz = 0.00256KzKztKdV2 (psf) (Eq 27.3-1)

– qz = 0.00256(0.85)(1.00)(0.85)(115)2 = 24.46 psf

– GCpi = ±±±± 0.18 (Enclosed building)(Figure 26.11-1)

– Transverse Wind Loading governs, by Inspection



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– Pressure on underside of roof overhang

» p = (24.46 psf)(0.85)(0.8) – (24.46 psf)(0.18)

» p = 12.23 psf (positive internal pressure)

» p = (24.46 psf)(0.85)(0.8) – (24.46 psf)(-0.18)

» p = 21.04 psf (negative internal pressure)

– Enter Figure 27.4-1 for pressures on windward roof

» h/L = 13.28’/60’ = 0.22 < 0.25

» Cp = 0.2 (Condition 1)

» Cp = -0.3 (Condition 2)

» Condition 1

– p = (24.46 psf)(0.85)(0.2) – (24.46 psf)(0.18)

– p = 0.24 psf (positive internal pressure)

– p = (24.46 psf)(0.85)(0.2) – (24.46 psf)(-0.18)

– p = 8.56 psf (negative internal pressure)



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» Condition 2

– p = (24.46 psf)(0.85)(-0.3) – (24.46 psf)(0.18)

– p = -10.64 psf (positive internal pressure)

– p = (24.46 psf)(0.85)(-0.3) –(24.46 psf)(-0.18)

– p = -1.83 psf (negative internal pressure)

» Combine Top & Bottom Pressures with Same Internal

Pressure Conditions

– Note: signs indicate pressure toward or away

from surface

– Change signs so (+) is up, globally

– Change signs so (-) is down, globally



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– Positive Internal Pressure

– povh = 12.23 psf - 0.24 psf = 11.99 psf

– povh = 12.23 psf + 10.64 psf = 22.87 psf

– overall povh is upward

– Negative Internal Pressure

– povh = 21.04 psf – 8.56 psf = 12.48 psf

– povh = 21.04 psf + 1.83 psf = 22.87 psf

– overall povh is upward

– Note: Net effect of internal pressures is zero so

that the total uplift on the overhang is the same.



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– Overall pressure on the Leeward Overhang is

calculated the same way, but uses the negative

pressure on the wall immediately adjacent to

the overhang for downward pressures on soffit.

– By inspection, total force on windward

overhang will control.



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Example-Roof Overhang


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• Roof Overhang (C & C) (Section 30.10)

• Equipment Building

–Unless otherwise listed, parameters are

identical to those for the MWFRS calculations

–Determine C & C loads for overhangs of roof

trusses, spaced at 2’-0” o.c.

• p = qh[(GCp) – (GCpi)] psf (Eq 30.10-1)

–Kd = 0.85 (Building C&C) (Table 26.6-1)

–All other parameters for qh are same as for

MFWRS

Example-Roof Overhang (C & C)


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– qh = 0.00256(0.85)(1.00)(0.85)(115)2 = 24.46 psf

• Enter Figure 30.4-2B for Roof Overhang C & C coefficients

• Determine Effective Wind Area of Studs

– Greater of Tributary Area or Effective Width

» Effective Wind Area Definition (Section 26.2)

– Greater of 2.0’ or (governs)

– Length/3 = 3’/3 = 1.00’

» Effective Wind Area: 3’x2’ = 6.00 ft2

» If Effective Wind Area > 700 ft2, use MFWRS loads

– Determine “a” distance (Figure 30.4-1 and 30.4-2A)

» Lesser of 10% of B = 0.10(30’) = 3’ and 0.4h =

0.4(13.28’) = 5.31’

– 3’ governs



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» Not less than the greater of 4% of B = 0.04(30’) = 1.2’

or 3’ (3’ controls)

» a = 3’ (equal to width of overhang; therefore, Zone 1

pressures are not applicable to any part of the

overhang)

– Zone 2: GCp = -2.2

» p = (24.46 psf)[(-2.2) – (0.18)] = -58.21 psf

– with positive internal building pressure

» p = (24.46 psf)[(-2.2) – (-0.18)] = -49.41 psf

– with negative internal building pressure

– Zone 3: GCp = -3.7

» p = (24.46 psf)[(-3.7) – (0.18)] = -94.90 psf

– with positive internal building pressure



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» p = (24.46 psf)[(-3.7) – (-0.18)] = -86.10 psf

– with negative internal building pressure

» For the overhang portion of the truss:

– Tributary Width = 2’

– Upward force on the entire truss end is:

– Zone 2: (-58.21 psf)(2’) = -116.42 plf

(upward)

– Zone 3: (-94.90 psf)(2’) = -189.90 plf

(upward)

– These are NOT the loads on the soffit material .



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• Section 30.10 states that the external coefficient for the

covering on the underside of the roof overhang (soffit) is the

same as the external pressure coefficient on the adjacent

wall surface, determined from Figure 30.4-1 or Figure 30.6-

1, as applicable.

– Use of the GCp with negative internal pressure yields the

greatest upward load on the material on the underside of

the overhang on the windward wall.

– Use of the GCp with positive internal pressure yields the

greatest downward load on the material on the

underside of the overhang on the leeward wall.

– For this building, assuming effective wind area is the

same as for the truss overhang:



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» Windward Wall Soffit Material (Figure 30.4-1)

– Zone 4 and Zone 5: GCp = +1.0

– p = (24.46 psf)[(1.0) –(-0.18)] = 28.86 psf

– acting upward

» Leeward Wall Soffit Material (Figure 30.4-1)

– Zone 4: GCp = -1.1

– p = (24.46 psf)[(-1.1) –(0.18)] = -31.31 psf

(acting downward)

– Zone 5: GCp = -1.4

– p = (24.46 psf)[(-1.4) –(0.18)] = -38.65 psf

(acting downward)



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• Chemical Silo (Other Structure MWFRS)

(Section 29.5)

• Silo Dimensions: h = 20’, D = 5.0’

• Welded Steel Tank: smooth sides, no ladder

• Roof Slope: 1:12 (conical)

–Maximum rise: 2.5 inches (consider

contribution to wind load, negligible)

• Unless otherwise listed, parameters for

calculation of qz are identical to those for the

MWFRS calculations for equipment building.

• F = qzGCfAf Lbs (Eq 29.5-1)

Example- Chemical SILO (MWFRS)


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– qz = 0.00256KzKztKdV2 (psf)

– Kz = 0.90 (Building C&C) (Table 29.3-1)

– Kd = 0.95 (Circular Tanks) (Table 26.6-1)

– G = 0.85 (Rigid Structure) (Section 26.9)

– qz = 0.00256(0.90)(1.00)(0.95)(115)2 = 28.95 psf

– Go to Table 29.5-1

» D/√ qz = 5’/Sqrt(28.95psf) = 0.93 < 2.5

» Go to bottom row

» h/D = 20’/5’ = 4.0

» Must interpolate between h/D=1.0 and h/D= 7.0

» Cf = 0.75

– F = (28.95 psf)(0.85)(0.75) Af = (18.45 psf)Af

– F = (18.45 psf)Af < (16psf)Af



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– Af = 5’x20’ = 100 ft2

– F = (18.45 psf)(100 ft2) = 1845 Lbs

» This is conservative, OR calculate F for increase in

pressure as height increases

– q15 = 0.00256(0.85)(1.00)(0.95)1152 = 27.34 psf

– q(15-20)= 0.00253(0.90)(1.00)(0.95) (115)2 = 28.95 psf

» For 0-15’: D/√ qz = 5’/Sqrt(27.34psf) = 0.96 < 2.5

» Cf = 0.75

– F0-15 = (27.34 psf)(0.85)(0.75)Af = 17.43 psf Af > 16 psf Af

– F0-15 = (17.43 psf)(5’)(15’) = 1307 Lbs.

– F15-20 = (18.45 psf)(5’)(5’) = 461 Lbs.

– Total F on Silo: 1307 Lbs + 461 Lbs = 1768 Lbs



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– Conservative OTM: (1845 Lbs)(10’) = 18,450 ft-lbs

– More Detailed OTM: (1307 Lbs)(15’/2)+(461

Lbs)(15’+5’/2) = 17,870 ft-lbs

» The taller the structure is, the more important it is to

use the stepped wind force approach.



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Questions


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