wind example claculations fema excerpt

7/27/2019 Wind Example Claculations FEMA Excerpt

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F. ExampleCalculationsDesign a CMU pier and ground anchor oundation or a manuactured home to be placed inan SFHA Zone AE having a ood velocity o 2 ps. The BFE is 9 eet and existing ground eleva-tion is approximately 7 eet. The ood depth is 2 eet and the reeboard is 1 oot, which yieldsa DFE depth o 3 eet. The manuactured home dimensions or these example calculations areshown in Figure F-1. The manuactured home is a single unit, 16 eet wide and 60 eet long

with a 30-degree gable roo with a 1-oot overhang. Roofng members are spaced 16 inches oncenter (o.c.). The manuactured home weighs 20 ps. Assume an NFPA 5000 soil classifcationo sot, sandy clay, or clay (allowable bearing pressure q

a=1,000 ps ; ultimate bearing pressure

qu

= 2,000 ps). Use ASCE 7 to calculate loads.

Foundation loads selected or this example o a manuactured home in an SFHA dier rom

those that may be ound in HUD standard 24 CFR 3280. Design loads in this example are in ac-cordance with ASCE 7-05 and other standards.

These example calculations assume transverse wind loads produce the controlling loading.Wind in the direction parallel to the roo ridge may produce greater loads or certain cases andmust be evaluated during fnal design.

FigureF-1.Manuactured

homedimensions.

ASCE 7-05

PROTECTING MaNufaCTuREd HOMEs fROM flOOds aNd OTHER HazaRds

A Multi-Hazard Foundation and Installation Guide

F-1


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F EXAMPLE CALCULATIONS

Step1: DetermineDesignCriteria

NORMALLOADS

DeadLoad(D)

D = 20 ps Given in the example statement

LiveLoad(L)

L is based on one- and two-amily dwellings

L = 40 ps

RooLiveLoad(Lr)

Lr= 20R

1R

2=20(1)(0.85)=17 ps

R1

= 1 or At 200 t2

At = 2(9.2 ft)(16 in) =24.5ft2

F = number o inches o rise per oot

F= 1ft tan 30 = 7 in

Note that the roo live load alls between the limits giv-en:

12 Lr 20

ENVIRONMENTALLOADS

WindLoading

Structure is a regular shape, located in a windborne debris region withterrain classifcation o Exposure C and surrounded by at terrain.

Mean roo height (h)

h = 3 t + 10 t + 0.5(4 t)

= 15 t

h < 16 t (least horizontal dimension)

Calculations are or a oundation system, which is a main wind orce re-sisting system (MWFRS).

VelocityPressures

Velocity pressures are determined using

Method 2: Analytical Procedure

ASCE 7-05

Table 4-1

ASCE 7-05

Section 6.2

Section 6.5

1ft12 in( )12 in

1ft( )

F-2 PROTECTING MaNufaCTuREd HOMEs fROM flOOds a Nd OTHER HazaRdsA Multi-Hazard Foundation and Installation Guide


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EXAMPLE CALCULATIONS F

(A simplifed alternative is to use ASCE 7, Section 6.4, Method 1. Windpressures are tabulated or basic conditions. The wind pressure must beadjusted or mean roo height and exposure category.)

Velocity Pressure Coefcient (qz)

qz= 0.00256K

zK

ztK

dV2I

Velocity pressure exposure coefcient evaluated at heightz(theheight above ground level in eet) (K

z)

Kz= 0.85

Topographical actor (Kzt)

Kzt=1 (assume a at surace)

Wind directionality actor (Kd)

Kd = 0.85

Basic Wind Speed (V)

V = 110 mph (3-second gust)

I = 1 (Category II building: Table 1-1 (ASCE 7))

Thereore, qz= 0.00256(0.85)(1)(0.85)(110)2(1)

= 23 ps

DesignPressuresorMWFRS

Internal Pressure Coefcient (GCpi

)

GCpi

= 0.18

External Pressure Coefcient (Cp)

h = Mean roo height, in eet

L = Horizontal dimension o building, in eet, measured par-allel to wind direction

B = Horizontal dimension o building, in eet, measured nor-

mal to wind direction

Table F-1 shows the External Pressure Coefcients calculated or thewindward, leeward and side walls. Computations o the External Pres-sure Coefcients or the windward and leeward roo are shown TableF-2.

Section 6.5.10

Eq. 6-15

Section 6.5.6

Table 6-3

Section 6.5.6

Section 6.5.4.4

Section 6.5.5

Table 6-1

Section 6.5.4

Section 6.5.11.1Figure 6-5

Section 6.5.11.2.1Figure 6-6

Figure 6-6



F-3


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TableF-1.ExternalWallPressureCoefcients

Surace WindDirection L/B Cp

Windward Wall n/a n/a 0.8

Leeward WallPerpendicular to

roo ridge-0.5

Side Wall n/a n/a -0.7

TableF-2.ExternalRooPressureCoefcients

Surace WindDirection h/L Cp

Windward Roo Perpendicular toroo ridge

-0.3

0.2

Leeward Roo -0.6

Foundation systems are considered rigid, thereore, G = 0.85.

Design Pressure (p)

The basic pressure equation (ASCE 7 6-17), which includes the internalpressure coefcient is as ollows:

p = qGCp qi(GCpi)However, this would only be used i designing individual componentswhose eective tributary area is equal to or greater than 700 s (ASCE7-05 6.5.12.1.3 and IBC 2006 1607.11.2.1). When determining loads onthe global structure (i.e., shear walls or oundation design), the inter-nal pressure components will act in equal and opposite directions onthe roo/oor and the leeward/windward walls, thereby algebraical-ly canceling each other. The resulting simplifed orm o the pressureequation is:

p = q x GCp

Table F-3 summarizes the design pressures calculated using this simpli-fed wind design pressure equation. Figure F-2 shows the application othese design pressures on the structure. For oundation design, internalpressures need not be considered since internal pressure on windwardwalls, leeward walls, oors, and roos cancel each other. For example,internal pressures acting on a windward wall are equal and opposite tothose acting on a leeward wall and the net orce on the oundation rominternal pressures is zero.

15 ft16 ft

=0.94

16 ft

60 ft=0.27

Figure 6-6

Figure 6-6

Section 6.5.8Eq. 6-17



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While internal pressures cancel, internal pressures or a partially en-closed building have been included in the example. This is to providean example o more general wind load calculations.

TableF-3.DesignPressuresorWindPerpendiculartotheRooRidge

SuraceDesignWindPressure

Calculationspressure

(ps)

Windward Wall p = 23 ps(0.85)(0.8) 15.7

Leeward Wall p = 23 ps(0.85)(-0.5) -9.8

Side Walls p = 23 ps(0.85)(-0.7) -13.7

Windward Roo

p = 23 ps(0.85)(-0.3) -5.9

p = 23 ps(0.85)(0.2) 4.0

Leeward Roo p = 23 ps(0.85)(-0.6) -11.8

MWFRSRooOverhangPressures

ASCE 7 only addresses the windward overhang, speciying the use o apositive pressure coefcient o C

p= 0.8. Acting on the bottom surace o

the overhang in combination with pressures acting on the top surace.For the leeward overhang, the coefcient or the leeward wall (C

p= -0.5)

could be used, but the coefcient has been conservatively taken as zero.

p = 23 ps(0.85)(0.8)

= 15.7 ps

pOH

= -5.9 ps 15.7 ps = -21.6 ps

A conservative simplifcation is to use the wind pressure acting away romthe roo case or uplit roo pressure simultaneously with the wind pres-sure toward the roo or the lateral roo pressure.

ASCE 7-05

Section 6.5.11.4.1



F-5


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FigureF-2.Maximum

uplitandlateralwind

loadsonroo.

SNOWLOADING

Ground Snow Load (pg)

pg

= 20 ps

Flat Roo Snow Load (p)

p

= 0.7CeC

tIp

g

p

= 0.7(1.0)(1.0)(1.0)(20)

= 14 ps

But not less than p=(I)p

g= 20 ps

ASCE 7-05

Section 7.2Figure 7-1

Section 7.3

Eq. 7-1



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Section 7.3.1

Table 7-2

Section 7.3.2

Table 7-3

Section 7.3.3

Table 7-4

Section 7.4

Eq. 7-2

Section 7.4.1

Figure 7-2

Section 7.6

Section 7.6.1

Figure 7-9

Eq. 7-3



F-7

Exposure Coefcient (Ce)

Ce

= 1.0 (partially exposed roo)

Thermal Factor (Ct)

Ct = 1.0Importance Factor (I)

I = 1.0 (Category II building: Table 7-4 (ASCE 7))

Sloped Roo Snow Load (ps)

ps

= Csp

= (1.0)(20 ps)

= 20 ps

Warm Roo Slope Factor (Cs)

Cs

= 1.0 (asphalt shingle not slippery)

Unbalanced Roo Snow Load (pu)

Since the roos eave to ridge distance 20 t, unbalanced uniorm snowloads shall be applied as ollows:

Pwindward

= 0.3 ps

= 6 ps

Pleeward.1

= ps

= 20 ps

Pleeward.2

= (hd)()/(S)

= (1.44 t)(16.6 pc)/(1.73)

= 18.2 ps

From the ridge toward the leeward eave a distance o:

x = (8/3)(hd)(S)

= 5.1 t

hd

= 1.44 t

= 0.13 pg + 14 30 pc= 16.6 pc


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FLOODLOADING

Hydrostatic Load (Fh)

I the manuactured home is elevated above the BFE onan enclosed oundation, venting must be provided in allmanuactured homes placed in a SFHA; the hydrostaticorces on either side o the oundation wall will cancel.However, the hydrostatic load is calculated because it isused in the hydrodynamic load calculation.

Fh

= PhH

Hydrostatic Pressure (Ph)

Ph

=H

Specifc Weight o Fresh Water ()

=62.4 pcf

Floodproofng Design Depth (H)

H = 2 t (base ood depth) + 1 t

Hydrodynamic Load

The hydrodynamic load is calculated by converting it to an equivalenthydrostatic load by increasing the ood depth. The increase in ooddepth is reerred to as d

h.

Drag Coefcient (Cd)

In the above equation, a value o 2.0 was assumedor C

d. This is a conservative estimate; the actual

value or Cd

could be anywhere between 1.2 and2.0.

Acceleration Due to Gravity (g)

g = 32.2 t/s2

with a hydrodynamic pressure o

Phydr

= (dh)= 62.4pcf(0.13ft) = 8.2psf

The equivalent hydrostatic load (Fh/ad

) taken into consid-eration the hydrodynamic load is :

Fh/ad

= Phydr

x H = 8.2 psfx3 ft= 24.6 plf

ASCE 7

Eq. 7-5CdV2

2gdh = = =0.13 ft

2.0(2ft/s)2

2(32.2ft/s)

FEMAs Coastal Construction Man-

ual(FEMA 55) recommends a valueo 2.0 or square or rectangular pilesand 1.2 or round piles.

For additional guidance regardingdrag coefcients, reer to Volume II

o the U.S. Army Corps o Engineers

Shore Protection Manual (USACE1984), FEMA 55, and theEngineeringPrinciples and Practices or Retro-

ftting Flood-Prone Residential Struc-tures(FEMA 259).

Note: A 1-oot reeboard is added

to the BFE depth to provide aprotection above the BFE; 3 eetbecomes the design depth or the

DFE.



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Since piers are 16 inches wide, the total hydrodynamic orce onthe pier is

CHECKSCOUR

Reference: Publication No. FHWA NHI 00-001,Evaluating Scour at Bridges, 4th Edition, May 2001,

Hydraulic Engineering Circular No. 18.

Ys

= 2.0 x (K1)x (K

2) x(K

3) x(K

4) x (a/Y

1)0.65 x F

r10.43

Y1

Where: Y

= Scour depth

Y1

= Flow depth directly upstream of pier

a = Pier width (ft.)

L = Pier length (ft)

Fr1

= Froude number

Fr1= V

1/(gY

1)1/2

Where V1

= Mean velocity of ow directly upstream of pier

g = acceleration due to gravity (32.2 feet/sec2)

lb

ft= 24.6 16 in 32 lbsper pier

1 ft

12 in

L = 16"

a = 8"

L = 16"



F-9


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K1

= Factor for pier nose shape. For square nose

K2

= Factor for Angle of attack . K2

= (cosine + (L/a) x sine )0.65

K3

= Factor for bed condition/. K3

= 1.1

K4

= Factor for armoring by bed material size.

Project parameters:

Flood low = 2 fps

Flood depth = 3 ft

Assume ood angle of attack = 0

So that:

K1

= 1.1 (Table 6.1)

K2

= [cosine 0o +(L/a)x sine 00]0.65 = [1.00 + (1.33/0.67) x 0]0.65 = 1.00

K3

= Factor for bed condition/. K3

= 1.1 (Table 6.3)

K4

= Factor for armoring by bed material size. K4

= 1.0 unarmored

Fr1

= V1/(gY

1)1/2 = 2/[32.2 x 3]1/2 =2/9.84 = 0.203

And

Ys = (2) x (1.1) x (1.0) x (1.1) x (1.0) x (0.67/2)0.65 x (0.203)0.43

Y1

Ys= (2.42) x (0.491) x (.504) = 0.6

Y1

Ys= (0.6) x (Y

1) = (0.6)x(3) = 1.8 ft

Scour protection or increased footing embedment required.



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Step2: SelectaDesignMethodologyandAssessLoad

CombinationsandFailureModes

Figure F-3 illustrates the loads applied to the manuactured home. Table F-4 lists the nomencla-

ture o the applied loads shown in Figure F-3.

TableF-4.LoadNomenclature

Nomenclature LoadDescription

D dead load

L live load

LR

roo live load

RH

horizontal reaction

RLV

leeward vertical reaction

RWV

windward vertical reaction

SB

balanced snow load

WH

horizontal wall wind pressure

WRH

roo horizontal wind pressure

FigureF-3.Loadingona

manuacturedhome.



F-11


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Nomenclature LoadDescription

WLRV

leeward roo vertical wind pressure

WWRV windward roo vertical wind pressure

Note that snow load governs over roo live load and wind downward load, and wind lateralload governs over earthquake lateral load. Load combinations or non-governing cases are notshown.

For the purposes o these calculations, the worst case wind load is taken to be perpendicular tothe roo ridge or all ailure modes. Wind in the direction parallel to the roo ridge may pro-duce greater loads or certain ailure modes.

UplitandDownwardFailureModeUplit ailure is a vertical orce phenomenon. The loads that act vertically are wind, snow, dead,and live loads. Table F-5 summarizes the loads that inuence uplit and downward ailure mode.Table F-6 assesses uplit and downward ailure load combinations. Note that uplit is based onMWFRS pressures or the global oundation design. Design o the connections to the ounda-tion may require components and cladding (C&C) pressures to be used.

TableF-5.VerticalLoadValues

LoadType

Totalloadactingonthestructureand,thereore,mustbesupportedbytheoundation

D D = [dead load per square oot][width o the manuactured home]

D = [20 ps][16 t]

D = 320 lbs per linear t o manuactured home length

L L = [live load per square oot][width o the manuactured home]

L = [40 ps][16 t]

L = 640 lbs per linear t o manuactured home length

Lr

Lr= [roo live load per square oot][width o the manuactured home]

Lr= [17 ps][18 t]

Lr= 306 lbs per linear t o manuactured home length

TableF-4.LoadNomenclature(continued)



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LoadType


W Maximum wind uplit loads occur or winds parallel to the roo ridge at the windward end.

W = WWRV

+WLRV

= [(vertical component roo wind pressures)(area roo)]/manuactured home length

W = [-17.6 ps][(9 t)(15 t)(2)] 0 t to 15 t +

[-9.8 ps][(9 t)(15 t)(2)] 15 t to 30 t +

[-5.9 ps][(9 t)(30 t)(2)] 30 t to 60 t

W = -10,584 lbs/60 t = -176 lbs per linear oot o manuactured home length (average)

In this case, vertical uplit loads are low and so this simplifcation is acceptable. However, toaccount or the unbalanced uplit i wind loads were higher, either overturning in this directionwould need to be considered, or the windward uplits conservatively made symmetrical about themiddle.

Maximum wind downward loads occur or wind perpendicular to the roo ridge; however, they aremuch less than, and do not govern over, roo live or snow loads.

S S = [snow pressure][horizontal projected roo area]

S = [20 ps][(9 t)]SW + [20 ps][(9 t)]SL

S = 360 lbs per linear t o manuactured home length

TableF-6.VerticalFailureModeASDLoadCombinations

LoadCombinations

4D + 0.75L + 0.75S

320 lbs + 0.75(640 lbs) + 0.75(360 lbs) = 1,070 lbs per linear t o manuactured home length

70.6D + W

0.6(320 lbs) - 176 lbs = 16 lbs per linear t o manuactured home length acting downward

Note that, or load combination 7, the 0.6 load actor should be applied to the dead load thatwould actually be present over the whole structure. Additions to the dead load tabulation suchas mechanical and miscellaneous or shingles should not be included in this value as they maynot be present in all areas or during a high-wind event and their inclusion would not be con-servative.

SlidingorShearingFailureMode

Sliding ailure is a lateral orce phenomenon. The loads that act laterally are wind and oodloads. Table F-7 summarizes the lateral loads and their values. Maximum lateral wind loads oc-

cur when the wind is perpendicular to the roo ridge. Note that lateral wind loads act on theoverall structure (i.e., oundation), whereas ood loads act on the individual piers. Table F-8gives the load combinations or sliding ailure. Once the number o piers is defned, the hy-drodynamic orces on these piers are to be added to load combination 4, and the oundationdesign will have to be checked to make sure it can resist the added hydrodynamic loads.

TableF-5.VerticalLoadValues(continued)



F-13


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TableF-7.LateralLoadValues

LoadType


W Maximum lateral wind loads occur or windsto the roo ridgeW = W

RH+ W

H=

[lateral roo pressures][roo height] + [wall pressures][wall height]

W = [4 ps (-11.8 ps)] (4.7 t)[(15.7 ps + 9.8 ps)(10 t)]

W = 329.3 lbs per linear t o manuactured home length

Fa

Hydrodynamic load per pier

Fa

= [hydrodynamic orce][pier length]

Fa

= 32.8 lbs per pier

Assume total o 9 piers x 2 rows or 1st iteration

Fa

= (32.8 lbs per pier)(9 piers per row)(2 rows)

Fa = 590.4 lbs / 60 t = 9.84 lbs per linear t o manuactured home length

TableF-8.SlidingLoadCombinations

LoadCombinations

5 W + 1.5Fa

329.3 lbs + 1.5(9.84 lbs) = 344 lbs per linear t o manuactured home length

Note: The vertical gravity loads are not considered to be conservative. Thus, the rictional re-

sistance o the ootings under the piers has been neglected. This component may be used inborderline situations at the discretion o the engineer.

OverturningFailureMode

Overturning ailure results rom loads that act on the whole structure and pivot about the bot-tom o the leeward pier. Dead, live, wind, and snow loads can all inuence the overturningmoment. Table F-9 summarizes the moments that aect overturning due to wind in this case.Table F-10 assesses the moment load combinations. Only the portions o the roo and oorlive loads that are over the part that cantilevers out past the leeward pier will contribute to theoverturning. Since this is the worst overturning case or each, only these conditions will be cal-

culated.



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TableF-9.MomentLoadValues

MomentType

Totalmomentaboutthebottomotheleewardoundationsupport(positivemomentiscounterclockwise)

D D = [dead load per square oot][home width][moment arm]D = [20 ps][(16 t)(4 t)]

D = +1,280 t-lbs per linear t o manuactured home length

L L1

= [live load per square oot][home width][moment arm]

L1

= [40 ps][(16 t)(4 t)]

L1

= 2,560 t-lbs per linear t o manuactured home length

L2

= [live load per square oot][cantilever width][moment arm]

L2

= [40ps][4 t][-1 t]

L2

= -160 t lbs per linear t o manuactured home length

Lr

Lr= [roo live load per square oot][roo width][moment arm]

Lr = [17 ps][4 t][1 t]L

r= -68 t-lbs per linear t o manuactured home length

W WIND PERPENDICULAR TO THE ROOF RIDGE

WWRV

= [vertical component roo wind pressures][roo width][moment arm]

WWRV

(-21.6 ps)(1 t)(12.5 t) + (-5.9 ps)(8 t)(8 t)

WWRV

= -648 t-lbs per linear t o manuactured home length

WLRV

= [vertical component roo wind pressures][roo width][moment arm]

WLRV

(-11.8 ps)(9 t)(-0.5 t)

WLRV

= +53 t-lbs per linear t o manuactured home length

WRH = [horizontal component roo wind pressures][roo height][moment arm]

WRH

= [4 ps (-11.8 ps)](-4.67 t)](15.3 t)

WRH

= -1,129 t-lbs per linear t o manuactured home length

WW+L

= [windward wall pressure + leeward wall pressure][homes height rom ground to rooeave][moment arm]

WW+L

= [15.7 ps + 9.8 ps](10 t)(-8 t)

WW+L

= -2,040 t-lbs per linear t o manuactured home length

Fa

Hydrodynamic load on piers

Fa

= [horizontal component][moment arm]

= (9.84 pl)(-3 t/2) = -15 t-lbs per linear oot o manuactured home length

S SB

= [balanced snow pressure][horizontal projected roo area][moment arm]

SB

= [20 ps][18 t][-4 t]

SB

= 1,440 t-lbs per linear t o manuactured home length



F-15


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TableF-10.OverturningLoadCombinations

MomentLoadCombinations(positivemomentiscounterclockwise)

6 D + 0.75W + 0.75L + 0.75Lr+ 1.5F

a(Partial live loading to produce max OT)

(1,280 t-lbs) + (0.75)(-648 t-lbs + 53 t-lbs 1,129 t-lbs 2,040 t-lbs) + (0.75)(-160 t-lbs)+(0.75)(-68 t-lbs) + (1.5)(-15 t-lbs) = -1,737 t-lbs per linear t o manuactured home length

D + 0.75W + 0.75L + 0.75S + 1.5Fa

(Full live and snow to produce max downward reaction)

(1,280 t-lbs) + (0.75)(-648 t-lbs + 53 t-lbs 1,129 t-lbs 2,040 t-lbs) + (0.75)(-2,560 t-lbs) +(0.75)(1,440 t-lbs) + (1.5)(-15 t-lbs) = 1,435 t-lbs per linear t o manuactured home length

7 0.6D + W + 1.5Fa

(0.6)(1,280 t-lbs) + (-648 t-lbs + 53 t-lbs 1,129 t-lbs 2,040 t-lbs) + (1.5)(-15 t-lbs)= -3,019 t-lbs per linear t o manuactured home length

Table F-11 summarizes the load combinations that govern or each o the three ailure modes.The maximum roo vertical and lateral load cases are assumed to act simultaneously as a con-servative simplifcation.

TableF-11.ASDLoadCombinationsorExampleProblem(loadsareinpoundsperlinearoot)

FailureModes

LoadCombinations

4 5 7

Uplit 1,070 lbs n/a1 15 lbs

Sliding n/a1

313 lbs n/a1

Overturning n/a1 n/a1 -979 t-lbs

1 Load combination does not govern.

Load combinations 1-3 do not govern. Load combination 6 does not comply with HUD 24 CFR 3280.

Step3: SelectFoundationTypeandMaterials

The example statement specifed a CMU pier and ground anchor oundation type. Since theood velocity is 2 ps, CMU piers must have surace bonded mortar that meets ASTM C887-79a

(2001) and ASTM C946-91 (2001) and maintain bonding between CMUs.

Step4: DetermineForcesatConnectionsandonFoundation

Components

CMU piers transer the compressive loads rom the manuactured home into ootings andthen into the ground. The masonry piers are not considered to provide any lateral or uplit



17/20


resistance. The governing load combination or downward orces is the vertical ailure mode(load combination 4), which produces a total downward orce rom the manuactured homeequal to

(downward orce)(length o manuactured home)

Thereore, (1,070 lbs)(60 t) = 64,200 lbs

This downward orce governs the number o ootings and, thereore, piers needed to transerthe downward load into the ground.

Following the braced masonry pre-engineered oundation design or ood velocities over 2 psspecifcation given in Chapter 10 o the use o a dry-stack 16-inch by 8-inch block pier with aminimum o an 1/4-inch thick surace bonded mortar and a 24-inch square, 10-inch deep oot-ing, calculate the number o ootings needed to adequately transer the downward loads to theground.

Required ooting area =

Consult the geotechnical engineer or the ultimate soil bearing capacity value. An approximatemethod to calculate the ultimate soil bearing capacity is to multiply the allowable soil bearingcapacity by a saety actor. The maximum pressure given in the NFPA 5000 Soil ClassifcationTable can also be used as the ultimate soil bearing capacity.

Required ooting area = = 64 ft2

Individual ooting area = (24 in x 24 in) = 4 t2

Number o ootings =

= = 16 ootings/piers

Thereore, provide 8 piers per side o the manuactured home

Pier spacing =

= = 8.6 t

The maximum spacing o the piers is set to 8 eet to provide eective oodborne debris protec-tion. To protect against oodborne debris, it is assumed that 1 pier will be lost due to oodbornedebris.

Minimum number o piers = +1= 8.5 piers, say 9 piers per side (i.e., 8 spaces at 7.5

eet). Thereore, the home will be supported by a total o 18 piers (9 piers on each side)

spaced at 7.5 eet.

comprehensive loadallowable soil bearing capacity

64,200 lbs

1,000 psf

1 ft2

144 in2( )total required footing areaindividual footing area

64 ft2

4 ft2

manufactured home length

(number piers per side-1)

60 ft

8-1

60 ft2

8 ft



F-17


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Lateral wind loads are resisted by the strapping and ground anchors. The fnal number o piersequals the initial guess; thereore, the lateral load on the piers does not have to be updated.

Calculate the number o anchors needed to resist sliding ailure.

The recommended design stiness o the anchors in Table 7-5 in this guide is given or 5-ootanchors installed at 45 degrees and axially loaded is 1,200 lb/in (Figure 7-4). The horizontalcomponent o the ground anchors strength is equal to

(1,200 lbs/in)(cos 45) = 848 lb/in

The manuactured home industry gives an allowable lateral manuactured home movement o3 inches. So the total lateral strength o a ground anchor is (3 in)(848 lbs) = 2,544 lbs.

Number o ground anchors needed =

Number o ground anchors needed = = 8 anchors per side

Calculate ground anchor spacing = = 8.5 t

The anchor strapping should attach into a wall stud; thereore, anchor spacing must be adjustedto 16-inch increments.

Both uplit and overturning ailure modes are resisted by the vertical strength o ground an-chors. The uplit orces will be resisted by all the ground anchors and the overturning moment

will be resisted only by the windward ground anchors.

For the worst uplit o the vertical ailure mode, load combination 7 (reer to Table F-6) governs.However, the maximum net uplit is 16 pl downward, which means that overturning will governthe uplit requirements.

For the overturning ailure mode, load combination 7 (reer to Table F-10) governs or windperpendicular to the roo ridge. Overturning moment is only resisted by the windward anchors.Thereore, the total vertical load each anchor will have to resist is

= 612 lbs per anchor

The vertical component o the anchor stiness equals

(1,200 lbs/in)(cos 45) = 848 lbs/in

The manuactured home industry gives an allowable vertical movement o 2 inches. This resultsin a vertical strength per anchor equal to

(2 in)(848 lbs/in) = 1,697 lbs per anchor

60 ft(8-1)

(979 ft-lbs)(60 ft)12 ft

8 anchors

(overturning moment)(length of home)moment arm

number of anchors per side=

lateral loadanchors lateral capacity

(313 lb)(60 ft)

2.544 lbs



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This is more than the strength needed by each anchor to resist the overturning moment.

The anchor strapping should attach into a wall stud and, thereore, anchor spacing must be ad-justed to 16-inch increments. Place anchors at each end o the home and space at 72 inches.

For the overturning case, the connection o the straps to the stud and the ground anchor em-bedment is based on MWFRS pressures. However, although it would likely not govern, to bethorough, the uplit only condition using C&C pressures should be checked or these two an-chorages (straps to studs and anchors in ground).

Step5: SpeciyConnectionsandFramingMethodsAlongwith

ComponentDimensionstoSatisyLoadConditions

The CMU pier and ground anchor oundation will consist o 16 dry-stack, 16-inch by 8-inch

block piers with a minimum o a 1/4-inch thick surace bonded mortar and 24-inch square,10-inch deep ootings. Ground anchors will be placed at 45-degree angles at each end o themanuactured home and spaced at 72 inches.

Step6: NoteAllDesignAssumptionsandDetailsonDrawings

Reer to pre-engineering oundation design drawings contained in Appendix H and specifca-tions presented in Chapter 10 herein as to how to adequately document assumptions and detaildrawings.



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wind example claculations fema excerpt

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