foundation design sample

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ENGINEERING PRINCIPLES AND PRACTICES or Retroftting Flood-Prone Residential Structures

E N G I N E E R I N G P R I N C I P L E S A N D P R A C T I C E S

APPENDIX C

Sample Design CalculationsTis appendix presents design examples o the retrotting techniques or elevation, dry oodproong, wetoodproong, and construction o a oodwall in a residential setting. Examples C1 through C5 are a set o examples that illustrate the elevation o a single-story home with a crawlspace. Example C6 demonstrateshow to size a sump pump or dry oodproong. Example C7 shows how to calculate both the net buoyancy orce exerted on a liquid propane tank in a wet oodproong scenario and the volume o concrete neededto oset the buoyancy orce. Te nal example, Example C8, demonstrates how to design a cantileveredoodwall to protect a residence subject to 3 eet o ooding. Please note that Examples C6 through C8 donot use the same example home presented in Examples C1 through C5, but are instead standalone examples.

Te analyses and design solutions presented in this appendix apply to the example problems only. A licenseddesign proessional should be engaged or actual projects involving the residential ood retrotting techniquespresented herein.


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C-2ENGINEERING PRINCIPLES AND PRACTICES or Retroftting Flood-Prone Residential Structures

APPENDIX C SAMPLE DESIGN CALCULATIONS

Elevation

Given or Obtained rom the Field Investigation:

Te owner o a single-story crawlspace home intends to elevate the structure to eliminate a repetitive ooding hazard. Her desire is to raise the structure one ull story (8 eet) and use the lower level or parking andstorage. She contracted with a local engineer to perorm the design. Te engineer’s investigation revealed theollowing inormation about the existing structure:

crawlspace home has 8-inch CMU ungrouted oundation (no vertical reinorcement);

the rst-oor elevation is 2 eet above the surrounding grade (which is level);

the property is located in a FEMA-designated oodplain (Zone AE) and the base ood elevation (BFE)

is 4 eet above ground level;

oodwater velocities during a base ood will average 6 eet per second;

oodwater ows parallel to the side elevation and impact the ront elevation;

oodwater debris hazard exists and is characterized as normal;

the structure is classied as a pre-Flood Insurance Rate Map (pre-FIRM) structure with no existing ood openings; and

local regulations require an additional 1 oot o reeboard above the 100-year ood elevation.




SAMPLE DESIGN CALCULATIONS APPENDIX C

Additional Inormation on Existing House:

wood-rame house 30 eet x 60 eet; and

gable roo with 4:12 slope.

Per International Residential Code (IRC) and American Society o Civil Engineers (ASCE) 7:

90 miles per hour Basic Wind Speed (3-second gust);

at open terrain surrounding house;

wind Exposure Category C, enclosed building;

seismic importance actor I e = 1.0:, S DS = 0.2; and

ground snow load o 20 lb/t2.

Extended oundation walls are proposed to be constructed o 8-inch-thick concrete masonry units. Teexisting ooting is 2 eet wide by 1 oot thick concrete reinorced with (3) #4 rebar continuously. op o ooting is 18 inches below grade; soil type is sti residual silts. New slab-on-grade will be 3½ to 4 inchesthick and set at grade.

Interior walls o the living area (elevated) are composed o 2x4 studs at 16 inches on center (o.c.) with plasteron each side. Exterior walls have 2x4 studs at 16 inches o.c., plaster on the inside, and sheathing and woodsiding on the exterior walls are insulated with berglass insulation.

First oor raming consists o 2x12s at 16 inches o.c. supported by the exterior long walls and a center support.Floor coverings are hardwood (oak) with a ¾-inch plywood suboor. Tere are 10 inches o insulationbetween the joists. A gypsum ceiling is planned or the proposed lower area.

Roo raming consists o pre-engineered wood trusses at 16 inches o.c. Te top chord consists o 2x6s andthe web and bottom chord consist o 2x4s. Te roo is berglass shingles with elt on ½-inch plywood. Teceiling is ½-inch plaster with ½-inch plywood backup. Tere are 16 inches o berglass insulation above theceiling.





EXAMPLE C1. CALCULATE VERTICAL LOADS

Given:

Per original Sample Calculation Statement:

• Property is located in a FEMA-designated oodplain (Zone AE) and is subject to a 100-yearood 4 t in depth above ground level;

• Local regulations require an additional 1 t o reeboard above the 100-year ood elevation;and

• Field investigation inormation and additional inormation on the site

Find:

1. Design ood elevation (DFE)

2. Floodproong design depth

3. otal vertical ood load due to buoyancy

4. Vertical loads on the house (excluding buoyancy orce)

5. otal structure dead weight

6. Is the total structural dead weight sufcient to prevent otation o the house rom the buoyancy orce during a ood event?

Solution or #1: o nd the DFE, use Equation 4-2:

5 t

NOE: Te DFE calculated includes reeboard.

Solution or #2: o nd the oodproong design depth over which ood orces will be considered,use Equation 4-3:

5 t

Solution or #3: Te buoyancy orce can be ound as ollows:

• Te calculation o buoyancy orces and comparison with structure weight is a criticaldetermination o this problem. While buoyancy o the rst oor is not an issue (since it is

elevated 5 t above the DFE), buoyancy o the entire structure (slab, oundation walls, andsuperstructure) must be checked i dry oodproong is being considered or the lower level. I buoyancy orces control, or i the slab and walls cannot withstand the hydrostatic and otherdesign ood orces, dry oodproong o the lower level is not applicable. Using Equation 4-6:

561.6 kips





EXAMPLE C1. CALCULATE VERTICAL LOADS (continued)

Solution or #4: Te vertical loads can be determined as ollows:

Calculate Structure Weight by Level

• abulate Dead Loads by Floor (based on ASCE 7-10, able C3-1)

Roo: 2x6 op Chord and 2x4 Web and BottomShingles – Asphalt – 1 layer: 2.0 lb/t2 Felt: 0.7 lb/t2

Plywood – 32/16–1/2 in.: 1.5 lb/t2 russes @ 16 in. o.c.: 5.0 lb/t2

First Floor Ceiling:Insulation – 16 in. o berglass: 8.0 lb/t2

1/2 in. plywood: 1.5 lb/t2

1/2 in. plaster and lath: 10.0 lb/t2 Misc., heating, electrical, cabinets: 2.0 lb/t2

First Floor:Oak Floor: 4.0 lb/t2

Suboor – ¾ in. plywood: 3.0 lb/t2

Joists (2x12): 4.0 lb/t2 Insulation – 10 in. berglass: 5.0 lb/t2

Misc., piping, electrical: 3.0 lb/t2

Gypsum ceiling – 1/2 in.: 2.5 lb/t2

Walls:Interior – wood stud, plaster each side: 20 lb/t2 Exterior – 2x4 @ 16 in. o.c., plaster insulation, wood siding: 18 lb/t2 Lower Level – 8 in. masonry, reinorcement at 48 in. o.c.: 46 lb/t2

• Now to determine the total weights by level

Roo: Using the roo overhang o 2 t

Surace Area =

Projected Area =

Shingles:Felt:Plywood:russ: 2,095 t2 (5lb/t2) = 10,475 lbs





EXAMPLE C1. CALCULATE VERTICAL LOADS (continued)

Gable end walls:

22,449 lbs or roo weight

First Floor Ceiling: Area =Insulation:Plywood:Plaster:Miscellaneous:

38,700 lbs or the rst oor ceiling

Walls:Exterior:

Interior:25,520 lbs or the walls

Calculating the subtotal or the roo, rst foor ceiling, and walls

First Floor Including Lower Level: Each o the components has the ollowing area: Area =

Floors:Oak Floor:

Suboor: Joists:Insulation:Miscellaneous:

34,200 lbs or the oors

Ceiling:= 4,500 lbs or the ceiling

Walls:Exterior:Interior:

Lower level above proposed dry oodprooed slab:

Weight o water per square oot o masonry wall:

Lower level below proposed dry oodprooed slab:





EXAMPLE C1. CALCULATE VERTICAL LOADS (concluded)

101,228 lbs or the walls

Footing: 31,536 lbs or ooting

Slab:89,100 lbs or the slab

Calculating the subtotal or the foor, ceiling, walls, ooting, and slab:

Solution or #5: Te total dead load o the structure can be ound by adding the two abovesubtotals:

347,233 lbs

Solution or #6: o determine i the dead load (structure weight) rom the house is sufcient toprevent overturning rom the buoyancy orce, compare the buoyancy orce to the structure weight:

347,233 lbs ≤ 561,600 lbs347 kips ≤ 562 kips N.G. (No Good)

Tereore, the structural weight is not enough to prevent oatation o house during designooding events. Additionally, in the case that the dead weight o the elevated structure could resist the buoyancy orces, the new slab would have to be designed to transer buoyancy loads to

exterior walls without cracking.

NOE: Buoyancy orces control the building (i dry oodprooed) during a ooding event unless structural measures, such as oor anchors or additional slab mass, or non-structural measures such as allowing the lower level to ood, are utilized to oset/equalize the buoyancy orces.

In our example, since buoyancy controls and the magnitude o the project represent a substantial improvement,the homeowner is required to allow the lower level to ood by incorporating vent openings in the oundationwall. While this action will equalize hydrostatic pressures on the oundation walls, hydrodynamic and ood-borne debris impact orces will still apply.

Section 5E.1.2.1 o FEMA echnical Bulletin 1-08 provides the ollowing guidance on ood openings: “A

minimum o two openings shall be provided on dierent sides o each enclosed area, having a total net area o not less than 1 square inch or every square oot o enclosed area subject to ooding.” Te enclosed area o the example building is 1,800 square eet (60 eet x 30 eet); thereore, a minimum o two ood openings with a minimum combined area o 1,800 square inches shall be installed 12 inches or less above grade.





EXAMPLE C2. CALCULATE LATERAL LOADS

Given:


• Floodwater velocities in the area o the house average 6 t/sec

• Floodwater ows parallel to ront elevation and impact side elevation

• Floodwater debris hazard exists and is characterized as normal

• Flood openings will be installed in the oundation walls

• V = basic wind speed = 90 mph

• Exposure Category C

• I = importance actor = 1.00 or residential construction

• S DS = 3.0 sec

Additional inormation or this example per Equation 4-13, able 4-6 and able 4-7:

• W = weight o debris = 1,000 lb

• C D = depth coefcient = 0.75

• C B = blockage coefcient = 0.6

• C str = building structure coefcient = 0.8

Find:

1. Lateral hydrostatic orce

2. Lateral hydrodynamic orces

3. Lateral debris impact orces

4. Lateral orces or wind perpendicular and parallel to the main ridge

5. Seismic orces

Solution or #1: o calculate lateral hydrostatic orces rom 5 t o water moving at 6 t/sec:

First, calculate hydrostatic orce rom 5 t o reestanding water using Equation 4-4.

780 lb/t acting at 1.67 t above ground surace





EXAMPLE C2. CALCULATE LATERAL LOADS (continued)

In the “additional inormation” provided in the description o the house, the top o ooting is 24 in.

below ground surace and the soil type is sti residual clay (S = 82 lb/t3

per able 4-3).

Calculate submerged soil and water orces per Equation 4-5.

lb/t acting at 0.75 t below ground surace

dh will be calculated in #2 to yield the orce due to owing water, using the equivalent hydrostaticmethod.

Finally, compute total lateral hydrostatic orce using Equation 4-4, ater removing the dh

term(which is being used in #2).

22 lb/t

Solution or #2: Since V < 10 t/sec, use Eq. 4-7 to convert hydrodynamic orce to equivalenthydrostatic orce

Determine drag coefcient C d by calculating and using able 4-5

0.70 t

Convert the equivalent head to equivalent hydrostatic orce.

From Equation 4-8:






218.4 lb/t acting at 2.5 t above ground surace

Now to calculate total orce due to ow velocity on the building ace (upstream)

6,552 lbs = 6.55 kips

Solution or #3: o calculate lateral debris impact loads, use Equation 4-11:

Te parameters in Equation 4-11 are briey discussed in Chapter 4 o this publication and discussedin greater detail in Chapter 8 o FEMA P-55 (Fourth Edition), Coastal Construction Manual (FEMA, 2011).

F i = 2,160 lbs = 2.16 kips

NOE: Since vents are being used to equalize the hydrostatic pressure, the wall will be subject to a net load equal to the combined hydrodynamic and impact loads. Te ability o the new oundation wall towithstand these orces is presented toward the end o Example C5.

Solution or #4: o calculate lateral orces or wind perpendicular and parallel to the main ridge:

Since the house is being elevated, wind pressures will be increased on the building. Depending upon

the amount o elevation, additional bracing o the roo or walls may be necessary.Reerence: International Residential Code (IRC) and ASCE 7

Basic Wind Speed has been determined to be 90 mph rom the 2012 IRC and veried with the localbuilding ofcial. Te method to determine lateral orces is the Directional Procedure per ASCE7-10. Because ASCE 7-10 procedures and load combinations are used in this sample problem, theequivalent ASCE 7-10 wind speed must be used or the same geographical region. Tat wind speed is115 mph (3-second gust).

From ASCE 7-10 Equation 27.3-1, calculate velocity pressure calculated at height z above ground(q z ):

Compute q z at two dierent heights:

• At

• At mean roo height






From ASCE 7-10 able 27.3-1, compute velocity pressure exposure coefcients (K z ) at heights listed

above:• For ,

• For , by linear interpolation

Use topographic actor, since house is surrounded by at, open terrain

From ASCE 7-10 able 26.6-1, use directionality actor, or buildings

24.5 lb/t2

26.0 lb/t2

Calculate Design Wind Pressures on Building Main Wind Force Resisting System, MWFRS ( p)using Equation 27.4-1 (ASCE 7-10)

From ASCE 7-10 Equation 27.3-1 use , the velocity pressure computed or windward walls calculated at wall height z 1 or z 2

2 above ground (lb/t )

From ASCE 7-10 Equation 27.3-1, use or all other walls and roo suraces(lb/t2)

From ASCE 7-10 Section 26.9.1, use gust eect actor G = 0.85 or rigid structures

From ASCE 7-10 Figure 27.4-1, compute external pressure coefcients (C p) or the ollowing scenarios:

• Perpendicular to the ridge, where:

a. For windward walls,

b. For leeward walls, ,

c. For windward roo, , , and -0.12 by linear interpolation

d. For leeward roo, , , by linearinterpolation






• Parallel to the ridge, where:

a. For windward walls,

b. For leeward walls, ,

c. For windward roo, , , or 0 to 20 t rom

windward edge, or 20 to 40 t rom windward edge

From Equation 27.3-1, use velocity pressure calculated at mean roo height, q h = q i = 26.0 lb/t2

From able 26.11-1, use internal pressure coefcients or enclosed buildings, GC pi = ± 0.18

MWFRS – Wind Perpendicular to Ridge

Walls: Windward:

12.0 lb/t 2 (inward)

21.3 lb/t 2 (inward)Leeward:

-15.7 lb/t 2 (outward)


Roo: Windward:





MWFRS – Wind Parallel to Ridge

Walls: Windward:






12.0 lb/t 2 (inward)




Roo: Windward, or distance rom leading edge:0 to 20 t


-15.2 lb/t 2 (outward)20 to 40 t

-15.7 lb/t 2 (outward)40 to 60 t







Solution or #5: o calculate the quantity and distribution o lateral seismic orces (base shear and

vertical distribution):Since the house is being elevated, the potential or seismic loading/overturning design loads will beincreased on the home. Depending upon the amount o elevation, additional bracing o the roo or walls may be necessary.

Reerence: ASCE 7-10 per Equivalent Lateral Force (ELF) procedurePer equation 12.8-1 in ASCE 7-10 and equation 12.8-2 in ASCE 7-10:

Where:

W = eective seismic weight

C S = seismic response coefcient

I = occupancy importance actor = 1 per ASCE 7-10 Section 11.5.1 and able 1.5-2

R = response modication actor

S DS = design spectral response acceleration parameter in the short period range

Now, per ASCE 7-10 able 12.14-1 and equations 12.8-11 and 12.8-12 in ASCE 7-10:R = 2 or ordinary reinorced masonry shear wall oundation

R = 6.54 or ramed walls with plywood

0.1

Efective seismic weights:

at roo/upper ceiling level






Recalculate without the eects o buoyancy on ooting weight and lower oundation walls

or conservative seismic condition.

Lower level masonry wall grouted at 48 in. o.c.: lbs orthe walls

Footing:

54,000 lbs or ooting

at rst oorlevel (without buoyancy orce on components below the slab)

= total eective seismic weight

38,093lbs

= height at roo/upper ceiling

= height at rst oor

1,560,042 t-lb

2,942,600 t-lb

0.35

0.65

13,332 lbs

24,760 lbs





EXAMPLE C2. CALCULATE LATERAL LOADS (concluded)

NOE: For summary, see below:

Lateral Forces Perpendicular to Long Direction

Seismic

Level

Height(t)h x

Level Weight(kips)

w x

Lateral Force(kips)

F x

Level Shear(kips) ∑ F x

2 18 83.97 1,511 13.3 13.3

1 10 303.03 3,030 24.8 38.1

TOTAL 4,541 38.1

Wind

Level

Wind Pressure(lb/t2)

P x

Area(t2)a x

Lateral Force(kips)

H x


2 19.4 300 5.8 5.8

1 37.0 1,080 40.0 45.8





EXAMPLE C3. LOAD COMBINATIONS

Lateral Forces Parallel to Long Direction

Seismic

Level

Height(t)h x

Level Weight(kips)

w x

Lateral Force(kips)

F x


2 18 83.97 1,511 13.3 13.3

1 10 303.03 3,030 24.8 38.1

TOTAL 4,541 38.1

Wind

Level

Wind Pressure(lb/t2)

P x

Area(t2)

a x

Lateral Force(kips)

H x

Level Shear(kips)

∑ F x

2 32.6 75 2.4 2.4

1 32.6 540 17.6 20.0

Given:

Per original Sample Calculation Statement:• Solutions to above Example C1 and C2• Field investigation inormation and additional inormation on the site

Find:1. I the structure can resist sliding using the most appropriate load combination allowable stress

design (ASD)2. I the structure can resist overturning 3. I the structure can resist uplit and buoyancy orces

Solution or #1: Check i the structure can resist sliding as ollows:

First, the most appropriate load combination or sliding must be ound using ASCE 7-10 Section2.4.1 or ASD, the load combinations are:

1. D 2. D + L3. D + (Lr or S or R )

4. D + 0.75L + 0.75(Lr or S or R )5. D + (0.6W + 0.7E )6a. D + 0.75L + 0.75(0.6W ) + 0.75(Lr or S or R )6b. D + 0.75L + 0.75(0.7E )+ 0.75S 7. 0.6D + 0.6W 8. 0.6D + 0.7E





EXAMPLE C3. LOAD COMBINATIONS (continued)

When a structure is located in a ood zone, the ollowing load combinations should be considered in

addition to the basic combinations on Section 2.4.1 o ASCE 7-10:• In Zone V or A Coastal Zone – 1.5 F a should be added to load combination 5, 6, and 7 and E

should be set equal to zero in Nos. 5 and 6

• In non-Coastal A Zones – 0.75 F a should be added to load combination 5, 6, and 7 and E should be set equal to zero in Nos. 5 and 6

Each possible building ailure mode (sliding, overturning, uplit, buoyancy) must be investigatedusing the most restrictive load combination.

• Controlling condition or sliding: Te sum o the orces in the horizontal direction mustbe LESS than the sliding resistance provided by the soil in order or building to not slide.

By inspection, load combination 7 is most restrictive sliding condition because dead loadis reduced and ood and wind loads are included or the direction parallel to the ridge. By inspection, load combination 7 is also the most restrictive load combination equation or thedirection perpendicular to the ridge, even though no ood load is involved in this direction.

• Weight o the structure used in sliding and overturning calculations must incorporate the useo ood vents. Te buoyancy orce or the slab and 5 t o wall must be subtracted rom thetotal weight o the structure:

Parallel to ridge direction: Te coefcient o soil riction is 0.3. Using load combination 7:

or the sliding orce

or the sliding resistance providing by the soilOK in parallel to ridge direction

Perpendicular to ridge direction: Te coefcient o soil riction is 0.3. Using load combination 7:where F a = 0 (ow and ood-borne debris directions are parallel to this

direction) and using the coefcient o soil riction 0.3

or the sliding orceor the sliding resistance providing by the soil

OK in perpendicular to ridge direction

Tereore, the structure can resist sliding.

Solution or #2: Check i the structure can resist overturning as ollows:

• Controlling condition or overturning: Te moments caused by the lateral orces must be lessthan the counter-moment provided by the building weight.





EXAMPLE C3. LOAD COMBINATIONS (concluded)

Te most likely overturning direction is along the short dimension o the building caused by seismic

orces. Te ASD actor or the seismic orce is 0.7 and the ASD actor or the dead weight o thebuilding is 0.6 per load combination 8.or building to resist overturning

Summing the moments about the pivot point:

OK, building weight keeps building rom overturning.

Tereore, the structure can resist overturning.

Solution or #3: Check i the structure can resist uplit and buoyancy orces as ollows:

• Controlling condition or uplit and buoyancy: Te dead weight o the building must resistthe uplit orces imposed by wind and buoyancy on the building. Te sum o the orces in thevertical direction must be greater than the buoyancy orce in order or the building to stay inthe ground. load combination 7 is most conservative. Buoyancy actored as ood load.

to resist uplit and buoyancy.

Per Example C1, the elevated structure is incapable o resisting buoyancy orces and ood openings are required.

NOE: For this example analysis, the existing structural components were assumed to be ade quate or the

loading conditions. However, the designer should check the existing truss-to-wall-connections, plywood roo diaphragm, upper level walls, and oor diaphragm or their ability to resist increased loadings.





EXAMPLE C4. CHECK OF EXISTING FOUNDATION DESIGN

Given:


• Solutions to above Examples C1, C2, and C3

• Field investigation inormation and additional inormation on the site

Find:

1. Adequacy o existing oundation design

Solution or #1: Te adequacy o the existing oundation design can be determined as ollows:

• All o the loads on the oundation must be calculated and tabulated using ASCE 7-10.Calculate loads along 60-oot exterior load bearing walls or highest ooting loading condition.

Snow load = S =

First ood live load= L =

Dead loads:

Roo:Shingles:Felt:Plywood:

russ:

Ceiling:Insulation:Plywood:Plaster:Miscellaneous: Wall (exterior):

Wall (interior):

First Floor:

Flooring:





EXAMPLE C4. CHECK OF EXISTING FOUNDATION DESIGN (concluded)

Suboor:

Joists:

Insulation:

Miscellaneous:

Ceiling:

Wall (exterior):

Footing:

Wall (interior):

New lower level wall:

NOE: A 20 lb/t 2

partition load is applied here; this approach is conservative due to the amount o interior walls in this building.

From our eld investigation it was determined that an allowable bearing pressure o 2,000 lb/t2 wasacceptable.

• Te total load on the oundation must be determined

otal load on oundation:

Most restrictive load combination is Eq.6a.

• Te total load on the oundation must be compared with the capacity o the existing oundation to determine adequacy

Te existing oundation is 2 t wide; thus, the bearing pressure or total loads is

Allowable, OK.

Tereore the existing oundation design is adequate.





EXAMPLE C5. NEW FOUNDATION WALL DESIGN

Given:


• Solutions to above Examples C1, C2, C3, and C4

• Assume seismic load condition controls design

• Field investigation inormation and additional inormation on-site as noted in previoussections

• Assume #4 reinorcing bars at 48 in. o.c. in solid grouted cores

• G =0.4E where E = modulus o elasticity per American Concrete Institute (ACI) 530-08Section 1.8.2.2

• Assume compressive strength o masonry, ' m = 2,000 lb/in.2

• ype M or S mortar per ACI 530-08 Section 1.8.2.2 E m = 900 ' m = 900 (2,000)= 1.8 x 106 lb/in.2

• With #4 @ 48 in. o.c. the equivalent solid thickness is 4.6 in. = 0.38 t

• M = 120 kips-t rom Example C2 (maximum moment split in hal due to 2 walls)

• V = 19.2 kips rom Example C2 (maximum shear split in hal due to 2 walls)

Find:

1. Design the oundation wall and connection to the ooting

2. Design top o wall connection. (Checking anchor bolts or pullout rom masonry

Solution or #1: o design the oundation wall connection to the ooting:

New Wall Design:

• Minimum wall reinorcement is #4 @ 48 in. As = 0.20/48 in.

• Next determine V m (shear strength provided by masonry) per ACI 530-08 Equation 3-22

• Moment is calculated rom Seismic Lateral Forces. Reer to the gure below or indicated

orces with associated moment arms. M

Vd A f ’ P mn= −

+4.0 1.75 0.25V m





Where P is equal to the average weight o the 30 t long ramed wall with gable end wall above. So,

2,924 kips- in.= 4.0 -1.75 11

19.1 kips)(360in.V m 20 in . )2 22,000lb/ in. +0.25 (5,670 lbs)

=

• Finding the resulting shear orce per ACI-530-08 Equation 3-23:

which is acceptable

Te resulting shear orce per ACI 530-08 Equation 2-25 is:

( )

• Determining the nominal shear strength and comparing to the in-plane shear capacity

Nominal Shear Strength

63.9 kips > 53.2 kips OK #4 @ 48 in. o.c. in-plane shear

(

( )

hor 9 h/r ≤ 99 h/r = 120/2.16 = 55.6 < 99r

OK

where: h =

≤ 9

EXAMPLE C5. NEW FOUNDATION WALL DESIGN (continued)

• Investigate long (60 t) wall or out-o-plane bending because axial load is also supported by this wall. Neglect impact load in this analysis.

reat wall as simple -beam 4 t wide and calculate axial pressure per load determined in ExampleC4. Section Properties ACI 580-08 Section 1.9 and






Axial Compressive Strength o Masonry per ACI-530-08 Equation 2-15. Maximum momentdetermined rom Simple Beam Load Combinations #1 and #5 rom Part 4 o Steel Construction Manual, Volume 1 (2nd Edition) using wind and hydrodynamic ood pressures ound in ExampleC2 and shown in the gure below:

Area (in.2) A A 2 I (I x )i (in.4)

(1.25)(48)=60 0.625 37.5 23.44 7.1 31.25

(8)(6.75)=54 3.81 205.74 783.87 205.03 988.90

Sum 114 243.24 1,020.15

Calculate section modulus or -beam section o wall






• Check combined bending and axial load using interaction equation per ACI 530-08 Equation2-13. Maximum Shear determined rom End Reactions per Simple Beam Load Combinations#1 and #5 rom Part 4 o Steel Construction Manual, Volume 1 (Second Edition) using the wind and hydrodynamic ood pressures ound in Example C2. Masonry shear stress ( v ) per ACI 530-08 Equation 2-23 and Equation 2-24:

OK

R Bottom

= + =( )(( . )( ) ( . )( )

( )( )[( )( )] lbs4

21 3 10

2

22 4 5

2 102 5 ) 762

OK • Reaction R op must be resisted by attachment o oor diaphragm to wall by bolts and wood sill

plate.

• Investigate pure bending o wall with maximum moment as determined at beginning o Section C5 and Section Properties per ACI 580-08 Section 1.9.

where M = 1,695 t-lbs or 20,340 in.-lbs, b = 48 in., d = 3.81 in.

and where and






So, the required As or M = 1,695 t-lb and 48 in. wide -beam is:

N.G., -1 -#4 BAR=0.20 in.2

Solution: Increase reinorcement to 5/8-in. diameter with

0.31 in.2 (1 -#5) OK

Tereore, use 1-#5 @ 48 in. o.c. to resist bending out-o-plane.

• Bolt Design

Shear at top o wall =

Check shear capacity o 0.5 in. A307 A.B. in Southern Pine #2 sill plate. Use National DesignSpecication or Wood (2005), able 11E. ry sill plate 2 in. x 6 in. so side member = 1-1/2 in., mainmember = 3 in., shear perpendicular to grain = 410 lbs. Modiy value with adjustment actors orconnections.

So

And,

Bolt spacing = , use 72 in. max. as required by IRC 2012 Section R403.1.6

Edge distance o bolt 4D =

Check bending o sill plate with bolts @ 6 t o.c.

M =

F b or Southern Pine #2 = 1,500 lb/in.2, b < F b OK

Check pullout o A.B. Uplit Force =

For anchors spaced 6 t o.c., uplit orce = 402 lb/bolt

Tereore, an 8 in. CMU wall with #5 @ 48 in. o.c. centered on grouted cell - 2,000 lb/in.2 masonry ( ’ m ) is acceptable.






Solution or #2: o design the top o wall connection, one must check anchor bolts or pullout

rom masonry as ollows:

ry ½ in. A307 anchor bolts @ 6 t o.c.

Uplit on bolt =402 lb/bolt

ry ½ in. A307 anchor bolt, area o bolt, Ab = 0.2 in.2

Edge distance, l be =

Embedment, l b = 7 in. Required minimum embedment or reinorced masonry IRC 2012 SectionR403.1.6

Reerence: ACI 530-08 Section 2.1.4 and ACI-530-08 Equation 2-10

Allowable load in tension:

B a = min ( )

Where:

Ab = Area o Anchor Bolt

' m = Compressive Strength o Masonry y = Yield Strength o Anchor Bolt

For this anchor bolt pattern,B a =OK

Allowable load in shear,

OK

Per ACI 530-08 Section 2.1.4.3.3, the combined ratio is

≤ 1





EXAMPLE C5. NEW FOUNDATION WALL DESIGN (concluded)

Tereore, a ½ in. A307 anchor bolt with an edge distance, l be = 3.56 in. and embedment,

l b = 15 in. is adequate.

Sample Bearing Wall Detail





Dry Floodproofng

EXAMPLE C6. SAMPLE CALCULATION FOR SUMP PUMP

Sump Pump Sizing Calculations

Given:

• Te drain line is 1.5 in. diameter steel pipe, which is 30 t long and includes one elbow, onegate valve, and one check valve (based on hydraulic engineering handbooks)

• Q = the ow rate o the sump pump = 20 gal/min

• V ps = Velocity o ow through the pipe in t/sec

• A pipe = Area o the discharge pipe cross-section in t

• Z = the elevation dierence between the bottom o the sump and the point o discharge,in t = 10 t

• h –pipe = head loss due to pipe riction (or this example use 2.92 t per 100 t)

• h –fttings = head loss through ttings, in t

• h –trans = head loss though transitions, in t

• K b

(elbow) = 0.63

• K b (gate valve) = 0.15

• K e (pipe entrance) = 0.5

• K o (sudden enlargement/outlet) = 1.0

• g = weight o gravity, 32.2 t/sec

• H = total head, in t

Find:1. Velocity converted rom gal/min to t/sec

2. Calculate the losses rom pipe ttings and pipe transitions3. Calculate the losses over the length o pipe4. Calculate the total dynamic head or the sump discharge

Solution or #1: o convert the velocity rom gal/min to t/sec:





EXAMPLE C6. SAMPLE CALCULATION FOR SUMP PUMP (concluded)

3.63 t/sec

Solution or #2: Te losses rom pipe ttings and pipes transitions can be calculated as ollows:

0.25 t

Solution or #3: o calculate the losses over the length o the pipe:

0.876 t

Solution or #4: o calculate the total dynamic head:

H = 11.772 t

NOE: Tereore select a pump capable o pumping 20 gal/min at 11.77 t o total head.





Wet Floodproofng

EXAMPLE C7. NET BUOYANCY FORCE ON A LIQUID PROPANE TANK

Given:

• F b = the net buoyancy orce o the tank, in lbs

• V t = the volume o the tank in gallons = 250 gal

• 0.134 is a actor to convert gal to t3

• = the specic weight o ood water surrounding the tank (generally 62.4 lb/t3 or resh

water and 64.1 lb/t3 or salt water)

• FS = the actor o saety to be applied to the computation, typically 1.3 or tanks

• W t = the weight o the tank = 670 lbs (empty – worst case scenario)

• V c = the volume o concrete required, in t3

• S c = the eective weight o concrete, typically 150 lb/t3

Find:

1. Te net buoyancy orce o the tank in pounds

2. Te volume o concrete required to oset the buoyant orce

Solution or #1: Te net buoyancy orce o the liquid propane tank can be ound as ollows:

= 2,048 lb

Solution or #2: Te volume o concrete required to oset the buoyant orce can be determined asollows:

Where:V c = 23.4 t 3





EXAMPLE C7. NET BUOYANCY FORCE ON A LIQUID PROPANE TANK (concluded)

NOE: o resist this buoyant orce, a slab o concrete with a volume, V c , is usually strapped to the tank to

resist the buoyant load.

Floodwalls and Levees

EXAMPLE C8. FLOODWALL DESIGN

Given:Site soil conditions based on clean dense sand:

• soil (unit weight o soil) = 120 lbs/t3

• S a (allowable soil bearing capacity) = 2,000 lbs/t2

• S (equivalent uid pressure o soil) = 78 lbs/t3

• C (coefcient o riction) = 0.47

• k p (passive soil pressure coefcient) = 3.69

• C s (cohesion) = 0

Local ood conditions:

• Fresh water ( w = 62.4 lbs/t3)

• Area o potential normal impact loading, C B = 0.2 (moderate upstream blocking), C Str = 0.8

• Expected ood velocity, V (t/sec)

Dimensional inormation

• H = 7.0 t

• D t = 4.0 t

• D = D h = 5.0 t

• t tg = 1.0 t

• B = 5.0 t

• Ah = 2.5 t





EXAMPLE C8. FLOODWALL DESIGN (continued)

• C = 1.5 t

• t wall = 1.0 t

Find:

1. Design a cantilever oodwall to protect a residence subject to 3 t o ooding.

Solution or #1: o design a cantilever oodwall to protect a residence subject to 3 t o ooding,ollow the eight step process or Floodwall Design in Chapter 5F:

Step 1: Assume wall height and ooting depth (see Figure 5F-15 in Chapter 5F)

• H = 7.0 t

• D t = 4.0 t

• D = D h = 5.0 t

• t tg = 1.0 t

Step 2: Determine dimensions (see Figure 5F-15 in Chapter 5F)

• B = 5.0 t

• Ah = 2.5 t

• C = 1.5 t

•

t wall = 1.0 t

Wall and ooting to be reinorced concrete having unit weight (S g ) o 150 lbs/t3.

Step 3: Calculate orces.Determine Lateral Forces:

Equation 4-4: =

Equation 4-5: =

Equation 4-7: =

Equation 4-8: =

=

Equation 5F-9: =






Since F n acts only at a single point, we will not include loading into the uniorm lateral oodwall

loading. Once the oodwall is sized, we will evaluate the wall perpendicular to ow to determineability to resist the impact loading. I necessary this wall will be redesigned to resist impact loads. Tisprocess will avoid overdesigning o the entire oodwall.

Equation 5F-12: =

Determine Vertical Forces:

Equation 5F-1: =

1Equation 5F-1: f D buoy2 = =γ w tB ( .0 5)(62. )4 4( ) ( )5 624lbs/ lf 2

Equation 5F-1: = 1,092 + =624 1,716 lbs/ lf

Equation 5F-2: =

Equation 5F-3: =

Equation 5F-4: =

Equation 5F-5: =

Equation 5F-6: =

Equation 5F-7: =

Equation 5F-8: =

Step 4: Check sliding.

Equation 5F-10: =

Equation 5F-11: =

Equation 5F-13: =

Equation 5F-14: = = 1.6 > 1.5 (recommended) OK or sliding

Step 5: Check overturning.






Equation 5F-15:

= 10,276 t-lbs/l

Equation 5F-16:

= 12,683 t-lbs/l

Equation 5F-17: = = 1.2 < 1.5 (recommended) No Good

ry increasing the ooting size to overcome the overturning moment. Assume B = 7.0 t; Ah = 4.0 t;and C = 2.0 t. Tis requires revision o Steps 3 and 4 or which the results are shown below. sta , di , dh , F sta , F p, w wall will not change. Recompute vertical orces.

Equation 5F-1: =

Equation 5F-1: =

Equation 5F-1: =

Equation 5F-2: =

Equation 5F-3: =

Equation 5F-4: =

Equation 5F-5: =

Equation 5F-6: =

Equation 5F-7: =

Equation 5F-8: =

Recheck Sliding

Equation 5F-10: =

Equation 5F-11: =






Equation 5F-13: =

Equation 5F-14: = = 1.79 > 1.5 (recommended) OK or sliding

Recheck Overturning

Equation 5F-15:

= 14,748 t-lbs/l

Equation 5F-16:

= 21,674 t-lbs/l

Equation 5F-17: = = 1.5 = 1.5 (recommended) OK

Step 6: Determine eccentricity.

Equation 5F-18: = 0.92 < OK

Step 7: Check soil pressure.

Equation 5F-19:

= 94.3 lbs/t2

= 673.3 lbs/t2 < 2,000 lbs/t2 OK

Step 8: Select reinorcing steel.

For steel in the vertical oodwall section:

Equation 5F-20 (note): = 2,583.7 t-lbs/l

Equation 5F-20: = 0.17 in.2/l




EXAMPLE C8. FLOODWALL DESIGN (concluded)

For top steel in the ooting section:

Equation 5F-20 (note): = 4,838.4 t-lbs/l


For bottom steel in the ooting section:

Equation 5F-20 (note): = 564.7 lb/in.2

Equation 5F-20 (note): = 1,274.2 t-lb/l


From American Concrete Institute Reinorced Concrete Design Handbook able 9a: Use #4 bars on14-inch centers in the vertical oodwall section, use #6 bars on 14-inch centers or the top steelin the ooting section, and use #3 bars on 14-inch centers or the bottom steel in the ooting

section or B = 7.0 t; Ah = 4.0 t; and C = 2.0 t.

NOE: Other ACI documents have similar inormation.

foundation design sample

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