handouts for 2 hr foundations for metai building systems

Alexander Newman, P.E.

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Foundations for Metal Building Systems

ASCE Web Seminar

Presented by Alexander Newman

Copyright © 2007--2012 Alexander Newman

All rights reserved.

Reproduction of this material without a written permission

of the copyright holder is a violation of the U.S. law

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Introduction

Some Reference Sources

A. Newman, “Foundation and Anchor Design Guide for Metal

Building Systems,” McGraw-Hill, 2013

A. Newman, “Metal Building Systems: Design and

Specifications,” 2nd ed. (McGraw-Hill, 2004)

2-day ASCE Seminar: Design and Strengthening of Shallow

Foundations for Conventional and Pre-engineered Buildings

ASCE Webinars Design of Anchor Bolts and Design of

Concrete Embedments for Shear and Tension


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Agenda Introduction to metal building systems

The main issues

Tie rods

Hairpins

Moment-resisting foundations

Trench footings, mats

Slab with haunch

Foundations for Quonset Hut-type buildings

Q&A

Introduction

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Introduction to Metal Building Systems

Two Main Types of MBS Frame-and-purlin types: Concentrated reactions on

foundations

Single-slope rigid frames

Multiple-span rigid frames

Tapered beam

Trusses

Quonset Hut-Style:

Distributed reactions

Arrival and Departure Airfield Control Facility, Pope AFB, NC (USACE)


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Introduction to Metal Building Systems

Frame-and-Purlin System

RIGID FRAME COLUMN

GIRT

FRAME WIDTH

CLEAR SPAN

RIGID FRAME

BRACING

ROOF SYSTEM

ROOF PURLIN

EAVE STRUT

ENDWALL FRAME

BAY SPACING

EAVE HEIGHT

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Main Issues

What Makes MBS Foundations Different From Conventional? Light weight => large net uplift

Large lateral reactions at columns


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Main Issues

Do Manufacturers Design Foundations?

MBMA Metal Building Systems Manual, Common Industry

Practices, Para. 3.2.2: Mfr is responsible only for providing

loads to Builder, not foundation or A.B. design.

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Main Issues

Foundations Designed Before the Building

Some say, “Foundation design is provided for bid purposes

only; the actual sizes to be determined by contractor using

similar details.” But… this introduces another party?

USACE TI 809-30, Metal Building Systems, Appendix


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Main Issues

Reliability and Redundancy vs. Cost

Establishing Size of Column Piers

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Main Issues

Estimating reactions:

1. Mfr’s tables (check code!) –

see App. D of MBS book

2. Specialized software

3. General analysis software

4. Use reference books

(e.g., Kleinlogel, Rigid Frame

Formulas, 1964)

5. Asking a mfr

With reaction uncertainty, conservative approach makes sense


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Main Issues

What Loads Are We Designing For?

MBMA Metal Building Systems Manual App. A3 refers to

sources suggesting that only 70% of total wind load on frame

needs to be considered in foundation design (but 100% for

anchor bolts)

IBC 2009 ASD Basic Load Combinations include:

0.6D + W + H (H = lateral earth pressure)

0.6D + 0.7E + H

IBC 2012 ASD Basic Load Combinations include:

0.6D + 0.6W + H

0.6(D + F) + 0.7E + H

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Main Issues

Wind and Seismic Loads + Dead Using ASD Basic Alternative Load Combinations

IBC 2009 (D + L + w W)

IBC 2012 (D + L + 0.6w W),

must use only 2/3 “of the minimum dead load likely to be in

place”

For seismic, ASCE 7 Sec. 12.13.4 allows foundation design for

75% of MOT at the base, but IBC-09 and -12 do not allow it with

Basic Alternative Load Combinations.


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Main Issues

1/3 Stress Increase with Wind & Seismic Loads? IBC-03 Para. 1616.1 allows soil stresses from wind and

seismic increased by 1/3 when alternate load combinations

used

IBC-06 does not?…

Para. 1605.3.1.1: No stress increase for basic load combs.

Para. 1605.3.2: For alternate, stress increase is allowed

“where permitted by the material chapter of this code or

the referenced standard.” [None given.] Cannot use

reduction of OT allowed by Sec. 12.13.4 ASCE 7-05, when

checking OT, sliding, bearing.

=> 1/3 stress increase is probably not allowed in IBC-06

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Main Issues

1/3 Stress Increase, Cont’d IBC-09 and IBC-12 allow it

Section 1806.1 specifically permits a one-third stress

increase for alternative basic combinations using ASD that

include wind or seismic loads.

Applies to vertical foundation pressures and lateral

bearing pressures in IBC Table 1806.2, Presumptive Load-

Bearing Values.

How about other allowable values determined by geotech.

investigation?


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Main Issues

Avoid Fixed-Base Columns

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Main Issues

Resisting Uplift

Ballast by soil

F.S. 1.5 for transient loads (Fig. 17 of NAVFAC 7.2),

or -- implied through load combinations?

- Count soil shear resistance? (Neglected in NAVFAC 7.2)

- Reduce in areas subject to flooding

NAVFAC 7.2, Figure 17


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Main Issues

Design Example: Proportion Footing for Uplift

Problem: Size the interior foundation of multi-span rigid

frame, using basic IBC-12 load combinations for:

Atrib = 60’ x 25’= 1500 ft2

Loads: D = 3 psf,

S (design roof snow) = 30 psf

0.6W uplift = 14 psf

Min. depth = 3’ below floor, Fp = 4000 psf

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Main Issues

Design Example, Cont’d

Solution:

Compute loads

D = 3 x 1500 = 4500 lbs = 4.5 kips

S = 30 x 1500 = 45,000 = 45 kips

0.6W = –14 x 1500 = –21,000 lbs = –21 kips

Total downward (D + S) = 4.5 + 45 = 49.5 kips

Total uplift (0.6D + 0.6W) = 0.6 x 4.5 – 21 = –18.3 kips


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Main Issues


Foundation weight, assuming ave. wt of soil, SOG, found. =

0.130 kips/ft3 x 3’ = 0.390 ksf

Net available soil pressure is 4.0 – 0.39 = 3.61 ksf

Arq for downward load is 49.5/3.61 = 13.71 ft2

=> Could use 3.7’ by 3.7’ footing for downward load

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Main Issues

Design Example, Cont’dFind min. size for uplift from 0.6Dmin, found + 0.6W = 0

Dmin, found = 18.3/0.6 = 30.5 kips, or 30.5/0.130 = 234.62 ft3 of

ave. weight of “ballast.”

If depth of footing = 3’, need min. square footing size of

(234.62/3)1/2 = 8.84’

Could use 8.0’ by 8.0’ footing, with depth = 234.62/(8)2 = 3.66’

Use 8.0’ x 8.0’footing, 3’- 8” deep

Size controlled by uplift!


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Main Issues

Resisting Horizontal Reactions

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Tie Rods

Tie Rods

Various designs


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Tie Rods

Common but Questionable Tie Rod Detail

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Tie Rods

Problems with Tie Rods Need mechanical connections and corrosion protection

ACI 318-08 and -11 Sec. 12.15.6: “Splices in tension tie

members shall be made with a full mechanical or full

welded splice…and splices in adjacent bars shall be

staggered at least 30 in.”

Sag under own wt. Turnbuckle tough to fit in sheath…

A problem to use at pits, trenches

Elongate under load… Use Ft <<0.6Fy


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Tie Rods Design Example: Tie Rod Elongation

Given: L = 120 ft, P = 36 kips, Fy = 60 ksi

If Ft = 36 ksi, Arq = 36/36 = 1.0 in2 or one #9 bar.

This can damage frame, finishes. (2) #9 bars would halve that.

If Ft = 24 ksi, Arq = 36/24 = 1.5 in2 or (2) #8 bars (A = 1.58 in2)

=> Can use (2) #9 bars to reduce elongation

AE

PLrod

sideeaatmovementorrod "9.0"79.1000,2900.1

1212036

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Tie Rods Tie Rods in Grade Beams

Determine bar area by controlling elongation


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Tie Rods

PT Tie Rods

Need concrete, or PT and

wind stresses are additive

(esp. @ base pl)

Cantilever-beam pier action

with passive pressure at

base

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Tie Rods

PT Tie Rods, Cont’d

Tendon placed in center of grade beam

Beam and pier placed together

12-16” below floor

Corrosion-resistant coating

A 36k compressive force in 16”x16” => fc = 0.141 ksi

To reduce upward pressure on beam, place on soft material

or cardboard


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Hairpins Hairpins: How They Work

Some take Ft in rebar = 24 ksi and in WWF, 20 ksi

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Hairpins

Hairpins, Cont’d

What if the slab is cut?

May be OK for smaller buildings

without joints or plumbing


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Hairpins Design Example: Hairpins in SOG

Find length of hairpins Lhair as follows:

Projected length of tensile crack ~ Lhair x 1.41 x 2 = 2.82 Lhair

For a given Awire can find tension capacity per ft of width, Tall

E.g., for WWF 6x6-W1.4x1.4 (old 66-1010) Awire = 0.028 in2/ft

Tall = 0.028 x 20 (ksi) x 2.82 Lhair = 1.58 x Lhair (kip)

If T = 10 kip, Lhair = 10/1.58 = 6.33 ft

For #4 bars @18” o.c., Abar = 0.13 in2/ft and

Tall = 0.13 x 24 (ksi) x 2.82 Lhair = 8.80 x Lhair (kip)

If T = 10 kip, Lhair = 10/8.8 = 1.13 ft (but use min 5 ft)

Add to that development length of hairpin bar or hook

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Hairpins Hairpin Design Example, Cont’d

Find hairpin bar size for T = 10 kip:

Ahair = 10(0.707)/24 ksi = 0.29 in2

Use # 5 hooked hairpins ~ 7 ft long


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Moment-Resisting Footings


Resist OT and sliding

Different proportions than in retaining walls

(+) and (-)’s

F.S. per CRSI

Design

Handbook:

Sliding 1.5

Overturning 2.0

At service loads

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Forces Acting on Moment-Resisting Foundations Can use Ka @ rotations > 0.1% H


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Moment-Resisting Footings Active and Passive Pressure

Rankine formulas use fluid analogy:

Pa = Ka (h) Ka = tan2 (45o - /2)

Pp = Kp (h) Kp = tan2 (45o + /2)

strengthshearfor

envelopefailure

s'Mohrfromfound

,frictionernalint

ofangle

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Active and Passive Pressure, Cont’d

Common values for drainable fill (sand, gravel free of fines):

For mixed grain sizes with fines, dense enough for low

permeability, all these increase. E.g.,

However, some codes provide design lateral loads directly…

00.333.0 pa KK

69.3271.035 pa KK


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Active and Passive Pressure, Cont’d

ASCE 7-10 Table 3.2-1 (ASCE 7-05 Table 3-1), Design Lateral

Soil Load gives lateral pressures directly

E.g, Well-graded and poorly graded clean gravel &

gravel/sand mixes (GW and GP); silty gravels: 35 psf/ft

But use 60 psf/ft for “relatively rigid walls, as when

braced by floors.” Basement walls extending < 8’ below

grade and supporting light floors are excepted.

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Soil Lateral Load per IBC

Backfill Material (per ASTM D 2487)

Unified Soil Classif’n

Design active pressure (psf/ ft of depth)

Design at-rest pressure (psf/ ft of depth)

Well- and poorly-graded clean gravels or sands; gravel-sand mixes

GW, GP, SW, SP

30 60

Silty sands, poorly graded sand-silt mixes SM 45 60

Clayey sands, poorly graded sand-clay mixes; mixture of inorganic silt and clay; inorganic clays of low to medium plasticity

SC,

ML-CL, CL

60 100

Notes:1. Table derived from IBC-12 Table 1610.1 (partial table is shown)2. Design loads are for optimum densities… add hydrostatic loading for saturated soil3. Can design basement wall extending < 8’ below grade and supporting flexible floor

systems for active pressure


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Soil Passive Resistance per IBCMaterial Lateral

bearing (psf/ft below natural grade)

Lateral sliding coefficient of friction

Lateral sliding resistance (psf)

Sandy gravel and/or gravel 200 0.35 --

Sand, silty sand, clayey sand, silty gravel, clayey gravel

150 0.25 --

Clay, sandy clay, silty clay, clayey silt, silt, sandy silt

100 -- 130 x contact area

Notes:1. Partial IBC-09, -12 Table 1806.2 is shown2. Total lateral sliding resistance = lateral bearing + lateral sliding3. Increases may be permitted if data submitted and approved, but max. sliding

resistance for clay, sandy clay, silty clay, clayey silt = ½ DL 4. Table values for lateral sliding resistance can be increased “by the tabular value

for each additional ft of depth to a max of 15 times the tabular value.”

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Soil Friction: Simplified Method

Fres = µ(W + Pv) F > F.S. x Ph

µ ~ 0.5 coefficient of friction, but perhaps not for slabs on

vapor barrier


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IBC Factors of Safety against OT and Sliding

IBC-09, -12: None for foundations, but Para. 1807.2.3

requires a factor of safety of 1.5 for both overturning and

sliding of retaining walls, without using the load

combinations of IBC Section 1605. Instead, 100% of the

nominal loads should be used (except 70% of seismic load);

consider some variable loads = 0.

Except use F.S. = 1.1 for EQ load

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Moment-Resisting Footings IBC Factors of Safety, Cont’d

SF against lateral sliding = (available soil resistance at

base)/(net lateral force applied to the retaining wall)

“Net” means passive pressure can be used?

CRSI Design Handbook: For retaining walls, F.S. against

sliding 1.5, overturning 2.0 at service loads


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Sliding Resistance With Shear Keys

General

method

Must be in

undisturbed

soil

NAVFAC DM-7.2

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Can Passive Pressure and Soil Friction be Combined? The problem

IBC-06 Para. 1804.3, Lateral Sliding Resistance:

The resistance of structural walls to lateral sliding shall be

calculated by combining the values derived from the

lateral bearing and the lateral sliding resistance shown in

Table 1804.2 unless data to substantiate the use of higher

values are submitted for approval.


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Combining Passive Pressure & Soil Friction, Cont’d IBC-09, IBC-12 Para. 1806.3.1, Combined Resistance:

The total resistance to lateral loads shall be permitted to be

determined by combining the values derived from the

lateral bearing pressure and the lateral sliding resistance

specified in Table 1806.2.

The previous para. prefaces this, “Where the presumptive

values of Table 1806.2 are used…”

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Design Procedure


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Moment-Resisting Foundations

MRF Design Procedure

Depends on Eccentricity

Rely on passive pressure for

sliding only

M = Fh x H + Mfix

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Moment-Resisting Foundations

The Pressure Wedge Method

When P is beyond kern


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Design Example: Moment-Resisting FoundationDesign the moment-resisting foundation for an industrial building framed with a metal building system. The primary framing consists of single-span rigid frames with pin-base columns. The frames have an eave height of 18 ft., span 80 ft., and are spaced 25 ft. on centers. The frost depth is 3.5 ft. Each column is supported by a 24-in square pedestal (pier), the top of which is 6 in. above the adjacent soil. A continuous foundation wall, of the same depth as the column footing, exists between the piers. The 6-in. slab on grade covers the interior of the building. The column vertical loads are applied at the center of the pier. The roof snow load is 30 psf. Use the frame reactions from the tables in the Appendix. Assume the following material properties:

Soil weight: 120 lbs/ft3

Concrete unit weight: 150 lbs/ft3

Allowable bearing pressure of soil: 3 ksf.

Concrete 28-day compressive strength: f’c = 4000 psi.

Assume soil is clean sand free of fines and use parameters from CRSI Design Handbook:

ϕ = 30o, Ka = 0.33, Kp = 3.00, µ = 0.55.

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Determine the design column reactionsThe following column reactions are used for an 80-ft wide frame: - Vertical: dead 4.8 kip, snow 30.9 kip- Horizontal: dead 2.9 kip, snow 21.8 kip -0.6xWind reactions on the right-side column, wind from right to left

(assume wind load = 0.6W for IBC load combinations) : Horizontal: 13.6 kip (inward)Vertical: 12.2 kip (uplift)

- 0.6Wind reactions on the left-side column, wind from right to left: Horizontal: 3.1 kip (outward); vertical: 8.4 kip (uplift)

Design the foundation for the right-side column for the following controlling load combinations:1. Dead + Snow load. Vertical: 4.8 + 30.9 = 35.7 kips (downward); horizontal: 2.9 + 21.8 = 24.7 kips (acting outward).2. Dead + 0.6Wind load from right. Vertical: 4.8 - 12.2 = - 7.4 kips (uplift); horizontal: 2.9 -13.6 = -10.7 kips (inward).


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Proportion the foundation Establish the foundation size to resist overturning, sliding, and uplift. The foundation size is determined by trial and error.

Case 1: Dead + Snow load P = 35.7 kips (downward), FH = 24.7 kip (outward).This case provides the largest horizontal force on the foundation. The force acts in the direction away from the building. Try a footing 9 ft. long, 4 ft. wide and 2 ft. thick, with 2 ft. by 2 ft. column pier.

W3

P

W1W2

Point of rotation

6”

1’-6”

2’-0”

4’-6”4’-6”

2’-0”3’-0” 4’-0”

1’-0”1’-0”

W4

Wo

e

FH

Al

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58ACI


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Example of a similar footing

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Trench Footings

Trench Footings

Need cohesive soils to support excavation


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Trench Footings Trench Footings, Cont’d

Some local codes require

forming

Best to form the top 6”

Uplift and horizontal

resistance similar to grade

beam (next)

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Mats

Mats


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Deep Foundations Deep Foundations

Pier and grade beam

Piles

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Slab with Haunch Slab with Haunch Common sizes: 24-30” deep, 24” wide

May be wider and deeper at columns

Not for areas of expansive soils or deep frost lines

May not be enough wt or passive pressure… Slab cracks!

“A crack will almost surely occur in the floor slab at the point where the

“grade beam” starts.” Metal Building Systems by Building Systems

Institute (an industry group), 1990.


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Forming Slab with a Haunch per USACE

USACE

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Slab with Haunch

Slab with Haunch: General Approach

Using weight of footing to counteract eccentricity of load

Works for light load, heavy footing, little lateral load


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Slab with Haunch

Slab with a Haunch in MBS For inward lateral load, weight of monolithically placed grade

beam is essential

Grade beam has to be reinforced to carry its weight

suspended

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Slab with Haunch

Slab with a Haunch in MBS, Cont’d

For outward lateral load, contribution of grade beam for

resisting OT is minimal… Need larger footing, rather than rely

on slab, or use ties in slab (next)

Authors of Metal Building Systems by BSI “do not recommend counting

on the slab shearing resistance for stability.”


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Slab with Haunch

Design Example: Slab with Haunch

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Foundations for Quonset Hut-Type Buildings

Photo: Capt. R. Vaira, USAF

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Foundation Reactions with or without Base

Moment


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Photo: Capt. R. Vaira, USAF

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Typical Foundation Types

Grade beam without footing

Wall foundation with footing

Downturned slab

Note: Do not place unbalanced fill until concrete is fully cured!


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Resisting Distributed Building Reactions

For vertical loads, start with minimum width and depth sized

as needed for uplift.

Use min. wall reinforcement to span over hard spots.

E.g., for 16” x 24” beam

Amin = 0.0025x16x24 = 0.96 in2 or 4 # 5 bars (2 top & 2 bot.)

Resist horizontal reactions by passive soil pressure, ties, or

rebars

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Design Example: Using Passive Pressure for Fh

Given: Soil is sand free of fines, density 115 #/cu. ft.

Fh = 500 #/ft

Solution:

Fh = 500 = ½(115)(h2)(3.00 – 0.33), from where

2 ft. of embedment seems enough…

00.3;33.0 pa KK

embedmentoffth .min8.1)33.000.3(115)2/1(

500


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…But Don’t Forget

Eccentricity of

Load

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Design Example: Resisting Horizontal Reactions

by Rebars or WWF in SOG

Given:

Fh = 1 kip/ft

f’c = 3000 psi,

fy = 60,000 psi,

Fp, allow = 4 ksf


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Using rebars (Ft = 24 ksi)

per ft of width

Can use #4 @18” (A = 0.133 in2)

Using WWF (Ft = 20 ksi)

Awire, rq = 1 kip/(20 ksi) = 0.05 in2/ft

Can use 6x6-W2.9xW2.9 (old style 66-66) with Awire = 0.058 in2/ft

But remember the caveats about slab cutting, joints,

trenches…

2042.024

)1(in

ksi

kipArq

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Q & A

[email protected]

handouts for 2 hr foundations for metai building systems

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