CH. 6 SOILS & FOUNDATIONS

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1 CH. 6 SOILS & FOUNDATIONS SOIL PROPERTIES Classified into four groups - Sands & gravels - Clays - Silts - Organics Subsurface Exploration Core borings: undisturbed samples of soil - Recovered bore samples are lab tested - Strength test for bearing capacity - Resistance to lateral pressure - Slope stability - Number of borings taken is determined by - Size of building - Suspected subsurface geological conditions - Local code - Typically a minimum of four bores unless wide variation is detected - Boring log: shows material, depth it was encountered, standard designation, moisture content, density - Standard penetration test (SPT): measures density of granular soils and consistency of some clays - 2 dia sampler is driven into the bottom of a bore hole by a 140lb hammer falling 30 - Number of blows (N) to drive cylinder 12 is recorded Test pits: trench dug for visual inspection of strata and direct collection of undisturbed samples - B/c open, practical limit on depth of 10 Location of each boring or test pit is shown on a plot plan and given and numbered corresponding to the boring log Tests requested by the architect and paid for by owner & are NOT part of contract documents Soil Types and Bearing Capacities Bearing capacities are given by code

2 Water In Soil Reduces load carrying capacity If present under one area is greater than other areas differential settlement may occur Hydrostatic pressure: - Difficulties - Additional loads - Waterproofing more difficult b/c pressure forces water into cracks or imperfections - Ways to minimize: - Slope ground away from building minimum of 1/4 per foot - Water from roofs and decks drained away from building with gutters and drainpipes, etc - Below grade: use drain tile or open web matting - Relieve pressure against floor slabs - Layer of gravel below slab - Use water proof membrane & drain tiles if pressure is significant Soil Treatment Increase bearing capacity or decrease settlement - Drainage: prevent hydrostatic pressure - Fill: if soil is unsuitable for building, undesirable material is removed and new fill brought in - Fill must be compacted - Controlled compaction requires moisture to lubricate the particles - There is an optimum relationship between fills density & optimum moisture content - Proctor test: method of determining optimum moisture content of fill where fill samples are tested in the lab to determine a standard for compaction - Specifications call for fill to be compacted between 90% to 100% of optimum Proctor density - Higher values for heavy loaded structures - Moisture content within 2% - 4% of optimum moisture content at time of compaction to be specified - Fill placed in 8 12 lifts with each being compacted before next placement - Compaction: existing soil compacted to provide base for construction - Densification: compaction of existing - Vibration - Dropping heavy weights - Pounding piles into ground & filling voids with sand - Surcharging: preloading of ground with fill material to cause consolidation and settlement of underlying soil. Once settlement has taken place, fill is removed - Suitable for large areas - Time & cost often preclude method of soil improvement Other Considerations Frost: foundation must be place below the frost line Expansive soils: clays (bentonite) expand when wet and shrink when dry - Foundations must be isolated from them - Piles or caisson foundation piers with concrete grade beam FOUNDATION SYSTEMS Spread Footings Spread footings: spread load over large area Wall footing: most common & under a continuous foundation wall that supports a bearing wall Independent column footing: supports one column Combined footing: two or more columns in situations where columns are spread to closely for separate ones or one is close to prop. line that a symmetrically loaded footing could not be poured - Strap footing or cantilever footing: variation if two columns are far apart Matt or raft footing: bearing is low or where loads are heavy in relation to soil pressures - One large footing as a two-way slab and supports columns above

3 Pile foundations When soil near grade is unsuitable for spread footings Transmit loads thru unstable soil to a more secure bearing with end bearing or side friction Piles: placed in groups or in a line under a bearing wall with building load transferred to them with pile caps - Pile is embedded from 4 6 into the pile cap - Either driven or drilled - Driven: timber steel or precast concrete - Drilled: called caissons also called piers - When two or more piles used to support one column, the centroid of the pile group is designed to coincide with the center of gravity of the column load If soil is soft a metal lining is used & removed as concrete is poured or left in place If soil pressure is not sufficient for drilled pier of normal dimensions, bottom is belled it increase area for bearing System used where expansive soils or clay are encountered near the surface - Grade beams are used & poured on carton forms that support the concrete during pouring but do not transmit any upward pressure from the soil Designing Footings Three primary factors to investigate - Unit loading: so allowable bearing pressure of soil is not exceeded and differential settlement in various parts of the structure is eliminated - Shear & bending: Two kinds of shear failure - Punching or two way shear: when column or pall load punches thru the footing - Flexural shear or diagonal tension: footing fail in bending when the lower surface cracks under flexural loading Simple spread footing acts like inverted beam with upward soil pressure as continuous load that is resisted by downward column load causes bending in upward direction which induces compression near the top of the footing and tension near the bottom - If tension great enough, tension reinforcement added near the bottom of the footing Area of spread footing determined by dividing total wall or column load + own weight + any soil on top of footing allowable soil bearing pressure - Then footing is designed for shear, moment and other loads with factored loads as required by American Concrete Institute AIC - These are safety factors to insure footing is of sufficient size and will resist all loads For foundations, the following formula is used to combine loads U = 1.4D + 1.7L U = required strength loads D = actual dead load L = actual live load When designing wall footings there are two critical sections to investigate - Face of the wall where bending moment is greatest - Distance (d) from the face of the wall footing where flexural shear is of most concern - However: critical two-way shear sections for columns footings is distance d/2 from the face of the wall

4 - Effective depth (d): distance from top of the footing to the centroid of the reinforcing steel - Concrete below the steel does not contribute any structural properties - d for masonry and concrete foundation walls is a little different & is measured from the face of wall For lightly loaded walls where footing width is not too great the bending is not as critical as shear that must be resisted by thickness of footing - Generally not economical to provide tension reinforcement in wall footings so width and thickness are designed to resist the wall load and shear forces using only strength of unreinforced concrete - However: longitudinal reinforcing is included (parallel to wall) for temperature reinforcing and to help span any intermittent weaker soil conditions For heavy loads or weak soil conditions, the footing width may become enough to require tension reinforcement - However: maximum allowable flexural shear governs design depth of wall footings Individual column footings are subject to two way action like flat slabs near columns as well as one way shear both types of shear must be calculated and the depth of the footing must be designed to resist - When both calculated, the greater shear of the two is used for design - Two locations where shear must be calculated illustrated above - For one way shear at distance (d) from the face of the column, the factored soil design pressure is calculated over the rectangular area indicated as abcd - Two way shear: soil design pressure is calculate over the area outside the square efgh indicated - Bottom reinforcing in both directions is usually required to resist the moment forces at the face of the column Example 6.1: find the required depth width and transverse reinforcing for the footing shown. The bottom of the footing is 5 below grade and carries a load per linear foot of 14,000lbf dead load, including the wall weight, and 7,000lbf live load on a 12 wide foundation wall. The concrete strength is 3000psi and the steel yield point is 60,000psi. soil tests have shown the allowable soil bearing pressure to be 3500psf

5 - Step one: consider a 1-0 long section of wall and footing - To find the footing width, divide the total load plus allowance for the weight of the footing and allowance for the soil on top of the footing by the allowable soil bearing pressure. - Estimate the footing width as 7 and depth as 12 - With concrete weighing about 150lbf/ft3 a 1-0 long section of footing weights 1050lbf - Soil weights about 100psff so soil weight is 4 times 100 or 400psf or 1200lbm for the 3 section on either side of the foundation wall - So the width of the footing is: b w = ( )/3500 b w = 6.99ft - Step two: to begin the footing design, ACI code requires the design soil pressure to be calculated based on factored loads according to: U = 1.4D + 1.7L U = (1.4)(14000) + (1.7)(7000) U = 31500lbf - Note: this excludes the weight of the footing and soil above the footing b/c the do not contribute to producing moment or shear in the footing - The design soil pressure is then equal to the factored design load divided by the area q s = U/A q s = 31,500/7 q s = 4500psf - Step three: assuming a footing depth of 12 check the flexural shear at the critical section since flexural shear almost always governs footing design. This section is at a distance (d) (effective depth of the footing) from the face of the wall - The distance is from the top of the footing to the centroid of reinforcing steel - ACI code requires a 3 clear dimension from steel to the bottom of a footing cast against earth - Use this plus an allowance (guess) of 1/2 for one half the diameter of the reinforcing steel BUT will probably be less than #8 bars but this gives an easy number of 3 1/2 to work with - Distance (d) is then = 8.5 or.708ft - Shear at this point is the distance to the end of the footing design soil pressure - Remember: this is a 1-0 long section of wall so units are in feet and pounds per inch V = (x d)q s V = ( )(4500) V = 10310plf - The ACI code limits one way shear on plain or reinforced concrete sections to a maximum of: V c = Max allowable shear = structural factor f c = compressive strength of concrete b w = width of footing d = effective depth V c = 2( f c )b w d V c = (.85)(2) 3000(12)(8.5) V c = 9498 lbf

6 - Actual shear of 10,310 lbf is more than the allowable of 9498lbf so the section needs to be revised - Try a 14 deep footing with d = = 10.5 Step four: find the moment at the face of the wall - The leg of the footing acts as an inverted cantilevered beam so moment is: M u = q s l 2 /s M u = [(4500)(3) 2 ]/2 M u = 20,250fl-lbf/ft Step five: Find the area of the steel required A s = M u /[ f y (d (a/2)] a = depth of rectangular stress block determined by a = A s f y /.85f c b - Since steel area is not known assume a value for a. (see ch 11) - Try 1 to begin A s = M u /[ f y (d (a/2)] A s = [(20250)(12)]/[(.90)(60000)(10.5 1/2)] A s =.45in 2 /ft of footing - There are several combinations of bar size and spacing that will satisfy this requirement - #5 8 spacing gives a steel area of.46in 2 /ft - so use this. Refer to table 11.1 for various bar spacing combinations Step six: find the development length required for the steel - This is the maximum length required to develop sufficient bond between steel and concrete - It is measured from face of wall to end of steel as shown below & is found by the formula: l d (fx /25 f c )d b - The equation assumes adequate concrete cover and bar spacing under normal conditions with = = = The area of a #5 bar is 31in 2 and its diameter is.625in l d = [(60000)(1)(1)(1)/ ].625 l d = 27.39in - The minimum length is 12 so governs - Since actual length is 36 3 or 33 there is sufficient nength of steel Step seven: find the longitudinal temperature reinforcement required - ACI requires at least.0018 times the area of the section to be steel so: A s = (.0018)(12)(14) A s =.3024in 2 - No 4 7 oc provide.34in 2 /ft (see table 11.1)

7 RETAINING WALLS Types of Retaining Walls Cantilever wall: resists forces by the weight of the structure & weight of the soil on the heel of base slab - Often with a key projecting from bottom to increase resistance to sliding - Most common type & constructed of reinforced concrete - Toe omitted if at property line or adjacent obstruction - Economically limited to tall Counterfort wall: similar to cantilever but a counterfort placed at distances equal or a bit larger than 1/2 the height - Counterforts: reinforced concrete webs act as diagonal braces Gravity walls: resist forces by own weight only - Non-reinforced concrete Forces on Retaining Walls Simplest case: force on wall results entirely from pressure of earth retained acting in a horizontal direction - Earth pressure increases proportionally with depth from the surface ranging from top to max at lowest depth Earth any point is given by: P = C o Wh C o = Coefficient of earth pressure - Coefficient of earth pressure depends on soil type, method of backfilling and compaction - Range: -.4 for uncompacted soils line sand and gravel - 1 for cohesive compacted soils - Formula is simplified so that C o is eliminated and weight of soil is considered to be equivalent to fluid weighing 30lb/ft 3 P = 30h - In diagram below pressure acts in a triangular form & the total pressure against the wall can be assumed to be acting thru centroid of the triangle or 1/3 the distance from base

8 - To find this value, find the pressure at the base and multiply by one half the height (the area of a triangle P = 1/2C o Wh 2 - Or if using the 30lb/ft 3 assumption P = 15h2 Example 6.2: what is the total pressure ona retaining wall 9.5 height - Assuming the spoil has an equivalent fluid weight of 30psf per foot of height, the total force acts at a point 9.5/3 or 3.17ft above the base P = (15)(9.5) 2 = 1354plf - Additional forces may act on retaining walls. Earth may slope upward from the top of the wall resulting in the total force acting thru the centroid of the pressure triangle but parallel to the slope - Additional loads: surcharges may result from driveways or other forces imposed on soil net to the wall - If ground becomes wet, additional pressure as a result, must be added to the soil pressure Design Considerations Failure: - As a whole by overturning or sliding - To prevent overturning or sliding, the resisting moment or forces that resist sliding are generally considered sufficient if a safety factor of Example: total dead load of wall + weight of earth backfill acting on footing of a cantilevered retaining wall should be at least 1.5x the overturning moment caused by earth pressure - To prevent sliding, friction between footing and surrounding soil and earth pressure in front off toe must be 1.5x the pressures tending to cause the wall to slide - Individual components fail such as arm or stem breaks due to excessive movement - To prevent individual components failure thickness, width and reinforcing of wall must be designed to resist the moment and shear forces induced