ENCE 4610 Foundation Analysis and Design. Combined Footings and Mat Foundations


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1 ENCE 4610 Foundation Analysis and Design Combined Footings and Mat Foundations
2 Mat Foundations A mat is continuous in two directions capable of supporting multiple columns, wall or floor loads. It has dimensions from 20 to 80 ft or more for houses and hundreds of feet for large structures such as multistory hospitals and some warehouses Ribbed mats, consisting of stiffening beams placed below a flat slab are useful in unstable soils such as expansive, collapsible or soft materials where differential movements can be significant (exceeding 0.5 inch). Topics for Mat Foundations o Floating Foundations o Coefficient of Subgrade Reaction o Loads and Moments on Combined Footings o Design of Mats This image cannot currently be displayed.
3 Conditions for Mat Foundations Structural loads require large area to spread the load Soil is erratic and prone to differential settlements Structural loads are erratic Unevenly distributed lateral loads Uplift loads are larger than spread footings can accommodate; weight of the mat is a factor here Mat foundations are easier to waterproof Example: Chase Tower, Houston, TX Mat foundation is 3 metres thick and bottomed at 19.2 m below street level This image cannot currently be displayed.
4 Floating Foundations Type of mat foundation that relies partially or entirely on the weight of the soil/water combination it displaces to support the structure above it A truly "floating" foundation exists where the weight of the soil removed is greater than or equal to the weight of the building that replaces it Although foundations can be made to float entirely, it many not be advisable due to heave or settlement due to changes in ground conditions Very useful for structures with hollow subterranean structures In this course, we will analyse floating foundations using the buoyancy method of analysis, i.e., comparing the weight of the structure to the weight of the soil displaced o We can do this either on a weight or pressure (weight/area) basis
5 This image cannot currently be displayed. Floating Foundations Basics of Buoyancy Method This image cannot currently be displayed. Buoyancy Method
6 Floating Foundation Example Given o Foundation as Shown o Foundation is 50 m long and 70 m wide o Sum of column and wall loads = 805 MN Find o Average bearing pressure of foundation o Increase in stress due to addition of foundation o Whether foundation will float This image cannot currently be displayed.
7 Floating Foundation Example Compute weight of soil displaced by foundation W s = (19)(50)(70)(8.7) = MN Compute displacement on pressure basis (same as total stress on base of foundation) p s = /3500 = MPa = kpa Compute change in pressure on foundation Δp = = kpa Foundation will float so long as bearing capacity or settlement analysis determines Compute weight of bottom of mat W f = (23.6)(50)(70)(1.8) = MN Compute total structure load on foundation W = = MN Compute area of foundation A = (50)(70) = 3500 m 2 Compute pressure of structure on foundation p = /3500 = MPa = kpa that kpa is an acceptable additional stress This image cannot currently be displayed.
8 Coefficient of Subgrade Reaction Nonrigid methods must take into account that both the soil and the foundation have deformation characteristics. These deformation characteristics can be either linear or nonlinear (especially in the case of the soils) The deformation characteristics of the soil are quantified in the coefficient of subgrade reaction, or subgrade modulus, which is similar to the modulus of elasticity for unidirectional deformation Definition of Coefficient of Subgrade Reaction k s = q δ k s = coefficient of subgrade reaction, units of force/length 3 (the units are the same as the unit weight, but not the significance!) q = bearing pressure δ = settlement
9 Determining the Coefficient of Subgrade Reaction Methods used to determine coefficient Use settlement techniques such as Terzaghi's consolidation theory, Schmertmann's or Hough s method, etc., and express the results in a k s value If using a pseudocoupled value, use values of k s in the centre of the mat which are half those along the perimeter This methodology has the potential of eliminating the problems described earlier while at the same time yielding values of k s which then can be used in a structural analysis of the mat with some degree of confidence Methods used to determine coefficient Plate load tests Test results must be adjusted between the shape of the loading plate and the actual shape of the foundation Adjustment must also be made for the size of the plate vs. the size of the foundation, and the influence of size on the depth of soil stress (see following slide) Attempts to make accurate adjustments have not been very successful to date Derived relationships between k s and E s Relationships developed are too limited in their application possibilities
10 Plate Load Test Common method of estimating the coefficient of subgrade reaction Load is applied to plate using reaction system Described in some detail in Murthy 13.2 Results in a loadsettlement curve of the plate
11 Coefficient of Subgrade Reaction Note nonlinear behaviour
12 Difficulties in Determining the Coefficient of Subgrade Reaction The position of the mat To model the soil accurately, ks needs to be larger near the edges of the mat and smaller near the centre Time With compressible (and especially cohesive compressible soils) mat settlement is a process which may take several years May be necessary to consider both short and long term cases Nonlinear nature of soil deformation makes unique value of k s nonexistent Width of the loaded area; wide mat will settle more than a narrow one because more soil is mobilised by a wide mat Depth of the loaded area below the ground surface Change in stress in the soil due to q is a smaller percentage of the initial stress at greater depths Shape of the loaded area: stresses beneath long, narrow loaded area is different from those below square loaded areas
13 Using Coefficient of Subgrade Reaction (Murthy 14.4) If results of plate load or other tests/methods are unavailable, use typical values of k 1 given in Table 14.1 Determine value of k s1 ( results from plate load test) o For sands, k s1 = k 1 for practical purposes o For clays, must correct for beam length of foundation Value of k s1 for clays: k k s s o Finite beams and foundations: 1 1 = = 2 k 3 2 k L + L + L L 0. 5 (SI (US Units, o Infinite beams and foundations: k s = 1 2 k 1 3 Units, L in meters) L in feet)
14 Using Coefficient of Subgrade Reaction (Murthy 14.4) Determine actual coefficient of subgrade reaction k s from k s1 o This basically upscales the results from the size of the plate load test to the actual size of the foundation o All of the limitations just discussed with this method need to be kept in mind when doing this Scale Factors k k s s o Sands = = k k s s 1 1 B + 2 B + 2 B o Clays 0.3 B (SI 3 k k = s s B Units, (US Units, 1 L in meters) L in feet)
15 Example of Using Coefficient of Subgrade Reaction Given Solution Structure to be supported on a 30 m wide by 50 m long mat foundation For medium sand, k 1 = 45 MN/m 3 For sand, k s1 = k 1 = 45 MN/m 3 Mat founded on medium sand Average bearing pressure is 120 kpa Find Settlement, using coefficient of subgrade reaction and assuming the following: Use the following equation to determine k s : k s = k sl B B Substituting B = 30 m, k s = 11.5 MN/m 3 2 Foundation is rigid relative to soil δ Substitute and solve for deflection: = q δ = q = 120 kpa k kn / m s k s 3 = 10.4 mm
16 Problems with the Coefficient of Subgrade Reaction Nonuniformity of the strata below the foundation Scaling the subgrade reaction to the actual size of the foundation With flexible foundations (and that includes mats and pavements) the coefficient of subgrade reaction is influenced by flexibility of foundation All of these factors make the solution of the example problem very preliminary
17 Coefficient of Subgrade Reaction Application of coefficient of subgrade reaction to larger mats Portions of the mat that experience more settlement produce more compression in the springs Sum of these springs must equal the applied structural loads plus the weight of the mat ΣP +W u D = qda = δk s f da
18 Structural Design of Combined Footings and Mats Structural design requires two analyses Strength Evaluate these requirements using factored loads and LRFD design methods Mat must have sufficient thickness T and reinforcement to safety resist these loads T should be large enough so that no shear reinforcement is required Serviceability Evaluate using unfactored loads for excessive deformation at places of concentrated loads, such as columns, soil nonuniformities, mat nonuniformities, etc. This is the equivalent of a differential settlement analysis Mat must be made thicker if this is a problem Example: Murthy, Ex. 14.4
19 Nonrigid Methods Nonrigid methods consider the deformation of the mat and their influence of bearing pressure distribution. These methods produce more accurate values of mat deformations and stresses These methods are more difficult to implement than rigid methods because of soilstructure interaction Winkler Methods Coupled Method PseudoCoupled Method MultipleParameter Method Finite Element Method
20 Winkler Methods The earliest use of these "springs" to represent the interaction between soil and foundation was done by Winkler in 1867; the model is thus referred to as the Winkler method The onedimensional representation of this is a "beam on elastic foundation," thus sometimes it is called the "beam on elastic foundation" method Mat foundations represent a twodimensional application of the Winkler method
21 Limitations of Winkler Method Soil springs do not act independently. Bearing pressure on one part of the mat influences both the "spring" under it and those surrounding it (due to lateral earth pressure) No single value of k s truly represents the interaction between the soil and the mat The independent spring problem is in reality the largest problem with the Winkler model Loadsettlement curves are not really linear; we must make a linear approximation to use the Winkler model Winkler model assumes that a uniformly loaded mat underlain by a perfectly uniform soil will uniformly settle into the soil. Actual data show that such a matsoil interaction will deflect in the centre more than the edges This is one reason why we use other methods (such as Schmertmann's or Hough s) to determine settlement
22 Coupled Method Ideally the coupled method, which uses additional springs as shown below, is more accurate than the Winkler method The problem with the coupled method comes in selecting the values of k s for the coupling springs
23 Multiple Parameter Method This method replaces the independentlyacting linear springs of the Winkler method with springs and other mechanical elements The additional elements define the coupling effects Method bypasses the guesswork involved in distributing the k s values in the pseudocoupled method; should be more accurate Method has not been implemented into software packages and thus is not routinely used on design projects
24 Finite Element Method Finite element method is used for structural analysis Mat is modelled in a similar way to other plate structures with springs connected at the nodes of the elements Mat is loaded with column loads, applied line loads, applied area loads, and mat weight Usually superstructure stiffness is not considered (conservative) Can be done but is rarely performed in practice Models the entire soilmat system in a threedimensional way In theory, should be the most accurate method
25 Other Considerations in Mat Foundations Total settlement "Bed of springs" solution should not be used to compute total settlement; this should be done using other methods Bearing capacity Mat foundations generally do not have bearing capacity problems With undrained silts and clays, bearing capacity needs to be watched Methods for spread footings can be used with mat foundations, including presumptive bearing capacities
26 Questions
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