Numerical Methods in Soil Mechanics 23.PDF

Numerical Methods in Soil Mechanics 23.PDF Numerical Methods in Geotechnical Engineering contains the proceedings of the 8th European Conference on Numerical Methods in Geotechnical Engineering (NUMGE 2014, Delft, The Netherlands, 18-20 June 2014). It is the eighth in a series of conferences organised by the European Regional Technical Committee ERTC7 under the auspices of the International Society for Soil Mechanics and Geotechnical Engineering (ISSMGE). The first conference was held in 1986 in Stuttgart, Germany and the series has continued every four years (Santander, Spain 1990; Manchester, United Kingdom 1994; Udine, Italy 1998; Paris, France 2002; Graz, Austria 2006; Trondheim, Norway 2010). Numerical Methods in Geotechnical Engineering presents the latest developments relating to the use of numerical methods in geotechnical engineering, including scientific achievements, innovations and engineering applications related to, or employing, numerical methods. Topics include: constitutive modelling, parameter determination in field and laboratory tests, finite element related numerical methods, other numerical methods, probabilistic methods and neural networks, ground improvement and reinforcement, dams, embankments and slopes, shallow and deep foundations, excavations and retaining walls, tunnels, infrastructure, groundwater flow, thermal and coupled analysis, dynamic applications, offshore applications and cyclic loading models. The book is aimed at academics, researchers and practitioners in geotechnical engineering and geomechanics.

Anderson, Loren Runar et al "FLOTATION" Structural Mechanics of Buried Pipes Boca Raton: CRC Press LLC,2000 Figure 23-1 Buried tank, 92D, tied down to a concrete anchor by straps, showing the buoyant uplift, Ww, of the empty tank (weight of tank neglected), and the effective resistance, Ws, of the soil wedge with 2-ft of soil cover The anchor and straps must prevent flotation Figure 23-2 12-kilogallon, 92D tank tied down to a slab, showing the net uplift, DW , the first-try location of hold down straps, and the moment diagram ©2000 CRC Press LLC CHAPTER 23 FLOTATION Buried tanks are used extensively at service stations Unfortunately, a groundwater table is not a valid reason to relocate service stations Therefore, when tanks are below the water table, holddowns, weights, etc., may be required to prevent flotation High soil cover can prevent flotation, but may not be economical Reinforced concrete pavement over a tank helps to resist flotation Holddowns require anchors — a concrete slab or deadmen When water table is a problem, soil at the bottom of the excavation is so wet that a concrete slab is used as a platform on which to work In some cases, two longitudinal footings (deadmen) may be adequate anchors The tank is tied to the anchors by straps See Figure 23-1 Tank D = 92 inches L = 35 ft Soil g d = 100 pcf = dry unit weight of soil, e = 0.5 = void ratio*, g b = 58.4 pcf = g d - g w/(1+e), j = 23o = soil friction angle from lab tests, q f = 56.5o = soil slip angle = 45o + j/2, H = 2.0 ft = soil cover * G = (1+e)100/62.4 = 2.4 = specific gravity of soil grains The specific gravity of most soil is in the range of 2.65 to 2.7 Therefore, this soil probably contains organic matter What force must be resisted by anchors? Holddowns Straps are generally used instead of cables and steel rods because straps minimize stress in the tank or pipe, and damage to the coating Questions arise, how many straps are needed, what size, and what spacing? Soil cover provides weight Anchors provide weight and soil resistance What is the soil resistance? Two mechanisms are: soil wedge and soil bearing capacity Soil Wedge If the embedment is granular and compacted, a floating tank must lift a soil wedge See Figure 23-1 If buoyant force of the tank exceeds the effective weight of the soil wedge, the anchors must restrain the difference See Chapter for effective unit weight of soil DW = Ww - Ws (23.1) where Ww = 2.881 kips/ft = buoyant uplift force per unit length of tank, Ws = 2.579 kips/ft = effective soil wedge (ballast) on top per unit length at q f = 45o + j/2, DW = 0.302 kips/ft to be restrained by straps For this 35-ft-long tank, The buoyant uplift force is 101 kips The resisting (ballast) force is 90 kips Neglecting the weight of the tank and the soil wedges at the ends of the tank, the force to be resisted by anchors is 11 kips Example Example 12,000-gallon tanks are common at service stations Suppose that a 12-kilogallon steel tank is buried under ft of soil cover How much anchorage is required if the tank is empty when the water table rises to or above the ground surface? For most analyses of granular embedments, engineers use a generalized soil wedge with slip plane slope of 2v:1h In Example 1, assuming the generalized soil wedge at effective (buoyant) unit weight, ©2000 CRC Press LLC Ww = 2.881 kips/ft = buoyant uplift force per unit length of tank, Ws = 2.257 kips/ft = effective generalized soil wedge (ballast) on top, DW = 0.624 kips/ft to be resisted by straps The buoyant uplift force is 101 kips The resisting soil wedge is 79 kips Neglecting soil wedges at the ends of the tank, the total uplift force to be restrained by the straps is 22 kips The conservatism (22 compared to 11) is justified because of conservative assumptions — in particular, the generalized soil wedge Figure 23-1 shows soil slip planes from the tank spring lines to the ground surface Tests show that planes are well established near the tank, but are not well established at the ground surface In fact, the "plane" may be more nearly a spiral cylinder that breaks out on the ground surface at a width less than the 15 ft shown For the generalized soil wedge (2v:1h), the break-out width at the surface is 13.5 ft Shearing stress of soil on pipe is neglected, weight of the tank is ignored, and soil liquefaction is not considered If there is any possibility of soil liquefaction, flotation is of concern The resisting soil force is 67 kips The total force on the anchors is 35(0.964) = 34 kips The force in each of four straps is 8.5 kips Design of Straps Continuing the example above, as a first try, use two slings (four straps) made of 0.25 x 1.25 hot rolled steel for which yield strength is 42 ksi and allowable is 21 ksi See Figure 23-2 If the force in each of the four straps is 8.5 kips, the resulting stress in each strap is s = 27 ksi — too high It would be prudent to increase the size or number of straps Try four straps, 0.25 x 1.75 The stress is s = 19.4 ksi This is acceptable unless fasteners are critical How are the straps attached to the slab? How are the straps tightened? How much prestress is required? Prestress reduces soil slip when the tank tries to float But prestress could cause flat spots on the bottom of the tank Prestress causes shearing stresses on the tank due to a "tight cinch" For double shear (both sides of the strap), shearing stress in the tank wall is, t = T/tD (23.2) Soil Bearing If a laborer sinks into the mud while walking on it, soil bearing capacity may be more critical than the soil wedge for resisting flotation Pressure on a laborer's shoe print is roughly ksf Just as a laborer's foot sinks into mud, so does a buoyant tank rise through mud It is assumed that the soil bearing capacity is the same — up or down where T = 8.5 kips tension in the strap, D = 92 inches, t = 0.187 = tank cylinder wall thickness (thin wall assumed) Substituting values, t = 494 psi Even if prestress should double the shearing stress, failure is improbable Example In the examples above, suppose that the soil is so poor that bearing capacity is only 250 lb/ft2 Soil resistance is Ws = (92D tank diameter)(250 lb/ft2) = 1.917 kips/ft Neglecting weight of the tank, Ww = 2.881 kips/ft, Ws = 1.917 kips/ft, DW = 0.964 kips/ft to be resisted by straps The buoyant uplift force is 101 kips ©2000 CRC Press LLC The next concern is longitudinal stress, especially at joints, due to bending of the tank when held down by straps Suppose straps are located tentatively at ft from the ends of the 35-ft-long tank See Figure 23-2 If, from the soil bearing case above, the resulting uplift force is DW = 964 lbs/ft, the moment diagram is maximum at B where M = 23.62 kip ft The maximum longitudinal stress, s , occurs at the top and bottom of the tank at B, where, s = M/(I/c) (23.3) Substituting values including I/c = pr2t; longitudinal stress is s = 230 psi Unless circumferential welds near B are bad, longitudinal stress is of no concern Safety factors are essential because soil-structure interaction is complex and variable In the above example, soil wedges at the ends of the tank are neglected Should they be included? Probably not if the soil wedges of Figure 23-1 are not true wedges but curve upward on a short concave section of a spiral A spiral is more probable if granular soil is loose If the soil is plastic or has more than about 10% fines, bearing capacity may be critical Bearing capacity must resist the vertical force on the slab (or deadman) that "pulls" it up through the soil For granular soils with less than 10% fines, the bearing capacity exceeds two kip/ft2, depending on degree of compaction For plastic soil or soil with a high fraction of fines or organic matter, the bearing capacity can be less Design of Anchors Anchors can be rock bolts, concrete slabs, reinforced concrete beams (deadmen), etc Slabs and deadmen are discussed in the following A typical slab is reinforced concrete approximately the same width and length as the tank, and a minimum of 8-inches thick See Figure 23-3 Deadmen can be two or more transverse beams on which the tank is positioned More generally, deadmen are two longitudinal beams located as shown in Figure 23-4 Holddown capability is the effective weight of the deadman (or slab) plus resistance of soil — either the effective weights of soil wedges (in this case, soil prisms) or the soil bearing capacity Following is the procedure for analysis ©2000 CRC Press LLC Soil Prism With shearing planes at 2v:1h, the soil lifted by the slab is the volume of prisms (partly dotted in Figure 23-3) times the effective unit weight of soil The soil is assumed to be granular and compacted Example Wedge Analysis Consider the 12 kilogallon tank, D = 7.667 ft diameter and L = 35 ft long, with soil cover H = ft The effective unit weight of the embedment, from the example above, is 58.4 lb/ft3 What is the safety factor against flotation? Try a slab See Figure 23-3 L = 35 ft = length, 2X = ft = width, t = inches = thickness The x 35-ft slab should be reinforced so that it can resist the concentrated forces of the straps Find the reaction to tension in the four straps The volume of each of the two soil prisms is 704 ft3 The effective weight of the two prisms is 2(704)(58.4) = 82 kips When added to the 79-kip effective weight of the generalized soil wedge on top of the tank, the resistance to flotation is 161 kips Because the buoyant uplift force is 101 kips, the safety factor against flotation is 165/101 = 1.6 Try longitudinal deadmen See Figure 23-4 Assume that the deadmen are as long as the tank, i.e., 35 ft, and that, a = 12 inches, b = 12 inches, r = 46 inches, h = 24 inches, j = 35 inches = (b+r/2), g = 58.4 pcf = effective (buoyant) unit weight of soil from the above examples Figure 23-3 Holddown straps to a slab, showing the soil prisms (partly dotted) that help to resist the uplift force of the straps Total strap resistance is the effective (buoyant) weights of the slab and soil prisms Figure 23-4 Holddown straps to longitudinal deadmen, showing (dotted) the prism shear surface AB and the spiral shear surface AC for less-than-select-soil ©2000 CRC Press LLC The area of the soil prism lifted by each deadman is roughly j(2r+h) = 28.2 ft2 Each prism is 35 ft long Therefore, the volume of each prism is 987 ft3 The effective weight of each prism is 115 kips For two prisms, this should be doubled However, in lessthan-excellent embedment, the soil shear surface is not plane AB, but is more nearly spiral AC, for which the volume to be lifted is roughly half of the prism Therefore the total effective weight of soil lifted by deadmen is 115 kips When added to the 79-kip effective weight of the soil wedge on top of the tank, resistance to flotation is 194 kips The buoyant uplift force of the tank is 101 kips The safety factor against flotation is 194/101 = 1.9 Try two longitudinal deadmen If the deadmen are 35 ft long and the cross sections are ft x ft, the soil bearing resistance plus buoyant weight of the deadmen is 2(35)(0.250) + 2(35)(0.081) = 23 kips Added to the 70-kip resistance of the soil cover, the total resistance is 93 kips Buoyant uplift force is 101 kips When empty, the tank will float in the mud One possible analysis of the resistance of deadmen might start with inversion of the classical soil bearing rationale As a deadman rises, a soil wedge forms on top and shoves soil outward in general shear This is explained in texts on soil mechanics The deadman “plows” its way up through the soil Soil Bearing Analysis PROBLEMS If the saturated soil is so poor that a laborer sinks into the mud, the tank and deadmen (or slab) could float upward through the mud Suppose the bearing capacity of the mud is 250 lb/ft2 Try a slab If the slab is 35 ft long by ft wide, half the area is effectively resisted by the soil bearing capacity Adding the buoyant weight of an 8-inch-thick concrete slab at 81 pcf, resistance to tension in the straps is (4 ft)(35 ft)(0.25 k/ft2) + 8(35)2/3)(0.081) = 35 + 15 = 50 kips Resistance of the soil cover is (8 ft)(35 ft)(250 lb/ft2) = 70 kips The sum of the two is 120 kips Buoyant uplift force is 101 kips The safety factor against flotation is 120/101 sf = 1.2 If there is any possibility of soil liquefaction, it would be prudent to specify select embedment ©2000 CRC Press LLC 23-1 A 20-kilogallon steel tank, 9-ft diameter, by 42ft length, is buried under H = ft of granular soil Design straps and a reinforced concrete slab ft x 42 ft x inches thick, to prevent flotation with a safety factor of 1.2 It is possible for a water table to reach the surface at a time when the tank is empty Assume effective unit weight of the submerged soil is 60 lb/ft3 a) What must be the number and size of steel straps if allowable tensile strength is 20 ksi? What must be the thickness of the slab to prevent flotation if unit weight of concrete is 142.2 lb/ft3? (What is the submerged unit weight of concrete?) ... strap), shearing stress in the tank wall is, t = T/tD (23.2) Soil Bearing If a laborer sinks into the mud while walking on it, soil bearing capacity may be more critical than the soil wedge... inversion of the classical soil bearing rationale As a deadman rises, a soil wedge forms on top and shoves soil outward in general shear This is explained in texts on soil mechanics The deadman “plows”... through the soil For granular soils with less than 10% fines, the bearing capacity exceeds two kip/ft2, depending on degree of compaction For plastic soil or soil with a high fraction of fines or