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

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

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 4 for effective unit

weight of soil

Example 1

12,000-gallon tanks are common at service stations

Suppose that a 12-kilogallon steel tank is buried

under 2 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?

Tank

D = 92 inches

L = 35 ft Soil

gd = 100 pcf = dry unit weight of soil,

e = 0.5 = void ratio*,

gb = 58.4 pcf = gd - gw/(1+e),

j = 23o = soil friction angle from lab tests,

qf = 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?

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 qf = 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 2 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,

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

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 2 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

Example 3

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.

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,

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

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 7 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 pr isms) or the soil bearing capacity Following

is the procedure for analysis

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 = 2 ft The effective unit weight of the embedment, from the example above, is 58.4 lb/ft3 What is the safety factor against flotation?

1 Try a slab See Figure 23-3

L = 35 ft = length, 2X = 8 ft = width,

t = 8 inches = thickness

The 8 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

2 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

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

less-than-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

Soil Bearing Analysis

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

1 Try a slab

If the slab is 35 ft long by 8 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

2 Try two longitudinal deadmen

If the deadmen are 35 ft long and the cross sections are 1 ft x 1 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

PROBLEMS

23-1 A 20-kilogallon steel tank, 9-ft diameter, by

42-ft length, is buried under H = 2 42-ft of granular soil Design straps and a reinforced concrete slab 9 ft x

42 ft x 9 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?)