Numerical Methods in Soil Mechanics 15.PDF

Numerical Methods in Soil Mechanics 15.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 "THRUST RESTRAINTS"

Structural Mechanics of Buried Pipes

Boca Raton: CRC Press LLC,2000

Figure 15-2 Impulse thrust, Qi, at a bend in a pipe due to change in direction of flow, showing a free-vector-diagram of the components Fx and Fy

CHAPTER 15 THRUST RESTRAINTS

A straight pipe section with end closures (valves or

caps) feels a longitudinal force when internal

pressure is applied In the case of static pressure,

the longitudinal force F is simply the internal

pressure times the area; i.e.,

F = PπD2/4

where

F = longitudinal thrust in the pipe,

P = internal pressure,

D = inside diameter = 2r

But suppose the pipe has gasketed joints between

the ends See Figure 15-1 Now it cannot resist the

force F Consequently, thrust restraints (thrust

blocks) must be supplied at the ends of the pipe

EVALUATION OF THRUST Q

Each thrust restraint must resist F without moving

enough to allow a joint to leak In addition to the

pressure, P, suppose that the fluid is moving in the

pipe due to pressure gradient, ∆P Force F is

increased by ∆F = ∆PπD2/4 For most buried

pipeline analyses, fluid friction, ∆F, is negligible

because the length between gaskets is short and is

easily resisted by soil friction on the pipe

But now, suppose that the gasketed pipe is not

straight A change in direction is introduced by an

elbow (or bend) The sidewise thrust Q at the elbow

is due to both pressure and the impulse of change in

direction of flow It is the vector sum of impulse and

pressure forces, Qi and Qp, on the fluid at the bend

Each is found separately

Impulse Thrust Qi

See free-vector-diagram, Figure 15-2

F = impulse force (vector),

Qi = thrust due to impulse,

Qi = vector sum of impulse forces Fx and Fy,

θ = offset angle of the bend,

v = average velocity of fluid flow in the pipe

Fx = axial force on the fluid at the elbow,

D = inside diameter,

ρ = mass density of the fluid, (v-vcosθ) = change in the x-components of velocity

as the column of fluid flows around the bend

Fx and Fy can be found by the principle "impulse

equals change in momentum." Both impulse and momentum are vector quantities Figure 15-2 shows

a free-body-diagram of a column of fluid (cross-hatched) The area is πD2/4, and the length is (vdt) Impulse is force times time dt, and change in momentum is mass times change in velocity In the x-direction;

Impulse = Fxdt Change in Momentum = (πD2/4)vdtρ(v-vcosθ) Equating impulse to the change in momentum in the x-direction,

Fx = π(Dv)2ρ(1-cosθ)/4

But Fx is only the x-component In a similar manner

by equating the y-component of impulse to change in momentum;

Fy = π(Dv)2ρ(sinθ)/4 From the free-vector-diagram of Figure 15-2, the resultant of Fx and Fy is,

2Qi = π(Dv)2ρsin(θ/2) (15.1)

The angle between Qi an d Fy is tan-1(Fx /Fy) = θ/2 Noting that the pipe is symmetrical about the Qi vector, Equation 15.1 could have been written directly, because the change in velocity in the Qi direction is simply 2vsin(θ/2)

Figure 15-3 Pressure thrust-Qp at a bend (elbow) in a pipe due to internal pressure, P, showing the free-vector-diagram for calculating Qp

Figure 15-4 Passive soil resistance on an elbow and on contiguous gasketed pipe sections showing how the soil envelope can provide thrust restraint

Pressure Thrust Qp

See Figure 15-3; where

D = inside diameter = 2r,

P = internal fluid pressure,

Qp = thrust due to internal pressure,

θ = offset angle of the bend (elbow)

A free-body-diagram of the elbow with pressurized

fluid contents is shown cross-hatched Neglecting

the small friction loss of flow around the bend, from

the free-vector-diagram,

2Qp = πD2Psin(θ/2) (15.2)

Qp is at an angle of θ/2 with the y-axis

Consequently, thrust, Q, is the sum, Qi+ Qp; i.e.,

2Q = πD2(P + v2ρ)sin(θ/2) (15.3)

where

v = average velocity of fluid flow,

ρ = mass density of the fluid,

θ = offset angle of the bend

Example

Find thrust-Q at a 90o elbow in a water pipe for

which,

θ = 90o,

D = 30 inches,

P = 200 psi = internal pressure,

v = 15 ft/second = flow velocity,

ρ = γw /g = mass density of water,

γ = 62.4 lb/ft3 = unit wt of water,

g = 32.2 ft/second2 = gravity

Substituting into Equation 15.1, Qi = 3 kips

Substituting into Equation 15.2, Qp = 200 kips

Combined, Q = 203 kips Impulse thrust, Qi is

usually neglected

If a large diameter pipe with high internal pressure

has an elbow with a large offset angle, θ, thrust-Q is

enormous

SPECIAL SECTIONS

Special sections redirect or alter flow Examples include elbows, wyes, tees, valves, reducers, caps, plugs, etc The following analyses for elbows can be applied to any special section In every case, thrust,

Q is the sum of impulse thrust, Qi, and pressure thrust, Qp

COMMON THRUST RESTRAINTS

1 Welded or Bolted Joints at Special Sections

In a pressurized pipe, at a gasketed elbow, Q must

be resisted by the soil or by a thrust restraint (thrust

block) For a welded elbow, Q is resisted by the pipe Two analyses of a welded elbow follow.

a) If the contiguous pipes are unrestrained and uncapped (like a garden hose), normal force, F, and

shearing force, S, act on the elbow Analysis is conservative because soil resistance reduces F and

S

σ = F/2πrt = average normal stress,

τ = S/2πrt = average shearing stress

From the equations of static equilibrium,

σ /P(r/t) = (1-cosθ) (15.4) NORMAL STRESS TERM

τ /P(r/t) = sinθ (15.5) SHEARING STRESS TERM

These stress terms are upper limits — twice the force-per-unit-area — to account for eccentricity of the F-force and redistribution of stresses The outside of a bend can stretch more than the inside Therefore, stresses are greater on the inside See Problem 15-12 Wall thickness is sometimes increased for elbows In general, greater wall thickness is not justified

b) If the contiguous pipes are restrained and capped, from the equations of equililbrium, longitudinal stress is,

σ = Pr/2t (15.6)

This is only half as great as circumferential stress,

and is independent of offset angle, θ A more

precise analysis would show that stress, σ , on the

inside of the bend is increased slightly as the offset

angle, θ, is increased Most pipes are ductile enough

that the material "plastic-flows" at yield, and does not

fail Moreover, soil friction resists thrust In

practice, contiguous pipes are seldom capped

Longitudinal stress is not critical for isotropic plain

steel and plastic pipes Of course, joints must be

adequate

For non-isotropic pipes (corrugated, ribbed, or

wrapped with fiberglas or wire), longitudinal strength

must be assured Neglecting impulse force and soil

resistance, for uncapped, unrestrained contiguous

pipes:

At elbows, for longitudinal design,

Pπr2(1-cosθ) = Aσf /sf (15.7)

where

A = area of longitudinal fibers,

σf = strength of the fibers

At valves or caps (not at bends) for design,

Pπr2 = Aσf /sf (15.8)

2 Embedment As Thrust Restraint

If thrust-Q is not large, the embedment is able to

develop adequate passive resistance It may not be

necessary to provide additional thrust restraint

Consider in Figure 15-4 the free-body-diagram of an

elbow and one section of pipe on each side The

joints are gasketed so the pipe can take no

longitudinal force Thrust-Q can be restrained only

by the soil bearing against the pipe The maximum

soil pressure bearing horizontally against the elbow

is passive resistance Px at the average depth of soil,

H + OD/2,

Px = (2H + OD)γ/2K

where

K = P/Px = (1-sinϕ)/(1+sinϕ),

ϕ = soil friction angle,

γ = unit weight of soil,

OD = outside diameter,

H = height of soil cover,

L = length of pipe section

The restraint capacity of soil against elbow is,

Qelb = (area) times Px where

(area) = (OD)Lelb,

Lelb = cord length (approximate) of elbow

from coupling to coupling as shown Multiplying (area) times Px,

Qelb = (2H + OD)γLelbOD/2K Added to this is the restraint capacity of the first section of pipe on each side of the elbow Full passive resistance of the soil would be developed at the elbow end of each section At the opposite end, each pipe section could rotate, because of the gasket But there would be no lateral movement Passive soil resistance would not be developed A crude, but reasonable and conservative assumption,

is that passive resistance varies linearly from Px at the elbow end to zero at the opposite end Due to soil supporting the two pipe sections, the component

of restraint in the direction of Q is,

Qsecs = (OD)LPxcos(θ/2)

or, substituting for Px,

Qsecs = (OD)L(2H + OD)γcos(θ/2)/2K Combining the thrust restraints provided by the elbow and the two pipe sections,

Restraint-Q = OD(2H + OD)γ[Lelb + Lcos(θ/2)]/2K

(15.9)

Rewriting Equation 15.3,

Thrust-Q = π(ID)2(P + v2ρ)sin(θ/2)/2

(15.10)

Equation 15.10 for thrust-Q was derived for a

horizontal bend For a vertical bend (in a vertical

plane), thrust-Q has a vertical component If soil

cover alone is to resist the upward component of

thrust-Q, then soil cover H must be great enough

that soil weight can hold the pipe down A

conservative restraint-Q for this vertical bend is,

Restraint-Q =

OD(2H + OD)γ[Lelb + Lcos(θ/2)]/2

(15.11)

This is the same as Equation 15.9 except that K is

eliminated For design, restraint-Q must be greater

than thrust-Q A safety factor should be included

3 Thrust Block as Thrust Restraint

Thrust blocks are the most common restraints in use

for pressurized gasketed pipes See Figure 15-5

Thrust blocks are usually concrete A reasonable

analysis for design starts with the

free-body-diagram Assuming a cubical block,

B = lengths of sides of the cube,

γ = unit weight of soil,

γc = unit weight of the thrust block,

jB = distance down to thrust-Q from the top of

the block,

K = (1-sinϕ)/(1+sinϕ),

ϕ = soil friction angle

Other data are shown on the sketch Friction on the

sides of the block is undependable and is

conservatively neglected

T wo modes of failure are considered: overturn

about point O, and slip The conditions under which

each mode controls are described by an example of

a cubical thrust block

Example — Assumptions

h = H/B = ratio of soil cover H to side B,

j = ratio of distance between top of block and

thrust-Q, to side B,

ϕ = 30° = soil friction angle,

K = 1/3 = (1-sinϕ)/(1+sinϕ),

γ = 120 pcf = unit weight of soil,

γc = 144 pcf = unit weight of concrete

Taking the sum of the moments of force about overturn fulcrum O,

Q/γB3 = (2h + 1.10)/(1-j) OVERTURN

(15.12) Taking the sum of the horizontal forces,

Q/γB3 = (3.577h + 2.193) SLIP

(15.13)

The dimensionless quantity Q/γB3 is the thrust block restraint number A table of values is shown as

Table 15-1 for typic al design based on the assumptions indicated

Overturn

In order to design a cubical thrust block with the typical soil properties assumed in the analysis above,

it is only necessary to guess a trial value for B from which values of h and j can be calculated Entering

Table 15-1 with h and j, the restraint number, Q/γB3

can be found in the overturn columns

For a soil unit weight of γ = 120 pcf, Q/B3 = (120 pcf)(restraint number)/sf Solve for B If not the same as the assumed B, using the new B recalculate values for h and j Enter Table 15-1 for a second trial solution of the restraint number from which a new value of B is calculated If this new B is unchanged, then the answer has been found If not, recycle the analysis with the new B

Slip The left of the two SLIP columns of Table 15-1

Figure 15-5 Free-body-diagram of a cubical thrust block.

Table 15-1 Values of cubical thrust block restraint number, Q/γB3, for concrete at 144 pcf and soil at 120 pcf and ϕ = 30o No safety factor is included

provides values for the restraint number from

Equation 15.13 The right of the two slip columns is

the minimum value of j at which slip is critical

No safety factor is included The analysis is so

conservative, that safety factors need not be large

Nevertheless, the risk of failure may warrant a

safety factor

One novel concept is the thrust pin designed to

conserve space See Figure 15-6 It can be located

inside the bend if necessary, tied with tendons

4 Tendons

Instead of thrust blocks or thrust pins, which restrain

the gasketed elbow by compression from outside the

bend, restraint is by tension tendons inside the bend

fastened to dead-men such as buried concrete

blocks, boulders, beams, pins, etc The tendons

could be rods, cables, wires, etc

Or, instead of tying tendons to dead-men, they could

be tied across the bend to corresponding pipe joints

on either side of the elbow See Figure 15-13

These "harp-strings" cannot resist the thrust without

other restraints such as longitudinal friction between

soil and pipes and soil bearing

PIPES ON STEEP SLOPES

The analysis of thrust restraints, for pipes on slopes,

is the same as above; but in addition, must include

longitudinal forces caused by gravity For most

pipes, the length tends to shorten when the pipe is in

service because internal pressure increases and

temperature decreases Therefore, it is good

practice to design restraints such that the pipe

shortens downhill Frictional resistance to

shortening is uphill and partially offsets the downhill

component of weight of the full pipe

The thrust restraint (anchor) is clamped to the pipe

uphill from it, and is slip-coupled to the pipe downhill

from it The downhill side is free to slip toward the

next anchor downhill Expansion joints allow slip

For short pipe sections, com-mon sleeve-type

couplings allow adequate slip

For most couplings, the pipe must be supported on both sides of the coupling to assure alignment Good backfill soil may provide alignment In the case of poor backfill, or pipe on piers, two yokes on each pier assure alignment as shown in Figure 15-7 Most couplings are not designed to resist longitudinal moment or transverse shear The allowable degree

of misalignment of a coupling is limited For example, the allowable misalignment (offset angle)

is about 3o for steel pipes 30 to 54 inch diameter Manufacturers of couplings should be consulted for restrictions and specific applications If the pipe is

on piers, couplings should not be located at midspans between piers

Particular care is required for large pipes on steep slopes because of the difficulties of installation as well as the additional loads on the thrust restraints

On slopes steeper than about 45°, the pipe is often placed on piers above ground The slope is too steep to excavate a trench, too steep to hold the pipe

in position for welding and backfilling, and too steep

to compact backfill Moreover, a pipe on a steep slope may feel the downhill drag from creep of surface soil (downhill freezing and thawing) and thus overload the anchor at the bottom of the slope Slopes steeper than 45° are usually rock outcrops that cannot be excavated without ripping or blasting

On steep slopes, or in inaccessible areas, construction might require lowering personnel, platforms, and equipment down the pipeline by cables Rock pins are drilled and grouted into place The pipe is laid downhill as the platform is lowered Economics often favor service from an overhead cable on towers, or from helicopters Helicopters are expensive (one to two thousand dollars per hour), but can place pipes and piers quickly if ground crews avoid delays Under some conditions, tunneling may be an option

Example The following example identifies some of the many problems associated with pipelines on steep slopes

Figure 15-6 Alternate concept of thrust restraint at an elbow in a pipeline provided by a thrust pin that conserves space and quantity of concrete The hole is bored, reinforcing steel is positionced, and concrete

is cast into the bored hole