Negative skin friction.In most situations, fill has to be placed to achieve necessary elevations. Weight of the fill would consolidate the clay layer.Generally the piles are driven after placing the fill layer.The settlement of the clay layer (due to the weight of the fill) would induce a downdrag force on the pile.If the piles were driven after full consolidation of the clay layer has occurred, no downdrag would occur.Unfortunately full consolidation may not be completed for years and many developers may not wait that long to drive piles.
There is little question that, if the elementary book of Tennekes and Lumley 1 were written today. 13 that presents a solution of this problem: we demonstrate that the scaling exponents of. The pumping rate is controlled manually using a. Tennekes and Lumley (1994) give an excellent summary of the scales of turbulence. These scales are not further treated here, except for the Kolmogorov scale λ 0. The latter is important, as it determines the size of flocs of cohesive sediment: the stresses that disrupt the flocs are in the viscous regime.
Skin friction in soft clay = α × c × perimeter surface area = 0.4 × 700 × π × d × L = 0.4 × 700 × π × 1 × 12 = 10,560 lbs ( 46.9 kN )Step 2: Find the skin friction from B to C:Skin friction in sandy soils = S = Kσ′ v × tan δ × A pS = skin friction of the pile; σ v′ = average effective stress along the pile shaft.Average effective stress along pile shaft from B to C = ( σ B + σ C)/2σ B = effective stress at B; σ C = effective stress at C.To obtain the average effective stress from B to C, find the effective stresses at B and C and obtain the average of those two values. Σ B = 100 × 4 + ( 100 − 62.4 ) × 8 = 700.8 lbs / ft. 2 ( 33.6 k P a ) σ C = 100 × 4 + ( 100 − 62.4 ) × 8 + ( 110 − 62.4 ) × 20 = 1452.8 lbs / ft. 5 k P a )Average effective stress along pile shaft from B to C = ( σ B + σ C)/2= (700.8 + 1452.8)/2 = 1076.8 lbs/ft.
2.Skin friction from B to C = Kσ′ v × tan δ × A p= 0.9 × 1076.8 × tan(25 °) × ( π × 1 × 20) = 28,407 lbsStep 3: Find the skin friction from C to D:Skin friction reaches a constant value at point “C”, 20 diameters into the bearing layer.Skin friction at point “C” = Kσ′ v × tan δ × A pσ′ v at point “C” = 100 × 4 + (100 − 62.4) × 8 + (110 − 62.4) × 20 = 1452.8 lbs/ft. 2Unit skin friction at point “C” = 0.9 × 1452.8 × tan25 ° = 609.7 lbs/ft. (29 kPa)Unit skin friction is constant from C to D. Skin friction in clay. SolutionThe Kolk and Van der Velde method depends on both effective stress and cohesion. Effective stress varies with the depth.
Hence, obtain the average effective stress along the pile length. Computation of skin friction in bored piles:.The same procedure used for driven piles could be used for bored piles, but with a lesser undrained shear strength.
The question is how much reduction should be applied to the undrained shear strength for bored piles?.Undrained shear strength may reduce as much as 50% due to the stress relief in bored piles. On the other hand, measured undrained shear strength is already reduced due to the stress relief occurring when the sample was removed from the ground. By the time, the soil sample reaches the laboratory, soil sample has undergone stress relief and the measured value already indicates the stress reduction.
Considering these two aspects, reduction of 30% in the undrained shear strength would be realistic for bored piles. Single pile in clay soil. SolutionStep 1: Find the end bearing capacity.End bearing capacity in clay soils = 9 cA.c = cohesion = 50 kN/m 2,N × c = 9,A = π × D 2/4 = π × 0.5 2/4 m 2 = 0.196 m 2Ultimate end bearing capacity = 9 × 50 × 0.196 = 88.2 kN.
(19.8 kips)Step 2: Find the skin friction.Ultimate skin friction = S u = α × c × A pUltimate skin friction = 0.75 × 50 × A pA p = perimeter surface area of the pile = π × D × L = π × 0.5 × 10 m 2.A p = 15.7 m 2Ultimate skin friction = 0.75 × 50 × 15.7 = 588.8 kN. (132 kips)Step 3: Find the ultimate capacity of the pileUltimate pile capacity = ultimate end bearing capacity + ultimate skin frictionUltimate pile capacity = 88.2 + 588.8 = 677 kN. (152 kips)Allowable pile capacity = ultimate pile capacity/FOSAssume a factor of safety of 3.0.Allowable pile capacity = 677/3.0 = 225.7 kN. (50.6 kips)Note: The skin friction was very much higher than the end bearing in this situation. Skin friction (groundwater considered).
SolutionUnlike sands, groundwater does not affect the skin friction in clayey soils.Step 1: Find the end bearing capacity.End bearing capacity in clay soils = 9 × c × A.c = cohesion = 35 kN/m 2,N c = 9,A = π × D 2/4 = π × 1 2/4 m 2 = 0.785 m 2Ultimate end bearing capacity = 9 × 35 × 0.785 = 247.3 kN. (55.5 kips)Step 2: Find the skin friction.Ultimate skin friction = S u = α × c × A pAdhesion factor α, is not given. Hannoush play in hindi full. Use the method given by American Petroleum Institute (API).α = 1.0, for clays with cohesion = 70 kN/m 2Since, the cohesion is 35 kN/m 2 interpolate to obtain the adhesion factor ( α). Pile in multiple layers. SolutionFind the skin friction at the top of the clay layer and bottom of the clay layer. Obtain the average of the two values.Step 1:At point A: σ′ = 3 × 17 = 51 kN/m 2; (1065 lbs/ft.
2)S u = 25 kN/m 2;S u/ σ′ = 25/51 = 0.50From the Kolk and Van der Velde table; α = 0.65At point B: σ′ = 6 × 17 = 102 kN/m 2.S u = 25 kN/m 2;S u/ σ′ = 25/102 = 0.25From the Kolk and Van der Velde table; α = 0.86Step 2:Ultimate skin friction at point A:f ult = α × S u = 0.65 × 25 = 16 kN/m 2Ultimate skin friction at point B:f ult = α × S u = 0.86 × 25 = 21 kN/m 2Assume the average of two points to obtain the total skin friction.Average = (16 + 21)/2 = 18.5 kN/m 2.Total skin friction = 18.5 × perimeter × length = 18.5 × ( π × 0.5) × 3 = 87 kN (19.6 kips)Summary of equations. Monteith, Mike H. Unsworth, in, 2013 9.1.2 Form DragIn addition to the force exerted by skin friction, which is a consequence of momentum transfer to a surface across the streamlines of flow, bodies immersed in moving fluids also experience a force in the direction of the flow as a result of the deceleration of fluid. This force is known as form drag because it depends on the shape and orientation of the body. Maximum form drag is experienced by surfaces at right angles to the fluid flow, and the force can be estimated by assuming that there is a point on the surface where the fluid is instantaneously brought to rest after being uniformly decelerated from a velocity V (remembering that Newton’s Second Law may be expressed as force = rate of change of momentum). If the initial momentum per unit volume of fluid is ρ V and the mean velocity during deceleration is V / 2, the rate at which momentum is lost from the fluid is ρ V × V / 2 = 0.5 ρ V 2. This is the maximum rate at which momentum can be transferred to unit area on the upstream surface of a bluff body and it is therefore the maximum pressure excess (pressure = force per unit area) that a fluid can exert in contributing to the total form drag over the body.
In practice, fluid tends to slip round the sides of a bluff body so that a form drag force smaller than 0.5 ρ V 2 is exerted on the upstream face. However, in the wake which forms downstream of a bluff body, the fluid pressure is less than in the free stream, and the associated suction force often makes an important additional contribution to the total form drag on a body.
(As examples, gulls diving for fish assume a “streamlined” shape that minimizes the wake to maximize their fall speed, and in the sporting world, golf ball manufacturers adopt complex patterns for the dimples on golf balls to reduce the wake and consequently to reduce the drag.) The total form drag on unit area is conveniently expressed as c f. 0.5 ρ V 2 where c f is a drag coefficient.
Hence c f is the fraction of the maximum bluff body pressure that a fluid can exert.In most problems, it is appropriate to combine skin friction and form drag to give a total force τ, usually the force on a unit area projected in the direction of the flow, i.e. 2 rl for a cylinder of radius r and length l in cross-flow and π r 2 for a sphere. The ratio τ / ( 0.5 ρ V 2 ) then defines the total drag coefficient c d, which for spheres and for cylinders at right angles to the flow lies between 0.4 and 1.2 in the range of Reynolds numbers between 10 2 and 10 5. Manufacturers of energy-efficient cars boast drag coefficients of about 0.25.As background for later discussion of mass and heat transfer, it should be noted that the diffusion of momentum in skin friction is analogous to the diffusion of gas molecules and of heat, provided the surface is parallel to the airstream. For such a surface, close relations may therefore be expected between r M, r H, and r V.
For a surface at right angles to the airstream however, there is no frictional force in the direction of flow. Friction will operate in all directions at right angles to the flow but the net sum of all these (vector) forces must be zero. In contrast, the net flux of heat or mass, which are scalar quantities, must be finite in the plane of the surface. In this case, r V and r H may be similar to each other but will be unrelated to the value of r M. Figure 46.15.
Concrete caisson with a W-section.The following parameters are given:Ultimate steel compressive strength = 36,000 psiUltimate concrete compressive strength = 3,000 psi Simplified design procedureSimplified design procedure is first explained. In this procedure, the composite nature of the section is ignored.Step 1: Structural design of the caissonAssume a diameter of 30 in. For the concrete caisson. Since, E (elastic modulus) of steel is much higher than concrete, a major portion of the load is taken by steel. Assume that 90% of the load is carried by steel.Load carried by steel = 0.9 × 1000 tons = 900 tonsAllowable steel compressive strength = 0.5 × 36,000 = 18,000 psi(NYC Building Code, Table 11.3 recommends a factor of safety of 0.5. Check your local building code for the factor of safety value).
Figure 46.16. I-beam.Step 2: Check whether this section fits inside a 30 in. Hole.Distance along a diagonal (Pythagoras theorem) = (16.36 2 + 17.54 2) 1/2 = 23.98 in.This value is smaller than 30 in. Hence, the section can easily fit inside a 30 in. Hole.Step 3: Compute the load carried by concrete.Concrete area = area of the hole − area of steel = 706.8 − 101 = 605.8.Allowable concrete compressive strength = 0.25 × ultimate compressive strength = 0.25 × 3000 = 750 psi(NYC Building Code, Table 11.3 recommends a factor of safety of 0.25.
1.3 The large eddies in a turbulent flow have a length scale t, a velocity scale v=u, and a time scale te-Yu. The smallest eddies have a length 26 Introduction scale a velocity scale u, and a time scale τ. Estimate the characteristic velocity vir) and the characteristic time tr) of eddies of size r, where r is any length in the range η. Gravograph drivers.
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Negative skin friction.In most situations, fill has to be placed to achieve necessary elevations. Weight of the fill would consolidate the clay layer.Generally the piles are driven after placing the fill layer.The settlement of the clay layer (due to the weight of the fill) would induce a downdrag force on the pile.If the piles were driven after full consolidation of the clay layer has occurred, no downdrag would occur.Unfortunately full consolidation may not be completed for years and many developers may not wait that long to drive piles.
There is little question that, if the elementary book of Tennekes and Lumley 1 were written today. 13 that presents a solution of this problem: we demonstrate that the scaling exponents of. The pumping rate is controlled manually using a. Tennekes and Lumley (1994) give an excellent summary of the scales of turbulence. These scales are not further treated here, except for the Kolmogorov scale λ 0. The latter is important, as it determines the size of flocs of cohesive sediment: the stresses that disrupt the flocs are in the viscous regime.
Skin friction in soft clay = α × c × perimeter surface area = 0.4 × 700 × π × d × L = 0.4 × 700 × π × 1 × 12 = 10,560 lbs ( 46.9 kN )Step 2: Find the skin friction from B to C:Skin friction in sandy soils = S = Kσ′ v × tan δ × A pS = skin friction of the pile; σ v′ = average effective stress along the pile shaft.Average effective stress along pile shaft from B to C = ( σ B + σ C)/2σ B = effective stress at B; σ C = effective stress at C.To obtain the average effective stress from B to C, find the effective stresses at B and C and obtain the average of those two values. Σ B = 100 × 4 + ( 100 − 62.4 ) × 8 = 700.8 lbs / ft. 2 ( 33.6 k P a ) σ C = 100 × 4 + ( 100 − 62.4 ) × 8 + ( 110 − 62.4 ) × 20 = 1452.8 lbs / ft. 5 k P a )Average effective stress along pile shaft from B to C = ( σ B + σ C)/2= (700.8 + 1452.8)/2 = 1076.8 lbs/ft.
2.Skin friction from B to C = Kσ′ v × tan δ × A p= 0.9 × 1076.8 × tan(25 °) × ( π × 1 × 20) = 28,407 lbsStep 3: Find the skin friction from C to D:Skin friction reaches a constant value at point “C”, 20 diameters into the bearing layer.Skin friction at point “C” = Kσ′ v × tan δ × A pσ′ v at point “C” = 100 × 4 + (100 − 62.4) × 8 + (110 − 62.4) × 20 = 1452.8 lbs/ft. 2Unit skin friction at point “C” = 0.9 × 1452.8 × tan25 ° = 609.7 lbs/ft. (29 kPa)Unit skin friction is constant from C to D. Skin friction in clay. SolutionThe Kolk and Van der Velde method depends on both effective stress and cohesion. Effective stress varies with the depth.
Hence, obtain the average effective stress along the pile length. Computation of skin friction in bored piles:.The same procedure used for driven piles could be used for bored piles, but with a lesser undrained shear strength.
The question is how much reduction should be applied to the undrained shear strength for bored piles?.Undrained shear strength may reduce as much as 50% due to the stress relief in bored piles. On the other hand, measured undrained shear strength is already reduced due to the stress relief occurring when the sample was removed from the ground. By the time, the soil sample reaches the laboratory, soil sample has undergone stress relief and the measured value already indicates the stress reduction.
Considering these two aspects, reduction of 30% in the undrained shear strength would be realistic for bored piles. Single pile in clay soil. SolutionStep 1: Find the end bearing capacity.End bearing capacity in clay soils = 9 cA.c = cohesion = 50 kN/m 2,N × c = 9,A = π × D 2/4 = π × 0.5 2/4 m 2 = 0.196 m 2Ultimate end bearing capacity = 9 × 50 × 0.196 = 88.2 kN.
(19.8 kips)Step 2: Find the skin friction.Ultimate skin friction = S u = α × c × A pUltimate skin friction = 0.75 × 50 × A pA p = perimeter surface area of the pile = π × D × L = π × 0.5 × 10 m 2.A p = 15.7 m 2Ultimate skin friction = 0.75 × 50 × 15.7 = 588.8 kN. (132 kips)Step 3: Find the ultimate capacity of the pileUltimate pile capacity = ultimate end bearing capacity + ultimate skin frictionUltimate pile capacity = 88.2 + 588.8 = 677 kN. (152 kips)Allowable pile capacity = ultimate pile capacity/FOSAssume a factor of safety of 3.0.Allowable pile capacity = 677/3.0 = 225.7 kN. (50.6 kips)Note: The skin friction was very much higher than the end bearing in this situation. Skin friction (groundwater considered).
SolutionUnlike sands, groundwater does not affect the skin friction in clayey soils.Step 1: Find the end bearing capacity.End bearing capacity in clay soils = 9 × c × A.c = cohesion = 35 kN/m 2,N c = 9,A = π × D 2/4 = π × 1 2/4 m 2 = 0.785 m 2Ultimate end bearing capacity = 9 × 35 × 0.785 = 247.3 kN. (55.5 kips)Step 2: Find the skin friction.Ultimate skin friction = S u = α × c × A pAdhesion factor α, is not given. Hannoush play in hindi full. Use the method given by American Petroleum Institute (API).α = 1.0, for clays with cohesion = 70 kN/m 2Since, the cohesion is 35 kN/m 2 interpolate to obtain the adhesion factor ( α). Pile in multiple layers. SolutionFind the skin friction at the top of the clay layer and bottom of the clay layer. Obtain the average of the two values.Step 1:At point A: σ′ = 3 × 17 = 51 kN/m 2; (1065 lbs/ft.
2)S u = 25 kN/m 2;S u/ σ′ = 25/51 = 0.50From the Kolk and Van der Velde table; α = 0.65At point B: σ′ = 6 × 17 = 102 kN/m 2.S u = 25 kN/m 2;S u/ σ′ = 25/102 = 0.25From the Kolk and Van der Velde table; α = 0.86Step 2:Ultimate skin friction at point A:f ult = α × S u = 0.65 × 25 = 16 kN/m 2Ultimate skin friction at point B:f ult = α × S u = 0.86 × 25 = 21 kN/m 2Assume the average of two points to obtain the total skin friction.Average = (16 + 21)/2 = 18.5 kN/m 2.Total skin friction = 18.5 × perimeter × length = 18.5 × ( π × 0.5) × 3 = 87 kN (19.6 kips)Summary of equations. Monteith, Mike H. Unsworth, in, 2013 9.1.2 Form DragIn addition to the force exerted by skin friction, which is a consequence of momentum transfer to a surface across the streamlines of flow, bodies immersed in moving fluids also experience a force in the direction of the flow as a result of the deceleration of fluid. This force is known as form drag because it depends on the shape and orientation of the body. Maximum form drag is experienced by surfaces at right angles to the fluid flow, and the force can be estimated by assuming that there is a point on the surface where the fluid is instantaneously brought to rest after being uniformly decelerated from a velocity V (remembering that Newton’s Second Law may be expressed as force = rate of change of momentum). If the initial momentum per unit volume of fluid is ρ V and the mean velocity during deceleration is V / 2, the rate at which momentum is lost from the fluid is ρ V × V / 2 = 0.5 ρ V 2. This is the maximum rate at which momentum can be transferred to unit area on the upstream surface of a bluff body and it is therefore the maximum pressure excess (pressure = force per unit area) that a fluid can exert in contributing to the total form drag over the body.
In practice, fluid tends to slip round the sides of a bluff body so that a form drag force smaller than 0.5 ρ V 2 is exerted on the upstream face. However, in the wake which forms downstream of a bluff body, the fluid pressure is less than in the free stream, and the associated suction force often makes an important additional contribution to the total form drag on a body.
(As examples, gulls diving for fish assume a “streamlined” shape that minimizes the wake to maximize their fall speed, and in the sporting world, golf ball manufacturers adopt complex patterns for the dimples on golf balls to reduce the wake and consequently to reduce the drag.) The total form drag on unit area is conveniently expressed as c f. 0.5 ρ V 2 where c f is a drag coefficient.
Hence c f is the fraction of the maximum bluff body pressure that a fluid can exert.In most problems, it is appropriate to combine skin friction and form drag to give a total force τ, usually the force on a unit area projected in the direction of the flow, i.e. 2 rl for a cylinder of radius r and length l in cross-flow and π r 2 for a sphere. The ratio τ / ( 0.5 ρ V 2 ) then defines the total drag coefficient c d, which for spheres and for cylinders at right angles to the flow lies between 0.4 and 1.2 in the range of Reynolds numbers between 10 2 and 10 5. Manufacturers of energy-efficient cars boast drag coefficients of about 0.25.As background for later discussion of mass and heat transfer, it should be noted that the diffusion of momentum in skin friction is analogous to the diffusion of gas molecules and of heat, provided the surface is parallel to the airstream. For such a surface, close relations may therefore be expected between r M, r H, and r V.
For a surface at right angles to the airstream however, there is no frictional force in the direction of flow. Friction will operate in all directions at right angles to the flow but the net sum of all these (vector) forces must be zero. In contrast, the net flux of heat or mass, which are scalar quantities, must be finite in the plane of the surface. In this case, r V and r H may be similar to each other but will be unrelated to the value of r M. Figure 46.15.
Concrete caisson with a W-section.The following parameters are given:Ultimate steel compressive strength = 36,000 psiUltimate concrete compressive strength = 3,000 psi Simplified design procedureSimplified design procedure is first explained. In this procedure, the composite nature of the section is ignored.Step 1: Structural design of the caissonAssume a diameter of 30 in. For the concrete caisson. Since, E (elastic modulus) of steel is much higher than concrete, a major portion of the load is taken by steel. Assume that 90% of the load is carried by steel.Load carried by steel = 0.9 × 1000 tons = 900 tonsAllowable steel compressive strength = 0.5 × 36,000 = 18,000 psi(NYC Building Code, Table 11.3 recommends a factor of safety of 0.5. Check your local building code for the factor of safety value).
Figure 46.16. I-beam.Step 2: Check whether this section fits inside a 30 in. Hole.Distance along a diagonal (Pythagoras theorem) = (16.36 2 + 17.54 2) 1/2 = 23.98 in.This value is smaller than 30 in. Hence, the section can easily fit inside a 30 in. Hole.Step 3: Compute the load carried by concrete.Concrete area = area of the hole − area of steel = 706.8 − 101 = 605.8.Allowable concrete compressive strength = 0.25 × ultimate compressive strength = 0.25 × 3000 = 750 psi(NYC Building Code, Table 11.3 recommends a factor of safety of 0.25.
1.3 The large eddies in a turbulent flow have a length scale t, a velocity scale v=u, and a time scale te-Yu. The smallest eddies have a length 26 Introduction scale a velocity scale u, and a time scale τ. Estimate the characteristic velocity vir) and the characteristic time tr) of eddies of size r, where r is any length in the range η. Gravograph drivers.