s u (mm) D r 1 0.1a D D r ) E p 5 Young s modulus of pile material (kn>mm 2 )

Transcription

1 11.1 Pile Load Tests 585 Pile settlement may increase with load to a certain point, beyond which the load settlement curve becomes vertical. The load corresponding to the point where the curve of Q versus s net becomes vertical is the ultimate load, Q u, for the pile; it is shown by curve 1 in Figure 11.3c. In many cases, the latter stage of the load settlement curve is almost linear, showing a large degree of settlement for a small increment of load; this is shown by curve in the figure. The ultimate load, Q u, for such a case is determined from the point of the curve of Q versus s net where this steep linear portion starts. One of the methods to obtain the ultimate load Q u from the load-settlement plot is that proposed by Davisson (1973). Davisson s method is used more often in the field and is described here. Referring to Figure 11., the ultimate load occurs at a settlement level (s u ) of s u (mm) D r 1 0.1a D D r b 1 Q ul A p E p (11.67) where Q u is in kn D is in mm D r 5 reference pile diameter or width ( 5 300mm) L 5 pile length (mm) A of pile cross section (mm p 5 area ) E p 5 Young s modulus of pile material (kn>mm ) Q u 0.1 D r D/D r Load, Q (kn) Q u L AE Eq. (11.67) Settlement, s (mm) Figure 11. Davisson s method for determination of Q u

2 586 Chapter 11: Pile Foundations The application of this procedure is shown in Example The load test procedure just described requires the application of step loads on the piles and the measurement of settlement and is called a load-controlled test. Another technique used for a pile load test is the constant-rate-of-penetration test, wherein the load on the pile is continuously increased to maintain a constant rate of penetration, which can vary from 0.5 to.5 mm>min (0.01 to 0.1 in.>min). This test gives a load settlement plot similar to that obtained from the load-controlled test. Another type of pile load test is cyclic loading, in which an incremental load is repeatedly applied and removed. In order to conduct a load test on piles, it is important to take into account the time lapse after the end of driving (EOD). When piles are driven into soft clay, a certain zone surrounding the clay becomes remolded or compressed, as shown in Figure 11.5a. This results in a reduction of undrained shear strength, c u (Figure 11.5b). With time, the loss of undrained shear strength is partially or fully regained. The time lapse may range from 30 to 60 days. For piles driven in dilative (dense to very dense) saturated fine sands, relaxation is possible. Negative pore water pressure, if developed during pile driving, will dissipate over time, resulting in a reduction in pile capacity with time after the driving operation is completed. At the same time, excess pore water pressure may be generated in contractive fine Pile Remolded zone Compressed zone Intact zone 1.5 D 0.5 D D diameter 0.5 D 1.5 D (a) c u Remolded zone Compressed zone Intact zone Sometime after driving Immediately after driving (b) Distance from pile Figure 11.5 (a) Remolded or compacted zone around a pile driven into soft clay; (b) Nature of variation of undrained shear strength (c u ) with time around a pile driven into soft clay

3 11.1 Pile Load Tests 587 sands during pile driving. The excess pore water pressure will dissipate over time, which will result in greater pile capacity. Several empirical relationships have been developed to predict changes in pile capacity with time. Skov and Denver (1988) Skov and Denver proposed the equation Q t 5 Q OED ca log a t t o b 1 1d (11.68) where Q t 5 pile capacity t days after the end of driving Q OED 5 pile capacity at the end of driving t 5 time, in days For sand, A 5 0. and t o days; for clay, A and t o days. Guang-Yu (1988) According to Guang-Yu, Q 1 5 (0.375S t 1 1)Q OED (applicable to clay soil) (11.69) where Q 1 5 pile capacity 1 days after pile driving S t 5 sensitivity of clay Svinkin (1996) Svinkin suggests the relationship Q t 5 1.Q OED t 0.1 (upper limit for sand) Q t Q OED t 0.1 (lower limit for sand) (11.70) (11.71) where t 5 time after driving, in days Example 11.9 Figure 11.6 shows the load test results of a 0-m long concrete pile ( 06 mm 3 06 mm) embedded in sand. Using Davisson s method, determine the ultimate load Q u. Given: E p kn>m. Solution From Eq. (11.67), s u D r 1 0.1a D D r b 1 Q ul A p E p

4 588 Chapter 11: Pile Foundations D r mm, D 5 06 mm, L 5 0 m 5 0,000 mm, A p 5 06 mm 3 06 mm 5 16,836 mm, and E p kn>m. Hence, s u 5 (0.01)(300) 1 (0.1) a b 1 (Q u)(0,000) (30)(16,836) Q u Q u The line s u (mm) Q u is drawn in Figure The intersection of this line with the load-settlement curve gives the failure load Q u kn. 800 Q u = 160 kn Q (kn) mm Settlement, s (mm) Figure Elastic Settlement of Piles The total settlement of a pile under a vertical working load Q w is given by s e 5 s e(1) 1 s e() 1 s e(3) (11.7)

5 where Elastic Settlement of Piles 589 s e(1) 5 elastic settlement of pile s e() 5 settlement of pile caused by the load at the pile tip s e(3) 5 settlement of pile caused by the load transmitted along the pile shaft If the pile material is assumed to be elastic, the deformation of the pile shaft can be evaluated, in accordance with the fundamental principles of mechanics of materials, as s e(1) 5 (Q wp 1jQ ws )L A p E p (11.73) where Q wp 5 load carried at the pile point under working load condition Q ws 5 load carried by frictional (skin) resistance under working load condition A p 5 area of cross section of pile L 5 length of pile E p 5 modulus of elasticity of the pile material The magnitude of j varies between 0.5 and 0.67 and will depend on the nature of the distribution of the unit friction (skin) resistance f along the pile shaft. The settlement of a pile caused by the load carried at the pile point may be expressed in the form: s e() 5 q wpd (1 m E s )I wp s (11.7) where D 5 width or diameter of pile q wp 5 point load per unit area at the pile point 5 Q wp >A p E s 5 modulus of elasticity of soil at or below the pile point m s 5 Poisson s ratio of soil I wp 5 influence factor < 0.85 Vesic (1977) also proposed a semi-empirical method for obtaining the magnitude of the settlement of s e(). His equation is s e() 5 Q wpc p Dq p (11.75) where q p 5 ultimate point resistance of the pile C p 5 an empirical coefficient Representative values of C p for various soils are given in Table

6 590 Chapter 11: Pile Foundations C p Table Typical Values of [from Eq. (11.75)] Type of soil Driven pile Bored pile Sand (dense to loose) Clay (stiff to soft) Silt (dense to loose) From Design of Pile Foundations, by A. S. Vesic. SYNTHESIS OF HIGHWAY PRACTICE by AMERICAN ASSOCIATION OF STATE HIGHWAY AND TRANSPORT. Copyright 1969 by TRANSPORTATION RESEARCH BOARD. Reproduced with permission of TRANSPORTATION RESEARCH BOARD in the format Textbook via Copyright Clearance Center. The settlement of a pile caused by the load carried by the pile shaft is given by a relation similar to Eq. (11.7), namely, s e(3) 5 Q ws pl D E s (1 m s )I ws (11.76) where p 5 perimeter of the pile L 5 embedded length of pile I ws 5 influence factor Note that the term Q ws >pl in Eq. (11.76) is the average value of f along the pile shaft. The influence factor, I ws, has a simple empirical relation (Vesic, 1977): L I ws (11.77) Å D Vesic (1977) also proposed a simple empirical relation similar to Eq. (11.75) for obtaining s e(3) : s e(3) 5 Q wsc s Lq p (11.78) In this equation, C s 5 an empirical constant 5 ( "L>D)C p (11.79) The values of for use in Eq. (11.75) may be estimated from Table C p Example The allowable working load on a prestressed concrete pile 1-m long that has been driven into sand is 50 kn. The pile is octagonal in shape with D mm (see Table 11.3a). Skin resistance carries 350 kn of the allowable load, and point bearing carries the rest. Use E p kn>m, E s kn>m, m s 0.35, and j50.6. Determine the settlement of the pile.

7 11.16 Laterally Loaded Piles 591 Solution From Eq. (11.73), From Table 11.3a for D mm, the area of pile cross section. A p cm, Also, perimeter p m. Given: Q ws kn, so From Eq. (11.7), Again, from Eq. (11.76), Q wp kn S e(1) 5 (Q wp 1jQ ws )L A p E p (350)(1) s e(1) m mm (0.105 m )( ) s e() 5 q wpd (1 m s)i E wp s ( )(0.85) m mm Hence, total settlement is s e(3) 5 Q ws pl D (1 m s)i E ws s L I ws Å D Å s e(3) 5 B R (1.168)(1) ( )(.69) m mm s e 5 s e(1) 1 s e() 1 s e(3) mm Laterally Loaded Piles A vertical pile resists a lateral load by mobilizing passive pressure in the soil surrounding it. (See Figure 11.1c.) The degree of distribution of the soil s reaction depends on (a) the stiffness of the pile, (b) the stiffness of the soil, and (c) the fixity of the ends of the pile. In general, laterally loaded piles can be divided into two major categories: (1) short or rigid piles and () long or elastic piles. Figures 11.7a and 11.7b show the nature of the variation of the pile deflection and the distribution of the moment and shear force along the pile length when the pile is subjected to lateral loading. We next summarize the current solutions for laterally loaded piles. Elastic Solution A general method for determining moments and displacements of a vertical pile embedded in a granular soil and subjected to lateral load and moment at the ground surface was given by Matlock and Reese (1960). Consider a pile of length L subjected to a lateral force Q g and

8 59 Chapter 11: Pile Foundations Deflection Shear Moment Q g M g (a) Loading Deflection Moment Shear Q g M g z (b) Figure 11.7 Nature of variation of pile deflection, moment, and shear force for (a) a rigid pile and (b) and elastic pile Q g M g x p z L (a) z (b) x x x M x V x p x z z z (c) z z Figure 11.8 (a) Laterally loaded pile; (b) soil resistance on pile caused by lateral load; (c) sign conventions for displacement, slope, moment, shear, and soil reaction

9 11.16 Laterally Loaded Piles 593 a moment M g at the ground surface (z 5 0), as shown in Figure 11.8a. Figure 11.8b shows the general deflected shape of the pile and the soil resistance caused by the applied load and the moment. According to a simpler Winkler s model, an elastic medium (soil in this case) can be replaced by a series of infinitely close independent elastic springs. Based on this assumption, where k 5 modulus of subgrade reaction pr 5 pressure on soil x 5 deflection k 5 pr(kn>m) x(m) The subgrade modulus for granular soils at a depth z is defined as k z 5 n h z (11.80) (11.81) where n h 5 constant of modulus of horizontal subgrade reaction. Referring to Figure 11.8b and using the theory of beams on an elastic foundation, we can write d x E p I p (11.8) dz 5 pr where E p 5 modulus of elasticity in the pile material I p 5 moment of inertia of the pile section Based on Winkler s model pr 5kx (11.83) The sign in Eq. (11.83) is negative because the soil reaction is in the direction opposite that of the pile deflection. Combining Eqs. (11.8) and (11.83) gives E p I p d x dz 1 kx 5 0 The solution of Eq. (11.8) results in the following expressions: Pile Deflection at Any Depth [x z (z)] (11.8) Q g T 3 M g T x z (z) 5 A x 1 B E p I x p E p I p (11.85) Slope of Pile at Any Depth [ u z (z)] Q g T M g T u z (z) 5 A u 1 B E p I u p E p I p (11.86)

10 59 Chapter 11: Pile Foundations Moment of Pile at Any Depth [M z (z)] M z (z) 5 A m Q g T 1 B m M g (11.87) Shear Force on Pile at Any Depth [V z (z)] V z (z) 5 A v Q g 1 B v M g T (11.88) Soil Reaction at Any Depth [ p9 z (z)] p z r(z) 5 A pr Q g T 1 B pr M g T (11.89) where A x, B x, A u, B u, A m, B m, A v, B v, A pr, and B pr are coefficients T 5 characteristic length of the soil pile system n h has been defined in Eq. (11.81) 5 Å 5 E pi p n h (11.90) When L $ 5T, the pile is considered to be a long pile. For L # T, the pile is considered to be a rigid pile. Table 11.1 gives the values of the coefficients for long piles (L>T $ 5) in Eqs. (11.85) through (11.89). Note that, in the first column of the table, Z 5 z T (11.91) is the nondimensional depth. The positive sign conventions for x z (z), u z (z), M z (z), V z (z), and p z r(z) assumed in the derivations in Table 11.1 are shown in Figure 11.8c. Figure 11.9 shows the variation of A x, B x, A m, and B m for various values of L>T 5 Z max. It indicates that, when L>T is greater than about 5, the coefficients do not change, which is true of long piles only. Calculating the characteristic length T for the pile requires assuming a proper value of n h. Table gives some representative values. Elastic solutions similar to those given in Eqs through for piles embedded in cohesive soil were developed by Davisson and Gill (1963). Their equations are Q g R 3 M g R x z (z) 5 A x r 1 B E p I x r p E p I p (11.9)

11 11.16 Laterally Loaded Piles 595 Table 11.1 Coefficients for Long Piles, k z 5 n h z Z A x A u A m A v A9 p From Drilled Pier Foundations, by R. J. Woodward, W. S. Gardner, and D. M. Greer. Copyright 197 McGraw-Hill. Used with permission of the McGraw-Hill Book Company. B x B u B m B v B9 p Table Representative Values of Soil n h kn>m 3 n h Dry or moist sand Loose Medium Dense 15,000 18,000 Submerged sand Loose Medium Dense ,000 and M z (z) 5 A m r Q g R 1 B m r M g (11.93) where A x r, B x, A m r, and B m r are coefficients. and R 5 É E p I p k (11.9)

12 596 Chapter 11: Pile Foundations Z max 1 A x Z z/t (a) Z max 1 B x B m Figure 11.9 Variation of A x, B x, A m, and with Z (From Matlock, H. and Reese, L. C. (1960). Generalized Solution for Laterally Loaded Piles, Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 86, No. SM5, Part I, pp With permission from ASCE.) Z z/t 3 5, 10 (b) The values of the coefficients Ar and Br are given in Figure Note that Z 5 z R (11.95) and Z max 5 L R (11.96)

13 11.16 Laterally Loaded Piles A m Z max Z z/t (c) 0 B m Z max Z z/t (d) Figure 11.9 (continued) The use of Eqs. (11.9) and (11.93) requires knowing the magnitude of the characteristic length, R. This can be calculated from Eq. (11.9), provided that the coefficient of the subgrade reaction is known. For sands, the coefficient of the subgrade reaction was given by Eq. (11.81), which showed a linear variation with depth. However, in cohesive

14 598 Chapter 11: Pile Foundations A x, A m Z max 5 Z 3 Z max 3 5 A x 5 (a) A m B x, B m , 5 Z max 1 3 Z max Z Figure Variation of A x, B x r, A m r, and B m r with Z (From Davisson, M. T. and Gill, H. L. (1963). Laterally Loaded Piles in a Layered Soil System, Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 89, No. SM3, pp With permission from ASCE.) 5 (b) B x B m

15 11.16 Laterally Loaded Piles 599 soils, the subgrade reaction may be assumed to be approximately constant with depth. Vesic (1961) proposed the following equation to estimate the value of k: Here, E s 5 modulus of elasticity of soil D 5 pile width (or diameter) m s 5 Poisson s ratio for the soil k E s D Å E p I p For all practical puroses, Eq. (11.97) can be written as k < E s 1 m s E s 1 m s (11.97) (11.98) Ultimate Load Analysis: Broms s Method For laterally loaded piles, Broms (1965) developed a simplified solution based on the assumptions of (a) shear failure in soil, which is the case for short piles, and (b) bending of the pile, governed by the plastic yield resistance of the pile section, which is applicable to long piles. Broms s solution for calculating the ultimate load resistance, Q u(g), for short piles is given in Figure 11.31a. A similar solution for piles embedded in cohesive soil is shown in Figure 11.31b. In Figure 11.31a, note that Q u(g) Q u(g) e L L D Restrained pile Free-headed pile 60 0 e D Q u(g) K p D 3 Ultimate lateral resistance, Restrained pile Free-headed pile 0 e L Ultimate lateral resistance, Q u(g) c u D Restrained pile Free-headed pile Length, L D Embedment length, L D (a) (b) Figure Broms s solution for ultimate lateral resistance of short piles (a) in sand and (b) in clay

16 600 Chapter 11: Pile Foundations K p 5 Rankine passive earth pressure coefficient 5 tan 5 1 fr (11.99) Similarly, in Figure 11.31b, c u 5 undrained cohesion < 0.75q u FS q u q u (11.100) where FS 5 factor of safety(5) q u 5 unconfined compression strength Figure 11.3 shows Broms s analysis of long piles. In the figure, the yield moment for the pile is where S 5 section modulus of the pile section F Y 5 yield stress of the pile material M y 5 SF Y (11.101) In solving a given problem, both cases (i.e., Figure and Figure 11.3) should be checked. The deflection of the pile head, x z (z 5 0), under working load conditions can be estimated from Figure In Figure 11.33a, the term h can be expressed as h 5 5 Å E p I p (11.10) The range of n h for granular soil is given in Table Similarly, in Figure 11.33b, which is for clay, the term K is the horizontal soil modulus and can be defined as n h K 5 pressure (kn>m ) displacement (m) (11.103) Also, the term b can be defined as b 5 KD Å E p I p (11.10) Note that, in Figure 11.33, Q g is the working load. The following is a general range of values of K for clay soils. Unconfined compression strength, q u kn>m K kn>m ,000 0, ,000 0, ,000

17 11.16 Laterally Loaded Piles Q u(g) K p D 3 Ultimate lateral resistance, e 0 D Restrained pile Free-headed pile ,000.0 Yield moment, M y (a) D K p Ultimate lateral resistance, Q u(g) c u D e 0 D 1 Restrained pile Free-headed pile M y Yield moment, c u D 3 (b) Figure 11.3 Broms s solution for ultimate lateral resistance of long piles (a) in sand (b) in clay

18 60 Chapter 11: Pile Foundations 10 Dimensionless lateral deflection, x z(z 0)(E p I p ) 3/5 (n h ) /5 Q g L Restrained pile Free-headed pile e.0 L Dimensionless length, L (a) 10 e 0. L 8 0. Dimensionless lateral deflection, x z(z 0)KDL Q g 6 Free-headed pile Restrained pile Dimensionless length, L (b) Figure Broms s solution for estimating deflection of pile head (a) in sand and (b) in clay

19 11.16 Laterally Loaded Piles 603 Example Consider a steel H-pile (HP ) 5 m long, embedded fully in a granular soil. Assume that n h 5 1,000 kn>m 3. The allowable displacement at the top of the pile is 8 mm. Determine the allowable lateral load, Q g. Let M g 5 0. Use the elastic solution. Solution From Table 11.1a, for an HP pile, I p m (about the strong axis) and let E p kn>m From Eq. (11.90), T 5 5 Å E p I p n h 5 5 ( )( ) m Å 1,000 Here, L>T 5 5> , Eq. (11.85) takes the form so the pile is a long one. Because M g 5 0, Q g T 3 x z (z) 5 A x E p I p and it follows that Q g 5 x z(z)e p I p A x T 3 At z 5 0, x z 5 8 mm m and A x 5.35 (see Table 11.1), so Q g 5 (0.008)( )( ) (.35)( ) kn This magnitude of Q g is based on the limiting displacement condition only. However, the magnitude of Q g based on the moment capacity of the pile also needs to be determined. For M g 5 0, Eq. (11.87) becomes M z (z) 5 A m Q g T According to Table 11.1, the maximum value of maximum allowable moment that the pile can carry is M z(max) 5 F Y I p d 1 A m at any depth is The

20 60 Chapter 11: Pile Foundations Let F From Table 11.1a, I p m Y 5 8,000 kn>m. and d m, so Now, I p d m 3 Q g 5 M z(max) A m T 5 ( )(8,000) kn (0.77)(1.16) Because Q g kn kn, the deflection criteria apply. Hence, Q g kn. Example 11.1 Solve Example by Broms s method. Assume that the pile is flexible and is free headed. Let the yield stress of the pile material, F y 5 8 MN>m ; the unit weight of soil, g518 kn>m 3 ; and the soil friction angle fr Solution We check for bending failure. From Eq. (11.101), From Table 11.1a, Also, S 5 I p 5 d 1 M y 5 SF y and M y ( ) 5 0. kn-m M y M y 5 D gk p D g tan 5 1 fr 5 0. (0.5) (18) tan From Figure 11.3a, for M the magnitude of Q u(g) >K p D 3 y >D gk p , g (for a free-headed pile with e>d 5 0) is about 10, so Q u(g) 5 10K p D 3 g510 tan (0.5)3 (18) kn

21 11.16 Laterally Loaded Piles 605 Next, we check for pile head deflection. From Eq. (11.10), so From Figure 11.33a, for hl 5 1.5, e>l 5 0 (free-headed pile): thus, and h5 Å 5 n h 1, E p I p Å ( )( m1 ) x o (E p I p ) 3>5 (n h ) >5 < 0.15 (by interpolation) Q g L Q g 5 x o(e p I p ) 3>5 (n h ) >5 0.15L hl 5 (0.86)(5) (0.008)3( )( ) 3>5 (1,000) >5 (0.15)(5) 5 0. kn Hence, Q g 5 0. kn (*15. kn). Example Assume that the 5-m long pile described in Example is a restrained pile and is embedded in clay soil. Given: c and K 5 5,000 kn>m 3 u kn>m. The allowable lateral displacement at the top of the pile is 10 mm. Determine the allowable lateral load. Given M y m g 5 0. Use Broms s method. Q g Solution From Example 11.1, M y 5 0. kn-m. So For the unrestrained pile, from Figure 11.3b, or M y c u D (100)(0.5) Q u(g) c u D < 65 Q u(g) 5 (65)(100)(0.5) kn

22 606 Chapter 11: Pile Foundations Check Pile-Head Deflection From Eq. (11.10), b5 Å KD E p I p 5 Å (5000)(0.5) ()( )( ) bl 5 (0.33)(5) From Figure 11.33b for bl , by extrapolation the magnitude of Q g 5 x z(z 5 0)KDL 8 x z (z 5 0)KDL Q g < 8 5 a b (5000)(0.5)(5) Hence, Q g kn(* 19.3 kn) kn Pile-Driving Formulas To develop the desired load-carrying capacity, a point bearing pile must penetrate the dense soil layer sufficiently or have sufficient contact with a layer of rock. This requirement cannot always be satisfied by driving a pile to a predetermined depth, because soil profiles vary. For that reason, several equations have been developed to calculate the ultimate capacity of a pile during driving. These dynamic equations are widely used in the field to determine whether a pile has reached a satisfactory bearing value at the predetermined depth. One of the earliest such equations commonly referred to as the Engineering News (EN) Record formula is derived from the work energy theory. That is, Energy imparted by the hammer per blow 5 (pile resistance)(penetration per hammer blow) According to the EN formula, the pile resistance is the ultimate load Q u, expressed as Q u 5 W Rh S 1 C (11.105) where W R 5 weight of the ram h 5 height of fall of the ram S 5 penetration of pile per hammer blow C 5 a constant

23 11.17 Pile-Driving Formulas 607 The pile penetration, S, is usually based on the average value obtained from the last few driving blows. In the equation s original form, the following values of C were recommended: For drop hammers, For steam hammers, C 5 5. mm if S and h are in mm C 5.5 mm if S and h are in mm Also, a factor of safety FS 5 6 was recommended for estimating the allowable pile capacity. Note that, for single- and double-acting hammers, the term W R h can be replaced by EH E, where E is the efficiency of the hammer and H E is the rated energy of the hammer. Thus, Q u 5 EH E S 1 C (11.106) The EN formula has been revised several times over the years, and other pile-driving formulas also have been suggested. Three of the other relationships generally used are tabulated in Table The maximum stress developed on a pile during the driving operation can be estimated from the pile-driving formulas presented in Table To illustrate, we use the modified EN formula: Q u 5 EW Rh S 1 C W R 1 n W p W R 1 W p In this equation, S is the average penetration per hammer blow, which can also be expressed as S 5 5. N (11.107) where S is in mm N 5 number of hammer blows per 5. mm of penetration Thus, Q u 5 EW R h (5.>N) 1.5 W R 1 n W p W R 1 W p (11.108) Different values of N may be assumed for a given hammer and pile, and may be calculated. The driving stress Q u >A p can then be calculated for each value of N. This Q u

24 608 Chapter 11: Pile Foundations Table Pile-Driving Formulas Name Modified EN formula Formula Q u 5 EW Rh W R 1 n W p S 1 C W R 1 W p where E 5 efficiency of hammer C 5.5 mm if the units of S and h are in mm W p 5 weight of the pile n 5 coefficient of restitution between the ram and the pile cap Typical values for E Single- and double-acting hammers Diesel hammers Drop hammers Typical values for n Cast-iron hammer and concrete piles (without cap) Wood cushion on steel piles Wooden piles Danish formula (Olson and Flaate, 1967) Q u 5 EH E EH E L S 1 É A p E p where E 5 efficiency of hammer H E 5 rated hammer energy E p 5 modulus of elasticity of the pile material L 5 length of the pile A p 5 cross-sectional area of the pile Janbu s formula (Janbu, 1953) Q u 5 EH E K u rs where K u r 5 C d 1 1 Å 1 1 lr C d C d W p W R lr 5 EH EL A p E p S

25 11.17 Pile-Driving Formulas 609 procedure can be demonstrated with a set of numerical values. Suppose that a prestressed concrete pile. m in length has to be driven by a hammer. The pile sides measure 5 mm. From Table 11.3a, for this pile, The weight of the pile is If the weight of the cap is.98 kn, then For the hammer, let Assume that the hammer efficiency is 0.85 and that n Substituting these values into Eq. (11.108) yields Q u 5 (0.85)( ) 5. N 1.5 Now the following table can be prepared: A p m A p Lg c 5 ( )(. m)(3.58 kn>m 3 ) kn W p kn Rated energy kn-m 5 H E 5 W R h Weight of ram 5. kn c. 1 (0.35) (0.08) d kip N 1.5 Q u A p Q u /A p N (kn) (m ) (MN/m ) Both the number of hammer blows per inch and the stress can be plotted in a graph, as shown in Figure If such a curve is prepared, the number of blows per inch of pile penetration corresponding to the allowable pile-driving stress can easily be determined. Actual driving stresses in wooden piles are limited to about 0.7f u. Similarly, for concrete and steel piles, driving stresses are limited to about 0.6f c r and 0.85f y, respectively. In most cases, wooden piles are driven with a hammer energy of less than 60 kn-m. Driving resistances are limited mostly to to 5 blows per inch of pile penetration. For concrete and steel piles, the usual values of N are 6 to 8 and 1 to 1, respectively.

26 610 Chapter 11: Pile Foundations 0 30 Q u /A p (MN/m ) Number of blows /5. mm (N) Figure 11.3 Plot of stress versus blows>5. mm. Example 11.1 A precast concrete pile m m in cross section is driven by a hammer. Given Maximum rated hammer energy kn-m Hammer efficiency Weight of ram kn Pile length 5.39 m Coefficient of restitution 5 0. Weight of pile cap 5.5 kn E p kn>m Number of blows for last 5. mm of penetration 5 8 Estimate the allowable pile capacity by the a. Modified EN formula (use FS 5 6) b. Danish formula (use FS 5 ) Solution Part a Q u 5 EW Rh S 1 C Weight of pile 1 cap 5 ( )(3.58 kn>m 3 ) kn W R 1 n W p W R 1 W p

27 11.18 Pile Capacity For Vibration-Driven Piles 611 Given: W R h kn-m Part b (0.8)( ) Q u Q all 5 Q u FS < 9.5 kn (0.) (55.95) kn Use E p kn>m Q u 5 EH E S 1 Ä EH E L A p E p EH E L (0.8)(0.67)(.39) 5 Å A p E p Å ( )( kn>m m mm ) Q u 5 (0.8)(0.67) < 1857 kn Q all < 6 kn Pile Capacity For Vibration-Driven Piles The principles of vibratory pile drivers (Figure 11.7e) were discussed briefly in Section 11.. As mentioned there, the driver essentially consists of two counterrotating weights. The amplitude of the centrifugal driving force generated by a vibratory hammer can be given as F c 5 mev (11.109) where m 5 total eccentric rotating mass e 5 distance between the center of each rotating mass and the center of rotation v5operating circular frequency Vibratory hammers typically include an isolated bias weight that can range from to 0 kn. The bias weight is isolated from oscillation by springs, so it acts as a net downward load helping the driving efficiency by increasing the penetration rate of the pile. The use of vibratory pile drivers began in the early 1930s. Installing piles with vibratory drivers produces less noise and damage to the pile, compared with impact driving. However, because of a limited understanding of the relationships between the load, the rate of penetration, and the bearing capacity of piles, this method has not gained popularity in the United States.

28 61 Chapter 11: Pile Foundations Vibratory pile drivers are patented. Some examples are the Bodine Resonant Driver (BRD), the Vibro Driver of the McKiernan-Terry Corporation, and the Vibro Driver of the L. B. Foster Company. Davisson (1970) provided a relationship for estimating the ultimate pile capacity in granular soil: In SI units, 0.76(H p ) 1 98(v p m>s) Q u (kn) 5 (v p m>s) 1 (S L m>cycle)(fhz) (11.110) where H p 5 horsepower delivered to the pile v p 5 final rate of pile penetration S L 5 loss factor f 5 frequency, in Hz The loss factor 1996): S L Closed-End Pipe Piles Loose sand: m>cycle Medium dense sand: m>cycle Dense sand: m>cycle for various types of granular soils is as follows (Bowles, H-Piles Loose sand: m>cycle Medium dense sand: m>cycle Dense sand: m>cycle In 000, Feng and Deschamps provided the following relationship for the ultimate capacity of vibrodriven piles in granular soil: Q u 5 3.6(F c 1 11W B ) v p c "OCR L E L (11.111) Here, F c 5 centrifugal force W B 5 bias weight v p 5 final rate of pile penetration c 5 speed of light m>min OCR 5 overconsolidation ratio L E 5 embedded length of pile L 5 pile length

29 11.19 Negative Skin Friction 613 Example Consider a 0-m-long steel pile driven by a Bodine Resonant Driver (Section HP ) in a medium dense sand. If H p horsepower, v p m>s, and f Hz, calculate the ultimate pile capacity, Q u. Solution From Eq. (11.110), Q u H p 1 98v p v p 1 S L f For an HP pile in medium dense sand, S L < m>cycle. So (0.76)(350) 1 (98)(0.0016) Q u kn ( )(115) Negative Skin Friction Negative skin friction is a downward drag force exerted on a pile by the soil surrounding it. Such a force can exist under the following conditions, among others: 1. If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will gradually consolidate. The consolidation process will exert a downward drag force on the pile (see Figure 11.35a) during the period of consolidation.. If a fill of granular soil is placed over a layer of soft clay, as shown in Figure 11.35b, it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pile. 3. Lowering of the water table will increase the vertical effective stress on the soil at any depth, which will induce consolidation settlement in clay. If a pile is located in the clay layer, it will be subjected to a downward drag force. z Clay fill H f Sand fill H f L L L 1 Sand z Clay Neutral plane (a) (b) Figure Negative skin friction

30 61 Chapter 11: Pile Foundations In some cases, the downward drag force may be excessive and cause foundation failure. This section outlines two tentative methods for the calculation of negative skin friction. Clay Fill over Granular Soil (Figure 11.35a) Similar to the b method presented in Section 11.1, the negative (downward) skin stress on the pile is f n 5 Krs o r tan dr (11.11) where Kr 5 earth pressure coefficient 5 K o 5 1 sin fr s o r 5 vertical effective stress at any depth z 5g f rz g f r 5 effective unit weight of fill dr 5 soil pile friction angle < fr Hence, the total downward drag force on a pile is Q n 5 3 H f 0 (pkrg f r tan dr)z dz 5 pkrg frh f tan dr (11.113) where H f 5 height of the fill. If the fill is above the water table, the effective unit weight, g f r, should be replaced by the moist unit weight. Granular Soil Fill over Clay (Figure 11.35b) In this case, the evidence indicates that the negative skin stress on the pile may exist from z 5 0 to z 5 L 1, which is referred to as the neutral depth. (See Vesic, 1977, pp. 5 6.) The neutral depth may be given as (Bowles, 198) L 1 5 (L H f) L 1 B L H f 1 g frh f R g frh f gr gr (11.11) where g f r and gr 5 effective unit weights of the fill and the underlying clay layer, respectively. For end-bearing piles, the neutral depth may be assumed to be located at the pile tip (i.e., L 1 5 L H f ). Once the value of L 1 is determined, the downward drag force is obtained in the following manner: The unit negative skin friction at any depth from z 5 0 to z 5 L 1 is where Kr 5 K o 5 1 sin fr s o r 5g f rh f 1grz dr fr f n 5 Krs o r tan dr (11.115)

31 11.19 Negative Skin Friction 615 L 1 L 1 Q n 5 3 pf n dz 5 3 pkr(g f rh f 1grz)tan dr dz (pkrg f rh f tan dr)l 1 1 L 1 pkrgr tan dr (11.116) If the soil and the fill are above the water table, the effective unit weights should be replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the downdrag zone to avoid this problem. A limited number of case studies of negative skin friction is available in the literature. Bjerrum et al. (1969) reported monitoring the downdrag force on a test pile at Sorenga in the harbor of Oslo, Norway (noted as pile G in the original paper). The study of Bjerrum et al. (1969) was also discussed by Wong and Teh (1995) in terms of the pile being driven to bedrock at 0 m. Figure 11.36a shows the soil profile and the pile. Wong and Teh estimated the following quantities: Fill: Moist unit weight, g f 5 16 kn>m 3 Saturated unit weight, g sat(f) kn>m 3 So g f r kn>m 3 and H f 5 13 m m f 16 kn/m 3 Fill 0 Axial force in pile (kn) m Groundwater table Fill sat ( f ) 18.5 kn/m 3 10 Pile D 500 mm 0 m Depth (m) 0 Clay 30 Rock (a) 0 (b) Figure Negative skin friction on a pile in the harbor of Oslo, Norway (Based on Bjerrum et al., (1969) and Wong and The (1995))

32 616 Chapter 11: Pile Foundations Clay: Kr tan dr < 0. Saturated effective unit weight, gr kn>m 3 Pile: L 5 0 m Diameter, D m Thus, the maximum downdrag force on the pile can be estimated from Eq. (11.116). Since in this case the pile is a point bearing pile, the magnitude of L m, and or Q n 5 (p)(kr tan dr)3g f 3 1 (13 )g f r(l 1 ) 1 L 1 pgr(kr tan dr) Q n 5 (p30.5)(0.) 3(16 3 ) 1 ( )(7) 1 (7) (p30.5)(9.19)(0.) 5 38 kn The measured value of the maximum Q n was about 500 kn (Figure 11.36b), which is in good agreement with the calculated value. Example In Figure 11.35a, let H f 5 m. The pile is circular in cross section with a diameter of m. For the fill that is above the water table, g f 5 16 kn>m 3 and fr 5 3. Determine the total drag force. Use dr 5 0.6fr. Solution From Eq. (11.113), with and Thus, Q n 5 pkrg fh f tan dr p 5p(0.305) m Kr 5 1 sin fr 5 1 sin dr 5 (0.6)(3) Q n 5 (0.958)(0.7)(16)() tan kn Example In Figure 11.35b, let H f 5 m, pile diameter m, g f kn>m 3, f, g sat(clay) kn>m 3 clay r 5 3, and L 5 0 m. The water table coincides with the top of the clay layer. Determine the downward drag force. Assume that dr 5 0.6f clay r.

33 11.0 Group Efficiency 617 Solution The depth of the neutral plane is given in Eq. (11.11) as Note that g f r in Eq. (11.11) has been replaced by g f because the fill is above the water table, so or Now, from Eq. (11.116), we have with and Hence, L 1 5 L 1 5 L H f L 1 (0 ) L 1 c (0 ) L 1 5. L 1 a L H f 1 1 g fh f gr 8.93; L m Q n 5 (pkrg f H f tan dr)l 1 1 L 1 pkrgr tan dr p 5p(0.305) m Kr 5 1 sin b g fh f gr (16.5)() ) d ()(16.5)() ( ) Q n 5 (0.958)(0.)(16.5)()3tan( )(11.75) 1 (11.75) (0.958)(0.)( )3tan( ) kn Group Piles 11.0 Group Efficiency In most cases, piles are used in groups, as shown in Figure 11.37, to transmit the structural load to the soil. A pile cap is constructed over group piles. The cap can be in contact with the ground, as in most cases (see Figure 11.37a), or well above the ground, as in the case of offshore platforms (see Figure 11.37b). Determining the load-bearing capacity of group piles is extremely complicated and has not yet been fully resolved. When the piles are placed close to each other, a

34 618 Chapter 11: Pile Foundations Section Pile cap Water table L d d Plan d d L d (b) B g L g d Number of piles in group n 1 n (Note: L g B g ) L g (n 1 1)d (D/) B g (n 1)d (D/) (a) (c) Figure Group piles reasonable assumption is that the stresses transmitted by the piles to the soil will overlap (see Figure 11.37c), reducing the load-bearing capacity of the piles. Ideally, the piles in a group should be spaced so that the load-bearing capacity of the group is not less than the sum of the bearing capacity of the individual piles. In practice, the minimum centerto-center pile spacing, d, is.5d and, in ordinary situations, is actually about 3 to 3.5D.

35 11.0 Group Efficiency 619 The efficiency of the load-bearing capacity of a group pile may be defined as h5 Q g(u) S Q u (11.117) where h5group efficiency Q g(u) 5 ultimate load-bearing capacity of the group pile Q u 5 ultimate load-bearing capacity of each pile without the group effect Many structural engineers use a simplified analysis to obtain the group efficiency for friction piles, particularly in sand. This type of analysis can be explained with the aid of Figure 11.37a. Depending on their spacing within the group, the piles may act in one of two ways: (1) as a block, with dimensions L g 3 B g 3 L, or () as individual piles. If the piles act as a block, the frictional capacity is f av p g L < Q g(u). [Note: p g 5 perimeter of the cross section of block 5 (n 1 1 n )d 1 D, and f av 5 average unit frictional resistance.] Similarly, for each pile acting individually, Q u < plf av. (Note: p 5 perimeter of the cross section of each pile.) Thus, h 5 Q g(u) 5 f av3(n 1 1 n )d 1 DL S Q u n 1 n plf av 5 (n 1 1 n )d 1 D pn 1 n (11.118) Hence, Q g(u) 5 B (n 1 1 n )d 1 D RS Q pn 1 n u (11.119) From Eq. (11.119), if the center-to-center spacing d is large enough, h.1. In that case, the piles will behave as individual piles. Thus, in practice, if h,1, then and if h$1, then Q g(u) 5hS Q u Q g(u) 5S Q u There are several other equations like Eq. (11.119) for calculating the group efficiency of friction piles. Some of these are given in Table It is important, however, to recognize that relationships such as Eq. (11.119) are simplistic and should not be used. In fact, in a group pile, the magnitude of f av depends on the location of the pile in the group (ex., Figure 11.38).

36 60 Chapter 11: Pile Foundations Table Equations for Group Efficiency of Friction Piles Name Converse Labarre equation Los Angeles Group Action equation Equation h51 B (n 1 1)n 1 (n 1)n 1 Ru 90n 1 n where u(deg) 5 tan 1 (D>d) h51 D pdn 1 n 3n 1 (n 1) 1 n (n 1 1) 1 "(n 1 1)(n 1) Seiler Keeney equation (Seiler and Keeney, 19) 11d h5b1 B 7(d 1) RBn 1 1 n 0.3 Rr 1 n 1 1 n 1 n 1 1 n where d is in ft 50 Sandy soil L = 18 D D = 50 mm Center pile Average skin friction, f (kn/m ) Border pile Corner pile D 3D Settlement (mm) Figure Average skin friction (f av ) based on pile location (Based on Liu et al., 1985)

37 11.1 Ultimate Capacity of Group Piles in Saturated Clay 61 3 Group efficiency, 1 D d d d d d D Figure Variation of efficiency of pile groups in sand (Based on Kishida and Meyerhof, 1965) Figure shows the variation of the group efficiency h for a group pile in sand (Kishida and Meyerhof, 1965). It can be seen that, for loose and medium sands, the magnitude of the group efficiency can be larger than unity. This is due primarily to the densification of sand surrounding the pile Ultimate Capacity of Group Piles in Saturated Clay Figure 11.0 shows a group pile in saturated clay. Using the figure, one can estimate the ultimate load-bearing capacity of group piles in the following manner: Step 1. Determine S Q u 5 n 1 n (Q p 1 Q s ). From Eq. (11.18), Q p 5 A p 39c u(p) Step. where c u(p) 5 undrained cohesion of the clay at the pile tip. Also, from Eq. (11.55), Q s 5S apc u DL So, S Q u 5 n 1 n 39A p c u(p) 1S apc u DL (11.10) Determine the ultimate capacity by assuming that the piles in the group act as a block with dimensions L g 3 B g 3 L. The skin resistance of the block is S p g c u DL 5S (L g 1 B g )c u DL Calculate the point bearing capacity: A p q p 5 A p c u(p) N c * 5 (L g B g )c u(p) N c *

38 6 Chapter 11: Pile Foundations Q g(u) (L g B g )c u L c u c u(1) L c u c u() c u c u(3) L g B g c u (p) N * c B g L g Figure 11.0 Ultimate capacity of group piles in clay Obtain the value of the bearing capacity factor N c * from Figure Thus, the ultimate load is S Q u 5 L g B g c u(p) N c * 1S (L g 1 B g )c u DL (11.11) Step 3. Compare the values obtained from Eqs. (11.10) and (11.11). The lower of the two values is Q g(u). N * c L g /B g L/B g Figure 11.1 Variation of N c * with L g >B g and L>B g

39 11.1 Ultimate Capacity of Group Piles in Saturated Clay 63 Example The section of a 3 3 group pile in a layered saturated clay is shown in Figure 11.. The piles are square in cross section (356 mm mm). The center-to-center spacing, d,of the piles is 889 mm. Determine the allowable load-bearing capacity of the pile group. Use FS 5. Note that the groundwater table coincides with the ground surface. Solution From Eq. (11.10), From Figure 11., c and c u() kn>m u(1) kn>m. For the top layer with c u(1) kn>m, From Table 11.10, a 1 < Similarly, SQ u 5 (3)() c (9)(0.356) (85.1) 1 (0.68)( )(50.3)(.57) d 1 (0.51)( )(85.1)(13.7) kn For piles acting as a group. SQ u 5 n 1 n 39A p c u(p) 1a 1 pc u(1) L 1 1a pc u() L c u(1) p a c u() p a < 0.85 a L g 5 (3)(0.889) m B g 5 ()(0.889) m G.W.T..57 m Clay c u 50.3 kn/m sat 17.6 kn/m m Clay c u 85.1 kn/m sat 19.0 kn/m mm Figure 11. Group pile of layered saturated clay

40 6 Chapter 11: Pile Foundations From Figure 11.1, N * c From Eq. (11.11), SQ u 5 L g B g c u(p) N * c 1S(L g 1 B g )c u DL 5(3.03)(.13)(85.1)(8.75) 1()( ) 3(50.3)(.57) 1 (85.1) (13.7) kn Hence, SQ u 5 1,011 kn. L g B g L B g SQ all 5 1,011 FS 5 1,011 < 3503 kn 11. Elastic Settlement of Group Piles In general, the settlement of a group pile under a similar working load per pile increases with the width of the group (B g ) and the center-to-center spacing of the piles (d). Several investigations relating to the settlement of group piles have been reported in the literature, with widely varying results. The simplest relation for the settlement of group piles was given by Vesic (1969), namely, s g(e) 5 É B g D s e (11.1) where s g(e) 5 elastic settlement of group piles B g 5 width of group pile section D 5 width or diameter of each pile in the group s e 5 elastic settlement of each pile at comparable working load (see Section 11.15) For group piles in sand and gravel, for elastic settlement, Meyerhof (1976) suggested the empirical relation s g(e) (mm) q"B gi N 60 (11.13)

41 11. Elastic Settlement of Group Piles 65 where q 5 Q g >(L g B g )(in kn>m ) (11.1) and L g and B g 5 length and width of the group pile section, respectively (m) N 60 5 average standard penetration number within seat of settlement ( <B g deep below the tip of the piles) I 5 influence factor 5 1 L>8B g > 0.5 (11.15) L 5 length of embedment of piles (m) Similarly, the group pile settlement is related to the cone penetration resistance by the formula S g(e) 5 qb gi q c (11.16) where q c 5 average cone penetration resistance within the seat of settlement. (Note that, in Eq. (11.16), all quantities are expressed in consistent units.) Example Consider a 3 3 group of prestressed concrete piles, each 1 m long, in a sand layer. The details of each pile and the sand are similar to that described in Example The working load for the pile group is 60 kn (3 3 3 Q all where Q all 5 50 kn as in Example 11.10), and d>d 5 3. Estimate the elastic settlement of the pile group. Use Eq. (11.13). Solution s e(g) 5 É B g D s e B g 5 (3 1)d 1 D 5 ()(3D) 1 D 5 7D 5 (7)(0.356 m) 5.9 m From Example 11.10, s e mm. Hence,.9 s e(g) 5 (19.69) mm É 0.356

42 66 Chapter 11: Pile Foundations 11.3 Consolidation Settlement of Group Piles The consolidation settlement of a group pile in clay can be estimated by using the :1 stress distribution method. The calculation involves the following steps (see Figure 11.3): Step 1. Let the depth of embedment of the piles be L. The group is subjected to a total load of Q g. If the pile cap is below the original ground surface, Q g equals the total load of the superstructure on the piles, minus the effective weight of soil above the group piles removed by excavation. Step. Assume that the load Q g is transmitted to the soil beginning at a depth of L>3 from the top of the pile, as shown in the figure. The load Q g spreads out along two vertical to one horizontal line from this depth. Lines aar and bbr are the two :1 lines. Step 3. Calculate the increase in effective stress caused at the middle of each soil layer by the load Q g. The formula is Dsr i 5 Q g (B g 1 z i )(L g 1 z i ) (11.17) B g L g Q g Groundwater table L a b 3 L Clay layer 1 Clay layer L 1 z V:1H Clay layer 3 V:1H L a Clay layer Rock b L 3 Figure 11.3 Consolidation settlement of group piles

43 11.3 Consolidation Settlement of Group Piles 67 Step. where Dsr i 5 increase in effective stress at the middle of layer i L g, B g 5 length and width, respectively of the planned group piles z i 5 distance from z 5 0 to the middle of the clay layer i For example, in Figure 11.3, for layer, z i 5 L 1 >; for layer 3, z i 5 L 1 1 L >; and for layer, z i 5 L 1 1 L 1 L 3 >. Note, however, that there will be no increase in stress in clay layer 1, because it is above the horizontal plane (z 5 0) from which the stress distribution to the soil starts. Calculate the consolidation settlement of each layer caused by the increased stress. The formula is De (i) Ds c(i) 5 B RH 1 1 e i o(i) (11.18) Step 5. where Ds c(i) 5 consolidation settlement of layer i De (i) 5 change of void ratio caused by the increase in stress in layer i e o(i) 5 initial void ratio of layer i (before construction) H i 5 thickness of layer i (Note: In Figure 11.3, for layer, H i 5 L 1 ; for layer 3, H i 5 L ; and for layer, H i 5 L 3.) Relationships involving De (i) are given in Chapter 1. The total consolidation settlement of the group piles is then Ds c(g) 5SDs c(i) (11.19) Note that the consolidation settlement of piles may be initiated by fills placed nearby, adjacent floor loads, or the lowering of water tables. Example 11.0 A group pile in clay is shown in Figure 11.. Determine the consolidation settlement of the piles. All clays are normally consolidated. Solution Because the lengths of the piles are 15 m each, the stress distribution starts at a depth of 10 m below the top of the pile. We are given that Q g kn. Calculation of Settlement of Clay Layer 1 For normally consolidated clays, Ds c(1) 5 c (C c(1)h 1 ) 1 1 e o(1) d log c s o(1) Ds (1) r 5 r 1Ds (1) r s o(1) r d Q g (L g 1 z 1 )(B g 1 z 1 ) kn>m ( )( )

44 68 Chapter 11: Pile Foundations Sand 16. kn/m 3 Groundwater table 10 m 15 m z V:1H Clay Clay (not to scale) Q g 000 kn Group pile Rock o(1) (1) o() () o(3), 9 m 7 m (3) 1 m V:1H Figure 11. Consolidation settlement of a pile group m 16 m m m Clay sat 18.0 kn/m 3 e o 0.8 C c 0.3 sat 18.9 kn/m 3 e o 0.7 C c 0. sat 19 kn/m 3 e o 0.75 C c 0.5 Pile group: L g 3.3 m; B g. m and So s o(1) r 5 (16.) 1 1.5( ) kn>m Ds c(1) 5 (0.3)(7) log c d m mm 13.8 Settlement of Layer As with layer 1, and Hence, Ds c() 5 C c()h log c s o() r 1Ds () d 1 1 e o() s o() r s s() r 5 (16.) 1 16( ) 1 ( ) kn>m Ds () r 5 Ds c() 5 (0.)() ( )(. 1 9) kn>m log c d m mm 181.6