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1 this document downloaded from vulcanhammer.info the website about Vulcan Iron Works Inc. and the pile driving equipment it manufactured Terms and Conditions of Use: All of the information, data and computer software ( information ) presented on this web site is for general information only. While every effort will be made to insure its accuracy, this information should not be used or relied on for any specific application without independent, competent professional examination and verification of its accuracy, suitability and applicability by a licensed professional. Anyone making use of this information does so at his or her own risk and assumes any and all liability resulting from such use. The entire risk as to quality or usability of the information contained within is with the reader. In no event will this web page or webmaster be held liable, nor does this web page or its webmaster provide insurance against liability, for any damages including lost profits, lost savings or any other incidental or consequential damages arising from the use or inability to use the information contained within. Visit our companion site This site is not an official site of Prentice-Hall, Pile Buck, or Vulcan Foundation Equipment. All references to sources of software, equipment, parts, service or repairs do not constitute an endorsement.

2 Critical Tension Stresses G. G. Goble c 8 When reinforced or prestressed concrete piles are driven in easy driving tension stresses are reflected from the pile tip back onto the, downward traveling compression stresses. This can produce resultant tension stresses in the pile that c.an be sufficiently large enough to cause tension cracking of the concrete. use of a "Wave Equation" analysis. This problem can be analyzed by If measurements are made they are usually made with transducers located near the pile top. However, the critical tension stress location will not be at the gage location so the magnitude of maximum tension stresses cannot be readily determined. this paper a procedure will be presented for determining the maximum _ tension stress in the pile given measurements at the pile top addi tion a procedure wi 11 be discussed for calculating maximum tension stress using a closed form solution of a continuous model of the pile Because excessive tension stresses are most likely to occur during easy driving, at times when the soil resistance is small in comparison to the maximum hammer force delivered, one can make a rough approximation of - the time and location of the maximum tension stress by consideration of the hammer force wave a1 one..c In Figure 1, hammer force pul se, ideal i zed for clarity, is shown in various stages of propagation. In In While the shape of this wave is not realistic, it embodies the important characteristics of observed hammer-induced waves: A rapid rise time to the force peak and a gradual decay with time of the force. The time from impact- to the - - maximum force is denoted by t P ' The ends of the rod are considered to be

3 free, and a wave is reflected from a free end as a force wave of the same shape but opposite sign. Hence, upon encountering the pile toe, the compression wave from the hammer is reflected in tension. any section after t = Llc, as shown in Figure lb, is given by the The stress at difference between the upward traveling tension wave and the downward traveling compression wave (the so-called tail). The maximum tension force will occur when the difference between these two is the greatest. However, to complicate matters, when the hammer wave returns to the pile top, it is reflected a second time from a free end, and becomes compression again. So, the location of the maximum tension stress is as shown in Figure Ic, a distance of t /2 from the pile top, where c is the speed of wave P propagation in the pile. This depth will be referred to as xc, the critical section. The compression force induced in the pile at the critical section and time (xc, t = 2L/c + t /2) serves to reduce the P maximum force by superposition. Examination of Figure lc shows that this force is F(ZL/c), where F, as defined above, is the input force at the pile top. force. This force value will be denoted by Ft, the so-called "tail" Close consideration will reveal that the use of a more realistic force wave wi 11 yield the same results. Now consider a more realistic case where top measurements are available as shown in Fig. 2. pile top are shown. The force and velocity measurements at the The force in the reflection wave is related to the measured force and velocity at a time 2L/c after impact, the time necessary for the original impact wave to travel the length of the pile, reflect.. and return to the measuring location. If we define

4 Fop Toe Coapression (a Tension (-) Fipure 1. t e n j, I,ncatian

5 at time 2L/c then the force in the tension reflection TR is (/2. C, As shown in Fig. 2, the tension force at any location x (which occurs at a time (2L-x)/c after impact) is the sum of the reflected tension force and the input wave measured at time 2x/c before 2L/c. If. the input wave exceeds the tension reflection force TRY then no net tension exists. This approach can be used for all locations along the pile so that a tension envelope, T(x), is obtained and from it the maximum tension. The tension envelope is given by superposition with the velocity input wave. Thus, tor L I \ Figure 2. Maximum Tension Example..

6 -C, The concepts used above can be used to obtain a direct solution to the problem without the necessity of a "Wave Equation" computer analysis. Referring again to Figure 1 i t can be seen that to solve the problem Fma,. and t must be known together with the shape of the "tail" of the curve. P In addition the reduction of the reflected tension by tip resistances must a1 so be known. This problem was solved by Parker (Reference) for ai r/steam hammers using a model shown in Figure 3. By writing the equations of motion of the masses and imposing a boundary condition relating the cushion spring force and the pile top velocity two third order differential equations are obtained. The solution can be of two forms. 1-L- Capbl ock [ I-, Pile Figure 3. Pile Model by Parker

7 c) The constants a, U, A, B, S, S,, P, Q are complicated in their definition. - Therefore, they have been tabulated for a variety of typical values of. dri vi ng sys tem parameters. in application this method is used by obtaining the values for FmaX and t from the tables in the Appendix. If t is known the location of the P P critical tension stress is known and therefore the associated time for calculating the force coming into the pile in the "tail" of the curve. The method.will be ill ustrated by a numerical example. The following nine-step procedure can be used to estiamte the maximum tension stress in a pile. Step 1 Establ i sh Driving System Constants Weights of Ram and Helmet (R and H) in Kips Stiffness of Capblock and Cushion (K and C) in Kips/Inch Vh = Hammer Impact Velocity ( ftlsec) where = 42 g h (eff) 2 g = Acceleration of Gravity (ft/sec ) h = Stroke (ft) eff = Hammer Efficiency c = Wave Speed where 2 E = Young's Modulus (kips/ft ) y = Weight 'Density of Pile (kips/cu ft) NOTE: Convert c to ft/msec for use with method

8 Step 2 Find Force Pulse Constants from the Appendix Find Critical Depth Step 4 Find Maximum Force Delivered by Hammer Fmax = (F/V) Vh Step 5 Find Ft, Hammer Force at Time Delay 2L/c td = ~L/c TYPE I Solution -a F~ = vh e td cos (wtd) + B sin (wtd)l f -. Step TYPE I1 Solution 6 Find Fs, Effect of Soil Resistance at Critical Section Use Table 1 Step 7 Find Fd, Effect of Soil Damping Fci = * Jc Fmax Step 8 Find Maximum Tension Force, Tmax Tmax=F ma x -Ft-Fs-Fd Step 9 Find Maximum Tension Stress - Tmax O t - A EXAMPLE t t. Situation An 80-foot long, 20-inch square HC pile has a prestress of 1.0 ksi. A Vulcan 020 Hammer has been chosen to assure that this pile can attain its..

9 C, highest ultimate capacity. To reach a sufficiently strong stratum, the piles must first pierce a thin, hard layer overlaying softer material. The geotechnical consultant predicts that this layer will exert a side force of 40 tons on the pile during driving and recommends a Case damping constant of Jc = 0.1. Problem Select a cushion made of 3/4-inch plywood sheets to prevent pile breakage. Solution Try a cushion of three sheets of 3/4-inch plywood. Step 1 Establ ish Driving System Constants Hammer: R = 20.0 kips Stroke = 36 inches Vh = 42 g h (eff) (Assume 80% efficiency) Helmet: H = 2.2 kips K = kips/inc Cushion: C A E = i- L A = 400 in2 (Pile top area) E = 30.0 ksi (Assumed for plywood) C = 7100 k/in Wave Speed: c = 5 Y L = 3(3/4") (p. 75) (25% shortening assumed)

10 -ci Step 2 Find Force Wave Constants NOTE : Since this cushion stiffness fa1 1s between the Type I and Type I1 solutions tabulated in the Appendix, values of Ft w i l l be found for the two closest cushion stiffnesses, and an interpolated value will then be used. C = 6000 C = 8000 a = SI= w = S;Z= A = P = -202 B = Q = 215 t = 2.8 P F/V = 73.2 Step 3 Find Critical Depth x = c t /2 C P = 11.12(2.8) /2 = 15.6 ft Step 4 Find Fmax F ma x = (F/V) Vh = (73.2) 12.4 = 908 kips

11 C Step 5 Find Ft - 2 L 2(80) = = 14.4 Msec td -c TYPE I Solution (C=6000) Ft = 12.4 e -"" 14.4 t-12.6 cos (.071 x 14.4) = 117 kips. TYPEIISolution (C-8000) sin (.071 x 14.4)) Ft = e x x 14.4) = 116 kips Interpolating for C=7100 Ft = kips Step 6 Find FS For a side force: Fs =!i R =!i (40 tons) = 20 tons = 40 kips Step 7 Find Fd Fd = Jc Fmax = 2(0.1) (908) = 182 kips Step 8 Find Tmax Fmax Fc Fs Fd T rnax 569 kips

12 C 1 at = 569/305 = 1.87 ksi occurring at approximately 15.6 feet below the top of the pile. Step 9 Find Maximum Tension Stress This stress exceeds the prestress, so a thicker cushion is required. Try nine sheets of 3/4-inch plywood. Step 1 Driving System Constants A1 1 constants remain the same except C. Step 2 Find Force Wave Constants Step 3 Find Critical Depth x = 11.12(5.9) /2 C = 31.1 ft Step 4 Find Fmax FmaX = (56)(12.4) = 094 kips Step 5 Find Ft td = unchanged = 14.4 Msec

13 TYPE I Solution - Ft = 12.4 e 14.4 i-12.7 cos (.I39 x 14.4) = 298 kips f sin (.I39 x 14.4)} Step 6 Find FS Same as above: Fs = 40 kips Step 7 Find Fd Fd = 2(0.1)(694) = 138 kips Step 8 Find Tmax 'max Ft -298 Step 9 Find Maximum Tension Stress O t = 218/305 = 0.71 ksi Occurring at approximately 31.1 feet below the pile top. The tension stress is less than the prestress, so the cushion is sufficient.

14 -C, - REFERENCE Parker, Eric J., "Tension Cracking in Concrete Piles Driven by Air/Stearn Hammers", Masters thesis, Department of Civi 1, Envi ronmental and Architectural Engineering, University of Colorado, 1979.

15 APPENDIX PILE TABLES Cons tan t Units a Msec

16 9 rn Y 4 ( p q r n ~ u - mommom mmmmmn mommmo mmommm l l - In. s 2 a a- - wawwww r-awwaw mmmmmm mmmmwm 0 0 Ulo w 0.(Y

17 RAM WEIGtiT = 3.0 KIPS TYPE I (TYPE 11) PICE SIZE 12 INCH SQUARE AREA = SO. IN. HELMET CUSHION (K/IN) E = KSI I = 51.8 E KSI I = 63.5 ALPHA OMEGA A B TP (F/V) (St) (S2) (PI (Q 1 ALPHA OMCGA (sf) (52) A 8 TP. (.F/V) (P 1 (0) H = I K I OOC..I , I

18 HELMET RAM WEIGHT = 3.0 KIPS TYPE 1 (TYPE 11) PILE SIZE 14 INCH SQUARE AREA = SO. IN. - CUSHl ON E KSX E KSX fk/in) X ALPHA OMEGA A 8 TP (F/V) ALPHA OMEGA A 8 TP- (St) (S2) (PI {Q) (St) (53) (P (0)

19 RAM WEIGHT = 5.0 KIPS HELMET PILE SIZE 10 INCH SQUARE AREA SQ. IN. CUSHION E = (H/lN) I OMEGA (52) KSI ALPHA (S1) OMEGA (S2) E m KSI

20 RAM WEIGHT = 5.0 KIPS TYPE I (TYPE 11) PILE SIZE 12 INCH SQUARE AREA SO. IN. HELMET CUSHION (H/IN) ALPHA OMEGA A 8 (St) (52) (PI (0) ALPHA (St) OMEGA i S 2)

21 HELMET RAM WEIGHT = 5.0 KlPS TYPE I (TYPE 11) PILE SIZE 14 INCH SQUARE AREA = IN. CUSHION E = KSI (K/IN) I E KSI 1 r 86.4 ALPHA (51) OMEGA (52) ALPHA OMEGA (st) (52)

22

23 HELMET RAM WEIGHT = 5.0 KIPS TYPE I (TYPE 11) PILE SIZE 18 INCH SOUARE AREA = SO. IN. CUSH 1 ON (&/IN) E = KSI I = ALPHA OMEGA A B fp (F/V) 151) Is21 (p) (0) A1 PHA (St OMEGA (S2)

24 RAM WEIGHT = 5.0 KIPS PILE SIZE 20 INCH SQUARE HC AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSHION (K/IN) E = KSI 1 * ALPHA (51) OMEGA (52) ALPHA (51 OMEGA (52)

25 RAM HEIGHT 10.0 KlPS TYPE 1 (TYPE 11) P.1LE SIZE 14 INCH SQUARE AREA IN. HELMET CUSHION (K/fN) E = KSI I * 70.5 ALPHA OMEGA A 8 (St) (S2) (PI (9) ALPHA (51 1 OMCGA (52)

26 HELMET RAM WEIGHT = 10.0 KIPS PILE SIZE 16 INCH SQUARE AREA = SQ. IN. CUSIIION (H/IN) TYPE I (TYPE 11) E a KSI I ALPHA OMEGA A 8 TP (F/V) (St) (S2) (P (01 ALPHA (Sf) OMEGA A 8 TP (F/V) (S2) (PI (0)

27 RAN WEIGHT = 10.0 HIPS PILE SIZE 18 INCH SQUARE AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSHION (K/IN) ALPHA OMEGA A 0 (S11 (S21 (P 1 (0 1 ALPHA (St 1 OMEGA (S21

28 RAM WEIGHT = 10.0 KIPS PILE SIZE 20 INCH SOUARE AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSHION (H/IN) E = KSI ALPHA OMEGA A 0 TP (F/v) (51) (52) ( p (0) ALPHA (St I I ,319 OMEGA (52).I I , I54, ,

29 RAM WEIGtIT = 10.0 KIPS PILE SIZE 20 INCH SQUARE HC AREA = SO. IN. TYPE 1 (TYPE 11) HELMET CUSH l QN (E / I N E = KSK I E KSI I = ALPHA CMEGA A 0 (51) (52) (P (0) ALPHA OMEGA (Sf) (S21

30 HELMET RAM WEIGHT = 10.0 KIPS,PILE SIZE 22 INCH SQUARE HC AREA = SO. IN. CUSHION (K/IN) E = KSI I TYPE I (TYPE 11) E KSI I ALPHA IS11 OMEGA A 8 (S2) (P 1 (0) ALPHA (Sf 1 OMEGA (52).I a

31 RAM WEIGHT = 15.0 KIPS PILE SIZE 16 INCH SQUARE AREA = SO. IN. TYPE I (TYPE If) HELMET CUSHION (K/IN) E = KSI 1 = ALPHA OMEGA A 8 TP (51) (S2) (PI. (0) ALPHA (St lo5.I50.I lo6.I51.I lo5.I50.I lo6-151.I96.205

32 W I- mn-ootm m ~ m m m - wm-om13 vmmmra 4a w - nommwn mmmmmm mmmonn mmnmmm a CCCC.-C -LC In l l l X Z I-... OQOOOO mmvam -mmoam

33 HELMET RAM WEIGHT = 15.0 KIPS PILE SIZE 20 INCH SQUARE AREA = SO. IN. CUSH I ON h KSI (K/IN).I ALPHA OMEGA A 0 TP (F/V) (51) (S2) (PI (0) ALPHA (St 1 TYPE I (TYPE 11) OMEGA (S2)

34 HELMET RAM WEIGHT = 15.0 KIPS PILE SIZE 22 INCH SQUARE AREA = SO. IN. E = KSI I ALPHA OMEGA A 8 TP (F/V) ISl) (S2) (PI (Q) TYPE I (TYPE X I ) E = KSI ALPHA OMEGA A 8 TP' (St) ('52) (PI (0)

35 ..,.'.". '..,....*A r..." ^.*.. <. -. I-... '..I ar-.. RAM WEIGHT = 15.0 KIPS PILE SIZE 20 INCH SQUARE HC AREA SO. IN. TYPE I (TYPE 11) HELMET CUSHION (K/IN) ALPHA OMEGA A 8 Tp (F/v) (51) (S2) ("1 (0) ALPHA (51 OMEGA (S2)

36 RAM WEIGHT = 15.0 KIPS PILE SIZE 22 INCH SQUARE HC AREA = SQ. IN. HELMET CUStII ON E = KSI (K/IN) TL PE i (TYPE 11) E KSI ALPHA OMEGA A 8 CSll. (S2) (P (0) ALPHA (Sf J , I OBS OMEGA (52),129, , I

37 RAM WEIGHT = 15.0 KIPS PILE SIZE 24 INCH SQUARE HC AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSHION (K/IN) ALPHA OMEGA A 8 (51) (S2) (PI (0) A1 PHA (St 1 OMEGA (S2)

38 RAM WEIGHT = 20.0 KIPS PILE SIZE 18 INCH SQUARE AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSHION (K/IN) E = KSI 1 = E = KSI I = ALPHA (51 OMEGA (52) ALPHA (St 1 OMEGA (S2)

39 RAM WEIGHT = 20.0 KIPS PICE SIZE 20 INCH SQUARE AREA = SO. IN. TYPE I (TYPE If) HELMET CUSH 1 ON (K/lN) ALPHA (51) OMEGA (52 1 OMEGA (S2)

40 HELMET RAM WEIGHT = 20.0 KIPS PILE SIZE 22 INCH SQUARE AREA = SO. IN. - CUSH 1 ON E = KSI (K/IN) ALPHA OMEGA A 8 TP (F/V) (S1) (52) (PI (0) A1 PHA (S.1) TYPE I (TYPE 11) OMEGA (S2) E KSI

41 RAM WEIGHT = 20.0 KIPS PILE SIZE 24 INCH SQUARE AREA = SO. IN. TYPE-I (TYPE 11) HELMET CUSHION (K/lN) ALPHA (51 1 OMEGA (S2) (F/V) ALPHA OMEGA A B T P (Sf) (52) (PI. (0) I I OOC.263

42 RAM WEIGHT =, 20.0 HIPS TYPE 1 (TYPE 111 PILE SIZE 20 INCH SQUARE HC AREA = SO. IN. HELMET CUSHION (K/IN) ALPHA OMEGA (51) (S2) ALPHA OMEGA A B (st) (S2) (P (0)

43 RAM WEIGHT = 20.0 KIPS PILE SIZE 22 INCH SQUARE HC AREA = SO. IN. HELMET CUSHION E = KS1 (K/IN) I TYPE I (TYPE It) E = HSI I = ALPHA OMEGA A 8 (51) (S2) (P 1 (0) ALPHA (St 1, I ,211, I OMEGA (S2).113.I53.I I80.I I I75.I13.I52.I I

44 RAM WEIGHT = 20.0 KIPS PILE StZE 24 INCH SQUARE HC AREA = SO. IN. TYPE I (TYPE 11) HELMET CUSH l ON (K/IN) E = KSI 1 = ALPHA (Sf OMEGA A (52) (PI ALPHA (St 1 OMEGA (52) t-' 0 vl

45 RAM WEIGHT = 30.0 KIPS PILE SIZE 18 INCH SQUARE AREA = SO. IN. TYPE I (TYPE If) HELMET E HSI I ALPHA OMEGA A B TP (F/V) ALPHA OMEGA A 8 TP (F/V) (St) (S2) ( p 1 (0) (St) (52) (PI (0)

46 -cc-t-wm... -0omgm (YNN--- I I I I I I... ooommm (YNN--- I I I I I I -A - m P o m ~ cn-m-mm Or--N-... -u OQOOOO -C(owmo -

47

48 HELMET CUSH 1 ON (K/IN) RAM WEIGHT = 30.0 KIPS PILE SIZE 24 INCH SQUARE AREA = SO. IN. E = KSI I TYPE I (TYPE 11) E YSI I - 253,s ALPHA OMEGA A 8 (51) (S2) (PI (0) ALPHA OMEGA (Sf) (52)

49 Y 8' w a 4 s ln. z XI ww- u-w -I- w-1 -A- en1 mmo An - m a... mom -..or-in - -.r-ao --mmmm FINNmFlo rc-m c I I I I I I I I I I I om... rnrn~r-mm --CC-C I I I I I

50 RAM WEIGHT = 30.0 KIPS PILE SIZE 22 INCH SQUARE HC AREA = SO. 1N. TYPE I (TYPE II) HELMET CUSHION (K/tN) E = KSI I = ALPHA OMEGA A 8 T P (51) (52) (PI (0) ALPHA OMECA A I3 (51) (52) (p) (9)

51 -?-lnm m-mn Or.--..,. mmfi 0.. r-om mma - n Yo? r-(co --.- I I I -mm UI m r- amp -uln I l l -1 a*.

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