ABSTRACT

Follow the same procedure as given for Case 1 to obtain values of Yg for the various loads Pt' The load deflection curve can be obtained from the calculated values as shown in Fig. 16.18. The measured values are also plotted. It is clear from the curve that there is a very close agreement between the two. The figure also gives the relationship between nh and Yg'

The data for the problem are taken from Matlock and Reese (1961). The pile is restrained at the head by the structure on the top of the pile. The pile considered is below the sea bed. The undrained shear strength c and submerged unit weights are obtained by working back from the known values of nh and T. The other details are

Pile diameter, d = 33 in, pipe pile EI = 42.35 x 1010 Ib-in2

c = 500 Ib/ft2

t = 150,000 lbs

( ) Mt - - T (b) T =[E nhI]5 , () (6 e)a ~T - 12.25 +1.078 T ' c Pe = Pt 1-O. 7T

Required (a) deflection at the pile head (b) moment distribution diagram

Solution Substituting the known values in Eq (16.30) and simplifying,

458 X 106 n =-----

Calculations 1. Assume e =0, P

From Eqs (d) and (b) nh =7.91b / in2 , T =140 in

From Eq. (a)

or

Therefore

M t -140 = -0.858 ~T 12.25 +1.078 x 140

e = 0.858 x 140 = 120 in

( e )1.5 ( 120)1.51+- = 1+- =10 d 33

Now from Eq. (d), nh =2.84Ib/in3, from Eq. (b) T =171.64 in After substitution in Eq. (a)

M t = -0.875, and e =0.875 x 171.64 =150.2 in ~T

( 150.2 )Pe = 1-0.67x-- x1.5x105 =62,205Ibs171.64

3. Continuing this process for a few more steps there will be convergence of values of nh, T and Pe. The final values obtained are

nh = 2.1lb / in3, T = 182.4 in, and Pe = 62,2461b

M t = -~e = -150,000 x 150.2 = -22.53 x 106 lb - in2

= 2.43 PeT3 = 2.43 x 62,246 x (l82.4)3 =2.17 in Yg EI 42.35 x 1010

Moment distribution along the pile may now be calculated by making use of Eq. (16.11) and Table 16.2. Please note that Mt has a negative sign. The moment distribution curve is given in Fig. 16.19. There is a very close agreement between the computed values by direct method and the Reese and Matlock method. The deflection and the negative bending moment as obtained by Reese and Matlock are

Y m

=2.307 in and Mt =-24.75 x 106 1b-in2

Pt=150,000 Ib~Mt S b d300 ,- ~e Deflected position.S 400 ~

M, Bending moment, (10)6, in-lb 025 -20 -15 -10 -5 0 5 10

Figure 16.19 Bending moment distribution for an offshore pile supported structure (Matlock and Reese, 1961)