Comparing Rigid Piles with Semi-Infinite Piles/Wave Equation Piles

In our last post A Tale of Two Coefficients of Restitution we discussed the development of equations of motion for impacting rams and piles that included a spring between them. This spring is bilinear using the coefficient of restitution. We showed that the behaviour of the ram and pile is the same with the spring in the middle as it was with classical momentum impact theory. What we learned was that the spring governed the force between the ram and the pile and thus the deceleration history of the ram and the force exerted on the pile.

Conventional wisdom has it that, when the dynamic formulae were first developed, the error due to the rigid pile assumption was not so great because piles were relatively short and wave effects were not as pronounced as they are with longer piles. The “longer piles” came with steel and concrete piles. But where is the transition point? This monograph attempts to answer that using a simplistic approach: the rigid model developed in the post A Tale of Two Coefficients of Restitution and the semi-infinite model as described in Closed Form Solution of the Wave Equation for Piles. Our method is simple: compute the pile stress and ram deceleration using the semi-infinite model and then determine the length of the pile at which both of these quantities are the same. The higher the pile length determined in this way, the less of a problem a rigid pile assumption is.

Rigid Pile Model

The schematic for the rigid ram model is shown below:

The development of the mathematical representation of this model is described in A Tale of Two Coefficients of Restitution. One thing that was left out was estimating the pile top stress; this is simply the force of the ram (or spring) divided by the pile head cross-sectional area.

Semi-Infinite Pile Model

The semi-infinite pile model is a way of modelling wave effects in the pile without having to model the entire pile but just the pile head. A schematic of the method is shown below.

The similarity in the simplification of the system between the two is easy to see. The rigid pile system models the pile as a rigid mass. The semi-infinite pile system models the pile as a dashpot based on the impedance of the pile. It is worth noting that neither system has a driving accessory mass. This isn’t realistic from a practical standpoint but it gives us an “apples to apples” way of comparing the two sytems. (In reality, Vulcan hammers that drove wood piles without a driving accessory did so without a cushion as well using the McDermid Base.)

The equations of motion are

$M{\frac {d^{2}}{d{t}^{2}}}{\it x_r}(t)+k\left ({\it x_r}(t)-{\it x_p}(t)\right )=0$ (1)

$Z{\frac {d}{dt}}{\it x_p}(t)-k\left ({\it x_r}(t)-{\it x_p}(t)\right )=0$ (2)

The solution for the displacement of the pile head is, from Closed Form Solution of the Wave Equation for Piles,

$x_p(t) = 1/2\,{\it V_0}\,\left (1-{e^{-{\it \alpha_0}\,t}}\left ({\frac {\sin({\it \alpha_0}\,{\it \beta_0}\,t)}{{\it \beta_0}}}+\cos({\it \alpha_0}\,{\it \beta_0}\,t)\right )\right ){{\it Z'}}^{-2}{{\it \alpha_0}}^{-1}$ (3)

where

$\alpha_0 = \frac{Z}{2 M Z'^2}$ (4)

$\beta_0 = \sqrt{4Z'^2-1}$ (5)

$Z' = \frac{Z}{\sqrt{kM}}$ (6)

It should be noted that there is a singularity when $Z' = 0.5$ and the circular functions change to hyperbolic ones when Z’ is below that value. More details on that can be found in Deeks (1992).

The displacement of the ram can be found by substituting Equation (3) into Equation (2) and solving for the ram displacement. Doing so (along with other algebra) yields

$x(r) =\frac{V_0}{Z'^2 \alpha_0} (1/4\,{\frac {{\it }\,\sin({\it \alpha_0}\,{\it \beta_0}\,t)}{{\it }\,{\it \beta_0}\,{e^{{\it \alpha_0}\,t}}}}-1/4\,{\frac {{\it }\,{\it \beta_0}\,\sin({\it \alpha_0}\,{\it \beta_0}\,t)}{{\it }\,{e^{{\it \alpha_0}\,t}}}}-1/2+1/2\,{\frac {{\it }\,\cos({\it \alpha_0}\,{\it \beta_0}\,t)}{{{\it }}{\it }\,{e^{{\it \alpha_0}\,t}}}})$ (7)

The deflection of the cushion spring is

$\Delta x = 1/4\,{\frac {{\it V_0}\,\left (-\sin({\it \alpha_0}\,{\it \beta_0}\,t)+{{\it \beta_0}}^{2}\sin({\it \alpha_0}\,{\it \beta_0}\,t)-2\,\cos({\it \alpha_0}\,{\it \beta_0}\,t){\it \beta_0}+2\,{e^{{\it \alpha_0}\,t}}{e^{-{\it \alpha_0}\,t}}\sin({\it \alpha_0}\,{\it \beta_0}\,t)+2\,{e^{{\it \alpha_0}\,t}}{e^{-{\it \alpha_0}\,t}}\cos({\it \alpha_0}\,{\it \beta_0}\,t){\it \beta_0}\right )}{{{\it Z'}}^{2}{\it \alpha_0}\,{\it \beta_0}\,{e^{{\it \alpha_0}\,t}}}}$ (8)

and that is at a maximum when

$t_{\Delta x_{min}} = \frac {\arctan (\beta_0)}{\alpha_0 \beta_0}$ (9)

At that point the force on both the pile head and the ram is at a maximum; the pile head experiences maximum force, and the ram maximum deceleration.

Steel Pile Case

Let us reconsider the example of the steel pile case from the post A Tale of Two Coefficients of Restitution. In that post we computed the maximum spring deflection, which in turn give us the maximum force on both ram and pile. The maximum pile stress is the maximum cushion spring force divided by the pile head area.

It should be evident that, with an increase in ram mass/weight, the natural frequency of the system will decrease and the maximum spring deflection will increase, along with the maximum ram deceleration and the pile head stress. The idea here is to establish the length of rigid pile at which the pile head stresses are the same as the semi-infinite pile. A “long” pile will indicate that a rigid pile model can be applied and a “short” pile will indicate that a semi-infinite/wave equation transmission line model should be applied. These are not very rigourous and you are invited to pursue this further.

In any case, if we use the same problem as we had in the previous post, applying semi-infinite pile theory as described earlier yields the following:

The pile impedance is 52,803 lb-sec/ft.
The hammer impedance is 93,793 lb-sec/ft
The impedance ratio Z’ = 0.563.
The two constants α₀ = 536 1/sec and β₀ = 0.517 (awfully close to the singularity!)
The maximum pile head force/cushion spring force is 423.9 kips.
The maximum ram deceleration is 84.7 g’s.
The maximum pile compressive stress is 14 ksi.

If we substitute the last result and back compute the length of pile to match it, we come up with a pile length of 9.2′! This indicates (but does not prove) that the transition from rigid pile model to wave model is very early; virtually all piles of this type are longer than this.

Wood Pile Case

We now consider a type of piles that have been touted as rigid for pile computational purposes: wood piles. In this case let us consider a wood pile that is 12″ at the head and 8″ at the butt. How long would such a pile have to be in order to give the same results for both theories:

As before we start with the semi-infinite pile model and compute the following;

The pile impedance is 11,997 lb-sec/ft.
The hammer impedance is 93,793 lb-sec/ft (unchanged)
The impedance ratio Z’ = 0.128. Because of this β₀ is imaginary (see Equation (5)) and the circular functions switch to hyperbolic. To accommodate this change we used Maple to solve the problem, as was the case with Closed Form Solution of the Wave Equation for Piles.
The two constants α₀ = 2359 1/sec and β₀ = 0.967i.
The maximum pile head force/cushion spring force is 129.3 kips.
The maximum ram deceleration is 25.8 g’s. This is the reason why Vulcan hammers which spend most of their lives driving wood piling last the longest.
The maximum pile compressive stress is 1.14 ksi.

The rigid pile weight needed to match this impact force/compressive stress is only 2.19′ long. It is safe to say that purely rigid piles do not exist either in theory or reality, and that some consideration must be made for the distributed mass and elasticity of the pile in determining its drivability and in estimating the SRD (soil resistance to driving) of the pile.