The Effect of Batter on Hammer Energy

Most people involved in the pile driving business know that, when you drive piles on a batter (angle) the energy from the hammer is reduced. But how much? This post attempts to answer that question, and show how Vulcan dealt with the issue.

Let’s start with single-acting hammers like the Warrington-Vulcan hammers: the rated striking energy is obviously

${\it E_r}={\it W_s}\,s$ (1)

where

• Er = rated striking energy, ft-lbs
• Ws = ram weight, lbs.
• s = stroke, ft.

When a hammer is put on a batter, its energy is reduced by two factors:

• The reduction in the fall through the gravity field. This is referred to as the “cosine effect” because the rated striking energy is multiplied by the cosine of the batter angle.
• The subtraction of the drag due to mechanical friction between the ram and the columns/ram tube. The drag is a product of the coefficient of friction of the ram riding on the columns/ram tube and the force of the ram on the frame, which should probably be referred to as the “sine effect.”

The effect of those two is shown by the equation

${\it E_{rd}}={\frac {{\it W_s}\,s}{\sqrt {1+{B}^{2}}}}-{\frac {{\it W_s}\,sfB}{\sqrt {1+{B}^{2}}}}$ (2)

where

• Erd = reduced rated striking energy on a batter, ft-lbs
• B = batter angle expressed as a fraction. Thus, a batter of 1:3 has B = 1/3, 1:4 has B = 1/4, etc. To use this the numerator has to be 1, thus a 9:12 batter has B = 1/1.333
• f = coefficient of friction

Using trigonometric identities eliminates the sines and cosines and reduces them to radicals.

Dividing Equation (2) by Equation (1) yields

$r_{diff} = {\frac {1-fB}{\sqrt {1+{B}^{2}}}}$ (3)

where rdiff = reduction ratio in energy.

So how did Vulcan deal with this problem? The only place where it did was in its Data Manual, which has been incorporated into the vulcanhammer.info Guide to Pile Driving Equipment. The relevant page for single acting hammers is shown below.

There are two significant differences between this formula and the one we derived. The first is that Vulcan incorporated “an assumed friction between leaders and hammer of 10%.” (What the friction between the leaders and the hammer has to do with this is hard to know, although keeping leaders greased during operation is important.) In any case this translates into a coefficient of friction f = 0.1. The second is that Vulcan’s formula expresses the portion of “plumb energy” as a percentage.

For the example case with a Vulcan #1 Hammer, if we substitute f = 0.1 and B = 1/3 into Equation (3), we get rdiff = 0.917, which is identical to Vulcan’s result.

The situation with differential acting hammers (or any hammer with downward assist) is more complicated. The rated striking energy for a plumb Super-Vulcan hammer is

${\it E_r}={\it W_s}\,s+s{\it A_{sp}}\,{\it p_{inlet}}$ (4)

where

• Asp = area of small piston sq.in.
• pinlet = inlet pressure at the hammer, psi

The reduced energy on a batter is thus:

${\it E_{rd}}={\frac {{\it W_s}\,s}{\sqrt {1+{B}^{2}}}}+s{\it A_{sp}}\,{\it p_{inlet}}-{\frac {{\it Ws}\,sfB}{\sqrt {1+{B}^{2}}}}$ (5)

Dividing the two equations as before yields

$r_{diff} = {\frac {{\it W_s}+{\it A_{sp}}\,{\it p+{inlet}}\,\sqrt {1+{B}^{2}}-{\it W_s}\,fB}{\sqrt {1+{B}^{2}}\left ({\it W_s}+{\it A_{sp}}\,{\it p_{inlet}}\right )}}$ (6)

Now we have a problem; the reduction depends upon the nature of the downward assist, or more specifically the relationship between the free drop energy and the energy derived from the downward assist. If we define frat = (Force due to downward assist)/(Force due to free drop) = (Asp pinlet)/(Ws) without effect of a batter, we obtain

$r_{diff} = {\frac {1+\sqrt {1+{B}^{2}}{\it f_{rat}}-fB}{\sqrt {1+{B}^{2}}\left (1+{\it f_{rat}}\right )}}$ (7)

For a single-acting hammer, frat = 0, and Equation (7) reduces to Equation (3) as it should.

Vulcan had a formula for this too, as can be seen below.

For a Vulcan 50C Hammer, frat = (55.917)(120)/(5000) = 1.341. Substituting this and the variables from the previous problem yields rdiff = 0.965, or a batter rated energy of 14,571, which is below Vulcan’s calculation.