Below are three photos taken of a Vulcan #1 hammer in Mentor, Ohio. It’s in fixed leaders with a moonbeam-style spotter. The hammer was made in Chicago, which means that it’s probably older than sixty years. Thanks to Ken Foster for sharing these great photos.
Most of our fluid mechanics offerings are on our companion site, Chet Aero Marine. This topic, and the way we plan to treat it, is so intertwined with the history of Vulcan’s product line that we’re posting it here. Hopefully it will be useful in understanding both. It’s a offshoot of Vulcan’s valve loss study in the late 1970’s and early 1980’s, and it led to an important decision in that effort. I am indebted to Bob Daniel at Georgia Tech for this presentation.
The basics of incompressible flow through nozzles, and the losses that take place, is discussed here in detail. The first complicating factor when adding compressibility is the density change in the fluid. For this study we will consider only ideal gases.
Consider a simple orifice configuration such as is shown below.
The mass flow through this system for an ideal gas is given by the equation
where
mass flow rate, 
throat area of orifice, 
adjusted throat area of orifice (see below,) 
upstream density, 
upstream pressure, psfa
downstream pressure, psfa
gravitational constant 
ideal gas constant or ratio of specific heats 
for air
gas constant 
upstream absolute temperature 
At this point we need to state two modifications for this equation.
First, we need to eliminate the density, which we can do using the ideal gas equation
Second, we should like to convert the mass flow rate into the equivalent volumetric flow rate for free air. Most air compressors (and our goal is to determine the size of an air compressor needed to run a test through this valve) are rated in volumetric flow of free air in cubic feet per minute (SCFM.) This is also the basis for the air consumption ratings for Vulcan hammers as well, both adiabatic and isothermal. This is accomplished by using the equation
Making these substitutions (with a little algebra) yields
In this article the coefficient of discharge 
We are basically considering the energy losses due to friction as an additional geometric constriction in the system.
One final–and very important–restriction on these equations is the critical pressure, given by the equation
The critical pressure is the downstream pressure for a given upstream pressure below which the flow is “choked,” i.e., the mass or volumetric flow rate will not increase no matter how much you either increase the upstream pressure or decrease the downstream pressure. This limitation, which was observed by Saint-Venant, is due to achieving the velocity of sound with the flow through the nozzle or valve. A more common way of expressing this is to consider the critical pressure ratio, or
As you can see, this is strictly a function of the ideal gas constant. It’s certainly possible to get around this using a converging-diverging nozzle, but most nozzles, valves or orifices are not like this, and certainly not a Vulcan 06 valve. We now turn to the analysis of this valve as an example of these calculations.
The first thing we should note is that pile driving equipment (except that which is used underwater) is designed to operate at sea level. Using this calculator and the standard day, free air has the following properties:
(or psfa)Now let’s consider the valve for the 06 hammer (which is identical to the #1 hammer.) A valve setting diagram (with basic flow lines to show the flow) is shown below.
Note the references to steam. Until before World War II most of these hammers (along with most construction equipment) was run on steam. With its highly variable gas constant and ability to condense back to liquid, steam presented significant analysis challenges for the designers of heavy equipment during the last part of the nineteenth century and the early part of the twentieth. For our purposes we’ll stick with air.
There are two cases of interest:
According to the vulcanhammer.info Guide to Pile Driving Equipment, the rated operating pressure for the Vulcan 06 at the hammer is 100 psig = 14,400 psfg = 16,516.22 psfa = 114.7 psia. For simplicity’s sake, we can consider the two cases as mirror images of each other. In other words, the upstream pressure in both cases is the rated operating pressure. This should certainly be the case during air admission into the hammer. For the exhaust, it should be true at the beginning of exhaust. Conversely, at the beginning of intake the downstream pressure should be atmospheric (or nearly so) and always so for exhaust.
From this and the physical characteristics of the system, we can state the following properties:
At this point calculating the flow in the valve should be a straightforward application of the flow equations, but there is one complicating factor: choked flow, which is predicted using the critical pressure ratio. For the case where 
The first is to fix the downstream pressure and then compute the upstream pressure with the maximum flow. In this case 
The reverse is to fix the upstream pressure and then to vary the downstream pressure. The critical downstream pressure is now 
We will concentrate on the latter case. If we substitute everything except the downstream pressure (expressed in psia,) we have
If 
As noted earlier, when air is first admitted into the cylinder the flow is constant. Once the critical pressure ratio is passed, the flow drops until the two pressures are equal.
It was this large volume of flow which prevented the use of the 06 valve (which could have been separated from the cylinder using a valve liner) in the valve loss study. The smaller DGH-100 valve was used instead.
It is interesting to note that the rated air consumption of the hammer is 625 cfm. This is lower than the instantaneous critical flow. Although on the surface it seems inevitable that the hammer will “outrun” the compressor, as a further complication the hammer does not receive air on a continuous basis but on an intermittent one. For much of the stroke the compressor is “dead headed” and no air is admitted into the cylinder from the compressor. To properly operate such a device, a large receiver tank is needed to provide the flow when it is needed. The lack of such large tanks on modern compressors is a major challenge to the proper operation of air pile hammers.
The introduction to this series is here. The first installment of the calculations is here.
Strength calculations assume that the inertial forces during impact are 150 times those of the weight.
We checked the rotor shaft strength in the optimal mode, i.e., when the impacting force direction formed a 90° angle with the direction of the blow. To simplify calculations consider that the forces act at one point. In the vertical place the shaft is loaded with impact inertia forces from the shaft weight and parts which are located on it.
where Q1 = inertial force from eccentric weight(s) and part of the shaft ahead of the eccentric.
Q2 = inertial force from the part of the shaft under the bearing.
Q3 = inertial force from the rotor weight and the middle part of the shaft.
A diagram of the shaft assembly is shown below.
A diagram of the beam forces in the vertical plane is shown below.
A diagram of the beam forces in the horizontal plane is shown below.
The forces which act on the shaft in the horizontal plane arise from the vibrating forces of the eccentrics.
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The reactions in the vertical plane are
The reactions in the horizontal plane are
The bending moment in the vertical plane in section A-A is
In section C-C it is
In section B-B it is
The bending moment in the horizontal plane in Sections A-A and C-C is
and for Section B-B
The sum of bending moments in Section A-A is
In Section B-B they are
and in Section C-C they are
The bending tension is calculated in the same way at all points.
For Section A-A
For Section B-B,
and Section C-C,
The tension in this section will be much less because the calculations do not take into account the force from the rotor shaft. Calculation of the shaft deflection will be done in Part C.
The calculations consider that the shaft is of uniform diameter, equal to 62 mm. In the vertical plane the deflection is equal to
where 
kg-cm
= axial inertial moment of cross-section of the shaft
E = spring modulus of shaft material = 2,000,000 kg/cm²
The deflection in the horizontal plane is equal to
The total deflection from horizontal and vertical moments is
In reality deflections will be smaller because we did not take into account the rotor forces.
The casing is subjected to loading tensions when the vibrator impacts on the pile cap. As the ram is located in the centre of the casing the critical sections are two perpendicular sections which are located at the planes of symmetry of the vibrator.
Let us determine the moment of resistance of the section which is shown in the drawing of bending tensions in this section, shown below.
This section is weakened by a hole for the ram but this weakness is compensated for by the local boss. So we do not take into account the hole and its boss.
The moment of inertia for the section relative to axis X-X is determined as
where 
= sum of inertial moments of the separate elements.
= sum of multiplication of squared distances from the mass centre of element ot the axis X-X by the area of the element.
The moment of resistance for this section is
The distance between the axes of the electric motors is 
mm. So the bending moment is equal to
The bending tension is equal to
Let us determine the bending tensions in the section perpendicular to the axis of the rotors. The section is shown in the drawing below.
To simplify the calculations consider the section of the casing is symmetrical and consists of two circles and two rectangles.
The inertial moment is equal to
The moment of resistance equals to
Let us now determine the bending moment considering that the load from the weight along the axis parallel to the rotor axis is distributed uniformly.
The bending tension is equal to
The maximum force for which spring deflection is required is P = 1000 kgf. The number of spring N = 2. The maximum deformation of the springs is f = 200 mm. The load for each spring is
As the springs are operating in relatively easy (not hard) conditions we can consider the permissible tension equal to 5500 kgf/cm². So the permissible tension per 1 kgf of load is equal to
The necessary spring stiffness is equal to
So we choose the spring with the following specifications:
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Average Diameter |
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Wire Diameter |
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Hardness of One (1) Turn |
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Number of Working Turns |
Npad = 14.5 |
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Total Number of Turns |
N = 21.5 |
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Tension per 1 kgf of Load |
A = 11.18 |
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Hardness of the whole spring |
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So the spring we have chosen meets all of the requirements.
Consider that the balanced part of the eccentrics (I and II; see diagram below) cancel each other.
So the coordinate of the center of mass of the rest of the eccentric (in the shape of a sector of a circle) is determined by the equation
The weight of the unbalanced part of the eccentric for a 1 cm thickness is equal to 1.7 kg. The eccentric moment of this eccentric is
The dynamic force of the eccentric is
The angular speed is 
rad/sec. The necessary eccentric moment of the eccentric is
The necessary total thickness of the eccentrics is
As during the determination of the eccentric moment it was increased a little, consider the thickness of the eccentrics equal to 80 mm.
This configuration of the eccentrics which we have come up with gives us an increase of its weight in comparison with the weight which is necessary to provide the required eccentric moment. So decreasing the moment of the rotary parts makes it easy to operate the motors.
The rotor shafts are mounted to spherical, double-row roller bearings No 3614 which have a coefficient of workability C = 330,000. The rotor weight Gb = 25 kgf. The eccentric weight is Gg = 28 kgf.
For the calculation of dynamic loads consider that the accelerations during impact are equal to 150 times the free weight.
As the shaft is symmetrical, each bearing is subjected to half the dynamic load
The shaft rotates at n = 950 RPM. Consider a factor of safety Kd = 1.5 and a dynamic load coefficient Kk = 1. The durability of the bearing “h” is determined as
Therefore, for 950 RPM, h = 160 hours.
The introduction to this series is here.
Moscow, 1963
Head of the Vibrating Machine Department L. Petrunkin
Head of Vibration Machine Construction: I. Friedman
Compiler: V. Morgailo and Krakinovskii
The impact-vibration hammer is intended for driving heavy sheet piles up to 30 cm in diameter as well as concrete piles 25 cm square up to a depth of 6 m for bridge supports and foundations.
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Parameter |
Value |
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Power N, kW |
9 |
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Blows per Minute Z |
475 |
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Revolutions per Minute |
950 |
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Ram mass |
650 |
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Force F, kgf |
5000 |
Impact velocity is determined:
where 
= fraction of natural frequency (without limiter) to force
frequency
i = fraction of the number of revolutions to the number of impacts
R’= coefficient of velocity recovering (assume R’=0.12)
In our case
therefore
rad/sec
kgf-sec²/m
Energy of blow is determined as
Power necessary to make impacts is
So that the impact-vibration hammer operates in the optimal mode while the gap is equal to zero, the spring suspension stiffness should meet the equation
where 
= stiffening coefficient = 1.1 to 1.3, assume 1.2
The upper springs are necessary to provide positive gaps, so their stiffness should be minimal to provide undisplaced operation the springs in the whole range of gap adjustment. Therefore
where Cb = stiffness of the upper springs
A = number of vibrations of the ram
a = maximum positive gap when the hammer is able to operate without danger of transferring into the impactless mode. When there is no limiter it is equal to the amplitude of vibrations
Assume a = 0.8.
where 
= coefficient which depends upon i and R’. Hammer coefficient of
velocity recovery may be increased up to R’ = 0.2. In this case = 7.1.
For calculation purposes let us assume A = 5.5. Now substitute the values into the formula
The bottom spring stiffness is then equal to
Now let us determine the maximum deformations. For upper and lower springs,
where b = negative gap. It is considered equal to “a” (maximum positive gap)
Assume 
.
Because of design considerations use four (4) upper and four (4) lower springs. The stiffness of one upper spring is
and the stiffness of one lower spring is
The material for the spring is “60 Sg” steel. The permissible tension in this steel is 
kgf/cm².
Tension per kgf of load is
According to the table of S.I. Lukowsky choose the spring as follows:
The stiffness of one turn 
and the number of working turns is
Assume 
turns. For this spring,
. The actual tensions in the spring are as follows:
and the total number of turns is
The full free height of the spring is
The distance between the support surfaces while the gap is equal to zero is
According to the table the closest value A = 4.24 corresponds to the spring with dimensions
The stiffness of one turn is equal to 
. The number of working turns is
Assume 10 turns.
The total number of turns is
The spring height in free position equals to
With this we begin a series of posts on the S-834 impact-vibration hammer, which the VNIIstroidormash institute in Moscow designed and produced in the early 1960’s. With the revived interest in Soviet and Russian technology, it’s a detailed look at how Soviet equipment designers came up with an equipment configuration. But it’s also a close-up view of how heavy machinery in general and pile driving equipment in particular is designed.
The impact-vibration hammer was a long-time interest for Soviet construction machinery institutes from 1954 to 1970. An overview of the history of this type of equipment in the Soviet Union is here. Since vibratory pile driving equipment was first developed in the Soviet Union, it’s also interesting to look at the entire subject; that overview is here.
The series is in three parts:
The specifications for the S-834 are here. What follows is an overview of the hammer itself and its general construction. We apologise for the poor quality of the scans.
In the posts that follow, the design calculations for the S-834 will be presented. In looking at the work of Soviet designers, it was tempting to revise the calculations. For one thing, although the metric system was introduced with the Russian Revolution, their implementation of the system is not really the “SI” system taught today, especially with the use of the kilogram-force. (That’s also true with many other Continental countries such as Germany and France.) For another, Russian technical prose can be very cryptic.
In the end, it was decided to reproduce the calculations pretty much “as they are,” with a minimum of revision. We apologise for the inconsistent sizing of the equations. Most of the transcription of this information was done in the 1990’s in Microsoft Word, and its conversion to HTML (for this format) in LibreOffice made the equations graphics (a good thing) but inconsistently sized the images (a bad thing.) This is one reason why we’ve migrated to LaTex for our newer technical productions online.
As with much of the Soviet material on vibration and impact-vibration pile driving, I am indebted to VNIIstroidormash’s L.V. Erofeev for the material itself and V.A. Nifontov for its translation.
All fluid flow in Vulcan hammers is regulated and directed by a valve. For most Vulcan hammers (the California series being a notable exception, the #5 is another) the valve is a Corliss type valve modified from those used in steam engines. Simple and reliable, it, like any other valve, is subject to losses as the air or steam passes through it. These are reflected in the mechanical efficiency of the hammer.
The losses due to air or steam flowing through the valve are generally not the most significant source of energy losses in a pile hammer. In the late 1970’s and early 1980’s, with the increase in sheer size of the hammers, these losses became of more concern. It was necessary to at least attempt to quantify these losses instead of using a “standard” back pressure value.
In May 1979 Vulcan contacted the Georgia Institute of Technology in Atlanta about using a Vulcan #1 series valve (like used in the #1, 06, etc.) in a test to determine the losses of air flowing through these valves. At this point a major problem was encountered: the air flow required to properly test the valve was too large for Georgia Tech’s equipment. Reaching out to Lockheed didn’t help either; they couldn’t do it. At this point Vulcan came up with an alternative: use the DGH-100 valve, which was a Corliss valve albeit much smaller, for the test. Making things easier was the fact that the DGH-100 used a small aluminium valve chest, which made the valve mounting simpler.
This proved feasible and Vulcan received a proposal from Brady R. Daniel at Georgia Tech for these tests. The valve was tested in two “configurations”:
The tests were run and the report was presented in October 1980. The immediate results were as follows:
The report also contained some suggestions for “streamlining” the design of the valve. These were not adopted, and the reason should be noted.
With the Corliss type valve, the Valve Port 1 is continuously pressurised, and this in turn forces the valve against the valve chest (or liner in the case of most newer Vulcan hammers.) With proper lubricant this seals the valve and further sealing (rings, seals, etc.) are unnecessary. This is a major reason why Vulcan hammers are as reliable as they are under the dire circumstances many operate. But that comes with a price. As with any design, there are trade-offs, and in this case the simplicity of the valve is traded off for efficiency. The simplest way to deal with this is to properly size the valve, and this was the main reason for the Valve Loss Study.
The Valve Loss Study is an interesting example of design analysis (others are here) which even an old product line like Vulcan’s can benefit from.
One of the things that was attempted in the TAMWAVE project is the use of cavity expansion theory to estimate soil set-up in cohesive soils. Doing this, however, brought some complications that need some explanation. Cavity expansion theory is basically the study of what happens when one body expands inside of another. When this takes […]
via TAMWAVE: Cavity Expansion Theory and Soil Set-Up — vulcanhammer.net
Vulcan had introduced its extractor line in the late 1920’s, after several design iterations. They had proven successful; for example, they were used in the construction of the original Tennessee Valley Authority systems of locks and dams. But, as is often the case with pile driving equipment, what contractors wanted could be summed up in one word: bigger.
In the extractor field, they got what they asked for: in 1954 Vulcan introduced the 1200A extractor, the largest in the and larger than any of the MKT “E” type extractors, their main competitor. To debut the line Vulcan did something completely different with its literature: it used a circus theme to emphasize its large size. You can see this below.
This may look silly today, but these days when we emphasize size, it’s completely different…when Vulcan came “down to earth” around the time it moved to Chattanooga, they put out this sheet, which shows all of the sizes and their specifications.
In the last post we looked at the SPW 2006 program analyze cantilever walls. In this post we will look at anchored walls, which are commonly seen with permanent works. The program, along with the example problem at hand, is here. Some instructions on the basic working of the program is here.
The problem we’ll be analyzing is once again from the BSC Piling Handbook (1984).
The soil profile input is similar to the cantilever wall except that it is necessary to put a layer boundary at the anchor point. Left and right side data, both tabular and graphic, are shown below.
The same comments re Kp, Kn, q and Dw apply here as they do to the cantilever wall. Note that SPW 2006 allows the entry of differing water table levels on each side of the wall. Note also in the original BSC diagram that a weep hole is installed in the sheeting. Proper drainage is essential for the relief of unbalanced hydrostatic forces.
In any case the one input not present with the cantilever wall is the anchorage. SPW 2006 does not have provision for anchor design. A brief summary of this is given below, from DM 7; more information on this is found in Sheet Pile Design by Pile Buck.
For our purposes we chose to specify a very stiff anchor, which renders the anchorage point essentially a fixed support, as shown at the right.
If one wants to consider the actual anchorage stiffness, it is necessary to determine the length, the cross-sectional area and the material to establish the stiffness, maximum stress and force, and the deflection at which plastic yielding takes place. If this is too much, a very stiff anchor is necessary; a very flexible anchor will render the calculations nearly useless. We have not included consideration of the flexibility of the soil bearing against the deadman; this further complicates the anchor input.
The graphical output (using PZ-22 sheeting) is shown below.
It is left as an exercise to show that the sheeting is adequate (or not) for the moment, using the same considerations as for cantilever walls.
Because of the soil-structure interaction, it is (in principle) unnecessary to apply Rowe’s moment reduction technique. That technique was developed to address deficiencies in the classical methods, which did not consider the interaction of the flexible sheeting with the relatively soft soil. Also, the technique of increasing the sheeting depth until a workable model is achieved is essentially a “free-earth” method. It is possible to apply the end conditions of the fixed earth (Blum’s or elastic line method) method; however, it can be tedious. Obviously SPW 2006 is happy to analyze sheeting lengths longer than the minimum required for geotechnical stability, and so if one wants to reduce the maximum moment (and thus the sheeting profile) a solution between the two can be found.
As an educational tool, SPW 2006 fits the bill nicely. It requires very few system resources and no installation. It has a very comprehensive and detailed input and uses soil-structure interaction methods which are becoming more common with retaining wall software. (OTOH, many engineers and owners are not “sold” on SSI, and prefer “classical” methods.)
For use in design, SPW 2006 simply lacks many of the convenience features that one expects with commercial software, and these can make using the program a time-consuming and mistake-prone business in a commercial environment. For those who want to graduate from strictly classical methods to SSI ones, it can be very useful for both training and as a check. But commercial use of this program is not recommended.
In our last post, we introduced the SPW 2006 sheet piling software, intended for educational purposes. The software can be downloaded here. In this installment we’ll look at its application to cantilever walls, i.e., those walls with no additional support other than the soil itself. These are used in temporary works. The file for this can be found with the software.
The problem is this one, taken from the BSC Piling Handbook, Fourth Edition (1984).
This is a fairly simple problem except that it has two different soil layers and properties. We’ll use the active and passive earth pressure coefficient values given in the example, although these can easily be computed from equations given in Verruijt or DM 7.
Based on this, the left side soil profile after input looks like this:
And the right side:
We note the following:
which is carried from the top downwards. It’s possible to vary that surcharge with depth; however, the program has no method of automatically computing variations in surcharge loading due to surface loads such as line and strip loads.
Selecting the proper case from the table at the bottom, the stroke can be computed as follows:
It is possible to be very precise with this calculation. For example, one could estimate the penetration below the dredge line 
In any case the soil profile looks like this:
The correspondence of the sections with the original problem is easily seen.
Now we select a sheeting length and a profile. We’ll select a length of 10 m (you will need to iterate from a short length, perhaps 6m and go upwards until you get a result that does not produce an error.) We’ll also start by assuming Profile #1 (Hoesch 95.) Running this yields the following beam diagrams:
The maximum moment is around 235 kN-m/m. But is this section suitable for this level of moment? The simplest way is to compute the maximum moment the sheeting section is capable of, and this can be done using the equation
Here 
Assuming that the sheeting is made of ASTM A572 Fr. 50 with an allowable stress of 220 MPa, for Hoesch 95 the maximum moment is as follows:
Obviously this is too light of a section for the moment level. This indicates that the EI of an acceptable section should be 
The solution printout is here.
In the next post, we will consider the case of an anchored wall.
![M_{max} = \frac{\sigma_{max}\left[EI \right]}{Eh}](https://s0.wp.com/latex.php?latex=M_%7Bmax%7D+%3D+%5Cfrac%7B%5Csigma_%7Bmax%7D%5Cleft%5BEI+%5Cright%5D%7D%7BEh%7D+&bg=ffffff&fg=333333&s=0 "M_{max} = \frac{\sigma_{max}\left[EI \right]}{Eh}")
