Those Pesky Kilogram-Force Units

Generally speaking, engineers educated in the U.S. must be educated in two units: the U.S. system (the Brits abandoned the Imperial system long ago) and the S.I. system, commonly called the “metric system.” I say commonly because they’re not really the same; countries that have been using the system the French came up with it during their Revolution (such as France itself, Germany and Russia) don’t always adhere to the S.I. system that Americans are taught.

The key problem is the unit of force. Americans have struggled with the difference between pound-force (lbf) and pound-mass (lbm,) throwing in the slug to try to solve the mess. But countries using a metric system have confused things as well. Engineers normally think that kilograms are restricted to being a unit of mass, but sad to say in many places we see kilograms used as a unit of force (sometimes it’s called a kilopond when it’s force, sometimes it isn’t.)

In a sense the confusion starts with daily life. We say that a person weighs 70 kilograms, and certainly any scale will measure the force on it, not the mass. But let’s consider something with a little more engineering sophistication but not much: the rated striking energy of a pile hammer.

Vulcan pile hammers are mostly single-acting. Their rated energy is the product of the force of gravity acting on the ram and the distance it drops before impacting the hammer cushion. In U.S. units, that usually means the weight of the ram (the force of gravity acting on its mass) in pounds-force multiplied by the drop in feet, for units of foot-pounds.

Vulcan exported a great deal of its output from the 1880’s onward, but the years it was really active in offshore oil platform construction saw a third of its output go outside of the country. That meant putting the specs into metric units as well. I say “metric” because the energy unit adopted was the kilogram-meter, where the weight of the ram in kilogram-force was multiplied by the drop in meters. This can be seen in the specification sheet below, in this case for the Vulcan 060 hammer.

When I came to Vulcan in 1978, having been schooled in the SI system, I thought this was wrong, and used the kN-m (kilonewtons being the SI units of force and meter length) or kJ (kilojoules being equal to kN-m.) So I changed the specifications accordingly. It wasn’t long before one of our Dutch customers complained that our specifications should be in “tonne-metres,” like our German competitor Menck. The “tonne” is the metric ton, or 1000 kg. It’s simply the kilogram-metre with the decimal point moved over three places.

That was the case for a long time. In Vulcan’s later years (the 1980’s and 1990’s) the SI specification was brought around to true SI units with the kJ as the unit of energy, as you can see below.

Offshore Specifications for 010-340 Hammers

So how do we deal with this problem? Let’s start by noting the following:

1 kilogram-force or kilopond = 9.80665 N
1 metric ton = 1000 kilogram-force = 9.80665 kN
1 N = 0.101972 kilogram-force
1 kN = 0.101972 metric ton

If some of those figures look suspiciously like the acceleration due to gravity of $~ 9.8 \frac{m}{sec^2}$ they should. Getting into using gravity in a kilogram-force system is just about as big of an ordeal as doing it in U.S. units, so let’s just look at a few simple examples to illustrate the difference.

Rated Striking Energy

Let’s go back to the 060 above. The 060 has a ram mass/weight of 27,273 kg or 27.273 metric tons. The stroke is 914 mm or 0.914 m. The energy is thus $27,273 \times 0.914 = 24,894\,kg-m$ or $27.273 \times 0.914 = 24.9\, tonne-metres$ . If we go to the SI system, we first convert the ram weight in metric tons to kN, thus $27.273 \times 9.80665 = 267.5\,kN$ , them multiply it by the stroke to yield $267.5 \times 0.914 = 244.5\,kN-m =244.5\,kJ$ .

Unit Weight of Soils

Most geotechnical calculations are static in nature, and some of the ones which aren’t are pseudo static (like the Mononobe-Okabe method.) But that hasn’t stopped the use of kilograms in computing unit weights of soil. An example of this is shown below, from Bengt Fellenius’ UniPhase program:

      ------------------------------------------------------------
      UniPhase 2.1                                     23 Sep 2019
      Project: HS1 Problem 5
      ------------------------------------------------------------
      Water density        :  1000.00  (kg/m^3)
      Solid density        :  2720.00  (kg/m^3)

      Degree of saturation :  89.00    (%)
      Water content        :  28.00    (%)
      Dry density          :  1465.73  (kg/m^3)
      Total density        :  1876.13  (kg/m^3)

      Saturated density    :  1926.86  (kg/m^3)
      Void ratio           :  0.86     (-)
      Porosity             :  46.11    (%)
      ------------------------------------------------------------

As shown here, it’s possible to do phase calculations in kilogram-force units as shown above. The problem then comes in applying them to effective stress calculations, which are usually done in SI units. Simplest way is to first note that the density of water is $1000 \frac{kg}{m^3}$ , or more simply put $1 \frac{tonne}{m^3}$ . To convert the total (or dry) density of the soil, move the decimal point three places to the left (converting it to metric tons) and then to convert to kN. So, for the total density $1.876 \times 9.80665=18.4\,\frac{kN}{m^3}$ .

It’s also worth noting that $1000\,\frac{kg}{m^3}=1\,\frac{g}{cm^3}$ , which you sometimes see in the old literature. That gets us into CGS territory, but that’s another post…

Bending Moment Problem

An example of a bending moment problem is shown in the calculations for the Soviet S-834 impact vibration hammer. The physical drawing for the beam (the reactions are at the bearings) is here:

The diagram for the vertical reactions is here:

Diagram for the vertical forces and reactions.

The vertical forces and locations are as follows:

$P_g = 1250\,kg$
$Q_1 = 4500\,kg$
$Q_2 = 450\,kg$
$Q_3 = 3300\,kg$
$l_1 = 9.2\,cm$
$l_2 = 19\,cm$
$l_3 = 5.7\,cm$

We’ll try to keep the calculations to a minimum; our objective is to illustrate the units. The original post has all the details. One thing you’ve probably noted is the use of cm, adding injury to insult in getting this to SI units.

Let’s find the bending moment and stress at the Section C-C for the vertical reactions at the centre of the shaft, where the diameter is 6.2 cm.

Because of symmetry, the reactions at the supports are both 4125 kgf, or 4.125 metric tons, or (4.125)(9.8) = 40.43 kN. The vertical moment is computed to be 48,500 kgf-cm, or 48.5 tonnes-cm; however, because of the influence of the horizontal forces, this is increased to 50 tonnes-cm. Converting the last moment, (50 tonnes-cm)(9.8 kN/tonnes)(1 m/100 cm) = 4.9 kN-m = 4.9 kJ.

Since the bending stress $\sigma = \frac{M}{S}$ , we need to compute the section modulus, which for a circle is $S = \frac{\pi d^3}{32}$ . Converting to metres and substituting, $S=\frac{\pi \times 0.062^3}{32} = 0.000023398\,m^3$ . (One of the real drawbacks of the SI system is its use of metres at times like this, as anyone who has been involved in sheet pile design will attest.) Thus the stress will be $\sigma = \frac{4.9}{0.000023398} = 209421\,kPa = 209.4\,MPa$ .

The original calculations yielded a result of $2080\,\frac{kgf}{cm^2}$ . We convert this as follows: (2.08 tonnes/sq.cm.)(9.8 kN/tonnes)(10,000 sq.cm./sq.m.) = 203,840 kPa = 203.8 MPa. The difference is because the Russians took a few, er, short-cuts in their calculations.

Soil Mechanics

Units in soil mechanics are a little different for soil mechanics than they are for machinery, because the distances involved are larger. For example, in U.S. units we speak of pounds/square inch while soil mechanics we speak of pounds/square foot (or kips/square foot.) Recently we published on a companion site Soil Mechanics by N.A. Tsytovich, a Russian text from the early 1970’s, and once again those pesky kilogram-force units appear there. So some explanation is necessary.

Tsytovich uses the $\frac{kg_f}{cm^2}$ as his principal unit of pressure and stress. If we convert to conventional SI units, 1 kg_f/cm² = 98.1 kPa. That, as it turns out, is nearly standard atmospheric pressure. Many correlations implicitly or explicitly use atmospheric pressure. For example, the “standard” correlation for SPT correction for effective stress is

$C_{N}=\sqrt{\frac{p_{atm}}{p_{0}}},\,C_{N}\leq2$

Using the kg_f/cm² units, we would only have to invert them and take the square root. Thus, if our effective stress is 4 kg_f/cm², then $C_{N}=\sqrt{\frac{1}{4}}=0.5$ .