Is that the beginning of roadway damage?

Solvers of the Five Axles puzzle discovered that differences in static axle-loadings for two types of trucks, C-rig and B-rig, cannot explain differences in roadway damage...

 Findings from Five-Axles Puzzle Topic Description C-rig B-rig Note 1. Roadway Damage More Less 1 2. Roadway Construction Technology Comparable 1 3. Weather Severity (erosion, freezing) Less More 4. Total Gross Weight (load limit) 40 tons 38 tonnes 2 5. Total Number of Axles 5 5 6. Peak Axle Load 18,970 lbs 32,500 lbs 7. Number of Wheels 18 12 8. Number of Duals vs Singles 8 vs 2 2 vs 10 9. Peak Wheel Load 4,740 lbs 8,140 lbs 10. Average Wheel Load (nominal) 4,444 lbs 6,666 lbs 3 11. Compression Stress per Contact Patch 45 psi 67 psi 4 12. Shear Stress per Contact Patch 10 psi 20 psi 5

Notes

1 Subjective observations by author (see text)

2 US limit 40 tons = 80,000 lbs; European limit 38 tonnes = 83,775 lbs

3 Equal gross weight for puzzle: 80,000 lbs divided by number of wheels per rig.

4 Estimated contact patch area 11 inches wide 9 inches long: 99 square inches.

5 Drive-wheels only: accelerating at 0.1 g (2.2 mph/sec), friction coefficient ≈ 1.

In either California or Brittany a freshly paved roadway may be considered 'perfect' with only gradual changes in elevation over long stretches.  As vehicles of various kinds come rolling along, dynamic forces are imposed on the roadway pavement by their wheels.  A typical segment of flat and level pavement is called upon to support the rather sudden application of wheel loadings: 950 lbs per wheel for a typical automobile, 4,444 lbs per wheel for a fully loaded C-rig and 6,666 lbs per wheel for a B-rig.  After a period of time (some 9 milliseconds at 60 mph), weight is just as suddenly relieved on that segment and moved on to another. At a given location, such impacts take place five times for each C-rig or B-rig that rolls over it.
Solvers may want to review how wheel-load, which is a vertical force, creates stress in the roadway materials underneath and necessarily results in strain.  Two types of stress are especially important: compression and shear. There are a half-dozen conventional failure theories, including stress (normal or shear), energy (strain or distortion), fracture or creep.
A roadway may start out in a 'perfect' condition when freshly paved but won't usually stay perfect.  For the Unsprung Secrets puzzle we will set aside environmental causes, especially water erosion of the subbase and icing conditions that can crack pavements.  Both may actually be more problematic in Brittany than in California.  Roadways naturally do start out with gentle humps and dips. With wear and tear these can become imperfections in many forms.  Our efforts must find out why damages are caused by C-rigs more than by B-rigsOur suspicion lies where the rubber meets to road...

Unsprung Mass

Since the earliest horse-drawn carriages, vehicles have been sprung -- 'spring-loaded' to protect their contents from bumps and potholes.  The vehicle's 'suspension system', which includes the heels and associated hardware, necessarily follow the vertical profiles of roadways.  Those parts of the suspension system are said to be unsprung, and minimizing unsprung mass is an important subject in assuring road-ability of vehicles in the presence of various imperfections.

Modern suspension systems assure safe-handling by drivers and comfortable riding for passengers.
An ideal
unsprung mass of zero would completely conceal vertical accelerations caused by irregularities in the roadway from the body of the vehicle and from people inside and from cargo in a truck.  That ideal would not impose dynamic forces on the roadway apart from the varying static loads described above.  A real suspension system on either a C-rig or a B-rig must accommodate non-zero unsprung mass.  Thus the suspension system transmits inertial stresses onto roadways, potentially creating and then aggravating defects in pavements, making small cracks larger and shallow pot-holes deeper.

Trucking regulations limit static axle-loadings but do not explicitly limit dynamic forces attributable to unsprung mass.  Let us explore that subject as it pertains to roadway damage.

Here is a convenient model for a truck wheel and prorated mechanical elements in the suspension system.  Using the sketch below, we find that
mU = wU / g slugs, where wU is the unsprung static weight in pounds and g is the acceleration of gravity (32.2 fpsps).

Legend for Unsprung Secrets Model.
• Static Wheel Load...includes tare weight and payload supported by suspension system
• Unsprung Static Weight.....includes wheel, tire, and pro-rata axle, suspension, brakes
• Pavement Force..............................includes vertical static loads and inertial forces
• Pressure Distribution........................profile of compression forces in civil structure
• Contact Patch........................area determined by Pavement Force and tire inflation
• Horizontal Momentum..............vector product of truck velocity and unsprung mass
The weight wU and therefore the mass mU for a given wheel and its associated hardware varies with the rôle being performed: [1] steering, [2] propelling, and [3] supporting the truck -- as can be identified in this sketch from the Five Axles puzzle...

Typical parameters from product literature for truck-tires will be used for modeling dynamic impacts on roadways: tire diameter 42 inches and weight 130 pounds.  The unsprung static weight
wU -- tire, rim, hub -- totals about 260 pounds, thus the unsprung mass mU 8 slugs.

Potential Energy

If roadway damage by trucks is to be blamed on onerous pounding by unsprung mass, one might reasonably suppose that vertical acceleration afforded by would apply.   Solvers of the Five Axles puzzle will notice that unsprung static weight wU of 260 pounds (lbs) is quite small compared to the static wheel-load fW pressing downward by the spring or, most commonly these days, from air-filled bellows in the suspension system...
C-rig:  fW = (wT + wL) / nW - uW 80,000 / 18 - 260 = 4,444 - 260 = 4,184 lbs
B-rig:  fW = (wT + wL) / nW - uW 80,000 / 12 - 260 = 6,666 - 260 = 6,406 lbs
The ratio fW / uW represents the acceleration aY of the unsprung mass if it were free to move vertically compared to the acceleration of a body falling under the influence of gravity alone, which is g = 32.2 fpsps.  In solving the Unsprung Secrets puzzle, we need to show that the five wheel-impacts on a given roadway segment are cumulatively 1.5 times worse for the C-rig compared to the B-rig.  However, the ratios of acceleration are the wrong way 'round for making that case: aY = 16.1 g for the C-rig and aY = 24.6 g for the B-rig. So much for potential energy in the impact comparison.
Kinetic Energy
The tractive effort of the prime mover in the tractor provides kinetic energy to the truck as a whole, including the unsprung mass.  We continue our analysis by postulating that the dynamic impact of the unsprung mass of a truck wheel is somehow proportional to that kinetic energy, some part of which gets to the roadway pavement.

Each truck-wheel has translational kinetic energy given by
mU vX 2 / 2 ft-lb.  Sophisticated solvers will want to take into account the fact that the wheel is also rotating at some angular velocity ωU radians-per-second, which is given by ωU = vX / rU, where rU is the radius of the wheel in feet.  The rotational kinetic energy is given by iU ωU 2 / 2 ft-lb, where iU is taken as the moment of inertia for a cylinder of radius rU and given by iU = mU rU 2 / 2 lb-ft/sec2.

The total kinetic energy of the wheel is the sum of its translational and rotational kinetic energies...
mU vX 2/2 + iU  ωU 2/2 = [mU vX 2  + (mU rU 2/2) ( vX / rU) 2] / 2 = 3 mU vX 2 / 4
A wheel with mass mU moves along the roadway at the same rectilinear velocity as the truck vX  feet-per-second.  Consider what occurs at the instant the unsprung wheel passes over a deflection in the roadway -- even a mild deflection.  From the diagram above, we see that, whereas 16 of the 18 wheels on the C-rig are configured as duals, only 4 of the 12 wheels on the B-rig are configured as duals.  Should solvers consider this to be one of the Unsprung Secrets?  Not so much.  Here's why...
The kinetic energy of the unsprung mass is proportional to the mass mU.  The combined mass for dual wheels will be twice that of a single wheel 2mU.  By the same token, however, the two contact patches in the dual configuration can be presumed to transfer energy equally and thus nullify differences in roadway impacts attributable to kinetic energy from a wheel configuration with doubling the mass.
The kinetic energy of a 260-lb truck wheel at 30 mph is about 11,600 ft-lbs.  At twice that speed, 60 mph, that figure quadruples to 46,500 ft-lbs. Oh, but kinetic energy is really a horizontal kind of thing.
What percentage of kinetic energy will get transferred to the roadway pavement?
Kinetic energy is a scalar quantity not a vectorNevertheless, to answer the question, we can avoid vector-mathematics (force, momentum, impulse), and we can also set aside local variations in tire pressure, mechanical distortions of contact patches, and flexing of sidewalls.  Here's how: We simply observe that in rolling over a deflection -- either a hump or a dip -- the wheel must follow a longer path than it would in continuing on the level roadway.
Here is a sketch of a roadway deflection.  It is exaggerated in the vertical direction.  One might expect a new roadway to have gentle deflections of no more than a few inches over a span of several feet.  Please excuse a statement of the obvious...

Imagine a truck wheel on either a C-rig or a B-rig rolling along on a straight roadway with a kinetic energy of 46,500 ft-lbs.  Suppose the wheel comes to a deflection in the roadway.  If deflected wheels are to keep up with undeflected wheels on the same truck, the time for traveling over the deflection tD must be the same as for wheels on straight roadway.  Accordingly, the average tangential velocity of deflected wheels must momentarily increase

 Sophisticated solvers avoid fallacies that lurk in averages by keeping time in the numerator.

A small deflection amounting to a 1-ft hump over a 100-foot span of the roadway will result in a change of kinetic energy by about 0.5%.  A 260-lb truck-wheel at 60 mph might then transfer 0.5% of its kinetic energy or 232 ft-lb to the pavement over the span of the hump. That's not an much of an impact, but -- hey, whatever it amounts to, we do not see any difference between the C-rig and B-rig in the transfer of kinetic energy via unsprung wheels.
By the way, the model transfers kinetic energy to the roadway not as a vertical impact but as a horizontal shear force along the pavement.  Hmm...
Pending Discovery
A truck's suspension system protects the roadway from damage by the truck as much as it protects the truck itself from an already damaged roadway.  For our solution to the Unsprung Secrets puzzle, we must look for a significant difference in the designs of the C-rig and the B-rig.  Our investigation of kinetic energy has given us a clue...

 The shortest distance between two points is a straight line.

Thus, we find ourselves trying to use the extra distance that must be traveled by an unsprung wheel in negotiating a roadway deflection.  But we are still looking for our solution.  Solvers need not despair.  A genuine discovery is at hand.  Exclamatory punctuation may be appropriate.

 What feature in the designs of the C-rig and B-rig accounts for the differences in roadway damage?