2015 AAR DP&FC Conference
Intermodal Weight Concentration
Norfolk Southern Railway
Matthew T. Hardin
Damage Prevention Commodity
Manager
Contents
• Definitions
• How to figure pounds per square foot
• Center of Gravity / Countering counterweights
• Container floor rating
• Why do we have failures
• How to distribute weight
• Current methods
Definitions
Center of Gravity
Is the average location of the weight of an object
Contact Area
When two objects touch, a certain portion of their
surface areas will be in contact with each other.
Surface Area
Is the sum of all the areas of all the shapes that
cover the surface of the object.
Linear Foot
Lineal feet, is a measure of length, one-foot length of
any long, narrow object.
Deflection
Action of deflecting or being deflected. The amount
by which something is deflected.
Force / Weight
Force is like water, it will find the weakest point.
Steel being such a harder material would actually
push against the force hence (reverse deflection)
or distribute the weight much better than softer
materials due to its density.
Force / Weight
Duct tape is like the force. It has a light side, a dark side, and it
holds the universe together....
How to Properly get LBS per Sqft
25,000 lbs per 10 linear feet (AAR rule), 300 lbs per square foot (NS IM rule circular). If you divide down the
runner chart in the AAR IM loading guide you get 304 lbs per square foot. Fact: the Holland sled can get a foot
print less than 294 lbs per square foot.
For everything else if starts to get a little gray.
First make sure you know how to get the pounds per square foot. You need to be able to measure the actual
surface area or contact area of the lading.
Contact Area
Box / square / rectangle (pallets, flat surface with equal sides) Width x length = sqft
Example 40” x 48” pallet weighting 2500 lbs; inches to feet 40/12 = 3.33 & 48/12 = 4
3.33 x 4 = 13.32 sqft; 2500lbs / 13.32 sqft = 187.68 lbs per sqft
Bottom of Paper Roll
Radius
Round / circular (paper rolls, other coils) Pi (3.14) x radius x radius = sqft
50” roll of paper weighting 7500lbs, inches to feet 50/12=4.16, radius is half of the diameter, = 2.08
3.14 x 2.08 x 2.08 = 13.58; 7500 lbs / 13.58 sqft = 552.28 lbs per sqft
Air tire Contact Point
Width
Length
Wheeled (forklift, tractor, scissor lift) Width of contact area of wheel x length of contact area of wheel
Wheel of tractor weighting 5000 lbs, 10 inches in width and 1 foot in length; inches to feet 10/12 = .833
.833 x 1 = .833 sqft, times four wheels = 3.33 sqft; 5000lbs / 3.33sqft = 1501.5 lbs per sqft
Forklifts and other like equipment has counterweights, axle weights will be different. Forklift
weighing 5000lbs rear axle could be distributing 3200 lbs to the rear two wheels verse the front axle
will only be supporting 1800 lbs. This would give you a different foot print on the rear two wheels
(1920.76 lbs per sqft on the rear two wheels) than the front two wheels (1080.43 lbs per sqft on the
front two wheels).
Not so Heavy end
Heavy end
Center of Gravity
If there is a 900lbs difference in the rear and front of a piece of lading or equipment what should be done?
Depends on a lot of items however the only way to distribute the weight is by removing the counterweights or
raising the heavy end of the equipment to move the center of gravity.
The center of gravity height is found using the rules of trigonometry and right triangles. Specifically, we are using
the Law of Tangents, and the Pythagorean Theorem. Sticking with the forklift example from last slide.
Here if the each rear wheel weighted 1600 lbs a piece (3200 total on rear axle) and each of the front wheels was
only 900 lbs a piece (1800 total on front axle) we would have to raise the rear axle over 11 inches higher than the
front to center the weight and distribute OR counter balance the counterweights.
Center of Gravity
Center of Gravity
So…. If the wheeled equipment is heavy and there is a large difference in axle weights it could look something like this.
Too Easy!!
Things to Think About
What are we Working with?
AAR M-930 and M-931 has a bunch of requirements for container making, however there is no
requirement for how many or how far part these are.
Container sub-floor, lateral braces
What are we Working with?
How containers test and certify floor rating.
Test Procedure A forklift was loaded to create a 24,000 lbs. load on the front axle. The forklift was
driven down the center of the floor mock-up for 3,000 complete cycles. . The forklift was then driven
as near as possible to the edge of the floor for an additional 2,000 cycles. A further 100 cycles were
driven down the edge with 30,000 lbs (125% overload).
24,000 lb front axle load
30,000 lb front axle load
What are we Working with?
What are we Working with?
What are we Working with?
Looking at one incident with forklifts which weight approximately 8,700 lbs a piece. The wheel that broke
through was exactly between the lateral braces. The container was only seven years old, was there a
repair job here? Or was the focused weight too much in the wrong place?
They tell me there was not a repair job in the floor and the container floor rating was 24,000 lbs.
Screw head showing
location of floor joist
10 inches
What are we Working with?
Small Contact
area and Weight
Container Floor
How did that Fail?
According to AAR M-930 “Flooring (Recommended Practice) shall be laminated hardwood or equivalent
composite material. The minimum strength properties must equal or surpass those of white oak.”
Floors are made with the minimum strength of White Oak which has an compression parallel to grain (crushing
strength) of 7440 psi
Floor plank fracture by bending.
These forklifts where a little more than 8700lbs with quite a small contact point.
Lateral crossbeam
2175 lbs per wheel
Container Floor
1.5”
I = 6h³ / 12
I = 12 (1.5)² / 12 = 3.38 in
10”
12”
M = Pd = (5” x 2175 lb)
M= 10,875 in-lb
(10,875 in-lb) (.75 in) / 3.38 in = 2413.09 PSI
White Oak strength 7440 psi, our one wheel that fell thru the floor 2413 psi
Are we Distributing the Weight?
Weight
distribute
on board /
runner
Weight
distribute
on floor
Could raise up the 2x6 and place the
brick under it easily. If weight was
distribute across the board then this
would not be possible.
A flat board/runner will distribute the weight
directly downward if no deflection. Too
distribute the weight across a runner the
board would have to have some opposite
deflection or arch. As shown..
Weight
distribute on
defective of
arch
Weight
distribute
on floor
Are we Distributing the Weight?
Ok, so I am now saying that a long runner under a piece of lading or pallet does not distribute weight as
commonly believed. What does the runner do and why does it seem to be working?
Pine wood board used as a runner is rated 4800
psi, doing the math shows that it should be able to
hold 4320 lbs in the same area between the lateral
braces.
Pine Wood runner
Container Floor
10”
We already figured out that the White Oak
floor is 7440 psi, the math says that in our
little area above it should support a max of
6696 lbs.
12”
Pine max weight PLUS white oak max weight, would mean
we would have to place more than 11,016 pounds here to
exceed the strength of the floor and pine runner together.
This is straight weight/pressure downward, if any
impact occurs it would greatly magnify the pressure on the
floor.
A 1G vertical impact would reduce capacity of the
floor by 50%!
Are we Distributing the Weight?
Ok, so I am now saying that a long runner under a piece of lading or pallet does not distribute weight as
commonly believed. What does the runner do and why does it seem to be working?
Pine wood board used as a runner is rated 4800
psi, doing the math shows that it should be able to
hold 4320 lbs in the same area between the lateral
braces.
Pine Wood runner
Container Floor
10”
We already figured out that the White Oak
floor is 7440 psi, the math says that in our
little area above it should support a max of
6696 lbs.
12”
Pine max weight PLUS white oak max weight, would mean
we would have to place more than 11,016 pounds here to
exceed the strength of the floor and pine runner together.
This is straight weight/pressure downward, if any
impact occurs it would greatly magnify the pressure on the
floor.
A 1G vertical impact would reduce capacity of the
floor by 50%!
AAR ILG Chart
AAR ILG Chart
25,000 lbs
AAR ILG Chart
20.4 ft
Long
25,000 lbs
6.0 ft Wide
Pallet
AAR ILG Chart
Runners
20.4 ft
long
25,000 lbs
AAR ILG Chart
Three 4x6s runners, 20.4’ long
Runners
20.4 ft
long
20.6 x .5 = 10.3 x three of them =
30.9 sq ft contact area.
25,000 lbs / 30.9 sg ft =
809.06 lbs per sqft
25,000 lbs / 20.4 linear feet =
1225.49 lbs per lineal foot
25,000 lbs
Are we Distributing the Weight?
Where we go next.
Thank You