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