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OOSC, Part II, Item 2. Coupling Devices, a.(1)(c) and (2)(b) - Cracks Through 20% or More Original Welds or Parent Metal

Entry DateFebruary 12, 2010
Issue Number10-005-VEH
NameMilan Orbovich
AgencyThe Public Utilities Commission of Ohio (PUCO)
Address180 East Broad StreetColumbus, OH, 43215United StatesMap It (opens in a new tab)
Phone614-466-0423
Fax(614) 466-5202
Email[email protected]
CommitteeVehicle Committee
StatusClosed
Summary of Issue

Clarification on determining 20% of parent metal or cracks in original welds. The "NOTE" under Part II of the North American Standard Out-of-Service Criteria (OOSC), Item 2. Coupling Devices, a.(1)(c), a.(2)(b) and b(5).

Justification or Need

Clarification on determining 20% of the parent metal and original welds referenced in the notes under Part II of the OOSC, Item 2. Coupling Devices. There are several ways to interpret this note. Examples are as follows: 1. If a crack exists in a 5th wheel mounting plate, is the length or width of the crack compared to 20% of the area of the plate (length X width), just the length or just the width of the mounting plate, or do you measure in the direction of the crack the length of parent metal and then the length of the crack and determine 20? 2. How is the 20% parent metal determined if a crack exists in a mounting angle iron? Is the crack compared to the area of the angle iron (width X length), just the length of the angle iron, just the width of the angle iron or does it depend on which way the crack is going? 3. When determining the 20% of the welds crack. Which of the following scenarios does it refer to? Example: There are 5 ? 6 inch welds on fifth wheel mounting plate on the left and right side of the truck. Do you determine the total length of the cracks in all 5 welds, determine the total length of all 5 welds and then divide the length of the cracks into the length of the welds? (3 inches of cracks total divided into 30 inches of weld is 10%) Do you look at each weld individually and divide a crack in one weld into the length of that weld? (3 inch crack in a 6 inch weld is 50 %) Do you determine that you have 2 welds with cracks out of the 5 welds and simply divide the number of cracked welds into the number of total welds (2 welds with cracks into 5 available welds would be 40%)

Request for Action

Both notes reference "...cracks through 20% or more original welds or parent metal? when dealing with angle irons, mounting plates and the upper coupler assembly. The intent of this note needs to be clarified in both instances and reworded to prevent the misinterpretation and intent of the note.