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COSmos
Vince Adams May 16, 2007 7:09 PM (in response to Troy Higgins)Hi Troy, the Multiplication Factor acts on your allowable stress. Therefore, using your example, if your allowable stress (yield stress?) was 10,000 and your max model stress was 5,000, you would have a SF of 2.0 with no multiplication factor. Now, if you use a multiplication factor of 4, CW would assume your allowable stress was 40,000 so, with the same 5,000 stress, your safety factor would be 8.0.
Does this explain it for you?
Vince 
COSmos
Ranga Narasimhan May 19, 2007 6:28 PM (in response to Troy Higgins)Hi,
what is the use of multiplication factor.. when my allowed stress itself is 10,000.. how does it make sense to set it as 40,000. In other words, in what situation i need to look after this feature?
thanks

COSmos
Vince Adams May 22, 2007 7:13 PM (in response to Ranga Narasimhan)Great question since I wasn't sure about why you might use this either. After talking to some of the people who help define this feature, it turns out the Multiplicaiton Factor is there for users who have a spec stating an allowable stress in terms of a % of yield or ultimate tensile. Some users have requested that the safety factor be built into the allowable value. In short, instead of requiring a safety factor of 2 on yield, some users want a safey factor of 1 on a reduced yield allowable.
Does that make sense? Can anyone suggest an alternate or more intuitive way to implement this?
Vince
COSmos
Peter Gillespie May 22, 2007 8:13 PM (in response to Vince Adams)From your description, it seems like wording such as "% of Yield / Ultimate Allowable" and a box that takes a percentage would be more clear.
We work with FOS, so I don't think I would personally use the box, but it would have taken awhile to figure out "Multiplication Factor" by myself.
COSmos
Troy Higgins Jul 27, 2007 12:45 PM (in response to Peter Gillespie)Well the way I do it is like this:
If I need a FOS of lets say 4, I apply 4 x the force and look for aFOS of 1.
Say I need to test a force of 50 lbs, I apply 100 which would be aFOS of 4, then If my results come out to 1 or above Im good.
COSmos
Pete Yodis Jul 27, 2007 1:33 PM (in response to Troy Higgins)That approach makes the assumption that you have a linear relationship/behavior between load applied and resutling stress. Just realize it and be careful, as it may not be the case for every problem and the assumption might not be valid.
Pete



