Is 3500 lb/in Referring to Force per Inch of Width?

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Discussion Overview

The discussion revolves around the interpretation of a failure load measurement of 3500 lb/in for a composite material, specifically CFRP (Carbon Fiber Reinforced Polymer). Participants are exploring the implications of this measurement in relation to tensile testing and the dimensions of the test piece, including its width and thickness.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the 3500 lb/in measurement indicates that the composite can withstand 3500 lb of force for every inch of width.
  • Another participant emphasizes the need for more information about the material properties to assess the validity of the 3500 lb/in figure.
  • A later reply suggests that the 3500 lb/in might refer to a single ply of CFRP rather than the entire composite piece with 12 plies.
  • There is speculation that if the failure load is indeed 3500 lb/in, then a 2-inch wide piece could theoretically fail at 7000 lb/in, although this is presented as a guess pending further information.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the interpretation of the 3500 lb/in measurement and whether it applies to the entire composite or just a single ply. No consensus is reached on the implications of this measurement.

Contextual Notes

Participants note the importance of understanding the material properties and the specific equations used in the spreadsheet to accurately interpret the failure load. There is an acknowledgment of missing information that could clarify the situation.

Corsan
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Hi all,
I have a spreadsheet which says failure will occur in my composite piece when a load of 3500 lb/in is applied.
If I were to tensile test this piece and the thickness has already been input (0.138 in) does this mean for every inch wide the piece can take 3500lb of force?
That seems high?

I'm confused about the units of measurement and if someone can clarify this I'd appreciate it,
Regards
 
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This is impossible to answer the general question without a complete description of what you are calculating.

For the particular number you got, it's impossible to say whether 3500 lbf is a sensible load to apply over a 1 x 0.138 inch area unless you tell us what material the test piece is made from.
 
AlephZero said:
This is impossible to answer the general question without a complete description of what you are calculating.

For the particular number you got, it's impossible to say whether 3500 lbf is a sensible load to apply over a 1 x 0.138 inch area unless you tell us what material the test piece is made from.

Ah right okay, its a 10" x 1" x 0.138" thick piece of CFRP (12 plies).

I'm planning on tensile testing unti failure and a spreadhseet I have from the internet says the failure load is 3500 lb/in but nowhere on this does it state the width/length of the piece so its obviously working it out from the ply properties (E1,E2...etc).

What I'm asking is, if that is the 'failure load' then does that mean that for every 1" width that the plies can take 3500 lb/in force?
 
I would have thought a 1 x 0.138 section of CFRP would take a lot more than that 3500lb. Even mild steel would do that.

Possibly your "3500 lb/in" is for one ply of CFRP, not for 12 ... ?
 
AlephZero said:
I would have thought a 1 x 0.138 section of CFRP would take a lot more than that 3500lb. Even mild steel would do that.

Possibly your "3500 lb/in" is for one ply of CFRP, not for 12 ... ?

Okay, I'll keep that in mind during the test.

Am I right in what I say though?
i.e. the 3500 lb/in refers to the width of the cross section, so if it were 2 inches wide the predicted failure would be 7000 lb/in?

Regards
 
Corsan said:
Am I right in what I say though?
i.e. the 3500 lb/in refers to the width of the cross section, so if it were 2 inches wide the predicted failure would be 7000 lb/in?

That seems a reasonable guess, but without seeing the complete spreadsheet (or at least the relevant equations) and understanding how it works, it is only a guess and I don't accept any responsibility if it turns out to be wrong!
 

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