Fire & Heat - Eurocode Parametric Fire Worked Example

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  • Опубліковано 1 гру 2024

КОМЕНТАРІ • 10

  • @brucezhang4278
    @brucezhang4278 3 роки тому +3

    I can't explain how lucky I am, this is EXACTLY what I need to learn. Thank you!

  • @alligodfreyukuni2975
    @alligodfreyukuni2975 5 місяців тому +1

    Great Explanation,

  • @LuziyaKambwela
    @LuziyaKambwela 5 місяців тому

    Are you able to make this lesson in a playlist. Fire design

    • @firesun
      @firesun  5 місяців тому

      A number of structural fire engineering videos are provided in this playlist: ua-cam.com/play/PLcXEIpaMKkuq2PknC_UsIErB1igWxPYQ2.html

  • @mileshermon9810
    @mileshermon9810 7 місяців тому

    so does this method not work if the enclosure is wood (as b is less than 400). If it is possible, how do you work it all out?

    • @firesun
      @firesun  7 місяців тому

      The reason it does not work if the enclosure is wood is because your HRR increases massively due to the contribution of the wood. The internal behaviour, burning duration, flame lengths, etc. will all be different from a timber compartment.

  • @satnamchana
    @satnamchana Рік тому

    If my Ted (equivalent time) is 35.51 mins so below my required resistance of R60 (60 mins) does that mean the beam fails?

    • @firesun
      @firesun  Рік тому

      It depends. If you calculate that your beam needs 35.5 minutes of equivalent resistance based on the room geometry, and you provide 60 minutes then you are fine. 60 minutes would normally represent the code requirement, whereas an equivalent design time of 35.5 min represents an updated requirement. Hence, they are requirements based on different assumptions, rather than one being pass or fail.

  • @cktong2711
    @cktong2711 2 роки тому

    At the 15:23, How to find Yq1=1.32? Thanks

    • @firesun
      @firesun  2 роки тому +1

      This comes from EN1991-1-2 where a factor based on the danger of activation is specified. Hence, it was taken from the code recommendations.