How to Calculate Reactions of a Cantilever Beam with a Uniformly Distributed Load

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  • Опубликовано: 7 ноя 2024

Комментарии • 37

  • @Eurocoded
    @Eurocoded  6 лет назад +9

    Like 👍🏻 share 👈🏻 and subscribe 🙌🏻

  • @soccerdogboy8946
    @soccerdogboy8946 4 года назад +4

    Thank you for the explanation! This was short and clear. Very helpful.

    • @Eurocoded
      @Eurocoded  4 года назад

      You are most welcome.

  • @clashroyale8683
    @clashroyale8683 5 лет назад +4

    Thank u sir
    Clear and effective explanation

  • @mbongogeoffred4424
    @mbongogeoffred4424 3 года назад +2

    Beautiful explanation sir, i was confusing how did u get the 5 metres because i self this since nine years ago after watching the video the second times i now recurred that moment is focused multiply perpendicular distance. Thanks 👍👍

  • @lisavanderwissel950
    @lisavanderwissel950 3 года назад +3

    How would you do this if the distributed load was a semicircle?

    • @Eurocoded
      @Eurocoded  3 года назад

      Calculate the area under the stress distribution and place the load at the centre of that area i.e. centre of the circle.

  • @AyandaMbhamali-yi8ee
    @AyandaMbhamali-yi8ee Год назад

    Where did you get that 5m

  • @leneay9
    @leneay9 5 лет назад +2

    thanks, clear and fast explanation!

  • @mukisajoshua2026
    @mukisajoshua2026 2 года назад +1

    Short and clear

  • @0therso
    @0therso 2 года назад

    Thanks for your explanation. Am I correct that the beam will fail when MA is greater than RA?

  • @Tranefine
    @Tranefine 4 года назад +3

    Thanks for the great explanation. Just one question: Am I correct that the direction of the bending moment depends on the direction of the load? So you could say, if there is a load negative to the Y-axis (as in your video), the Momentum M_A is anticlockise. If the load is in positive Y-direction, M_A is clockwise?

    • @Eurocoded
      @Eurocoded  4 года назад +1

      Tranefine you’re welcome. Watch this video where I have given a detailed answer 👉🏽 ruclips.net/video/6SJiDkYkOLs/видео.html

    • @Eurocoded
      @Eurocoded  4 года назад +1

      Tranefine let me know if the video answered your question 👍🏽

    • @Tranefine
      @Tranefine 4 года назад +1

      @@Eurocoded Thank you, sir! Unfortunately, not everything has been clear so I commented straight under the linked video. Hopefully, you can help me there!

    • @Eurocoded
      @Eurocoded  4 года назад +1

      Tranefine check out my comment under the linked video. Hopefully that will clear things up for you.

    • @Tranefine
      @Tranefine 4 года назад

      Eurocoded - it was definitely helpful, thank you!

  • @puritykhamala7518
    @puritykhamala7518 2 года назад

    Question please where is the 5m coming from.... distance ya RA... haven't got tht please......

  • @accadoe3584
    @accadoe3584 3 года назад +1

    Greetings Sir, have you done a video on frame reactions?

  • @TonyMrBoring
    @TonyMrBoring 5 лет назад +3

    Shouldn't the reaction Ra force in the y direction be negative?

    • @Eurocoded
      @Eurocoded  5 лет назад

      Hi Miss - 100kN force is acting downwards. Therefore, R_A reaction should be in the opposite direction (i.e. upwards). R_A is drawn as acting upwards. Hence, R_A is positive.

  • @ravishankaryadav1008
    @ravishankaryadav1008 2 года назад

    Sir at the fixed support A ,can we take RA support downward and moment MA clockwise. Please reply

  • @NdambomaAkem-vb2ug
    @NdambomaAkem-vb2ug Год назад

    Sir please how did u have 5 meters

  • @nicholaspeak3335
    @nicholaspeak3335 4 года назад +1

    when you have converted the UDL to point load and placed it in the centre of its load span. Can it be placed at any point on that span? and be added to another point load that would occur at 'C'?

    • @Eurocoded
      @Eurocoded  4 года назад +2

      nicholas peak In this example, converted UDL is applied 5m from A. If you offset this point to a different location (example: to C) you will then need to counter the change in moment resulting from shift of converted UDL. Then you can add the converted UDL to another point load applied at C.
      Example: Imagine the same beam arrangement with another point load of 50kN applied at C.
      If you shift the converted UDL to C then the lever arm becomes 7m when taking moments about point A, resulting in 700kNm bending moment about A. But in reality, it should be 500kNm (5m x 100kN). Therefore, we can add an external anti-clockwise bending moment of 200kNm to counter the additional 200kNm (700-500) resulting from shifting the converted ULD. Then you can take the sum of two point loads at C as 150kN.
      I don’t recommend above method for this sort of beam arrangement. It would be much simpler to look at equilibrium with two point loads applied at two different locations. However, sometimes you will need to adopt above method depending on the arrangement.

  • @chaminiwickramage7392
    @chaminiwickramage7392 2 года назад +1

    Thank you sir ,,

  • @jonm3131
    @jonm3131 5 лет назад +2

    Why is moment not equal to 0 about A?

    • @Eurocoded
      @Eurocoded  5 лет назад

      If you are referring to the moment resulting from 100kN, it is not equal to 0 because the lever arm is not 0. Moment of a force = force × lever arm. In this example both the force and the lever arm are not equal 0.

    • @hilbertdavid4058
      @hilbertdavid4058 4 года назад

      @@Eurocoded why Ma is CCW?

  • @joyprokash6757
    @joyprokash6757 2 года назад +1

    Thanks

  • @charithaheshan1048
    @charithaheshan1048 5 лет назад +1

    Are you sri lankan?

    • @Eurocoded
      @Eurocoded  5 лет назад

      Hi Charitha - you’re the second one to ask on RUclips. Looks like I sound Sri Lankan on my videos. 🇱🇰👍🏻

  • @bunadz683
    @bunadz683 2 года назад +1

    ❤🇩🇿