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Homework checkover, Beams

  1. May 25, 2012 #1
    Just wondered if someone was able to give the homework the once over as im not feeling confident i have this right.

    1. I have a simply supported beam 10metres in length, supported at 2 metres in from both ends. there is a 5Kn U.D.L along the entire length of the beam and downward forces of 10Kn 3 metres in from the LHS and 20Kn at the far RHS of the beam

    Determine the vertical reactions at the supports.

    Ok, first of all i worked out the U.D.L and its position. Total U.D.L= 10(length of beam) x 5 (Load)= 50Kn. This force is then added to the beam as a downward force of 50Kn in the middle of the distance between the 2 supports.

    Moments about A:
    (force x distance from support Ra) = (Support Rb x distance between the 2 supports)
    (10x1)+(50x3)+(20x8) = (Rbx6)
    10+150+160 = 6Rb
    320 = 6Rb
    Rb = 320/6
    Total upward force at support Rb = 53.3Kn

    Moments about B:
    (force x distance from support Rb) = (support Ra x distance between the 2 supports)
    (20x2)+(Rax6)=(10x5)+(50x3)
    40+6Ra=50+150
    6Ra=50+150-40
    6Ra=160
    Ra=160/6
    Total upward force at support Ra = 26.7Kn

    To check the answer,
    upward forces = 53.3Kn +26.7Kn=80Kn
    downward forces = 10Kn+50Kn+20Kn=80Kn

    I'm pretty happy i have this right as my upward forces are the same as my downward forces putting the beam in equilibrium.

    I have other Q's but ill do 1 at a time.
     
  2. jcsd
  3. May 25, 2012 #2
    Sketch the shear force diagram for the beam.

    Ok, downward forces are Negative and upward forces Positive. Because there is a UDL involved the diagram will have a slant to it.
    Working from the far LHS:
    @ LHE Q (let the shear force @ Y0 be Q0 and so on) = 0
    @ 1M Q1= -1x5 udl (distance from LHE x force) = -5
    @ 2M Q2 = -2x5 udl = -10
    Include Ra = -2x5 udl+26.7 = +16.7
    @ 3M Q3 = -3x5 udl+26.7 = +11.7
    Include concentrated load of 10Kn = -3x5 udl+26.7-10 = -1.7
    @ 4M Q4 = -4x5 udl+26.7-10 = -3.3
    @ 5M Q5 = -5x5 udl+26.7-10 = -8.3
    @ 6M Q6 = -6x5 udl+26.7-10 = -13.3
    @ 7M Q7 = -7x5 udl+26.7-10 = -18.3
    @ 8M Q8 = -8x5 udl+26.7-10 = -23.3
    Include Rb = -8x5 udl+26.7-10+53.3 = +30
    @ 9M Q9 = -9x5 udl+26.7-10+53.3 = +25
    @ 10M Q10 = -10x5 udl+26.7-10+53.3 = +20
    Include concentrated load of 20Kn = -10x5 udl+26.7-10+53.3-20 = 0

    I'm happy i have this right as the end result is 0.

    sorry for the essays..
     
  4. May 25, 2012 #3
    Calculate the bending moments along the beam @ 1M intervals.

    working from left to right

    Bending moment @ far LHE = Ma = 0
    @ 1M from far LHE M1 = -M due to udl
    @ 1M = -(5x1x0.5) = -2.5
    @ 2M = -(5x2x1) = -10
    @ 3M = -(5x3x1.5) + (26.7x1) = +4.2
    @ 4M = -(5x4x2) + (26.7x2)-(10x1) = +3.4
    @ 5M = -(5x5x2.5) + (26.7x3)-(10x2) = -2.4
    @ 6M = -(5x6x3) + (26.7x4)-(10x3) = -13.2
    @ 7M = -(5x7x3.5) + (26.7x5)-(10x4) = -29
    @ 8M = -(5x8x4) + (26.7x6)-(10x5) = -49.8
    @ 9M = -(5x9x4.5) + (26.7x7)-(10x6)+(53.3x1) = -22.3

    State the position and magnitude of the maximum bending moment in the beam.
    -49 @ 2M from the far RHE
     
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