1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Have i done this right? Simply Supported Beams

  1. Oct 17, 2012 #1
    ƩƩ1. The problem statement, all variables and given/known data

    I am having issues finding Ra and Rb using the Sum of the momemnts, it's ƩMCW = MACW.

    I need to find the position of the maximum and minimum bending moments.

    The Supported beams in question is 3.5m long. Starting from the left and moving towards the right: 1m in and then a 2KN load, then 1.5m further in and then another load of 3KN hits down and then an extra 1m when it meets the end.

    Like this: 1m, 2KN, 1.5m, 3KN, 1m.
    Ra Rb

    Looking at my notes for some example i did (didn't finish the notes, must of forgot and now paying the price), example i did:

    3m, 4KN, 4m, 3KN.
    Ra Rb

    And here are the notes:

    ƩMCW = MACW

    (4x3) + (3X7) = Rb x 12

    12 + 21 =Rb x 12

    33/12 = Rb =2.75KN

    ƩFup = ƩFdown

    Ra + Rb = 4+3

    Ra + 2.75 = 7

    Ra = 7 - 2.75

    Ra = 4.25KN

    In the Engineering book i brought: Engineering Science by Mike Tooley and Lloyd Dingle, there is a beam example:

    2m, 20KN, 3m, 5KN, 2m, 30KN, 2m, 15KN.
    Ra Rb

    He did this: 2x20 + 5x5 + 7x30 + 9x15 = 8xRb

    Using this example and my own, i attempted to answer the question to find Ra and Rb. The very first and poorly drawn supported beam at the top.

    2. Relevant equations

    Direct Stress σ = F/A = Force/Area n/m squared

    Sheer stress ζ = F/A = Force/Area n/m squared

    Strain ε = Δ x L/L = Change in length/Original length.

    Modulus of elasticity E = σ/ε = stress/strain N/M squared.

    3. The attempt at a solution

    Here is my attempt at the very first diagram (the 3.5 over all length)

    (2x1) + (2.5x3) = 9.5

    Rb x 3.5 = 9.5 x 3.5 = 33.25

    2+7 = Rb x 3.5

    9.5/3.5 = 2.71KN

    Rb = 2.71KN

    ƩFup = ƩFdown

    Ra + Rb = 2 + 3 = 5

    Ra + 2.71 = 5 - 2.71 = 2.5 KN

    Ra = 2.5KN

    So Rb = 2.71 KN and Ra = 2.5KN

    What i don't undertand is on my previous attempt, Ra = 2.286 KN and Rb = 2.714KN

    Where have i gone wrong? And what do i need to do after finding Ra and Rb?
     
  2. jcsd
  3. Oct 17, 2012 #2
    Found Ra. How did i miss that.

    My first notes were correct. Rb = 2.714kn and Ra = 2.286kn

    I did the calculations wrong.
     
    Last edited: Oct 17, 2012
  4. Oct 17, 2012 #3
    Ok, i need help, how do i plot the sheer force diagram?

    I did a rough example:

    +

    +2.28 for 1m then with the 2KN i come down to .28 for 2.5m and with the 3KN i come down to -2.72 for 3.5m and then back up to 0 inbetween the plus and minus.
    -

    Is this rough version ok?
     
  5. Oct 17, 2012 #4
    You can always check your own reactions by taking moments about ANY point you haven't used to get the initial result. Your SFD seems ok to me, But people use various sign conventions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Have i done this right? Simply Supported Beams
  1. Simply supported beam (Replies: 4)

Loading...