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Homework Help: Load factors

  1. Mar 26, 2012 #1
    hi, not sure if im posting in the right section, had a good look through the search option but cannot find what im looking for, probably because im being stupid..

    im working on some simple beam calculations, im a studying architecture, however im trying to get my head round load factors..

    if i have a 2 storey building and i want to remove a facing wall say 5.0 in length this is how i would calculate the load on the beam

    Dead load
    1. area of wall it will carry in this case 5m x 3m high wall x 4.4kn/m2 (load for 240mm brick wall with plaster.

    2. floor 5m x 2.5 (half of floor distribution onto beam) x 2.05kn/m2

    3. roof 5m x 2m x 1.19kn/m2

    then the imposed loads on the roof

    my question is, im trying to input this data onto CSC Tedds calculation software and im confused as to what loading information i should input .. i would assume i would put total dead and total imposed loads by multiplying the width by height by load of material which i have tried..

    however it fails the beam size that has been approved manually. i believe im doing something wrong with the data input as Tedds has load factors of 1.6 and 1.4 which i did not use any load factors when working out manually.. can somone guide me as to what the load factors for dead and imposed means? do i need to use these if i have worked out the total udl with the example above?

    thank you so much in advance
     
  2. jcsd
  3. Mar 26, 2012 #2
    its cool guys i figured it out.. thanks
     
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