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Thread: Proper beam selection

  1. #1
    Associate Engineer
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    Proper beam selection

    I am trying to determine the specific size and type of beam needed for a project. I have searched the internet and was unable to come up with the answer. I have a basement wall that is beginning to crack due to soil loads from the clay soil in my location. The best data I could find about lateral force on walls from soil load says to design the wall with the ability to withstand a force of 15.71 pounds per square foot, per foot of depth which obviously would translate into 31.42 P/SF at a depth of 2 feet ect. Now for the specifics; the concrete walls are 6 inches thick and 9 feet tall measured from the basement floor to the top of the wall. The soil level on the outside of the wall is roughly 1 foot from the top of the 9 foot wall. I would like to install small I, W, or S beams vertically on the inside of the wall roughly every 8 feet and secure them at the top and bottom in order to combat the force of the soil load acting on the outside of the wall. But as previously stated I cannot find anything that would help me determine the best size and type of beam for this project. Any help would be greatly appreciated.
    Norm

  2. #2
    Technical Fellow Kelly_Bramble's Avatar
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    There will not be a simple chart vs wall support reference webpage.. not that simple.

    First, determine the beam or structural section you think might work..

    https://www.engineersedge.com/struct...pes_menu.shtml

    Determine the Area moment of Inertia

    https://www.engineersedge.com/standa...properties.htm

    Then determine the Yield strength of the material used in the structural shape

    https://www.engineersedge.com/materi...d_strength.htm

    https://www.engineersedge.com/manufa...s_strength.htm

    Determine the desired Factor of Safety (FOS) relative to the yield strength - I recommend 1.5 or so..

    https://www.engineersedge.com/analys...ety-review.htm

    Figure out the loading configuration then select the calculator that best fits.

    https://www.engineersedge.com/beam_calc_menu.shtml

    Calculate the Deflection, stress and so on and compare to the structural shape specifications and FOS.

    Repeat until you're satisfied you have the right structural support member.
    Last edited by Kelly_Bramble; 04-23-2016 at 08:32 AM.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  3. #3
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    First of all I would like to thank you for taking the time to answer my question. I’m no stranger to using mathematical formulas but I’m not very well acquainted with some of the engineering terminology or formula applications. I have never tried to get this technical with any of my previous projects. I do however want to have a mathematical solution that I feel is sound enough before I proceed. All of the links you supplied look great and would be perfect if I knew how to properly apply the equations. The best I can tell, the equation for deflection in link you supplied is for uniform loading across the entire length of the beam. This would not be a proper loading scenario for my situation. I was hoping to be able to use a calculator like the one here, which gives 66 different loading scenarios. The one that best fits the load on the beam in my case is #52.
    In this case they are just looking for E, I, q, Mo, M1, and M2 which would give me the max deflection and end slope. Is that correct?

  4. #4
    Technical Fellow Kelly_Bramble's Avatar
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    The best I can tell, the equation for deflection in link you supplied is for uniform loading across the entire length of the beam.
    No, did you actually look at the links?

    In this case they are just looking for E, I, q, Mo, M1, and M2 which would give me the max deflection and end slope.
    - Yes, that's it - knowing the resultant stresses would be helpful. Also there are not any equations associated so verification of the calculations is challenging.

    This calculator is close... Are you sure this is the loading configuration?

    Beam supported One End, Pin Opposite End and Triangular Distributed Load


    but I’m not very well acquainted with some of the engineering terminology or formula applications.

    All of the links you supplied look great and would be perfect if I knew how to properly apply the equations.
    Then you should hiring a local licensed engineer to determine what your requirements are and to ensure that the job is done correctly.
    Last edited by Kelly_Bramble; 04-24-2016 at 07:52 AM.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  5. #5
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    Quote Originally Posted by Kelly Bramble View Post
    No, did you actually look at the links?


    Yes, I did look at the links at length. I apologize for the fact that I don't speak "engineer" but after deciding to try the calculations for a W4X13 beam I couldn’t decide what value I was supposed to be looking for in your second link. I saw the Ixx and Iyy which I suppose is inertia for the x and y directions but there were no headings for “area moments of inertia”. I saw the yield strength for ASTM A36 was 250 but was unable to work past the “I” problem. I’m not trying to build a sky scraper I was just trying to look for a way to make myself feel better about my choice of steel for a DIY reinforcement other than “yep, that looks like it should hold”. This may be an over simplistic way of thinking but all I really need to know is the max load capacity for beam “X” over a span of 9 feet. I can calculate the variations in load from there within tolerances close enough for this project. I looked for 1 ½ days for a table on beam capacities but never found one. I understand you can’t give specific engineering advice but if you know where I can locate a table that shows the max load capacity for a W4X13 beam just that much would be greatly appreciated.
    Thank you for your time!
    Norm

  6. #6
    Lead Engineer Cake of Doom's Avatar
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    There is a nifty little formula for working out the approximate minimum second moment of area required: KWL^2 = Ixx required.

    K is derived from the deflection limit you need to achieve. Say for this instance you want a limit of L/500 this would be 500/161.3 = K =3.1.

    W will be the total load the steel will take.

    L is length.

    I've found that this works better for metric units but there is no reason that it shouldn't work with imperial units. As deflection checks are of the "not greater than" type I wouldn't base the entire design from it but it's handy for finding a starting point. That being said, it would still be worth having a local, licensed professional take a look at it. There may be more going on than just the cracking issues.

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