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  1. #41
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    Chuck,

    That's a long span, it's gonna be a monster. I don't have time to run the numbers right now, but if it were me, I'd engineer up a portable column which "clipped" via bolts to the bottom flange of your beam and then used an acme pressure screw to provide positive contact with the concrete. Think about the column-jacks you've probably seen supporting house I-beams in a basement.

    In other words, a portable column, which you would use as necessary to reduce the span of the beam to minimum and thus increase its capacity. Even better would be to have 2 columns of such design, they could then be used as a pair to minimize the loading span anywhere in the building.

    My .02

    -Matt

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    I have same type of question, the existing beam is 2x4" hollow aluminum spanning 10', I want to hang a yoga swing, how to test if it will hold my weight (125#)slightly swinging?

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    You will have to supply the wall thickness before any calculations can be run, also the orientation. Max dimension up or sideways?
    Tom

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    Whatever you do, load test the swing with at least twice the anticipated weight before use. A sling/seat/pole that gives way suddenly can lead to impacts or falls that can damage your spine or break a limb. You can use bags of sand or buckets of water, and remember to "bounce" them a bit to allow for kinetic loading, not just static.

    A picture of the beam and how it's attached to the structure will be a help. Be sure the mounting hardware is tight and cannot loosen under use. A low-strength Loctite will stop nuts or screws from getting loose.

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    Quote Originally Posted by mark thomas View Post
    Machinery's Handbook is confusing because it's just a reference, not a how-to -- it assumes the reader has studied structural engineering and just needs to look up the formulas and constants. But calculating deflection of a point load on a simple beam is extremely easy. There are only three things you need from the book: 1) the modulus of elasticity (known by the symbol "E") of the material, 2) the moment of inertia of the beam shape (symbol "I"), and 3) the formula to calculate the deflection for your configuration.

    All of these are in the chapter "Strength of Materials". Look at the section with tables for "Steel Wide-Flange Sections". Wide-flange is the standard construction shape that most people colloquially call "I beam". I-beams are actually a different and mostly obsolete shape but still available, as someone else pointed out. Wide-flange beams are known by a designation like "W6x25", "W14x136", etc. The first number is the height and the second number is the weight per foot. There are typically around a dozen different weights for each height. The weights, and structural properties, vary because the width and the thickness of the flanges vary for each size within a given height. The Wide-Flange table in the Handbook lists the full dimensions of each size, along with the moment of inertia for each shape.

    Moment of Inertia (I) is too complicated to explain and not necessary to understand beyond knowing that it is a number that reflects how rigid a given cross section is. For example, a solid round is more rigid than a tube of the same diameter, and has a different I (the specific value which of course varies, depending on diameter and wall thickness) For your purposes, you just need to find out the I value for the beams you might choose to use. These are in the Handbook tables.

    Modulus of elasticity (E) is a very simple concept. It is merely a measure of how much a material stretches under load. Every linear elastic material has a constant E, and you just need to know what it is and plug it into the formulas. (A linear elastic material is one that will stretch a given amount per given load. In other words if you applied n pounds of tension to a wire and it stretched x inches, and you apply 2n pounds and it stretches 2x, it is linear.) Steel is linear elastic, and its E value is 2.9E7. It varies slightly depending on alloy, but not enough to be relevant here.

    Then there are tables titled "Stresses and Deflections in Beams", which has a series of drawings showing beam and load configurations. For example, a simple beam with a single load in the middle; a simple beam with a single load at some point that is not the middle; a beam with a uniforming distributed load; etc. The pictures are largely self-explanatory. Foe each configuration, the table lists formulas to calculate stress and deflection at any given point along the length of the beam, as well as handy shortcut formulas for common things, like maximum deflection at the center of a beam.

    So, with I and E (which describe the material and the shape of the beam), plus your parameters of length and load, you have all the ingredients to do the calcs. Note that for a simple point load at the center of a beam, the maximum deflection is simply the formula:

    W * length^3 / 48 * E * I

    So if I pick a W6x12, its moment of inertia (I value) is 21.7. Plugging your parameters in, gives us this:

    3000 * 108^3 / 48 * 2.9E7 * 21.7 = 0.125"

    3,000 pounds on the center of a 108" long W6x12 will cause it to sag 1/8" in the middle.


    So that's how you interpret what you saw in Machinery's Handbook.

    Here's the formula applied to the example Forrest cited from some handbook:

    I of S5x10 = 12.3 (I-beams are designated "Snxn")

    3800 * 120^3 / 48 * 2.9E7 * 12.3 = 0.384"

    His book said it would sag .4", so we're basically dead-on here.


    In closing, I would note that your question was what size beam is "needed to support", but the need must be defined. Probably you mean "need not to break and fall on your head." But note that no material starts out rigid and then just breaks. First, metals flex under load and return to prior shape when the load is removed; this is the elastic range. Second, when the load is high enough, the material flexes under load but does not return to prior shape, i.e. it bends. Third, when the load is really high, the material may break.

    3,000 pounds is not a lot of weight relative to Wide-Flange beams, and even small ones will not bend, let alone break under a static load (3,000 pounds dropping 50' and hitting a W4x13 would be a different story of course.) So in many applications, deflection is the design constraint, not failure. For example, the Uniform Building Code specifies minimum building floor joist sizes not to a safety standard, but to a comfort standard -- people don't like to feel the floor under them flex, even if it is perfectly safe.

    So in summary, something from an S5x10 to a S7x15.3 (like John has) would be in order. A middle ground is a S6x12 which is twice as strong as a S5x10, and about 60% as strong as the 7 incher. None of these pose a failure risk in a relatively static load, home shop environment -- they will just vary in the amount of flex.
    For some reason the system is not allowing me to leave new posts yet being a new user, but I am impatient and want to get some information now, so I am forced to piggyback on this old post being similar to what I want to ask only on a different scale. You seem to really know your stuff and the mathematic equations required to determine what I need. What I have is a 40x40x25' tall shop that has all this extra height we would like to build a loft that we could use as a large studio apartment / office. I would like to know what it would take beam wise and their sizes to make a 40' span like that. I would consider adding a vertical support or 2 in the center if absolutely required and if it meant a drastic savings on materials and so forth. But to start I would like to try and estimate the project from the expensive end and we were to remain the free spanning. I was told according to the 2015 IRC for the live load we would need 50psf uniform and 1000psf concentrate with a floor deflection live load of L360, with downstairs being a garage with light manufacturing that is 150psf uniform and 2000psf concentrate. It asked about the collateral loads but I wasn't sure if that was a different term for any of the about code weights or not. I wanted a downstairs height of 15', which would give us 2' two make the floor while still having 8' ceiling height at the eaves upstairs. Since we don't have any height restrictions to the floor thickness itself, instead of all the beams meeting perpendicular with metal plates and bolts keeping them on the same plain. We had planned it would be easier to install if we ran the 2 main beams north and south on either side of the shop then stacking the cross beams running east and west on top of the first beans instead of connecting them to the side of each beam having to drill a 1000 holes with connecting plates. Instead if we stacked the cross beams we had also planned to weld each beam connection along with gussets on each joint to help with lateral loads. I should mention those main beams running 40' north and south do not need to be free spanning since they are along the walls so any vertical supports required would be out of the way. We also considered a 3rd main beam running north and south in the center of the shop so that the cross beams would only need to span 20' instead. I had thought this might be cheaper by having 3 main beams where only 1 makes the actual 40' span while the cross members were only a 20' span instead, then if absolutely necessary or if a major cost savings adding a vertical support or two down the center beam in the middle of the shop. The foundation itself is 6" thick fiber reinforced, while 18" around the perimeter the foundation drops down to 12" thick for the footing.

    Even though the shop is a good size with some heavier equipment the upstairs would basically just be where I do my upholstery work and may be an apartment I could use for time to time. So I just wanted it set up as a big studio where the heaviest piece of equipment would be a 3-400lb sewing machine, few tables, desk, computer, and etcetera. If possible I would like the loft to be free spanning so that it does not interfere with the shop down stairs. I am hoping even though it is a 40' span that being a lighter duty use would mean that vertical supports in the center could be avoided. I have emailed 15+ engineers in my state all of which have chosen to not reply. I don't know if that is due to my project being ridiculous or not large enough being a single floor design to capture their attention. For now I am just trying to determine the type of materials that would be required to make a free spanning floor to code so that we can put together an estimate of what this project would cost. Then I can determine if the cost savings are significant enough that it would be worth working around a post or two in the shop.

    If there are any engineers out there interested in a side job please contact me at [email protected]. Once I can determine a cost and all that is involved I planned to talk to the city to pull actual permits. At that time it will probably help to have some drawings that are official so the city engineer can sign off that the floor will support local codes. But beforehand I need to put together a cost by determining what would be involved and which way we would approached the project construction wise by figuring out what it would require to make sure we even have the funds available to take on this task. I am more than capable of the construction, but the math and formulas you guys use as engineers is way over my head. So I would definitely need some sort of a design to follow before we can begin construction to be sure exactly where all the beams should meet and how. But I am hoping someone may be able to help put together a rough estimate by letting me know what size beams I will need and how many so I can begin to source supplies and what they will cost.

    Thanks for taking the time to read this and in advance for anyone who offers any real help. It will be in Everett, WA 98204 if that makes a difference. [email protected]

  6. Likes Ronald S. liked this post
  7. #46
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    Quote Originally Posted by mark thomas View Post
    Machinery's Handbook is confusing because it's just a reference, not a how-to -- it assumes the reader has studied structural engineering and just needs to look up the formulas and constants. But calculating deflection of a point load on a simple beam is extremely easy. There are only three things you need from the book: 1) the modulus of elasticity (known by the symbol "E") of the material, 2) the moment of inertia of the beam shape (symbol "I"), and 3) the formula to calculate the deflection for your configuration.

    All of these are in the chapter "Strength of Materials". Look at the section with tables for "Steel Wide-Flange Sections". Wide-flange is the standard construction shape that most people colloquially call "I beam". I-beams are actually a different and mostly obsolete shape but still available, as someone else pointed out. Wide-flange beams are known by a designation like "W6x25", "W14x136", etc. The first number is the height and the second number is the weight per foot. There are typically around a dozen different weights for each height. The weights, and structural properties, vary because the width and the thickness of the flanges vary for each size within a given height. The Wide-Flange table in the Handbook lists the full dimensions of each size, along with the moment of inertia for each shape.

    Moment of Inertia (I) is too complicated to explain and not necessary to understand beyond knowing that it is a number that reflects how rigid a given cross section is. For example, a solid round is more rigid than a tube of the same diameter, and has a different I (the specific value which of course varies, depending on diameter and wall thickness) For your purposes, you just need to find out the I value for the beams you might choose to use. These are in the Handbook tables.

    Modulus of elasticity (E) is a very simple concept. It is merely a measure of how much a material stretches under load. Every linear elastic material has a constant E, and you just need to know what it is and plug it into the formulas. (A linear elastic material is one that will stretch a given amount per given load. In other words if you applied n pounds of tension to a wire and it stretched x inches, and you apply 2n pounds and it stretches 2x, it is linear.) Steel is linear elastic, and its E value is 2.9E7. It varies slightly depending on alloy, but not enough to be relevant here.

    Then there are tables titled "Stresses and Deflections in Beams", which has a series of drawings showing beam and load configurations. For example, a simple beam with a single load in the middle; a simple beam with a single load at some point that is not the middle; a beam with a uniforming distributed load; etc. The pictures are largely self-explanatory. Foe each configuration, the table lists formulas to calculate stress and deflection at any given point along the length of the beam, as well as handy shortcut formulas for common things, like maximum deflection at the center of a beam.

    So, with I and E (which describe the material and the shape of the beam), plus your parameters of length and load, you have all the ingredients to do the calcs. Note that for a simple point load at the center of a beam, the maximum deflection is simply the formula:

    W * length^3 / 48 * E * I

    So if I pick a W6x12, its moment of inertia (I value) is 21.7. Plugging your parameters in, gives us this:

    3000 * 108^3 / 48 * 2.9E7 * 21.7 = 0.125"

    3,000 pounds on the center of a 108" long W6x12 will cause it to sag 1/8" in the middle.


    So that's how you interpret what you saw in Machinery's Handbook.

    Here's the formula applied to the example Forrest cited from some handbook:

    I of S5x10 = 12.3 (I-beams are designated "Snxn")

    3800 * 120^3 / 48 * 2.9E7 * 12.3 = 0.384"

    His book said it would sag .4", so we're basically dead-on here.


    In closing, I would note that your question was what size beam is "needed to support", but the need must be defined. Probably you mean "need not to break and fall on your head." But note that no material starts out rigid and then just breaks. First, metals flex under load and return to prior shape when the load is removed; this is the elastic range. Second, when the load is high enough, the material flexes under load but does not return to prior shape, i.e. it bends. Third, when the load is really high, the material may break.

    3,000 pounds is not a lot of weight relative to Wide-Flange beams, and even small ones will not bend, let alone break under a static load (3,000 pounds dropping 50' and hitting a W4x13 would be a different story of course.) So in many applications, deflection is the design constraint, not failure. For example, the Uniform Building Code specifies minimum building floor joist sizes not to a safety standard, but to a comfort standard -- people don't like to feel the floor under them flex, even if it is perfectly safe.

    So in summary, something from an S5x10 to a S7x15.3 (like John has) would be in order. A middle ground is a S6x12 which is twice as strong as a S5x10, and about 60% as strong as the 7 incher. None of these pose a failure risk in a relatively static load, home shop environment -- they will just vary in the amount of flex.
    For some reason the system is not allowing me to leave new posts yet being a new user, but I am impatient and want to get some information now, so I am forced to piggyback on this old post being similar to what I want to ask only on a different scale. You all seem to really know your stuff and the mathematic equations required to determine what I need. What I have is a 40x40x25' tall shop that has all this extra height we would like to build a loft that we could use as a large studio apartment / office. I would like to know what it would take material / beam wise and their sizes to make a 40' span like that. I would consider adding a vertical support or 2 in the center if absolutely required and if it meant a drastic savings on materials and so forth. But to start I would like to try and estimate the project from the expensive end if we were to remain the free spanning. I was told according to the 2015 IRC for the live load we would need 50psf uniform and 1000psf concentrate with a floor deflection live load of L360, with downstairs being a garage with light manufacturing that is 150psf uniform and 2000psf concentrate. It asked about the collateral loads but I wasn't sure if that was a different term for any of the above info about code weights or not. I wanted a downstairs height of 15', which would give us 2' of height space to build the floor while still having 8' ceiling height at the eaves upstairs. Since we don't have any height restrictions to the floor thickness itself, instead of all the beams meeting perpendicular with metal plates and bolts keeping them on the same plain. We had planned it would be easier to install if we ran the 2 main beams north and south on either side of the shop then stacking the cross beams running east and west on top of the first beans instead of connecting them to the side of each beam and having to drill a 1000 holes with connecting plates. Instead if we stacked the cross beams we had also planned to weld each beam connection along with gussets on each joint to help with lateral loads. I should mention those main beams running 40' north and south do not need to be free spanning since they are along the walls so any vertical supports required would be out of the way. We also considered a 3rd main beam running north and south in the center of the shop so that the cross beams would only need to span 20' instead. I had thought this might be cheaper by having 3 main beams where only 1 makes the actual 40' span while the cross members were only a 20' span instead, then if absolutely necessary or if a major cost savings adding a vertical support or two down the center of the 40' beam in the middle of the shop. The foundation itself is 6" thick fiber reinforced, while 18" around the perimeter the foundation drops down to 12" thick for the footing.

    Even though the shop is a good size with some heavier equipment downstairs the upstairs would basically just be where I do my upholstery work and may be an apartment I could use from time to time. So I just wanted it set up as a big studio where the heaviest piece of equipment would be a 3-400lb sewing machine, few tables, desk, computer, and etcetera. If possible I would like the loft to be free spanning so that it does not interfere with the shop down stairs. I am hoping even though it is a 40' span that being a lighter duty use would mean that vertical supports in the center could be avoided. I have emailed 15+ engineers all of which have chosen to not reply or bow out being uncertified in Washington. I don't know if that is due to my project being ridiculous or not large enough being a single floor design to capture their attention. For now I am just trying to determine the type of materials that would be required to make a free spanning floor to code so that we can put together an estimate of what this project would cost. Then I can determine if the cost savings are significant enough that it would be worth working around a post or two in the shop.

    If there are any engineers out there interested in a side job please contact me at [email protected]. Once I can determine a cost and all that is involved I planned to talk to the city to pull actual permits. At that time it will have to submit drawings in triplicate that the city engineer can sign off that the floor will support local codes. But beforehand I need to put together a cost by determining what would be involved and which way we would approached the project construction wise to make sure we even have the funds available to take on this task. I am more than capable of the construction, but the math and formulas you guys use as engineers is way over my head and I don't intend to attempt them myself. I know where my expertise lie and I intend to stay in my lane. So I would definitely need some sort of a design / blueprints to follow before we can begin construction to be sure exactly where all the beams should meet and how. But I am hoping someone may be able to help put together a rough estimate for now by letting me know what size beams I will need and how many so I can begin to source supplies and what they will cost.

    Thanks for taking the time to read this and thanks in advance for anyone who takes the time to offer any real help. It will be in Everett, WA 98204 if that makes a difference, and city advises we follow the 2015 IRC building codes. [email protected]

  8. #47
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    Sir Forrest,

    If 5"x3" 10.0 lb per ft. How do you measure the weight of 6 meter height?
    Could you please educate me?

  9. #48
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    Not an engineer but we have a similar building we built...

    Ours is 30x60 2 level.

    Mezzanine joists are made from rolled 12 gauge galvanized steel into 4 x 12 C section.

    Each bay has one on each end, building is 5 bays so 4 I beams.

    Original had single post in center but we wanted drive through so changed to 2 posts.

    Loading is north of 100, maybe 150 per square ft.

    Your floor joists as well as floor matters, we used 1.125 48 inch span rated subfloor with 16 in centers so no problem with loading.

    Forget about free span as the cost will be great and walls may need different support.

    You will need a permit to be proper and that will require engineered drawings.

    Check with the building manufacturer before doing anything else.

    Sent from my SAMSUNG-SM-G930A using Tapatalk

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    Sir Mark,

    It is very interesting to read your explanation but I don't have much time to read everything. Besides, I am not familiar to the symbols...What is the max load weight capacity of column/beam? If I will build my house of 3 storey with 36 square meter per floor, What are the suitable number,size,thick of column/beam to be used? Consider, thickness of slab is 3 inches, 160kg/m2. Could you please show me your computation.

    Your response is highly appreciated.

    Regards,
    Ronald S.

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    Quote Originally Posted by Ronald S. View Post
    Sir Mark,

    It is very interesting to read your explanation but I don't have much time to read everything. Besides, I am not familiar to the symbols...What is the max load weight capacity of column/beam? If I will build my house of 3 storey with 36 square meter per floor, What are the suitable number,size,thick of column/beam to be used? Consider, thickness of slab is 3 inches, 160kg/m2. Could you please show me your computation.

    Your response is highly appreciated.

    Regards,
    Ronald S.
    Your answer requires much work.

    Please try over at Eng-Tips Engineering Forums

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    I know this is old, but when I was looking for the same answers, I found it was easiest to see what commercially built cranes were available and see what they had used for the construction. Probably cheating, but it was sure easy.


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