# How to calculate maximum weight before a length of box section bends

February 9, 2015 2:25 PM Subscribe

I have a length of steel box section, 80mm x 80mm, 3.6mm thick which is 4m in length. I want to put something onto it, in the centre that weighs 900kg.
I want to know how to calculate whether this is possible but am unsure how to do it. I really want to know the maximum amount of weight I can put on before it breaks.

I believe that an actual engineer will come along to explain this better, but you might be able to figure it out from the data on this site:

http://www.engineeringtoolbox.com/square-hollow-structural-sections-hss-d_1478.html

and the information on this site:

http://www.engineersedge.com/strength_of_materials.htm

But you should note that if this suspended especially overhead at any point you need to add an 8:1 safety factor or working load limit to this (i.e. if it breaks at 800kg you can only load it to 100kg). There are also two different points at which there is an unsafe amount of weight on an object, one is when it breaks and the other is well before that when it deforms and no longer returns to it's original state after it has been loaded.

posted by edbles at 2:42 PM on February 9, 2015

http://www.engineeringtoolbox.com/square-hollow-structural-sections-hss-d_1478.html

and the information on this site:

http://www.engineersedge.com/strength_of_materials.htm

But you should note that if this suspended especially overhead at any point you need to add an 8:1 safety factor or working load limit to this (i.e. if it breaks at 800kg you can only load it to 100kg). There are also two different points at which there is an unsafe amount of weight on an object, one is when it breaks and the other is well before that when it deforms and no longer returns to it's original state after it has been loaded.

posted by edbles at 2:42 PM on February 9, 2015

Response by poster: Again i'm not even sure! I want to know whether the beam can safely hold the 900kg thing. I guess before it breaks.

The width of the thing on top will be 2 metres.

posted by aqueousdan at 2:47 PM on February 9, 2015

The width of the thing on top will be 2 metres.

posted by aqueousdan at 2:47 PM on February 9, 2015

This is a textbook engineering problem, but you need a little more information to solve it. The biggest piece of missing information is how the ends of the beam are being held up - are you resting it on top of two supports, or welding it into place? Are there more than two supports?

Second piece of information needed is whether this is a static or dynamic load. If you're just laying a box of bricks on top of this beam, that's a static load. If it's something like a winch or a tire swing or whatever, then that's a dynamic load.

The deflection calculator brainmouse linked to is a good first approximation, but your box is large compared to the beam you're resting it on, so you can't really assume it's a point mass like that calculator does.

Answer those questions and then I'll see what I can do for you.

posted by backseatpilot at 3:31 PM on February 9, 2015 [2 favorites]

Second piece of information needed is whether this is a static or dynamic load. If you're just laying a box of bricks on top of this beam, that's a static load. If it's something like a winch or a tire swing or whatever, then that's a dynamic load.

The deflection calculator brainmouse linked to is a good first approximation, but your box is large compared to the beam you're resting it on, so you can't really assume it's a point mass like that calculator does.

Answer those questions and then I'll see what I can do for you.

posted by backseatpilot at 3:31 PM on February 9, 2015 [2 favorites]

There's one more piece of data you'll need: Exactly what type of steel is it, and how has it been formed? All steel will

posted by clawsoon at 3:33 PM on February 9, 2015

*deflect*at approximately the same rate, but the point at which it*breaks*can vary widely depending on things like carbon content, alloying, and heat treatment. As you can see from the Yield Strength column in this table, the strength of steel can vary by almost a factor of ten.posted by clawsoon at 3:33 PM on February 9, 2015

Also, what's holding the beam up? It would be a shame to calculate that the beam is strong enough for the load and then end up pulling down your roof or something. 900kg is a lot.

posted by clawsoon at 3:54 PM on February 9, 2015

posted by clawsoon at 3:54 PM on February 9, 2015

Actually, it won't work; it's so far from working that a back of the envelope calculation kills it. The limiting factor is the cross sectional area of the top of the beam, which is only 80 * 3.6 = 266 square millimeters, or 0.446 square inches. You are applying a torque to each half of the beam of 900 kg * 2 meters, or 1984 lb * 6.5 ft, or about 12,800 foot-pounds. The box section must resist this torque with an arm distance of only the height of the box section, 80 mm or 0.26 feet This results in a compressive force of 49,000 pounds or about 110,000 PSI on the cross section of the top of the box section.

There are some errors in this back of the envelope calculation but mild steel is generally good for 40,000 PSI and good hardened steel rarely over 80,000. It's not worth the bother to check deflection and buckling because there is no way it gets any better. In this case it doesn't actually matter how the beam is mounted, since there are no mounts which could hold it rigid against so much torque any more than the center of the beam could maintain its integrity.

posted by localroger at 4:04 PM on February 9, 2015 [4 favorites]

There are some errors in this back of the envelope calculation but mild steel is generally good for 40,000 PSI and good hardened steel rarely over 80,000. It's not worth the bother to check deflection and buckling because there is no way it gets any better. In this case it doesn't actually matter how the beam is mounted, since there are no mounts which could hold it rigid against so much torque any more than the center of the beam could maintain its integrity.

posted by localroger at 4:04 PM on February 9, 2015 [4 favorites]

An additional note -- it

posted by localroger at 4:10 PM on February 9, 2015

*would*be worth doing real calculations if the beam was a solid 80mm rectangle instead of a box section. It would still be marginal especially with respect to safety factors but between the extra strength and protection from buckling, it would probably hold. Of course in that case your 4 meter beam would weigh around 100 kg all by itself.posted by localroger at 4:10 PM on February 9, 2015

Localroger, that the 900kg will be spread over 2m would mean rather less of a moment arm, yes? Not near my textbooks to check for myself...

posted by notsnot at 4:52 PM on February 9, 2015

posted by notsnot at 4:52 PM on February 9, 2015

Supposing this is a real-world problem and not a sneaky way to get engineers to do your homework for you, I'd carefully reevaluate your attitude toward risk management if you think putting 900kg on a 80mm square steel tube section spanning a 4m gap is a good idea.

Which is to say I wouldn't stand under it even if it held— you probably want a safety factor of 2, maybe 3 if people are going to be anywhere nearby, so you're looking at a beam with a load rating of 1800-2700kg to be safe.

posted by a halcyon day at 5:42 PM on February 9, 2015 [1 favorite]

Which is to say I wouldn't stand under it even if it held— you probably want a safety factor of 2, maybe 3 if people are going to be anywhere nearby, so you're looking at a beam with a load rating of 1800-2700kg to be safe.

posted by a halcyon day at 5:42 PM on February 9, 2015 [1 favorite]

And I'm betting it dies by web buckling long before the moment calculations take hold.

posted by rudd135 at 7:31 PM on February 9, 2015

posted by rudd135 at 7:31 PM on February 9, 2015

Yeah if it weren't so obviously hosed I'd go next to the column buckling formulas. But there's no need, it can't hold.

If the 900kg load is 2m long along the length of the beam, assuming it is itself absolutely rigid, it cuts the stress exactly in half (leaves 1m instead of 2m as the moment arm on each side) which still leaves it over the line for most normal steels and over any sense of safety margin for something you'd want to use.

posted by localroger at 7:44 PM on February 9, 2015

If the 900kg load is 2m long along the length of the beam, assuming it is itself absolutely rigid, it cuts the stress exactly in half (leaves 1m instead of 2m as the moment arm on each side) which still leaves it over the line for most normal steels and over any sense of safety margin for something you'd want to use.

posted by localroger at 7:44 PM on February 9, 2015

You...need to hire an engineer. Seriously, please don't ask the internet for advice on rigging a 900kg load.

posted by d. z. wang at 8:35 PM on February 9, 2015

posted by d. z. wang at 8:35 PM on February 9, 2015

Response by poster: Sorry the ends will be welded. The whole thing will be like a table structure... 4 uprights joined to each other at the top. kind of like in this image

The load will weigh 1800kg but lay across the two longest lengths, hence the 900kg on each one part.

For the back of an envelope calculation can you give me an idea of what I am looking to calculate? Ultimate yield strength?

posted by aqueousdan at 3:36 AM on February 10, 2015

The load will weigh 1800kg but lay across the two longest lengths, hence the 900kg on each one part.

For the back of an envelope calculation can you give me an idea of what I am looking to calculate? Ultimate yield strength?

posted by aqueousdan at 3:36 AM on February 10, 2015

IANYME.

You need to calculate the bending stress and compare that to your yield strength. But you should also check shear stresses, and check your legs for compressive strength and buckling.

I think the bending load is the most significant for this setup, though. For a point load (worst-case, but easiest to calculate) I calculate a little over 650 MPa in bending. That's not the principal stress, just tensile/compressive load from bending.

Can you add 2 more legs in the middle?

posted by Huffy Puffy at 5:19 AM on February 10, 2015

You need to calculate the bending stress and compare that to your yield strength. But you should also check shear stresses, and check your legs for compressive strength and buckling.

I think the bending load is the most significant for this setup, though. For a point load (worst-case, but easiest to calculate) I calculate a little over 650 MPa in bending. That's not the principal stress, just tensile/compressive load from bending.

Can you add 2 more legs in the middle?

posted by Huffy Puffy at 5:19 AM on February 10, 2015

The formula for bending stress is

Stress = (bending moment) * (height) / (area moment of inertia)

Height is the height from the neutral axis, so .04 meters at the top (which is the worst case). The other things you should know how to calculate if you want to solve this problem.

posted by Huffy Puffy at 5:23 AM on February 10, 2015

Stress = (bending moment) * (height) / (area moment of inertia)

Height is the height from the neutral axis, so .04 meters at the top (which is the worst case). The other things you should know how to calculate if you want to solve this problem.

posted by Huffy Puffy at 5:23 AM on February 10, 2015

There are multiple things to calculate; an actual engineer could tell you which ones are likely to be most important.

In your long horizontal beams, there's:

As for ultimate yield strength vs. yield strength, stick with yield strength. Ultimate yield strength happens after deformation has occurred, which you don't want.

posted by clawsoon at 6:18 AM on February 10, 2015 [1 favorite]

In your long horizontal beams, there's:

- maximum stress from bending in the top and bottom cross sections of the beam, which localroger already showed was way too high
- shear stress in the sides ("web") of the beam, mentioned by rudd135 and Huffy Puffy

- Euler buckling, which depends on the length of the legs
- compressive stress (the easiest to calculate, and the only one likely to be well within safety margins)

- stress in your corner welds
- weakness added by any additional welding or any holes you're drilling (what are you using to prevent the load from sliding?)
- load on the floor or ground; 1800 kg resting on four vertical legs could be enough to punch through some floors, or cause one side to sink down a few millimeters and your 1800 kg load to slide off, which would be bad

As for ultimate yield strength vs. yield strength, stick with yield strength. Ultimate yield strength happens after deformation has occurred, which you don't want.

posted by clawsoon at 6:18 AM on February 10, 2015 [1 favorite]

*Which is to say I wouldn't stand under it even if it held— you probably want a safety factor of 2, maybe 3 if people are going to be anywhere nearby, so you're looking at a beam with a load rating of 1800-2700kg to be safe.*

posted by a halcyon day at 8:42 PM on February 9 [+] [!]

posted by a halcyon day at 8:42 PM on February 9 [+] [!]

Again if this is overhead you need a minimum of a 5:1 safety factor if it's a static load, 8:1 if it moves.

posted by edbles at 6:52 AM on February 10, 2015 [2 favorites]

The reason that trusses simplify things so much, BTW, is that they only involve compression and tension along the axis of the columns (to a good-enough approximation). As soon as you get into bending, the calculations become much more complicated.

posted by clawsoon at 7:58 AM on February 10, 2015

posted by clawsoon at 7:58 AM on February 10, 2015

Another question for clarification: Will the load be pressing down all across its 2m on the beam, or only at the load's ends (the 1m and 3m marks)?

http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowRectangle.aspx

http://www.tech.plym.ac.uk/sme/desnotes/buccalc.htm

http://pages.jh.edu/~virtlab/bridge/bridge.htm

...I calculated (and please double-check my calculations! please! this is not engineering advice!) that a truss like this, built out of your square steel tubing, would give you a safety factor of around 24:1:

All of the other concerns about whether your floor would be able to hold it continue to apply.

posted by clawsoon at 8:44 AM on February 10, 2015

*If*the load is only pressing down at the 1m and 3m marks, and*if*you're willing to consider a truss, the calculations become much simpler. Using these calculators:http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowRectangle.aspx

http://www.tech.plym.ac.uk/sme/desnotes/buccalc.htm

http://pages.jh.edu/~virtlab/bridge/bridge.htm

...I calculated (and please double-check my calculations! please! this is not engineering advice!) that a truss like this, built out of your square steel tubing, would give you a safety factor of around 24:1:

450kg 450kg load load | | v v <---2m---> __________ ^ /\ /\ | / \ / \ 2m / \ / \ | / \ / \ v /________\/________\ <---2m---><---2m--->If you change any of the dimensions, you'll have to re-calculate. A change in the dimensions of the truss will change both the forces produced (they get rapidly higher as the top-to-bottom height is made smaller) and the buckling strengths of the columns (they get rapidly weaker as they get longer).

All of the other concerns about whether your floor would be able to hold it continue to apply.

posted by clawsoon at 8:44 AM on February 10, 2015

And, again, if your load is mostly in the centre, it adds bending forces and completely changes the situation (for the worse). Don't use my calculations anyway - I'm not an engineer - but definitely don't use them if the loads are anything other than 450kg point loads right at the joints.

posted by clawsoon at 9:11 AM on February 10, 2015

posted by clawsoon at 9:11 AM on February 10, 2015

If you already have this length of box section, put it on a concrete floor with a brick under each end. Then take your ~100kg body and stand in the middle and jump up and down on it.

Bouncy, isn't it?

Don't even

posted by flabdablet at 10:50 AM on February 10, 2015 [1 favorite]

Bouncy, isn't it?

Don't even

*think*about putting a ton on that.posted by flabdablet at 10:50 AM on February 10, 2015 [1 favorite]

Here's a more direct answer to your back-of-the-envelope question:

What localroger did, more-or-less, was use simple lever equations. Pretend you took a 2m length of the beam and attached the bottom at one end to a hinge that was embedded in a wall. If you pushed up on the other end of the beam with 900kg of force, how much force would the top of the beam, directly above the hinge, put on the wall?

Since this is a simple lever, the force is going to be multiplied by the ratio of distances. The distance from where you're pushing to the fulcrum is 2m; the distance from the fulcrum to the point where the force is being applied to the wall is 80mm. That means the force is going to be multiplied, as per the lever equation, by 2m/80mm = 25 times. So your 900kg of force is multiplied to 22500kg.

That's the first step. What you really need to find, though, is how much stress you're creating. Stress is basically the same as pressure; it's just force divided by area. If that force is spread over a wide area, the stress is low, and a low-strength material can resist it. If the force is concentrated in a small area, the stress is high, and a high-strength material is required. When "yield strength" is specified, it's always in units of stress: Force divided by area. The SI standard unit for stress is the pascal, which is one newton of force spread over one square meter. You'll typically see strengths quoted in megapascals (MPa).

So the next thing we have to do is convert our mass in kilograms to a force in newtons. This is simple; just multiply by 9.8 meters per second squared, which is the acceleration due to gravity at earth's surface:

22500 kg * 9.8 m/s^2 = 220500 N.

(A 900kg mass on the moon would produce much less force and could be held up by a much weaker beam.)

Finally, we divide that force by the area. In a hollow beam that's bending like this, the vast majority of the compressive force is going to be concentrated in the top surface of the beam. The cross-sectional area of the top, where most of the force is concentrated, is 80mm*3.6mm = 288mm^2.

Now we're ready for the final calculation: Force/area.

220500 N / 288 mm^2 = 765 625 000 pascals = 765 MPa

That's how much stress we're putting on the steel. It's higher than the yield strength of average steel, so we know that the beam won't hold.

Google is handy for calculations like this, since it'll convert most of your units for you.

The actual bending equation is more complicated than that; this just gives you a rough estimate. However, this rough estimate also tells you what you can do to reduce the maximum stress on the top and bottom surface of your beam:

- Make the distance you're spanning shorter. If it was 3m instead of 4m, that would reduce the stress by 25% or so.

- Use a stronger material. If you had a material with a yield strength of 5,000 MPa or so, you'd meet your safety factor requirement. (This material probably does not exist.)

- Use a thicker-walled beam. If the walls of the beam were 7.2mm thick instead of 3.6mm, you'd almost cut the stress in half.

- Increase the distance between the top and bottom of the beam. This is the most practical solution in most cases. It explains the shape of I-beams, of bridge trusses, of crane arms, and of engineered joists. However, this presents its own engineering challenges; if you make it too tall and thin, it'll bow sideways when force is applied and collapse; the webbing needs to be correctly designed and oriented to take the shear stress; the webbing needs to be properly attached so that it transmits the forces.

All that said, I'd still recommend a truss where no bending forces are present at all.

posted by clawsoon at 7:40 PM on February 10, 2015 [1 favorite]

What localroger did, more-or-less, was use simple lever equations. Pretend you took a 2m length of the beam and attached the bottom at one end to a hinge that was embedded in a wall. If you pushed up on the other end of the beam with 900kg of force, how much force would the top of the beam, directly above the hinge, put on the wall?

Since this is a simple lever, the force is going to be multiplied by the ratio of distances. The distance from where you're pushing to the fulcrum is 2m; the distance from the fulcrum to the point where the force is being applied to the wall is 80mm. That means the force is going to be multiplied, as per the lever equation, by 2m/80mm = 25 times. So your 900kg of force is multiplied to 22500kg.

That's the first step. What you really need to find, though, is how much stress you're creating. Stress is basically the same as pressure; it's just force divided by area. If that force is spread over a wide area, the stress is low, and a low-strength material can resist it. If the force is concentrated in a small area, the stress is high, and a high-strength material is required. When "yield strength" is specified, it's always in units of stress: Force divided by area. The SI standard unit for stress is the pascal, which is one newton of force spread over one square meter. You'll typically see strengths quoted in megapascals (MPa).

So the next thing we have to do is convert our mass in kilograms to a force in newtons. This is simple; just multiply by 9.8 meters per second squared, which is the acceleration due to gravity at earth's surface:

22500 kg * 9.8 m/s^2 = 220500 N.

(A 900kg mass on the moon would produce much less force and could be held up by a much weaker beam.)

Finally, we divide that force by the area. In a hollow beam that's bending like this, the vast majority of the compressive force is going to be concentrated in the top surface of the beam. The cross-sectional area of the top, where most of the force is concentrated, is 80mm*3.6mm = 288mm^2.

Now we're ready for the final calculation: Force/area.

220500 N / 288 mm^2 = 765 625 000 pascals = 765 MPa

That's how much stress we're putting on the steel. It's higher than the yield strength of average steel, so we know that the beam won't hold.

Google is handy for calculations like this, since it'll convert most of your units for you.

The actual bending equation is more complicated than that; this just gives you a rough estimate. However, this rough estimate also tells you what you can do to reduce the maximum stress on the top and bottom surface of your beam:

- Make the distance you're spanning shorter. If it was 3m instead of 4m, that would reduce the stress by 25% or so.

- Use a stronger material. If you had a material with a yield strength of 5,000 MPa or so, you'd meet your safety factor requirement. (This material probably does not exist.)

- Use a thicker-walled beam. If the walls of the beam were 7.2mm thick instead of 3.6mm, you'd almost cut the stress in half.

- Increase the distance between the top and bottom of the beam. This is the most practical solution in most cases. It explains the shape of I-beams, of bridge trusses, of crane arms, and of engineered joists. However, this presents its own engineering challenges; if you make it too tall and thin, it'll bow sideways when force is applied and collapse; the webbing needs to be correctly designed and oriented to take the shear stress; the webbing needs to be properly attached so that it transmits the forces.

All that said, I'd still recommend a truss where no bending forces are present at all.

posted by clawsoon at 7:40 PM on February 10, 2015 [1 favorite]

I finally tracked down the exact equation for maximum stress in the outer fibres of the beam assuming that the loads are concentrated point loads both at the same distance from their end of the beam. The equation is:

σ

where:

σ

P = load at each point (450kg * 9.8m/s

a = distance from the beam support (1m, in this case)

and:

Z = 2I/H

I = (H

where:

Z = section modulus

I = moment of inertia about bending axis (or second moment of area)

H = outer size of square tube (0.080m)

h = inner size of square tube (0.0728m)

posted by clawsoon at 11:38 AM on February 13, 2015

σ

_{max}= Pa/Zwhere:

σ

_{max}= maximum stress (this is the number you compare to the material's yield strength, after making the correct safety factor adjustment)P = load at each point (450kg * 9.8m/s

^{2}= 4410N, in this case)a = distance from the beam support (1m, in this case)

and:

Z = 2I/H

I = (H

^{4}- h^{4}) / 12where:

Z = section modulus

I = moment of inertia about bending axis (or second moment of area)

H = outer size of square tube (0.080m)

h = inner size of square tube (0.0728m)

posted by clawsoon at 11:38 AM on February 13, 2015

Here's one more example with the actual engineering equations and some explanation of what they're doing.

In this example, we'll calculate the maximum stress on a hollow rectangular beam that's like yours except that it's twice as tall. Same wall thickness, same width.

We have to use consistent units. I'll go with meters, newtons, and pascals.

The first thing we calculate is I, which is the moment of inertia about the bending axis (also called the second moment of area). This tells us how much resistance to bending a beam will have based solely on its dimensions.

I = b × h

where b is the width and h is the height. (I don't know why they use b for the width, but they always do.) In this example, b = 0.080m and h = 0.160m (since this beam is twice as tall).

This equation is for a solid rectangle. To calculate for our hollow rectangle, we calculate the same equation for the missing metal in the middle and subtract it. So, for the outside dimensions:

I

For the inside dimensions:

I

For the hollow beam:

I = I

This gives us an abstract indication of how much a beam shaped and sized like this will bend. The second thing we calculate is Z, the section modulus, which gives us an abstract indication of how much the outer fibres of the beam will be squished or stretched when the beam bends. The amount of squishing or stretching is formally called "strain". Z is I divided by half the height, i.e. the distance from the centre to the outermost fibres:

Z = I / (0.5 × h)

In this case:

Z = 0.000005664m

If we collapse the equations for Z and I for a solid rectangle together, we can see that there's a relationship between them and our lever intuition above:

Z = b × h

b×h is the area, so this becomes:

Z = A × h / 6

If Z is higher, the maximum stress from a given load will be lower, as we'll see below. So, from the equation Z = A×h/6, we can see that maximum stress will go down as 1) the area resisting the force goes up and 2) the height of the beam goes up. (What's the 6 in there for? That's calculus, and it would make this explanation much more complicated.) So this tells us, just like the lever intuition tells us, that we want to put the top and bottom of the beam as far apart as we reasonably can.

Note that Z=A×h/6 is

We are now ready to calculate the maximum stress that will occur in the beam, which we can compare to the yield strength of steel to see if we've met our safety factor. For combinations of arbitrarily placed loads, calculus is necessary. However, the calculus yields simple equations for simple situations, so we'll use one of those. In this case, we'll use the equation for 900kg placed at the center of the beam.

σ

where σ is the stress, P is the load (in newtons, not kilograms!), L is the length of the beam, and Z is what we calculated above.

P = 900kg × 9.8m/s

σ

Hopefully I've done the calculations correctly; please double-check my work. If I have, this tells you that a hollow beam twice as tall as yours would hold the load, but not safely, since 124MPa is about two-thirds of the stress that the weakest steel can resist.

Remember, of course, that this is for a hollow beam

That doesn't tell you what the stress is going to be on your legs; calculating that is a whole 'nother world of pain. (Pain which is once again much simplified by making everything into triangular trusses.)

posted by clawsoon at 12:01 PM on February 14, 2015

In this example, we'll calculate the maximum stress on a hollow rectangular beam that's like yours except that it's twice as tall. Same wall thickness, same width.

We have to use consistent units. I'll go with meters, newtons, and pascals.

The first thing we calculate is I, which is the moment of inertia about the bending axis (also called the second moment of area). This tells us how much resistance to bending a beam will have based solely on its dimensions.

I = b × h

^{3}/ 12where b is the width and h is the height. (I don't know why they use b for the width, but they always do.) In this example, b = 0.080m and h = 0.160m (since this beam is twice as tall).

This equation is for a solid rectangle. To calculate for our hollow rectangle, we calculate the same equation for the missing metal in the middle and subtract it. So, for the outside dimensions:

I

_{outside}= 0.080m × (0.160m)^{3}/ 12 = 0.000027307m^{4}For the inside dimensions:

I

_{inside}= 0.0728m × (0.1528m)^{3}/ 12 = 0.000021643m^{4}For the hollow beam:

I = I

_{outside}- I_{inside}= 0.000027307m^{4}- 0.000021643m^{4}= 0.000005664m^{4}This gives us an abstract indication of how much a beam shaped and sized like this will bend. The second thing we calculate is Z, the section modulus, which gives us an abstract indication of how much the outer fibres of the beam will be squished or stretched when the beam bends. The amount of squishing or stretching is formally called "strain". Z is I divided by half the height, i.e. the distance from the centre to the outermost fibres:

Z = I / (0.5 × h)

In this case:

Z = 0.000005664m

^{4}/ (0.5 × 0.160m) = 0.0000708m^{3}If we collapse the equations for Z and I for a solid rectangle together, we can see that there's a relationship between them and our lever intuition above:

Z = b × h

^{3}/ (12 × 0.5 × h) = b × h^{2}/ 6 = b × h × h / 6b×h is the area, so this becomes:

Z = A × h / 6

If Z is higher, the maximum stress from a given load will be lower, as we'll see below. So, from the equation Z = A×h/6, we can see that maximum stress will go down as 1) the area resisting the force goes up and 2) the height of the beam goes up. (What's the 6 in there for? That's calculus, and it would make this explanation much more complicated.) So this tells us, just like the lever intuition tells us, that we want to put the top and bottom of the beam as far apart as we reasonably can.

Note that Z=A×h/6 is

*only*good for solid rectangles. For hollow rectangles, you have to go through all the steps above, first calculating both values of I and subtracting before calculating Z.We are now ready to calculate the maximum stress that will occur in the beam, which we can compare to the yield strength of steel to see if we've met our safety factor. For combinations of arbitrarily placed loads, calculus is necessary. However, the calculus yields simple equations for simple situations, so we'll use one of those. In this case, we'll use the equation for 900kg placed at the center of the beam.

σ

_{max}= P × L / (4 × Z)where σ is the stress, P is the load (in newtons, not kilograms!), L is the length of the beam, and Z is what we calculated above.

P = 900kg × 9.8m/s

^{2}= 8820Nσ

_{max}= 8820N * 4m / (4 × 0.0000708m^{3}) = 124576271 pascals = 124 MPaHopefully I've done the calculations correctly; please double-check my work. If I have, this tells you that a hollow beam twice as tall as yours would hold the load, but not safely, since 124MPa is about two-thirds of the stress that the weakest steel can resist.

Remember, of course, that this is for a hollow beam

*twice the height of yours*.That doesn't tell you what the stress is going to be on your legs; calculating that is a whole 'nother world of pain. (Pain which is once again much simplified by making everything into triangular trusses.)

posted by clawsoon at 12:01 PM on February 14, 2015

BTW, if you're tempted to ignore the "factor of safety", you might appreciate den Hartog's preferred term, "factor of ignorance". We don't know if there's a hairline crack hidden inside your beam that'll fatally weaken the tension side, or if there were manufacturing defects, or exactly how much the welding will weaken it, or whether someone will unthinkingly lean on the load after it's up there, or whether it will accidentally get soaked with its own weight of water or snow, or shaken by the tremors of a far-off earthquake, or bounced when whatever's holding it up crumbles a bit.

We also don't know how much the engineering equations reflect the real world; they are useful simplifications, but that's all.

So, yeah... in that 8:1 or 5:1 safety factor is all of our ignorance about the present and inability to predict the future.

posted by clawsoon at 1:06 PM on February 14, 2015

We also don't know how much the engineering equations reflect the real world; they are useful simplifications, but that's all.

So, yeah... in that 8:1 or 5:1 safety factor is all of our ignorance about the present and inability to predict the future.

posted by clawsoon at 1:06 PM on February 14, 2015

« Older Vending at an indie wedding expo: What can I... | What's great about the Twin Cities? Newer »

This thread is closed to new comments.

bendsor before itbreaks? You say it both ways, and obviously those are very different things.This page has a deflection calculator that should show you have far it will bend given a certain load (you'll have to do some unit conversion) and free vs. fixed ends. There are other variables, though - like the distribution of your load (how wide is the 900kg thing, and therefore how wide will the load be spread?).

posted by brainmouse at 2:36 PM on February 9, 2015 [1 favorite]