Steel question

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I'd like to know the strength of unequal angle steel.
Length is 4 metre, cross section is 6mm x 75mm x 50 mm. The short side is vertical.
What sort of weight would this take in the middle if supported both ends?
It can be a small amount of flex, but not more than say 5mm.
 
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In theory, around 100kg central point load would make it deflect 5mm. But that is assuming it can be held rigidly vertical, which would be almost impossible. The self-weight will also make it deflect - prob. around a further 5mm. Angles are not intended to be used as beams, because of the torsion effect.
 
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In theory, around 100kg central point load would make it deflect 5mm. But that is assuming it can be held rigidly vertical, which would be almost impossible. The self-weight will also make it deflect - prob. around a further 5mm. Angles are not intended to be used as beams, because of the torsion effect.
It's not actually for a lintel.
The proposed use is for a perimeter support for a modern log cabin. So the load is spread across the length and there would be a support in the middle.
To backtrack on the story. We have a log cabin that was left by previous owners. It was badly installed with unsuitable bearers. The floor is constructed of 3x2 timbers and floorboards. To support the cabin walls, I want to place lengths of the angle steel around the perimeter, bolt it to the 3x2, then using jacks raise the whole structure and correct the bearers underneath.
I would also use short lengths of the angle to spread the jacking load.
The cabin company has said the total weight of the cabin is around 1400kg. so each side will be around 1/4 of that, ie 450kg.
I'm not so concerned with some deflection since the final stage will be to place supports along the length. But I don't want it to flex beyond repair.
 
How to compute the deflectifon of a beam

First you need to know:

The length of the beam in meters, L
The weight supported by the beam in newtons, W
(If you know the mass supported in kg, multiply by 9.8 (gravity) to get the weight in newtons)
The elasticity, or Young Modulus, of the material in N/m^2, E. For steel this is about 200 x 10^9.

You also need to know a parameter called the second moment of area, I; this is a function of the dimensions and shape of the cross-section of the beam. There are standard formulas for things like rods, tubes, solid and hollow bars etc. Here's a calculator that will work it out for your angle: http://www.calcresource.com/moment-of-inertia-angle.html . Entering your dimensions, it tells me that I = 1.48 X 10^-7. (There are separate values for x and y, and I'm a bit confused about which we need; but I think that your orientation is less rigid than if it were rotated by 90 degrees, so I've chosen the smaller value.)

Finally there is a constant that depends on how the beam is supported and loaded; if it's supported at the ends by resting on something and loaded evenly the value is 5/384.

Putting that all together, the deflection d is

d = 5/384 x W x L^3 / ( E x I )

In your case, that's

5/384 x 450x9.8 x 4^3 / ( 200x10^9 x 1.48x10^-7 )

= 0.124m

That's quite a lot more than the 5mm you were hoping for!

One thing to note is that the deflection increases very quickly as the length of the beam increases - it increases with the cube of the length. So if you halved the length of the beam, e.g. by using 2 2m sections, the deflection would decrease 8-fold to about 15mm. But you can do better than that; if you have a single 4m piece of steel that's supported at the middle it's more rigid than two separate 2m pieces - I think by a factor of about 2.5 (the 5/384 parameter becomes 0.0052). So that would get you quite close to the 5mm you're looking for.

(Edit - oops, I forgot that the weight is also halved in these cases, further reducing the deflection.)

Of course, I'm not a structural engineer, and what a structural engineer (or rather the software he uses) can do is also quantify other aspects such as torsion - as tony1851 mentions. Maybe long before it has deflected by 124mm it will twist, unless something is holding it back.
 
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The 3x2 joists will, I am hoping, prevent the steel from twisting. The very nature of the log cabin is such that the 'logs' have their own structural strength.
I really don't think I will leave the steel completely unsupported aside from the ends. It will have probably two extra supports along the lengths by the time I am finished.
Basically, because the supporting bearers were so badly installed, the whole cabin is sloping down. So I need to re-support it independent of the old bearers, remove old bearers and replace.
 
endecotp, I was just trying to follow your calc and I'm not sure that your value of 124mm deflection is correct.

You seem to have calculated deflection for the angle under its own weight only, but I work this out to be just over 6mm.

I'll work in mm and Newtons for all variables to reduce risk of error...

Weight of angle (50x75x6 UA) = 0.055N/mm (5.6kg/m)
Length of angle = 4000mm
Modulus of elasticity (steel) = 205000N/mm2
Second moment of area (weak axis) = 148000mm4

So for a simply supported beam:

(5/384) x ((0.055x4000x4000^3)/(205000x148000)) = 6.04mm

Anyway, on its weak axis the angle isn't really capable of carrying any load when spanning 4m.
 
You seem to have calculated deflection for the angle under its own weight only

No, why do you think that? I don't even know the weight of the steel itself. I'm using 450kg, which is 1/4 the weight of the shed as in post #5.

Your calculation, however, does seem to be for deflection under its own weight, and I believe your answer.
 
No, why do you think that? I don't even know the weight of the steel itself. I'm using 450kg, which is 1/4 the weight of the shed as in post #5.

Your calculation, however, does seem to be for deflection under its own weight, and I believe your answer.

Oops, sorry, completely misread :oops:. And didn't read post 5 at all...this what happens when I am supposed to be doing proper work...

So isn't the deflection more like 490mm...or am I being dense again??
 
No, length is cubed.
It's only to the 4th power if you have weight-per-unit-length; that is multiplied by the length to get total weight.
 
For those regularly doing deflection calcs for steel beams, there is a simpler equation which I have found quick and less prone to mistakes through having lots of 000s;

(for udl) Δ max = 5 x W x L³ / 384 x 2.05 x I

W is in kN
L is in m
I is in cm^4, taken directly from the tables.
 

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