Load resistance of steel I-beam 120: vertical vs horizontal orientation (Wx 54.7, Wy 7.4)

Question

Please answer this general question: Does a steel I-section laid on the side (like the letter H) and supported at its ends have more than 7 times less load resistance than when laid normally? I obtained this knowledge from the data of the I-section 120, where Wx = 54.7 and Wy = 7.4 Please do not...

roman2361 · Answer

there is no such thing as a load-bearing capacity is the wall bending strength, squeezing strength and twisting strength for freely supported I-sections, there is only bending and shearing Bending Mmax = ql ^ 2/8 shear Qmax = ql / 2 When loading with continuous vertical forces because that's what you mean Mmax

akm97 · Answer

Thank you for your answer. For example, I mean such an I-section and I was wrong, because I meant a 160 - for it I gave the values Wx and Wy. One is 7 times bigger than the other. I do not understand, how to understand that 7 times is a cliche, there is more or less?

roman2361 · Answer

The W index is not the same as the resistance to bending, for example Understand Added after 1 [minutes]: it's really not that simple

akm97 · Answer

I understand that, but you can say whether the flexural strength of a horizontal I-section is 7 times lower, or is this value lower? I am asking, out of pure curiosity, because I have such an I-section 160 with a length of about 2.7 m under my load (less than 100 kg), it bends quite strongly, not from the load itself, but when I jump on it.

janek1815 · Answer

It bends when you put it on its side (with your foot vertical). When you put it on your foot, I can't believe it will bend. I have a hall made of 140 I-sections, the span between the walls is 6.5 m, this is where the roof rests and nothing bends. The ends of your I-section are not fixed, which also causes more deflection.

akm97 · Answer

I know that when I put it on my foot, it won't bend, because I tried it. My point was for someone to say that if the cross-sectional strength factor differs 7 times, then the difference in load resistance is the same. I know that I do not use scientific terms, but I meant such practical knowledge.

_jta_ · Answer

I do not know what the dimensions of this I-beam are (or maybe such as here ?) - if the width 's' is much smaller than the height 'h', such a difference is possible. For s = h and of the same thickness, I have the impression that the ratio of the bending strength and the force needed for a specific deflection in both directions should be about 2, with s = h / 2 strength about 4, the force for deflection about 8. But this is for flat feet, and they are wider in the middle, which makes the difference even more.

akm97 · Answer

I will measure it carefully tomorrow, but the dimensions seem to be correct. The feet are therefore probably a little more than twice as narrow. But there are two and they are actually widened in the middle.

_jta_ · Answer

This difference of several times in strength (and even greater deflection) looks quite reasonable.

roman2361 · Answer

Gentlemen no offense but you write total nonsense, it cannot be described in such a general way. You wanted to find out if the endurance is 7x greater with 7x greater Wx to you, so you write that it is not. It takes some complicated calculations to figure it out, but it will be less than 7x for sure Depends on how the back is attached and a few other things When it comes to deflection, you have basic formulas, what are the symbols, don't ask because one pulls the other

akm97 · Answer

I wrote these unnecessary nerves that I am a layman in this topic, so I will not do the calculations anyway. If you would like to make these calculations for me, I would be grateful and provide the necessary data. Greetings

_jta_ · Answer

The resistance of the I-beam to bending towards the flange (i.e. how the flanges lie horizontally and the crossbar vertically, i.e. bend downwards, if something is loaded on it) can be quite simply calculated: crossbeam height X (flange cross-section area + 1/6 of the cross-bar cross-sectional area) ) X material strength. Worse with the resistance to bending in the perpendicular side (i.e. as the cross-section is the letter H with a fairly wide bar) - you need to calculate the integral zx ^ 2 * g (x) * dx within the range from -s / 2 to s / 2, where x is the height above the plane of the crossbar (negative means that we are under it), ag (x) is the thickness of the foot at height x, and the result is multiplied by (4 / s) X the material strength (as we integrate one foot and the other is the same). Let us say that the foot has the shape of a symmetrical pentagon, obtained by cutting the arms of an isosceles triangle with a small height g1: in the middle it is still g1, the base is s wide, at the ends of the base there are sides with heights g2 perpendicular to it, resulting from trimming then there are sides connecting them to the vertex. Then g (x) = g1-2 * | x | * (g1-g2) / s; you need to integrate x ^ 2 * g1-2 * x ^ 3 * (g1-g2) / s from 0 to s / 2 and multiply the result by 2; integration results in g1 * s ^ 3 / 24- (g1-g2) * s ^ 3/32 = (4 * g1-3 * g1 + 3 * g2) * s ^ 3/96 = (g1 + 3 * g2) * s ^ 3/96; the final strength result is: material strength * (g1 + 3 * g2) * s ^ 2/12. Consider the case of g2 = 0, s = h / 2. Then the first result (divided by the strength of the material so as not to write it everywhere) will be g1 * s * h / 2 = g1 * s ^ 2 (not counting the strength provided by the crossbar), and the second g1 * s ^ 2/12 - that is 12 times smaller (if you count the bar, it will come out 16). Second case: g2 = g1, s = h; then (without the crossbar strength) the first result will be g1 * s ^ 2, the second will be g1 * s ^ 2/3. It seems that in the drawing (and in the table of dimensions) h was counted together in the thickness of the feet, and for calculating the strength, h should be taken without this thickness (and this is approximately) - in this way it is taken into account that we lose some strength due to unequal deformation of the feet.

roman2361 · Answer

buddy _jta_ please stop misleading people, you're counting like algebra I am a civil engineer with several years of practice, so I cannot accept these calculations with ease The formula for calculating the deflection for the case of a free-supported beam loaded with a continuous constant load (the simplest case) is the last case in the link I have cited f = 5/384 * (ql ^ 4 / EJ) q = value of continuous load (KN / m) l = spread E = Young's modulus of elasticity w / z from the steel grade, e.g. 20500MPa, as I remember well for St3x ( and just J = the moment of inertia (x, y) which is responsible for the arrow w depending on the x and y axis and depending on the profile (aha and types of I-sections there are several min. IP IPN IPE HEA HEB) so each has completely different moments of inertia. To count correctly, you still need to convert units to cm, we enter the Young's modulus in MPA so for this I-beam. what you gave is probably IP you have calculations: deflection in the xx axis, i.e. the web vertically f = 5/384 * (ql ^ 4 / EJ) Permanent load from IP 160 only with a span of 6.0m weight of 1 running meter = 15.8 kg / m = 0.158 KN / m q = 0.158 KN / m = 0.00158 KN / cm l = 6 m = 600 cm E = 20500 MpaJx = 935 cm4 fx = 0.139102648 cm = 1.39mm y-axis deflection, i.e. the web horizontally q = 0.158 KN / m = 0.00158 KN / cm l = 6 m = 600 cm E = 20500 Mpa Jy = 54.7 cm4 fy = 2.377714362 cm = 23.77 mm fx / fy = 17.09323583 that is 17x greater deflection in the yy axis than in the xx axis wx / out = 7.8 now you get it ?? the deflection depends on Jx or Jy the indicator here has nothing to do calculations made according to the Polish standard: PN-B-03200: 1990

akm97 · Answer

Thank you, colleague Roman2361. If I would like to use the formula for the third or fourth link provided by you, then how and on what scale to insert the force F (weight conversion). PS Thank you also my colleague _jta_

roman2361 · Answer

I also throw in the calculation for the ULS calculation, i.e. beam dimensioning: dimensioning, i.e. checking the ULS of the ultimate limit state in the xx axis M = ql ^ 2/8 Permanent load from IP 160 only with a span of 6.0m weight of 1 running meter = 15.8 kg / m = 0.158 KN / m q = 0.158 KN / m l = 6 m fd = 21.5 KN / cm2 for ST3X steel for t M / alpha * fd = 1.322790698 cm3 required in x Wxmax / Wx = 0.011305903 1% i.e. the cross-section will transfer the load with a margin of 99% (it is only a foreign beam of the same beam - for imaging) dimensioning for the Wy axis yy q = 0.158 KN / m = l = 6 m = fd = 21.5 KN / cm2 Wy = 15 cm3 M = 0.711 KNm 71.1 KNcm this way of supporting the beam secures torsion because the flange will not buckle, alpha = 1.0 ULS = Wy max> M / alpha * fd = 3.306976744 cm3 required wy Dimax / Wy = 0.220465116 22% i.e. the cross-section will transfer the load with a margin of 78% I hope this will give you a rough idea of how it is designed It's really not worth joking around with it, best done by a permissions designer. greetings

roman2361 · Answer

force is concentrated force in KN, 1KN = 100kg there is no scale, there is an individual

_jta_ · Answer

And I wrote about strength (i.e. the moment of force beyond which an inelastic deformation occurs), not about deflection. The deflection for these I-sections differs more than the strength, by the factor h / s - i.e. Ix / Iy = Wx / Wy * h / s - but I wrote about it in # 8. The formulas I derived in # 13 allow me to calculate Wx and Wy to see if there are any errors in the table - I counted this to show that this data may be correct, which you questioned in # 11 It takes some complicated calculations to figure it out, but it will be less than 7x for sure - and it came out 7.8 ...

roman2361 · Answer

and it came out 7.8 ... what did you do ??? for IP 160 Wx / O = 7.8 (it was always like that - because the tables say so) akm97 asked if the load capacity is also greater 7.8x, so it is not, full stop. if you get it then come on. Why do you check these indicators, why did Mr. Bogucki and Zyburtowicz publish it in the tables so as not to count. All polytechnics are based on this, you think they check :) Added after 9 [minutes]: And I wrote about strength (i.e. the moment of force beyond which an inelastic deformation occurs), not about deflection. right, you confuse the concepts and then you wonder that you do not understand What you are writing now is the theory of elasticity. Theory!!! checked in laboratory conditions, which is used, for example, to determine the Young's modulus, i.e. after applying the force F, the cross-section becomes plastic, i.e. deformed in such a way that it does not return to zero. The founder of the post asked about: load resistance He probably means this unfortunate beam that he has at home and is free-propelled, and I don't think he was going to protect it from torsion.

_jta_ · Answer

Why do you check these indicators, why did Mr. Bogucki and Zyburtowicz publish it in the tables so as not to count. All polytechnics are based on this, you think they check Sometimes it is worth making a conversion to make sure what a given indicator means ... Of course, you can read it in the book where it is described, but sometimes there are mistakes in them - maybe you remember a movie, probably entitled Exam, like a professor gives the student a task to count, leaves, the student checks the answer in the book, corrects the mark, and the professor returns, looks at the result, and says there is a typing error in the book? I used to count formulas from a book, my results did not match what was printed, I checked them again and found that there must be a mistake in the book; then I looked at two other editions of this book, the patterns were consistent with mine, only one typesetter made a mistake, and the correction did not catch (my friend had three consecutive editions, the error was in the middle one). And I remember a situation where many engineers claimed that these were the results of research (it was about the characteristics of the engine), who wrote to me that it was described in the book - I looked at this book, at the beginning there was information that these were the results of a simplified theory - others probably skipped the introduction, didn't notice, or maybe thought it didn't affect the results - and I just knew that the results of this theory are very different from reality. Unfortunately, many views repeated by experts are not true - because hardly anyone checks what he repeats. And what do you mean by the load capacity if it is not 7.8 times greater?

roman2361 · Answer

load capacity ULS - ultimate limit state I gave the formula above you do not agree with this ??????? Added after 3 [minutes]: proposes to pick up PN or Eurocode Everything is there

akm97 · Answer

I have one more question for my friend roman 2361 or someone else. Can the formulas provided by roman2361 in the link (in particular the third and fourth formulas) be applied to the HEA and HEB beams, substituting only the appropriate values of J. Because I understand that the E factor remains constant, ie 20500 Mpa?

_jta_ · Answer

roman2361 You write in an incomprehensible way. Even if pick up the PN . First of all, there are many Polish Standards - much more than you can read - and you need to know which one; perhaps you and your colleagues only use one, and for you PN means just this one, but for others it is not. Second, you may have the appropriate standard handy, but someone else should be shown where to find it. Third, such a standard contains a lot of information and it would be nice to say which ones are correct - in the sense that they are what you mean. Besides, you use a lot of markings without explaining what they mean (maybe you are used to these markings, maybe your colleagues know them and you don't need to explain them, but here is a different situation) - and sometimes you write incorrectly, e.g. in # 14 mass of 1 running meter = 15.8 kg / m = 0.158 KN / m - probably the mass of 1 running meter of the tee is 15.8 kg (not kg / m!), and its weight (this is not same as mass, do not write '=') is 15.8 kG = 155 N = 0.155 kN (not KN!). I have the impression that the correct result can only miraculously come out of such calculations, so I think I would prefer a designer with permissions who counts like this wouldn't do it for me.

roman2361 · Answer

Read it again what you wrote. Keep banging your head against the wall. Thank you for your attention.

akm97 · Answer

Thank you for the answers to both _jta_ and roman2361, but could anyone answer the question from the 2 posts above?

Load resistance of steel I-beam 120: vertical vs horizontal orientation (Wx 54.7, Wy 7.4)

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