How to do a steel beam deflection calculation

[@Iffy50]

This is a great video. I calculated deflection last week at work and I compared my answer in SI and Imperial units to be sure they matched. They did not. I switched from meters to mm and the Young's Modulus used here and everything worked out.

[@emmadavies7434]

Interesting and very useful to know thank you 😊

[@masroormalik7964]

Excellent work

[@asrajo5728]

thanks for sharing sir

[@geoffreyworley5853]

Excellent clear and concise

[@StructuralEngineerCalcs]

Thank you for the kind words! I'm glad you found it clear & concise.

[@samirel-nagi6978]

Hi Thank you for these informative, refreshing structural analysis videos. It will be much appreciated if you cover analysis of masonry walls with one and two openings subject to lateral wind loads. Please try to cover most scenarios.

[@StructuralEngineerCalcs]

Thank you for your positive feedback on the videos and for your suggestion. We appreciate your interest in our content & we take all suggestions seriously. Analysis of masonry walls with one & two openings subject to lateral wind loads is an important topic in structural engineering & it's definitely worth covering it. We will consider creating videos on this topic in the future and will keep our viewers informed about new content. We appreciate your support & look forward to providing you with more informative & refreshing structural analysis videos.

[@ahmedaweysmohamed6378]

it was useful. thanks alot

[@StructuralEngineerCalcs]

Glad to hear that

[@oseghaleengineer2135]

Interesting! Appreciated. Pls how do you get the beam member sizing supporting a slab load over a span of 17m? I mean the classification of I-beam for such span?

[@AbuOmar1970h]

helpful

[@rockervipro]

Nice video thanks! From the table of properties I get a second moment of area (257x152x60) of 25.72 x 10^7 mm4. In your video you seem to have used the one of (257x152x52) which is 21.59 x 10^7. Could you review this?

[@chitpaul]

Start from where you got the imposed load from, 4.9kN/m .?

[@dragosionutbratosin3291]

the right answer is 12.205mm, because the second moment for the UB in the example 25500 cm4

[@StructuralEngineerCalcs]

The answer is correct please review your calcs.

[@sudhakarpersonal1926]

Hello for steel beam, Rafters, and columns all deflection check is span/360. Or it will change. Please let me know.

[@arvindpatil7222]

Thank you so much this Just wanted to know, at last the length was divided by 360 to check whether the deformation is acceptable or not What is 360 ?? Is it standard value to be considered while accepting the deformation?

[@edmarferreirajunior724]

It is a pattern defined by codes. The beam span must be divided by 360. And depending on the calculation method adopted, such as LRFD (Load and Resistance Factor Design method), there are different values ​​for each type of use, such as floor beams, roof beams, etc.

[@user-yw2xx8jv3o]

L/360 is for brittle finishes I.e. plaster.

[@pogarda001]

from the Structural engineer's pocket book third edition, the second moment of area UB 457x152x60 is 25500cm^4, it can't be 21.5x10^7

[@ayopreciousgold3213]

what then do you think the correct answer should be? please im asking for my self. currently working on this and I am struggling to understand where the young modulus is gotten from.

[@Eren-dq4uj]

@@ayopreciousgold3213 the youngs modulus is a material property. If you know the metal you also know the E.

[@GaloAalo]

Why load 4.9 is not multiply with factor load?

[@JPcostarica27]

Where span/360 came from? Is that a formula for maximun accepted deflection no matter beam type, just considering the lenght?

[@newsherts1259]

Where does 384 come from?

[@bart6978]

Can you calculate the defelction on 2,5 m from the left side? How?

[@Iffy50]

There are tables that show deflections for different loading scenarios. Just look up "deflection tables". Be sure you are using the correct end conditions.

[@wafikhwijeh4006]

Please.. How could you calculate I for I beam section? I know the equation of it is b*h3/12 ?

[@edmarferreirajunior724]

You must use the parallel axis theorem, also known as Huygens-Steiner theorem. The total moment of inertia will be the sum of the moments of inertia of the 3 rectangles that form the section of the I-beam.

[@user-yw2xx8jv3o]

steel Blue Book, you won't need to work it out your self.

[@MrMuhammadhafizuddin]

I got the answer is 1.447mm...please correct me if i wrong....one more question, can i use this formula to calculation Square Hollow Section?

[@StructuralEngineerCalcs]

Please review your calcs. You can also use this formula to check the deflection for any simply supported section.

[@user-yw2xx8jv3o]

you have got your units mixed up.

[@kyleharris3592]

The million dollar question is how did you get the second moment of area as 21500cm4 and in the design tables it is clearly 25500cm4 for a 457 x 152 x 60

[@kazisohail6798]

W is factord load?

[@urdogdanii1206]

W is unfactored for serviceability checks

[@sajjanbharti6896]

Thanks and regards🎂🎂🍰🍥💝🗽🍥🍰

[@akashsorathiya3078]

why we neglected self weight of beam ????

[@edmarferreirajunior724]

Because, unless you are designing a beam for fabrication on demand, all you need is just a reference number for a necessary minimum moment of inertia, in order to search, in a standard catalog of ready to use steel beams, those whit a moment of inertia slightly superior to that found in calculation. So you take the self wheight of the catalog beam and redo the same calculation. In general, the self wheigt is not significant in final result.

[@Irishfan]

I would have preferred to see this in US Standard units instead of metric.

[@AMJB100000000]

I reckon it would be less deflection in the real world 27mm seems like alot

[@jerinsan9078]

Too much use full, please do video about column...