How to design a steel column using an easy approach.

[@andreipopovcean9494]

Thanks! This is one of the most understable and practical engineering channel on KZbin.

[@gulfadventurevlogs]

Thanks. Great Explanation. easy to understand. great job.

[@emmadavies7434]

Great topic guys well done!.one of the most easiest ways I've ever come across cheers!.

[@jacksonphillips6276]

Great Channel.

[@user-ov4dw3yr5d]

Thank you for all you very informative videos. 👍

[@dengineerslifevlog5534]

Thanks for this topic...

[@shahedsiddiki1165]

It's a good video but we want more about column design

[@Pgan803]

Excellent

[@boy-br2tr]

Can you sample a building design from roof down to foundation structual analysis?..

[@cnjrq]

I have a complete tutorial on how loads are distributed from roof to foundation via columns. You can watch example 2 in my recent tutorial on steel columns kzbin.info/www/bejne/joLMoaafba-YaLM

[@vesaltorabi9]

You are amazing!!

[@zakwanabo9265]

Thank you very much

[@goodmind6446]

Thanks Sir ! want to see video on stability analysis and design of steel structure.

[@cnjrq]

I have lecture number 23 to 26 planned for stability analysis of steel buildings. I'm on lecture 10 at the moment and you'll see three vidoes per week uploaded. Please feel free to follow this playlist kzbin.info/aero/PLOQ_D0oq27oCKwuVHk-mgE0SRIGpOpSVu

[@nadinbu257]

thank you

[@nirshunarayanyadav5297]

Very nice explanation

[@Beginner2Creator]

Great topic : let me ask you, what about slenderness more than 3 ?

[@igirithankgod5395]

Thank you Sir for this easy approach. I have my actual stress as 157N/mm^2, is the column suitable?

[@ashutoshsuthar3753]

So core topics, please make video regarding design of crane grider on which the crab travels

[@maggysantana3878]

Just looking it few time and I don’t understand where you get the 350knm and the 300 Knm... from where you get the 4000mm ? Can you please do the example step by step... and show how you find those actor I mentioned to understand the math clear please.? Thanks.

[@telestruct]

the design DL=350KN & LL=300KN are obtained from structural analysis. the selected column height of 4m is just an example.

[@harigovind2968]

can u make a video on design of inclined cantilever columns

[@IvanIvanov-lo6lv]

Please explain for 1.35 and 1.5 in design load. What do they mean, when to use them.

[@sajadhussain6897]

they are used to increase the load to induce factor of safety

[@AS-tq2mn]

They're safety factors used to increase the load. You use 1.35 for permanent loads (which is the weight of the structure and anything else that's part of the build up) and 1.5 for variable loads (which is the load you consider from people using the structure, or any other load that comes and goes/isn't constant like snow).

[@dopareel]

nice

[@suhaibkhan7356]

How 100 n/mmsq compressive stress assume

[@rijvanansari6502]

Same question

[@user-ov4dw3yr5d]

If I may request a vlog for alum. Curtain wall and steel brackets structural calcs.? Thanks in advance.

[@sakally6392]

cool

[@youubik]

Just throw in that wacking great safety factor at the end and its doesn't matter what you calculate it to

[@saddammb5633]

How did you consider 100 N/MM2 as compressive stress?

[@Kal-El_000]

That was initial trial stress

[@burhanalibaig2533]

@@Kal-El_000 which book can i refer to for this calculations?

[@Kom-pks]

Hi, Anyone know the formula for calculating 'non-dimensional slenderness' for SS400 ? i seem to couldnt find it anywhere, or any keyword to search would be great too Thank you!

[@cnjrq]

Slenderness (Lambda) = (Lcr/iz)x(1/(93.9e), where e = SQRT(235/fy). Fy is yeild stress, Lcr is buckling lenggth, iz is radius of gyration in minor axix. You can follow the proceodure in my recent video, if it helps you kzbin.info/www/bejne/q5Wmn3SDaLeYpLM

[@ashleyrandall6582]

The 203 x 203 x 46 works. However can anyone tell me which codes this example is going according to ?

[@vinothkumar-qq1gf]

As per 203 x 203 x 46Kg channel I got an actual stress value is 55.90 N/mm2. Is it correct?

[@faithoko-ukoni5234]

No, you get about 156.89N/mm2

[@unitvector-dq1gb]

Trying 203×203×46 as you ask : This had properties, A = 58.8 cm i ( radius of gyration) = 5.11 cm The yield stress (fy) for this type of steel = 275 Mpa or N/mm² , it's the same. And the effective length is 4 m or *4000mm • Non-dimensional slenderness of the section is L/86i So 4000mm/86×5.11×10mm = 0.91 • The reduction factor X from the curve is approximately 0.68 • The buckling stress is : X × fy 0.68×275= 187Mpa • Actual strees is : Design load/ Area of section 922.5×10³ N / 58.8×10² mm = 156.9 Mpa < 187 Mpa ✓ok Susceptive to correct .

[@charlesa1000]

hi can you suggest a column for 8m length

[@civilking5437]

Wrong. The reduction factor should be 0.58 not 0.68

[@sandeepguleria8214]

So it failed actually

[@victorgetsbetter]

here right on time 😂

[@cooprice0237]

This is a good example, but that's it. Stop assuming, explain these assumptions.

[@bkingtz1]

Who will help me calculate what type of columns and beams etc I need to make 7.5m L x 6m w and 4.5m H, cover/garage for my car lift? I live in country where there is snow, also no other weight on the roof except snow. I am just going to cover it with thermo panels. And what is necessary for column foundation (concrete and rebar etc type and spec)

[@hudumasaud8295]

As an engineer, do it first. Then submit for vetting

[@bkingtz1]

My 🧠 hurts 😵‍💫