COMPRESSION MEMBERS IN STEEL STRUCTURES || SOLVED NUMERICAL || DESIGN OF COMPRESSION MEMBERS

[@amitakki7594]

Mam u r a very good teacher of civil engineer. Thanks mam

[@amitakki7594]

Again thanks ❤❤❤❤🌹🌹🌹

[@adarshmishra7729]

OUTSTANDING LECTURE ....

[@avinashverma9024]

Amazing explanation

[@Jay_94]

awesome mam, this video helped to me a lot.

[@yanirajorgeundo1648]

Very good 👍

[@mastangori2236]

Nce mam

[@tachopacho5929]

Truely helpful maam

[@sunilteronofficial4112]

Good job 👍

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Thank u mam😊😊

[@MK2Vlogs]

Good going....keep up the good work👍

[@nabanitadaimary735]

Thank you so much for appreciating

[@tachopacho5929]

All revised done

[@lensitikhir3984]

Q1
Soln:KL=2.6m
Slenderness ratio= 50.38
h/bf= 1.6>1.2
tf=12.7mm < 40mm
Therefore buckling is about Yaxis
Where buckling class is b
fcd=193.506N/mm²
Pd= 2025.23KN

[@tachopacho5929]

U r wrong

[@tachopacho5929]

U r wrong

[@mudenrai6711]

An excellent content as always,ma'am.😊😊

[@nabanitadaimary735]

Means a lot 😊& that lifts me up to make more such contents. Thank you

[@mdnoorali2879]

Good Voice and thank you 😊

[@nanziadawasangma9380]

Ma'am for Practise Q 1, I got
KL=2.6m
Slenderness ratio=50.388
Buckling is about y axis with buckling class b
Design Compressive stress (fcd) =193.50 N/mm²
Design compressive strength (Pd)=2025.17 KN

[@nabanitadaimary735]

Good

[@nabanitadaimary735]

Note: In the interpolation part the (+ 206) is missing however the calculation is correct including the (+206)

[@laradomicky7808]

Q.1
Design strength (pd) =2025.066KN
Which is greater than the imposed load, hence the design is safe.
Q.2
Since the type of column is not given, the dimensions of the member, rxx,ryy,slenderness ratio are unknown.
Assuming an fcd =100N/mm^2
Area required= imposed load/fcd (950×10^3/100) =9500mm^2
Choosing a suitable section for the design I.e ISWB550 @112.5kg/m
(A)provided =143334mm^2.
Area provided > area required, hence ok.
Along y-y axis
Interpolating from table 9 (b)
Fcd provided= 136.98 N/mm^2
Pd =1086.41KN
Along x-x axis
Interpolating from table 9(b)
Fcd (provided) = 121.44 N/mm^2
Pd= 1740.72KN
Since the design load along both axis is > the imposed load, hence the design is safe.

[@nabanitadaimary735]

Right but assume the slenderness ratio n w.r.t. Get the fcd. Dont assume fcd

[@laradomicky7808]

Ok Mam

[@karticjaryal3883]

Help a lot , bt I'm quite slow to understand ! So please can you be a bit slow on it !

[@limekaswu7658]

Q1.
KL=2.6m
Slanderness ratio=50.38
Buckling about y axis with buckling class b
Fcd=193.50
Pd= 2025.17 KN
Q2.
Slenderness ratio is not known
So, assuming an fcd=100 N/mm^2
Area required=imposed load/fcd
=950*10^3/100
=9500mm^2
Choosing suitable section for a Design ie ISWB550@122.5 kg/m
A provided=143334mm^2
Area provided>area required
Hence ,it is ok
Along x-x axis
Interpolating from table 9(b)
Fcd( provided)=121.44N/mm^2
Pd=1740.72 KN
Since the design load along x axis is greater then the imposed load
Hence,the design is safe

[@limekaswu7658]

Along y-y axis
Interpolating from table 9(b)
Fcd (provided)=121.44N/mm^2
Pd=1740.72 KN
Hence the design load along x axis >the impose load hence it is safe

[@nabanitadaimary735]

Correct design

[@nabanitadaimary735]

But assume slenderness ratio not fcd

[@limekaswu7658]

Okay
Thank you ma'am

[@dipashribharali4938]

INTERPOLATING PART IS SOME WHERE NOT CORRECTLY WRITTEN I THINK ,OTHERWISE GOOD ONE.

[@nabanitadaimary735]

Dear in the interpolation part the (+ 206) is missing but the calculation is correct including the (+206).

[@akshaytamta8463]

Make a video on question 2

[@nabanitadaimary735]

ok sure

[@akshaytamta8463]

Thanks for reply. Means a lot to me

[@akshaytamta8463]

So mam when will you upload this video??

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Please come back

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Where to find pdf?

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Madam thoda jor laga ke bolia