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How to taken the valie of modulis of elasticity

Sir minimum thickness is what?

18:53 sir why did you take b1 =75

15:10 In question it was given that the Load acting on the simply supported beam is 21 kn, so the value of factor load in calculating deflection should be 31.5 by including one value of Length

Why don't we use 400 N/mm2 if we want to calculate for bolt bearing strength?

Sir but here the area of base plate is greater than area of concrete padestal 157500>155555.55 Any problem?

How do we know what e is

Thank you sir u literally cleared all my problems

must be 10,5 m2 x 8,75 kN/m2 = 91,875 kN and then 91,875/5 meter = 18,375 kN/m on beam SB.👌👌👍👍

The answer is wrong must be 18.375 kN/m on SB

Thanks sir 👍

Thanks a lot sir very useful video

From which textbook u explained, I mean terminologies are from which textbook?

what about the the segments whuch are not completely under teh strip but some of the portion is being covered ?

you did not tell about the value of zp provided

Ye Truck ka horn q dal diya hai😅

Thanks for explanation, very useful.

upar upaar bataya. detail me kuch nhi bataya 😢

184 kaha se Aiya Sir

Very well explained

Tooo good ❤❤❤

Sir please make videos on step by step mix designs according to FAA and ICAO

step jump kar kar ke sikha rahe ho

step 5 me Zprovided 348 kaha se aaya aur Z required 234 kaha se aaya

sir bht mistake hain calculation part me

Plzz give IS code page nos also

How you get 97.40 % please exapalin

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How tension and compression is developed.....

How to determine the diameter if load is given...

Nice explanation

Sir can u explain in shear strength formula what is the other area in th formula which u have considered as zero directly .

Fub is 5×100×0.6= 300 Sir

good and perfect!

Better than my professor thanks...

Good explanation my concept is very clear about this topic ❤❤

I believe we can calculate the load from the slabs S1 and S2 to the beam AB by getting the area of the two trapezoidal portions which is 2 Nos x 1/2 x (5+2) x 1.5 = 10.5 m2. We then multiplied the total area of 10.5 m2 x 8.75 kN/m2 and you get a total load of 91.875 kN to be carried by the beam AB. However using the UDL method (?) as you did we get a sum of 23.12 kN/m which when multipled by the length of the beam, we get a higher value of 115 kN. Further to this, we must add the Wall Loads to the beam, the self-weight of the Beam and then add the factor to be able to truely use this information in day to day job. Please go through that difference between the UDL and the area method. I have done this with another example, again the area method (which is nearer to Etabs calculation and result) has a lower value when doing the UDL.

Sorry I have only one liked buttom 🥹❤❤❤ thankyou thankyou so much keep itt

Thank you so much sir

P ki value

Very bad

Sir how to remember all the value

Tiada sarikata susa faham.🥴

Sir in step two for Design load you have denoted by (Pu) and in the last step you have denoted (Pu) as applied load . which is correct?

One of best video in DSS and this subtopic...

Excellent

Why √3

Extremely well explained

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thank you sir, the content is very informative.....