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PROJECT 2 NAME OF THE PROJECT: Design & Analysis…
MD ABU ATHAR
updated on 07 Sep 2022
PROJECT 2
NAME OF THE PROJECT: Design & Analysis of a Hostel building.
LOCATION OF THE PROJECT: Guwahati, Assam.
Aim:- To frame the structural design and analysis of a hostel building for the given architectural drawings provided.
Introduction:-
The flow of the project may be in the way that are as follows
The given Architectural drawings
Procedure:-
Open the E-tabs software and choose to create a new model.
Assume thickness of the wall=250 mm
Ht of the wall =3.45 m
Density of AAc block=6kn/m^3
Thickness of plastering both sides assumed to be=25mm
Density of plastering=8kn/m^3
Brickwall load = load imparted by the brickwork+load imparted by both sides plastering
Calculation of the Brickwall loads=height of wall X thickness of wall X density of AAC block+Height of plastering on both sides X density of plastering X thickness of plastering=3.45X0.25X6+0.025X2X8X3.45=6.55KN/m
For a Residential building the code of practice for live load are as follows,
All rooms And Kitchen=2 Kn/m^2
Toilet and Bathrooms=2 Kn/m^2
Terrace live load=1 Kn/m^2
Partition wall load=3.5 Kn/m
Corridoor,passsages,Stircase,store rooms,Balconies=2 Kn/m^2
Ast max for Longitudnal Reinforcement:-
Beam B34 Maximum Lonitudnal Beam Reinforcement requirement,
Maximum Ast requirement for shear reinforcing for beam B29.
Using the above reinforcement area results we have to design and detail the beam Reinforcement.
Astmax at top= 1428mm2
Ast bottom R/f=1289 mm2
Formulae for the anchorage length as per IS 13920
Anchorage length=Ld+10db, Where Ld is development length and db is nothing but the diameter of the bar for which we are calculating the anchorage length.
Ld can be calculated as, Ld = φ σst/4τ bd
In the model,section properties arte provided are concrete of M30 grade while steel rebar for longitudnal reinforcement of HYSD415 are provided.
τ bd value from the IS 456:2000 ,we know=2.72
so,Ld for T-20 rebar,
Ld=((20x500)/(4x2.72))=
Ld=919 mm ,but as per IS 13290 add 10db in the development length so,
anchorage length=919+200=1119 mm.
similarly Ld for T12 rebar = 551mm, anchorage length=551+120=671 mm.
Now,calculations for providing the shear R/f
the depth of column as 500mm,
As per IS 13290,from the end support upto 2d of the beam confined shear reinforcement shall be provided.
=2d, d=500-25-20= 455mm,2d=2x455=910 mm
The area of shear R/f from the etabs at the support is equal to 1531 mm2.
let's provide 2L-T8 for the shear R/f.
one round of 2L-T8 counts for 100 mm2,so,for 1531 mm2
Total numbers required are=1531/100=15.31, let's take=16 No.s
For spacing we have
Spacing=(910/15)=56.875 mm, let's consider 70 mm c/c
let's provide 2L-T8 for the shear R/f.
one round of 2L-T8 counts for 100 mm2,so,for 1114 mm2
Total numbers required are=1114/100=12,
For spacing we have
Spacing=(1114/12)=92.83 mm, let's consider 110 mm c/c.
As per Is 13290 spacing should be ≤d/2.
Column Reinforcement:-
1. Longitudinal Reinforcement
As obtained from the ETABS output, the maximum Ast required for column and the value for the same is of 4071mm^2. Hence, if we provide 8 bars of 25mm dia and 2 bars of 20mm dia.
Ast = 8x491 + 2x314 = 4553 > 4071 . Hence OK.
2. Shear Reinforcement
Provide 8mm dia Lateral Ties.
Formulae for the anchorage length as per IS 13920
Anchorage length=Ld+10db, Where Ld is development length and db is nothing but the diameter of the bar for which we are calculating the anchorage length.
Ld can be calculated as, Ld = φ σst/4τ bd
In the model,section properties arte provided are concrete of M30 grade while steel rebar for longitudnal reinforcement of HYSD415 are provided.
τ bd value from the IS 456:2000 ,we know=1.875 (for the deformed bars in compression)
so,Ld for T-32 rebar,
Ld=((25x500)/(4x1.875))
Ld=1667 mm.
Anchorage length=Ld+10db=1667+25*10=1917mm
Similarly for 20mm bars=1534 mm
From there suggessions we have,
L0 is nothing but a particular distance from the top of the beam or bottom of the beam into the column upto which confined reinforcing is suggested by the codal provisions.
So, L0 is considered to be 650mm
2.Confinement spacing nbot more than,
So, 100 mm is considered.
Note:-As we are getting the same area of shear reinforcement requirement all along the height of the building. It suggests that the shear reinforcement is not required but, IS code has suggested to provide the minimal shear reinfocements in that perspective so,
Cover in the column is considered to be =40mm
and for the anchorage length in the foundation cover considered is=50 mm
Thus, using the above values the following detailed drawing was prepared in AutoCAD for the column height from building base upto the first slab.
Part B:-
Design and analysis of substructure:-
Required Data for Design and Analysis:-
Let's see for A1 Grid position,
Preliminary calculations
Unfactoreed load=287 Kn.
Dead load Breakup=287/2=143.5 Kn.
Live load Breakup=287/2=143.5 Kn.
Point Spring of Pile= A X E /L=
area of one pile=(πxd^2)/4=(πx500^2)/4=196349.5 mm^2
E=5000 X Fck^0.5=5000x30^0.5=27387N/mm2
L=8000
Point Spring of Pile= A X E /L=196349.5X27387/8000=672176 N/mm=0.672KN/m
Number of piles required= unfactored load with self weight addition/individual pile capacity=287X1.1/280=1.12 Nos.
To distribute evenly, Let's consider 2 piles for symmetric placement of piles under pile cap.
Similarly,the no of piles required for other grid locations are also calculated and a reference model for the same is erected in the Auto-Cad software.
Use the play button available above the model space to perform the analysis of the model,
After the analysis the model reflected in the model space as,
Check for settlement-
Punching Shear check,
As per the provision this value should not cross 1,if it crosses depicts that it has failed in punching shear check.
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