Challenge

Project 1

AAyyappa Adapureddy8 Feb 2023

Q1). Design a multi-storey Residential Building located in Bangalore using STAAD Pro Connect Edition. The specification must be as follows:

Building floors	G + 6 + R
Plan dimension	B = 25m , L = 16m
Storey Height	h = 3.5m
Soil	Hard soil
Use	Residential Purpose
Safe Bearing Capacity	180 kN/m2
Maximum slab thickness	150mm

Assume suitable sections for structural elements. Follow IS456, IS1893 and IS 875

Prepare a Design Basis Report for the project
Create structural model using STAAD Pro Connect Edition
Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
Design slab element using MS – Excel
Design beam and column element using STAAD RCDC. Attach the design calculation report.
Design Foundations using STAAD Foundation.
Prepare bar bending schedule for beam and column using STAAD RCDC
Attach necessary sketches and drawings wherever required.
All stipulations, assumptions and design parameter must adhere to Indian Standards.

AIM:

To design a multi-storey Residential Building located in Bangalore using STAAD Pro Connect Edition.

INTRODUCTION:

A beam is a structural element that primarily resist loads applied laterally to the beams axis.
Beams are used to support the weight of floors ceilings and roofs of abuilding and to transfer the load to a vertical load bearing element of the structure.
Beams are flexural members the beam takes to load by bending and flexural members carry bending moment and shear force.The design of beams is for bending moment,Shear force,Torsion.
Clumn is a compression member the effective length of which exceeds three times the least lateral dimension.
Columns are structural elements used primarily to support compressive loads.
Column transmit the loads into foundation.
The columns are different shapes like rectangle,circular square,octagonal etc.

PROCEDURE:

DBR - Design Basis Report

Step 1) Given data

Building floors	G + 6 + R
Plan dimension	B = 25m , L = 16m
Storey Height	h = 3.5m
Soil	Hard soil
Use	Residential Purpose
Safe Bearing Capacity	180 kN/m2
Maximum slab thickness	150mm

Step 2) Creating Geometry of the Model

i) Create the geometry of the model

ii) As it is Residential Building and it has following dimensions

iii) L = 16m

iv) B = 25 m

v) H = 30 m

vi) Once geometry is created assign support condition as Fixed

Step 3) Material Specification

i) Assume the size of beams - 0.300 x 0.450 m

ii) Assume the size of columns - 0.500 x 0.500 m

Step 4) Loads

Calculation of Loads:

1) Calculation of Dead Load - IS 875 part 1

i)Dead load on Floor

i) Dead load due to slab = 0.150 x 25

= 3.750 Kn/m2

ii) Partitions = 1.5 Kn/m2

iii) Floor Finish = 2.5 Kn/m2

iv) Miscellaneous = 1 Kn/m2

v) Total Floor Load = 8.75 Kn/m2

ii) Dead load on Roof

i) Dead load of Roof Slab = 3.75 Kn/m2

ii) Insulaions and Water Proofing = 0.5 Kn/m2

iii) MEP Services = 0.5 Kn/m2

iv) Miscellaneous = 0.5 Kn/m2

v) Total Roof Load = 5.25 kn/m2

2) Live Load - IS 875 part 2

i) For Residential Building LL = 3 Kn/m2

ii) For Roof = 1.5 Kn/m2

3) Calculation of Wind load

i) Design Wind Speed

We know that

Vz = Vb x k1 x k2 x k3 x k4

where ,

Vb=Basic wind speed (m/sec)

K1=Risk coefficient (Table 1 of IS 875 (Part 3)-2015)

K2 = Terrain Roughness and height factor

K3=Topography factor (Clause 6.3.3 of IS 875 (Part 3)-2015)

K4 =Importance factor for Cyclic Region

AS Vb = 33 m/s ( Darjeeling ) Is 875

K1 = 1

K2 = 1.15 ( Ht for 30m) ( Category 1)

K3 = 1

K4 = 1

Therefore ,

Vz = 33 x 1 x 1.15 x 1 x 1

= 37.95 m/s

ii) Wind Pressure = Pz

Pz = 0.6 x Vz ^2

= 0.6 x 37.95 x 37.95

= 864.121 N

= 0.864 Kn

Similarly we can calculate for different heights

Height (m)	k2	Vz (m/s)	Pz	Pz in kN
3.5	1.05	34.65	720.37	0.720
7	1.05	34.65	720.37	0.720
10.5	1.06	35.2	743.424	0.743
14	1.08	35.64	762.12	0.762
17.5	1.10	36.3	790.61	0.790
21	1.125	37.13	827.18	0.827
24.5	1.13	37.29	834.32	0.834
28	1.15	37.95	864.121	0.864

4) Calculation of Seismic load

1) Design Base Shear

VB = Ah x W

where ,

Ah = Horizontal Coefficient of acceleration

W = Seismic Weight

ii) Ah can be calculated by using formula

Ah = Z/2 x Sa/g x I/R

As given building in location is in Bangalore -

a) Seismic Zone (Z) - Zone II ( 0.10 )

b) Soil condition or Soil type (Sa) - Hard Soil

c) Importance Factor - 1 ( For Residential Building )

d) Response Reduction Factor - 5 ( SMRF)

i) Now we have to calculate the Sa/g ( Spctral Acceleraion cpefficient) which depends on Type of soils .

ii) For hard Soil -

Sa/g = 1 + 15 T .........................0.00 < T < 0.10

2.50 ...............................0.10 < T < 0.40

1 / T ..............................0.40 < T < 4.00

iii) First we have to calculat Time peroiod in x and y direction (20x 30m)

Tx = 0.09hd">0.09h√d0.09hd

= 0.09X304">0.09X3040.09X304

Tx = 0.68 sec

Ty = 0.09hd">0.09h√d0.09hd

= 0.09X305">0.09X3050.09X305

Ty = 0.54 sec

Considered Max time period i.e Tx = 0.68 sec

Sa / g = 1/T

= 1.47

So put all the values in Ah equation

Ah = 0.10/2 x 1.47 x 1/5

Ah = 0.0147

3) Calculation of Seismic weight

i) Area of each Floor = 16 x 25

= 400 m^2

ii) Due to Dead Loads , Typical Floor = DL = 8.75 Kn/m^2

So Total Dead load of all floors = 8.75 x 400

= 3500 Kn

iii) Due to Dead Loads ,Roof = DL of roofs = 5.25 Kn/m^2

So Total Dead load of all floors = 5.25 x 400

= 2100 Kn

iv) Due to Live Loads , Typical Floor = LL = 3 Kn/m^2 ( Is 875 part 2)

But as per IS 1893:2016 If Live load is greater than 3 Kn/m^2 we have to reduced LL by 50%

Therefore reduced it by 50%

LL = 1.5 Kn/m^2

Total Live load for all floors = 1.5 x 600

= 600 KN

v) Due to Live Loads , Roof - AS per IS 1893 : 2016 Live load due to roof is neglected in calculation of seismic weight .

vi) Total Seismic Weight = W = DL + LL + DL

= (3500+ 600) x 7 + 2100

= 30800 Kn

4) Design Base Shear VB

VB = Ah x W

= 0.0147 x 30800

= 452.76 Kn

Storey level	Wi	hi	Wihi2	ki	Lateral Force
Storey level	(kN)	(m)			Qx	Qz
Roof	2100	28	3214400	0.31	140.035	140.035
6th	4100	24.5	2461025	0.24	108.66	108.66
5th	4100	21	1808100	0.17	76.96	76.96
4th	4100	17.5	1255625	0.12	54.33	54.33
3rd	4100	14	803600	0.07	31.70	31.70
2nd	4100	10.5	452025	0.044	19.92	19.92
1st	4100	7	200900	0.019	8.60	8.60
GF	4100	3.5	50225	0.005	2.26	2.26
		∑	10245900	1	442.47	442.47

Step 5)

i)WL + X direction

ii) WL -X direction

iii) WL +Z direction

iv) WL -Z direction

v) DL and LL

Step 6) Design of slab element

Step 1 ) Data

Span Shorter Direction (Clear)	ly	=	4	m
Span Longer Direction (Clear)	lx	=	5	m
Live Load on the Slab	LL	=	3	kN / m2
Compressive stength of concrete	fck	=	25	N / mm2
Yield strength of steel	fy	=	415	N / mm2
Unit weight of concrete	Ƴc	=	25	kN / m3
Unit weight of floor finish 100 mm	Ƴ	=	22	kN / m3
Clear concrete cover		=	25	mm
Bearing of slab	B	=	250	mm

Step 2)

Span to effective depth ratio	l/d	=	32
Minimum effective depth	d	=	125	mm
Overall depth	D	=	156	mm
Provide Overall depth	D	=	160	mm
Dia of bars for short direction	Φ	=	12	mm
Dia of bars for long direction	Φ	=	12	mm
Effective Depth	d	=	120	mm

Step 3) Loading

i) Dead Load of the slab (DL) = 3.75 Kn/m2

ii) Super Dead Load = 5 Kn/m2

iii) Live load (LL) = 3Kn/m

Total Load on the slab (TL) = 5+3.75+3

= 11.75 Kn/m

Factored Load = 1.5 x 11.75

= 18 Kn/m

ii) Effective span = ly + d = 5000 + 120 = 5.120 m

= lx + d = 4000 + 120 = 4.120

take the minimum value

Therefor ,

Effective length ( Leff) = 4.120 m

iii) Ly / Lx

= 5.120 / 4.120

= 1.2 < 2 ....................................Two way Slab

iv) BM Coefficient as per IS 456

Considered Two adjacent Edges Discontinues (Ly /Lx = 1.2 )

a) For Shorter Span

αx">ααx">xαx">αx = 0.060

αy">αyαy= 0.045

b) For Longer Span

αx">αxαx = 0.047

αy">αyαy = 0.035

Step 4) Bending Moment

Mx = αx">αxαx X w X lx^2

= 0.060 x 18 x 4.120 x 4.120

= 18.33 Kn-m /m ( Negative Moment )

My = αy">αyαy X w X lx^2

= 0.045 x 18 x 4.120 x 4.120

= 13.75 Kn-m /m ( Positive moment )

Mx = αx">αxαx x w x lx^2

= 0.047 x 18 x 4.120 x 4.120

= 14.36 Kn- m /m

My = αy">αyαy x w x lx^2

= 0.035 x 18 x 4.120 x 4.120

= 10.69 Kn- m /m

Considered Max valus of Mx and My

i) Shear Force - Vu = Wl /2

= 18 x 4.120 / 2

= 37.08 Kn

Step 5) To check the Effective depth

Mu,lim = 0.138 fck bd^2

18.3 x 10^6 = 0.138 x 25 x 1000 x d^2

d = 72.83mm

d prov > d req ...................................Ok

Step 6) Depth of slab for shear force

i) τc,max = 3.1 N/mm^2

ii) Tv = Vu / bd

= 37.08 x 10^3 / (1000 x 120 )

= 0.30

iii) For τc

Assume Pt % = 0.25

τc = 0.36

k = 1.3

τc = k x τc

= 1.3 x 0.36

= 0.468

Tv < τc < τc,max .....................Safe

Step 7 ) Determination of Ast

For Negative moment

Mu = 0.87fyAstd(1-Astfy/fckbd)

18.33 x 10^6 = 0.87 x 415 x Ast x 120 ( 1 - Ast / 25 x 1000x 120)

Ast = 302 mm^2

For Positive Moment

13.75 x 10^6 = 0.87 x 415 x Ast x 120 ( 1 - Ast / 25 x 1000x 120)

Ast = 236 mm^2

Step 8) Min Distribution Steel

Astmin = 0.12bd

= 0.12 x 1000 x 120

= 144 mm^2

Step 8) Selection of reinforcing bars

i) Area of bars (Top steel) = 302 mm^2

So provide 12 mm dia bars

Spacing = (Ast of dia / Total Ast) x 1000

= 374

= 300 mm

So Provide 12mm dia @ 300mm c/c

ii) Area of bars (Bottom steel) = 236mm^2

Spacing = (Ast of dia / Total Ast) x 1000

= 478.81

= 300 mm

So Provide 12mm dia @ 300mm c/c

5) Design of beam and column element using STAAD RCDC. Attach the design calculation report.

Step 1)

i) Once model is run successfully

ii) Click on post processing mode

iii) We can see the results , reactions , Bending moment etc

iv) We can Design beam and Column in RCDC software as well

v) So Click on Advance concrete design

vi) Click on Design of Beam first

vii) Enter the Project details as above

viii) And Click on Create Project

Step 2)

i) Once you click on Create a New Project

ii) It will take some time and to read the all data of structure

iii) Once process is complete you will see the Layout of the beam

Step 3)

i) Click on Setting

ii) Click on General and Reinforcement

iii) Enter the data - as shpwn below

Step 4)

i) Click on Setting - Rebar Curtailment

ii) If you want you can make the changes otherwise let it be the same

Step 5)

i) Click on Setting - Details and Drawings

ii) Make the necessary changes

iii) Click on Okay .

Step 6)

i) Once setting is done

ii) Click on Design and Click on Auto Design

iii) All the load cases which we have defined will come

iv) Select the type of load

v) Click on Okay

vi) Click on Add from Ananlysis option for adding Load combinations

vii) Click on Okay

Step 7)

i) Once you click Okay

ii) Software will design the beams

iii) As we can see all the beams are in Green colour

iv) So all the beams are passed

v) And We can see the design also

Step 8) Beam Design Calculation Report

Step 9) - Preparing BBS

i) For BBS

ii) Click on BBS

iii) Click on Generate BBS

iv) We can select All beams or we can select group wise

v) Select B1-B2-B3-B4

vi)Click on Generate

Step 10)

i) Click on Report

ii) Click on Design Summary

Similarly Follow the same procedure for Column Also:

i) Create a New Project

As we can see grids are automatically created with column mark

iDesi gn Foundations using STAAD Foundation.

Step 1)

i) Open the model

ii) Click on Staad Foundation

iii) Create a Job

iv) Enter concrete and reinforcement details

v) Enter Soil property

vi) Click on Design

vii) Once design is completed you will see the details of footing

viii) Design output of footing

RESULTS:

The multi-story residential building located in banglore has been designed and analysed.

i) DBR is prepared from Step 1 to Step 4 ( Input Details )

ii) Structural Model has been created and Analyze it successfully

iii) DL, LL, WL and EQL load are calculated in above steps as per IS 875 standards

iv) Slab element is designed and calculated in above steps

v) Beam and Column is designed in RCDC and Calculation sheet is ptovided in above steps

vi) BBS is also prepared

vii) Foundation is designed in Staad Foundation

Project 1

Ship work like this
in your own course.

Talk to a career counsellor

Ship work like thisin your own course.

Talk to a career counsellor

Ship work like this
in your own course.