STRUCTURAL

ANALYSIS AND

DESIGN

OF THE RESIDENTIAL BUILDING

Mrs. SAKKAL KUMARI LAMA

IMADOL -9, LALITPUR

Prepared by:

JAN, 2015

NOTATIONS

SYMBOLS DESCRIPTION

Ac : Area of concrete

Ag : Gross area of section

Ast : Area of tensile steel

Asc : Area of compression steel

Asv : Area of vertical stirrups

b : Width of beam or shortest dimension of column

D : Overall depth of a beam or slab

d : Effective depth of a beam or slab

fck : Characteristic compressive strength of concrete

fsc : Compressive stress in steel corresponding to a strain 0.002

fst : Tensile stress in reinforcement

fy : Characteristic yield strength of steel

Ld : Development length of bar

L : Length of column or span of beam

Lo : Anchorage length of bar

lo : Distance between points of zero moments

Leff : Effective length of beam or column

Mu : Factored design moment

Mu,lim : Limiting moment of resistance

Mux : Factored design moment along x-x Axis

Muy : Factored design moment along y-y Axis

Muxl : Maximum uniaxial moment capacity with axial load, bending about x-x axis

Muyl : Maximum uniaxial moment capacity with axial load, bending about y-y Axis

I : Moment of inertia of a section

lx,ly : Moment of inertia about X and Y axis

P : Axial load

Pu : Factored design axial load

p : Percentage of compression reinforcement

pc : Percentage of compression reinforcement

Sv : Spacing or vertical stirrups

V : Shear force

Vu : Factored shear force

Vus : Design strength of shear reinforcement

Xu : Depth of neutral six at the limit state of collapse

Xu max : Maximum depth of neutral axis

Z : Lever arm

σabc : Permissible stress in concrete in bending compression

σabc : Permissible stress in steel in compression

σst : Permissible stress in steel in tension

σsv : Nominal shear stress

τ v : Design bond stress

τ bd : Maximum shear stress in concrete with shear reinforcement

Ф : Diameter of bar

A st,req : Required area of tensile steel

A sc,req : Required area of compression steel

A st,Pro : Provided area of tensile steel

A sc,Pro : Provided area of compression steelSv, prov : Provided spacing of vertical stirrups

Distribution of Chapter

This report has been broadly categorized into five chapters, Summary of each chapter are mention below:

Chapter 1: Introduction

Chapter 2: Preliminary load calculation and design

In this chapter, the preliminary load calculation is done and every element is designed for a particular section. We generally deal with the design of every structural element of particular floor like roof, typical floor, first floor and basement floor. Structural arrangements is done with necessary computations that are performed for the vertical load calculation, preliminary design of the structure elements, seismic load calculation and the different load combinations that are used.

Chapter 3: Modeling and Structural Analysis

This chapter deals with the modeling techniques with SAP2000 that is followed by the analysis of the different structural members. This includes the steps that are performed for the vertical load calculation, preliminary design of the structure elements, seismic load calculation and the different load combinations that are used. inputs given and outputs obtained in the process, the time period calculation and storey drift of the building.

Chapter 4: Structural Design

It deals with the earthquake resistance design of beams, columns, slabs, shear walls and footings considering limit state of collapse and serviceability, their comparison with the provided ones and locating the areas of insufficient designs. The result is compared with the results obtained from the proposed program

Chapter 5: Structural Detailing and Drawings

The various structural detailing and drawings of the different members as obtained from their respective design are listed in this chapter.

CHAPTER 1: INTRODUCTION

INTRODUCTION

Building Type: Residential Building, Located in Dhakshi, Chandragiri, Kathmandu. Concrete Mix: M25 in columns and M20 beams, slab, staircase.

Steel Type: Tor steel Fe 500

BUILDING DESIGN PARAMETERS

Structural System : No. of Storey :

RCC Space Frame Basement + 3

Height of Building = 8.535 m

Floor Height : 2.845 m for each floors

Type of Sub-Soil : Medium Soil (Zone II)

Size of column = 304.8 X 304.8 mm

Size of beam Main Beam 230 mm X 355 mm (including slab depth)

Secondary Beam 230 mm X 300 mm

Depth of slab = 125 mm

Depth of waist slab of staircase = 125 mm

Depth of Footing = 610 mm

Density of RCC = 25 kN/m3

Density of PCC = 20 kN/m3

Density of Brick = 19 kN/m3

CHAPTER 2 : PRELIMINARY LOAD CALCULATION AND DESIGN

Load calculation is done based on IS 875: Part 1, 1987 Dead loads are calculated on the basis of unit weights.Earthquake forces are calculated in accordance with IS 1893: 2002The loads on beams supporting solid slabs spanning in two directions at right angles and supporting uniformly distributed loads may be assumed to be in accordance with fig.below.

B

45'A

45'

ALoad in this shaded area to be carried by Beam 'A'

BLoad in this shaded areato be carried by Beam 'B'

Fig:- Load carried by supporting beams.

Wind and earthquake loads are not assumed to occur simultaneously. However wind load is not critical in Kathmandu so earthquake loads have been taken for analysis. The earthquake loads have been calculated using the Equivalent Static Approach or Seismic Coefficient Method of Design and then checked through Dynamic Analysis.

Dead Loads:

Dead loads of slabs

Slab thickness: 125 mm 3.125 kN/m2

Stair waist slab thickness: 125 mm 3.125 kN/m2

Live Load:

Rooms 2 kN/m2

Kitchens 3 kN/m2

Balconies 3 kN/m2

Stairs 3 kN/m2

Floor Finishes

The floor finishes is assumed to be of 1 kN/m2

Wall Load:Storey Height 2.8448 m Density of Brick= 19 KN/m3

Size of beam 0.3556 mHeight ofParapet(4")= 1 m

Wall loads

Types of Wall ( " )Noopening

10%opening 20% opening

30%opening

40%opening

9 10.81 9.73 8.65 7.57 6.494 4.81 4.32 3.84 3.36 2.88

Parapet 1.69 1.52 1.35 1.18 1.01

Partition Load:The load due to internal wall partitions has been taken as 0.95 kN/m2 which

shall be distributed uniformly to the frames(as per real brick partitioned load)

Preliminary Design :

1. Slab depthd = Le/(1.35*23) = 4978/(1.75*27)

= 105 mm Let effective cover be 20 mmThen Overall Depth, D = 125 mm

2. BeamDepth =(L/10 to L/15) = L/15 = 4978/15 = 331.87 mm ~ 335mm Width =(D/3 to 2D/3) = D/2 = 2*335/3 = 230 mm

3. ColumnCritical Column is the one that is bounded by a maximum slab area. i.e. 74.8 m2

LOAD COMBINATIONS The following combinations are considered according to IS 1893:2002 which is taken as default from SAP provided for Indian Code

Lateral Load CalculationThe characteristic (intensity, duration, etc) of seismic ground vibration

expected at any location depends upon the magnitude of earthquake, its depth of focus, distance from the epicenter, characteristic of the path through which the seismic waves travel, and the soil strata on which the structure stands. The random earthquake ground motion, which causes the structure to vibrate, can be resolved in any three mutually perpendicular directions. The predominant direction of ground vibration is usually horizontal. Earthquake is not likely to occur simultaneously with wind.

Seismic Coefficient Method

The Indian Standard IS 1893:2002 contains provisions for both the static analysis and the dynamic analysis of buildings. Static analysis using equivalent lateral force procedure is restricted to regular buildings having height less than 40 m and irregular buildings having height less than 12 m in seismic Zone V(Cl.7.8.1). At the core of seismic analysis, is the use of response spectra plot as given in figure 2 of IS1893:2002, in which the spectral acceleration is plotted for wide range of fundamental natural period of the structures. For the static analysis, the static forces in the structure are derived from the design seismic base shear (Vb) given by:

Vb AhW (Cl. 7.5.3)

Where, W is the seismic weight of the building equal to full dead load plus fraction of live load, and Ah, is the design horizontal seismic coefficient, given by:

A ZI Sa (Cl. 6.4.2)

h2R

g

Adopted coefficients for Seismic coefficient method are as follows.

Z = 0.36, Seismic zone factor (Zone V) (Cl. 6.4.2, table-2)RSa/g

==

Importance factor (Cl. 6.4.2, table-6)5, Response reduction factor (Cl.6.4.2, table-7)Average response acceleration coefficient depending upon soil typeand fundamental natural period (Ta) (Cl.6.4.2, figure 2)

The approximate fundamental natural period of vibration (Ta), in seconds, of a moment resisting frame building without brick infill panels may be estimated by the empirical expression:Ta = 0.075 h0.75 for RC frame building

= 0.085 h0.75 for steel frame buildingWhere,h = Height of building, in m. This excludes the basement storey, where basement walls are connected with the ground floor deck or fitted between the building columns. But it includes the basement storey, when they are not so connected. (Cl.7.6.1)

Similarly, the approximate fundamental natural period of vibration (Ta), in seconds, of all other buildings, including moment-resisting frame buildings with brick infill panels, may be estimated by the empirical expression:Ta =0.09*h/√d Where,h= Height of building in m as defined in Cl.7.6.1d= Base dimension of the building at the plinth level in m, along the considered direction of the lateral force.

Distribution of Design Force

The design base shear (Vb) computed in Cl.7.5.3 shall be distributed along the height of the building as per the following expression:

Qi=Vb*

Where,

Qi = Design lateral force at floor i,

Wi = Seismic weight of floor i,

hi = Height of floor i measured from base, and

n = Number of storeys in the building is the number of levels at which the masses are located.

BASE SHEAR CALCULATION

In calculating the seismic weight of floors, the full dead load at each storey is considered, while the imposed load is considered as following criteria,

If IL3 kN/m2 -consider only 50% of IL If IL < 3 kN/m2 -consider only 25% of IL

The lateral force in each storey level is calculated using seismic coefficient method and then checked through dynamic analysis. These lateral forces thus calculated was then applied in both X and Y direction at the centre of mass of each floor element accordingly.

TABLE: Auto Seismic - IS1893:2002LoadPat Dir PercentEcc ZCode SoilType I R TUsed C WeightUsed BaseShear

Text Text Unitless Text Text Unitless Unitless Sec Unitless KN Neqx X 0.05 0.36 II 1 5 0.696 0.075 3801.83127 267.4233eqy Y 0.05 0.36 II 1 5 0.6261 0.075 3801.83127 297.3018

CHAPTER 3 : MODELLING AND STRUCTURAL ANALYSIS

A three-dimensional mathematical model of the physical structure should be used that represents the spatial distribution of the mass and stiffness of the structure to an extent that is adequate for the calculation of the significant features of its dynamic response. Further, structural models are required to obtain member forces and structure displacements resulting from applied loads and any imposed displacements or P-Delta effects. Thus, the essential requirements for the analytical model are the inclusion of sufficient details of geometry, material, loading and support such that it reflects the near-true behavior of the physical structure. For the structural modeling of the present building, SAP software was used. SAP is a general purpose structural analysis and design program with applications primarily in the building industry - commercial buildings, bridges and highway structures, industrial structures, chemical plant structures, dams, retaining walls, turbine foundations, culverts and other embedded structures, etc.A 3D beam elements having 12 DOFs with 6 DOFs at each node were used for modeling the beams and columns in the building, while surface elements were used to model the shear-wall. Surface element is essentially a three-dimensional super element developed from the assemblage of several shell elements by statically condensing the inactive DOFs, and using stiff fictitious beams to enforce the compatibility at the boundary of super elements.The in-plane stiffness of the floor systems of most building structures are extremely high compared to the stiffness of framing members. As a result, the in-plane deformations of beams can often be neglected, and columns, braces and walls connected to a given diaphragm will be constrained to move as one single unit in the lateral directions. This property is widely used in structural analysis to reduce the size of the system equations of buildings with such rigid floor types. When the “Rigid Diaphragm” option is selected for a given floor in any finite element (FE) based program, a transformation of coordinates and degrees of freedom is carried out to arrive at a system equation that allocates only three in-plane degrees of freedom for that particular diaphragm.

The structural analysis and design was carried out using relevant Indian Codes of Practice and is generally in accordance with Nepal building Code. The earthquake loading was carried out using Indian Standard. Soil performance factor was based on the geo technical investigation carried out at the site. The structural design of foundation, column, beam and slab was based on IS 456. Also, the system has been designed to meet the ductility requirements of IS 13920.

Y ZX

Figure: Axial Force Diagram (Typical) UDCON-2, 1.5 (DL+LL)

3D Model of the Building for Analysis

Figure: Shear Force Diagram (Typical) UDCON-2, 1.5 (DL+LL)

Figure: Bending Moment Diagram (Typical) UDCON-2, 1.5 (DL+LL)

The figure above shows the Axial Force, Shear Force and Bending Moment Diagram typical along 3-3 Axis.

DEFLECTION AND STORY DRIFTIn order to control deflection of structural elements, the criteria given in clause

23.2 of IS:456:2000 is proposed to be used.To control overall deformation due to earthquake load, the criteria given in clause7.11 of IS 1893 :2002 is applied. The story drift in any story due to the minimum specified design lateral force, with partial load factor of 1.0 shall not exceed 0.004 times the story height.

Due to EQx

FloorDisplacement (mm)

Storey drift (mm) .004*h

Fourth 12.444 2.153 11.39Third 10.291 3.396 11.39Second 6.895 3.729 11.39First 3.166 3.166 11.39Ground 0 0 0

Due to EQy

FloorDisplacement (mm)

Storey drift (mm)

.004*h (mm)

Fourth 9.778 2.273 11.39Third 7.505 3.575 11.39Second 3.93 3.7714 11.39First 0.1586 0.1586 11.39Ground 0 0 0

The results shows that the drift are within the permissible limit.

Figure: Joint Labeling at Ground level

SOME OUTPUT RESULTS FROM MODELLING

1. Reinforcement Output for Beams and Column s

TABLE: Joint ReactionsJoint OutputCase CaseType F3Text Text Text KN

7 UDCON2 Combination 471.87813 UDCON2 Combination 586.28619 UDCON2 Combination 392.86725 UDCON2 Combination 260.96931 UDCON2 Combination 598.32137 UDCON2 Combination 1033.14843 UDCON2 Combination 670.91750 UDCON2 Combination 526.27962 UDCON2 Combination 740.48368 UDCON2 Combination 442.28674 UDCON2 Combination 295.47186 UDCON2 Combination 353.32792 UDCON2 Combination 129.386

CHAPTER 4 : STRUCTURAL DESIGN

The design of all structural elements is done using ‘Limit State Method’. All relevant Limit State must be considered in design to ensure adequate safety and serviceability.

The structural elements are designed for the worst combination of the loads. Designs of foundations have been done with respect to the joint reactions given in the table above for the Load combination 1.5(DL+LL).

The manual calculations for foundation, staircase and slab have been given below.

DESIGN OF ISOLATED FOUNDATION (JOINT-74)

Grade of Concrete (fck) = 20 MpaGrade of Steel (fy) = 500 MPaLonger Side of Rectangular Column (Lc) = 300 mmShorter Side of Rectangular Column (Bc) = 300 mmSafe Bearing Capacity of Soil (SBC) = 150 KN/m3Unfactored Axial Load From Column (P) = 196.9807 KNCover to rebar center (dc)= 50 mm

AREA OF FOUNDATION BASE

Area of Foundation Base Required (Areq) = 1.444525 m2Length of Foundation Base Requied (Lreq) = 1.201884 m

Provide Length of Foundation (L) = 1.9 mArea of Foundation Base Provided (A) = 3.61 m2

NET SOIL PRESSURE

Net Soil Pressure at Ultimate Loads (qu) = 81.84792 KN/m2= 0.081 N/mm2

DEPTH OF FOUNDATION

Bending Moment about an axis passing through the face of the column:

BM = 49.76354 KNm

Calculation of Effective Depth Required

1) From Flexure

Effective Depth (d) = 97.41465 mm

2) One-Way Shear

The critical section is at a distance d from the column face

Assume Percentage of Steel (pt) = 0.21 %

Design Shear Strength of Concrete (Tc) = 0.32 N/mm2

Effective Depth Required (d) = 160.794 mm

3) Two-Way (Punching) Shear

ks = 1Tc = 1.118034 N/mm2

Design Shear Strength of Concrete (T'c) = 1.118034 N/mm2

Effective Depth Required (d) = 140.4667 mm

Depth Required from Flexure = 97.41465 mmDepth Required from One-Way Shear = 160.794 mmDepth Requierd from Two-Way Shear = 140.4667 mm

Effective Depth of Foundation Required (dreq) = 160.794 mm

Provide Depth of Foundation (d) = 300 mm

Total Foundation Depth (D) = 350 mm

DESIGN OF FLEXURAL REINFORCEMENT

Factored Bending Moment at Column Face (BM) = 49.76354 KNm

Area of Steel Required = 387.9302

Size of Bar Provided = 12 mm

Area of Each Bar (a) = 113 mm2

Spacing Required = 291.3926 mm

Spacing Provided = 150 mm

Provide 12 mm dia. Bars @ 150 mm C/C (both ways)

Area of Steel Provided (Ast) = 753.6 mm2

Percentage of Steel (pt) = 0.132211 %

Design of strap footing 4C (ID 68) & 3C (ID 62)ST1Working Load carry by Working Load carry by

column A (PA)= column B(PB)=

294.85733 KN493.65533 KN

Factored load (PAu)= Factored load (PBu)=

442.286 KN740.48 KN

sizeof column A

size of column B

x-dir A y-dir A x-dir B y-dir B

300 mm300 mm300 mm300 mm

c/c distance between column A & B (x')= 4.75 m consdering;unit weight of soil (We)= 20 KN/m2Angle of repose (φ) = 30 degreesafe bearing of capacity of soil(sqc)= 150 KN/m2characteristic strength of concrete(fck)= 20 N/mm2characteristic strength of steel(fy)= 500 N/mm2

1. Depth of foundation:

h= 𝑄𝑄𝑜𝑜/𝑊𝑊𝑒𝑒 ∗ ((1−𝑠𝑠𝑖𝑖𝑛𝑛𝜑𝜑)/(1+𝑠𝑠𝑖𝑖𝑛𝑛φ))^2= 0.833 ≅ 1 m

The base of foundation is located at the depth of 1m below which soil is not subjected to seasonal volumn

changes caused by alternate wetting and drying;

assuming selfweight of footing=10% of (PA+PB)Total service load= (PA+PB)+10% of (PA+PB)

= 867.3639333 KNRequired area of footing (Af)= 5.78 m2

or; L1B1 +L2B2 = 5.78 m2 …….(i)

where; L1,B1 :- length and width of footing under exterior column A L2,B2 :- length and width of footing under interior column B considering ; B1=B2=B= 1.674 mL1+L2 = 3.454 m …….(ii)

c/c distance between column A & B (x)= 4.75 mc.g. of column loads from the axis of column A (x')=

x'= 2.97 mThus, the distance of c.g. of column loads from the left edge column A (x")=

(𝑃𝑃𝐵𝐵∗𝑥𝑥)/(𝑃𝑃𝐴𝐴+𝑃𝑃𝐵𝐵) m

distance of the centroid of the whole area from the left edge of column A; 3.124

L1B1*L1/2+L2B2*( x-dirA/2+x)=(L1B1+L2B2)*x" …….(iii)

(L1^2)/2+L2*( (x-dirA)/2+x)=(L1+L2)*x"(L1^2)/2+L2*( (x-dirA)/2+x)= 10.790(L1^2)/2+ 4.9 L2 = 10.79substituting the value of L1 as L1= 3.454 -L2

L2^2 + 2.891 L2 + -9.65 = 0L2 = 1.980 m

L1 = 1.474 mconsider ; L1= 1.674 m L2 = 1.67 m

The effective thickness of footing may be determined by considering singly reinforced balanced section;

Pu = net upward ultimate soil pressure= 211.037 KN/m2

The width of beam is considered as the max width of pedestal = 350 mm Cantilever projection of the footing slab = 0.662 mMaximum bending moment per meter width = 92.49 KNm Equating Mu,lim to Mu;

k fck b d^2 = 92.49 k= 0.133 ( for Fe= 500 N/mm2 )

d= 186.46 mmProviding 12 mm φ at a clearcover = 50 mm Overall depth (D) = 242.46 mmProvided overall depth (D) = 350 mm d= 294 mm

Mu/bd2 = 1.070Percentage of steel required

pt % 0.263 % <Ast = 774.54 mm2

pt,lim = 0.758 % ( for Fe= 500 N/mm2 )

Spacing of 12 mm φ bar 146.01 mm c/c

Provided spacing of 12 mm φ bar @ 150 mm c/c

Actual area of steel provided = 753.96 mm2percentage of steel provided (Pt% provided) = 0.256 % Distribution steel = 0.12% of bd

= 352.8 mm2spacing of 12 mm dia required 320.56 mm c/c provided spacing of 12 mm dia 150 mm c/c Check for one way shear;The critical section for one-way shear is taken at a distance of effective depth from the face of beam. Shear force (V) = 77.66 KNNominal shear stress (𝜏𝜏𝑣𝑣) = 0.264 N/mm2

percentage of steel provided (Pt% provided) = 0.256 %permissible shear stress (𝜏𝜏𝑐𝑐)

= 0.363 N/mm2

Therefore; SAFE

Design of strap beamThe maximum bending moment for the strap beam will occur where the shear force is zero. Let , zero shear force occur at a distance x''' from left edge of column A;

rebar percent not OK!

2. Footing dimensions:

3. Design of footing slab

a. Thickness of the footing based on moment:

x''' = 1.252 m (from left edge) x''' = 2.0960 m (form right edge)Maxiumum B.M = -210.52 KNm Maxiumum B.M = -664.97 KNm

Then, Maximum Bending Moment (Mu,max) = -664.9704 KNm ConsiderThe width of beam is considered as the max width of pedestal = 350 mm Equating Mu,lim to Mu,

Mu = k fck b d^2 ……. (iv) k= 0.133 ( for Fe= 500 N/mm2 )

d = 845 mm

let , 20 mm ∅ at a clearcover of = 50 mm

overall depth = 905.14 mmProvide an overall depth of = 762 mmThus, effective depth = 702 mm

𝑀𝑀𝑢𝑢/( 3.855

pt % = 1.327 % <Therefore, Ast = 3260.11 mm2

pt,lim = 0.758 % ( for Fe= 500 N/mm2 )

Provide 3 20

mm∅+ 2 16 mm∅

Provded Ast = 1344.56 mm2 provided pt% = 0.547 %

S.F. at distance 'd' from the face of the column A = -88.30 KNS.F. at distance 'd' from the face of the column B = 386.50 KN

Nominal shear stress at column A = 0.36 N/mm2Nominal shear stress at column B = 1.57 N/mm2

percentage of steel provided (pt%) = 0.547 %permissible shear stress (𝜏𝜏𝑐𝑐) = 0.528 N/mm2

nominal shear reinforcement is enough at column Ashear reinforcement is require at column BSpacing shear reinforcement ;

2 - legged stirrups of dia 8 mm ∅Sv = 615.07 mm c/c min of 0.75d = 526.5 mm & 450 mm

min. shear reinforcement criteria259.26 mm c/c

Provide shear stirrups of 2- legged stirrups of 8mm dia @ 200mmc/c

Check for shear

PA=442.286 KNPA+PB= 1182.769PB = 740.483 KNx =4.75 m

x''= 3.12 mY

X

m 0.35

X𝐿𝐿1 = 1.674 m 𝐿𝐿2 = 1.67 mY

2-L-stirrups of 8mmφ @200 mm c/c

3-20 mm dia +2-20 mm dia

3-16mm dia

B = 1.674

fig: strap beam footing

PA = 442.286 KN PB = 740.483 KN

353.28 KN/m 353.28 KN/m444.79

149.0985 149.098552.99 (+)

(-)

-389.29 -295.69fig: shear force diagram(KN)

-210.518 -664.97-179.06

-124.80

3.97

fig: Bending mo ment diagramK Nm)

123.747

350 3-20 mm dia +mm 2-16 mm dia

3-16mm dia

762 mm

fig: section @ X-X

3-20 mm dia+12mm dia @ 150 mm c/c bothways 2-16 mm dia

3-16mm dia762 mm

350 mm

150 omm

fig: section @Y-Y

Design Of Staircase

Height of Riser (R) = 189.6 mm Grade Of Concrete M 20 Tread Width (T)= 254 mm Grade Of Steel Fe 500Width of Support = 228.6 mm Clear cover = 20 mm Width of Stairs = 1066.8 mm Bar Diameter Assumed = 12 mm Width of Landing = 1066.8 mm Unit Weight of Concrete (Y) = 25Effective Span 〖

(𝐿𝐿〗_𝑒𝑒 = 2870 mm Unit Weight of Marble (

𝑌𝑌)_=𝑚𝑚26.5

Let,

𝐿𝐿_𝑒𝑒𝑓𝑓𝑓𝑓/

𝑑𝑑

= 23

Unit Weight of Screed ( 𝑌𝑌)_= 24

or, d = 124.78 mmadopt d = 124.78 mm

Overall Depth (D) = 124.79 + cover +

= 150.783 mm

Load Calculations

(A) Load on Flight

ф/2

(i) Self Weight = Y*b*D

= Y* D √(𝑅𝑅^2+ 𝑇𝑇^2 ) /𝑇𝑇

= 4.70 KN/ 𝑚𝑚^2

(ii) Weight of Steps =

1/*2 Y * (

𝑅𝑅∗𝑇𝑇)/𝑇𝑇

= 2.37 KN/𝑚𝑚^2

(iii) Floor Finish = 30 mm thick marble + 25mm thick screed

= 𝑌𝑌_𝑚𝑚*0.03

+𝑌𝑌_𝑠𝑠 *0.025

= 1.395 KN/𝑚𝑚^2(iv) Live Load = 3 KN/𝑚𝑚^2

Total Load = 11.47 KN/𝑚𝑚^2Factored Load = 1.5 * 11.47

= 17.20 KN/𝑚𝑚^2

= 1.85 𝑚𝑚SFD

(B) Load on Landing

(i) Self Weight = Y * D = 3.76957 KN/𝑚𝑚^2(ii) Floor Finish (same as in flight ) = 1.395 KN/𝑚𝑚^2(iii) Live Load = 3 KN/𝑚𝑚^2

Total Load = 8.16457 KN/𝑚𝑚^2Factored Load = 1.5 * 8.16457

= 12.25 KN/𝑚𝑚^217.20

12.25 12.25

𝐴𝐴 𝐵𝐵1.07 𝑚𝑚 0.7 𝑚𝑚 1.07𝑚𝑚

𝑅𝑅_𝐴𝐴 𝑅𝑅_𝐵𝐵

BMDHere,𝑅𝑅_𝐴𝐴𝑅𝑅_𝐵𝐵

= 35.728 KN= 35.728 KN

Let, point of zero SF occurs at distacnc x from A, then𝑉𝑉_𝑥𝑥 = 0

x = 1.85 𝑚𝑚 Hence, Maximum moment at x is,

M = 32.28 37.494

Depth Check

M = 0.138*𝑓𝑓_𝑐𝑐𝑘𝑘*b* 𝑑𝑑^2d = 104.71 mm < 124.783 mm ok

Adopt d = 124 mm

Hence D = 150 mm

Area Of Steel

M = 0.87 *𝑓𝑓_𝑦𝑦* d *𝐴𝐴_𝑠𝑠 *(1 − (𝑓𝑓_𝑦𝑦 𝐴𝐴_𝑠𝑠𝑡𝑡)/(𝑓𝑓_𝑐𝑐𝑘𝑘∗𝑏𝑏∗𝑑𝑑))or, 598.4427141 = 𝐴𝐴_𝑠𝑠 - 0.00019 〖𝐴𝐴_𝑠𝑠𝑡𝑡〗^2or, 𝐴𝐴_𝑠𝑠 = 682.32 〖𝑚𝑚𝑚𝑚〗

Spacing = (1000∗π/4∗ ^2)/ф 𝐴𝐴_𝑠𝑠𝑡𝑡= 165.67 mm= 125 mm (Adopt)

Hence Provide 12 mm bar at the spacing of 125 mm c/c

𝐴𝐴_𝑠𝑠 Provided = 904.32

For Secondary Reinforcemet,Assume diameter of distribution bar = 10 mm

Required

𝐴𝐴_𝑠𝑠

= 0.12% of (b*d)

= 158.74 〖𝑚𝑚𝑚𝑚〗^2

Spacing =

(1000∗π/4∗ ^2)/ф 𝐴𝐴_𝑠𝑠𝑡𝑡

Hence Provide 10 mm bar at the spacing of 150 mm c/c

= 494.52 mm= 150 mm (Adopt)

Check For Shear

𝑉𝑉_𝑢𝑢

= 35.728

Nominal Shear Stress ( �

�𝜏)

_𝑣𝑣=

𝑉𝑉_𝑢𝑢/(𝑏𝑏∗𝑑𝑑)

= 0.27

Percentage Of Steel = (100∗𝐴𝐴_𝑠𝑠𝑡𝑡)/(𝑏𝑏∗𝑑𝑑)Check For Serviciability Requiremet

Effective Depth Provided (d) = 124Basic (L/d) ratio = 23Actual (L/d) ratio = 23.15Modification Factor Required = 1.01Steel Stress at Service Load (

)𝑓𝑓=_ 8.75

Percentage of Steel = 0.79Modification Factor = 2.00Permissible (L/d) = 46 > 23.15 Safe in deflectionMinimum Effective Depth = 62.3913 < 124 Safe

= 0.79 %

For M20 and 0.59% steel, 𝜏𝜏_𝑐𝑐 = 0.45

Shear Strength for slab = k * ( k = 1.25) 𝜏𝜏_𝑐𝑐=

0.56 > 0.27 OK

205

Design of Slab

Here the aspect ratio Ly/Lx is less than 2 hence the slab is designed as a two way slab.

Lx Ly3.43 3.66

Hence designed as two way slab

safe

safe

safe

safe

Check for Deflection

The effective depth provided ###### mm From figure 4 of I.S 456:2000 modification factor isModification factor 2.00Required depth under deflection consideration 72.35 mm

HENCE SAFE

F:\USER\RAM\3990\TWO WAY SLAB FINAL.xls RR/ST 12/10/2015

Span Moment Coefficient

Mu kN.m

Mu/bd² N/mm

2Ptreq Ast

reqd mm²

Min Ast mm²

Dia of bar mm

Spacing mm

Ast pro mm²shorter x neg 0.050 5.38 0.48 0.12 120.13 127.2 8 150 335.10

x pos 0.040 4.32 0.38 0.12 95.94 127.2 8 150 335.10

longer y neg 0.047 5.08 0.45 0.12 113.20 127.2 8 150 335.10y pos 0.035 3.78 0.34 0.12 83.70 127.2 8 150 335.10

Two Adjacent Edge DisContinuous Basic dimensions of slab

=

Basic Ly/Lx ratio = 1.067 <2

Clear cover to reinforcement d' = 15 mmProvided overall depth D = 125.00 mmEffective depth Diameter of bar

d

==

106.00 mm8 mm

Select Grade of Concrete fck = N/mm²Select Grade of Steel fy = N/mm²

Load calculation :

Dead load of the slab DL = 3.125 kN/m²Floor finish(Roof finish) FF = 1 kN/m²Live load LL = 2 kN/m²

Total load TL = 6.125 kN/m²

Moment and Area of Steel calculations: