Design of Isolated StairCase
GEOMETRY
Flight SpanStart LandingL1=1.2m
Steps SpanL2=3m
End LandingL3=1.2m
Number of StepsNr=8n
Riser R=197mm
T T=375mm
Total Design Flight SpanL=5.4mL=L1+L2+L3
Flight WidthFw=1.2 m
Flight Height GainedFh=1.575 m
Steps Slope=ATAN(1.575/3)Sa=0.4834RadiansATAN(Fh/L2)
Staircase Geometry=
Design Parameters
Partiall Safety Factor for LoadsLimit State of Collapse (DL+LL)gl=1.5ConstantCl.36.4.1 of IS 456-2000
Partiall Safety Factor for Steelgs=1.15ConstantCl.36.4.1 of IS 456-2000
Partiall Safety Factor for Concretegc=1.5ConstantCl.36.4.1 of IS 456-2000
Elasticity of SteelEs=200000N/mm^2Cl 5.6.3 of IS 456-2000
Elasticity of ConcreteEc=25000N/mm^2Cl 6.2.3.1 of IS 456-2000
Strength of Concretefck=25N/mm^2
Strength of Steelfy=415N/mm^2
Weight of Concretewc=25 kN/Cum
INPUT SPECIFICATIONS
Clear Cover to Rebarsdc=20 mm
Main Bar Diadb=12 mm
Slab Breadth ConsideredB=1000mmAssmed 1 m wide strip
Assumed Slab ThicknessDg=150mm
Effective Depth=150-20-12/2D=124 mm
DESIGN LOAD & MOMENTS
Floor Finishes=1 kN/SqmIS 875
Live Load=3 kN/SqmIS 875
Total Self Weight=0.15/Cos(0.4834)*25ws1=4.235kN/m
Total Steps Weight=8*0.197*0.375/2*25/3ws2=2.461 kN/m
Load No: 01
Loading PatternL Ty=UniformLoadPt = 1; UDL = 2; TDL = 3; VDL= 4
Distributed Loadw=10.696kN/m
Start Locationa=1.2m
Length of Spreads=3m
DirectionDirection=270Deg
Increasing/Decreasing Rate=None1: None, 2: Increasing, 3: Decreasing
Equation exponent Eq=0For example Eq=1 for y=mx+b AND Eq=3 for y=cx^3
Load No: 11
Loading PatternL Ty=UniformLoadPt = 1; UDL = 2; TDL = 3; VDL= 4
Distributed Loadw=7.75kN/m
Start Locationa=0m
Length of Spreads=1.2m
DirectionDirection=270Deg
Increasing/Decreasing Rate=None1: None, 2: Increasing, 3: Decreasing
Equation exponent Eq=0For example Eq=1 for y=mx+b AND Eq=3 for y=cx^3
Load No: 21
Loading PatternL Ty=UniformLoadPt = 1; UDL = 2; TDL = 3; VDL= 4
Distributed Loadw=7.75kN/m
Start Locationa=4.2m
Length of Spreads=1.2m
DirectionDirection=270Deg
Increasing/Decreasing Rate=None1: None, 2: Increasing, 3: Decreasing
Equation exponent Eq=0For example Eq=1 for y=mx+b AND Eq=3 for y=cx^3
Bending Moments, Shear Forces and Deflections with Any Combination of Loads
Inputs: Beam Specifications
Length of BeamL=5.4m5.4 m
Section BreadthB=1000mm1 m
Secton DepthD=150mm0.15 m
Modulus of ElasticityE=25000N/mm^225000000 kN/m^2
Moment of InertiaIz=281250000mm^40.00028125 m^4
Inputs Loads:
Loading and Reactions Diagram=
Shear Force=
Results
Support ReactionIterationR1=25.344kN25.344 kN
Support ReactionSum of All loads - R1R2=25.344N25.344 kN; R2 = P - R1
FEM at support AIteration MethodMab=0kNm
FEM at support BIteration MethodMba=0kNm
Max Sagging MomentIteration MethodMaxMx+=36.866kNmOccurs at: 2.65
Net Sagging MomentThe DifferenceNetMx+=36.866kNm MaxBM3 - (MaxBM1 + (MaxBM2 - MaxBM1) / L.Value * a)
Bending Moment=
Max Sagging MomentIteration MethodMx+=36.866kN-mOccurs at: 2.65
Deflection Diagram=15.56 mm occurs at 2.75
Factored Moment=1.5*10.696*5.4^2/8Mu=24457445 N-mmWL^2/8 * Conversion to N-mm from KN-m
STRESS BLOCK & MOMENT OF RESISTANCE
Ultimate Strain in Steel=(415/1.15/200000)+0.002esu=0.0038Constant((fy / gs / Es) + 0.002
Limiting Depth of Nuetral Axis=(0.0035 / (0.0035 +0.0038)) *124Xu=59.417mm(0.0035 / (0.0035 + esu)) * D OR 0.48*D
Parabolic Depth from Origin/Neutral Axis=0.002*59.42/ 0.0035X1=33.95mm0.002 * Xu / 0.0035
Total area Constant of the Stress Block=(((2/3/1.5)*25*59.42*1000)-((2/3/1.5)*25*33.95/3*1000))/25/59.42/1000k1=0.36Constant0.364 * fck * x * b ; This can be taken as 0.36 for all Practical Purposes -cl 37.1 of IS 456-2000 {(((2 / 3 / gc) * fck * X * B) - ((2 / 3 / gc) * fck * X1 / 3 * B))}
Distance of the Centre of Compression from the Compression edge=(59.42 - (((2/3/1.5) * 25*59.42*1000*(59.42/2) - ((2/3/1.5) *25*1000*33.95^ 2 /12))/(0.36*25*59.42*1000))) /59.42k2=0.416Constant0.416 OR say 0.42 as per IS 456 clause 37.1c ((X - (((2 / 3 / gc) * fck * X * B * (X / 2) - ((2 / 3 / gc) * fck * B * X1² / 12)) / (k1 * fck * X * B))) / X)
Moment of Resistance by Concrete Section=0.36*25*59.42*1000*(124-(0.42*59.42))MRc=53.08N-mm k1 * fck * xu * B * (D - (k2 * xu)) / 10&sup6;
Coefficient of Moment of Resistance =53.082/(25*1000*15376)k=0.138Constant0.138 for fe 415 (= Mu/ fck * B * D²)
MINIMUM EFFECTIVE DEPTH REQUIRED
Ultimate Strain in Steel=(415/1.15/200000)+0.002esu=0.0038Constant((fy / gs / Es) + 0.002
Limiting Depth of Nuetral Axis=(0.0035 / (0.0035 +0.0038)) *124Xu=59.417mm(0.0035 / (0.0035 + esu)) * D OR 0.48*D
Parabolic Depth from Origin/Neutral Axis=0.002*59.42/ 0.0035X1=33.95mm0.002 * Xu / 0.0035
Total area Constant of the Stress Block=(((2/3/1.5)*25*59.42*1000)-((2/3/1.5)*25*33.95/3*1000))/25/59.42/1000k1=0.36Constant0.364 * fck * x * b ; This can be taken as 0.36 for all Practical Purposes -cl 37.1 of IS 456-2000 {(((2 / 3 / gc) * fck * X * B) - ((2 / 3 / gc) * fck * X1 / 3 * B))}
Distance of the Centre of Compression from the Compression edge=(59.42 - (((2/3/1.5) * 25*59.42*1000*(59.42/2) - ((2/3/1.5) *25*1000*33.95^ 2 /12))/(0.36*25*59.42*1000))) /59.42k2=0.416Constant0.416 OR say 0.42 as per IS 456 clause 37.1c ((X - (((2 / 3 / gc) * fck * X * B * (X / 2) - ((2 / 3 / gc) * fck * B * X1² / 12)) / (k1 * fck * X * B))) / X)
Moment of Resistance by Concrete Section=0.36*25*59.42*1000*(124-(0.42*59.42))MRc=53.08N-mm k1 * fck * xu * B * (D - (k2 * xu)) / 10&sup6;
Coefficient of Moment of Resistance =53.082/(25*1000*15376)k=0.138Constant0.138 for fe 415 (= Mu/ fck * B * D²)
Minimum Depth of Section Required=Sqrt(24457445.434/ (0.138*25*1000))Dr=84.197mmUsing formula Mu = k * fck * B * D²
DEPTH PROVIDED IS ADEQUATE
Ultimate Strain in Steel=(415/1.15/200000)+0.002esu=0.0038Constant((fy / gs / Es) + 0.002
Limiting Depth of Nuetral Axis=(0.0035 / (0.0035 +0.0038)) *124Xu=59.417mm(0.0035 / (0.0035 + esu)) * D OR 0.48*D
Parabolic Depth from Origin/Neutral Axis=0.002*59.42/ 0.0035X1=33.95mm0.002 * Xu / 0.0035
Total area Constant of the Stress Block=(((2/3/1.5)*25*59.42*1000)-((2/3/1.5)*25*33.95/3*1000))/25/59.42/1000k1=0.36Constant0.364 * fck * x * b ; This can be taken as 0.36 for all Practical Purposes -cl 37.1 of IS 456-2000 {(((2 / 3 / gc) * fck * X * B) - ((2 / 3 / gc) * fck * X1 / 3 * B))}
Distance of the Centre of Compression from the Compression edge=(59.42 - (((2/3/1.5) * 25*59.42*1000*(59.42/2) - ((2/3/1.5) *25*1000*33.95^ 2 /12))/(0.36*25*59.42*1000))) /59.42k2=0.416Constant0.416 OR say 0.42 as per IS 456 clause 37.1c ((X - (((2 / 3 / gc) * fck * X * B * (X / 2) - ((2 / 3 / gc) * fck * B * X1² / 12)) / (k1 * fck * X * B))) / X)
Reqd Steel Percentage %=1.645pt=1.645%By trials; pt = 1.625 produces lower Moment of Resistance than 598.166636961428
Moment of Resistance= (1000*(124 ^ 2) * ((415/1.15) * (1.645/100) * (1-(0.416*415) / 1.15) / (0.36*25) * (1.645/100)))/10 ^ 6MR=24.87kN-m;(B * (D²) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100)))/106
Steel Required=(1.64*1000*124) / 100Ast=2039.576mm²Ast = (pt * B * D) / 100
Spacing Required: #12@440c/c1000/(0/ (3.142*12²/4)) ,0) /10,0) *10Spacing=440mmB / (Ast / (PI * dia² / 4)) & rounded to nearest 10 mm
BEAM GEOMETRY
Design Parameters
0System of UnitsSys=SI
1UnitSystems.Length Units L=metre
2UnitSystems.Force Units F=Newton
3Concrete Strengthfck=25N/mm^2Characteristic Strength
4Concrete safety factorgc=1.5Cl 36.4.1 of IS 456-2000
5Concrete Young's ModulusEc=25000N/mm^2Cl 6.2.3.1 of IS 456-2000
6Steel Strengthfy=415N/mm^2
7Steel safety factorgs=1.15Cl.36.4.1 of IS 456-2000
8Steel Young's ModulusEs=200000N/mm^2Cl 5.6.3 of IS 456-2000
9Member UnitSystems.LengthL=5.7m
10Member BreadthB=250mm
11Member DepthDg=400mm
12Clear Coverdc=20mm
13Min UnitSystems.Area Tension SteelAstm=111mm^2
14Min UnitSystems.Area Compr SteelAscm=111mm^2
15Min Tension Steel %ptt=0.12%
16Min Compression Steel %ptc=0.12%
17Weight of Concretewc=25kN/m^3
18Weight of Steelws=785kN/m^3
19Load Safety Factorgl=1.5
20Main bar DiabDia=20mm
21Stirrup Steel UnitSystems.Stressfys=250N/mm^2
22Stirrup bar Dias_bar=8mm
23Stirrup Spacingsv=450mm
24Min SpacingminSv=50mm
25Flange widthbf=250mm
26Span/d ratio limitLim L=10
27Creep Coefficientcft=1.6
28Beam TypebTy=Simply Supported
29Crack Width LimitWcrL=0.3mm
Bending Moments, Shear Forces and Deflections with Any Combination of Loads
Inputs: Beam Specifications
Length of BeamL=5.7m
Section BreadthB=250mm0.25mm
Secton DepthD=370mm0.37mm
Modulus of ElasticityE=0.1m100m
Moment of InertiaIz=1055270833.33333mm^40.00105527083333333 m^4
Inputs Loads:
Loading and Reactions Diagram for Support Beam=
Shear Force for Support Beam=
Results
Support ReactionIterationR1=36.541kN36.541 kN
Support ReactionSum of All loads - R1R2=39.81N39.81 kN; R2 = P - R1
FEM at support AIteration MethodMab=0kNm
FEM at support BIteration MethodMba=0kNm
Max Sagging MomentIteration MethodMaxMx+=31.048kNmOccurs at: 2.95
Net Sagging MomentThe DifferenceNetMx+=31.048kNm MaxBM3 - (MaxBM1 + (MaxBM2 - MaxBM1) / L.Value * a)
Bending Moment for Support Beam=
Max Sagging MomentIteration MethodMx+=31.048kN-mOccurs at: 2.95
Deflection Diagram for Support Beam=1031573.35 mm occurs at 2.85
DESIGN FLEXURE
UnitSystems.Stress BLOCK
Eff depth=400-30-20/2d=370mmde = Dg - dc - barDia/2
Ultimate strain in steel=(415/1.15/200000/)+0.002esu=0.0038Constant((fy / gs / Es) + 0.002
Limiting depth of N-A=(0.0035 / (0.0035 +0.0038)) *370Xu=177mm(0.0035 / (0.0035 + esu)) * d OR 0.48d
Parabolic depth=0.002*177/ 0.0035X1=101mm0.002 * Xu / 0.0035
Area stress block=(((2/3/1.5)*25*177*250)-((2/3/1.5)*25*101/3*250))/25/177/250k1=0.36Constant0.364 * fck * x * b ; == 0.36 cl 38.1 of IS 456-2000 {(((2 / 3 / gc) * fck * X * B) - ((2 / 3 / gc) * fck * X1 / 3 * B))}
Stress CG Quotient from compr edge=(177 - (((2/3/1.5) * 25*177*250*(177/2) - ((2/3/1.5) *25*250*101^ 2 /12))/(0.36*25*177*250))) /177k2=0.416Constant0.416 OR say 0.42 as per IS 456 clause 37.1c ((X - (((2 / 3 / gc) * fck * X * B * (X / 2) - ((2 / 3 / gc) * fck * B * X1 ^ 2 / 12)) / (k1 * fck * X * B))) / X)
MR of Concrete=0.36*25*177*250*(370-(0.416*177))/10^6MRc=118,174,647N-mm k1 * fck * xu * B * (d - (k2 * xu))
UnitSystemsStress Block=
Min Eff d REQD
MR Coefft=118,174,647/(25*250*136900)k=0.138Constant0.138 for fe 415 (= Mu/ fck * B * de^2)
Factored UnitSystems.Moment=31,047,634*1.5Mu=46,571,451Nmmwith safety load factor: 1.5
Min Depth=Sqrt(46,571,451/ (0.138*25*250))dr=232mmUsing formula Mu = k * fck * B * d^2
DEPTH IS ADEQUATE
Reqd Steel Percentage %=0.42pt=0.42%By trials; pt = 0.4 produces lower Moment of Resistance than 46.571
Moment of Resistance= 250*(370^2) * ((415/1.15 ) * (0.42/100 ) * (1-(0.416 * (415/1.15) / (0.36*25) * (0.42/100))))/10^6MR=48.24kN-mB * (d ^ 2) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100))))
Steel Required=0.42*250*370/100Ast=388.5mm^2Ast = pt * B * d / 100
#20mm bars required=388.5/ (PI() *20^ 2 / 4)Nr=2Nr
DESIGN SHEAR
Shear UnitSystems.Force factored=1.5*36541.3333333333Vu=54812Ncl 40 IS 456
Shear UnitSystems.Stresstv=0.593N/mm^2
beeta Constant= 0.8 * 25/ (6.89 * 0.679)beeta=4.273Constant8 * fck / (6.89 * pt)
Shear strength of concrete=(0.85 * Sqrt(0.8 *25)* (Sqrt(1 + 5 *4.273) - 1)) / (6 * 4.273)tc=0.553N/mm^20.85 * Sqrt(0.8 * fck * (Sqrt(1 + 5 * beeta) - 1)) / (6 * beeta)
Shear reinforcement required since0.553<0.593
Stirrups Provided#8@450 mm c/c=
SLENDERNESS CHECKS
The beam passes slender checkspan: 5700is less than the lower of 60 *250 and 250 *250^ 2/370=
DEFLECTION CHECKS
Service UnitSystems.MomentMs=31,047,634
UnitSystems.Area of Steel RequiredAsReq=389
UnitSystems.Area of Steel ProvidedAsProv=628
Service UnitSystems.Stress =0.58 * 415fs=241N/mm^258% of yeild UnitSystems.Stress fy
Gross UnitSystems.Moment of Inertia=250*(400)^ 3 / 12Igr=1,333,333,333
UnitSystems.Moment Constant=1031573.35*(25000*1333333333)MCon=34,385,778,324,736,900,000reverse calculation from elastic deflection
SimplySupportedSpan/effDepth RatioL/d=20Clause 42 and 23.2.1 of IS 456
Tension Steel % modification factor=1 / (0.225 + 0.00322 * 149.1+ 0.625 * Log10(0.679))F1=1.667Limited to 2.0
Compression Steel % factor=1.6 *0.12/ (0.12 + 0.275)F2=1Range 1.0 to 1.5
Flange modification factor=0.8 + 2 / 7 * (250/250- 0.3)F3=1
Calculated modification factor=20*1.667*1*1S/d=33.34Computed final span/eff d ratio
Span to eff depth ratio is with in limits
SERVICEABILITY CHECKS Annex F of IS 456
Dist from compression face=400a=400mma=Dg >> Clause 43.1 & 26.3.2
Initial fy strain @ level considered=1/628e1=0.00159e1 = 1 / Ast.Value
Ave fy strain @ level considered= 0.00159-((250*(400-0)*(400-0))/(3 *628*200000*(370-0)))em=0.0013e1 - ((b.Value * (Dg.Value - Xu.Value) * (a.Value - Xu.Value)) / (3 * Ast.Value * Es.Value * (D.Value - Xu.Value)))
Eff cover corner=20+20/ 2effCover=30mmeffCover = dc.Value + m_bar.Value / 2
Dist surface of bar =Sqrt(2 * 30^ 2)-20 / 2acr=32.43mmSqrt(2 * effCover ^ 2) - m_bar / 2
Crack width at mid span bottom corner=3 *32.43*0.0013/ (1 + (2 *32.43-20)/(400-0))Wcr=0.114mm Annex-F of IS 456
Dist surface of bar=20dc=20mm= dc
Crack width at mid span soffit=3 *20*0.0013/ (1 + (2 *20-20)/(400-0))Wcr=0.074mm Annex-F of IS 456
The crack width is with in limits