design stair

7/30/2019 Design Stair

1/25

Torsion design

TORSION DESIGN FOR RECTANGULAR BEAM SECTIONUSI NG ESCP 2 1983 CODE TORSION ON GI RDER B FB3

B

B= 0.3 Uef=1.6 Mean perimeter (Calculated value)

equiva D= 0.6 Aef=0.1375 Equi valent hollow section area (Calculated value)

dw= 0.55

D d=0.575 bw=hef = bw1/5 0.05 bw=

C=25

bw =B-2*250.25

bw1

Equi valent hollow section

W=11.45 Distiributed cantiliverd load TRd=38.9583 Limiting value of ultimate torque (Calculated value)

L= 2 Length of cantiliverd Tc=8.4975 Torsional resistance of concrete (Calculated value)

P=3.64 Concenterated load Tsd=26.54

fcd=11.3333 fyk=300 Vc=58.6328

ftcd=1.03 fyd=260.87 Vsd=204.01 (GIven value)

The lim iti ng values of torsional & shear resistance are mul tipl ied by the fol lowing reductions .

(a) torsion

t= 1/(1+((Vsd/VRd)/(TRd/Tsd))2)^.5= 0.66799

TRd= 26.0238

TRd>=Tsd OK!The section is enough

(a) shear

v= 1/(1+((Tsd/TRd)/(Vsd/VRd))2) .5= 0.74417

VRd= 43.6327

The values of torsional & shear r esistance of section are mul tipl ied by the foll owing reductions .

(a) torsion

tc= 1/(1+((Vsd/Vc)/(Tsd/Tc))2) .5 = 0.66799

Tc= 5.67625

(a) shear

vc= 1/(1+((Tsd/Tc)/(Vsd/Vc))2)^.5 = 0.74417

Vc= 43.6327

Reinf orcement design

Tef=Torsional resistance of reinforcement 20.86375

Tef=2*Aef*fyd*As/s As 58.1656018 Cross-sectional area of stirrupps

S=spacin 200 mm

Tef=2*Aef*fyd*Al/uef Al 465.324815 Cross-sectional area of longitudnal reinforcement.

Distiributed equally for each corner.


2/25

Tors ion d


3/25

RECTANGULAR BEAM DESIGN(singly reinforcement)

Mspan

Beam deph d>=(.4+.6*fyk/400)LeB/a 0.219583 a= 24 (EBCS 2

D=d+35= 0.254583 calculated value use D= 0.55 d=

LeB=Lengith of beam = 6.2 useB= 0.25

M01 57.45 KNm out put from sap analysis Material p

M12=57.45 KNm out put from sap analysis Use conce

Lspan=86 KNm out put from sap analysis fcd

Msup= M11-M120 Paxial=0 KN

msup/M11= 0


4/25

Asm= 696.19 From moment

Asp= Paxl/fyd 0 From axial load

As design= 696.19

No of reinforcement 16 = 3.4636 No of reinforcement 14


No of reinforcement 24 = 1.54024

SO USE 216 +220

Shear reinforcement

Vc= .25fctdk1k2bwd 39.9648

where k1=(1+50)==1.0 1.075

Vs=Vsd-Vc=64.3552 =2Avdfyd/S S=Spacing

S=2Avdfyd/Vs

Spacing for 8 = 212.8135 Spacing for 12 = 480.958


SO USE 8c/c140


5/25

M11

M12

Longitudnal Beam

ABLE 5.1)

0.525

roperty

ret=C 25 Reinforcement Fyk 300

11.16667 fyd 260.87

s

= 2.91336

= 3.97042


6/25

= 4.52072

= 6.16098


7/25

RECTANGULAR BEAM DESIGN(singly reinforcement)

Mspan

Beam deph d>=(.4+.6*fyk/400)LeB/a 0.2125 a= 24 (EBCS 2

D=d+35= 0.2475 calculated value use D= 0.45 d=

LeB=Lengith of beam = 6 useB= 0.25

M01 60 KNm out put from sap analysis Material p

M12=60 KNm out put from sap analysis Use conce

Lspan=17 KNm out put from sap analysis fcd


msup/M11= 0


8/25

Asm= 158.4 From moment

Asp= Paxl/fyd 0 From axial load

As design= 158.4



No of reinforcement 24 = 0.35044

SO USE 216 +220

Shear reinforcement

Vc= .25fctdk1k2bwd 35.362

where k1=(1+50)==1.0 1.175


S=2Avdfyd/Vs



SO USE 8c/c140


9/25

M11

M12

Longitudnal Beam

ABLE 5.1)

0.425

roperty

ret=C 25 Reinforcement Fyk 300

11.16667 fyd 260.87

s

= 3.7861

= 5.15981


10/25

= 1.02857

= 1.40177


11/25

Beam design

RECTANGULAR BEAM DESIGN M11(singly reinforcement) M12

Longitudnal Beam

Mspan

Beam deph d>=(.4+.6*fyk/400)LeB/a 0.2125 a= 24 (EBCS 2 TABLE 5.1)

D=d+35= 0.2475 calculated value use D= 0.55 d= 0.525

LeB=Lengith of beam = 6 useB= 0.25

M01 126.07 KNm out put from sap analysis Material property

M12=126.07 KNm out put from sap analysis Use conceret=C 25 Reinforcement Fyk

Lspan=205 KNm out put from sap analysis fcd 11.16667 fyd 260.87


msup/M11= 0


12/25

Beam des

SO USE 8c/c140


13/25

300


14/25

Beam desgn

RECTANGULAR BEAM DESIGN M11(Dubly Reinforcement) M12

Longitudnal Beam

Mspan

Beam deph d>=(.4+.6*fyk/400)LeB/a 0.2125 a= 24 (EBCS 2 TABLE 5.1)


LeB=Lengith of beam = 6 useB= 0.25 d'= 0.25

M01 80 KNm out put from sap analysis Material property

M12=65 KNm out put from sap analysis Use conceret=C 25 Reinforcement Fyk

Lspan=45 KNm out put from sap analysis fcd 11.16667 fyd 260.8696


msup/M11= 0.1875 .0176 Doubly reinforcement

Ks= 4.65 Ks1= 0.56 (FROM ESCP-2 TABLE 33)

Asm= 1225.91 From moment d'/d= 0.909091

Asp= Paxl/fyd 38.33333 From axial load

As design= 1264.24 Tensile reinforcement

As design= 147.636 Compression reinforcement

No of reinforcement 16 = 6.2898 No of reinforcement 14 = 8.20937

No of reinforcement 20 = 4.0262 No of reinforcement 12 = 11.188

SO USE 416 +220

Span rein for cement

MspanD=52.5 KNm Paxial=10 KN

Kd= 0.01898


15/25

Beam desgn

SO USE 8c/c140


16/25

Beam desgn

300


17/25

TORSION DESIGN FOR RECTANGULAR BEAUSING ESCP 2 1983 CODE

B

B=0.3 Uef=1.9

D=0.75 Aef=0.175

dw=0.7

D d=0.725 bw=

hef = bw1/5 0.05 bw=

C=25

bw =B-2*250.25

bw1

Equivalent hol low section

Length of beam 3.6

W=12 Distiributed cantiliverd load TRd=49.58333

L=1.7 Length of cantiliverd Tc=10.815

P=12 Concenterated load Tsd=49.572

fcd=11.33333 fyk=300 Vc=53.90602

ftcd=1.03 fyd=260.8696 Vsd=49.01

The limiti ng values of torsional & shear resistance are mul tiplied

(a) torsion

t= 1/(1+((Vsd/VRd)/(TRd/Tsd))2) .5= 0.98089

TRd= 48.63581


(a) shear

v= 1/(1+((Tsd/TRd)/(Vsd/VRd))2) .5= 0.194562

VRd= 10.48806

The values of torsional & shear resistance of section are mul tipli ed

(a) torsion

tc= 1/(1+((Vsd/Vc)/(Tsd/Tc))2) .5 = 0.98089

Tc= 10.60833


18/25

(a) shear

vc= 1/(1+((Tsd/Tc)/(Vsd/Vc))2)^.5 = 0.194562

Vc= 10.48806

Reinforcement design (from torsion)

Tef=Torsional resistance of reinforcement

Tef=2*Aef*fyd*As/s Cross-secti



SO USE 8c/c = 150

Tef=2*Aef*fyd*Al/uefAl= 810.8154641 Cross-sectiDistiribute

Shear reinforcement (f rom shear force)

Vc= 10.4881

where k1=(1+50)==1.0 0.875


S=2Avdfyd/Vs



SO USE 8c/c = 150

(Over all shear reinforcement)

When shear force &torsion acts together Tc=0

When considering combined shear & torsion of it is necessery ADD th

Av+t/s=Av/s+2At/s

A(v+t)/s = 0.52858

As= 0.528585

Spacing 100 for As 52.8585




19/25

SO USE 12c/c = 120


20/25

SECTIONTORSION ON BEAM

Mean perimeter (Calculated value)

Equivalent hol low section area (Calculated value)

Limiting value of ultimate torque (Calculated value)

Torsional resistance of concrete (Calculated value)

(GIven value)

y the following reductions .

by the foll owing reductions .


21/25

38.96367

onal area of stirrupps

onal area of longitudnal reinforcement.equally for each corner.

(f rom shear force)

e stirrups


22/25

TORSION DESIGN FOR RECTANGULAR BEAM SECTIONUSI NG ESCP 2 1983 CODE TORSI ON ON BEAM

B

B= 0.6 Uef=2.2 Mean perimeter (Calculated value)

D= 0.6 Aef=0.3025 Equi valent hollow section area (Calculated value)

dw=0.55

D d=0.575 bw=

hef = bw1/5 0.11 bw=

C=25

bw =B-2*250.55

bw1

Equi valent hollow section

Length of beam 6

W= 15.3 Distiributed cantiliverd load TRd=188.558 Limiting value of ultimate torque (Calculated value)

L=2.5 Length of cantiliverd Tc= 41.1279 Torsional resistance of concrete (Calculated value)

P=5 Concenterated load Tsd=162.188

fcd=11.3333 fyk=300 Vc=100.164

ftcd=1.03 fyd=260.87 Vsd=71.5 (GIven value)

The limi ting values of torsional & shear resistance are multiplied by the fol lowing reductions .

(a) torsion

t= 1/(1+((Vsd/VRd)/(TRd/Tsd))2)^.5= 0.98401

TRd= 185.543


(a) shear

v= 1/(1+((Tsd/TRd)/(Vsd/VRd))2)^.5= 0.17812

VRd= 17.8412

The values of torsional & shear r esistance of section are mul tipl ied by the fol lowing reductions .

(a) torsion

tc= 1/(1+((Vsd/Vc)/(Tsd/Tc))2)^.5 = 0.98401

Tc= 40.4702


23/25

(a) shear

vc= 1/(1+((Tsd/Tc)/(Vsd/Vc))2)^.5 = 0.17812

Vc= 17.8412

Reinforcement design (from torsion)

Tef=Torsional resistance of reinforcement 121.717

Tef=2*Aef*fyd*As/s Cross-sectional area of stirrupps



SO USE 14c/c = 200

Tef=2*Aef*fyd*Al/uefAl= 1696.66514 Cross-sectional area of longitudnal reinforcement.

Distiributed equally for each corner.

Shear reinforcement (fr om shear f orce)

Vc= 17.8412

where k1=(1+50)==1.0 1.025

Vs=Vsd-Vc= 53.6588 =2Avdfyd/S S=SpacingS=2Avdfyd/Vs



SO USE 8c/c = 200

(Over all shear reinforcement) (fr om shear force)

When shear force &torsion acts together Tc=0

When considering combined shear & torsion of it is necessery ADD the stirrups

Av+t/s=Av/s+2At/s

A(v+t)/ s = 0.95007

As= 0.95007




SO USE 12c/c = 120


24/25

STAIR DESIGN

STAR CASE DESGN- 2

L1

H1 (H1/L1)= 0.6667

= 30 0.5236

Angle o inclination

L

stair depth d>=(.4+.6*fyk/400)Le/a 0.17 a= 25 (EBCS 2 TABLE 5.1)


Lex=shorter span of slab 1.5 L=length of the stair 5 L1= 2.7

Use conceret=C25 Reinforcement Fyk= H1= 1.8

Dead load =

Conceret 5 Conceret unit weight=25Conc tri 2 Riser 0.16

Screed = 0.345 Screed unit weight =23 (EBCS 1TABL E 2.8)

Marble 0.405 Tile unit weight =27

W1= 7.75

W1/COS = 8.94893

L ive load =

Category Specific use =0.5/Fyk = 0.00167LL= 3 KN/M2

es gn oa =

Design load = 1.3*DL+1.6*L 16.4336 Design moment span=WL^2/8 51.355 For simply support

DL= 16.4336

Reinforcement design=

Bottom reinforcement Span

Kd= 0.026 Ks= 4.13 (FROM ESCP-2 TABLE 27)

As= 1146.47

Asmin= bd= 308.333

As design 1146.47

Spacing for 8 43.7867 Spacing for 10 68.47127 Spacing for 12 98.5637

Spacing for 14 134.326 Spacing for 16 175.321

SO USE 16C/C=150

Page 24


25/25

STAIR DESIGN

r from analysis OK!

design stair

Documents