Symmeterical loaded free standing stair case

Computing loads on flight and landing

Flight

D.l =Assuming a 250 thick waist slab 0.3

R = 0.344 mm 0.1667 R

= 10.836 KN/m

= 2.0004 KN/m

= 1.89 KN/m

= 14.7264 KN/m

= 6.3 KN/m

= = 30.7

landing

= 11.25 KN/m

= 2.25 KN/m

= 13.5 KN/m

= 15 KN/m

= KN/m = 42.9 KN/m

Given Data35

460

= 1.575 m H1 = H2 = 2.5 m

= 4.2 m E/G = 2.35

= 30.7 KN/m 30.79 = 0.859

= 42.9 KN/m 30.79 = 0.5118

= 1.875 m 30.79 = 0.738

B' = 3.75 m 30.79 = 0.262

Self wieght of waist slab(W1) + Self weight of step(W2) +S.F Load(W3)

W1

W2

W3

P1

P2

P(1+2) 1.4*P1 +1.6*P2

W3

W4

P3

P4

P(3+4) 1.4*P1 +1.6*P2

fcu = N/mm2

fy = N/mm2

B1

L1

P(1+2) cos Φ =

P(4+5) sin Φ =

L3 cos 2Φ =

sin 2Φ =

= 200 mm

Determining the moment of inertia

= 1575 mm

= = 1.05E+09

= = 6.51E+10

= = 6.62E+10

m = = 0.016125

n = = 0.006753

f = = 2.175 m

= 0.0136694604

= 0.0092083082

= 0.0024555081

= 4.1772186857

From Equilibrium and method of least work = Factored load on Flight A = Factored load on Flight B

R1 = R2 = 0

RA = RD = = 354 KN

Ha = Hd = 369.5 KN

Design moments

= = -703.1 KNm

= = -107.1532 KNm

= = 401.9 KNm

Reinforcment

hs

B1

Ix B1hs3/12 mm4

Iy hsB13/12 mm4

Ip (B1*hs*(B12*hs

2))/12 mm4

Ix/Iy

(E/G)*Ix/Ip

B' - B1

K1

K2 N/mm2

K3

K4

P(1+2) P(1+3)

B1*L1*P(1+2) +0.5*L3*B'*P(4+5)

H1/K4{B1*L1*P(1+2) +0.5*L3*B'*P(4+5)(0.333*L1+0.25*H1)} =

Mx1 = Mx2 0.5*[{-2H1*HA+L1*RA+0.5*L3*B'*P(4+5)}(L1+H1)]

My1 = My2 -f((K3HA/2K2)

Mz1 = Mz2 0.5*f*HA

Flexural Reinforcement

d =

123 mmso lets say d = 250 mmhs = 280 mm

Z = 237.5 mm

moment of resistance =b = 1575 mmd = 250 mm

l

= 1550.4 KN.m

Reinforcement sizes and areas that can be used hereBar type Dia in mm AreaT20 20 314T25 25 491T32 32 804T40 40 1260

Required steel area for design moment =

6775

USING 25 DIA BARS = 9

spacing of provided bars = 154 mm

Provide T25@150mm c/c both top and bottom

Reinforcement in the

Mx1 = Mx2 =0.85fcubd2

=0.85*35*1575*d2

√{Mx1/0.85fcub}

0.45fcubd2

M/(0.95fyz)

mm2

= 401.9 KNm b = 280 mmd = 1545 mmz = 1467.75 mm

Required steel area for design moment =

= 627

Provided area of steel for reinforcement = 628

Number of bars = 8

Spacing provide for the bars = 554.857143

provide T32@550mm cc in close rings

Design for Shear Reinforcement Design

V = 354 KN

v = 0.9

1.73 = 1.21

400/d = 1.6 = 1.13

1.4 = 1.120.53

= = 0.82

˂ v ˂Hence shear reinforcement in the form of closed loop rings are required

=

1493 Provide 2108

Number of links = 19

Sv = 220

Provide T12@200 mm c/c

Mz1 = Mz2

M/(0.95fyz)

mm2

mm2

V/bvd = N/mm2

100As/bvd = N/mm2 (100As/bvd)1/3

(400/d)1/4

fcu/25 = (fcu/25)1/3

0.79/γm =

vc (0.79/γm)* [(100As/bvd)1/3*(400/d)1/4(fcu/25)1/3]

vc 0.8√fcu

Asv = V/(0.95*fyv)

vbd/0.95fyv

Asv = mm2 Asv = mm2

Torsional Reinforcement Design

Let

= 280 mm

= 1575 mmT = 107.16 KN.m

= 2T

= 1.85

25 30 35 400.33 0.37 x 0.4

4 4.38 n 5

= 0.385

= 4.69

= 2.75 ˂

hence

˂ ˂

Extra link reinforcement required

=where

= 220 mm

= 1815 mmFor additional Torsional link reinforcement

0.8381830213

hmin

hmax

vt

h2min(hmax-hmin/3)

vt N/mm2

vtmin

vtu

vtmin

vtu

vd vt + v = vtu

vtmin vd vtu

Asv T/{(0.8X1Y1*(0.87*fyv)}

X1

Y1

Asv/sv ≥ T / (0.8*(x1*y1*(0.87fyv)

≥

= minimum of

then C)200 mm

= 220 mm

Therefore = 184.400265

Hence providing T8@ 140 mm c/c in landing

Sv a) X1

b)Y1/2

Sv

Asv mm2