Download - Free Standing Staircase 11
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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 =
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= 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
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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
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= 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
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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)
≥
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= 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
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