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Stress 6 5 4 3 2 1 0.000
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Typical concrete stress-strain curve, with short-term loading.
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אאE٣ J١F Φ Weight mm Kg/m 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38
0.222 0.395 0.617 0.888 1.21 1.58 2.00 2.47 2.98 3.55 4.17 4.83 5.55 6.31 7.13 7.99 8.90
1 .283 .503 .785 1.13 1.54 2.01 2.54 3.14 3.80 4.52 5.31 6.16 7.07 8.04 9.08 10.2 11.3
2 .566 1.01 1.57 2.26 3.08 4.02 5.09 6.28 7.60 9.04 10.6 12.3 14.1 16.1 18.2 20.4 22.6
Area of Cross-Section 3 4 5 6 .848 1.13 1.41 1.70 1.51 2.01 2.51 3.02 2.36 3.14 3.93 4.71 3.39 4.52 5.65 6.79 4.62 6.16 7.70 9.24 6.03 8.04 10.1 12.1 7.63 10.2 12.7 15.3 9.42 12.6 15.7 19.8 11.4 15.2 19.0 22.8 13.6 18.1 22.5 27.1 15.9 21.2 25.5 31.9 18.5 24.6 30.8 37.5 21.2 28.3 35.3 42.4 24.1 32.2 40.2 48.3 27.2 36.3 45.4 54.5 30.6 40.9 50.7 61.2 33.9 45.2 56.5 67.8
in cm2 7 8 1.98 2.26 3.52 4.02 5.50 6.28 7.92 9.05 10.8 12.3 14.1 16.1 17.8 20.4 22.0 25.1 26.6 30.4 31.7 36.2 37.2 42.5 43.1 49.3 49.5 56.6 56.3 64.3 53.6 72.6 71.4 81.6 79.1 90.4
9 2.54 4.52 7.07 10.2 13.9 18.1 22.9 28.3 34.2 40.7 47.0 55.4 63.6 72.4 81.7 91.8 102
10 2.83 5.03 7.85 11.3 15.4 20.1 25.4 31.4 38.0 45.2 53.1 61.6 70.7 80.4 90.8 102 113
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R.C slab
(plaster)
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Thickness of beam (t) = span / 8-10
For simple beams choose t = span / 8
For continuous beams ,, t = span / 10
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i.e. O.w. of beam = 1.0 x 0.2 x (0.6 – 0.1) x2.5 = 0.25 t/m L٠{٢٥Zאאא Wאא
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γ wall = 1.8
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- ٢١ -
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Live loadsאאZL
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(2-2)
אאאא אאE٣ Wאאאא
U = 1.4 D + 1.7 (E + L)
(2-3)
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(2-1)אאU
- ٢٢ -
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Own weight of slab = Ws =1×1× t ×γ c = t × γ c
אאE
= 0.1× 2.5 = 0.25 t/m2 = 2.5 kN / m2 Flooring (א) = 0.15 t / m2 = 1.5 kN / m2 Live Loads (L.L.) (א = )א0.20 t/m2 = 2.0 kN / m2 Total Loads (wt) ( = )اﻟﺤﻤﻞ اﻟﻜﻠﻲ0.25+ 0.15+ 0.20 = 0.60 t/m2 = 6.0 kN/m2
W = 6.0 x (3.5/2) = 10.5 kN/m
B1אאE B2אאE
W = 6.0 x 3.5 = 21 kN/m
- ٢٣ -
אא
٢٥١
א
אאאא
א
ﺑﻼﻃﺔ
L=8,0 m
B2 آﻤﺮة
B1 آﻤﺮة
L=3,5 m
L=3,5 m
E٢{٢F
א B2 B1 א S1 א א א א א W ٢ ١٢Zא K א א א E٣ J٢F
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Wא
Own weight of slab = Ws =1×1× t ×γ c = t × γ c
אאE
= 0.12× 2.5 = 0.30 t/m2 = 3.0 kN / m2 Flooring (א) = 0.20 t / m2 = 2.0 kN / m2 Live Loads (L.L.) (א = )א0.30 t/m2 = 3.0 kN / m2 Total Loads (wt) ( = )اﻟﺤﻤﻞ اﻟﻜﻠﻲ0.30+ 0.20+ 0.30 = 0.80 t/m2 = 8.0 kN/m2
W = 8.0 x (4.0/2) = 16.0 kN/m - ٢٤ -
B1אאE
אא
٢٥١
א
אאאא
א
B2אאE
W = 8.0 x(4+3)/2 = 28 kN/m 1
4.5 m
٢
٣
4.5 m
4.5 m
٤
A
3.0 m
B2
B
S1 4.0 m
C
B1 9m
E٣{٢F
- ٢٥ -
א אאאא
אא אא
٣
אא
٢٥١
א
אאאא
א
Wאא א א אא א א א א
Kאאאאא א،א א א א א א א
Kאאא אאא
Wאא Wאאאאא
Kאאא E١
א א א א E٢ Kאאאא
א א א א א א E٣
אא
Wאא Wאאאאא
KאאאאE١
KאאאE٢
KאאאE٣
KאאאאאאאאE٤ Wאאאא • Wאא Kאא - ٢٦ -
אא
٢٥١
א
אאאא
א
אאאא Structural Analysisfor Slabs and Beams
WאאW١ J٣ Wאאאאא
אא: אא J١
א א (Beams) אא (slabs) א
WאאK(Footings)א(Columns)
Slabsא
Beamsא א
Girdersא א Columns א
Footings א
KאאאאE١ J٣Fא
Over hanging beam
Simple slab or beam
continous slab or beam
(Column)
Simple frame
(Column)
אאאE١ J٣F - ٢٧ -
אא
٢٥١
א
אאאא
א
אאWאא J٢
אא אא א א، א אאWKאאא
KKאאאא
T،،،Wא
KאאאE٢ J٣FאKUL،
U
L
U - section
L - section
T circular sec
T - section square sec
Rectangular sec
KאאאE٢ J٣F
WאאאאW٢ J٣ א Statically System א א א א א א Kא א א
Kאאא
EאאFאא W W١ J٢ J٣ E٣ J٣F (Knife Edge Support) א א א א
K
- ٢٨ -
אא
٢٥١
א
אאאא
א
d
L
Knife edge support
E٣ J٣F
WאאW٢ J٢ J٣
אא אא (Linear Elastic Theory)אא א א א א א א א Kאא א
W
Three Moment EquationאאE
Virtual Work MethodאאאE Moment DistributionאE
א א Eא אF א א אא٠{٤،אא٠{٨(lo)אא(d)א i.e
d / l0 < 0.8 for simple beams. d / l0 < 0.4 for continues beams. (Deep Beam)אא אאא
Kאאא(Plastic Analysis)א
W(Effective Span l0)אאאאW٣ J٣ WאאאE
Wאאאאאאא l01אאא
l02 = l0 × 1.05 אא١{٠٥ - ٢٩ -
• •
אא
٢٥١
א
אאאא
א
l03 = l0 + d
•
E٤ J٣FאאאאZd
d
L L
1
L 2 = 1.05 L
C L
C L
אאאE٤ J٣F
WאאאE
אFאEאFאאאאE١ WאאאאאE(Rigid Connections)
l01 = Centerline to Centerline of Supportsא• א l02 = 1.05 × l0אא١{٠٥ •
אאFאאאאאE٢
אאאאא KEאא– Kאאאא
Or
L03 = Distance from Centerline to Centerline L03 =l0 +d
WאאאE
Wאאאאאאא (l01)אא - ٣٠ -
•
אא
٢٥١
א
אאאא
א
Wאא
•
L02 = l0 + d2
E٥ J٣Fאא
C
L
d1 d2
L L 1
C
L
E٥ J٣F
:(Straining Actions)אאW٤ J٣ (Normal Forces) אא(Straining Actions)אא אא(Bending Moments)א(Shear Forces)א א Kא אאאא
Kאאאאאאאאא
(N)אא אאאאאא א אאאאא (M) א (Q)א
Kאאא
Wאא(D.L + L.L )אא
Z(N)אא
:Simple Slab or Simple BeamEאאFE١ - ٣١ -
אא
٢٥١
א
אאאא
א
•
Wאא(Q)א Q = K q × wt × L
(3-1)
E J٦ J٣Fאא٠{٥Z (shear factor) אZK q
Uniformly Distributed Load (D.L +L.L) אאZwt
Span of Slab or beam אאZL
Wאא(M)אא • 2
M = w × L / Km
(3-2)
E J٦ J٣Fאא
٨Z(moment factor)אZKm
w t/m Loads L
wL 2 K q = 0.5
Q=
S.F.D
Q B.M.D
K m= 8 2
wL M= 8
E٦ J٣F WOver Hanging Slab or BeamאאE٢
Wאא(D.L + L.L )אא W(Q)א - ٣٢ -
אא
٢٥١
א
אאאא
א
(D.L+L.L) אא א א א א KKKK (D.L only) אא Eא אאF אא KE J٧ J٣Fאא(D.L + L.L)אאאאא i.e.
Q = K q × wt × L
(3-3)
،EאF٠{٥Z (shear factor) אZK q W
Uniformly Distributed Load Span of Slab or beam
(D.L +L.L) אאZW t
אאZL W(M)אא
א א א א E א א (D.L) אא EאאF אא (D.L+L.L)
KE J٧ J٣F
- ٣٣ -
אא
٢٥١
א
אאאא
א
w1 = D.L + L.L (t/m)
w
D.L
w
Loads L
L
1
L
w1L 2
1
wL1 S.F.D
Kq = 0.5 wL1 w1L 2
M -v e
B.M.D
Case of max +ve M w1L2 8 w t/m (D.L)
w1 t/m
w1(D.L + L.L) Loads
M -ve =
w1L21 2 B.M.D
Case of max -ve M 2
wL 8
E٧ J٣F
W
2
2
Max. M +ve. = wt×L /8 – w × L1 /2
(3-4)
(D.L+L.L)אאאאאאZwtW (D.L. only)אאאאZw
אאZL1 אאZL
- ٣٤ -
אא
٢٥١
א
אאאא
א
אא א א א E E J٧ J٣F אא (D.L+L.L.)אEאאFאא (D.L.)
Max. M-ve = w × L12/2
(3-5)
Continues Slab or Beam
E٢
א א א א א אא א Kאאאא٪٢٠א
WאאאאאW W(M)אא •
2
M = w×L / Km
(3-7)
E٨ J٣Fאא(moment factor)אZKm
אאאZL (Q)א •
Q = K q × wt × L
(3-6)
E٩ J٣Fאא (shear factor) אZK q
(D.L+L.L)אאאאאאZwt
אאאZL
L1אL،٪٢٠א،אא
Where,
L = (L1 + L2) / 2 L1 < 1.2 L2
and, wt = (wt1 + wt2) / 2 or, L2 < 1.2 L1 - ٣٥ -
WL2
אא
٢٥١
א
אאאא
א
w t/m or w1
w t/m or w2 Loads
L or L
L or L
2
1
-9
-24
-24
+11 wL 24
( Moment factor)
+11 w L2 9
2
Km
w L2 24 B.M.D
w L2 11
w L2 11
אE٨ J٣F
0.4
0.6
0.6
0.4 K q ( Shear factor)
0.6 wL 0.4 wL S.F.D 0.4 wL 0.6 wL
אE٩ J٣F
WאאאאW (Q)א
Q = K q × wt × L
(3-8)
•
E١٠ J٣Fאא٠ (shear factor) אZK q - ٣٦ -
אא
٢٥١
א
אאאא
א
(D.L+L.L)אאאאאאZw t
Effective Span of Slab or beam אאאZL W(M)אא • 2
M = w×L / Km
(3-9)
E١٢ J٣FE١١ J٣Fאא(moment factor)אZKm EffectiveSpan of Slab or beam אאאZL
L ،٪٢٠ א ،א א W
Where,
WL2L1א
L = (L1 + L2) / 2 L1 < 1.2 L2
0.45
and, wt = (wt1 + wt2) / 2 or, L2 < 1.2 L1
0.6
0.5
0.5
0.5
0.5
0.5
Kq
אאאאאE١٠ J٣F -24
-10 +10
-12 +12
-12 +12
Km ( Moment factor)
אאE١١ J٣F
- ٣٧ -
אא
٢٥١
א
אאאא
א
-24
-10 +12
-12 +16
-12 +16
Km ( Moment factor)
אאאאE١٢ J٣F
WאאאW٥ J٣ אאאאE١ J٣Fאאאא WאאאKאא
WאEאFEאאF E EE٦ J٣FאאFאאאא Kאאא
Wאא E
F א א א א א E٧ J٣אאFEא
אאא
WEאFE
E١٠ J٣Fאאאאאא
א אאKאאאא E١٢ J٣FE١١ J٣Fא
- ٣٨ -
אא
٢٥١
א
אאאא
א
W٦{٣
KE٢ J٢FאאאאW١
Wא
W = 0.6 t/m2 Loads m 3.0
wL2 24
m 3.0
wL2 24
--v M e
+v e M
B.M.d
+v e M
loads & B.M at section s-s
Bending Moments: 2 × 0 .6 × 3 .5 2 w l 1 t M +ve = 11 = 11 = 0.668mt / m
אא
M −ve
× = w l t
9
2 1
0 .6 × 3 .5 2 = = 0 .816 mt / m אא 9
- ٣٩ -
:٢ ﻣﺜﺎل
אא
٢٥١
א
אאאא
א
.أوﺟﺪ ﻋﺰوم اﻻﻧﺤﻨﺎء ﻋﻠﻰ اﻟﻜﻤﺮة اﻟﺘﺎﻟﻴﺔ ذات ﺑﺤﺮﻳﻦ و ﻳﺆﺛﺮ ﻋﻠىﻬﺎ ﺣﻤﻞ ﻣﻮزع ﺑﺎﻧﺘﻀﺎم 3.5 t/m C
A
B 6m
6m
Wא
M + ve M −ve
× =w l t
11
× = w l t
9
3.5 × 6 2 = 11 .45 mt / m 11
אא
3 .5 × 6 2 = = 14 .0 mt / m 9
אא
2 1
2 1
=
Q A = 0 .4 × wt × l1 = 0 .4 × 3 .5 × 6 = 8 .4 mt / m
Q A = 0 .6 × wt × l1 = 0 .6 × 3 .5 × 6 = 12 .6 mt / m
C وAאא Bאא :٣ ﻣﺜﺎل
.أوﺟﺪ ﻋﺰم اﻻﻧﺤﻨﺎء ﻋﻠﻰ اﻟﻜﻤﺮة اﻟﺘﺎﻟﻴﺔ ذات ﺑﺤﺮ واﺣﺪ و ﻳﺆﺛﺮ ﻋﻠىﻬﺎ ﺣﻤﻞ ﻣﻮزع ﺑﺎﻧﺘﻀﺎم 1.8 t/m B
A 7m
- ٤٠ -
אא
٢٥١
א
אאאא
א
Wא M + ve
× =w l t
8
2 1
אא
1 .8 × 7 2 = = 11 .025 mt / m 8
Q A = + 0 .5 × wt × l1 = 0 .5 × 1 .8 × 7 = + 6 .3mt / m Q B = − 0 .5 × wt × l1 = −0 .5 × 1 .8 × 7 = − 6 .3mt / m
Aאא Bאא
W١ E٢ J٢F א B2 B1 א א א א א KE٢אF
W٢ E٣ J٢Fא B2 B1א S1 אא אא KE٢אF
- ٤١ -
א
אא
א א
٤
אאא
٢٥١
א
אא
א
Wאא אא א א א א EאF א א אא א ،אאא אא
אאאאא Kאאא
Kאאאא
Wאא Kאאאאא E١
KEאאאFאאאאאE٢
א א א א אא א E٢ Kאא
אא א א א א א א א E٣ Kאא
אאאאאאאא E٤ Kא
Wאא ٪١٠٠אאאאאאאאאא Wאאאא • Wאא Kאא
- ٤٢ -
אאא
٢٥١
א
אא
א
אאאW١ J٤
Wאאא Jא Jא Solid SlabsאאE١ אאW(Hollow Block Slabs)אאE٢ KE١{٤ Fאא אאאאW(Flat Slabs)אאE٣ F אא KE٢{٤ Waffle SlabsאאאאE٤ Lift SlabsאאE٥ Pre-SlabsאאE٦
- ٤٣ -
אאא
٢٥١
א
אא
א
אא WE١ J٤F
אא WE٢ J٤F
- ٤٤ -
אאא
٢٥١
א
אא
א
אא WE٣ J٤F א א (Solid Slabs) א א א Kאאאאא
אאאFאאאאאא אא א(Beams)אאאאאKEKKKKKא
Kא(girders) Wאאא .אאא
Kא
E
E
W
Wאאאאא٢{٤ W
אאאאאא
i.e. r = L / B ≥ 2
E٤ J٤אאFאאאא
the slab is one way slab אאB،אאL،אrW
- ٤٥ -
אאא
٢٥١
א
אא
א
1m
d/2 B d/2
L
E٤ J٤F
אFאא א אאE٤ J٤Fא
F אא KEאאא א
KאאאאאאאאE
א א א אא א א א א Kאאאא
F א אא א א א א
WEא(Deflection)
t min.. =B/30
א E١
t min.. =B/40
א E٣
t min.. =B/35 אא E٢
Kאא אאאאאאB
- ٤٦ -
אאא
٢٥١
א
אא
א
Wאאא٣{٤
W
אאאא،אאאאא K?٢?אאאאאא،א i.e. L / b ≥ 1.0 ≤ 2.0 אZbאZL
J٤FאאKאאאא
K א א אאא אKE٥
אאאאאא
אאאKאא
אאאאא
Kא אאאאאאא
Wאאא
אאאאE٥ J٤F
KL/b אE
KאאאאאאאE
αEאאFאאא
βEאאFאאא
- ٤٧ -
אאא
٢٥١
א
אא
א
Wאאאא t min.. =b/35 א E
t min. =b/45 אא E
Kb
Wאאאאאאא٤ J٤ אא(Hook’s Law) (Bernoulli theorem)
א א א א K א א
KWאאאא
(Pure Bending Moment or Pure א א א ١{٤{٤
Flexure) א •
(Bond slip )EFאא •
א א א א א א א א • א א
א (Plane cross sections) א • א KE٦ J٤FאאKא fc
c
x
3 C
x
Z=d -x 3
d
T
R.C sec
s Strain diagram
E٦ J٤F - ٤٨ -
fs n
Stress diagram
אאא
٢٥١
א
אא
א
Wאאאאאא
(4-1)
d
=
k
1
M b
Wאאא
EאFאאZb אאZd
(f c) א א Z k1 KE١ J٤FאאK(f s)אא
KאאאZM
Wאא
(4-2)
A
s
=
M k 2 × d
(M)אאאZAs אאZd
א(f c)אאZK2 KE١ J٤FאאK(f s)א
KאאאZM
- ٤٩ -
אאא
٢٥١
א
אא
א
אE١ J٤F
Fs=2200
Fs=2000
Fs=1600
Fs=1400
Fs=1200
Fs=1000
fs
fc α β k1 k2 α β k1 k2 α β k1 k2 α β k1 k2 α β k1 k2 α β k1 k2
45 50 .403 .428 .866 .857 .357 .330 866 857 .360 .385 .880 .872 .374 .345 1058 1048 .325 .349 .892 .884 .391 .360 1248 1237 .297 .319 .901 .894 .408 .375 1441 1430 .273 .909 .402 1818 .254 .915 .415 2013
55 .452 .849 .308 849 .407 .864 .322 1037 .372 .876 .335 1227 .340 .887 .347 1419 .292 .803 .371 1898 .273 .909 .383 2000
60 .474 .842 .289 842 .428 .857 .301 1029 .391 .870 .313 1217 .360 .880 .324 1408 .310 .897 .346 1793 .290 .903 .357 1987
65 .494 .835 .273 835 .448 .851 .284 1021 .411 .863 .295 1208 .379 .874 .305 1398 .328 .891 .328 1782 .307 .898 .334 1976
70 .512 .829 .259 829 .467 .844 .269 1013 .429 .857 .279 1200 .396 .868 .288 1389 .344 .885 .306 1771 .323 .892 .315 1962
75 .529 .824 .247 524 .484 .839 .256 1007 .446 .851 .265 1192 .413 .862 .274 1380 .360 .880 .290 1769 .338 .887 .298 1952
80 .545 .818 .237 818 .500 .833 .245 1000 .462 .846 .253 1185 .429 .857 .261 1371 .375 .875 .276 1750 .353 .882 .283 1940
90 .574 .809 .219 809 .529 .823 .226 988 .491 .836 .233 1171 .458 .847 .240 1367 .403 .866 .252 1731 .380 .873 .269 1921
95 .588 .804 .211 804 .543 .819 .218 983 .504 .832 .224 1165 .471 .843 .230 1349 .416 .861 .242 1723 .393 .869 .248 1912
100 .600 .800 .204 800 .555 .815 .210 978 .517 .826 .216 1162 .484 .839 .222 1342 .429 .857 .233 1715 .406 .865 .239 1903
105 .612 .796 .193 796 .567 .811 .204 973 .529 .824 .209 1154 .496 .835 .214 1335 .440 .853 .225 1706 .417 .861 .230 1894
- ٥٠ -
אאא
٢٥١
א
אא
א
Diagonal Tensionאא٢{٤{٤ Horizontal אאאVertical Shearאאאאא
KShear
Wqאא
(4-3)
q =
Q 0 . 87 × bd
KאאZq
Kאאאא ZQ
WאאאW٣ J٤ J٤ WאW Eא א א F א א א (١ Kאאא٪٠{٢٥
EאאF אאאא (٢
אאK אאE 1 F 5
KE٥F
E٪ ٢٠ F א א א א א (٣ אא ٥L١ אאא א
אאא ٤L ١אאא
Kאא
אא אא (٤
אאE٥Fא ٢٠
K١٠א
K٢{٠١{٥א (٥
E ٣L ١F א א א א (٦ Kאאאא - ٥١ -
אאא
٢٥١
א
אא
א
K٨ (٧
WאW אאאא٢L٤٠٠אאא Kאאאא
Kא
Kא
E١
W
E٢
E٣
WאW K ١٠אא K١٥א
WאWא ١{٠٥ א א א א א KאאKאא
- ٥٢ -
אאא
٢٥١
א
אא
א
W ٤{٤{٤ א א KE٧ J٤F א אא א א א WE١F K
Wא
Dead loadsWאאא
١٠Z(t)אא
A
One - way slab
m 2.25
L=6
m
A
2
w = 0.6 t/m = 6 KN/m Loads
i.e.
M
m
B.M.d
2.25
E٧ J٤F Own weight of slab =1×1× t ×γ c = t × γ c = 0.1× 2.5 = 0.25 t/m2 = 2.5 KN / m2 Flooring (א) = 0.15 t / m2 = 1.5 KN / m2 Live Loads (L.L.) (א = )א0.20 t/m2 = 2.0 KN / m2 = 0.60 t / m2 = 6.0 KN / m2
Total Loads (wt) (א)א Bending Moments (א)
- ٥٣ -
אאא
٢٥١
א
אא
א
Section A-A: M = wt ×L2/8 = 0.6 × (2.25)2 / 8 = 0.380 m.t /m Assume f c =60 kg / cm2
(f cu = 250 kg / cm2)
f s = 1400 kg/cm2
And
(mild steel 37)
i.e.
From table (4-1)
k 1= 0.313
and k 2 = 1217
Eb=100cm.F١{٠אEdFאא d = k1
M b
=
0 . 313
0 . 380 × 100000 100
= 6.10 cm.
א א K ٨ Z EtF א א א
K
I.e. take t min = 8.0 cm.
d act. = t – cover = t – (1.5: 2.0 cm) = 8.0 – 1.5 = 6.5 cm.
Wא
As main =
M × 10 5 k2× d
=
0.380 × 10 5 1217 × 6.5
= 4.8 cm2/m
Choose 7Ø10 mm/m (5.5cm2/m) Check: As min = 0.25% Ac = 0.25 / 100 ×10×100 = 2.5 cm2 i.e. As main chosen is okay. As secondary = 0.20 A s main = 0.20 × 4.8 = 0.96 cm2 /m Choose As secondary = 5 Ø 8 mm/m (2.51 cm2)
- ٥٤ -
אאא
٢٥١
א
אא
א
WE٢F א א אא א א א א
אKE٨ J٤Fאא אא ٣{٠٠ KE٢L٢٠٠ZאאFא א
r=
L 6 = =2 b 3
Wא (One way slab) Loads on Slab:
assume t s=10 cm.
O.w. of slab = 0.1 ×2.5 = 0.25 t/m2 = 2.5 KN/m2 Flooring
= 0.15 t/m2 = 1.5 KN/m2
L.L.
= 0.20 t/m2 = 2.0 KN/m2
Total load wt =
= 0.60 t/m2 = 6.0 KN/m2 3.0 m
3.0 m
W d.L + L.L Loads m 3.0
m 3.0
S
S wL2 24
--v M e
+ve M
wL2 24
m 6.0 B.M.d
+ve M
loads & B.M at section s-s
m L = 3.0
E٨ J٤F
- ٥٥ -
m L = 3.0
אאא
٢٥١
א
אא
א
Bending Moments: 2 × wt l1 = 0.6 × 9 = 0.49 mt / m M +ve = 11 11 2 w × l 0.6 × 9 = 0.6 mt / m M −ve = t 1 = 9 9 (Cases of א٢L٤٠٠אא W
K(case of total loads)אאloading) Assume f c =60 kg / cm2
(f cu = 250 kg / cm2)
f s = 1400 kg/cm2
(mild steel 37—24/35)
And
i.e. k 1= 0.313
From table (4-1) and
k 2 = 1217
Eb=100cm.F١{٠אEdFאא
d = k1
M b
=
0 . 313
0 . 6 × 100000 100
= 7.74 cm.
Take t = 10.0 cm d act. = t – cover = t – (1.5: 2.0 cm) = 10.0 – 1.5 = 8.5 cm.
Wאאא
A main = s1
M +ve ×10 k2 × d
5
0 . 49 × 10 5 = = 4.74 cm2/m 1217 * 8 . 5 Choose 7 Ø 10 mm/m (5.5cm2/m)
Check: As min = 0.25% Ac = 0.25/100 ×10×100 = 2.5 cm2 i.e. As1 main chosen is okay. As1 secondary = 0.20 As1 main = 0.20× 4.74 = 0.948 cm2 /m Choose As1 secondary = 5 Ø 8 mm/m (2.51 cm2)
- ٥٦ -
אאא
٢٥١
א
אא
א
Wאאא M −ve ×105 As2 main = k2 × d
=
0.6 ×105 = 5.8 cm2 / m 1217× 8.5
Choose 8 Ø 10 mm/m (6.28cm2/m) Check: As min = 0.25% Ac = 0.25/100 ×10×100 = 2.5 cm2 i.e. As2 main chosen is okay. As2 secondary = 0.20 As2 main = 0.20× 5.8 = 1.16 cm2 /m Choose As2 secondary = 5 Ø 8 mm/m (2.51 cm2)
אE٢Fאאאא
KE٩ J٤F
KאאאאWE٩ J٤F Wאאאאאא٥{٤ WE١٠.٤Fא
אאאאאא E١ K٥
٧٠אאאKאא E٢ K٥א
- ٥٧ -
אאא
٢٥١
א
אא
א
WE١١{٤FאאאאWאא E٣ Kאאאא • Kאאאא א • Kאאא
١٠ J٤
١١ J٤
- ٥٨ -
אאא
٢٥١
א
אא
א
אE١٢{٤FאWא
אאאא
KאאאK
אאאאWאא
אאא א٦ KE١٠{٤Fאא
אאאא٦٤אאא Kאאא
אאאאאאא
KאאאאKאאאא
KאW١٢{٤
- ٥٩ -
א
אאא
אאא
٥
אא
٢٥١
א
אאא
א
Wאא אאאאאאא אא
א א א א א ،א א
א אאאא אאאKא
Kאאאא
Wאא אKאאאאאא E١
א א א א א א א א E٢ Kאאאאא
Kאאאאאאאא E٣
אאאא א E٤
Kאאאאאא
Kאא ٪١٠٠אאאאאאאאאא Wאאאא Wאא Kאא
אאאאאא
- ٦٠ -
אא
٢٥١
א
אאא
א
W١{٥ Wאאאאאא
א אאאאא E١ אא א K א א
K٢٠٢٠WKKא
KE١ J٥FאF٢٥٢٥
T 12 cm b
20 cm
25 cm 12 cm b
b< T
12 cm b
25 cm b< 25 cm
20 cm b< 20 cm
E١ J٥F
אאאאאא E٢ אאאאKא Kאא
אאאאאאא E٣ Kאא
א אא אא א E٤ Kאאא
Kא E٥
Kאאאאאאאא
- ٦١ -
אא
٢٥١
א
אאא
א
Wאאאאא٢{٥
Wאאאא Main girders אא אE١ Secondary beamsאא אE٢
Eא Fא אאאא F א אא א א א אא
KE٢ J٥FאאEאאא
4.00 m
4.00 m
Beam
0.3
Seconadry
0.2
m Seconadry
3.00
m
Beam Main
0.3
Beam m
m
3.00
m
E٢ J٥F
Fא א אא KEא אא א
Kא
K E٣ J٥F א א א א ،א
א (G1) א א (B1→B2→B3) א
KC4אB1אאאאK - ٦٢ -
אא
٢٥١
א
אאא
א
א K E٣ J٥F א א א א א
אאאאאאאG1א ،C1אא G1אאאB3،B2،B1
KאאאC4،C3،C2
4.00
4.00
4
G1 C3
C4 3 3.00
m 2
C5
B2 1
3.00
m
B3 B1
C2
C1
E٣ J٥F
WאאאאW١ J٢ J٥
א Eא א א אא א F א E١ (Embedded beams) א (Dropped beams) א א
Kאא
אא (Loop of loading)E ٢
Kאאאאאאא
אאאא Kאאאאאאא
EאFאEFאE J٤ J٥Fא א aאאא d c b a א
Kאאא
KE J٤ J٥FאאאאאאE J٤ J٥Fא cbaאאE J٤ J٥Fאא - ٦٣ -
אא
٢٥١
א
אאא
א
אKאא Dא e d Kאאא
א א Eא אF א א
א ،E J٥ J٥F ، E J٥ J٥F א K E J٥ J٥F א א אא KE J٥ J٥Fאאא
m
m
m D
S
m D
S
m D
m D
m
b
c
a
d
S
m
b
c
a
d
S
m S
m D
m D
e m D
m
m S
D
m D
S
E٤ J٥F
- ٦٤ -
m S
אא
٢٥١
א
אאא
א
X m
X m
ks ac Cr
Y
m
Y
Cr ac ks
X
m
Y
m
m
E٥ J٥F
Wאאאא ٢.٢.٥
אאאEאאFאאא
א אאאאאאאאא
א א א א K אאאא
אאאאא Kאאא
אאאאאאE٥ J١Fא אאאאאאאאא
Kא
Wאאא٣{٥
Wאאאא
אאאKאאאא EאFאאא - ٦٥ -
(١ (٢
אא
٢٥١
א
אאא
א
EאFאאאאא
(٣
א א א אא א א א
אא E٧{٦FE٦ J٥F אא
Kאאאאאאא
EF א K EF א K
אאאאא
א א א אא E٥{٧ F א
KE١ J٥F
1m
d/2
B
L
KאאאאאאאאאE٦ J٥F
- ٦٦ -
d/2
אא
٢٥١
א
אאא
א
L Y
X
45°
Beam A
2X
Beam B
KאאאאאאאE٧ J٥F E١ J٥F
אאאאאאאβ،αא
L/2x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 α 0.667 0.725 0.769 0.803 0.829 0.853 0.870 0.885 0.897 0.908 0.917 β 0.500 0.554 0.582 0.615 0.642 0.667 0.688 0.706 0.722 0.637 0.750
W
KאאאאאZW KאאZL KZX W
(Equivalent coefficient for moment)אאZα (Equivalent coefficient for shear)אאZβ
W
- ٦٧ -
אא
٢٥١
א
אאא
א
EאאאFאאאZα ×w ×x אאאא א Kאא
EאאאFאאאZβ ×w ×x
א א F אא א א KEαאאא
KE٢ J٥FאאאאW
E٢ J٥F
0.5 0.667
L/2 L
1
0.5
0.5
2
=
From table
3
0.5 0.667
4
0.5 0.333
5
=
0.5
6
=
From table
7
Take = 1/2x for one trabizoidal
E١ J٥F אאאא E٨ J٥Fא אKאאE J٨ J٥FאKE J٨ J٥F
W
KE٣L١{٢אF٢٠אאאא
WאKE J٨ J٥FאאK٣{٠Zאאא - ٦٨ -
אא
٢٥١
א
אאא
א
KB2אאB1אאאא
Kאאאאאא 1
2
3
4
A
A
m 6.00
B
B
6.00
m
C
C 5.00 m
5.00m
4.00 m
1
2
3
4
E J٨ J٥F
1
2
4
3 G2
A
A S2 m 6.00 B2
S2
S1
B
B2
m
B
G1
m B2 6.00
S2
B1
1
B1
0.4 m
S1
5.00 m
G2 4.00 m
C
0.2
2
C 5.00 m 3
E J٢ J٥F
- ٦٩ -
B2
S2
4
E١
E٢
אא
٢٥١
א
אאא
א
h2 ( clear height )
E J٢ J٥F
א
B2،B1אאאW
E٩ J٥FאאB1אא
2.00
G2
2.50
2.00
G1
2.50
G2
6.00 m
6.00 m
B1 w t t/m G1
G2
G2
B1 E٩ J٥F - ٧٠ -
אא
٢٥١
א
אאא
א
Wt on beam B1 = loads from slab + o. w. of beam Loads on slab:
Assume t s = 12 cm = 0.12 m O.W. of slab = 0.12 × 2.5 = 0.30 t/m2 = 3.0 KN / m2 O.W. of flooring
= 0.15 t/m2 = 1.5 KN / m2
L.L
= 0.30 t/m2 = 3.0 KN / m2
Total load on slab
= 0.75 t/m2 = 7.5 KN / m2 Slab s1
is 4 × 6 m
r1 = 6/4 = 1.5 α1 = 0.853
and
β1 = 0.667 (from table 5-1) Slab s2
is 5 × 6 m
r2 = 6/5 = 1.2 α1 = 0.769
and
β1 = 0.582 (from table 5-1)
Beam B1 has two equal span = 6.0 m Assume And
b =20 cm
t = span / 10 = 60 cm
W
WאB1אא(working loads)א W α = total working load for moment
= 0.853 × 0.75 × 4/2 + 0.769 × 0.75 × 5/2 + 0.2 (0.6 – 0.12) × 2.5 = 2.96 t/m
WאB1אא(working loads)א W β = total working load for shear
= 0.667 × 0.75 × 4/2 + 0.582 × 0.75 × 5/2 + 0.2 (0.6 – 0.12) × 2.5 = 2.33 t/m
KאB1אאאאאW Kאאא B2אא
KאG1(Main girder)אEאFא
E١٠ J٥FאאB2אא - ٧١ -
אא
٢٥١
א
אאא
א
2.50
2.50 col
col
col
6.00 m
6.00 m
B2 w t/m t col
col
col
B2
E١٠ J٥F Loads on slab: Assume t s = 12 cm = 0.12 m O.W. of slab = 0.12 × 2.5 = 0.30 t/m2 O.W. of flooring = 0.15 t/m2 L.L = 0.30 t/m2 Total load on slab = 0.75 t/m2 Slab s2 is 5 × 6 m r2 = 6/5 = 1.2 and β1 = 0.582 α1 = 0.769 Beam B2 has two equal span = 6.0 m Assume b =20 cm And t = span / 10 = 60 cm Loads from wall: = 3.0 – 0.6 = 2.4 mאאH wall = W wall = γ b × t w + o.w. of plaster = 1.2× 0.2 + 0.05 = 0.29 t/m2
W
WEא)B2אא(working loads)א W α = total working load for moment
= 0.769 × 0.75 × 5/2 + 0.2 (0.6 – 0.12) × 2.5 + 0.29 × 2.4 = 2.38 t/m
WEאFB2אא(working loads)א
- ٧٢ -
אא
٢٥١
א
אאא
א
W β = total working load for shear = 0.582 × 0.75 × 5/2 + 0.2 (0.6 – 0.12) × 2.5 + 0.29 × 2.4 = 2.38 t/m
:B2،B1אאאאW
J٥FE١١ J٥FאאEאFB1אE١ E١٢
2.33 t/m Load for shear 6.00 m
0.4
6.00 m
0.6 0.6
0.4
K ( Shear factor) q
8.4 t 5.6 t S.F.D Q=K * w L q
B
5.6 t 8.4 t
E١١ J٥F
2.96 t/m
6.00 m
-24
6.00 m
-9 +11
-4.44
Load for moment
9.7
2 w L= 11.84 9
-24 K m ( moment factor)
+11
- 4.44
9.7
E١٢ J٥F - ٧٣ -
2 B.M.d M= wL /Km
אא
٢٥١
א
אאא
א
WEאFB2אE٢
W β (for shear) = 1.74 t/m W α (for moment) = 2.22 t/m
KB2אאאאאW١
WDesign of R.C. Sections of Beamsאאא٤{٥
א א א א א א א
א Kאא א א K א
א K א א א א KאאאאאאE١٣ J٥Fא
E١٣ J٥F
א(Deflection)אאA-BאE١٤ J٥Fא
אא אKאא
WEB-BאFאאEA-AאF
- ٧٤ -
אא
٢٥١
א
אאא
א
E١٤ J٥F
אאאאא
אE١٥ J٥FאאKE١٥ J٥FE١٥ J٥Fא
E١
א K א א א א
אאאKא T א
E١٥ J٥Fא
אאאאא
KE١٥ J٥Fאא
b
A-A
B-B
- ٧٥ -
A-A T
E٢
אא
٢٥١
א
אאא
א
E١٥ J٥F
Wאאאאאא
E١٥ J٥אאFאאאE
E١٥ J٥אאFTאאTE Lא (L-shaped)L אא K(L-Section)>
Wאאאא١{٤{٥
אאאאאאאא Wאאאא
אא (Linear) א
E١
אאא א
E٢
KE١٦ J٥F
א אאאאאאא
א f
c
א א Kf sאאאא
- ٧٦ -
אא
٢٥١
א
אאא
א
ft
d
+
d
fc Stress diagram
Strain diagram
E١٦ J٥F
א א א א
E٣
Wאאdאא
E٤
Kאא(Q)אאאא(M)
(5-1) d =
k
1
M b
W
אא(Depth of R.C. Sec.)אאאZd
EE١٧ J٥F&E١٦ J٥FאאF
(f c) א א Z k1 KאאאאE١ J٤Fאא(f s)אא
KאאאZB
f s,bf c f cuE٥ - ٧٧ -
אא
٢٥١
א
אאא
א
f cu = 250 kg/cm2
WאW
f c = 80 kg/cm2
,
So, k1 = 0.253
and f s = 1400 kg/cm2
,
k 2= 1185
E١ J٤FאאאdאאE٦ WאאאAsא E٧
(5-2)
A
s
=
k
M × d 2
From figure (5-21)
t = d +d′ Where; d′ = 3 + Ø /2 + (2.5 + Ø /2) × (n-1)
KאZnW
W٢٠ZØ،אZ(n)אא d′ = 3 + Ø /2 ≈ 4 cm. KE١٧ J٥FאאKא
fc
t
d
+
d
fs / n
b
E١٧ J٥F
- ٧٨ -
אא
٢٥١
א
אאא
א
EאאאאF LTאאא٢{٤{٥ LאTאIא∏אKאאא KE١٨ J٥אFLאTאאאא
א א א אא Kאאאא
א K T א א א א א
(Floor) Kא אא KLאא
אא אא (B)אא
WLאT
WאאאאאBאא T
B = 12 ts + b Or
B = L/3 (for simple beams)
Or
B = L/4.5 (for continuous beams)
E
(5-3)
LE
B = 4.5 t s + b Or
B = L/6 (for simple beams)
(5-4)
Or
B = L/9 (for continuous beams)
t sK٨KאZ
LKאאZ
- ٧٩ -
אא
٢٥١
א
אאא
א
M
M B
B ts
d
ts d
As
As
bO T - section
bO L - section
E١٨ J٥F
אאא(Maximum B)אאW KE١٩ J٥F
- ٨٠ -
אא
٢٥١
א
אאא
א
C
L
C
L
S
S
S
6.30 L sec
m
T sec bo
C
L
L sec
C
L
E١٩ J٥F
WLTא
WT
EאFא (Z)א
E١
אF א (Z) א
E٢
KאאE٢٠ J٥F
KTאאKE٢٠ J٥FE
- ٨١ -
אא
٢٥١
א
אאא
א
B
B z
ts
ts
z d
t
t
bO
bO
E٢٠ J٥F
Wאא (neutral axis distance (z))אא
Z = 0 . 14
M B
(5-5)
א אאא BK Bא (r)א
WKאאBr
(5-6)
Br= r × B
B/b0&ts/zאK(r)אE٣ J٥Fא
- ٨٢ -
אא
٢٥١
א
אאא
א
(Br=r.B) (r)אE٣ J٥F B/b0 2.0 2.5 3.0 3.5 4.0 5.0
1.0 1.0 1.0 1.0 1.0 1.0 1.0
0.9 1.0 .99 .99 .99 .99 .99
0.8 .98 .98 .97 .97 0.97 0.97
0.7 .96 .95 .94 .94 .93 .93
ts/z 0.6 .92 .90 .89 .89 .88 .87
0.5 .88 .85 .83 .82 .81 .80
0.4 .82 .78 .76 .74 .73 .71
0.3 .76 .70 .67 .65 .63 .61
0.2 .68 .62 .57 .54 .52 .49
Wאא d =
k
M 1
B
r
:k1אW f c′= 0.75fc
T = d + d′
Kאאא f cW
K(t)אא
WאTאאא
A
s
=
k
M × d 2
Kf s , (f c′)אE١ J٤Fאk2W
אא (shear stress)אא (Check)א Kאא
Wא אLTאאאE١ Kאאא - ٨٣ -
אא
٢٥١
א
אאא
א
אא(L)אE٢ אאאEאאFא Kאאא٪١٥
: Shear stressאאא٥{٥
(q) א א א א (shear forces (Q)) א
אא(Diagonal Tension)(T)(shear stress) KE٢١ J٥Fאא
Wאאqא
q =
Q 0 . 87 × b × d
Kg / cm2
T q q
q q
T E٢١ J٥F
Wא
K٢L٧אאא E١
١٨ E F E٢ KאאאאאHאאאK٢L - ٨٤ -
אא
٢٥١
א
אאא
א
:אאאאא١{٥{٥
אאאאאאאא
Wא
KE١٣ J٥١١ J٥FאאK(Shear force diagram) (S.F.D)א אE١ WאאאאאE٢
q
max
=
Q 0 . 87 × b × d
q maxK (Linearity)א Q qW K(Face of column)אא
K(7kg/cm2)אאאאאאE٣ KE٢٢ J٥Fאאא
Q q
Q
By concrete
S.F.D
st
7 kg/cm q = 1: 1 q st 3 2 max
Ab
q max
E٢٢ J٥F
- ٨٥ -
אא
٢٥١
א
אאא
א
Kq max½⅓אאאאאאE٤ I.e.
q stirrups =(⅓: ½) q max =
n × Ast ×
f
s
b× s
K(No. of branches of stirrups)אאאZnW
K(Area of bar of one branch of stirrups)אZA st
(A st = 0.503 cm2)W٨א
K٢L١٤٠٠אאZf s (breadth of beam)א Zb
KאZs
Wאא(A sb)אאE٥
A
sb
A
=
b
f
Where;
× b
s
A b = area from diagram of shear stress.
- ٨٦ -
אא
٢٥١
א
אאא
א
K٦{٥
KאאE١ J٥FאאB1א
Wא
Design data: L= Effective Span =6.0 m ton Q max. +ve = 8.4 ton Q max. – ve = 5.6 M max. –ve = 11.84 t.m. M max. +ve = 9.7 t.m. Assume f cu = 250 kg/cm2 kg/cm2 f c = 90 f s =1400 kg/cm2 8.4 ton Therefore: Q Design = At Sec. 2-2 M –ve = 11.84 t.m. t.m. At Sec. 1-1 M +ve = 9.7 From table k1 = 0.233 k2 = 1171 Design of Sec. 2-2 (Slab lies in Tension zone) So design as a rectangular Sec. d =
k
M
1
2
× 10 b
5
= 0 . 233
11 . 84 × 10 20
Take:
5
= 56 . 7 cm
t = 60 cm d act. = 56 cm
A
s
=
M × 10 k ×d 2
2
act
5
=
11 . 84 × 10 1171 × 56
5
= 18 . 06
cm
2
Take 6 Ø 20 mm (18.8 cm2) Check on section 1-1 (slab is in compression zone) So design as a T section B (breadth of flange) is taken the least of: B = 12 t s + b = 12 × 10 + 20 = 140 cm Or B = L/4.5 = 600/4.5 = 133.33 cm Or B = ¢ :¢ =450 cm I.e. B =133.33 cm
- ٨٧ -
אא
٢٥١
א
אאא
א
And
Z = 0 .14
M
+ ve
B
9 .7 × 10
= 0 .14
133 .33
6
= 11 .94 cm
Z > t s > 10 cm So: the sec. is actually T sec. f c′ = 0.75 f c = 0.75× 90 = 67.5 kg/cm2 Take f c′ = 70 kg/cm2 I.e. k1= 0.279 k2= 1200 d act.= 56 cm
As at sec .1 =
M × 10 k ×d
5
1
2
act
=
9 .7 × 10 5 = 14 .43 cm 2 1200 × 56
Take 5 Ø 20 mm (15.7 cm2)
8.4 t
sec 2
q
255 q
max
8.62 kg/cm
Ab
st
7 kg/cm
E٢٣ J٥ Check of shear:
q
2 max .
=
8 . 4 × 10 3 = 8 . 62 kg / cm 2 > 7 kg/cm2 0 . 87 × 20 × 56
Take stirrups 2 branches 5 Ø 8 mm/m - ٨٨ -
45°
3.6 m
x 2.4 m
5.6 t
אא
٢٥١
א
אאא
א
q
st
A
=
bent
n × Ast ×
=
f
s
b× s
=
2 × 0.503×1400 1 1 = 3.52 = ( : ) q 20× 20 3 2 2 max.
A ×b = × b A f f
= (45 ×
b
b
s
8.62 − 3.52 + 7 − 3.52 20 2 )× = 2.28 cm 2 1400
s
Take 2 Ø 20 mm (6.28 cm2 for more safety) KB1אאE٢٤ J٥Fאא
٧{٥
- ٨٩ -
אא
٢٥١
א
אאא
א
Kא
B1
B1אE٢٤ J٥F
2Φ18
٦Φ٢٠
آﺎﻧﺎتΦ
5Φ20
٢Φ٢٠
ﻓﻲ وﺳﻂ اﻟﻜﻤﺮة
ﻓﻮق اﻟﺮآﻴﺰة
B1אאWE٢٥ J٥F
Wאאאאאא WאאE١
K٣٥٠H٣٠{٤H٣٠{٨
KאאאאאאאE٢ - ٩٠ -
אא
٢٥١
א
אאא
א
א٢٨אאאאE٣ ٢L٢٥٠
K٢٤אאאE٤
KEאאF٣٥L٢٤אאאאE٥
Kאא٤L١א٥L١אאאE٦ Kאא٧L١אאאE٧ K٣٥אאאE٨
אאEFאאאאאאאE٩
Kאא٤L١א
אאאאE١٠
E٣٥L٢٤Fא
אא
אא
L٨Ø٥
١٦Ø٢
L٨Ø٥
١٨Ø٢
٢٠Ø٢ ٢٠Ø٢ ٢
L١٠Ø٦
٢٠Ø٢
٢٥Ø٣ ٢٢Ø٣ ٣
L١٠Ø٦
٢٥Ø٢
٢٢Ø٥ ٢٢Ø٥ ٤
- ٩١ -
J
١٦Ø٢ ١
א
אא
א א
٦
אא אא
٢٥١
א
א
Wאא אאאאאאאא
א Kא אא א א ،א
Kאאאאאאא Wאא Kאאא E١
Kאאאאא
E٢
Kאאא
E٣
Wאא ٪١٠٠אאאאאאאאאא Kאאאא • Wאא Kאא
אאאאאא
- ٩٢ -
אא אא
٢٥١
א
א
Wאא١{٦ Kאאאאאאאאא א א א א א א
א א א K א א
Kא
א א א K א Kאאאאא
KאאE١ J٦FאאKאאא
W(Braced and un-braced column) אאא١{١{٦ אאאאאאאאא
א א (Shear walls)א א א
אא אKאK Kאאאא
WE٢ J٦FאW
אאאא (Shear wall A and B)א Y אאאא
Eא F ٥אK ٨ ← ١ X א KXא(un-braced)Yא(Braced) - ٩٣ -
אא אא
٢٥١
א
א
E٢ J٦F Wאא٢{١{٦ (λ i = He/ i)א(λ b=He/b)אאאא KE١ J٦Fאאאא
WKאאאאZiW
i = 0.3b for rectangular section i = 0.25 D for circular section KאאאאD،אאאbW
- ٩٤ -
אא אא
٢٥١
א
א
אאאE١ J٦F א אאא אא א λ i λb λb ٥٠ ١٢ ١٥ ٣٥
٨
١٠
Kאאאאא• Wאא٢{٦ Wאאא Kאאא E١
א F Kא א אא א
(٢
KEא Wאאא١{٢{٦
Wאאאאאא
KאEאאFאאא E١ KE٣ J٦FאאK(Y)א(X)אאאא
(B-3) א אא א Eא א F א E٢ K(A)אאאאא
- ٩٥ -
אא אא
٢٥١
א
א
E٣ J٦F Wאאאא(B-3)אאPא אE٣
KEAאאאFא אE
Kאאאא E
KאאאאאאE Kאאא E
Therefore; total load P of each floor is equal to: P = W slab ×A + weight of beams + weight of walls + own weight of column - ٩٦ -
אא אא
٢٥١
א
א
Where; W slab = t s × 2.5 + weight of flooring + Live Loads. If Weight of flooring = 150 kg/m2
t s = 12 cm,
and Live Loads = 300 kg/m2
So, W slab = 0.12 × 2.5 + 0.15 + 0.3 = 0.75 t/m2 Weight of beams = b × t × 2.5 × ∑ L beams
KE٢٠ZFאZ b W K(t = span / 10: 14) אZ t
אאא (A)אאאאאאאZ ∑ L beams K
Weight of walls = γ wall × h wall ×t wall × ∑ L walls KאZγ wallW
KEFאאאZ h wall
KEאאאFאZ t wall
א א F א א אא אא א Z ∑ L
walls
KEא
For example; If γ wall = 1.2 t/m3; t wall = 0.2 m
;
∑ L walls = 6.0 m ; h wall = 2.4 m
;
And, own weight of plaster = 50 kg/ m2 = 0.05 t/m2 i.e., Weight of wall = (1.2 × 0.2 + 0.05) × 2.4 × 6.0 = 4.176 tons / floor Own weight of column / floor = b c × t c × 2.5 × h c Where;
E٠{٦F٠{٦←٠{٢ZאZb c ٢{٠←٠{٢٥ZאZt c
- ٩٧ -
אא אא
٢٥١
א
א
אאZh c
WPcאאא Pc = total vertical load on column = N × Pc/floor KאאZNW KאאאאאאZPc/floor
W
א Eא א א א F א א א ٪ ١٠ ± א
א KאאאאאKא Wאא٣{٦
אאEFאא
K א א EF Ke min.אא
e min = 0.05 t or 20 mms (Whichever is bigger)
א א א א א א Wאאאא
P = fc0 × Ac + 0.44 f y ×A sc
(6-1)
W
KEE٤ J١FאאKf cu FZfc0
KאאאZAc
KאאאאאZf y KאאאאZA sc
A sc = 1% ×AcWאµ = A sc/Ac = 1%א
E١ J٦Fאאאא א W Kא
- ٩٨ -
אא אא
٢٥١
א
א
אאW٤ J٦ א ٪ ٠{٨ א א א א א א א E١ אא אא٪ ٠{٦א
KE١ J٦Fאאא(λ i)א(λ b)
א ٪ ١ א א א א א א א E٢ Kאאא٪١{٢א
אאאאאאE٣ Wאא
Kאאאא٪٤ • Kאאאא٪٥ •
Kאאאא٪٦ •
KאE٤ K١٢אE٥
K٢٠אאאאאאE٦
K ٣٠אאא E٧ אאאK ٢٥
KE٤ J٦F א א F ١٥ א א א א א
KEF٦אאאאא
E٤ J٦F - ٩٩ -
אא אא
٢٥١
א
א
WאאאאאאE٨ KE١٥FE
K٢٠אאאE
K٨¼E٩
KאאאאאאE١٠
א K ٣ K ٨ א E١١ אאא
Kאאא١٠١٠אאא
Kאא٤٠אאE١٢
- ١٠٠ -
אא אא
٢٥١
א
א
W٥{٦
WאK١٠٠ZאאאW١ א٣Zא• א
K(36/52)אא• א
K٢L٢٥٠Z(f cu)א •
Assume:
א
µ (total steel ratio) = 1% ; f co = 60 kg/cm2
(for f cu =250 kg/cm2)
And
f y = 3600 kg/cm2
i.e.,
applying in equation (6-1)
So
100×1000 = 60×Ac + 0.44×0.01×Ac×3600 = 75.84 Ac Ac = 1318.56 cm2
b = breadth of column = 25 cm; Take 1318 . 56 = 52 . 74 cm 25
So
t =
Take
t = 60 cm A sc = 1% ×25×60 = 15 cm2 Choose 8 Ø 16 mm
- ١٠١ -
אא אא
٢٥١
א
א
W٢
WאK150Zאא א4Zא• א
K(36/52)אא• א
K٢L300Z(f cu)א • Assume:
א
µ (total steel ratio) = 1% ; f co = 70 kg/cm2
(for f cu =300 kg/cm2)
And
f y = 3600 kg/cm2
i.e.,
applying in equation (6-1)
So
150×1000 = 70×Ac + 0.44×0.01×Ac×3600 = 85.84 Ac Ac = 1747.43 cm2
d = diameter of column = 50 cm; Take So
Ac =
π × (50 )2 4
= 1962 . 5 cm
A sc = 1% ×1962.5 = 19.62 cm2 Choose 8 Ø 18 mm =20.4 cm2
- ١٠٢ -
2
אא אא
٢٥١
א
א
Wאא
٦{٦
K١ אE٥ J٦Fא
אE٥ J٦F
٨Φ١٨
ﺳﻢ٥٠
אאאE٦ J٦F
- ١٠٣ -
אא אא
٢٥١
א
א
W3
WאאE٦FאאאE٧ J٦Fא (C5 ; C6)٦W٥אאא
אאאK (C5 ; C6) ٦W ٥א
Kא
E١
E٢
אאאE٧ J٦F
א
W
(L.L = 200 kg/m2)א• א
KאאE٨ J٦Fאא •
٧٠ZאאאK٢{٩٠Zאא • - ١٠٤ -
אא אא
٢٥١
א
א
EאאFA5א(C5)٥אא
EאאFA6א(C6)٦אא
E٨ J٦F For column (C5): • Slabs: W slab = t s× γ c + flooring + L.L = 0.1×2.5 +0.15 + .200 = 0.6 t/m2 Area (A5) = (3.0 + 1.0) ×(3.5/2) = 7 m2 ∑ loads from slabs = 0.6×7 + (0.12 -0.1) × 2.5× (3.5/2) ×3 = 4.463 tons - ١٠٥ -
אא אא
٢٥١
א
א
• Beams: ∑ L (of Beams) = (3.5/2) + 1 +3 +1 = 6.75 ms O.W. of Beams = 0.2× (0.7 – 0.1) ×2.5 = 0.3 t/m ∑ loads of Beams = 6.75 ×0.3 = 2.025 tons. • Walls: γ wall = 1.2 t/m3 O.W. of wall = 0.2 ×1.2 +0.05 (Plaster) = 0.29 t/m2 ∑ L (of Walls) = 1 + 3 + (3.5/2) +1 = 6.75 ms h wall = 2.9 – 0.6 = 2.3 ms ∑ loads of walls = 6.75 ×0.29×2.3 = 4.5 tons. • Columns:
assume column dimension = 20 × 60 cm O.W. of column = 0.2 ×0.6 ×2.5 ×3 = 0.9 ton Total load P on column C5 = (Pc5) Pc5 (per one floor) = 4.463 + 2.025 + 4.5 + 0.9 =11.888 tons Pt = 11.888 × 6 = 71.328 tons = 72.0 tons For column (C6):
• Slabs: W slab = t s× γ c + flooring + L.L = 0.1×2.5 +0.15 + .200 = 0.6 t/m2 Area (A6) = (2.5/2 + 2.0) × (3.5/2 + 3.5/2) = 3.25 × 3.5 = 11.375 m2 ∑ loads from slabs = 0.6×11.375+ (0.12 -0.1)×2.5×(3.5/2)×3.25 =
7.11 tons
• Beams: ∑ L (of Beams) = (2.5/2) + 2 +3.5/2 +2.5/2 +3.5/2 = 8.0 ms O.W. of Beams = 0.2× (0.7 – 0.1) ×2.5 = 0.3 t/m - ١٠٦ -
אא אא
٢٥١
א
א
∑ loads of Beams = 8.0 ×0.3 = 2.4 tons. • Walls: γ wall = 1.2 t/m3 O.W. of wall = 0.2 ×1.2 +0.05 (Plaster) = 0.29 t/m2 ∑ L (of Walls) = 8.0 ms h wall = 2.9 – 0.6 = 2.3 ms ∑ loads of walls = 8.0 ×0.29×2.3 = 5.336 tons. • Columns:
assume column dimension = 20 × 70 cm O.W. of column = 0.2 ×0.7 ×2.5 ×3 = 1.05 ton Total load P on column C5 = (Pc6) Pc6 (per one floor) = 7.11 + 2.4 + 5.336 + 1.05 =15.896= 16 tons Pt = 16 × 6 = 96 tons Design of column C5: Assume: f c0 = 60 kg/cm2
for
f y = 3600 kg/cm2 So;
f cu = 250 kg/cm2; and A sc= 1 % Ac
Pc5 = Ac × fc0 + 0.44× A sc× f y
72×1000 = 60 × Ac + 0.44×0.01×Ac×3600 = 60 Ac +12.32 Ac = 75.84 Ac Ac = 72000/ 75.84 = 949.3 cm2 Take b = 20 cm i.e.
t = 949.3 /20 = 47.4 cm
Take column section = 20 × 50 cm Area of required steel reinforcement (A sc) = 1/100 × 20×50 = 10 cm2 Take 6 Ø16 mm (A sc = 12.1 cm2) Use stirrups 5 Ø 8 mm/m
. C5 ( واﻟﺬي ﻳﻮﺿﺢ ﻗﻄﺎﻋًﺎ ﻋﺮﺿﻴًﺎ ﻓﻲ اﻟﻌﻤﻮد٩-٦) اﻧﻈﺮ اﻟﺸﻜﻞ رﻗﻢ
- ١٠٧ -
אא אא
٢٥١
א
א
C5אE٩ J٦F Design of column C6: Assume: f c0 =60 kg/cm2 f y = 3600 kg/cm2 So;
f cu = 250 kg/cm2;
for
and A sc= 1 % Ac
Pc6 = Ac × fc0 + 0.44× A sc× f y
96×1000 = 60 × Ac + 0.44×0.01×Ac×3600 = 60 Ac +12.32 Ac = 75.84 Ac Ac = 96000/ 75.84 = 1265.8 cm2 Take i.e.
b = 20 cm
t = 1265.8 /20 = 63.3 cm ≈ 70 cm Take column section = 20 × 70 cm
Area of required steel reinforcement (A sc) = 1/100 × 20×70 = 14 cm2 Take 8 Ø 16 mm (A sc = 16.1 cm2) Use stirrups 5 Ø 8 mm/m
KC6א אE١٠ J٦Fאא
- ١٠٨ -
אא אא
٢٥١
א
א
C6אE١٠ J٦F
- ١٠٩ -
אא אא
٢٥١
א
א
WאאE٨ J٦Fא א (C1-C2-C3-C4-C7) ٧– ٤– ٣– ٢– ١א אא
אK ٧٠ZאאאK ٢{٩٠Zאאא
E١
Kא (C1-C2-C3-C4-C7) ٧– ٤– ٣– ٢– ١א (36/52)אאאאKאא
E٢
K٢L300Z(f cu)א
Kאאאאא Kאא
- ١١٠ -
E٣
א
אאאא
אאא א
٧
١٠٣
אא
٢٥١
א
אאאא
א
אאאאWאא א• א א א א אא א א א א
אאאאאאא א א ،אא
Kאא Wא• א
אאאא
Kאאאאאאא
א א א א אא א א אKאאא
Kא
א א א א א א א Kאאאאא
E١
E٢ E٣
E٤
Wאאא • ٪١٠٠אאאאאאאאאא Kאאא• א • Wאא • Kאא Kאאא Kאא
- ١١١ -
אא
٢٥١
א
אאאא
א
WאאW١ J٧ אKאאאאאא א K א א א א
אאאאאאאא Kאאאאא
Shallow א א א א
א א אF א א א K foundation
א ،אאא Eא
KאאאKDeep foundationא
،אאא אאKאאאא
Wא א אא א א Kא א
KאאאאאאאאKא
אFאאאאא א אאKEאא،אא
Wאאא
אאK (Strip footings)אא
(١
Kאאאאאאא
אאK(Spread footings)אא
(٢
K (Combined footings) א א
(٣
Kאאא
KאאאK،א K
- ١١٢ -
אא
٢٥١
א
אאאא
א
K א K (Raft foundations) א א
(٤
Kאאאאאא،
WאאאאאW٢ J٧ אאאאאאאא Kא
،אאאאאא אאאאאא
Kא(D f)א(q all.)אא
P
T
=
γ 1−
P a
×
q
D
(7-1)
f
all
KאZPW
KאאאZPT
KE٣L٢٠Z٣L٢ZFאאZγ a KאZD f
(Gross allowable bearing stress of the soil)אאאZq all
אאאאאא
KאאPTא
א Kא א אא
א א אא E١ J٧F א K א א
Kא
Two Way Footings אאאE - ١١٣ -
אא
٢٥١
א
אאאא
א
One Way FootingsאאאE
אאאאאE١ J٧F
אאאKאאאא
אא אא Kא אKאאאאאKאאאא WKאאא
(L - l) = (B - b)
(7-2)
Wאאא
Or
A = L × B = PT/q all
(7-3)
q all = P T / (L × B)
(7-4) - ١١٤ -
אא
٢٥١
א
אאאא
א
א B ، LE3 J٧F،E2 J٧F
KE١ J٧FאPTKE٥Fא
אאאE١ J٧FאאPT K(B,L)אאPTK
אאאE2 J٧F
אא אאא
Kאאאאאאאא
Kא א א א א א א א
KאKE٢ J٧F
Wאא Section a -a:
M
P = A
⎡ ⎛ L − l ⎞2 ⎟ / 2 + ⎢b ⎜ ⎢⎣ ⎝ 2 ⎠
(B − b )⎛⎜ L − l ⎞⎟ ⎝ 2 ⎠
- ١١٥ -
2
⎤ / 3⎥ ⎥⎦
אא
٢٥١
א
אאאא
א
(7-5) M
(2
P 24 × A
=
)(L
B + b
− l
)2 Section a′- a′:
(7-6) M
=
P 24 × A
(2
L + l
)(B
− b
)2 Wאא Section a -a:
(7-7)
(7-8)
M
=
P L
(L
−
l
M
=
P B
(B
−
b
)2
/ 8
)2
/ 8
א א KE6 J٧F ، E5 J٧F א א אא Kאאאאאא
א אאאא
KאאאKאאא
WQ bא
Fn=P/A
(7-9)
WאאQ b For section a -a: Q b =¼ (B + b) (L - l) × F n
(7-10) For section a′- a′: - ١١٦ -
אא
٢٥١
א
אאאא
א
Q b =¼ (L + l) (B - b) × F n
(7-11) Punching Depth אאW١ J٢ J٧
(L × B) א א (l × b) א א
אאK(Direct Shear)אאאאאא Kאאאאא
K٢L٨q Pאאאאא
Kאאא אאאא
KE3 J٧Fאאאא אא
EאאFא˚٤٥א
WאאKאא
Q
2 ⎞ ⎛ 2 ⎞ ⎛ = P − Fn ⎜ b + d ⎟ × ⎜ l + d ⎟ P 3 ⎠ ⎝ 3 ⎠ ⎝
(7-12)
Wאאאא
qP =
QP 2 d (l + b + 1 .33 d )
- ١١٧ -
(7-13)
אא
٢٥١
א
אאאא
א
E3 J٧F WW3 J٧ ١
K ١{٠אK ١٠٠ ٤٥ × ٤٥
E١٨ø٨אF٢L١{٧٥א
א
E٤ J٧F - ١١٨ -
אא
٢٥١
א
אאאא
א
P
T
Footing Dimensions:
= 1 −
γ
P a
× D q all
= f
100 = 112 . 9 ton 2 ×1 1 − 17 . 5
A = 112.9 / 17.5 = 6.452 m2
B=
A = 2.54 = 2.55m Concrete Sections:
M=
P (2B + b)(L − l )2 = 100 2 (5.1 + 0.45)(2.55 − 0.45)2 = 15.683m.t 24 × A 24 × (2.55)
Q b =¼ (B + b) (L - l) × F n =¼ (2.55 + 0.45) (2.55 -0.45) × 15.38 = 24.221 ton
Q But
2 ⎞ ⎛ 2 ⎞ ⎛ = P − Fn ⎜ b + d ⎟ × ⎜ l + d ⎟ P 3 ⎠ ⎝ 3 ⎠ ⎝ 15 . 683 × 10 5 = 49 cm (45 + 20 )
d = 0 . 313 ×
Take d =50 cm so, Q P = 100 −
qP
&
t = 55 cm
100 ⎛ 2 2 ⎞ ⎞⎛ 0 . 45 0 . 5 0 . 45 + × 0.5 ⎟ = 90.564ton + × ⎟ ⎜ ⎜ 2 3 3 2.55 ⎝ ⎠ ⎠⎝
QP 90 . 564 × 10 3 = 5 . 78 kg / cm = = 2 d (l + b + 1 . 33 d ) 2 × 50 (45 + 45 + 66 . 7 )
2
M 15.68 × 10 5 2 As = k 2 × d = 1217 × 50 = 25.75cm Choose 13 Ø 16 mm
q
b
=
Q 24.221× 1000 = = 8.57 < 10kg / cm2 0.87 × d × ∑ o o o o 0.87 × 50 × (13π × 1.6) Footing is 2.55 ×2.55 ×0.55 (13 Ø 16 in each direction)
Wאא ٤ J٧ - ١١٩ -
אא
٢٥١
א
אאאא
א
א א א א א א E٥ J٧F א א Kא
E٥ J٧F
- ١٢٠ -
אא
٢٥١
א
אאאא
א
W٢
١٥٠ K ١٦ Ø ١٢E٧٥ × ٣٠F
אאאK ٢L ٢{٠אא KEq p = 8 kg/cm2אאאFKא
Wא
A = P / F all.net = 150 / 20 = 7.5 m2 Take B=2.5 m;
i.e. (L - l) = (B - b) L = (2.5 – 0.30) + 0.75 = 2.95 m Choose L = 3.0 m , B = 2.5 m i.e. L x B = 3 x 2.5 = 7.5 m2
E٦ J٧F
- ١٢١ -
אא
٢٥١
א
אאאא
א
M a−a =
P (2 B + b )(L − l )2 = 150 (5 . 3 )(2 . 25 )2 = 22 . 36 t .m 24 × A 24 × 7 . 5
M a ′− a ′ =
P (2 L + l )(B − b )2 = 150 (6 . 75 )(2 . 2 )2 = 27 .225 t .m 24 × A 24 × 7 . 5
Q b a-a =¼ (B + b) (L - l) × P/A = ¼ × 150/7.5(2.8) (2.25) = 31.50 ton Q b a′-a′ =¼ (L + l) (B - b)× P/A = ¼ × 150/7.5(3.75) (2.2) = 41.25 ton Concrete Sections; d = 0 . 313 ×
22 . 360 × 10 5 = 66 . 16 cm (30 + 20 )
t = 70 cm
M 22.360 × 105 2 As = k 2 × d = 1217 × 65 = 28.24cm M 27.225 × 10 5 2 As′ = k 2 × d = 1217 × 63 = 35.5cm
Choose 15 Ø 16 mm
Choose 18 Ø 16 mm
Q 41.25 × 103 qb = 0.87 × d × ∑ o o o o = 0.87 × 63 × (18π × 1.6) = 8.36 < 10kg / cm2 (Safe) Q p = 150-20(0.3+0.44)(0.75+0.44) = 132.4 ton
QP 132 . 4 × 10 3 qP = = = 5 . 04 < 8 kg / cm 2 ( Safe ) 2 d (l + b + 1 . 33 d ) 2 × 65 (75 + 30 + 97 )
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⎡ P = 1.5 × 5.0⎢
P =
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WאאאאE٢ א אאאאאאא
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א א F א א א א א K X אאאEToeFא X =
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Kp = P = W × H × Ka
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Pa1=1.5x(1/3)x7.5=3.75 t/m
Pa2=(1/2) x1.7x(7.5)2x (1/3)=15.94 t/m Pp=(1/2) x1.7x(1.5)2x (3)=5.74 t/m
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Wall Stability Wאאא
ΣM/Toe Wאא ΣM/Toe = (34.177x3.2) + (5.138x1.75) + (2.997x1.48) + (7.313 x2.25)+ (5.74x0.50)-(3.75x3.75)-(15.94x2.5) = 142.118 – 53.913 = 86.22 t.m/m.
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e = B − X = 4.5 − 1.78 = 0.47 p B = 4.5 0.75 2
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ΣM/base = ΣV x e = 49.625 x 0.47 = 23.324 tm/m. إﺟﻬﺎد ﻋﻨﺪ اﻟﻘﺪم 49.625 23.324 × 6 + = 11.03 + 6.91 = 17.94t / m 2 2 4.5 (4.5)
f toe =
إﺟﻬﺎد ﻋﻨﺪ اﻟﻜﻌﺐ f heel =
49.625 23.324 × 6 − = 11.03 − 6.91 = 4.12t / m 2 2 4.5 (4.5)
W(Sliding)אאאאא Driving Force = Pa1 + Pa2 = 3.75 + 15.94 =19.69 t/m Resisting Force = Pp + V tan Φ = 5.74 + 49.625 x 0.5774 = 34.39 t/m.
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WEarth Pressure Wא
PD = (1.5 x1/3) + (1.7 x 1.725 x 1/3) = 0.5 + 0.977 = 1.477 t/m2. PC = (1.5 x1/3) + (1.7 x 3.45 x 1/3) = 0.5 + 1.955 = 2.455 t/m2. PB = (1.5 x1/3) + (1.7 x 5.15 x 1/3) = 0.5 + 2.918 = 3.418 t/m2. PA = (1.5 x1/3) + (1.7 x 6.85 x 1/3) = 0.5 + 3.881 = 4.381 t/m2.
Shear ForceWא VD = (0.5 x 1.725) + (0.977 x 1.725 x 1/2) =0.862 + 0.843 = 1.705 t. VC = (0.5 x 3.45) + (1.955 x 3.45 x 1/2) =1.725 + 3.372 = 5.097 t. - ١٣٩ -
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VB = (0.5 x 5.15) + (2.918 x 5.15 x 1/2) =2.575 + 7.513 = 10.089 t.
VA = (0.5 x 6.85) + (3.881 x 6.85 x 1/2) =3.425 + 13.292 = 16.717 t.
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t.m. M C =
0.5 × (3.45) 2 ⎛ 1.955 × 3.45 3.45 ⎞ +⎜ × ⎟ = 2.975 + 3.878 = 6.853 2 2 3 ⎠ ⎝
t.m. M B =
0.5 × (5.15) 2 ⎛ 2.918 × 5.15 5.15 ⎞ +⎜ × ⎟ = 6.630 + 12.900 = 19.530 2 2 3 ⎠ ⎝
t.m. M A =
0.5 × (6.85) 2 ⎛ 3.881 × 6.85 6.85 ⎞ +⎜ × ⎟ = 11.730 + 30.356 = 42.086 2 2 3 ⎠ ⎝
Section A: Assume And
f c =60 kg / cm2
(f cu = 250 kg / cm2)
f s = 1400 kg/cm2
(mild steel 37—24/35)
i.e. k 1= 0.313
From table (4-1) and
k 2 = 1217
Eb=100cm.F١{٠אEdFאא
d = k1
M b
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42 , 086 × 100000 100
= 64.21 cm.
Take t = 70.0 cm d act. = t – cover = 70.0 – 5 = 65.0 cm.
Wאא M ×105 As = k2 × d
=
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Choose 12 Ø 24 mm/m (54.24cm2/m) - ١٤٠ -
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Section B: d = 0 . 313
19 . 530 × 100000 = 43 . 74 cm. 100
t = 60.0 cm d act. = t – cover = 60.0 – 5 = 55.0 cm. M ×105 As = k2 × d
=
19.55×105 = 29.20 cm2/m 1217× 55
Choose 12 Ø 20 mm/m (39.6 cm2/m)
Section C: 6 . 853 × 100000 = 25 . 91 cm. 100
d = 0 . 313
t = 50.0 cm d act. = t – cover = 50.0 – 5 = 45.0 cm.
As =
M ×105 k2 × d
=
6.853×105 = 12.51 cm2/m 1217× 45
Choose 6 Ø 20 mm/m (19.8 cm2/m) Check: As min = 0.25% Ac = 0.25/100 ×70×100 = 17.5 cm2
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(References) אא
1) BS 8110 – 1985
2) CP 110 , 1972 3) Jack C. McCormack ˝Design of Reinforced Concrete˝ Fourth Edition Copyright © 1998 by Addison Wesley Longman, Inc. 4) James G. Macgregor ˝Reinforced Concrete – Mechanics and Design˝ Third Edition © 1997 Prentice-Hall, Inc. אאאא˝אK E٥ Kא Jא˝א K١٩٩٥א١٩٨٩˝אאאא ˝אE٦
7) M. Hillal ˝Fundamentals of reinforced and pre-stressed concrete˝ Dar ElNashr Cairo 1992. 8) Reynolds, C.E and Steadman, J.C ˝Reinforced Concrete Designers˝ Handbook 9th Edition. -1981. 9) Robert Park and William L. Gamble. ˝Reinforced Concrete Slabs˝ Second Edition John Wiley & Sons Inc 2000. 10) Shaker El-Behairy, ˝Reinforced Concrete Design Handbook˝ Part I , II and III Fifth Edition Dar-El-Nashr-Cairo 1998.
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