ﻴﻘﻊ ﺍﻟﻤﺸﺭﻭﻉ ﻓﻲ ﻀﺎﺤﻴﺔ ﻗﺩﺴﻴﺎ ﻭﻴﺘﺄﻟﻑ ﻤﻥ ﻜﺘﻠﺔ ﻭﺍﺤﺩﺓ ﺘﺘﺄﻟﻑ ﻤﻥ ﺃﺭﺒﻌﺔ ﻁﻭﺍﺒﻕ ﻤﺘﻜﺭﺭﺓ ﻤﻊ ﺍﻟﻘﺒﻭ. ﻴﺘﺄﻟﻑ ﺍﻟﻤﺸﺭﻭﻉ ﻤﻥ ﺃﺭﺒﻌﺔ ﻁﻭﺍﺒﻕ ﻤﺘﻜﺭﺭﺓ ﻓﻲ ﻜل ﻁﺎﺒﻕ ﻴﻭﺠﺩ ﺃﺭﺒﻌﺔ ﺸﻘﻕ ﺴﻜﻨﻴﺔ. ﺘﻡ ﺍﺴﺘﺨﺩﺍﻡ ﺒﻼﻁﺎﺕ ﻫﻭﺭﺩﻱ ﺘﻌﻤل ﺒﺎﺘﺠﺎﻩ ﻭﺍﺤﺩ ﻓﻲ ﺍﻟﺒﻼﻁﺎﺕ ﺍﻟﻁﻭﺍﺒﻕ ﺍﻟﻤﺘﻜﺭﺭﺓ ﺃﻤﺎ ﺒﻼﻁﺔ ﺴﻘﻑ ﺍﻟﻘﺒﻭ ﻓﻘﺩ ﺍﺴﺘﺨﺩﻤﻨﺎ ﻓﻴﻬﺎ ﺒﻼﻁﺔ ﻤﺼﻤﺘﺔ ﺘﻌﻤل ﺒﺎﺘﺠﺎﻫﻴﻥ ﻭﻴﻐﻁﻰ ﺴﻘﻑ ﺍﻟﻁﺎﺒﻕ ﺍﻷﺨﻴﺭ ﺃﻴﻀﺎﹰ ﺒﻼﻁﺔ ﻤﺼﻤﺘﺔ ﺘﻌﻤل ﺒﺎﺘﺠﺎﻫﻴﻥ ﻋﻠﻰ ﺍﻟﻘﺴﻤﻴﻥ ﺍﻟﻤﺎﺌل ﻭﺍﻷﻓﻘﻲ ﻭﻗﻤﻨﺎ ﺒﺘﻭﺯﻴﻊ ﺠﺩﺭﺍﻥ ﺍﻟﻘﺹ ﺒﺸﻜل ﻤﻨﺎﺴﺏ ﻟﻤﻘﺎﻭﻤﺔ ﺍﻟﻘﻭﻯ ﺍﻷﻓﻘﻴﺔ )ﺯﻻﺯل( ﻭﺘﻡ ﺍﺨﺘﻴﺎﺭ ﻨﻭﻉ ﺍﻷﺴﺎﺴﺎﺕ ﺃﺴﺎﺴﺎﺕ ﻤﻨﻔﺭﺩﺓ. ﺘﻡ ﺍﻋﺘﻤﺎﺩ ﺍﻟﻜﻭﺩ ﺍﻟﻌﺭﺒﻲ ﺍﻟﺴﻭﺭﻱ ﻜﻤﺭﺠﻊ ﺃﺴﺎﺴﻲ ﻋﻨﺩ ﺤﺴﺎﺏ ﺍﻟﻤﻨﺸﺄ ﻭﻓﻕ ﻁﺭﻴﻘﺔ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻘﺼﻭﻯ ﻭﺘﻡ ﺍﻋﺘﻤﺎﺩ ﺍﻟﻤﻭﺍﺼﻔﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ: ﺍﻟﻤﻘﺎﻭﻤﺔ ﺍﻻﺴﻁﻭﺍﻨﻴﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﻠﻴﺘﻭﻥ f C′ = 20 N / mm 2 ﺤﺩ ﺍﻟﺴﻴﻼﻥ ﻟﻔﻭﻻﺫ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻁﻭﻟﻲ ﻋﺎﻟﻲ ﺍﻟﻤﻘﺎﻭﻤﺔ f y = 420 N / mm 2 ﺤﺩ ﺍﻟﺴﻴﻼﻥ ﻟﻔﻭﻻﺫ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻌﺭﻀﻲ f yp = 240 N / mm 2 ﻗﺩﺭﺓ ﺘﺤﻤل ﺍﻟﺘﺭﺒﺔ 3 kg / cm 2 -ﻭﻗﺩ ﺃﺨﺫﻨﺎ ﺍﻷﺒﻌﺎﺩ ﺒـ mﻭﺍﻟﺤﻤﻭﻻﺕ ﺒـ KN / m ′
ﺤﻤﻭﻻﺕ ﺍﻟﺤﻴﺔ .3 KN/m2 a = a ′ = 5cmﻤﺴﺎﻓﺔ ﺘﻐﻁﻴﺔ ﺤﺩﻴﺩ ﺍﻟﺘﺴﻠﻴﺢ.
ﻨﺄﺨﺫ ﺍﻟﺤﺎﻻﺕ ﺍﻷﺨﻁﺭ ﻤﻥ ﺃﺠل ﺘﺤﺩﻴﺩ ﺴﻤﺎﻜﺔ ﺍﻟﺒﻼﻁﺔ ﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺒﻼﻁﺎﺕ ﺍﻷﻜﺒﺭ ﺃﺒﻌﺎﺩﺍﹰ ﻤﻥ ﺸﺭﻭﻁ ﺍﻟﻜﻭﺩ ﻟﺘﺤﺩﻴﺩ ﺴﻤﺎﻜﺔ ﺍﻟﺒﻼﻁﺔ ﻤﻥ ﺍﻟﺒﻨﺩ ١-٤-٣-٧ﻤﻥ ﺃﺠل ﺒﻼﻁﺔ ﻭﺍﻋﺘﻤﺎﺩ ﺍﻟﺠﻭﺍﺌﺯ ﺍﻟﻤﺨﻔﻴﺔ: ﻤﺴﺘﻤﺭ ﻤﻥ ﻁﺭﻑ ﻭﺍﺤﺩ L 3750 = = 187.5 m 20 20
= ht
ﺠﺎﺌﺯ ﻤﺴﺘﻤﺭ ﻤﻥ ﻁﺭﻑ ﻭﺍﺤﺩ ﻭﻓﻕ ﺍﻟﻤﺤﻭﺭ 5-5 4200 =4 18
ﻨﻌﺘﻤﺩ ﺴﻤﺎﻜﺔ ﺍﻟﺒﻼﻁﺔ .26 cm
= ht
ﺴﻤﺎﻜﺔ ﺒﻼﻁﺔ ﺍﻟﺘﻐﻁﻴﺔ . t f 1 S min t f = max 10 ⇒ t f = 6 cm 50 mm
ﺴﻨﻌﺘﻤﺩ ﺍﻟﻘﻴﻤﺔ b w = 150 mmﺒﺤﻴﺙ ﺃﻥ ﺍﻟﻌﺭﺽ ﺍﻷﺩﻨﻰ ﻟﻠﻌﺼﺏ 1 h = 87 m = max 3 t 100 m
ﺴﻨﻌﺘﻤﺩ ﺍﻟﻘﻴﻤﺔ ﻟﻠﺘﺒﺎﻋﺩ ﺒﻴﻥ ﺍﻷﻋﺼﺎﺏ .S = 500 m
ﻭﻗﺩ ﺘﻡ ﺘﺴﻠﻴﺢ ﺒﻼﻁﺔ ﺍﻟﺘﻐﻁﻴﺔ ﺘﺴﻠﻴﺤﺎﹰ ﺇﻨﺸﺎﺌﻴﺎﹰ ﻋﻠﻰ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ: ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻤﺘﻌﺎﻤﺩ ﻤﻊ ﺍﻷﻋﺼﺎﺏ `5 φ 6 / m ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻤﻭﺍﺯﻱ ﻟﻸﻋﺼﺎﺏ ﺒﻠﻭﻜﺔφ 6 / O6/ / `5O6/m /
0,15
0,35
0,15
0,18
0,32
0,18
b w min
R 1 :ﺠﻤﻠﺘﻪ ﺍﻹﻨﺸﺎﺌﻴﺔ ﺠﺎﺌﺯ ﻤﺴﺘﻤﺭ ﻋﻠﻰ ﺃﺭﺒﻊ ﻓﺘﺤﺎﺕ
3,75
3,2
3
3,45
: ﺍﻟﻭﺯﻥ ﺍﻟﺫﺍﺘﻲ ﻟﺒﻼﻁﺔ ﺍﻟﺘﻐﻁﻴﺔ-
[
g 1 = t f * γ = 0.06 × 5 = 1.5 kN / m 2
g2 =
(b1 + b 2 )
] : ﻭﺯﻥ ﺍﻷﻋﺼﺎﺏ-
(h t
− tf )
γ S
2 0.15 + 0.18 (0.26 − 0.06) 25 = 1.65 kN / m 2 = 2 0.5
[
] : ﻭﺯﻥ ﺍﻟﺒﻠﻭﻙ-
g3 =
[
0.1 × 5 = 1 kN / m 2 0.5
] : ﺤﻤﻭﻻﺕ ﺍﻟﺘﻐﻁﻴﺔ-
[
g 4 = 2.5 kN / m 2
]
g u = 1.5 (g 1 + g 2 + g 3 + g 4 ) = 9.975 kN / m 2
:ﺍﻟﺤﻤﻭﻻﺕ ﺍﻟﻤﺼﻌﺩﺓ ﻋﻠﻰ ﺍﻟﻤﺘﺭ ﺍﻟﻁﻭﻟﻲ g u × 0.5 = 4.98 ≈ 5 kN / m`
P = 3 kN / m 2
ﺤﻤﻭﻻﺕ ﺤﻴﺔ ﻤﺼﻌﺩﺓ ﻋﻠﻰ ﺍﻟﻤﺘﺭ ﺍﻟﻁﻭﻟﻲ Pu = 3 × 1.8 × 0.5 = 2.7 kN / m`
ﺇﻥ ﺍﻟﻤﺴﺎﻨﺩ ﺍﻟﻌﺭﻴﻀﺔ ﺘﺅﺜﺭ ﺒﺸﻜل ﻜﺒﻴﺭ ﻋﻠﻰ ﻗﻴﻡ ﺍﻟﻌﺯﻭﻡ ﺍﻟﺴﺎﻟﺒﺔ ﻟﺫﻟﻙ ﻨﻘﻭﻡ ﺒﺘﺩﻭﻴﺭ ﺍﻟﻌﺯﻭﻡ ﺍﻟـﺴﺎﻟﺒﺔ
:ﻭﻓﻕ ﺍﻟﻌﺯﻡ ﺍﻟﺴﺎﻟﺏ ﺍﻟﻤﺨﻔﺽ ﺍﻟﺘﺼﻤﻴﻤﻲ
MS = M −
R.B 8
ﺍﻋﺘﻤﺎﺩ ﻋﻠﻰ ﻋﺭﺽ ﺍﻟﺠﻨﺎﺡ ﻤﺴﺎﻭﻴﺎﹰ ﺍﻟﺘﺒﺎﻋﺩ ﺒﻴﻥ ﻤﺤـﺎﻭﺭT ﺘﺤﺴﺏ ﺍﻟﻤﻘﺎﻁﻊ ﻭﺍﻟﻭﺴﻁ ﺒﺸﻜل ﺤﺭﻑ . ﻭﻋﺭﺽ ﺍﻟﺠﺴﺩ ﻤﺴﺎﻭﻴﺎﹰ ﺍﻟﻌﺭﺽ ﺍﻟﻭﺴﻁﻲ ﻟﻠﻌﺼﺏS ﺍﻷﻋﺼﺎﺏ b1 + b 2 = ﺍﻟﻌﺭﺽ ﺍﻟﻭﺴﻁﻲ ﻟﻠﻌﺼﺏ 2
ﻓﻲ ﺤﺎﻟﺘﻨﺎb w =
500cm 60cm
150 + 180 = 165 m 2
ht=260cm
200cm
165cm
T d = 230 ⇒ A S min = 0.9 µ S max = 0.5 ×
b w .d = 0.81 cm 2 fy
f 455 . C = 0.014 630 + f y f y
A S max = µ S max .d.b w = 5.27 cm 2
:ﺇﻥ ﺃﻜﺒﺭ ﻗﻭﺓ ﻗﺹ ﻴﺘﺤﻤﻠﻬﺎ ﺍﻟﻤﻘﻁﻊ ﻋﻨﺩ ﺍﻟﻤﺴﺎﻨﺩ ﺍﻷﻋﺼﺎﺏ ﻫﻲ
[
t cu = t u = 0.23 f C′ = 1.03 N / mm 2 t max = 0.65 f C′ = 2.9 N / m`
]
Vu = 24.1 N / mm 2 tu = =
Q × 10 3 0.85 b w .d
24.1 × 10 3 = 0.75 N / m` 0.85 × 165 × 230
ﺇﺫﻥ ﻴﻠﺯﻡ ﺘﺴﻠﻴﺢ ﻗﺹ ﺇﻨﺸﺎﺌﻲt u < t cu n as =
0.35 .b w .S fy
⇐ φ 6 ﻨﺨﺘﺎﺭ ﺇﺴﻭﺍﺭﺓ ﺒﻔﺭﻋﺘﻴﻥ ﻗﻁﺭ S=
2 π 62 240 × = 23.5 4 0.35 × 165 d = 23 cm 30 cm
ﺍﻷﺼﻐﺭS max φ 6 / 200 m
b w = 25 cm ﺒﻌﺭﺽH = 50 cm ﻨﺄﺨﺫ ﺴﻘﻭﻁ ﺍﻟﺠﻭﺍﺌﺯ ﺍﻟﻤﺤﻴﻁﻴﺔ
5-5
:ﺠﺎﺌﺯ ﻤﺴﺘﻤﺭ ﺴﺘﺔ ﻓﺘﺤﺎﺕ
1
2.93
3.183
3.18
2.93
1
:ﻤﻘﻁﻌﻪ ﺒﺸﻜل ﻤﺴﺘﻁﻴل
260
1000
ﻭﺯﻥ ﺫﺍﺘﻲg 1 = (h t − t f ) γ − ( ﻭﺯﻥ ﺍﻷﻋﺼﺎﺏ+ )ﻭﺯﻥ ﺍﻟﺒﻠﻭﻙB 0.825 + 0.5 = (0.26 − 0.06) × 25 − × 1 = 2.35 [kN / m`] 0.5
ﻭﺯﻥ ﺍﻟﺠﺩﺍﺭg 2 = 0.85 × 3.24 × 0.1 × 18 = 4.96 kN / m`
:ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻷﻭﻟﻰ R 1 ﺭﺩ ﻓﻌل ﺍﻟﻌﺼﺏ
g3 =
R g1 0.5
=
29.9 = 59.8 [kN / m`] 0.5
:ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻟﺜﺎﻨﻴﺔ g4 =
R g2 0.5
= 78.22 [kN / m`]
:ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻟﺜﺎﻟﺜﺔ g5 =
R g4 0.5
= 64.2 [kN / m`]
ﺒﺎﻗﻲ ﺍﻟﻔﺘﺤﺎﺕ ﺒﺎﻟﺘﻨﺎﻅﺭ
P1 =
R P1 = 40.84 [kN / m`] 0.5
P2 =
R P2 = 49.8 [kN / m`] 0.5
P3 =
R P4 = 41.6 [kN / m`] 0.5
ﺒﺎﻟﻨﺴﺒﺔ ﻟﺠﻤﻴﻊ ﺍﻟﻤﻘﺎﻁﻊ
A S min = 0.9
b w .d 0.9 × 1000 × 230 = = 492.85 cm 2 fy 420
260
1000
:ﻨﻔﺭﺽ A S max = 0.5 A Sb ⇒ y max = 0.5 y b A max = 0.5
455 20 × × 100 × 26 = 26.826 630 + 420 420
A S max = 26.825 cm 2
tu =
τ = 1.83 N / m` 0.85 × 1000 × 230
t u max = 0.65 f C′ = 2.91 N / m` t cu = 0.23 f C′ = 1.03 N / m`
ﻨﺤﺘﺎﺝ ﺇﻟﻰ ﺘﺴﻠﻴﺢ ﻗﺹ ﺤﺴﺎﺒﻲ t st = τ u − t ou = 1.83 − 0.72 = 1.4 t st .S.b w = n.a S .f y n=6
φ8 ⇒ S =
6 π 82 × 240 = 101.88 m 4 × 1000 × 1.11
300 S max = = 230 d 8 φ / 20 cm
2-2
2
1
2,7
3,25
:ﻤﻴﺘﺔ 0.825 + 0.5 × 0.8 = 1.88 kN / m` 0.5
( = ﻭﺯﻥ ﺫﺍﺘﻲ0.26 − 0.06) × 0.8 × 25 −
= ﻭﺯﻥ ﺠﺩﺍﺭ0.85 × 3.24 × 0.1 × 18 = 4.96 kN / m` = ﺤﻤﻭﻟﺔ ﻤﻴﺘﺔ ﻤﻨﻘﻭﻟﺔ ﻤﻥ ﺍﻷﻋﺼﺎﺏ ﻤﻥ ﺍﻟﻔﺘﺤﺔ ﺍﻷﻭﻟﻰ = ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻟﺜﺎﻨﻴﺔ
30.57 = 61.14 kN / m` 0.5
52.5 = 105.4 [kN / m`] 0.5
= ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻷﻭﻟﻰ
20.57 = 41.14 [kN / m`] 0.5
= ﻓﻲ ﺍﻟﻔﺘﺤﺔ ﺍﻟﺜﺎﻨﻴﺔ
29.9 = 59.8 [kN / m`] 0.5
.ﺭﺴﻡ ﻤﻐﻠﻔﺎﺕ ﺍﻟﻌﺯﻭﻡ ﻭﻗﻭﻯ ﺍﻟﻘﺹ ﻭﺭﺩﻭﺩ ﺍﻷﻓﻌﺎل
ﺒﺎﻋﺘﺒﺎﺭ ﺠﻤﻴﻊ ﺍﻟﺒﻼﻁﺎﺕ ﻤﺴﺘﻨﺩﺓ ﻋﻠﻰ ﻤﺴﺎﻨﺩ )ﺠﺩﺭﺍﻥ ﺃﻭ ﻜﻤﺭﺍﺕ( ﻋﻠﻰ ﺤﻭﺍﻓﻬﺎ ﺍﻷﺭﺒﻌﺔ ﻓﺈﻥ ﺫﻟـﻙ ﻴﺘﻡ ﺒﺤﺴﺎﺏ ﺩﺭﺠﺔ ﺍﺴﺘﻁﺎﻟﺔ rﺤﺴﺏ ﺍﻟﻌﻼﻗﺔ: m1 .L1 m 2 .L 2
=r
ﺤﻴﺙ: : m 2 , m1ﺘﺴﺎﻭﻱ 0.87ﻟﻠﻔﺘﺤﺎﺕ ﺍﻟﻁﺭﻓﻴﺔ ﻭﺘﺴﺎﻭﻱ 0.76ﻟﻠﻔﺘﺤﺎﺕ ﺍﻟﺩﺍﺨﻠﻴﺔ : L1ﺍﻟﺒﻌﺩ ﺍﻟﻜﺒﻴﺭ ﻭ : L 2ﺍﻟﺒﻌﺩ ﺍﻟﺼﻐﻴﺭ ﻭﺍﻵﻥ ﺘﻜﻭﻥ ﺍﻟﺒﻼﻁﺔ ﻤﺼﻤﺘﺔ ﺒﺎﺘﺠﺎﻩ ﻭﺍﺤﺩ ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﺴﺘﻁﺎﻟﺘﻬﺎ rﺃﻜﺒﺭ ﻤﻥ .2 ﺍﻟﺒﻼﻁﺎﺕ ﻤﺴﺘﻨﺩﺓ ﺍﺴﺘﻨﺎﺩ ﻤﺘﻤﺎﺜل ﻤﻥ ﺍﻟﻁﺭﻓﻴﻥ ﻨﺤﺴﺏ ﺩﺭﺠﺔ ﺍﻻﺴﺘﻁﺎﻟﺔ: m1 .L1 3.75 = = 1.2 m 2 .L 2 3.1 m .L 3.2 ﺠﻤﻴﻊ ﺍﻟﺒﻼﻁﺎﺕ ﻋﺎﻤﻠﺔ ﺒﺎﺘﺠﺎﻫﻴﻥ ﺤﻴﺙ ﺩﺭﺠﺔ ﺍﻻﺴﺘﻁﺎﻟﺔ ﺃﺼﻐﺭ ﻤﻥ = 1.16 ⇒ 2 = r= 1 1 m 2 .L 2 2.75 m .L 4.2 = 1.22 = r= 1 1 m 2 .L 2 3.45 = r1
)4.20 + 3.45 + 0.76 (4.20 + 3.45 = 9.62 cm 140
=
اﻟﻤﺤﯿﻂ 140
اﻟﻤﻜﺎﻓﺊ
=t
)3.60 + 0.76 (420 × 2 + 360 = 9.1 cm 140
=t
)375 + 310 + 0.76 (375 + 310 = 8.61 cm 140
=t
)320 + 375 + 0.76 (375 + 320 = 8.74 cm 140
=t
ﺤﻤﻭﻟﺔ
ﺤﻤﻭﻟﺔ
ﺍﻟﺒﻌﺩ
ﺍﻟﺒﻌﺩ
ﻤﻜﺎﻥ
ﺍﻻﺘﺠﺎﻩ
ﺍﻻﺘﺠﺎﻩ
ﺍﻟﺼﻐﻴﺭ
ﺍﻟﻜﺒﻴﺭ
ﺍﻟﺒﻼﻁﺔ
ﺍﻟﻘﺼﻴﺭ
ﺍﻟﻁﻭﻴل
L2
L1
ﻭﺭﻗﻤﻬﺎ
6.14
3.14
0.45
0.23
1.2
310
375
S1
9
1.64
0.66
0.12
1.6
275
375
S2
6.14
3.14
0.45
0.23
1.2
320
375
S3
5.7
3.55
0.42
0.26
1.14
320
320
S4
6.7
3.79
0.44
0.26
1.16
275
320
S5
3.68
5.46
0.27
0.40
0.90
310
320
S6
7.23
2.46
0.53
0.18
1.31
360
420
S7
6.7
3.4
0.44
0.25
1.16
310
360
S8
5.46
3.69
0.40
0.27
1.12
320
360
S9
6.28
3
0.46
0.22
1.22
345
420
S10
6.7
2.73
0.49
0.2
1.27
310
345
S11
6.4
3
0.47
0.22
1.23
320
345
S12
α2
α1
r
:ﻨﻘﻭﻡ ﺒﺤﺴﺎﺏ ﺍﻟﺤﻤﻭﻻﺕ ﻋﻠﻰ ﻤﺘﺭ ﻤﺭﺒﻊ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﺒﻼﻁﺔ
g 1 = 0.12 × 25 = 3 kN / m 2
g = 2.5 kN / m 2
g u = 1.5 (g 1 + g 2 ) = 1.5 (3 + 2.5) = 8.25 kN / m 2
Pu = 3 × 1.8 = 5.4 kN / m 2
Pu = 8.25 + 5.4 = 13.65 kN / m 2
ﺴﻴﺘﻡ ﺤﺴﺎﺏ ﺍﻟﺒﻼﻁﺔ ﺒﻁﺭﻴﻘﺔ ﺍﻟﺸﺭﺍﺌﺢ ﻤﻥ ﺍﻟﻜﻭﺩ ﺍﻟﻌﺭﺒﻲ ﺍﻟﺴﻭﺭﻱ ﻭﺫﻟﻙ ﻋﻥ ﻁﺭﻴﻕ ﺤﺴﺎﺏ ﻋﺯﻭﻡ ﺍﻻﻨﺤﻨﺎﺀ ﻟﺸﺭﺍﺌﺢ ﺒﺎﻻﺘﺠﺎﻫﻴﻥ ﺤﻴﺙ: ﺍﻷﺤﻤﺎل ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻁﻭﻴل w 1 = α1 .w
ﺍﻷﺤﻤﺎل ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻘﺼﻴﺭ w 2 = α 2 .w
ﺤﻴﺙ α 2 ,α1ﻫﻤﺎ ﻤﻌﺎﻤﻼﻥ ﺘﺅﺨﺫ ﻗﻴﻤﻬﻤﺎ ﻤﻥ ﺍﻟﺠﺩﻭل ) (٧-٨ﺜﻡ ﻗﻤﻨﺎ ﻤﻥ ﺃﺠل ﺤـﺴﺎﺏ ﻋـﺯﻭﻡ ﺍﻻﻨﺤﻨﺎﺀ ﻤﻥ ﺍﻟﺸﺭﺍﺌﺢ ﻭﺒﺎﻻﺘﺠﺎﻫﻴﻥ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺤﺎﺴﺏ ﺒﻌﺩ ﺃﻥ ﺘﻡ ﺇﻴﺠﺎﺩ ﻗﻴﻡ ﺍﻟﻤﻌـﺎﻤﻠﻴﻥ d 1 , d 2ﻟﺠﻤﻴـﻊ ﺍﻟﺒﻼﻁﺎﺕ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺒﺎﻟﺠﺩﻭل ﺍﻟﺴﺎﺒﻕ.
6.14
9
6.14
2,75
3,1 8.76
8.536
4.455
7.9
5T10
4T10 5T10
ﺘﺴﻠﻴﺢ ﺠﻤﻴﻊ ﺍﻟﻤﻘﺎﻁﻊ 5 T 10
7.389
5T10 5T10
4T10 5T10
2-2
2,75
3,2
:ﺃﺒﻌﺎﺩ ﺍﻟﺠﺎﺌﺯ ﺍﻷﻭﻟﻴﺔ
ht
bw`
L L ⇐ ht = − 10 12 h t = 500 mm b w = 250 mm
:ﻭﺯﻥ ﺫﺍﺘﻲ g 1 = 1.5 b w .h t .γ g 1 = 1.5 × 0.25 × 25 (0.5 − 0.12 ) = 3.56 kN / m`
B = 2.3 kN / m` ﻤﻊ ﻁﻴﻨﺔ ﻤﻥ ﺍﻟﻭﺠﻬﻴﻥ10 m ﺍﻟﺠﺩﺍﺭ ﺴﻤﺎﻜﺘﻪ
ﺍﻻﺭﺘﻔﺎﻉ ﺍﻟﺤﺭH = H S − L = 3500 − 500 = 300 L = 3200 L 3200 = = 0.93 < 1 1.15 H 1.15 × 3000
ﺇﺫﺍﹰ ﺤﻤﻭﻟﺔ ﺍﻟﺠﺩﺍﺭ ﺘﻭﺯﻉ ﺒﺸﻜل ﻤﺜﻠﺜﻲ g m = α.β × 0.87 Lﺤﻤﻭﻟﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻌﺯﻡ 2 × 2.3 × 0.87 × 3.2 = 4.27 3
=
g 5 = 0.5 × 2.3 × 0.87 × 3.2 = 3.2ﺤﻤﻭﻟﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻘﺹ 0.87L 60°
60°
ﺍﻟﺒﻼﻁﺎﺕ ﻭﻓﻕ ﺨﻁﻭﻁ ﺍﻨﻜﺴﺎﺭﻫﺎ ﺒﺸﻜل ﺸﺒﻪ ﻤﻨﺤﺭﻑ ﻭﻤﺜﻠﺙ ﺒﺯﺍﻭﻴﺔ ﺍﻨﻜﺴﺎﺭ ٤٥ﺩﺭﺠﺔ L2 2 2.75 × = × 8.25 = 2.75 2 3 2
(g n )m = α.g n .ﺤﻤﻭﻟﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻌﺯﻡ
L2 2.75 × = 0.5 × 8.25 `= 5.42 kN / m 2 2
(g n )S = B.g nﺤﻤﻭﻟﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻘﺹ
q L2 2
2 ,7 5
3 ,2
2.75 = 4.95 2 2.75 = 3.11 kN 2
× (Pn )m = × 5.4ﺤﻤﻭﻟﺔ ﺤﻴﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻌﺯﻡ 2 3
× (Pn )S = 0.5 × 5.4ﺤﻤﻭﻟﺔ ﺤﻴﺔ ﻤﻜﺎﻓﺌﺔ ﻟﻠﻘﺹ
ﺇﻥ ﺃﻜﺒﺭ ﻗﻭﺓ ﻗﺹ ﻴﺘﺤﻤﻠﻬﺎ ﺍﻟﺒﻴﺘﻭﻥ ﻤﻥ ﻤﺴﺎﻨﺩ ﺍﻟﺠﻭﺍﺌﺯ ﺍﻟﺭﺌﻴﺴﺔ ﺍﻟﺴﺎﻗﻁ:
]
[
t cu = t u = 0.23 f C = 0.23 20 = 1.03 N / mm 2
ﻭﻫﻭ ﺍﻹﺠﻬﺎﺩ ﺍﻟﻤﺴﻤﻭﺡ ﺒﺎﻟﺒﻴﺘﻭﻥ Q ⇒ Q = 1.03 × 0.85 × 250 × 460 × 10 −3 = 100.68 0.85 b w .d
= tu
ﺇﺫﺍﹰ ﺃﻱ ﻗﻭﺓ ﻗﺹ ﻋﻨﺩ ﺍﻟﻤﺴﺎﻨﺩ ﺃﺼﻐﺭ ﻤﻥ Q = 100.68 kNﺘﺤﺘﺎﺝ ﺇﻟﻰ ﺘﺴﻠﻴﺢ ﻋﺭﻀﻲ ﺇﻨﺸﺎﺌﻲ ﻓﻘﻁ φ 8 / 20 cm
ﻭﻗﺴﻡ ﺃﻓﻘﻲ ﻴﻘﻊ ﻋﻠﻰ ﻤﺴﺘﻭﻱ ﻋﻠﻰ16 m ﺒﻼﻁﺔ ﺍﻟﺴﻘﻑ ﺘﻘﺴﻡ ﺇﻟﻰ ﻗﺴﻤﻴﻥ ﻗﺴﻡ ﻤﺎﺌل ﻭﻋﻠﻰ ﻤﺴﺘﻭﻱ ﻟﺫﻟﻙ ﺘﻡ ﺩﺭﺍﺴﺔ ﺒﻼﻁﺔ ﺍﻟﺴﻘﻑ ﺒﻼﻁﺔ ﻤﺼﻤﻤﺔ ﻋﺎﻤﻠﺔ ﺒﺎﺘﺠﺎﻫﻴﻥ ﻭﻋﻠﻰ ﻗﺴﻤﻴﻥ ﺍﻷﻭل ﻤﺎﺌل14.80 m .ﻭﺍﻟﺜﺎﻨﻲ ﺃﻓﻘﻲ
.ﻨﻘﻭﻡ ﺒﺤﺴﺎﺏ ﺍﻟﺤﻤﻭﻻﺕ ﻋﻠﻰ ﻤﺘﺭ ﻤﺭﺒﻊ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﺒﻼﻁﺔ
g 1 = 0.12 × 25 = 3 kN m 2
g = 2.5 kN m`
g u = 1.5 (g 1 + g 2 ) = 1.5 (3 + 2.5) = 8.25 kN m`
Pu = 3 × 1.8 = 5.4 kN m`
Pu = 8.25 + 5.4 = 13.85
9.89
48.7
49.7
1.2
6.02 11.5 5.9
ﻤﺭﻜﺯﺓ = 8.25 × 6.02 = 49.605 kNﺍﻟﻘﻭﺓ ﺍﻟﻜﻠﻴﺔ ﺍﻟﻤﻴﺘﺔ ﻤﺼﻌﺩﺓ ﺘﺤﻠل ﺇﻟﻰ ﺍﻟﻤﺭﻜﺒﺘﻴﻥ: ﺍﻷﻓﻘﻴﺔ 9.89 kN ﺍﻟﺸﺎﻗﻭﻟﻴﺔ 48.7
ﻨﺤﻭل ﺍﻟﻘﻭﺓ ﺍﻟﻤﺭﻜﺯﺓ ﺇﻟﻰ ﻤﻭﺯﻋﺔ ﺒﺎﻨﺘﻅﺎﻡ: ﻗﻤﺔ ﺍﻟﻘﻭﺓ ﺍﻟﻤﻭﺯﻋﺔ ﻋﻠﻰ ﺍﻟﻤﺘﺭ ﺍﻟﻁﻭﻟﻲ 48.7 `= 8.26 kN m 5.9
= gu
= 2.7 × 6.02 = 16.25 kNﺍﻟﺤﻤﻭﻟﺔ ﺍﻟﺤﻴﺔ ﺍﻟﻜﻠﻴﺔ ﺍﻟﻤﺼﻌﺩﺓ ﺘﺤﻠل ﻭﻓﻕ ﺯﺍﻭﻴﺔ ﻤﻴﻼﻥ ﺍﻟﺴﻘﻑ ﺍﻟﻤﺭﻜﺯﻱ ﺇﻟﻰ ﻤﺭﻜﺒﺘﻴﻥ: ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﺸﺎﻗﻭﻟﻴﺔ ﺍﻟﻤﺭﻜﺯﺓ16.254 × cos 11.5 = 15.93 kN : ﺍﻟﻤﺭﻜﺒﺔ ﺍﻷﻓﻘﻴﺔ ﺍﻟﻤﺭﻜﺯﺓ16.254 × sin 11.5 = 3.3 kN :
ﻨﺤﻭل ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﺸﺎﻗﻭﻟﻴﺔ ﺍﻟﻤﺭﻜﺯﺓ ﺇﻟﻰ ﻤﻭﺯﻋﺔ: 15.93 `= 2.7 kN m 5.9
= Pu
ﻤﻜﺎﻥ
ﺤﻤﻭﻟﺔ
ﺤﻤﻭﻟﺔ
ﺍﻻﺘﺠﺎﻩ
ﺍﻻﺘﺠﺎﻩ
ﺍﻟﻁﻭﻴل
ﺍﻟﻘﺼﻴﺭ
0.23
0.45
3.14
6.14
0.66
1.64
9
3.14
6.14 5.7
ﺍﻟﺒﻌﺩ
ﺍﻟﺒﻌﺩ
ﻭﺭﻗﻤﻬﺎ
ﺍﻟﻜﺒﻴﺭ
ﺍﻟﺼﻐﻴﺭ
L1
L2
S1
375
310
1.2
S2
375
275
1.6
0.12
S3
375
320
1.2
0.23
0.45
S4
320
320
1.14
0.26
0.42
3.55
S5
320
275
1.16
0.26
0.44
3.79
6.7
S6
320
310
0.90
0.40
0.27
5.46
3.68
S7
420
360
1.31
0.18
0.53
2.46
7.23
S8
360
310
1.16
0.25
0.44
3.4
6.7
S9
360
320
1.12
0.27
0.40
3.69
5.46
S10
420
345
1.22
0.22
0.46
3
6.28
S11
345
310
1.27
0.2
0.49
2.73
6.7
S12
345
320
1.23
0.22
0.47
3
6.4
ﺍﻟﺒﻼﻁﺔ
r
α1
α2
ﺴﻴﺘﻡ ﺤﺴﺎﺏ ﺍﻟﺒﻼﻁﺔ ﺒﻁﺭﻴﻘﺔ ﺍﻟﺸﺭﺍﺌﺢ ﻤﻥ ﺍﻟﻜﻭﺩ ﺍﻟﻌﺭﺒﻲ ﺍﻟﺴﻭﺭﻱ ﺫﻟﻙ ﻋﻥ ﻁﺭﻴﻕ ﺤﺴﺎﺏ ﻋﺯﻭﻡ ﺍﻻﻨﺤﻨﺎﺀ ﻟﺸﺭﺍﺌﺢ ﺒﺎﻻﺘﺠﺎﻫﻴﻥ ﺤﻴﺙ: ﺍﻷﺤﻤﺎل ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻁﻭﻴلW1 = a 1 W :
ﺍﻷﺤﻤﺎل ﺒﺎﻻﺘﺠﺎﻩ ﺍﻟﻘﺼﻴﺭW2 = a 2 W :
ﺤﻴﺙ a 1ﻭ a 2ﻫﻤﺎ ﻤﻌﺎﻤﻼﻥ ﺘﺅﺨﺫ ﻗﻴﻤﻬﻤﺎ ﻤﻥ ﺍﻟﺠﺩﻭل ) (7-8ﺜﻡ ﻗﻤﻨﺎ ﻤﻥ ﺍﺠل ﺤﺴﺎﺏ ﻋﺯﻭﻡ ﺍﻻﻨﺤﻨﺎﺀ ﻤﻥ ﺍﻟﺸﺭﺍﺌﺢ ﻭﺒﺎﻻﺘﺠﺎﻫﻴﻥ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺤﺎﺴﺏ ﺒﻌﺩ ﻤﺎ ﺘﻡ ﺇﻴﺠﺎﺩ ﻗﻴﻡ ﺍﻟﻤﻌﺎﻤﻠﻴﻥ a 1ﻭ a 2ﻟﺠﻤﻴﻊ ﺍﻟﺒﻼﻁﺎﺕ ﻜﻤﺎ ﻫﻭ ﻤﻭﻀﺢ ﺍﻟﺠﺩﻭل ﺍﻟﺴﺎﺒﻕ.
ﺤﺴﺎﺏ ﺍﻟﺘﺴﻠﻴﺢ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻁﻭﻴل ﻋﻠﻰ ﺃﺴﺎﺱ ﻤﻘﻁﻊ ﻤﺴﺘﻁﻴل ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﻓﻴﻨﺘﺞ ﺘﺴﻠﻴﺢ ﺍﻟﺒﻼﻁﺎﺕ ﻓﻲ ﺍﻻﺘﺠﺎﻫﻴﻥ. 5 T 10 m :
120
1000
. ﺠﻤﻠﺘﻪ ﺍﻹﻨﺸﺎﺌﻴﺔ ﺠﺎﺌﺯ ﻤﺴﺘﻤﺭ ﻋﻠﻰ ﻤﻨﺤﻨﻴﻴﻥ1-1 ﺍﻟﺠﺎﺌﺯ ﻭﻓﻕ ﺍﻟﻤﺤﻭﺭ
h = 500 m L L ht = ⋅ ⇒ t 10 12 b W = 250 m
g 1 = 0.25 × 0.5 × 25 × 1.5 = 4169 kN m`
B = 3 kN m` ﻤﻊ ﻁﻴﻨﺔ ﻋﻠﻰ ﺍﻟﻭﺠﻬﻴﻥ150 m :ﺴﻤﺎﻜﺔ ﺍﻟﺠﺩﺍﺭ
( ﻭﺯﻥ ﺠﺩﺍﺭ ﻤﻜﺎﻓﺊ ﻟﻠﻌﺯﻡg n )n = × 3 × 0.87 × 3.1 = 5.39 2 3
( ﻭﺯﻥ ﺠﺩﺍﺭ ﻤﻜﺎﻓﺊ ﻟﻠﻘﺹg n )S = 0.5 × 3 × 0.87 × 3.1 = 4.05
( ﻭﺯﻥ ﺒﻼﻁﺔ ﻤﻜﺎﻓﺊ ﻟﻠﻌﺯﻡg u )n = × 8.25 × 2 3
3.1 = 8.52 kn m` 2
( ﻭﺯﻥ ﺒﻼﻁﺔ ﻤﻜﺎﻓﺊ ﻟﻠﻘﺹg u )S = 0.5 × 8.25 ×
3.1 = 6.39 kN m` 2
.ﻨﺎﺘﺠﺔ ﻋﻥ ﺍﻨﻜﺴﺎﺭ ﺍﻟﺒﻼﻁﺎﺕ ﻓﻘﻁ
(Pn )n
=
2 3.1 × 2.7 × = 2.79 kN m` 3 2
(Pn )S = 0.5 × 2.7 × 3.1 = 2.1 kN 2
m`
ﺤﺴﺎﺏ ﺍﻟﺘﺴﻠﻴﺢ ﻭﻗﺩ ﺘﻡ ﺤﺴﺎﺏ ﺍﻟﺘﺴﻠﻴﺢ ﻋﻠﻰ ﺃﺴﺎﺱ ﻤﻘﻁﻊ ﻤﺴﺘﻁﻴل ﺃﺒﻌﺎﺩﻩ 38 × 25ﻋﻠﻰ ﺠﻤﻊ ﺍﻟﻤﻨﺎﻁﻕ ﻤﻨﺎﻁﻕ ﺍﻟﻌﺯﻡ ﺍﻟﻤﻭﺠﺏ +ﻤﻨﺎﻁﻕ ﺍﻟﻌﺯﻡ ﺍﻟﺴﺎﻟﺏ.
0.9 ﻨﺴﺒﺔ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﺩﻨﻴﺎ = b ⋅ d ρy
= MS
0.9 × 380 × 250 = 203.6 m m 420
= MS
ﻨﺴﺏ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻌﻅﻤﻰ ﺤﺴﺏ ﺍﻟﻜﻭﺩ ﺍﻟﻌﺭﺒﻲ ﺍﻟﺴﻭﺭﻱ ﺤﺴﺏ ﺍﻟﺒﻨﺩ ٥-٢-٩ﺍﻟﻌﻼﻗﺔ ):(9-5 455 b⋅d 630 + ρ y
= ASb
455 20 × 380 × 250 = 26.4 cm 2 360 + 420 420
=
ﻨﺤﺴﺏ ﺍﻟﺤﻤﻭﻟﺔ ﺍﻟﻤﻁﺒﻘﺔ ﻋﻠﻰ ﺍﻟﻌﻤﻭﺩ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ: ) N u = K e (1.5 G + 1.8 P
ﺤﻴﺙ: : K eﻋﺎﻤل ﺍﻟﺘﻜﺎﻓﺅ ﻴﺄﺨﺫ ﺍﻟﻘﻴﻡ ﺍﻟﻭﺍﺭﺩﺓ ﻓﻲ ﺍﻟﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﻭﺒﺎﻋﺘﺒﺎﺭ ﻭﺠﻭﺩ ﺒﻼﻁﺎﺕ ﻜﺎﺒﻭﻟﻴـﺔ ﺒﺎﻟﻤـﺴﻘﻁ ﺤﺴﺏ ﺍﻟﺠﺩﻭل )٢-٨ﺏ( ﺼﻔﺤﺔ ) (١٨٧ﺒﺎﻟﻜﻭﺩ ﺍﻟﻌﺭﺒﻲ ﺍﻟﺴﻭﺭﻱ. ﻤﻭﻗﻊ ﺍﻟﻌﻤﻭﺩ
ﻭﺴﻁﻲ
ﻁﺭﻓﻲ
ﺭﻜﻨﻲ
ﺍﻟﻁﺎﺒﻕ ﺍﻷﺨﻴﺭ
1.1
1.3
1.4
ﺍﻟﻁﺎﺒﻕ ﺘﺤﺕ ﺍﻷﺨﻴﺭ
1
1.2
1.3
ﺒﺎﻗﻲ ﺍﻟﻁﻭﺍﺒﻕ
1
1.1
1.2
ﺍﻟﻁﺎﺒﻕ
ﻭﺘﺤﺴﺏ ﻁﺎﻗﺔ ﺘﺤﻤل ﺍﻟﻌﻤﻭﺩ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ:
]
[
N u = ξ.Ω f C .A ′C + f y .A S
ﺤﻴﺙ: : ξﻋﺎﻤل ﻴﺄﺨﺫ ﺒﻌﻴﻥ ﺍﻻﻋﺘﺒﺎﺭ ﺍﻟﻼﻤﺭﻜﺯﻴﺔ ﻭﻴﺅﺨﺫ ﻤﺴﺎﻭﻴﺎﹰ 0.8ﻟﻠﺘﺴﻠﻴﺢ ﺍﻟﻁﻭﻟﻲ : Ωﻤﻌﺎﻤل ﺘﺨﻔﻴﺽ ﻁﺎﻗﺔ ﺘﺤﻤل ﺍﻟﻤﻘﻁﻊ ﺍﻟﻌﺭﻀﻲ ﻭﻴﺴﺎﻭﻱ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﻀﻐﻁ ﺍﻟﻤﺭﻜﺯﻱ 0.7
ﻭﺍﻵﻥ ﻴﻤﻜﻨﻨﺎ ﺤﺴﺎﺏ ﻤﺴﺎﺤﺎﺕ ﻤﻘﻁﻊ ﺍﻟﺒﻴﺘﻭﻥ ﺍﻟﻼﺯﻤﺔ ﻟﻜل ﻋﻤﻭﺩ ﺒﻌﺩ ﺘﻌﻴﻴﻥ ﺍﻟﺤﻤﻭﻟﺔ ﺍﻟﻤﻁﺒﻘﺔ ﻋﻠﻴﻪ ﻭﺒﻔﺭﺽ ﻨﺴﺒﺔ ﺘﺴﻠﻴﺢ ﺒﻴﻥ ﺍﻟﺩﻨﻴﺎ ﻭﺍﻟﻌﻅﻤﻰ ) (0.01-0.025ﻭﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻔﻀﻠﺔ ) (0.015ﻤﻥ ﺍﻟﻌﻼﻗﺔ: Nu A ′C = 0.8 × 0.7 (0.85 f C + f y .u )
ﻭﻗﺩ ﺘﻡ ﺘﻘﺴﻴﻡ ﺍﻷﻋﻤﺩﺓ ﺇﻟـﻰ ﺃﺭﺒﻌـﺔ ﻤﺠﻤﻭﻋـﺎﺕ ﺤـﺴﺏ ﺤﻤـﻭﻻﺕ ﺍﻷﻋﻤـﺩﺓ ﻟﻬـﺎ ﻤﺘﻘﺎﺭﺒـﺔ ) .( C 4 , C 3 , C 2 , C1
ﺤﻴﺙ ﺃﻥ ﺠﻤﻴﻊ ﺍﻷﻋﻤﺩﺓ ﻤﺴﺘﻁﻴﻠﺔ ﺍﻟﺸﻜل ﻴﺘﻡ ﺍﻟﺘﺤﻘﻕ ﻤﻥ ﺃﻥ ﺍﻷﻋﻤﺩﺓ ﻗﺼﻴﺭﺓ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ: L0 40 ≤ ⇒ = 11.55 h 12 3.2 = 10.66 ≤ 11.55 0.3
ﺠﺩﻭل ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻷﻭﻟﻰ C 1ﻨﻭﻉ ﺍﻟﻌﻤﻭﺩ ﻁﺭﻓﻲ ﻤﻭﻗﻊ
ﺤﻤﻭﻟﺔ ﺍﻟﻌﻤﻭﺩ
ﺃﺒﻌﺎﺩ ﺍﻟﻌﻤﻭﺩ
ﺍﻟﻌﻤﻭﺩ
ton
)h (cm) b (cm
ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﺤﺴﺎﺒﻲ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻤﺨﺘﺎﺭ cm2
AS
ﺍﻟﻘﺒﻭ
143.42
55
25
13.28
12 T 12
ﺃﺭﻀﻲ
135.1
55
25
12.21
12 T 12
ﺃﻭل
101.3
50
25
8.62
8 T 12
ﺜﺎﻨﻲ
67.5
50
25
6.3
6 T 12
ﺜﺎﻟﺙ
33.7
45
25
5.6
6 T 12
ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺜﺎﻨﻴﺔ C2ﻨﻭﻉ ﺍﻟﻌﻤﻭﺩ ﺭﻜﻨﻲ ﻤﻭﻗﻊ
ﺤﻤﻭﻟﺔ ﺍﻟﻌﻤﻭﺩ
ﺃﺒﻌﺎﺩ ﺍﻟﻌﻤﻭﺩ
ﺍﻟﻌﻤﻭﺩ
ton
)h (cm) b (cm
ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﺤﺴﺎﺒﻲ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻤﺨﺘﺎﺭ cm2
AS
ﺍﻟﻘﺒﻭ
76
40
25
7.05
6 T 14
ﺃﺭﻀﻲ
57.4
40
25
5.8
6 T 12
ﺃﻭل
55.6
30
25
5.6
6 T 12
ﺜﺎﻨﻲ
37.1
35
25
4.06
6 T 12
ﺜﺎﻟﺙ
18.5
35
25
3.75
6 T 12
ﺍﻟﻤﺠﻤﻭﻋﺔ ﺍﻟﺜﺎﻟﺜﺔ C3ﻨﻭﻉ ﺍﻟﻌﻤﻭﺩ ﻭﺴﻁﻲ ﻤﻭﻗﻊ
ﺤﻤﻭﻟﺔ ﺍﻟﻌﻤﻭﺩ
ﺃﺒﻌﺎﺩ ﺍﻟﻌﻤﻭﺩ
ﺍﻟﻌﻤﻭﺩ
ton
)h (cm) b (cm
ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﺤﺴﺎﺒﻲ ﺍﻟﺘﺴﻠﻴﺢ ﺍﻟﻤﺨﺘﺎﺭ cm2
AS
ﺍﻟﻘﺒﻭ
204.1
60
25
26
12 T 16
ﺃﺭﻀﻲ
177.6
55
25
19.82
10 T 16
ﺃﻭل
133.2
50
25
6.03
6 T 12
ﺜﺎﻨﻲ
88.8
50
25
6.25
6 T 12
ﺜﺎﻟﺙ
44.5
45
25
5.63
6 T 12
ﺍﻟﻤﺠﻤﻭﻋﺔ Cnﺃﺒﻌﺎﺩ ﺍﻟﻌﻤﻭﺩ 25 × 80ﻓﻲ ﺍﻟﻘﺒﻭ 8 × 14ﺒﺎﻗﻲ ﺍﻟﻁﻭﺍﺒﻕ 10 T 12
ﺍﻷﺤﻤﺎل ﺍﻟﺤﻴﺔ + 0.25ﺍﻷﺤﻤﺎل ﺍﻟﻤﻴﺘﺔ = W ﻭﺯﻥ ﺍﻟﻁﺎﺒﻕ ﺍﻟﻤﺘﻜﺭﺭ W = 4000 kN
ﺍﻟﻭﺯﻥ ﺍﻟﻜﻠﻲ ﻟﻠﻤﺒﻨﻰ W = 20000 kN
V ﻫﻲ ﻤﺠﻤﻭﻉ ﺍﻟﻘﻭﻯ ﺍﻟﻘﺎﺼﺔ ﺍﻷﻓﻘﻴﺔ ﺍﻟﻤﻁﺒﻘﺔ ﻋﻠﻰ ﺍﻟﻤﺒﻨﻰ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻤﺩﺭﻭﺱ ﻋﻨﺩ ﻤﻨﺴﻭﺏ ﺍﺘﺼﺎل ﺍﻷﺴﺎﺱ ﻤﻊ ﺍﻟﻤﻨﺸﺄ V = [Z.I.K.C.S].W
ﺤﻴﺙ: :Zﻤﻌﺎﻤل ﺯﻟﺯﺍﻟﻴﺔ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻤﺩﺭﻭﺴﺔ :Iﻤﻌﺎﻤل ﺃﻫﻤﻴﺔ ﺍﻟﻤﻨﺸﺄ :Kﻤﻌﺎﻤل ﺍﻟﺴﻠﻭﻙ ﺍﻟﻼﻤﺭﻥ ﻟﻠﻤﻨﺸﺄ :Cﻨﺴﺒﺔ ﺍﻟﺘﺴﺎﺭﻉ ﺍﻟﺯﻟﺯﺍﻟﻲ /ﺍﻟﺘﺴﺎﺭﻉ ﺍﻷﺭﻀﻲ :Sﻤﻌﺎﻤل ﺍﻟﺘﺭﺍﺒﻁ ﻭﺍﻟﻁﻨﻴﻥ ﺒﻴﻥ ﺍﻟﻤﻨﺸﺄ ﻭﺘﺭﺒﺔ ﺍﻟﺘﺄﺴﻴﺱ Z.I.K.C.S = λ
ﺩﻤﺸﻕ ﺘﻘﻊ ﻓﻲ ﺍﻟﻤﻨﻁﻘﺔ ﺍﻟﻨﻁﺎﻕ )(4
ﺍﻟﻘﻴﻤﺔ
ﺍﻟﻌﺎﻤل
0.4
Z
ﺍﻟﺠﻤﻠﺔ ﺍﻹﻨﺸﺎﺌﻴﺔ ﺍﻟﻤﻘﺎﻭﻤﺔ ﻟﻸﺤﻤـﺎل
ﺍﻟﺯﻟﺯﺍﻟﻴﺔ ﻫﻲ ﺠﻤﻠﺔ ﺠﺩﺍﺭﻴﺔ ﻓﻘﻁ ﻭﻻ
1.3
K
ﺘﻭﺠﺩ ﻤﺴﺎﺤﺔ ﺇﻁﺎﺭﻴﺔ
1.2
ﺍﻨﻅﺭ ﺍﻟﻔﻘﺭﺓ )(1-3
ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻘﺼﻴﺭ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﺍﻟﻁﻭﻴل
1.0
I C S
C ﻭﻫﻭ ﺍﻟﻨﺴﺒﺔ ﺒﻴﻥ ﺍﻟﺘﺴﺎﺭﻉ ﺍﻟﻨﺎﺠﻡ ﻋﻥ ﺍﻟﺯﻟﺯﺍل /ﺍﻟﺘﺴﺎﺭﻉ ﺍﻷﺭﻀﻲ 1 15 T
=C
ﺤﻴﺙ: :Tﻫﻭ ﺍﻟﺩﻭﺭ ﺍﻷﺴﺎﺴﻲ ﻻﻫﺘﺯﺍﺯ ﺍﻟﻤﻨﺸﺄ ﻭﻟﺤﺴﺎﺏ Tﺘﻭﺠﺩ ﻁﺭﻴﻘﺘﻴﻥ: ﺘﻘﺭﻴﺒﻴﺔ ﺘﺠﺭﻴﺒﻴﺔ. ﻭﺃﺨﺭﻯ ﺩﻗﻴﻘﺔ.ﻭﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺘﻘﺭﻴﺒﻴﺔ ﻨﺠﺩ ﺃﻨﻪ: 0.09 h n D
=T
ﺤﻴﺙ: : h nﺍﺭﺘﻔﺎﻉ ﺍﻟﻤﻨﺸﺄ ﺍﻟﻤﺩﺭﻭﺱ ﻤﻥ ﺍﻟﻘﺎﻋﺩﺓ ﻭﺤﺘﻰ ﺃﻋﻠﻰ ﻤﻨﺴﻭﺏ )ﺍﻟﻤﻨﺴﻭﺏ (h :Dﺒﻌﺩ ﺍﻟﻤﻨﺸﺄﺓ ﻓﻲ ﺍﺘﺠﺎﻩ ﺍﻟﻘﻭﻯ ﺍﻟﺯﻟﺯﺍﻟﻴﺔ ﺍﻟﻤﻁﺒﻘﺔ OK
D = 14.50 ⇒ T = 0.41 > 0.3 D = 20.8 ⇒ T = 0345 1 =C = 0.104 15 0.41
= 0.11
1 15 0.345
=C
ﺤﻤﻭﻻﺕ ﻋﻠﻰ ﺍﻟﻭﺍﺠﻬﺔ ﺍﻟﻘﺼﻴﺭﺓ V = Z.I.K.C.S
= 0.4 × 1.2 × 1.3 × 1.5 × 0.11 × 20000 = 2059.2 kN
ﺤﻤﻭﻻﺕ ﻋﻠﻰ ﺍﻟﻭﺍﺠﻬﺔ ﺍﻟﻁﻭﻴﻠﺔ V = 0.4 × 1.2 × 1.3 × 1.5 × 0.104 × 20000
= 1946.88 kN
XR =
∑ K .x ∑K jy
j
m
jk
j=1
:ﺤﻴﺙ Y ﻋﻥ ﺍﻟﻤﺤﻭﺭJ ﻫﻲ ﺒﻌﺩ ﻤﺭﻜﺯ ﺍﻟﻘﺹ ﻟﻠﺠﺩﺍﺭ: X j X ﻋﻥ ﺍﻟﻤﺤﻭﺭJ ﻫﻲ ﺒﻌﺩ ﻤﺭﻜﺯ ﺍﻟﻘﺹ ﻟﻠﺠﺩﺍﺭ: Yj Y XJ Ky
m
YR =
∑K j=1
jx
.y j J
m
∑K j=1
Y
C.g
jx
X
CR
X
Vy ex=1.04
e y = Ye = 0.079 ⇐ e x = X R = 0
Y
e y = YR (min ) = 0.05 L1 = 0.725
C.g
e x = X R (min ) = 0.05 L 2 = 1.04 Fyj =
K jy
Vy +
m
∑K j=1
jy
∑ (K
K jy .x j
m
j=1
CR
jy
2 j
)
.MT
2
.x + K jx .y j
ey=0.725
Vx X
n
V = Ft + ∑ Fi
Fn
F6
i=n
i =1
Fn-1
T < 0.7 ⇒ Ft = 0 V − Ft = V Fi =
(V − Ft ).w i .h i n
∑ w .h i =1
i
i=1
F1 i
:ﺤﻴﺙ ﺃﻱ ﺍﻟﺤﻤﻭﻟﺔ ﺍﻟﺸﺎﻗﻭﻟﻴﺔ ﺍﻟﻤﺭﻜﺯﺓ ﻋﻨﺩ ﺍﻟﻤﻨﺴﻭﺏ ﺃﻭ ﺍﻟﻨﺎﺠﻤﺔ ﻓﻘﻁ ﻋـﻥ ﻭﺯﻥi ﻭﺯﻥ ﺍﻟﻤﻨﺴﻭﺏ: w i .ﻫﺫﺍ ﺍﻟﻤﻨﺴﻭﺏ . ﻋﻥ ﻤﻨﺴﻭﺏ ﺃﺴﺎﺱ ﺍﻟﻤﻨﺸﺄi ﺍﺭﺘﻔﺎﻉ ﺍﻟﻤﻨﺴﻭﺏ: h i
ﺭﻗﻡ ﺍﻟﻁﺎﺒﻕ
wi
hx
w x .h x
fi
1
4000 3.5
1400
137.28
2
4000
2800
274.65
3
4000 10.5 4200
400.84
4
4000
5000
500.12
5
4000 17.5
700
637.4
7 14
ﺭﻗﻡ ﺍﻟﻁﺎﺒﻕ
wi
hx
w x .h x
fi
1
4000 3.5
1400
129.79
2
4000
2800
259.58
3
4000 10.5 4200
380.37
4
4000
5600
510.16
5
4000 17.5 7000
600.96
7 14
21000 = ﻤﺠﻤﻭﻉ
:ﻭﺒﺎﻟﺘﺎﻟﻲ ﺘﻜﻭﻥ ﺤﺼﺔ ﻜل ﺠﺩﺍﺭ ﻜﻤﺎ ﻴﻠﻲ
Y
Y
XJ Y
Ky
Vx
C.g
J C.g
X
CR
X
CR
X
ey=0.725
Vy ex=1.04
Fjx =
K jx
Vx +
m
∑K j=1
jx
∑ (K
K jx .y j
m
j=1
jy
2 j
2
.x + K jx .y j
)
MT
Fyj =
K jy
Vy +
m
∑K j=1
jy
∑ (K
K jy .x j
m
j=1
jy
2 j
2
.x + K jx .y j
)
MT
Vy × e x : e x = 1.04 Fxj Vx
=
K jx
×1 +
m
∑K j=1
jx
∑ (K
K jx .y j
m
j=1
jy
2 j
2
.x + K jx .y j
)
ex
Fyj Vy
=
K jy
×1 +
m
∑K j=1
jy
∑ (K
K jy .x j
m
j=1
jy
2 j
2
.x + K jx .y j
)
ex
1 ﺍﻟﺠﺩﺍﺭ
ﺍﻟﻁﺎﺒﻕ
NDL
2 ﺍﻟﺠﺩﺍﺭ NLL
MS
5
197.88 46.3 231.4 334.2 86.8
186
4
395.66 92.6 725.5 668.46 173.6 669.67 688.54 186
449.6
3
593.64 138.9 1326.1 1002.6 260.4 435.7 1032.8 279
116.9
2
791.52 185.2 2037
179.4
1
989.4 231.5 2781.7 1671
ﺍﻟﻁﺎﺒﻕ
NLL
MS
NDL
3 ﺍﻟﺠﺩﺍﺭ
NDL
NLL
344.27 93
224.8
336.8 347.2 670.87 1377.1 372
4 ﺍﻟﺠﺩﺍﺭ
434 915.37 1721.4 462
5 ﺍﻟﺠﺩﺍﺭ
MS
2450.3
6 ﺍﻟﺠﺩﺍﺭ
NDL
NLL
MS
NDL
NLL
MS
NDL
NLL
MS
5
184.2
41.67
213.85
184.2
43.67
175.7
304.47
71.6
321.56
4
368.4
83.34
270.39
368.4
87.34
222.6
608.94
143.2
915.5
3
552
125.01
501.39
552
131.61 420.62
913.41
214.8 1711.3
2
736
166.7
773.17
736
174.68
1217.88 286.4 2645.6
1
`920
208.32 1065.34
920
218.35 901.05 1522.35
ﺍﻟﻁﺎﺒﻕ
655.5
358
3649.2
(1) ﺍﻟﺠﺩﺍﺭ
(2) ﺍﻟﺠﺩﺍﺭ
(3) ﺍﻟﺠﺩﺍﺭ
ﺍﻟﺘﺭﻜﻴﺏ
ﺍﻟﺘﺭﻜﻴﺏ
ﺍﻟﺘﺭﻜﻴﺏ
0.8 (1.5 C + 1.8 P + 1.8 × 1.1 M S )
0.8 (1.56 + 1.8 P + 1.8 × 1.1 N S )
0.8 (1.56 + 1.8 P + 1.8 × 1.1N S )
Nu
Mu
Nu
Mu
Nu
Mu
5
195.344
458.08
336.8
330
513.41
404.7
4
390.61
1305.92
673.8
665.4
1028.82
809.4
3
586.03
2387
1010.78
784.28
1540.22
2104.2
2
781.38
3666.6
1347.52
1207.56
2053.6
3230.13
1
976.72
5007.12
1684.4
1647.66
2567.04
4410.7
ﺍﻟﻁﺎﺒﻕ
(4) ﺍﻟﺠﺩﺍﺭ
(5) ﺍﻟﺠﺩﺍﺭ
(6) ﺍﻟﺠﺩﺍﺭ
ﺍﻟﺘﺭﻜﻴﺏ
ﺍﻟﺘﺭﻜﻴﺏ
ﺍﻟﺘﺭﻜﻴﺏ
0.8 (1.5 C + 1.8 P + 1.8 × 1.1 M S )
0.8 (1.56 + 1.8 P + 1.8 × 1.1 M S )
0.8 (1.56 + 1.8 P + 1.8 × 1.1M S )
Nu
Mu
Nu
Mu
Nu
Mu
5
252.32
213.8
255.92
316.3
303.84
578.8
4
504.66
486.7
511.84
400.68
607.68
1647.94
3
756.96
902.5
767.76
757.12
911.52
3080.3
2
1009.3
1391.7
1623.68
1179.92
1215.36
4762.1
1
1261.5
1917.61
1279.6
1621.9
1519.2
5658.60
–
ﺍﻟﺘﺴﻠﻴﺢ
N
M
b (cm)
L (cm)
ﺭﻗﻡ ﺍﻟﺠﺩﺍﺭ
ﺒﺎﻟﻭﺴﻁ
ﺃﻋﻤﺩﺓ ﻤﺨﻔﻴﺔ
5 T 12/m`
12 T 18
976.8
5007.12
25
415
W1
5 T 12/m`
12 T 18
1620
1647.66
25
295
W2
5 T 12/m`
12 T 18
2567.04
4410.7
25
395
W3
5 T 12/m`
12 T 18
1261.6
1917.61
25
330
W4
5 T 12/m`
12 T 18
1279.6
1621.9
25
330
W5
5 T 12/m`
12 T 18
1519.2
6568.6
25
510
W6
–
ﺍﻟﺘﺴﻠﻴﺢ
N
M
b (cm)
L (cm)
ﺭﻗﻡ ﺍﻟﺠﺩﺍﺭ
ﺒﺎﻟﻭﺴﻁ
ﺃﻋﻤﺩﺓ ﻤﺨﻔﻴﺔ
5 T 10 /m`
12 T 16
390.61
1305.92
25
415
W1
5 T 10 /m`
12 T 16
673.8
665.4
25
295
W2
5 T 10 /m`
12 T 16
1026.82
809.4
25
395
W3
5 T 10 /m`
12 T 16
504.6
486.7
25
330
W4
5 T 10 /m`
12 T 16
511.84
400.68
25
330
W5
5 T 10 /m`
12 T 16
607.68
1647.94
25
510
W6
[3.18] m ﻨﺄﺨﺫ ﺍﻻﺭﺘﻔﺎﻉ ﺍﻟﻁﺎﺒﻘﻲ ﻤﺴﺎﻭﻱ 15.9 [cm] ﺩﺭﺠﺔ ⇐ ﺍﺭﺘﻔﺎﻉ ﺍﻟﺩﺭﺠﺔ ﺍﻟﻭﺍﺤﺩﺓ ﻤﻥ ﺍﻟﻤﻘﻁﻊ ﺍﻟﻤﻌﻤﺎﺭﻱ٢٠ ﺃﺨﺫﻨﺎ
b s = 30 [cm] ﻨﺄﺨﺫ ﻋﺭﺽ ﺍﻟﺩﺭﺠﺔ ﻤﺴﺎﻭﻴﺎﹰ
ﻤﻥ ﺸﺭﻭﻁ ﺍﻟﻜﻭﺩ
0.63 t=
3m
3630 = 181.5 → 145.2 mm → t = 15 cm 20 − 25
g1 =
25 t 25 × 0.15 × 1 × 1 = = 4.24 kN / m 2 Ο cos α cos 27.9
(
)
g2 =
hS 0.159 γ= × 25 = 1.98 2 2
g 3 = 25 × 0.15 = 3.75 kN / m 2
g4 =
1.5 1.5 = = 1.5 kN / m 2 b 1
g 5 = 2.5
P = 4 kN / m 2 Pu = 1.8 × 4 = 7.2 kN / m 2
g u1 = 1.5 (4.24 + 1.98 + 2.5 + 1.5) = 15.33 kN / m 2
g u 2 = 1.5 (2.5 + 3.75) + 1.8 × 4 = 16.575 kN / m 2
7.2 9.375
15.3
0.63
3m 21.82
10.29
ﻨﺄﺨﺫ ﺘﺴﻠﻴﺢ ﺜﺎﻨﻭﻱ5 φ12 / m` ﻭﻨﺄﺨﺫ ﺍﻟﺘﺴﻠﻴﺢ ﻋﻨﺩ ﺍﻟﻭﺜﺎﻗﺔ10 φ12 / m` ﻨﺄﺨﺫ ﺘﺴﻠﻴﺢ ﺍﻟﺸﺎﺤﻁ
0.15 × 25 × 0.63 = 2.36 kN / m`
2.5 × 0.63 = 1.56 kN / m`
4 × 0.63 = 2.52 kN / m`
30.08 [kN / m`] ﻤﻴﺕ
15.85 [kN / m`] ﺤﻲ
5 6 .3 5
2 .5
1.5 (2.36 + 1.56) + 1.8 (2.52 ) + 30.08 + 15.85 = 56.346 kN 6 ×16 :ﻨﺄﺨﺫ ﺘﺴﻠﻴﺢ ﺍﻟﻤﻴﺩﺓ
1.53
0.63
3m
h S = 15.9 [cm] ﻨﺄﺨﺫ ﺍﺭﺘﻔﺎﻉ ﺍﻟﺩﺭﺝ ﻤﻥ ﺍﻟﻤﺴﻘﻁ ﺍﻟﻤﻌﻤﺎﺭﻱ b S = 30 cm ﻭﻋﺭﺽ ﺍﻟﺩﺭﺠﺔ
:ﺘﺤﺩﻴﺩ ﺴﻤﺎﻜﺔ ﻓﻲ ﻫﺫﻩ ﺍﻟﺠﻤﻠﺔ ﻤﻥ ﺸﺭﻭﻁ ﺍﻟﻜﻭﺩ t=
5160 = 258 → 206.4 ⇒ t = 200 mm 20 → 25
g1 =
t 0.2 .l.l.γ = × 1 × 1 × 25 = 5.66 kN / m` cos α 0.884
g2 =
hS 0.159 .γ = × 25 = 1.98 kN / m 2 2 2
g3 =
1.5 1.5 = = 1.5 kN / m 2 b 1
g 4 = 2.5 kN / m 2 g u = 17.46 kN / m`
P = 4 kN / m 2 ⇒ Pu = 4 × 1.8 = 7.2 kN / m 2
g 1′ = t.γ = 0.20 × 25 = 5 kN / m 2
g ′2 = 2.5 kN / m 2
g u = 11.25 kN / m 2
P = 4 kN / m 2
Pu = 4 × 1.8 = 7.2 kN / m 2
b = 1m
d = 0.18 ⇒ 5×12 ﻭﺘﺴﻠﻴﺢ ﺍﻟﻅﻔﺭ6 × 12 ﻴﺼﺒﺢ ﺘﺴﻠﻴﺢ ﺍﻟﺸﺎﺤﻁ
:ﺍﻟﺩﺭﺝ ﻋﻨﺩ ﺍﻟﻁﺎﺒﻕ ﺍﻷﺨﻴﺭ
0.63
1.6
3m
5X16
.ﺘﺴﻠﻴﺢ ﺍﻟﻨﺼﻑ ﺍﻵﺨﺭ ﻤﻥ ﺍﻟﻤﻴﺩﺓ ﻓﻲ ﻤﻨﺘﺼﻑ ﺍﻟﻁﺎﺒﻕ
16X 6
γ = 20 kN / m 2 ϕ = 25 Ο C = 0.3 Kg / cm 2
. ﻭﻨﺤﺴﺏ ﺍﻟﺤﻤﻭﻻﺕ1 m ﻨﺄﺨﺫ ﺸﺭﻴﺤﺔ ﻤﻥ ﺍﻟﺠﺩﺍﺭ ﺒﻌﺭﺽ : K i ﻋﻨﺩ ﺤﺴﺎﺏ ﻀﻐﻁ ﺍﻟﺘﺭﺒﺔ ﻋﻠﻰ ﺠﺩﺍﺭ ﺍﻟﻘﺒﻭ ﻭﻨﺩﺨل ﻤﻌﺎﻤل ﻀﻐﻁ ﺍﻟﺘﺭﺒﺔ ﻓﻲ ﺤﺎﻟﺔ ﺍﻟﺭﺍﺤﺔ K i = 1 − sin ϕ = 1 − sin (25) = 0.6
t = 20 cm ﺒﻔﺭﺽ
G 1 = 0.2 × 4.68 × 25 = 36 kN / m`
G 2 = 23.4 kN / m`
G 3 = 40 kN / m 2
Q = a S .H.K i 40
= 20 × 4.68 × 0.6 = 55.6
+
36kN/m
M (−max ) =
Q.H 2 = 85.52 kN.m / m` 15
M (+max ) =
Q.H 2 = 38.1 kN.m / m` 33.6
M − = 12.4 kN.m / m` M + = 33.3 kN.m / m`
M (+max ) G 1 = 0.2 × 1.9 × 25 = 14.25 kN / m` N = 14.25 + 23.4 = 37.65
e=
M = 0.9 > 0.3 t = 0.09 ()ﻻﻤﺭﻜﺯﻴﺔ ﻜﺒﻴﺭﺓ N
eS = e +
t − a =1 2
M O = 37.65 × 0.99 = 37.63 kN.m / m` C=T= AS =
Mo 37.63 × 10 3 = = 100.12 kN / m` 0.85 d 0.85 × 270
T = 709.1 mm 2 > A S min = 6 cm 2 0.55 f y
8 T14 / m`
M (−max ) M − = 12.4 kN.m / m` N = 59.43 kN / m`
e=
M = 0.21 > 0.3 t = 0.09 m ()ﻻﻤﺭﻜﺯﻴﺔ ﻜﺒﻴﺭﺓ N
e S = 21 +
30 − 3.5 = 32.4 kN / m` 2
M o = 59.43 × 0.324 = 19.23 C=T=
19.23 × 10 3 = 83.8 kN / m` 0.85 × 270
T 83.8 × 10 3 AS = = = 363.53 > A S m` 0.55.f y 0.55 × 420 A S = 8 T12 / m` ﻨﺨﺘﺎﺭ
ﺒﺩﻭﻥ ﺘﺼﻌﻴﺩq = 128 t ﺃﻜﺒﺭ ﺤﻤﻭﻟﺔ ﻋﻠﻰ ﺍﻟﻌﻤﻭﺩ : ﺍﻟﻤﺴﺎﺤﺔ ﺍﻟﻤﻁﻠﻭﺒﺔ2.5 Kg / cm 2 ﻭﺍﻹﺠﻬﺎﺩ ﺍﻟﻤﺴﻤﻭﺡ ﺒﺎﻟﺘﺭﺒﺔ1.5 m ﺒﻔﺭﺽ ﻋﻤﻕ ﺍﻟﺘﺄﺴﻴﺱ A=
Pt 128 = = 5.15 m 2 qa 25
L = 1.5 >>> β = 1.8 m, L = 2.7 m β
V=
0.65 × 1.8 × 128 = 30.8 ton 1.8 × 2.7
M=
30.8 × 0.05 = 10 ton 2
d M 10 × 10 5 d = C× = 0.264 = 37 cm ⇒ d = 40 cm β 30 + 20
10 = 27.8 ton 0.9 × 0.4
C=
FC =
2C 2 × 27.8 × 128 = = 19.3 Kg / cm 2 < 80 Kg / cm 2 k.d.β 0.4 × 180 × 0.4
d = 40 cm ﻨﺄﺨﺫ ﺍﻻﺭﺘﻔﺎﻉ
V= d=
0.315 × 128 = 8.3 ton 1.8 × 2.7
3.8 × 10 3 = 10.5 cm < 40 cm 110 × 0.9 × 8
AS =
M 10 × 10 5 = = 13.8 cm 2 ⇒ 10 T 14 j * d * f S 0.9 × 2000 × 40
V=
0.5 × 2.7 × 128 = 35.5 ton 1.8 × 2.7
M=
35.5 × 0.5 = 8.8 ton 2
AS =
8.8 × 10 3 = 10.9 cm 2 ⇒ 8 T 14 0.9 × 2000 × 40
L = 2.7 m ، B = 1. m ﺃﺒﻌﺎﺩ ﺍﻷﺴﺎﺱ
h = 45 cm ﻭﻟﻜﻥ ﺍﻻﺭﺘﻔﺎﻉ ﺍﻟﻜﻠﻲd = 40 cm ﺍﻻﺭﺘﻔﺎﻉ ﺍﻟﻔﻌﺎل 15 cm ﺒﻴﺘﻭﻥ ﻨﻅﺎﻓﺔ ﺍﺭﺘﻔﺎﻉ A S = 10 T 14 ﺍﻟﺤﺩﻴﺩ ﺍﻟﻤﻭﺍﺯﻱ ﻟﻼﺘﺠﺎﻩ ﺍﻟﻁﻭﻴل A S = 8 T 14 ﺍﻟﺤﺩﻴﺩ ﺍﻟﻤﻭﺍﺯﻱ ﻟﻼﺘﺠﺎﻩ ﺍﻟﻘﺼﻴﺭ
