( Estimating Reinforcement (steel
▪ Total amount of steel = mass of steel × length of steel for 1 m length
▪ Mass of 1 m length of steel for each bar diameter is as listed in table below:
▪ An extend and bend length must be added to the main bar length both side
Leb=16 db , for both side Leb=2(16db)=32db
▪ An over lab length (Lo) must be added to the reinforcement length (LR) when LR
when LR is greater than the max bar length (Lb) which is (12m)
Lo = 25 db to 40 db and Lo ≥ 0.3 m
Where db is the diameter of bar (m)
Number of over labs= LR/12
Number of stirrups=Length of main steel reinforcement/ Distance between stirrups +1
▪ An extend (Le) and bend length (Lb) should be added to the stirrup-crosssectional length
▪ Le= 6db and Le ≥ 0.1 m
▪ Lb= 4 db
❖ Mass of steel (Ms)(kg) can be estimated using : Ms=LRD^2
/162
where D= diameter of bar (mm)
❖ Overlap for dowel = 30 db to 50 db
❖ Extend for dowel = 12 db
❖ Bend for dowel = 4db
Example /
A strip foundation of 40 m length has a cross – section as shown below. compute
the amount of steel reinforcement . Concrete cover is 5 cm.
So/
❖ Main reinforcement
LR=40 m
Number of overlaps = 40/12=3.33=4
Lϕ16=40+4×0.3=41.2
Lϕ12= lϕ16 =41.2m
Lϕ16=3(lϕ16)=3(41.2 )=123.6 m
Mϕ16=123.6 m ×1.55 kg/m =191.52 kg
or
Mϕ16= 123.6 (16)^2/162 = 195.3 kg
Lϕ12= 2 (lϕ12 ) =2 (41.2) = 82.4m
Mϕ12= 82.4m ×0.995 kg/m =81.99 kg
or Mϕ16= 82.4 (12)^2/162 = 73.24 kg
❖ Stirrups
Lsϕ10=2 (0.6-2(0.05)+0.4-2(0.05))+0.3 =1.9 m
where 0.05 is the cover and 0.3=Le+Lb
Note:0.3=(6×0.01)×2+(4×0.01)×5
Note:0.3=(6×0.01)×2+(4×0.01)×5
Number of stirrups (Ns)= 40/0.25+1=161
L ϕ10 = 1.9 ×161=305.9 m
M ϕ10= 305.9m ×0.567 kg/m=173.45 Kg
Mϕ10= 305.9 (10)^2/162 = 188.82 kg
Example /
Estimate the amount of steel reinforcement for the foundation shown below (cover 7 cm)?
Sec 1-1❖
LR(ϕ16&12) each side = 2.5 – 2×0.07 = 2.36 m ( LR<12 ) no over labs
No of bars each side(ϕ16&12) = 2.36/0.3+ 1 = 8.87 bars = 9 bar
lϕ16 = 2.36 +32 (0.016) = 2.872 m
Lϕ16 = 2 (9) (2.872) = 51.696 m
Mϕ16= 51.696 m ×1.55 kg/m = 80.13 kg
Lϕ12= 2.36 +32(0.012) = 2.74 m
Lϕ12 = 2(9) (2.74) = 49.4 m
Mϕ12 = 49.4 m+0.995 = 49.15 kg
:Dowels
Dowel length ( LD = overlap + bend + extend + (H- cover)
Where :
Overlap = 25 db to 40 db
Bend = 4 db
Extend = 12 db
(40+4+12=56)
H= foundation depth
LD ϕ25= 56(0.025 ) + (0.4-0.07) = 1.73 m
LD ϕ25 = 4 (LD) 4×1.73 →LD = 6.92 m
Mϕ25 = 6.92 m × 3.97 kg/m = 27.49 kg
Total amount of Sec 1-1
Mϕ16 = (80.13) kg
Mϕ12 = (49.15) kg
Mϕ25= (27.470 kg
❖ Sec 2-2
Along 10 m:
LR10(ϕ16&12)= 10-2(0.07) → LR= 9.86 m
LR20(ϕ16&12)= 20-2 (0.07) → LR= 19.86 m
Lϕ16= 9.86 + 32 (0.016) = 10.372 ˂ 12 m no overlap
Lϕ12= 9.86 + 32 (0.012) = 10.244 ˂ 12 m no overlap
No. of bars (ϕ16&12)= LR 20/distance+1=19.86/0.3+1= 67.2
= 68 bars
Lϕ16= 68(lϕ16 ) →Lϕ16 = 68 (10.37) → Lϕ16 = 705.2 m
Mϕ16 = 705.2 m × 1.55 kg/m = 1093 kg
Lϕ12 = 68(lϕ12)→ Lϕ12 = 68 (10.244) → Lϕ12= 686.3 m
Mϕ12= 686.3 m× 0.995 ks/m = 682.9 kg
Along 20 m:
LR(ϕ16&12)= 20-2 (0.07)=19.86 m
LR10(ϕ16&12)= 10-2(0.07) → LR= 9.86 m
lϕ16= 19.86+32 (0.016)=20.37≥ 12 m overlap
No. of overlap =20.37/12=1.69=2
lϕ16 = 20.37+2 (40×0.016) = 21.65 m
lϕ12 = 19.86+32 (0.012) = 20.24 m ≥ 12 m
No. of overlap =20.24/12= 1.68 =2
lϕ12 = 20.24 +2 (40×0.012) = 21.20 m
No. of bars(ϕ16&12) = 9.86 /0.3 +1= 33.87 ≈ 34 bars
Lϕ16= 34 (21.65) →Lϕ16 = 736 m
Mϕ16 = 736 m ×1.55 kg/m = 1141 kg
Lϕ12 = 34×(21.20) →Lϕ12=720.8 m
Mϕ12 = 720.8 m × 0.995 kg/m = 717.2 kg
Dowel :
LD ϕ25= 56(0.025)+(0.4-0.07)=1.73 m
LD ϕ25= 2 × 4(1.73)=13.84 m
Mϕ25 = 13.84 m × 33.97 = 54.9 kg
Total amount of sec 2-2
Mϕ16 = 1093+ 1141=2234 kg
Mϕ12 = 682.9 +717.2 = 1400 kg
Mϕ25 = 54.9 kg
❖ Total amount of steel ( sec 1-1 & 2-2 )
Mϕ12 = 98.3 +1400 = 1498.3 kg
Mϕ16 = 160.3 + 2234 = 2394.3 kg
Mϕ25=54.9+54.9=109.8 kg
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