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Shear Strength in Two-Way Floor Systems
Similar to spread footing with concentrated column load
Wide-Beam Action: nominal strength in usual case where no shear
reinforcement is
Vn Vc 0.53 fc bw d
Two-Way Action: cracking occur around column with periphery b0 at
distance d / 2 outside column. Vn is the smallest of
4
Vn Vc 0.27 2
fc b0 d
c
ACI Formula (11-35)
sd
Vn Vc 0.27
2 fc b0 d
b0
ACI Formula (11-36)
Vn Vc 1.06 fc b0 d
ACI Formula (11-37)
SLAB_90
where
b0 = perimeter of critical section at distance d /2 outside column*
c = ratio of long side to short side of column
s = 40 for interior columns, 30 for edge columns and 20 for
corner columns
การวัดระยะ d /2 เพือ่ หาเส้ นรอบรู ป b0:
1) ขอบหรือมุมของเสา นา้ หนักบรรทุกเป็ นจุด หรือพืน้ ทีแ่ รงปฏิกริ ิยา
2) ตาแหน่ งที่เปลีย่ นความหนาในแผ่นพืน้ เช่ น ขอบของหมวกเสา(Column
capital) หรือแป้นหัวเสา(Drop panel)
SLAB_91
Minimum size of square columns for Vc 1.06 fc bw d
d/2
d/2
d/2
s
s
b0 4(s d )
If s 4d ,
40
2
b0 / d
Interior column
b0 3s 2d
If s 4.33d,
30
2
b0 / d
Edge column
b0 2s d
If s 4.5d,
20
2
b0 / d
Corner column
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Example 17: Investigate the shear strength in wide-beam and two-way actions in
the flat slab design example for an interior column with no bending moment to be
transferred
Solution: (a) Wide-beam action. Check section 1-1 and 2-2 in long direction
1
2
2.10 m
3.10 m
Equivalent
square
column
1.33 m
avg d = 16 cm
Drop panel
6.0 m
6.0 m
avg d = 24 cm
2.59 m
2.50 m
1
8.0 m
2
8.0 m
Vu 1,658 6.0 3.1 30,839 kg
Vu 1,658 6.0 2.59 25,765 kg
Vn Vc 0.53 210 600 16 73,732 kg
Vn 62,672 kg > Vu
OK
Vn 0.85 73,732 62,672 kg > Vu OK
It will be rare that wide-beam (one-way) action will govern.
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(b) Two-way action.
Check section 1-1 circular at d/2 = 12 cm from edge of column capital
and section 2-2 at d/2 = 8 cm from edge of drop panel
Section 2-2
1.50 m
1.74 m
2.26 m
6.0 m
6.0 m
Section 1-1
2.66 m
8.0 m
8.0 m
Vu 1.658(6 8 0.25 1.742 ) 1.4 0.08 2.4(2.1 2.5 0.25 1.742 )
8 cm drop panel
75.64 0.77 76.41 ton (Section 1-1)
Vu 1.658(6 8 2.26 2.66) 69.62 ton (Section 2-2)
SLAB_94
Shear strength at section 1-1:
b0 = (174) = 546.6 cm, b0/d = 546.9/24 = 22.8
Since b0/d > 20, and c = 1, ACI Formula (11-36) controls.
40
2 fc b0 d 1.01 fc b0 d
22.8
Vn Vc 0.27
0.85 1.01 210 546.6 24 / 1,000 163.2 ton
Shear strength at section 2-2:
b0 = 2(226+266) = 984 cm, b0/d = 984/16 = 61.5
Since b0/d > 20, and c = 1, ACI Formula (11-36) controls.
40
2 fc b0 d 0.716 fc b0 d
61.5
Vn Vc 0.27
0.85 0.716 210 984 16 / 1,000 138.9 ton
Shear reinforcement is not required at this interior location.
SLAB_95
Example 18: Investigate the shear strength in wide-beam and two-way actions in
the flat plate design example for an interior column with no bending moment to be
transferred. Note that f’c = 280 ksc.
avg d = 13 cm 1
30 cm
Section 2-2
4.0 m
4.0 m
2.17 cm
40 cm
43 cm
53 cm
1
5.0 m
5.0 m
(a) Wide-beam action. Assuming d = 16-3 = 13 cm
Vu 1,048 4.0 2.17 / 1,000 9.10 ton
Vn Vc 0.53 280 400 13 / 1,000 46.12 ton
Vn 0.85 46.12 39.20 ton > Vu OK
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(b) Two-way action.
Vu 1.048(4 5 0.43 0.53) 20.72 ton
b0 = 2(43+53) = 192 cm, b0/d = 192/13 = 14.8
Since b0/d < 20, and c = 1, ACI Formula (11-37) controls.
Vn Vc 1.06 fc b0 d
0.85 1.06 280 192 13 /1,000 37.63 ton
Shear reinforcement is not required at this interior location.
SLAB_97
Shear Reinforcement in Flat Plate Floors
where no column capitals and drop panels, shear reinforcement is
frequently necessary.
Nominal strength when bar reinforcement is used,
Vn Vc Vs 0.53 fc b0 d
Av fy d
s
1.59 fc b0 d
Section
at which
v n 0.53 fc
Double U-stirrup
Stirrup
SLAB_98