Transcript No Slide Title

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)
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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)
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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.
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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