Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Shear in slab without beams z

Drop Panels (ACI 9.5.3.2 and 13.3.7.1):

Prof. Dr. Qaisar Ali

80

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Shear in slab without beams z

Column Capital:

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Shear in slab without beams z

Minimum depth p of slab in case of shear reinforcement to be provided as integral beams or bent bars: z

ACI 11.12.3 requires the slab effective depth d to be at least 6 in., but not less than 16 times the diameter of the shear reinforcement.

z

When bent bars and integral g beams are to be used, ACI 11.12.3.1 reduces ΦVc by 2

82

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Calculation of Punching shear demand (Vu): 25'-0"

Critical perimeter: d = h – 1 = 9″ bo = 4(c+d) = 4(14+9) = 92″ Tributary y area ((excluding g area of bo): At = (25×20) – (14+9)2/144 = 496.3 ft2 wu = 0.3804

C4

25'-0" C3

25'-0" C3

25'-0" C3

C4 20'-0"

C2

C1

C1

C1

C2 20'-0"

C2

C1

C1

C1

C2

kip/ft2

Vu = wu × At = 189 kip Prof. Dr. Qaisar Ali

20'-0" C4

C3

C3

C3

C4

83

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Calculation of Punching shear capacity (ΦVc): √ (fc′)bod=√(4000)×92×9/1000=52 k ΦVc is least of:

25'-0" C4

25'-0" C3

25'-0" C3

25'-0" C3

C4 20'-0"

Φ4√ (fc′)bod = 156 k (2 + 4/βc) √ (fc′)bod = 312 k

C2

C1

C1

C1

C2

{(αsd/bo +2} √ (fc′)bod = 307 k Therefore,

20'-0"

C2

C1

C1

C1

C2

C3

C4

ΦVc = 156 k < Vu (190 k) , N.G

20'-0" C4

C3

C3

84

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 01): Drop panels y

In drop panels, the slab thickness in the vicinity of the columns is increased to increase the shear capacity (ΦVc = Φ4√ (fc′)bod) of concrete.

y

The increased thickness can be computed by equating Vu to ΦVc and simplifying the resulting equation for “d” to calculate required “h”.

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 01): Drop panels Equate Vu to ΦVc: 25'-0"

Vu = ΦVc 189 = 0.75 × 4 √ (fc′) × 92 × d

C4

25'-0" C3

25'-0" C3

25'-0" C3

C4

25/6 = 4.25′

d = 10.82″ Therefore, h = d+1≈ 12″ This gives 2″ drop panel.

20'-0"

20/6 = 3.5′ C2

C1

C1

C1

C2

According to ACI, minimum

20'-0"

thickness of drop panel = h/4 = 10/4 = 2.5″, which governs.

C2

C1

C1

C1

C2

Drop Panel dimensions: 25/6 ≈ 4.25′; 20/6 ≈ 3.5′

20'-0" C4

Prof. Dr. Qaisar Ali

C3

C3

C3

C4

86

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 02): Column Capitals y

Occasionally, the top of the columns will be flared outward, as shown in figure. This is known as column capital.

y

This is done to provide a larger shear perimeter at the column and to reduce the clear span, ln, used in computing moments.

y

ACI 6.4.6 requires that the capital concrete be placed at the same time as the slab concrete. As a result, the floor forming becomes considerably more complicated and expensive.

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 02): Column Capitals Equate Vu to ΦVc: Vu = ΦVc

25'-0" C4

25'-0" C3

25'-0" C3

25'-0" C3

C4

190 = 0.75 × 4 √ (fc′) × bo × 9

20'-0"

bo = 111.26″ C2

C1

C1

C1

C2

Now,

20'-0"

bo = 4 (c + d) 111.26 = 4(c + 9)

C1

C2

C1

C1

C2

Simplification gives,

20'-0"

c = 18.8 ≈ 19″ C4

C3

C3

C4

C3

88

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 02): Column Capitals c = 19″

According to ACI code, θ <

2.5″

45o

y = 2.5/ tanθ

θ y

Let θ = 30o, then y ≈ 4.35″ For θ = 20o, y ≈ 7″

capital

14″ column

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 03): Integral Beams

Vertical stirrups For 4 sides, total

Vertical stirrups are used in

stirrup area is 4

conjunction with supplementary horizontal bars radiating outward

times individual 2

lv

legged stirrup area

in two perpendicular directions from the support pp to form what are termed integral beams contained entirely within the slab thickness. In such a way, critical perimeter is increased

Horizontal bars

Increased critical perimeter

90

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 03): Integral Beams

bo = 4R + 4c Prof. Dr. Qaisar Ali

91

10

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 03): Integral Beams y

ΦVc = 156 kips

y

When integral beams are to be used, ACI 11.12.3 reduces ΦVc by 2. Therefore ΦVc = 156/2 = 78 kips

y

Using 3/8″ Φ, 2 legged (0.22 in2), 4 (side) = 4 × 0.22 = 0.88 in2

y

Spacing (s) = ΦAvfyd/ (Vu – ΦVc) s = 0.75 × 0.88 × 60 × 9/ (190 – 78) = 3.18 ≈ 3″

y

Maximum spacing allowed d/2 = 6/2 = 3″ controls.

92

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 03): Integral Beams y

Four #5 bars are to be provided in each direction to hold the stirrups. We know minimum bo = 111.26″

y

bo = 4R + 4c1 ........ (1)

y

R = √ (x2 + x2)

y

From figure, x = (3/4)(lv – c1/2), therefore,

y

R = √ (2) x, and eqn. (1) becomes,

y

bo = 4√ (2) x + 4c1 bo = 4√ (2){(3/4)(lv – c1/2)} + 4c1 Or bo = 4.24lv – 2.12c1 + 4c1= 4.24lv + 1.88c1

y

Prof. Dr. Qaisar Ali

Therefore lv ≈ 20″

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Example: Calculate the shear capacity of slab at 14″ column C1 of the 10″ flat p plate shown. y Design for shear (option 03): Integral Beams details. lv = 20″ ≈ 24″ or 2′

94

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

DDM Limitations: z

For slabs with beams between supports on all sides (ACI 13.6.1.6):

0.2 ≤ α1l22/α2l12 ≤ 5.0 α = EcbIb / EcsIs Where, Ecb = Modulus of elasticity of beam concrete Ecs = Modulus of elasticity of slab concrete Ib = Moment of inertia of beam section Is = Moment of inertia of slab section Prof. Dr. Qaisar Ali

95

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

DDM Limitations: z

Explanation of Ib and Is:

α = EcbIb / EcsIs

96

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Example on α calculation z

hf = 7″, hw = 18″, bw = 12″

z

Effective flange width z

bw + 2hw = 48″, bw + 8hf = 68, 48″ governs

z

Ib = 33060 in4 OR,

z

IT-section ≈ 2Irectangle section & IL-section ≈ 1.5Irectangle section

z

Ib = 2 × 12 × 243/12 = 27648 in4

z

Is = (10 + 10) × 12 × 73/12 = 6860 in4

z

α = Ib/Is = 33060/ 6860 = 4.82

Prof. Dr. Qaisar Ali

20′ 20′

25′

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Longitudinal Distribution of Static Moments

Prof. Dr. Qaisar Ali

98

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Longitudinal Distribution of Static Moments

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Column Strip Moments z

ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to assign moments to column strip.

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Column Strip Moments z

ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to assign moments to column strip.

101

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Column Strip Moments z

ACI tables 13.6.4.1, 13.6.4.2 and 13.6.4.4 of the ACI are used to assign moments to column strip.

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Middle Strip Moments z

The remaining moments are assigned to middle strip in accordance with ACI 13.6.6.

z

Beams between supports shall be proportioned to resist 85 percent of column strip moments if Îą1l2/l1 {Where l2 shall be taken as full span length irrespective of frame location (exterior or interior)} is equal to or greater than 1.0 (ACI 13.6.5.1).

103

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Graph A4 z

Lateral distribution of longitudinal moments can also be done using Graph A.4 (Design of Concrete Structures, Nilson 13th Ed)

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

In graph A.4, l2 shall be taken

as

full

span

length irrespective of frame location (exterior or interior).

105

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Example on graph A4: z

Find the lateral distribution to column strip of positive and interior negative moments using graph A4. Take

Prof. Dr. Qaisar Ali

z

l2/l1 = 1.3

z

Îąl2/l1 > 1. 1

106

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Example on graph A4 z

l2/l1 = 1.3

z

αl2/l1 > 1

65 % of interior positive negative longitudinal longitudinal moment moment will go towill column go to column strip strip 107

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Torsional Stiffness Factor (βt) z

In the presence of an exterior beam, all of the exterior negative factored moment goes to the column strip, and none to the middle strip, unless the beam torsional stiffness is high relative to the flexural stiffness of the supported slab.

z

Torsional stiffness factor βt is the parameter accounting for this effect. βt reflects the relative restraint provided by the torsional resistance of the effective transverse edge beam.

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Torsional Stiffness Factor (Î˛t) 25'-0"

25'-0"

25'-0"

25'-0"

For a considered frame, the transverse edge beam

20'-0"

provides restraint through its torsional resistance. 20'-0"

20'-0"

109

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Prof. Dr. Qaisar Ali

Torsional Stiffness Factor (Î˛t)

110

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt z

Where walls are used as supports along column lines, they can be regarded as very stiff beams with an α1l2/l1 value greater than one.

z

Where the exterior support consists of a wall perpendicular to the direction in which moments are being determined, determined βt may be taken as zero if the wall is of masonry without torsional resistance.

z

βt may be taken as 2.5 for a concrete wall with great torsional resistance that is monolithic with the slab.

111

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt: z

Prof. Dr. Qaisar Ali

βt can be calculated using the following formula:

112

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt: z

Where, Ecb = Modulus of elasticity of beam concrete; Ecs = Modulus of elasticity of slab concrete C = cross-sectional constant to define torsional properties x = shorter overall dimension of rectangular part of cross section, in. y = longer overall dimension of rectangular part of cross section, in. Is = Moment of inertia of slab section spanning in direction l1 and having width bounded by panel centerlines in l2 direction.

Prof. Dr. Qaisar Ali

113

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt: z

C for βt determination can be calculated using the following formula.

y2

y2 x2

x1

x1

y1 Prof. Dr. Qaisar Ali

x2

y1

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): For determination of E-W frame exterior negative moment distribution to column strip, find βt for beam marked. Take slab depth = 7″ and Ecb = Ecs. 25'-0"

25'-0"

25'-0"

25'-0"

20'-0"

Exterior edge beam

20'-0"

(12″ × 24″)

20'-0"

115

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is) z

hw ≤ 4hf = 17″

Calculation of C:

7″ 24″

12″ Prof. Dr. Qaisar Ali

116

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is) z

Calculation of C: C = {1 – 0.63×12/24}{123 ×24/3} + {1 – 0.63×7/17}{73 ×17/3} = 10909 in4

y2 = 17″

2 y1= 24″

x2 = 7″

1 x1 =12″

Prof. Dr. Qaisar Ali

117

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is) z

Calculation of C: C = {1 – 0.63×12/17}{123 ×17/3} + {1 – 0.63×7/29}{73 ×29/3} = 8249 in4

y2 = 17″ + 12″ = 29″

2 y1= 17″ Prof. Dr. Qaisar Ali

x2 = 7″

1 x1 =12″

118

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is) z

Calculation of C: Therefore, C = 10909 in4

119

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): βt = EcbC/(2EcsIs) = C/ (2Is) z

Calculation of Is: Is = bhf3/12 = (20 × 12) × 73/12 = 6860 in4 25'-0"

25'-0"

25'-0"

25'-0"

20'-0"

b 20'-0"

20'-0"

Prof. Dr. Qaisar Ali

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

Lateral Distribution of Longitudinal Moments z

Determination of βt (Example): z

βt = C/ (2Is) = 10909/ (2 × 6860) = 0.80

121

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method z

Additional Requirements for Slab with Beams z

βt = 0.8strip 90 % of exterior negative moment goes to column

Lateral Distribution of Longitudinal Moments z

Once βt is known, exterior negative moment in column strip can be found. For,

z

l2/l1 = 1.3

z

αl2/l1 > 1 and βt = 0.8

Prof. Dr. Qaisar Ali

122

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Minimum thickness for two way slab: z

For 0 0.2 2 ≤ αm ≤ 2: fy ⎞ ⎛ ln ⎜⎜ 0.8 + ⎟⎟ 200,000 ⎠ h= ⎝ 36 + 5β (α m − 0.2 )

But not less than 5 in. fy in psi. z

For αm > 2: fy ⎞ ⎛ ⎟ ln ⎜⎜ 0.8 + 200,000 ⎟⎠ h= ⎝ 36 + 9β

But not less than 3.5 in. fy in psi. 123

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Two Way Slabs (General) y Minimum thickness for two way Slab: h = Minimum slab thickness without interior beams. ln = length of clear span in direction that moments are being determined, measured face-to-face of supports. β = ratio of clear spans in long to short direction of two-way slabs. αm = average value of α for all beams on edges of a panel.

z

For αm < 0.2, use the ACI table 9.5 (c).

Prof. Dr. Qaisar Ali

124

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Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

Direct Design Method y Special Reinforcement at exterior corner of Slab z

The reinforcement at exterior ends of the slab shall be provided as per ACI 13.3.6 in top and bottom layers as shown.

z

The positive and negative reinforcement in any case, should be of a size and spacing equivalent to that required for the maximum positive moment (per foot of width) in the panel.

125

Prof. Dr. Qaisar Ali

Department of Civil Engineering, N-W.F.P. University of Engineering and Technology Peshawar

The End

Prof. Dr. Qaisar Ali

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