Slabs and Flat Slabs
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Slabs and Flat Slabs
Lecture 5 19th October 2016
Contents – Lecture 5
• Designing for shear in slabs  including punching shear • Detailing – Solid slabs • Flat Slab Design – includes flexure worked example • Exercise  Punching shear
EC2 Webinar  Autumn 2016
Lecture 5/1
Designing for shear in slabs
• When shear reinforcement is not required – e.g. usually one and twoway spanning slabs
• Punching shear – e.g. flat slabs and pad foundations
Shear
There are three approaches to designing for shear: • When shear reinforcement is not required e.g. usually slabs • When shear reinforcement is required e.g. Beams, see Lecture 3 • Punching shear requirements e.g. flat slabs
The maximum shear strength in the UK should not exceed that of class C50/60 concrete
EC2 Webinar  Autumn 2016
Lecture 5/2
Shear resistance without shear reinforcement
Without Shear Reinforcement
EC2: Cl. 6.2.2
Concise: 7.2
VRd,c = [0.12k(100 ρ l fck)1/3 + 0.15σcp] bwd
(6.2.a)
with a minimum of VRd,c = (0.035k3/2fck1/2 + 0.15 σcp) bwd
(6.2.b)
where:
k
= 1 + √(200/d) ≤ 2.0
ρl
= Asl/bwd ≤ 0.02
Asl
= area of the tensile reinforcement,
bw
= smallest width of the crosssection in the tensile area [mm]
σ cp
= NEd/Ac < 0.2 fcd [MPa] Compression +ve
NEd
= axial force in the crosssection due to loading or prestressing [in N]
Ac
= area of concrete cross section [mm2]
EC2 Webinar  Autumn 2016
Lecture 5/3
Shear – vRd,c

Concise Table 7.1 or 15.6
vRd,c resistance of members without shear reinforcement, MPa As Effective depth, d (mm) (bd) % ≤200 225 250 275 300 350 400 450 500
0.25 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.41 0.40
0.50 0.59 0.57 0.56 0.55 0.54 0.52 0.51 0.49 0.48
0.75 0.68 0.66 0.64 0.63 0.62 0.59 0.58 0.56 0.55
1.00 0.75 0.72 0.71 0.69 0.68 0.65 0.64 0.62 0.61
1.25 0.80 0.78 0.76 0.74 0.73 0.71 0.69 0.67 0.66
1.50 0.85 0.83 0.81 0.79 0.78 0.75 0.73 0.71 0.70
1.75 0.90 0.87 0.85 0.83 0.82 0.79 0.77 0.75 0.73
2.00 0.94 0.91 0.89 0.87 0.85 0.82 0.80 0.78 0.77
k
2.00 1.94 1.89 1.85 1.82 1.76 1.71 1.67 1.63
Table derived from: vRd,c = 0.12 k (100ρI fck)(1/3) ≥ 0.035 k1.5 fck0.5 where k = 1 + √(200/d) ≤ 2 and ρI = As/(bd) ≤ 0.02 Note: This table has been prepared for fck = 30. Where ρI exceeds 0.40% the following factors may be used:
fck factor
25 28 32 35 40 45 50 0.94 0.98 1.02 1.05 1.10 1.14 1.19
600 0.38 0.47 0.53 0.59 0.63 0.67 0.71 0.74 1.58
750 0.36 0.45 0.51 0.57 0.61 0.65 0.68 0.71 1.52
Shear in Slabs
Most slabs do not require shear reinforcement
∴Check VEd < VRd,c
Where VRd,c is shear resistance of members without reinforcement
vRd,c = 0.12 k (100 ρI fck)1/3 ≥ 0.035 k1.5 fck0.5
Where VEd > VRd,c, shear reinforcement is required and the strut inclination method should be used
Howto Compendium p21
EC2 Webinar  Autumn 2016
Lecture 5/4
Punching shear
Punching shear symbols
ui = ith perimeter. u1 = basic control perimeter at 2d u1* = reduced basic control perimeter u0 = column perimeter d = average effective depth k = coeff. depending on column shape –see Table 6.1 W1 = a shear distribution factor – see 6.4.3(3)
Punching Shear
EC2:Cl. 6.4
Concise: Figure 8.3
Punching shear does not use the Variable Strut inclination method and is similar to BS 8110 methods
• The basic control perimeter is set at 2d from the loaded area • The shape of control perimeters have rounded corners
2d 2d u1 2d u1 u1 2d
bz
by
• Where shear reinforcement is required the shear resistance is the sum of the concrete and shear reinforcement resistances.
EC2 Webinar  Autumn 2016
Lecture 5/5
Punching Shear
EC2: Cl. 6.4.3 & 6.4.4
6.4.3 (2)
When calculating vRd,c: 6.4.4 (1)
Punching Shear
The applied shear stress should be taken as: vEd = β VEd/ui d where β =1 + k(MEd/VEd)u1/W1
For structures where: • lateral stability does not
depend on frame action • adjacent spans do not differ
by more than 25% the approximate values for β shown may be used:
EC2 Webinar  Autumn 2016
Lecture 5/6
Punching Shear
Where the simplified arrangement is not applicable then β can be calculated
z
For a rectangular internal column
with biaxial bending the following
2d
simplification may be used:
c1
y
β = 1 + 1.8{(ey/bz)2 + (ez/by)2}0.5
where by and bz are the dimensions of the control perimeter
c2 2d
For other situations there is plenty of guidance on determining β given in Cl 6.4.3 of the Code.
Punching shear control perimeters Basic perimeter, u1
EC2: Cl. 6.4.2
Near to an edge
Concise: Figure 8.4
Near to an opening
Concise: Figure 8.6
EC2 Webinar  Autumn 2016
Lecture 5/7
Punching Shear Reinforcement
EC2: Cl. 6.4.5
Concise: Figures 12.5 & 12.6
Outer control perimeter
Outer perimeter of shear reinforcement
kd
A
0.75d 0.5d
0.75d 0.5d
1.5d (2d if > 2d from
A column)
Outer control perimeter kd
The outer control perimeter at which shear reinforcement is not required, should be calculated from:
uout,ef = βVEd / (vRd,c d)
The outermost perimeter of shear reinforcement should be placed at a distance not greater than kd ( k = 1.5) within the outer control perimeter.
Section A  A u1
uout
Punching Shear Reinforcement
EC2: Cl. 6.4.5
Concise; Figure 8.10
Where proprietary systems are used the control perimeter at which
shear reinforcement is not required, uout or uout,ef (see Figure) should be calculated from the following expression:
uout,ef = βVEd / (vRd,c d)
uout,ef
uout
> 2d
2d 1,5d
d 1,5d
d
EC2 Webinar  Autumn 2016
Lecture 5/8
Punching Shear Reinforcement
EC 2: Cl. 6.4.5, Equ 6.52
Concise: 8.5
Where shear reinforcement is required it should be calculated in
accordance with the following expression: vRd,cs = 0.75 vRd,c + 1.5 (d/sr) Asw fywd,ef (1/(u1d)) sinα
(6.52)
Asw = area of shear reinforcement in each perimeter around the col.
sr
= radial spacing of layers of shear reinforcement
α
= angle between the shear reinforcement and the plane of slab
fywd,ef
= effective design strength of the punching shear reinforcement, = 250 + 0.25 d ≤ fywd (MPa.)
d
= mean effective depth of the slabs (mm)
Max. shear stress at column face,
vEd = βuVdEd ≤ vRd,max = 0.5 ν fcd 0 EC2 Equ 6.53
Punching Shear Reinforcement
EC 2: Cl. 6.4.5 (3), Equ 6.53
Concise: 8.6
Max. shear stress at column face, the u0 perimeter
vEd = βuVdEd ≤ vRd,max = 0.5 ν fcd 0
c1 and c2 are illustrated in Concise Figure 8.5
EC2 Webinar  Autumn 2016
Lecture 5/9
Punching Shear Reinforcement
Check vEd ≤ 2 vRdc at basic control perimeter ( NA check)
Note: UK NA says ‘first’ control perimeter, but the paper* on which this guidance is based says ‘basic’ control perimeter
The minimum area of a link leg (or equivalent), Asw,min, is given by the
following expression:
Asw,min (1.5 sinα + cosα)/(sr st) ≥ (0,08 √(fck))/fyk
EC2 equ 9.11
Asw,min ≥ (0,053 sr st √(fck)) /fyk For vertical links
*FRASER, AS & JONES, AEK. Effectiveness of punching shear reinforcement to EN
199211:2004. The Structural Engineer ,19 May 2009.
Punching shear Worked example
From Worked Examples to EC2: Volume 1 Example 3.4.10
EC2 Webinar  Autumn 2016
Lecture 5/10
Lecture 5 19th October 2016
Contents – Lecture 5
• Designing for shear in slabs  including punching shear • Detailing – Solid slabs • Flat Slab Design – includes flexure worked example • Exercise  Punching shear
EC2 Webinar  Autumn 2016
Lecture 5/1
Designing for shear in slabs
• When shear reinforcement is not required – e.g. usually one and twoway spanning slabs
• Punching shear – e.g. flat slabs and pad foundations
Shear
There are three approaches to designing for shear: • When shear reinforcement is not required e.g. usually slabs • When shear reinforcement is required e.g. Beams, see Lecture 3 • Punching shear requirements e.g. flat slabs
The maximum shear strength in the UK should not exceed that of class C50/60 concrete
EC2 Webinar  Autumn 2016
Lecture 5/2
Shear resistance without shear reinforcement
Without Shear Reinforcement
EC2: Cl. 6.2.2
Concise: 7.2
VRd,c = [0.12k(100 ρ l fck)1/3 + 0.15σcp] bwd
(6.2.a)
with a minimum of VRd,c = (0.035k3/2fck1/2 + 0.15 σcp) bwd
(6.2.b)
where:
k
= 1 + √(200/d) ≤ 2.0
ρl
= Asl/bwd ≤ 0.02
Asl
= area of the tensile reinforcement,
bw
= smallest width of the crosssection in the tensile area [mm]
σ cp
= NEd/Ac < 0.2 fcd [MPa] Compression +ve
NEd
= axial force in the crosssection due to loading or prestressing [in N]
Ac
= area of concrete cross section [mm2]
EC2 Webinar  Autumn 2016
Lecture 5/3
Shear – vRd,c

Concise Table 7.1 or 15.6
vRd,c resistance of members without shear reinforcement, MPa As Effective depth, d (mm) (bd) % ≤200 225 250 275 300 350 400 450 500
0.25 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.41 0.40
0.50 0.59 0.57 0.56 0.55 0.54 0.52 0.51 0.49 0.48
0.75 0.68 0.66 0.64 0.63 0.62 0.59 0.58 0.56 0.55
1.00 0.75 0.72 0.71 0.69 0.68 0.65 0.64 0.62 0.61
1.25 0.80 0.78 0.76 0.74 0.73 0.71 0.69 0.67 0.66
1.50 0.85 0.83 0.81 0.79 0.78 0.75 0.73 0.71 0.70
1.75 0.90 0.87 0.85 0.83 0.82 0.79 0.77 0.75 0.73
2.00 0.94 0.91 0.89 0.87 0.85 0.82 0.80 0.78 0.77
k
2.00 1.94 1.89 1.85 1.82 1.76 1.71 1.67 1.63
Table derived from: vRd,c = 0.12 k (100ρI fck)(1/3) ≥ 0.035 k1.5 fck0.5 where k = 1 + √(200/d) ≤ 2 and ρI = As/(bd) ≤ 0.02 Note: This table has been prepared for fck = 30. Where ρI exceeds 0.40% the following factors may be used:
fck factor
25 28 32 35 40 45 50 0.94 0.98 1.02 1.05 1.10 1.14 1.19
600 0.38 0.47 0.53 0.59 0.63 0.67 0.71 0.74 1.58
750 0.36 0.45 0.51 0.57 0.61 0.65 0.68 0.71 1.52
Shear in Slabs
Most slabs do not require shear reinforcement
∴Check VEd < VRd,c
Where VRd,c is shear resistance of members without reinforcement
vRd,c = 0.12 k (100 ρI fck)1/3 ≥ 0.035 k1.5 fck0.5
Where VEd > VRd,c, shear reinforcement is required and the strut inclination method should be used
Howto Compendium p21
EC2 Webinar  Autumn 2016
Lecture 5/4
Punching shear
Punching shear symbols
ui = ith perimeter. u1 = basic control perimeter at 2d u1* = reduced basic control perimeter u0 = column perimeter d = average effective depth k = coeff. depending on column shape –see Table 6.1 W1 = a shear distribution factor – see 6.4.3(3)
Punching Shear
EC2:Cl. 6.4
Concise: Figure 8.3
Punching shear does not use the Variable Strut inclination method and is similar to BS 8110 methods
• The basic control perimeter is set at 2d from the loaded area • The shape of control perimeters have rounded corners
2d 2d u1 2d u1 u1 2d
bz
by
• Where shear reinforcement is required the shear resistance is the sum of the concrete and shear reinforcement resistances.
EC2 Webinar  Autumn 2016
Lecture 5/5
Punching Shear
EC2: Cl. 6.4.3 & 6.4.4
6.4.3 (2)
When calculating vRd,c: 6.4.4 (1)
Punching Shear
The applied shear stress should be taken as: vEd = β VEd/ui d where β =1 + k(MEd/VEd)u1/W1
For structures where: • lateral stability does not
depend on frame action • adjacent spans do not differ
by more than 25% the approximate values for β shown may be used:
EC2 Webinar  Autumn 2016
Lecture 5/6
Punching Shear
Where the simplified arrangement is not applicable then β can be calculated
z
For a rectangular internal column
with biaxial bending the following
2d
simplification may be used:
c1
y
β = 1 + 1.8{(ey/bz)2 + (ez/by)2}0.5
where by and bz are the dimensions of the control perimeter
c2 2d
For other situations there is plenty of guidance on determining β given in Cl 6.4.3 of the Code.
Punching shear control perimeters Basic perimeter, u1
EC2: Cl. 6.4.2
Near to an edge
Concise: Figure 8.4
Near to an opening
Concise: Figure 8.6
EC2 Webinar  Autumn 2016
Lecture 5/7
Punching Shear Reinforcement
EC2: Cl. 6.4.5
Concise: Figures 12.5 & 12.6
Outer control perimeter
Outer perimeter of shear reinforcement
kd
A
0.75d 0.5d
0.75d 0.5d
1.5d (2d if > 2d from
A column)
Outer control perimeter kd
The outer control perimeter at which shear reinforcement is not required, should be calculated from:
uout,ef = βVEd / (vRd,c d)
The outermost perimeter of shear reinforcement should be placed at a distance not greater than kd ( k = 1.5) within the outer control perimeter.
Section A  A u1
uout
Punching Shear Reinforcement
EC2: Cl. 6.4.5
Concise; Figure 8.10
Where proprietary systems are used the control perimeter at which
shear reinforcement is not required, uout or uout,ef (see Figure) should be calculated from the following expression:
uout,ef = βVEd / (vRd,c d)
uout,ef
uout
> 2d
2d 1,5d
d 1,5d
d
EC2 Webinar  Autumn 2016
Lecture 5/8
Punching Shear Reinforcement
EC 2: Cl. 6.4.5, Equ 6.52
Concise: 8.5
Where shear reinforcement is required it should be calculated in
accordance with the following expression: vRd,cs = 0.75 vRd,c + 1.5 (d/sr) Asw fywd,ef (1/(u1d)) sinα
(6.52)
Asw = area of shear reinforcement in each perimeter around the col.
sr
= radial spacing of layers of shear reinforcement
α
= angle between the shear reinforcement and the plane of slab
fywd,ef
= effective design strength of the punching shear reinforcement, = 250 + 0.25 d ≤ fywd (MPa.)
d
= mean effective depth of the slabs (mm)
Max. shear stress at column face,
vEd = βuVdEd ≤ vRd,max = 0.5 ν fcd 0 EC2 Equ 6.53
Punching Shear Reinforcement
EC 2: Cl. 6.4.5 (3), Equ 6.53
Concise: 8.6
Max. shear stress at column face, the u0 perimeter
vEd = βuVdEd ≤ vRd,max = 0.5 ν fcd 0
c1 and c2 are illustrated in Concise Figure 8.5
EC2 Webinar  Autumn 2016
Lecture 5/9
Punching Shear Reinforcement
Check vEd ≤ 2 vRdc at basic control perimeter ( NA check)
Note: UK NA says ‘first’ control perimeter, but the paper* on which this guidance is based says ‘basic’ control perimeter
The minimum area of a link leg (or equivalent), Asw,min, is given by the
following expression:
Asw,min (1.5 sinα + cosα)/(sr st) ≥ (0,08 √(fck))/fyk
EC2 equ 9.11
Asw,min ≥ (0,053 sr st √(fck)) /fyk For vertical links
*FRASER, AS & JONES, AEK. Effectiveness of punching shear reinforcement to EN
199211:2004. The Structural Engineer ,19 May 2009.
Punching shear Worked example
From Worked Examples to EC2: Volume 1 Example 3.4.10
EC2 Webinar  Autumn 2016
Lecture 5/10
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