This document provides details of the analysis and design of a flat slab foundation according to BS8110:Part 1:1997. It includes the slab geometry, material properties, loading details, and calculations for the design of reinforcement in the sagging and hogging bending moments for internal and edge spans in the x-direction. Reinforcement areas are calculated and reinforcement arrangements are selected to satisfy design requirements. Deflection checks are also performed.

1. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Date
Calc. by
Dr. C. Sachpazis
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Chk'd by
Date
18/01/2014
App'd by
Date
FLAT SLAB DESIGN TO BS8110:PART 1:1997
Slab geometry
Span of slab in x-direction;
Spanx = 7200 mm
Span of slab in y-direction;
Spany = 7200 mm
Column dimension in x-direction;
lx = 400 mm
Column dimension in y-direction;
ly = 400 mm
External column dimension in x-direction;
lx1 = 250 mm
External column dimension in y-direction;
ly1 = 250 mm
Edge dimension in x-direction;
ex = lx1 / 2 = 125 mm
Edge dimension in y-direction;
ey = ly1 / 2 = 125 mm
Effective span of internal bay in x direction;
Lx = Spanx – lx = 6800 mm
Effective span of internal bay in y direction;
Ly = Spany – ly = 6800 mm
Effective span of end bay in x direction;
Lx1 = Spanx – lx / 2 = 7000 mm
Effective span of end bay in y direction;
Ly1 = Spany – ly / 2 = 7000 mm
ex
B
C
Span x
Span x
lx
ey
1
A
lx
l y1
Sp
an
y
lx1
ly
Sp
an
y
2
ly
3
h
Slab details
Depth of slab;
h = 250 mm
1

2. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
2
Characteristic strength of concrete;
fcu = 35 N/mm
Characteristic strength of reinforcement;
fy = 500 N/mm
Characteristic strength of shear reinforcement;
fyv = 500 N/mm
Material safety factor;
γm = 1.15
2
2
Cover to bottom reinforcement;
c = 20 mm
Cover to top reinforcement;
c’ = 20 mm
Loading details
2
Characteristic dead load;
Gk = 7.000 kN/m
Characteristic imposed load;
Qk = 5.000 kN/m
2
Dead load factor;
γG = 1.4
Imposed load factor;
γQ = 1.6
Total ultimate load;
Nult = (Gk × γG) + (Qk × γQ) = 17.800 kN/m
Moment redistribution ratio;
βb = 1.0
Ratio of support moments to span moments;
i = 1.0
2
DESIGN SLAB IN THE X-DIRECTION
SAGGING MOMENTS
End bay A-B
Effective span;
L = 7000 mm
Depth of reinforcement;
d = 200 mm
Midspan moment;
m = (Nult × L ) / (2 × (1 + √(1 + i)) ) = 74.823 kNm/m
Support moment;
m’ = i × m = 74.823 kNm/m
2
2
Design reinforcement
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m / (d × fcu) = 0.053
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 187.3 mm
Area of reinforcement designed;
2
As_des = m / (z × fy / γm) = 919 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 919 mm /m
2
Provide 20 dia bars @ 150 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 2094 mm /m
PASS - Span reinforcement is OK
Check deflection
Design service stress;
fs = 2 × fy × As_req / (3 × As_prov × βb) = 146 N/mm
2
2
2
2
Modification factor;
k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.545
Allowable span to depth ratio;
0.9 × 26 × k1 = 36.151
Actual span to depth ratio;
L / d = 35.000
2

3. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
PASS - Span to depth ratio is OK
Internal bay B-C
Effective span;
L = 6800 mm
Depth of reinforcement;
d = 202 mm
Midspan moment;
m = (Nult × L ) / (2 × (√(1 + i) + √(1 + i)) ) = 51.442 kNm/m
Support moment;
m’ = i × m = 51.442 kNm/m
2
2
Design reinforcement
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m / (d × fcu) = 0.036
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 191.9 mm
2
Area of reinforcement designed;
As_des = m / (z × fy / γm) = 617 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 617 mm /m
2
2
Provide 16 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1005 mm /m
PASS - Span reinforcement is OK
Check deflection
2
Design service stress;
fs = 2 × fy × As_req / (3 × As_prov × βb) = 204 N/mm
Modification factor;
k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.601
Allowable span to depth ratio;
0.9 × 26 × k1 = 37.469
Actual span to depth ratio;
L / d = 33.663
2
2
2
PASS - Span to depth ratio is OK
HOGGING MOMENTS – INTERNAL STRIP
Penultimate column B3
Consider the reinforcement concentrated in half width strip over the support
Depth of reinforcement;
d’ = 200 mm
Support moment;
m’ = 2 × i × m = 149.646 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
2
K = m’ / (d’ × fcu) = 0.107
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 172.5 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 1996 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1996 mm /m
2
Provide 20 dia bars @ 150 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 2094 mm /m
PASS - Support reinforcement is OK
3

4. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Internal column C3
Consider the reinforcement concentrated in half width strip over the support
Depth of reinforcement;
d’ = 200 mm
Support moment;
m’ = 2 × i × m = 102.884 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
2
K = m’ / (d’ × fcu) = 0.073
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 182.1 mm
2
Area of reinforcement required;
As_des = m’ / (z × fy / γm) = 1300 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1300 mm /m
2
2
Provide 20 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1571 mm /m
PASS - Support reinforcement is OK
HOGGING MOMENTS – EXTERNAL STRIP
Penultimate column B1, B2
Consider one and a half bays of negative moment being resisted over the edge and penultimate column
Width of span;
B = 7200 mm
Edge distance;
e = 125 mm
Depth of reinforcement;
d’ = 200 mm
Support moment;
m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 158.265
kNm/m
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m’ / (d’ × fcu) = 0.113
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 170.5 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 2134 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 2134 mm /m
2
Provide 20 dia bars @ 125 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 2513 mm /m
PASS - Support reinforcement is OK
Internal column C1, C2
Consider one and a half bays of negative moment being resisted over the edge and penultimate column
Width of span;
B = 7200 mm
Edge distance;
e = 125 mm
Depth of reinforcement;
d’ = 200 mm
Support moment;
m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 108.810
kNm/m
4

5. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Lever arm;
Date
Chk'd by
Date
18/01/2014
App'd by
Date
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m’ / (d’ × fcu) = 0.078
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 180.9 mm
2
Area of reinforcement required;
As_des = m’ / (z × fy / γm) = 1383 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1383 mm /m
2
2
Provide 20 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1571 mm /m
PASS - Support reinforcement is OK
Corner column A1
Depth of reinforcement;
d’ = 206 mm
Total load on column;
S = ((Spanx / 2) + ex) × ((Spany / 2) + ey) × Nult = 247 kN
Area of column head;
A = lx × ly1 = 0.100 m
Support moment;
m’ = S × (1 – (Nult × A / S) ) / 2 = 99.639 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
1/3
2
2
K = m’ / (d’ × fcu) = 0.067
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 189.3 mm
2
Area of reinforcement required;
As_des = m’ / (z × fy / γm) = 1211 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1211 mm /m
2
2
Provide 16 dia bars @ 150 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1340 mm /m
PASS - Support reinforcement is OK
Edge column A2, A3
Depth of reinforcement;
d’ = 202 mm
Total load on column;
S = Spanx × (Spany / 2 + ey) × Nult = 477 kN
Area of column head;
A = lx1 × ly = 0.100 m
Support moment;
m’ = S × (1 – (Nult × A / S) ) / 5.14 = 78.476 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
1/3
2
2
K = m’ / (d’ × fcu) = 0.055
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 188.8 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 956 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 956 mm /m
2
Provide 16 dia bars @ 175 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1149 mm /m
5

6. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
PASS - Support reinforcement is OK
Between columns 1-2, 2-3
Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom
reinforcement.
Area of reinforcement required;
2
As_req = Asx1 / 2 = 1047 mm /m
Provide 16 dia bars @ 150 centres - 'U' bars with 1600 mm long legs
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1340 mm /m
PASS - Edge reinforcement is OK
Distribution reinforcement
Provide 12 dia bars @ 300 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 377 mm /m
DESIGN SLAB IN THE Y-DIRECTION
SAGGING MOMENTS
End bay 1-2
Effective span;
L = 7000 mm
Depth of reinforcement;
d = 220 mm
Midspan moment;
m = (Nult × L ) / (2 × (1 + √(1 + i)) ) = 74.823 kNm/m
Support moment;
m’ = i × m = 74.823 kNm/m
2
2
Design reinforcement
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m / (d × fcu) = 0.044
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 208.6 mm
2
Area of reinforcement designed;
As_des = m / (z × fy / γm) = 825 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 825 mm /m
2
2
Provide 20 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1571 mm /m
PASS - Span reinforcement is OK
Check deflection
2
Design service stress;
fs = 2 × fy × As_req / (3 × As_prov × βb) = 175 N/mm
Modification factor;
k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.579
Allowable span to depth ratio;
0.9 × 26 × k1 = 36.942
Actual span to depth ratio;
L / d = 31.818
2
2
2
PASS - Span to depth ratio is OK
Internal bay 2-3
Effective span;
L = 6800 mm
6

7. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Depth of reinforcement;
d = 222 mm
Midspan moment;
m = (Nult × L ) / (2 × (√(1 + i) + √(1 + i)) ) = 51.442 kNm/m
Support moment;
m’ = i × m = 51.442 kNm/m
2
2
Design reinforcement
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m / (d × fcu) = 0.030
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 210.9 mm
2
Area of reinforcement designed;
As_des = m / (z × fy / γm) = 561 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 561 mm /m
2
2
Provide 16 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1005 mm /m
PASS - Span reinforcement is OK
Check deflection
2
Design service stress;
fs = 2 × fy × As_req / (3 × As_prov × βb) = 186 N/mm
Modification factor;
k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.798
Allowable span to depth ratio;
0.9 × 26 × k1 = 42.062
Actual span to depth ratio;
L / d = 30.631
2
2
2
PASS - Span to depth ratio is OK
HOGGING MOMENTS – INTERNAL STRIP
Penultimate column C2
Consider the reinforcement concentrated in half width strip over the support
Depth of reinforcement;
d’ = 220 mm
Support moment;
m’ = 2 × i × m = 149.646 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
2
K = m’ / (d’ × fcu) = 0.088
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 195.7 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 1758 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1758 mm /m
2
Provide 20 dia bars @ 150 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 2094 mm /m
PASS - Support reinforcement is OK
Internal column C3
Consider the reinforcement concentrated in half width strip over the support
Depth of reinforcement;
d’ = 220 mm
7

8. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Support moment;
m’ = 2 × i × m = 102.884 kNm/m
Lever arm;
Date
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
2
K = m’ / (d’ × fcu) = 0.061
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 204.0 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 1160 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1160 mm /m
2
Provide 20 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1571 mm /m
PASS - Support reinforcement is OK
HOGGING MOMENTS – EXTERNAL STRIP
Penultimate column A2, B2
Consider one and a half bays of negative moment being resisted over the edge and penultimate column
Width of span;
B = 7200 mm
Edge distance;
e = 125 mm
Depth of reinforcement;
d’ = 220 mm
Support moment;
m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 158.265
kNm/m
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m’ / (d’ × fcu) = 0.093
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 194.1 mm
Area of reinforcement required;
2
As_des = m’ / (z × fy / γm) = 1875 mm /m
2
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1875 mm /m
2
Provide 20 dia bars @ 150 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 2094 mm /m
PASS - Support reinforcement is OK
Internal column A3, B3
Consider one and a half bays of negative moment being resisted over the edge and penultimate column
Width of span;
B = 7200 mm
Edge distance;
e = 125 mm
Depth of reinforcement;
d’ = 220 mm
Support moment;
m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 108.810
kNm/m
Lever arm;
2
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
K = m’ / (d’ × fcu) = 0.064
8

9. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 203.0 mm
2
Area of reinforcement required;
As_des = m’ / (z × fy / γm) = 1233 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 1233 mm /m
2
2
Provide 20 dia bars @ 200 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1571 mm /m
PASS - Support reinforcement is OK
Edge column B1, C1
Depth of reinforcement;
d’ = 222 mm
Total load on column;
S = (Spanx / 2 + ex) × Spany × Nult = 477 kN
Area of column head;
A = ly1 × lx = 0.100 m
Support moment;
m’ = S × (1 – (Nult × A / S) ) / 5.14 = 78.476 kNm/m
Lever arm;
K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2
1/3
2
2
K = m’ / (d’ × fcu) = 0.045
Compression reinforcement is not required
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 210.1 mm
2
Area of reinforcement required;
As_des = m’ / (z × fy / γm) = 859 mm /m
Minimum area of reinforcement required;
As_min = 0.0013 × h = 325 mm /m
Area of reinforcement required;
As_req = max(As_des, As_min) = 859 mm /m
2
2
Provide 16 dia bars @ 175 centres
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1149 mm /m
PASS - Support reinforcement is OK
Between columns A-B, B-C
Around the perimeter between the column heads provide a minimum of 50% of the required end span bottom
reinforcement.
Area of reinforcement required;
2
As_req = Asy1 / 2 = 785 mm /m
Provide 16 dia bars @ 200 centres - 'U' bars with 1600 mm long legs
Area of reinforcement provided;
2
2
As_prov = π × D / (4 × s) = 1005 mm /m
PASS - Edge reinforcement is OK
PUNCHING SHEAR
Corner column A1
Design shear transferred to column;
Vt = ((0.45 × Spanx) + ex) × ((0.45 × Spany) + ey) × Nult = 202 kN
Design effective shear transferred to column;
Veff = 1.25 × Vt = 252 kN
Area of tension steel in x-direction;
Asx_ten = Ascorner = 1340 mm /m
Area of tension steel in y-direction;
Asy_ten = Ascorner = 1340 mm /m
Column perimeter;
uc = lx1 + ly = 650 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
2
2
9

10. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 1.811 N/mm
App'd by
Date
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 1292 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 1731 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.707 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 0.911 N/mm
Shear reinforcement required at perimeter;
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 119 mm
vc < v <= 1.6 × vc
2
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 1613 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 2161 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.707 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.730 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 16 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 1934 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 2592 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.707 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.609 N/mm
v < vc no shear reinforcement required
Penultimate edge column A2
Design shear transferred to column;
Vt = ((0.45 × Spanx) + ex) × (1.05 × Spany) × Nult = 453 kN
Design effective shear transferred to column;
Veff = 1.4 × Vt = 634 kN
Area of tension steel in x-direction;
Asx_ten = Asx_edge = 1148 mm /m
2
10

11. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
App'd by
Date
2
Area of tension steel in y-direction;
Asy_ten = Asy1e = 2094 mm /m
Column perimeter;
uc = (2 × lx1)+ ly = 900 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.292 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2184 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 3588 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.757 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.356 N/mm
1.6 × vc < v <= 2 × vc
Shear reinforcement required at perimeter;
Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 947 mm
2
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2826 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 4628 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.756 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.048 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 372 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 3468 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 5669 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.756 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 0.854 N/mm
Shear reinforcement required at perimeter;
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 154 mm
vc < v <= 1.6 × vc
2
11

12. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
App'd by
Date
Shear reinforcement at a perimeter of 3.75d - (803 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4110 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 6710 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.755 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.721 N/mm
v < vc no shear reinforcement required
Internal edge column A3
Design shear transferred to column;
Vt = ((0.45 × Spanx) + ex) × Spany × Nult = 431 kN
Design effective shear transferred to column;
Veff = 1.4 × Vt = 604 kN
Area of tension steel in x-direction;
Asx_ten = Asx_edge = 1148 mm /m
2
2
Area of tension steel in y-direction;
Asy_ten = Asye = 1570 mm /m
Column perimeter;
uc = (2 × lx1)+ ly = 900 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.135 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2184 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 2989 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 1.292 N/mm
Shear reinforcement required at perimeter;
Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 945 mm
1.6 × vc < v <= 2 × vc
2
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2826 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 3862 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
12

13. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
Nominal design shear stress at perimeter;
App'd by
Date
2
v = Veff / (u × d) = 0.998 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 365 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 3468 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 4734 mm2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.814 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 159 mm
Shear reinforcement at a perimeter of 3.75d - (803 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4110 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 5607 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.711 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.686 N/mm
v < vc no shear reinforcement required
Penultimate edge column B1
Design shear transferred to column;
Vt = (1.05 × Spanx) × ((0.45 × Spany) + ey) × Nult = 453 kN
Design effective shear transferred to column;
Veff = 1.4 × Vt = 634 kN
Area of tension steel in x-direction;
Asx_ten = Asx1e = 2513 mm /m
Area of tension steel in y-direction;
Asy_ten = Asy_edge = 1148 mm /m
2
2
Column perimeter;
uc = lx + (2 × ly1) = 900 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.292 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2184 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 4066 mm
2
13

14. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.789 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.356 N/mm
1.6 × vc < v <= 2 × vc
Shear reinforcement required at perimeter;
Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 789 mm
2
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2826 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 5241 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.788 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.048 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 331 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 3468 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 6416 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.788 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.854 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 104 mm
Shear reinforcement at a perimeter of 3.75d - (803 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4110 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 7592 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.787 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.721 N/mm
v < vc no shear reinforcement required
14

15. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Penultimate central column B2
Design shear transferred to column;
Vt = (1.05 × Spanx) × (1.05 × Spany) × Nult = 1017 kN
Design effective shear transferred to column;
Veff = 1.15 × Vt = 1170 kN
Area of tension steel in x-direction;
Asx_ten = Asx1e = 2513 mm /m
Area of tension steel in y-direction;
Asy_ten = Asy1e = 2094 mm /m
2
2
Column perimeter;
uc = 2 × (lx + ly) = 1600 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.417 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4168 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 9601 mm2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.847 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.312 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 5452 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 12559 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.847 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.003 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 382 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 6736 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 15516 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
15

16. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
App'd by
Date
2
vc = 0.847 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.812 N/mm
v < vc no shear reinforcement required
Internal central column B3
Design shear transferred to column;
Vt = (1.05 × Spanx) × Spany × Nult = 969 kN
Design effective shear transferred to column;
Veff = 1.15 × Vt = 1114 kN
Area of tension steel in x-direction;
Asx_ten = Asx1i = 2094 mm /m
2
2
Area of tension steel in y-direction;
Asy_ten = Asye = 1570 mm /m
Column perimeter;
uc = 2 × (lx + ly) = 1600 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.254 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4168 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 7636 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.249 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 5452 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 9988 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 0.955 N/mm
Shear reinforcement required at perimeter;
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 418 mm
vc < v <= 1.6 × vc
2
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 6736 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
16

17. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
As_ten = 12340 mm
App'd by
Date
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.773 N/mm
v < vc no shear reinforcement required
Internal edge column C1
Design shear transferred to column;
Vt = Spanx × ((0.45 × Spany) + ey) × Nult = 431 kN
Design effective shear transferred to column;
Veff = 1.4 × Vt = 604 kN
Area of tension steel in x-direction;
Asx_ten = Asxe = 1570 mm /m
Area of tension steel in y-direction;
Asy_ten = Asy_edge = 1148 mm /m
Column perimeter;
uc = lx + (2 × ly1) = 900 mm
2
2
(Library item: Flat slab shear map C1)
d = h – c - φp = 214 mm
Average effective depth of reinforcement;
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.135 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2184 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 2989 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 1.292 N/mm
Shear reinforcement required at perimeter;
Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 945 mm
1.6 × vc < v <= 2 × vc
2
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 2826 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 3862 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 0.998 N/mm
Shear reinforcement required at perimeter;
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 365 mm
vc < v <= 1.6 × vc
2
Shear reinforcement at a perimeter of 3.00d - (642 mm)
17

18. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 3468 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 4734 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.712 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.814 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 159 mm
Shear reinforcement at a perimeter of 3.75d - (803 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4110 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 5607 mm2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.711 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.686 N/mm
v < vc no shear reinforcement required
Internal central column C2
Design shear transferred to column;
Vt = Spanx × (1.05 × Spany) × Nult = 969 kN
Design effective shear transferred to column;
Veff = 1.15 × Vt = 1114 kN
Area of tension steel in x-direction;
Asx_ten = Asxe = 1570 mm /m
Area of tension steel in y-direction;
Asy_ten = Asy1i = 2094 mm /m
2
2
Column perimeter;
uc = 2 × (lx + ly) = 1600 mm
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.254 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4168 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 7636 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.249 N/mm
18

19. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
App'd by
Date
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 872 mm
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 5452 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 9988 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
2
Nominal design shear stress at perimeter;
v = Veff / (u × d) = 0.955 N/mm
Shear reinforcement required at perimeter;
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 418 mm
vc < v <= 1.6 × vc
2
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 6736 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 12340 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.785 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.773 N/mm
v < vc no shear reinforcement required
Internal column C3
Design shear transferred to column;
Vt = Spanx × Spany × Nult = 923 kN
Design effective shear transferred to column;
Veff = 1.15 × Vt = 1061 kN
Area of tension steel in x-direction;
Asx_ten = Asxi = 1570 mm /m
Area of tension steel in y-direction;
Asy_ten = Asyi = 1570 mm /m
Column perimeter;
uc = 2 × (lx + ly) = 1600 mm
2
2
Average effective depth of reinforcement;
d = h – c - φp = 214 mm
Maximum allowable shear stress;
vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm
Design shear stress at column perimeter;
v0 = Veff / (uc × d) = 3.099 N/mm
2
2
PASS - Maximum concrete shear stress not exceeded at column perimeter
Shear reinforcement at a perimeter of 1.50d - (321 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 4168 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 6544 mm
2
19

20. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Date
Chk'd by
Date
18/01/2014
App'd by
Date
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.746 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 1.190 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 834 mm
Shear reinforcement at a perimeter of 2.25d - (482 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 5452 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 8560 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.746 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.910 N/mm
vc < v <= 1.6 × vc
Shear reinforcement required at perimeter;
2
Asv_req = (v - vc) × u × d / (0.95 × fyv) = 403 mm
Shear reinforcement at a perimeter of 3.00d - (642 mm)
Length of shear perimeter;
u = uc + (2 × (kx × ky) × k × d) = 6736 mm
Area of tension steel at shear perimeter;
As_ten = (ky × (px + (kx × k × d)) × Asy_ten) + (kx × (py + (ky × k × d)) ×
Asx_ten)
As_ten = 10576 mm
2
Design concrete shear stress;
1/3
1/3
1/4
vc=(min(fcu,40)/25) ×0.79×min(100×As_ten/(u×d),3) ×max(400/d,1) /1.25
2
vc = 0.746 N/mm
Nominal design shear stress at perimeter;
2
v = Veff / (u × d) = 0.736 N/mm
v < vc no shear reinforcement required
CURTAILMENT OF REINFORCEMENT
Internal column
Radius of circular yield line;
1/2
r = (lx × ly / π)
1/3
× (1.05 × Spanx × 1.05 × Spany / (lx × ly))
= 1601
mm
Minimum curtailment length in x-direction;
lint_x = Max(r + 12 × D, 0.25 × Spanx) = 1841 mm
Minimum curtailment length in y-direction;
lint_y = Max(r + 12 × D, 0.25 × Spany) = 1841 mm
Corner column
Radius of yield line;
ly))
r = (lx1 × ly / π)
1/2
× ((0.45 × Spanx + ex) × (0.45 × Spany + ey)/ (lx1 ×
1/3
r = 863 mm
Minimum curtailment length in x-direction;
lcorner_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm
20

21. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Calc. by
Dr. C. Sachpazis
Minimum curtailment length in y-direction;
Date
Chk'd by
Date
18/01/2014
App'd by
Date
lcorner_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm
Edge columns
Radius of yield line in x-direction;
ly))
r = (lx1 × ly / π)
1/2
× ((0.45 × Spanx + ex) × (1.05 × Spany) / (lx1 ×
1/3
r = 1130 mm
Minimum curtailment length in x-direction;
ledge_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm
Radius of yield line in y-direction;
r = (lx × ly1 / π)
1/2
× ((0.45 × Spany + ey) × (1.05 × Spanx) / (lx ×
1/3
ly1))
r = 1130 mm
Minimum curtailment length in y-direction;
ledge_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm
21

22. Job Ref.
Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Date
Calc. by
Dr. C. Sachpazis
Chk'd by
Date
18/01/2014
B
A
ex
1
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
ey
Span x
n
l y1
e
s
n
Date
C
Span x
l x1
App'd by
f
s
r
0.2 x Spany
s
r
Span y
c
a
b
q
e
x
2
j
j
Span y
x
f
ly
0.5 x Spany
p
l
q
d
p
g
h
3
k
m
k
q
0.2 x Spanx
lx
0.5 x Spanx
When the effective span in the x direction, Lx, is greater than the effective span in the y direction, Ly, the
reinforcement in the outer layer is assumed to be that in the x direction otherwise it is assumed to be that in the y
direction.
22

23. Flat Slab Analysis & Design, In accordance with
BS8110:PART 1:1997
Project:
Section
Civil & Geotechnical Engineering Consulting Company for
Structural Engineering, Soil Mechanics, Rock Mechanics,
Foundation Engineering & Retaining Structures.
Tel.: (+30) 210 5238127, 210 5711263 - Fax.:+30 210 5711461 Mobile: (+30) 6936425722 & (+44) 7585939944, costas@sachpazis.info
Sheet no./rev. 1
Civil & Geotechnical Engineering
GEODOMISI Ltd. - Dr. Costas Sachpazis
Date
Calc. by
Dr. C. Sachpazis
Job Ref.
Chk'd by
18/01/2014
Date
App'd by
Date
REINFORCEMENT KEY
2
a = 20 dia bars @ 150 centres - (2094 mm /m);
2
b = 16 dia bars @ 200 centres - (1005 mm /m)
2
c = 20 dia bars @ 200 centres - (1570 mm /m);
2
d = 16 dia bars @ 200 centres - (1005 mm /m)
2
e = 20 dia bars @ 125 centres - (2513 mm /m);
2
f = 20 dia bars @ 200 centres - (1570 mm /m)
2
g = 20 dia bars @ 150 centres - (2094 mm /m);
2
h = 20 dia bars @ 200 centres - (1570 mm /m)
2
j = 20 dia bars @ 150 centres - (2094 mm /m);
2
k = 20 dia bars @ 200 centres - (1570 mm /m)
2
l = 20 dia bars @ 150 centres - (2094 mm /m);
2
m = 20 dia bars @ 200 centres - (1570 mm /m)
2
n = 16 dia bars @ 150 centres - (1340 mm /m)
2
p = 16 dia bars @ 175 centres - (1148 mm /m);
2
q = 16 dia bars @ 150 centres - (1340 mm /m)
2
r = 16 dia bars @ 175 centres - (1148 mm /m);
2
s = 16 dia bars @ 200 centres - (1005 mm /m)
2
Distribution bars = 12 dia bars @ 300 centres - (377 mm /m)
Shear reinforcement is required - Refer to output above for details.
23