Sachpazis: Flat slab design to bs8110 part 1-1997

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

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by


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

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by


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


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

Lever arm;

2

K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2

K = m / (d × fcu) = 0.036
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 191.9 mm
2


As_des = m / (z × fy / γm) = 617 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1005 mm /m

Check deflection
2




k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.601


0.9 × 26 × k1 = 37.469


L / d = 33.663

2

2

2

HOGGING MOMENTS – INTERNAL STRIP
Penultimate column B3
Consider the reinforcement concentrated in half width strip over the support
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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 172.5 mm

2

As_des = m’ / (z × fy / γm) = 1996 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 2094 mm /m
PASS - Support reinforcement is OK
3

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by


Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

Internal column C3
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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 182.1 mm
2


As_des = m’ / (z × fy / γm) = 1300 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1571 mm /m

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


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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 170.5 mm

2

As_des = m’ / (z × fy / γm) = 2134 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 2513 mm /m

Internal column C1, C2
Width of span;

B = 7200 mm

Edge distance;

e = 125 mm


d’ = 200 mm

Support moment;

m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 108.810

kNm/m
4

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 180.9 mm
2


As_des = m’ / (z × fy / γm) = 1383 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1571 mm /m

Corner column A1

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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 189.3 mm
2


As_des = m’ / (z × fy / γm) = 1211 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1340 mm /m

Edge column A2, A3

d’ = 202 mm


S = Spanx × (Spany / 2 + ey) × Nult = 477 kN


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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 188.8 mm

2

As_des = m’ / (z × fy / γm) = 956 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 1149 mm /m
5

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

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.

2

As_req = Asx1 / 2 = 1047 mm /m

Provide 16 dia bars @ 150 centres - 'U' bars with 1600 mm long legs

2

2

As_prov = π × D / (4 × s) = 1340 mm /m
PASS - Edge reinforcement is OK

Distribution reinforcement

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


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

Lever arm;

2

K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2

K = m / (d × fcu) = 0.044
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 208.6 mm
2


As_des = m / (z × fy / γm) = 825 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1571 mm /m

Check deflection
2




k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.579


0.9 × 26 × k1 = 36.942


L / d = 31.818

2

2

2

Internal bay 2-3
Effective span;

L = 6800 mm
6

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by


Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date


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

Lever arm;

2

K’ = 0.402 × (βb – 0.4) – 0.18 × (βb – 0.4) = 0.176
2

K = m / (d × fcu) = 0.030
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d = 210.9 mm
2


As_des = m / (z × fy / γm) = 561 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1005 mm /m

Check deflection
2




k1 = min(0.55+(477N/mm -fs)/(120×(0.9N/mm +(m/d ))),2) = 1.798


0.9 × 26 × k1 = 42.062


L / d = 30.631

2

2

2

HOGGING MOMENTS – INTERNAL STRIP
Penultimate column C2

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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 195.7 mm

2

As_des = m’ / (z × fy / γm) = 1758 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 2094 mm /m

Internal column C3

d’ = 220 mm
7

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by


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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 204.0 mm

2

As_des = m’ / (z × fy / γm) = 1160 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 1571 mm /m

HOGGING MOMENTS – EXTERNAL STRIP
Penultimate column A2, B2
Width of span;

B = 7200 mm

Edge distance;

e = 125 mm


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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 194.1 mm

2

As_des = m’ / (z × fy / γm) = 1875 mm /m
2


As_min = 0.0013 × h = 325 mm /m



2


2

2

As_prov = π × D / (4 × s) = 2094 mm /m

Internal column A3, B3
Width of span;
B = 7200 mm
Edge distance;

e = 125 mm


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

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 203.0 mm
2


As_des = m’ / (z × fy / γm) = 1233 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1571 mm /m

Edge column B1, C1

d’ = 222 mm


S = (Spanx / 2 + ex) × Spany × Nult = 477 kN


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
z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 210.1 mm
2


As_des = m’ / (z × fy / γm) = 859 mm /m


As_min = 0.0013 × h = 325 mm /m



2

2


2

2

As_prov = π × D / (4 × s) = 1149 mm /m

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.

2

As_req = Asy1 / 2 = 785 mm /m

Provide 16 dia bars @ 200 centres - 'U' bars with 1600 mm long legs

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

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



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


u = uc + (2 × (kx × ky) × k × d) = 1613 mm



Asx_ten)
As_ten = 2161 mm

2

1/3

1/3

1/4

2

vc = 0.707 N/mm

2

v = Veff / (u × d) = 0.730 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 1934 mm



Asx_ten)
As_ten = 2592 mm

2

1/3

1/3

1/4

2

vc = 0.707 N/mm

2

v = Veff / (u × d) = 0.609 N/mm

v < vc no shear reinforcement required
Penultimate edge column A2

Vt = ((0.45 × Spanx) + ex) × (1.05 × Spany) × Nult = 453 kN


Veff = 1.4 × Vt = 634 kN


Asx_ten = Asx_edge = 1148 mm /m

2

10

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Date

Calc. by

Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

App'd by

Date

2


Asy_ten = Asy1e = 2094 mm /m

Column perimeter;

uc = (2 × lx1)+ ly = 900 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.292 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 2184 mm



Asx_ten)
As_ten = 3588 mm

2

1/3

1/3

1/4

2

vc = 0.757 N/mm

2

v = Veff / (u × d) = 1.356 N/mm
1.6 × vc < v <= 2 × vc


Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 947 mm

2


u = uc + (2 × (kx × ky) × k × d) = 2826 mm



Asx_ten)
As_ten = 4628 mm

2

1/3

1/3

1/4

2

vc = 0.756 N/mm

2

v = Veff / (u × d) = 1.048 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 3468 mm



Asx_ten)
As_ten = 5669 mm

2

1/3

1/3

1/4

2

vc = 0.756 N/mm

2


v = Veff / (u × d) = 0.854 N/mm



vc < v <= 1.6 × vc
2

11

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Date

Calc. by


Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

App'd by

Date


u = uc + (2 × (kx × ky) × k × d) = 4110 mm



Asx_ten)
As_ten = 6710 mm

2

1/3

1/3

1/4

2

vc = 0.755 N/mm

2

v = Veff / (u × d) = 0.721 N/mm

Internal edge column A3

Vt = ((0.45 × Spanx) + ex) × Spany × Nult = 431 kN


Veff = 1.4 × Vt = 604 kN


Asx_ten = Asx_edge = 1148 mm /m

2

2


Asy_ten = Asye = 1570 mm /m

Column perimeter;

uc = (2 × lx1)+ ly = 900 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.135 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 2184 mm



Asx_ten)
As_ten = 2989 mm

2

1/3

1/3

1/4

2

vc = 0.712 N/mm

2


v = Veff / (u × d) = 1.292 N/mm



1.6 × vc < v <= 2 × vc
2


u = uc + (2 × (kx × ky) × k × d) = 2826 mm



Asx_ten)
As_ten = 3862 mm

2

1/3

1/3

1/4

2

vc = 0.712 N/mm

12

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014


App'd by

Date

2

v = Veff / (u × d) = 0.998 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 3468 mm



Asx_ten)
As_ten = 4734 mm2
1/3

1/3

1/4

2

vc = 0.712 N/mm

2

v = Veff / (u × d) = 0.814 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 4110 mm



Asx_ten)
As_ten = 5607 mm

2

1/3

1/3

1/4

2

vc = 0.711 N/mm

2

v = Veff / (u × d) = 0.686 N/mm

Penultimate edge column B1

Vt = (1.05 × Spanx) × ((0.45 × Spany) + ey) × Nult = 453 kN


Veff = 1.4 × Vt = 634 kN


Asx_ten = Asx1e = 2513 mm /m


Asy_ten = Asy_edge = 1148 mm /m

2

2

Column perimeter;

uc = lx + (2 × ly1) = 900 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.292 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 2184 mm



Asx_ten)
As_ten = 4066 mm

2

13

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

1/3

1/3

1/4

2

vc = 0.789 N/mm

2

v = Veff / (u × d) = 1.356 N/mm
1.6 × vc < v <= 2 × vc



2


u = uc + (2 × (kx × ky) × k × d) = 2826 mm



Asx_ten)
As_ten = 5241 mm

2

1/3

1/3

1/4

2

vc = 0.788 N/mm

2

v = Veff / (u × d) = 1.048 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 3468 mm



Asx_ten)
As_ten = 6416 mm

2

1/3

1/3

1/4

2

vc = 0.788 N/mm

2

v = Veff / (u × d) = 0.854 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 4110 mm



Asx_ten)
As_ten = 7592 mm

2

1/3

1/3

1/4

2

vc = 0.787 N/mm

2

v = Veff / (u × d) = 0.721 N/mm


14

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

Penultimate central column B2

Vt = (1.05 × Spanx) × (1.05 × Spany) × Nult = 1017 kN


Veff = 1.15 × Vt = 1170 kN


Asx_ten = Asx1e = 2513 mm /m


Asy_ten = Asy1e = 2094 mm /m

2
2

Column perimeter;

uc = 2 × (lx + ly) = 1600 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.417 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 4168 mm



Asx_ten)
As_ten = 9601 mm2
1/3

1/3

1/4

2

vc = 0.847 N/mm

2

v = Veff / (u × d) = 1.312 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 5452 mm



Asx_ten)
As_ten = 12559 mm

2

1/3

1/3

1/4

2

vc = 0.847 N/mm

2

v = Veff / (u × d) = 1.003 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 6736 mm



Asx_ten)
As_ten = 15516 mm

2

1/3

1/3

1/4


15

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Date

Calc. by

Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

App'd by

Date

2

vc = 0.847 N/mm

2

v = Veff / (u × d) = 0.812 N/mm

Internal central column B3

Vt = (1.05 × Spanx) × Spany × Nult = 969 kN


Veff = 1.15 × Vt = 1114 kN


Asx_ten = Asx1i = 2094 mm /m

2

2


Asy_ten = Asye = 1570 mm /m

Column perimeter;

uc = 2 × (lx + ly) = 1600 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.254 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 4168 mm



Asx_ten)
As_ten = 7636 mm

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2

v = Veff / (u × d) = 1.249 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 5452 mm



Asx_ten)
As_ten = 9988 mm

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2


v = Veff / (u × d) = 0.955 N/mm



vc < v <= 1.6 × vc
2


u = uc + (2 × (kx × ky) × k × d) = 6736 mm



Asx_ten)
16

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Date

Calc. by

Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

As_ten = 12340 mm

App'd by

Date

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2

v = Veff / (u × d) = 0.773 N/mm

Internal edge column C1

Vt = Spanx × ((0.45 × Spany) + ey) × Nult = 431 kN


Veff = 1.4 × Vt = 604 kN


Asx_ten = Asxe = 1570 mm /m


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



vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.135 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 2184 mm



Asx_ten)
As_ten = 2989 mm

2

1/3

1/3

1/4

2

vc = 0.712 N/mm

2


v = Veff / (u × d) = 1.292 N/mm



1.6 × vc < v <= 2 × vc
2


u = uc + (2 × (kx × ky) × k × d) = 2826 mm



Asx_ten)
As_ten = 3862 mm

2

1/3

1/3

1/4

2

vc = 0.712 N/mm

2


v = Veff / (u × d) = 0.998 N/mm



vc < v <= 1.6 × vc
2

17

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date


u = uc + (2 × (kx × ky) × k × d) = 3468 mm



Asx_ten)
As_ten = 4734 mm

2

1/3

1/3

1/4

2

vc = 0.712 N/mm

2

v = Veff / (u × d) = 0.814 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 4110 mm



Asx_ten)
As_ten = 5607 mm2
1/3

1/3

1/4

2

vc = 0.711 N/mm

2

v = Veff / (u × d) = 0.686 N/mm

Internal central column C2

Vt = Spanx × (1.05 × Spany) × Nult = 969 kN


Veff = 1.15 × Vt = 1114 kN


Asx_ten = Asxe = 1570 mm /m


Asy_ten = Asy1i = 2094 mm /m

2

2

Column perimeter;

uc = 2 × (lx + ly) = 1600 mm


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.254 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 4168 mm



Asx_ten)
As_ten = 7636 mm

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2

v = Veff / (u × d) = 1.249 N/mm

18

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Date

Calc. by

Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

App'd by

Date

vc < v <= 1.6 × vc

2



u = uc + (2 × (kx × ky) × k × d) = 5452 mm



Asx_ten)
As_ten = 9988 mm

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2


v = Veff / (u × d) = 0.955 N/mm



vc < v <= 1.6 × vc
2


u = uc + (2 × (kx × ky) × k × d) = 6736 mm



Asx_ten)
As_ten = 12340 mm

2

1/3

1/3

1/4

2

vc = 0.785 N/mm

2

v = Veff / (u × d) = 0.773 N/mm

Internal column C3

Vt = Spanx × Spany × Nult = 923 kN


Veff = 1.15 × Vt = 1061 kN


Asx_ten = Asxi = 1570 mm /m


Asy_ten = Asyi = 1570 mm /m

Column perimeter;

uc = 2 × (lx + ly) = 1600 mm

2
2


d = h – c - φp = 214 mm


vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm


v0 = Veff / (uc × d) = 3.099 N/mm

2

2


u = uc + (2 × (kx × ky) × k × d) = 4168 mm



Asx_ten)
As_ten = 6544 mm

2

19

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis

Date

Chk'd by
Date

18/01/2014

App'd by

Date

1/3

1/3

1/4

2

vc = 0.746 N/mm

2

v = Veff / (u × d) = 1.190 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 5452 mm



Asx_ten)
As_ten = 8560 mm

2

1/3

1/3

1/4

2

vc = 0.746 N/mm

2

v = Veff / (u × d) = 0.910 N/mm
vc < v <= 1.6 × vc


2



u = uc + (2 × (kx × ky) × k × d) = 6736 mm



Asx_ten)
As_ten = 10576 mm

2

1/3

1/3

1/4

2

vc = 0.746 N/mm

2

v = Veff / (u × d) = 0.736 N/mm

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

lcorner_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm
20

Job Ref.

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



Calc. by

Dr. C. Sachpazis


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

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

ledge_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm

21

Job Ref.

BS8110:PART 1:1997
Project:

Section

Date

Calc. by

Dr. C. Sachpazis

Chk'd by
Date

18/01/2014

B

A
ex
1

Sheet no./rev. 1



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

BS8110:PART 1:1997
Project:

Section

Sheet no./rev. 1



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

Sachpazis: Flat slab design to bs8110 part 1-1997

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Sachpazis: Flat slab design to bs8110 part 1-1997