Download - Sachpazis: Flat slab design to bs8110 part 1-1997

GEODOMISI Ltd. - Dr. Costas Sachpazis 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, [email protected]

Project: Flat Slab Analysis & Design, In accordance with BS8110:PART 1:1997

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Section

Civil & Geotechnical Engineering Sheet no./rev. 1

Calc. by

Dr. C. Sachpazis

Date

18/01/2014

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1

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

Slab details

Depth of slab; h = 250 mm

Spanx Spanx

lx

ly

h

ly1lx1

1

2

3

A B C

Span

y

Span

y

lx

ly

ey

ex






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Date

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2

Characteristic strength of concrete; fcu = 35 N/mm2

Characteristic strength of reinforcement; fy = 500 N/mm2

Characteristic strength of shear reinforcement; fyv = 500 N/mm2

Material safety factor; γm = 1.15

Cover to bottom reinforcement; c = 20 mm

Cover to top reinforcement; c’ = 20 mm

Loading details

Characteristic dead load; Gk = 7.000 kN/m2

Characteristic imposed load; Qk = 5.000 kN/m2

Dead load factor; γG = 1.4

Imposed load factor; γQ = 1.6

Total ultimate load; Nult = (Gk × γG) + (Qk × γQ) = 17.800 kN/m2

Moment redistribution ratio; βb = 1.0

Ratio of support moments to span moments; i = 1.0

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 × L2) / (2 × (1 + √(1 + i))2) = 74.823 kNm/m

Support moment; m’ = i × m = 74.823 kNm/m

Design reinforcement

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

K = m / (d2 × 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; As_des = m / (z × fy / γm) = 919 mm2/m

Minimum area of reinforcement required; As_min = 0.0013 × h = 325 mm2/m

Area of reinforcement required; As_req = max(As_des, As_min) = 919 mm2/m

Provide 20 dia bars @ 150 centres

Area of reinforcement provided; As_prov = π × D2 / (4 × s) = 2094 mm2

/m

PASS - Span reinforcement is OK

Check deflection

Design service stress; fs = 2 × fy × As_req / (3 × As_prov × βb) = 146 N/mm2

Modification factor; k1 = min(0.55+(477N/mm2-fs)/(120×(0.9N/mm2

+(m/d2))),2) = 1.545

Allowable span to depth ratio; 0.9 × 26 × k1 = 36.151

Actual span to depth ratio; L / d = 35.000






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Dr. C. Sachpazis

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PASS - Span to depth ratio is OK

Internal bay B-C



Midspan moment; m = (Nult × L2) / (2 × (√(1 + i) + √(1 + i))2) = 51.442 kNm/m



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

K = m / (d2 × fcu) = 0.036


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






/m


Check deflection



+(m/d2))),2) = 1.601




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)2 = 0.176

K = m’ / (d’2 × fcu) = 0.107


z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 172.5 mm

Area of reinforcement required; As_des = m’ / (z × fy / γm) = 1996 mm2/m





/m

PASS - Support reinforcement is OK






Job Ref.

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Date

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Internal column C3




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

K = m’ / (d’2 × fcu) = 0.073


z = min((0.5 + √(0.25 – (K / 0.9))), 0.95) × d’ = 182.1 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


Support moment; m’ = m × i ×(e + B + B / 2) / ((0.5 × B) + (0.2 × B) + e) = 158.265

kNm/m

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

K = m’ / (d’2 × fcu) = 0.113


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






/m


Internal column C1, C2






kNm/m






Job Ref.

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Calc. by

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Date

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5

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

K = m’ / (d’2 × fcu) = 0.078


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






/m


Corner column A1


Total load on column; S = ((Spanx / 2) + ex) × ((Spany / 2) + ey) × Nult = 247 kN

Area of column head; A = lx × ly1 = 0.100 m2

Support moment; m’ = S × (1 – (Nult × A / S)1/3) / 2 = 99.639 kNm/m

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

K = m’ / (d’2 × fcu) = 0.067


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






/m


Edge column A2, A3


Total load on column; S = Spanx × (Spany / 2 + ey) × Nult = 477 kN

Area of column head; A = lx1 × ly = 0.100 m2

Support moment; m’ = S × (1 – (Nult × A / S)1/3) / 5.14 = 78.476 kNm/m

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

K = m’ / (d’2 × fcu) = 0.055


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






/m






Job Ref.

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Calc. by

Dr. C. Sachpazis

Date

18/01/2014

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6


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; As_req = Asx1 / 2 = 1047 mm2/m

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


/m

PASS - Edge reinforcement is OK

Distribution reinforcement



/m

DESIGN SLAB IN THE Y-DIRECTION

SAGGING MOMENTS

End bay 1-2



Midspan moment; m = (Nult × L2) / (2 × (1 + √(1 + i))2) = 74.823 kNm/m



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

K = m / (d2 × fcu) = 0.044


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






/m


Check deflection



+(m/d2))),2) = 1.579




Internal bay 2-3







Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

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Midspan moment; m = (Nult × L2) / (2 × (√(1 + i) + √(1 + i))2) = 51.442 kNm/m



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

K = m / (d2 × fcu) = 0.030


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






/m


Check deflection



+(m/d2))),2) = 1.798




HOGGING MOMENTS – INTERNAL STRIP

Penultimate column C2




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

K = m’ / (d’2 × fcu) = 0.088


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






/m


Internal column C3








Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

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

K = m’ / (d’2 × fcu) = 0.061


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






/m


HOGGING MOMENTS – EXTERNAL STRIP

Penultimate column A2, B2






kNm/m

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

K = m’ / (d’2 × fcu) = 0.093


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






/m


Internal column A3, B3






kNm/m

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

K = m’ / (d’2 × fcu) = 0.064






Job Ref.

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Dr. C. Sachpazis

Date

18/01/2014

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






/m


Edge column B1, C1


Total load on column; S = (Spanx / 2 + ex) × Spany × Nult = 477 kN

Area of column head; A = ly1 × lx = 0.100 m2

Support moment; m’ = S × (1 – (Nult × A / S)1/3) / 5.14 = 78.476 kNm/m

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

K = m’ / (d’2 × fcu) = 0.045


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

Area of reinforcement required; As_req = Asy1 / 2 = 785 mm2/m

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


/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 mm2/m

Area of tension steel in y-direction; Asy_ten = Ascorner = 1340 mm2/m

Column perimeter; uc = lx1 + ly = 650 mm

Average effective depth of reinforcement; d = h – c - φp = 214 mm






Job Ref.

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Calc. by

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Date

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10

Maximum allowable shear stress; vmax = min(0.8 × √(fcu), 5) = 4.733 N/mm2

Design shear stress at column perimeter; v0 = Veff / (uc × d) = 1.811 N/mm2

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 mm2

Design concrete shear stress;

vc=(min(fcu,40)/25)1/3×0.79×min(100×As_ten/(u×d),3)1/3×max(400/d,1)1/4/1.25

vc = 0.707 N/mm2

Nominal design shear stress at perimeter; v = Veff / (u × d) = 0.911 N/mm2

vc < v <= 1.6 ×××× vc

Shear reinforcement required at perimeter; Asv_req = (v - vc) × u × d / (0.95 × fyv) = 119 mm2




Asx_ten)

As_ten = 2161 mm2



vc = 0.707 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 2592 mm2



vc = 0.707 N/mm2


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


Area of tension steel in x-direction; Asx_ten = Asx_edge = 1148 mm2/m






Job Ref.

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Calc. by

Dr. C. Sachpazis

Date

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11

Area of tension steel in y-direction; Asy_ten = Asy1e = 2094 mm2/m

Column perimeter; uc = (2 × lx1)+ ly = 900 mm








Asx_ten)

As_ten = 3588 mm2



vc = 0.757 N/mm2


1.6 ×××× vc < v <= 2 ×××× vc

Shear reinforcement required at perimeter; Asv_req = 5 × ((0.7 × v) - vc) × u × d / (0.95 × fyv) = 947 mm2




Asx_ten)

As_ten = 4628 mm2



vc = 0.756 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 5669 mm2



vc = 0.756 N/mm2


vc < v <= 1.6 ×××× vc







Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

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12




Asx_ten)

As_ten = 6710 mm2



vc = 0.755 N/mm2



Internal edge column A3

Design shear transferred to column; Vt = ((0.45 × Spanx) + ex) × Spany × Nult = 431 kN


Area of tension steel in x-direction; Asx_ten = Asx_edge = 1148 mm2/m

Area of tension steel in y-direction; Asy_ten = Asye = 1570 mm2/m

Column perimeter; uc = (2 × lx1)+ ly = 900 mm








Asx_ten)

As_ten = 2989 mm2



vc = 0.712 N/mm2


1.6 ×××× vc < v <= 2 ×××× vc





Asx_ten)

As_ten = 3862 mm2



vc = 0.712 N/mm2






Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

13


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 4734 mm2



vc = 0.712 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 5607 mm2



vc = 0.711 N/mm2



Penultimate edge column B1

Design shear transferred to column; Vt = (1.05 × Spanx) × ((0.45 × Spany) + ey) × Nult = 453 kN


Area of tension steel in x-direction; Asx_ten = Asx1e = 2513 mm2/m

Area of tension steel in y-direction; Asy_ten = Asy_edge = 1148 mm2/m

Column perimeter; uc = lx + (2 × ly1) = 900 mm








Asx_ten)

As_ten = 4066 mm2






Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

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14



vc = 0.789 N/mm2


1.6 ×××× vc < v <= 2 ×××× vc





Asx_ten)

As_ten = 5241 mm2



vc = 0.788 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 6416 mm2



vc = 0.788 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 7592 mm2



vc = 0.787 N/mm2








Job Ref.

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Date

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Penultimate central column B2

Design shear transferred to column; Vt = (1.05 × Spanx) × (1.05 × Spany) × Nult = 1017 kN


Area of tension steel in x-direction; Asx_ten = Asx1e = 2513 mm2/m

Area of tension steel in y-direction; Asy_ten = Asy1e = 2094 mm2/m

Column perimeter; uc = 2 × (lx + ly) = 1600 mm








Asx_ten)

As_ten = 9601 mm2



vc = 0.847 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 12559 mm2



vc = 0.847 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 15516 mm2








Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

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vc = 0.847 N/mm2



Internal central column B3

Design shear transferred to column; Vt = (1.05 × Spanx) × Spany × Nult = 969 kN


Area of tension steel in x-direction; Asx_ten = Asx1i = 2094 mm2/m

Area of tension steel in y-direction; Asy_ten = Asye = 1570 mm2/m









Asx_ten)

As_ten = 7636 mm2



vc = 0.785 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 9988 mm2



vc = 0.785 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)






Job Ref.

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Date

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As_ten = 12340 mm2



vc = 0.785 N/mm2



Internal edge column C1

Design shear transferred to column; Vt = Spanx × ((0.45 × Spany) + ey) × Nult = 431 kN


Area of tension steel in x-direction; Asx_ten = Asxe = 1570 mm2/m

Area of tension steel in y-direction; Asy_ten = Asy_edge = 1148 mm2/m

Column perimeter; uc = lx + (2 × ly1) = 900 mm

(Library item: Flat slab shear map C1) Average effective depth of reinforcement; d = h – c - φp = 214 mm







Asx_ten)

As_ten = 2989 mm2



vc = 0.712 N/mm2


1.6 ×××× vc < v <= 2 ×××× vc





Asx_ten)

As_ten = 3862 mm2



vc = 0.712 N/mm2


vc < v <= 1.6 ×××× vc








Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

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18



Asx_ten)

As_ten = 4734 mm2



vc = 0.712 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 5607 mm2



vc = 0.711 N/mm2



Internal central column C2

Design shear transferred to column; Vt = Spanx × (1.05 × Spany) × Nult = 969 kN


Area of tension steel in x-direction; Asx_ten = Asxe = 1570 mm2/m

Area of tension steel in y-direction; Asy_ten = Asy1i = 2094 mm2/m









Asx_ten)

As_ten = 7636 mm2



vc = 0.785 N/mm2







Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

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Date App'd by Date

19

vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 9988 mm2



vc = 0.785 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 12340 mm2



vc = 0.785 N/mm2



Internal column C3

Design shear transferred to column; Vt = Spanx × Spany × Nult = 923 kN


Area of tension steel in x-direction; Asx_ten = Asxi = 1570 mm2/m

Area of tension steel in y-direction; Asy_ten = Asyi = 1570 mm2/m









Asx_ten)

As_ten = 6544 mm2






Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

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20



vc = 0.746 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 8560 mm2



vc = 0.746 N/mm2


vc < v <= 1.6 ×××× vc





Asx_ten)

As_ten = 10576 mm2



vc = 0.746 N/mm2



CURTAILMENT OF REINFORCEMENT

Internal column

Radius of circular yield line; r = (lx × ly / π)1/2 × (1.05 × Spanx × 1.05 × Spany / (lx × ly))

1/3 = 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; r = (lx1 × ly / π)1/2 × ((0.45 × Spanx + ex) × (0.45 × Spany + ey)/ (lx1 ×

ly))1/3

r = 863 mm

Minimum curtailment length in x-direction; lcorner_x = Max(r + 12 × D, 0.2 × Spanx) = 1440 mm






Job Ref.

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Minimum curtailment length in y-direction; lcorner_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm

Edge columns

Radius of yield line in x-direction; r = (lx1 × ly / π)1/2 × ((0.45 × Spanx + ex) × (1.05 × Spany) / (lx1 ×

ly))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 ×

ly1))1/3

r = 1130 mm

Minimum curtailment length in y-direction; ledge_y = Max(r + 12 × D, 0.2 × Spany) = 1440 mm






Job Ref.

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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.

Spanx

a b

c

d

s s s

q

q

q

e f

j

k

p

r

g h

l

m

j

k

e f

p

n

rn

x x

Spanx

Span

ySpan

y

0.5 x Spanx

lx

0.5 x Spany

l y

A B C

1

2

3

0.2 x Spanx

ex

0.2 x Spany

ey

lx1

ly1






Job Ref.

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Date

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23

REINFORCEMENT KEY

a = 20 dia bars @ 150 centres - (2094 mm2/m);

b = 16 dia bars @ 200 centres - (1005 mm2/m)

c = 20 dia bars @ 200 centres - (1570 mm2/m);

d = 16 dia bars @ 200 centres - (1005 mm2/m)

e = 20 dia bars @ 125 centres - (2513 mm2/m);

f = 20 dia bars @ 200 centres - (1570 mm2/m)

g = 20 dia bars @ 150 centres - (2094 mm2/m);

h = 20 dia bars @ 200 centres - (1570 mm2/m)

j = 20 dia bars @ 150 centres - (2094 mm2/m);

k = 20 dia bars @ 200 centres - (1570 mm2/m)

l = 20 dia bars @ 150 centres - (2094 mm2/m);

m = 20 dia bars @ 200 centres - (1570 mm2/m)

n = 16 dia bars @ 150 centres - (1340 mm2/m)

p = 16 dia bars @ 175 centres - (1148 mm2/m);

q = 16 dia bars @ 150 centres - (1340 mm2/m)

r = 16 dia bars @ 175 centres - (1148 mm2/m);

s = 16 dia bars @ 200 centres - (1005 mm2/m)

Distribution bars = 12 dia bars @ 300 centres - (377 mm2/m)

Shear reinforcement is required - Refer to output above for details.

Download - Sachpazis: Flat slab design to bs8110 part 1-1997

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