flat slab design to bs8110

1

GEODOMISI Ltd. - Dr. Costas SachpazisCivil & Geotechnical

Engineering Consulting Company for Structural

Engineering, Soil Mechanics, Rock Mechanics, Foundation

Engineering & Retaining Structures.

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

Job Ref.

Section

Civil & Geotechnical EngineeringSheet no./rev. 1

Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Spanx = 7200 mmSpany = 7200 mmlx = 400 mmly = 400 mmlx1 = 250 mmly1 = 250 mmex = lx1 / 2 = 125 mmey = ly 1 / 2 = 125 mmLx = Spanx - lx = 6800 mmLy = Spany - ly = 6800 mmLxi = Spanx - lx / 2 = 7000 mmLyi = Spany - ly / 2 = 7000 mm

h = 250 mm

FLAT SLAB DESIGN TO BS8110:PART 1:1997

Slab geometrySpan of slab in x-direction;Span of slab in y-direction;Column dimension in x-direction;Column dimension in y-direction;External column dimension in x-direction; External column dimension in y-direction; Edge dimension in x-direction;Edge dimension in y-direction;Effective span of internal bay in x direction; Effective span of internal bay in y direction; Effective span of end bay in x direction; Effective span of end bay in y direction;Slab details

Depth of slab;

2

SAGGING MOMENTS

End bay A-B

Effective span;

Depth of reinforcement;

Midspan moment;

Support moment;

Design reinforcementLever arm;

Area of reinforcement designed; Minimum

area of reinforcement required; Area of

reinforcement required;

Provide 20 dia bars @ 150 centres

Area of reinforcement provided;

Check deflection

Design service stress;

Modification factor;

Allowable span to depth ratio;

Actual span to depth ratio;





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Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

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Characteristic strength of concrete;

Characteristic strength of reinforcement;

Characteristic strength of shear reinforcement;

Material safety factor;

Cover to bottom reinforcement;

Cover to top reinforcement;

Loading details

Characteristic dead load;

Characteristic imposed load;

Dead load factor;

Imposed load factor;

Total ultimate load;

Moment redistribution ratio;

Ratio of support moments to span moments;

DESIGN SLAB IN THE X-DIRECTION

fcu = 35 N/mm2 fy =

500 N/mm2 fyv =

500 N/mm2 Ym =

1.15 c = 20 mm c'

= 20 mm

Gk = 7.000 kN/m2

Qk = 5.000 kN/m2

Yg = 1.4 Yq = 1.6

Nult = (Gk x yg) + (Qk x yq) = 17.800 kN/m2 Pb =

1.0 i = 1.0

L = 7000 mmd = 200 mmm = (Nult x L2) / (2 x (1 + V(1 + i))2) = 74.823

kNm/m m' = i x m = 74.823 kNm/m

K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176 K = m / (d2 x fcu) = 0.053

Compression reinforcement is not required

z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d = 187.3 mm As_des = m / (z x fy / Ym) = 919 mm2/m As_min = 0.0013 x h = 325 mm2/m As_req = max(As_des, As_min) = 919 mm2/m

As_prov = n x D2 / (4 x s) = 2094 mm2/mPASS - Span reinforcement is OK

fs = 2 x fy x As_ req / (3 x As_ prov x Pb) = 146 N/mm2 ki = min(0.55+(477N/mm2-fs)/(120x(0.9N/mm2+(m/d2))),2)

= 1.545 0.9 x 26 x ki = 36.151 L / d = 35.000

3





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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal bay B-C

Effective span;


Midspan moment;

Support moment;

Design reinforcement

Lever arm;






Check deflection





PASS - Span to depth ratio is OK

L = 6800 mm d = 202 mmm = (Nult x L2) / (2 x (V(1 + i) + V(1 + i))2) = 51.442

kNm/m m' = i x m = 51.442 kNm/m

K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176 K = m / (d2

x fcu) = 0.036


z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d = 191.9 mm As_des = m / (z x fy / ym) = 617 mm2/m As_min = 0.0013 x h = 325 mm2/m As_req = max(As_des, As_min) = 617 mm2/m


fs = 2 x fy x As_req / (3 x As_prov x Pb) = 204 N/mm2 ki = min(0.55+(477N/mm2-fs)/(120x(0.9N/mm2+(m/d2))),2) = 1.601 0.9 x 26 x ki = 37.469 L / d = 33.663


HOGGING MOMENTS - INTERNAL STRIP Penultimate column B3

Consider the reinforcement concentrated in half width strip over the support

Depth of reinforcement; d' = 200 mmSupport moment; m' = 2 x i x m = 149.646 kNm/mLever arm; K = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.107

Compression reinforcement is not required z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d' = 172.5 mm Area of reinforcement required; As_des

= m' / (z x fy / ym) = 1996 mm2/mMinimum area of reinforcement required; As_min = 0.0013 x h = 325 mm2/mArea of reinforcement required; As_req = max(As_des, As_min) = 1996 mm2/mProvide 20 dia bars @ 150 centresArea of reinforcement provided; As prov n x D2 / (4 x s) = 2094 mm2/m

PASS - Support reinforcement is OK

4





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Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal column C3

Consider the reinforcement concentrated in half width strip over the supportDepth of reinforcement; d' = 200 mmSupport moment; m' = 2 x i x m = 102.884 kNm/mLever arm; K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.073




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 mmEdge distance; e = 125 mmDepth of reinforcement; d' = 200 mmSupport moment; m' = m x i x(e + B + B / 2) / ((0.5 x B) + (0.2 x B) + e) = 158.265

kNm/mLever arm; K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.113




Internal column C1, C2


5

kNm/m

6





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Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Lever arm;

Area of reinforcement required;

Minimum area of reinforcement required;




Corner column A1


Total load on column;

Area of column head;

Support moment;Lever arm;






Edge column A2, A3




Support moment;

Lever arm;






K = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176 K = m' / (d'2 x fcu) = 0.078


z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d' = 180.9

mmAs_des = m' / (z x fy / ym) = 1383 mm2/mAs_min = 0.0013 x h = 325 mm2/mAs_req = max(As_des, As_min) = 1383 mm2/m

As_prov = n x D2 / (4 x s) = 1571 mm2/m


d' = 206 mmS = ((Spanx / 2) + ex) x ((Spany / 2) + ey) x Nult = 247 kN A = lx x lyi = 0.100 m2

m' = S x (1 - (Nult x A / S)1/3) / 2 = 99.639 kNm/m K = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176 K = m' / (d'2 x fcu) = 0.067

Compression reinforcement is not required z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d' = 189.3 mm As_des = m' / (z x fy / ym) = 1211 mm2/m As_min = 0.0013 x h = 325 mm2/m As_req = max(As_des, As_min) = 1211 mm2/m

As_prov = n x D2 / (4 x s) = 1340 mm2/mPASS - Support reinforcement is OK

d' = 202 mmS = Spanx x (Spany / 2 + ey) x Nult = 477 kN A = lxt

x ly = 0.100 m2

m' = S x (1 - (Nult x A / S)1/3) / 5.14 = 78.476

kNm/m K' = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2

= 0.176 K = m' / (d'2 x fcu) = 0.055


As_ prov = n x D2 / (4 x s) = 1149 mm2/m

7





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Job Ref.

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

L = 7000 mm

d = 220 mm

m = (Nuit x L2) / (2 x (1 + V(1 + i))2) = 74.823 kNm/m m' = i x

m = 74.823 kNm/m

K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176 K = m / (d2 x

fcu) = 0.044


z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d = 208.6 mm

As_des = m / (z x fy / ym) = 825 mm2/m As_min = 0.0013 x h =

325 mm2/m As_req = max(As_des, As_min) = 825 mm2/m

As_prov = n x D2 / (4 x s) = 1571 mm2/m

PASS - Span reinforcement is OK

fs = 2 x fy x As_req / (3 x As_prov x Pb) = 175 N/mm k1 =

min(0.55+(477N/mm2-fs)/(120x(0.9N/mm2+(m/d2))),2) = 1.579 0.9 x 26

x k = 36.942 L / d = 31.818


L = 6800 mm


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/mProvide 16 dia bars @ 150 centres - 'U' bars with 1600 mm long legsArea of reinforcement provided; As prov n x D2 / (4 x s) = 1340 mm2/m

PASS - Edge reinforcement is OK

Distribution reinforcement Provide 12 dia bars @ 300 centresArea of reinforcement provided; As prov n x D2 / (4 x s) = 377 mm2/m

DESIGN SLAB IN THE Y-DIRECTION

SAGGING MOMENTS

End bay 1-2

Effective span;Depth of reinforcement;Midspan moment;Support moment;Design reinforcementLever arm;

Area of reinforcement designed; Minimum area of reinforcement required; Area of reinforcement required;Provide 20 dia bars @ 200 centresArea of reinforcement provided;

Check deflectionDesign service stress;Modification factor;Allowable span to depth ratio;Actual span to depth ratio;

Internal bay 2-3Effective span;

8





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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date


Midspan moment;

Support moment;

Design reinforcement

Lever arm;






Check deflection





d = 222 mmm = (Nutt x L2) / (2 x (V(1 + i) + V(1 + i))2) = 51.442 kNm/m m' = i x m = 51.442 kNm/m

K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176 K = m / (d2

x fcu) = 0.030


z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d = 210.9 mm As_des = m / (z x fy / ym) = 561 mm2/m As_min = 0.0013 x h = 325 mm2/m As_req = max(As_des, As_min) = 561 mm /m


fs = 2 x fy x As_ req / (3 x As_ prov x Pb) = 186 N/mm2 ki = min(0.55+(477N/mm2-fs)/(120x(0.9N/mm2+(m/d2))),2) = 1.798 0.9 x 26 x ki = 42.062 L / d = 30.631


HOGGING MOMENTS - INTERNAL STRIP

Penultimate column C2

Consider the reinforcement concentrated in half width strip over the supportDepth of reinforcement; d' = 220 mmSupport moment; m' = 2 x i x m = 149.646 kNm/mLever arm; K = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.088



PASS - Support reinforcement is OKInternal column C3

Consider the reinforcement concentrated in half width strip over the support Depth of reinforcement; d' = 220 mm

9





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Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Support moment; m' = 2 x i x m = 102.884 kNm/mLever arm; K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.061




HOGGING MOMENTS - EXTERNAL STRIP

Penultimate column A2, B2


kNm/mLever arm; K = 0.402 x (pb - 0.4) - 0.18 x (pb - 0.4)2 = 0.176

K = m' / (d'2 x fcu) = 0.093


= m' / (z x fy / ym) = 1875 mm2/mMinimum area of reinforcement required; As_min = 0.0013 x h = 325 mm2/mArea of reinforcement required; As_req = max(As_des, As_min) = 1875 mm2/mProvide 20 dia bars @ 150 centresArea of reinforcement provided; As_prov = n x D2 / (4 x s) = 2094 mm2/m


Internal column A3, B3


kNm/mLever arm; K' = 0.402 x (Pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176

10

K = m' / (d'2 x fcu) = 0.064

11





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Job Ref.

Section

Civil & Geotechnical Engineering Sheet no./rev. 1Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Edge column B1, C1




Support moment;

Lever arm;






Corner column A1

Design shear transferred to column;

Design effective shear transferred to column;

Area of tension steel in x-direction;

Area of tension steel in y-direction;

Column perimeter;

Average effective depth of reinforcement;


z = min((0.5 + V(0.25 - (K / 0.9))), 0.95) x d' = 203.0 mm Area of reinforcement required; As_des = m' / (z x fy / ym) = 1233 mm2/mMinimum area of reinforcement required; As_min = 0.0013 x h = 325 mm2/mArea of reinforcement required; As_req = max(As_des, As_min) = 1233 mm2/mProvide 20 dia bars @ 200 centresArea of reinforcement provided; As prov n x D2 / (4 x s) = 1571 mm2/m


d' = 222 mmS = (Spanx / 2 + ex) x Spany x Nuit = 477 kN A = lyi x lx = 0.100 m2

m' = S x (1 - (Nutt x A / S)1/3) / 5.14 = 78.476 kNm/m K' = 0.402 x (pb - 0.4) - 0.18 x (Pb - 0.4)2 = 0.176 K = m' / (d'2 x fcu) = 0.045


As_ prov = n x D2 / (4 x s) = 1149 mm2/mPASS - 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; As_req = Asy1 / 2 = 785 mm2/mProvide 16 dia bars @ 200 centres - 'U' bars with 1600 mm long legsArea of reinforcement provided; As_prov = n x D2 / (4 x s) = 1005 mm2/m

PASS - Edge reinforcement is OK

PUNCHING SHEAR

Vt = ((0.45 x Spanx) + ex) x ((0.45 x Spany) + ey) x Nut = 202

kNVeff = 1.25 x Vt = 252 kNAsx_ten = Ascorner = 1340 mm /m

Asy_ten = Ascorner = 1340 mm /m

uc = lx 1 + ly = 650 mmd = h - c - ^p = 214 mm

12





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Dr. C. Sachpazis

Date

18/01/2014

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Vt = ((0.45 x Spanx) + ex) x (1.05 x Spany) x Nuit = 453 kN Veff =

1.4 x Vt = 634 kN

Asx_ten = Asx_edge = 1148 mm /m

Maximum allowable shear stress; Vmax = min(0.8 x V(fcu), 5) = 4.733 N/mm2

Design shear stress at column perimeter; v0 = Veff / (u x 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 x(kx x ky) x k x d) = 1292 mmArea of tension steel at shear perimeter; As_ten = (ky x(px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

As_ten = 1731 mm2

Design concrete shear stress;vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

vc = 0.707 N/mm2

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

vc < v <= 1.6 x vc

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

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x(kxx ky) x k x d) = 1613 mmArea of tension steel at shear perimeter;As_ten = (ky x(px+ (kx x k x d)) x Asy_ten) + (kx x(py+ (ky xk xd)) x

Asx_ten

)

As_ten = 2161 mm2


vc = 0.707 N/mm2


vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u = uc + (2 x(kxx ky) x k x d) = 1934 mmArea of tension steel at shear perimeter;As_ten = (ky x(px+ (kx x k x d)) x Asy_ten) + (kx x(py+ (ky xk xd)) x

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;Design effective shear transferred to

13

column; Area of tension steel in x-direction;

14





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Dr. C. Sachpazis

Date

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Area of tension steel in y-direction; Column

perimeter;


Maximum allowable shear stress;

Design shear stress at column perimeter;

As_ ten: 3588 mm

Asy_ten = Asy1e = 2094

mm2/m uc = (2 x lx1)+ ly = 900 mm d = h - c - = 214 mm

VMAX = min(0.8 x V(fCU), 5) = 4.733 N/mm2 VO = VEFF / (uC x d) = 3.292 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 + (2x (kxx ky) x k x d) = 2184 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) xA

sx_ten)

Design concrete shear stress;Vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.757 N/mm2


1.6 x vc < v <= 2 x vc

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

mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky)x k x d) = 2826 mmArea of tension steel at shear perimeter; As_ten = (kyx (px + (kx x

k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) xA

sx_ten)

As_ten = 4628 mm2


vc = 0.756 N/mm2


vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 3468 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x

k x d)) x Asy_ten) + (kx x (py + (ky x k x

d)) xA

sx_ten)

As_ten = 5669 mm2


vc = 0.756 N/mm2


15

vc < v <= 1.6 x vc


16





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Job Ref.

Section


Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal edge column A3





Column perimeter;




Shear reinforcement at a perimeter of 3.75d - (803 mm)Length of shear perimeter; u = Uc + (2 x (kx x ky) x k x d) = 4110 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x

d)) x

Asx_ten

)

As_ten = 6710 mm2


vc = 0.755 N/mm2



Vt = ((0.45 x Spanx) + ex) x Spany x Nult = 431 kNVeff = 1.4 x Vt = 604 kNAsx_ten = Asx_edge = 1148 mm /m

Asy_ten = Asye = 1570 mm2/m

uc = (2 x lx1)+ ly = 900 mmd = h - c - ^p = 214 mmVmax = min(0.8 x V(fcu), 5) = 4.733 N/mm2

VO = VEFF / (uC X d) = 3.135 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 x (kx x ky) x k x d) = 2184 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x

d)) xA

sx_ten)

As ten = 2989 mm2

Design concrete shear stress;Vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.712 N/mm2


1.6 x vc < v <= 2 x vc

Shear reinforcement required at perimeter; Asv_req = 5 x ((0.7 x v) - vc) x u x d / (0.95 x fyv) = 945

mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) xk x d) = 2826 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten

)

As_ten = 3862 mm2

17


vc = 0.712 N/mm2

18





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Job Ref.

Section


Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Penultimate edge column B1





Column perimeter;




Nominal design shear stress at perimeter; v = Ver / (u x d) = 0.998 N/mm2

Vc < v <= 1.6 x Vc


Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u = uc + (2 x(kx x ky) xk x d) = 3468 mmArea of tension steel at shear perimeter;As_ten = (ky x(px + (kx xk x d)) x Asy_ten) + (kx x(py+ (ky xk xd)) x

Asx_ten

)

As_ten = 4734 mm2


vc = 0.712 N/mm2


vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 3.75d - (803 mm)Length of shear perimeter; u = uc + (2 x(kx x ky) xk x d) = 4110 mmArea of tension steel at shear perimeter;As_ten = (ky x(px + (kx xk x d)) x Asy_ten) + (kx x(py+ (ky xk xd)) x

Asx_ten)As_ ten = 5607 mm2


vc = 0.711 N/mm2



Vt = (1.05 x Spanx) x ((0.45 x Spany) + ey) x Nuit = 453 kN Veff = 1.4 x Vt = 634 kNAsx ten = Asx1e = 2513 mm /mAsy_ten = Asy_edge = 1148 mm /m

uc = lx + (2 x ly1) = 900 mmd = h - c - = 214 mmvmax = min(0.8 x V(fcu), 5) = 4.733 N/mm2

v0 = VEFF / (uC X d) = 3.292 N/mm 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 x (kx x ky) x k x d) = 2184 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x

d)) x

Asx_ten

)

As ten = 4066 mm2

19





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Section


Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

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Design concrete shear stress;Vc=(min(fcu,40)/25)1/3x0.79xmm(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.789 N/mm2

Nominal design shear stress at perimeter; v = Vf / (u x d) = 1.356 N/mm2

1.6 x Vc < v <= 2 x Vc

Shear reinforcement required at perimeter; Asv_req = 5 x ((0.7 x v) - vc) x u x d / (0.95 x fyv) = 789

mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 2826 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x

Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten

)

As ten — 5241 mm

Design concrete shear stress;vc—(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

vc — 0.788 N/mm2

Nominal design shear stress at perimeter; v — Vf / (u x d) — 1.048 N/mm2

vc < v <= 1.6 x vc

Shear reinforcement required at perimeter; Asv_req — (v - vc) x u x d / (0.95 x fyv) — 331 mm2

Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u — uc + (2 x (kx x ky) x k x d) — 3468 mmArea of tension steel at shear perimeter; As_ten — (ky x (px+ (kx x k x d)) x


Asx_ten

)

As_ten — 6416 mm2


vc — 0.788 N/mm2


vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 3.75d - (803 mm)Length of shear perimeter; u — uc + (2 x (kx x ky) x k x d) — 4110 mmArea of tension steel at shear perimeter; As_ten — (ky x (px + (kx x k x d)) x


Asx_ten

)

As_ten — 7592 mm2


vc — 0.787 N/mm2


20


21





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

Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Penultimate central column B2


Design effective shear transferred to column; Area of tension steel in x-

direction;


Column perimeter;

Average effective depth of reinforcement; Maximum allowable shear

stress;


As_ten

: 9601 mm2

Vt = (1.05 x Spanx) x (1.05 x Spany) x Nult = 1017 kN Veff = 1.15 x Vt =

1170 kN

Asx ten = Asx1e = 2513 mm /m

Asy_ten = Asyte = 2094 mm /m uc = 2 x (lx + ly) = 1600 mm d = h - c - =

214 mm

VMAX = MIN(0.8 x V(FCU), 5) = 4.733 N/MM2 VO = VEFF / (UC x D) = 3.417 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 x (kx x ky) x k x d) = 4168 mm

Area of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

Design concrete shear stress;

Vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.847 N/mm2


vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 5452 mm


Asx_ten)

As_ten = 12559 mm2


vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

vc = 0.847 N/mm2


vc < v <= 1.6 x vc




Asx_ten)

As_ten = 15516 mm2



22





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Job Ref.

Section


Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal central column B3Design shear transferred to column;

Design effective shear transferred to column



Column perimeter;


Maximum allowable shear stress;Design shear stress at column perimeter;

As_ ten: 7636 mm

Vc = 0.847 N/mm2


v < Vc no shear reinforcement required

Vt = (1.05 x Spanx) x Spany x Nuit = 969 kNVeff = 1.15 x Vt = 1114 kNAsx_ten = Asxii = 2094 mm2/m

Asy ten — Asye — 1570 mm /muc — 2 x (lx + ly) — 1600 mmd — h - c - — 214 mmvmax — min(0.8 x V(fcu), 5) — 4.733 N/mm2

vO — VEFF / (uC X d) — 3.254 N/mm 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 x (kx x ky) x k x d) — 4168 mmArea of tension steel at shear perimeter; As_ten — (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten

)


vc — 0.785 N/mm2

Nominal design shear stress at perimeter; v — Veff / (u x d) — 1.249 N/mm2

vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u — uc + (2 x (kx x ky) x k x d) — 5452 mmArea of tension steel at shear perimeter; As_ten — (ky x (px + (kx x

k x d)) x Asy_ten) + (kx x (py + (ky x k x

d)) x

Asx_ten

)

As_ten — 9988 mm2


vc — 0.785 N/mm2

Nominal design shear stress at perimeter; v — Veff / (u x d) — 0.955 N/mm2

vc < v <= 1.6 x vc


Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u — uc + (2 x (kx x ky) x k x d) — 6736 mm

23

Area of tension steel at shear perimeter; As_ten — (ky x (px+ (kx xk x d)) x


Asx_ten

)

24

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

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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal edge column C1


Design effective shear transferred to column Area of tension steel in x-

direction;


Column perimeter;(Library item: Flat slab shear map C1)

Maximum allowable shear stress;d = h - c - = 214 mm

As_ten = 12340 mm2


Vc=(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.785 N/mm2


v < Vc no shear reinforcement required

Vt = Spanx x ((0.45 x Spany) + ey) x Nult = 431

kN Veff = 1.4 x Vt = 604 kN Asx_ten = Asxe = 1570

mm2/m Asy_ten = Asy_edge = 1148 mm /m uc = lx +

(2 x ly1) = 900 mm Average effective depth of

reinforcement;

vmax = min(0.8 x V(fcu), 5) = 4.733 N/mm2

Design shear stress at column perimeter; vo = Veff / (uc x d) =

3.135 N/mm2

PASS - Maximum concrete shear stress not exceeded at column perimeter


Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 2184 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

As ten = 2989 mm2



vc = 0.712 N/mm2


1.6 x vc < v <= 2 x vcShear reinforcement required at perimeter; Asv_req = 5 x ((0.7 x v) - vc) x u x d / (0.95 x fyv) = 945 mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 2826 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

As_ten = 3862 mm2



vc = 0.712 N/mm2


vc < v <= 1.6 x vcShear reinforcement required at perimeter; Asv_req = (v - vc) x u x d / (0.95 x fyv) = 365 mm2

25


26




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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

VO — VEFF / (UC X D) — 3.254 N/MM PASS - Maximum concrete shear stress

not exceeded at column perimeter Shear reinforcement at a perimeter of 1.50d - (321 mm)

Internal central column C2



direction;


Column perimeter;


stress;


Vt — Spanx x (1.05 x Spany) x Nuit — 969 kN

Veff — 1.15 x Vt — 1114 kN

Asx_ten — Asxe — 1570 mm2/m

Asy ten — Asy1i — 2094 mm /m

uc — 2 x (lx + ly) — 1600 mm

d — h - c - ^p — 214 mm

vmax — min(0.8 x V(fcu), 5) — 4.733 N/mm2

Length of shear perimeter;

Area of tension steel at shear perimeter;

Asx_ten)


Length of shear perimeter; u = Uc + (2 x (kx x ky) x k x d) = 3468 mm


Asx_ten)

As ten — 4734 mm


Vc—(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc — 0.712 N/mm2


Vc < v <= 1.6 x VcShear reinforcement required at perimeter; Asv_req — (v - Vc) x u x d / (0.95 x fyv) — 159 mm2

Shear reinforcement at a perimeter of 3.75d - (803 mm)Length of shear perimeter; u — uc + (2 x (kx x ky) x k x d) — 4110 mm

Area of tension steel at shear perimeter; As_ten — (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

As_ten — 5607 mm2


vc—(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

vc — 0.711 N/mm2



u — uc + (2 x (kx x ky) x k x d) — 4168 mm

27

2

Nominal design shear stress at perimeter;

vc — 0.785 N/mm v — Veff / (u x d) —

1.249 N/mm2

As_ten — (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

As_ten — 7636 mm2

vc—(min(fcu,40)/25)1/3x0.79xmin(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

28




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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Internal column C3



direction;


Column perimeter;


stress;


As_ten

: 6544 mm

Vc < v <= 1.6 X VcShear reinforcement required at perimeter; Asv_req = (v - Vc) X u X d / (0.95 X fyv) = 872 mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 X (kx X ky) X k X d) = 5452 mm

Area of tension steel at shear perimeter; As_ten = (ky X (px + (kx X k X d)) X Asy_ten) + (kx X (py + (ky X k X d)) X

Asx_ten)

As_ten = 9988 mm2



vc = 0.785 N/mm2

Nominal design shear stress at perimeter; v = Vf / (u X d) = 0.955 N/mm2

vc < v <= 1.6 X vcShear reinforcement required at perimeter; Asv_req = (v - vc) X u X d / (0.95 X fyv) = 418 mm2

Shear reinforcement at a perimeter of 3.00d - (642 mm)Length of shear perimeter; u = uc + (2 X (kx X ky) X k X d) = 6736 mm

Area of tension steel at shear perimeter; As_ten = (ky X (px + (kx X k X d)) X Asy_ten) + (kx X (py + (ky X k X d)) X

Asx_ten)

As_ten = 12340 mm2



vc = 0.785 N/mm2

Nominal design shear stress at perimeter; v = Vf / (u X d) = 0.773 N/mm2


Vt = Spanx X Spany X Nult = 923 kN

Veff = 1.15 X Vt = 1061 kN

Asx_ten = Asxi = 1570 mm2/m

Asy_ten = Asyi = 1570 mm2/m

uc = 2 X (lx + ly) = 1600 mm

d = h - c - ^p = 214 mm

vmax = min(0.8 X V(fcu), 5) = 4.733 N/mm2

VO = VEFF / (UC X D) = 3.099 N/MM 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 X (kx X ky) X k X d) = 4168 mmArea of tension steel at shear perimeter; As_ten = (ky X (px + (kx X k X d)) X Asy_ten) + (kx X (py + (ky X k X d)) X

Asx_ten)

29




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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

CURTAILMENT OF REINFORCEMENT

Internal column

Radius of circular yield line; mm

Minimum curtailment length in x-direction; Minimum curtailment

length in y-direction;

Corner column

Radius of yield line;

ly))1/3

Minimum curtailment length in x-direction;


Vc=(min(fcu,40)/25)1/3x0.79xmm(100xAs_ten/(uxd),3)1/3xmax(400/d,1)1/4/1.25

Vc = 0.746 N/mm2


Vc < v <= 1.6 x VcShear reinforcement required at perimeter; Asv_req = (v - vc) x u x d / (0.95 x fyv) = 834 mm2

Shear reinforcement at a perimeter of 2.25d - (482 mm)Length of shear perimeter; u = uc + (2 x (kx x ky) x k x d) = 5452 mmArea of tension steel at shear perimeter; As_ten = (ky x (px + (kx x k x d)) x Asy_ten) + (kx x (py + (ky x k x d)) x

Asx_ten)

As_ten = 8560 mm2



vc = 0.746 N/mm2


vc < v <= 1.6 x vcShear reinforcement required at perimeter; Asv_req = (v - vc) x u x d / (0.95 x fyv) = 403 mm2



Asx_ten)

As_ten = 10576 mm2



vc = 0.746 N/mm2



r = (lx x ly / n)1/2 x (1.05 x Spanx x 1.05 x Spany / (lx x ly))1/3 = 1601

lint_x = Max(r + 12 x D, 0.25 x Spanx) = 1841 mm lint_y = Max(r + 12 x D,

0.25 x Spany) = 1841 mm

r = (lx1 x ly / n)1/2 x ((0.45 x Spanx + ex) x (0.45 x Spany + ey)/ (lx1 x r = 863 mm

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

30





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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

Minimum curtailment length in y-direction;

Edge columns

Radius of yield line in x-direction;

ly))1/3

Minimum curtailment length in x-direction; Radius of yield line in y-

direction;

ly1))1/3

Minimum curtailment length in y-direction;

lcorner_y = Max(r + 12 x D, 0.2 x Spany) = 1440 mm

r = (lx1 x ly / n)1/2 x ((0.45 x Spanx + ex) x (1.05 x Spany) / (lx1 x

r = 1130 mm

ledge_x = Max(r + 12 x D, 0.2 x Spanx) = 1440 mm r = (lx x ly1 / n)1/2 x ((0.45 x Spany + ey) x (1.05 x

Spanx) / (lx x

r = 1130 mm

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

31





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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

Date App'd by Date

0.5 x Spanx ->|0.2 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.

32





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Section


Calc. by

Dr. C. Sachpazis

Date

18/01/2014

Chk'd by

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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.

flat slab design to bs8110

Documents

xdirectioncolumn dimension

xdirectionedge dimension

gk x yg qk x yq

nult x l2

n x d2

d2 x fcu

momentsdesign slab

mmspan of slab