L1675 – 32 Lansdowne Road
MOMENT FRAME 1
ANALYSIS
Tedds calculation version 1.0.05
Element UDL loads
Geometry
Eleme
nt
Load case
Type
FB3
10
7
8
9
10
7
8
9
6
3
C3
5
UGB4
2
Position
LGB7
C2
Dead
Dead
Dead
Dead
Live
Live
Live
Live
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
Ratio
Start
0
0
0
0
0
0
0
0
Load Orientatio
n
End
1
1
1
1
1
1
1
1
(kN/m)
6.8
111.8
16.8
95
4.3
22.2
8.4
13.8
GlobalZ
GlobalZ
GlobalZ
GlobalZ
GlobalZ
GlobalZ
GlobalZ
GlobalZ
C1
4
1
Reactions
Load combination: 1.4D + 1.6I (Service)
Node
Load combination: 1.4D + 1.6I (Strength)
Load combination: 1.4D + 1.6I (Service)
1
6
Load combination: 1.2D + 1.2I +1.2W (Strength)
Load combination: 1.2D + 1.2I +1.2W (Service)
Node
Node loads
5
2
3
Fx
(kN)
-2.9
2.9
Load case
Wind
Wind
Wind
Fz
(kN)
399
400.5
Moment
My
(kNm)
0
0
Fz
(kN)
367.7
431.8
Moment
My
(kNm)
0
0
Fz
(kN)
290.8
380.5
Moment
My
(kNm)
0
0
Load combination: 1.2D + 1.2I +1.2W (Service)
Load combination: 1.0D + 1.4W
Node
Force
Force
X
(kN)
13.1
5.3
5.3
1
6
Moment
Z
(kN)
0
0
0
(kNm)
0
0
0
Force
Fx
(kN)
-14.5
-9.2
Load combination: 1.0D + 1.4W
Node
1
6
Force
Fx
(kN)
-19.3
-13.9
;
Page 46 of 72
Pringuer-James Consulting Engineers Ltd
5
4
6
2
1
3
L1675 – 32 Lansdowne Road
Page 47 of 72
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
BEAM UGB4
BEAM FB3
STEEL MEMBER DESIGN (BS5950)
STEEL MEMBER DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Section details
Section type;
Section details
UKC 305x305x118 (Corus Advance);
Steel grade;
S275
Section type;
Classification of cross sections - Section 3.5
Tensile strain coefficient;
UKC 254x254x132 (Corus Advance);
Section classification;
Plastic
Fv = 365 kN;
Design shear resistance;
Py,v = 600.1 kN
Tensile strain coefficient;
 = 1.02;
Design shear force;
Shear capacity - Section 4.2.3
Fv = 524 kN;
Moment capacity high shear;
Mc = 515.1 kNm
Design bending moment;
M = 220 kNm;
Mb / mLT = 481.3 kNm
Buckling resistance moment;
Moment capacity high shear;
Mb = 461.8 kNm;
Mc = 471.2 kNm
Mb / mLT = 461.8 kNm
PASS - Buckling resistance moment exceeds design bending moment
Compression members - Section 4.7
Fc = 13 kN;
Compression resistance;
Pcx = 3742.8 kN
Design compression force;
Fc = 5 kN;
PASS - Compression resistance exceeds design compression force
Fc = 13 kN;
Compression resistance;
Pcy = 2987.7 kN
Design compression force;
Compression members with moments - Section 4.8.3
Compression resistance;
Pcx = 4082.1 kN
PASS - Compression resistance exceeds design compression force
Fc = 5 kN;
PASS - Compression resistance exceeds design compression force
Compression resistance;
Pcy = 3051.1 kN
PASS - Compression resistance exceeds design compression force
Compression members with moments - Section 4.8.3
Fc / (A  py) + M / Mc = 0.498
Comp.and bending check;
Fc / (A  py) + M / Mc = 0.468
PASS - Combined bending and compression check is satisfied
Member buckling resistance - cl.4.8.3.3.2
Buckling resistance checks;
Py,v = 672.2 kN
Buckling resistance moment - Section 4.3.6.4
Mb = 481.3 kNm;
Compression members - Section 4.7
Comp.and bending check;
Design shear resistance;
PASS - Design shear resistance exceeds design shear force
PASS - Buckling resistance moment exceeds design bending moment
Design compression force;
Plastic
Moment capacity - Section 4.2.5
M = 255 kNm;
Buckling resistance moment - Section 4.3.6.4
Design compression force;
Section classification;
Shear capacity - Section 4.2.3
Moment capacity - Section 4.2.5
Buckling resistance moment;
S275
Shear capacity - Section 4.2.3
PASS - Design shear resistance exceeds design shear force
Design bending moment;
Steel grade;
Classification of cross sections - Section 3.5
 = 1.02;
Shear capacity - Section 4.2.3
Design shear force;
TEDDS calculation version 3.0.04
PASS - Combined bending and compression check is satisfied
Member buckling resistance - cl.4.8.3.3.2
Fc / Pcx + mx  M / Mc  (1 + 0.5  Fc / Pcx) = 0.499
Buckling resistance checks;
Fc / Pcy + mLT  MLT / Mb = 0.534
Fc / Pcx + mx  M / Mc  (1 + 0.5  Fc / Pcx) = 0.468
Fc / Pcy + mLT  MLT / Mb = 0.478
PASS - Member buckling resistance checks are satisfied
Page 48 of 72
PASS - Member buckling resistance checks are satisfied
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
BEAM LGB7
COLUMN C1
STEEL MEMBER DESIGN (BS5950)
STEEL MEMBER DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Section details
Section type;
UKC 203x203x46 (Corus Advance);
Steel grade;
Section details
S275
Section type;
UKC 254x254x89 (Corus Advance);
Steel grade;
S275
Classification of cross sections - Section 3.5
Tensile strain coefficient;
 = 1.00;
Section classification;
Classification of cross sections - Section 3.5
Compact
Tensile strain coefficient;
Shear capacity - Section 4.2.3
Design shear force;
Fv = 66 kN;
Design shear resistance;
Py,v = 241.4 kN
 = 1.02;
Section classification;
Plastic
Fv = 60 kN;
Design shear resistance;
Py,v = 426.3 kN
Shear capacity - Section 4.2.3
PASS - Design shear resistance exceeds design shear force
Design shear force;
PASS - Design shear resistance exceeds design shear force
Shear capacity - Section 4.2.3
Shear capacity - Section 4.2.3
Moment capacity - Section 4.2.5
Design bending moment;
M = 81 kNm;
Moment capacity low shear;
Mc = 136.8 kNm
Moment capacity - Section 4.2.5
Design bending moment;
Buckling resistance moment - Section 4.3.6.4
Buckling resistance moment;
Mb = 90.8 kNm;
Mb / mLT = 90.8 kNm
M = 140 kNm;
Moment capacity low shear;
Mc = 324.3 kNm
Buckling resistance moment - Section 4.3.6.4
PASS - Buckling resistance moment exceeds design bending moment
Buckling resistance moment;
Mb = 323.7 kNm;
Mb / mLT = 323.7 kNm
PASS - Buckling resistance moment exceeds design bending moment
Compression members - Section 4.7
Design compression force;
Fc = 575 kN;
Compression resistance;
Pcx = 2899.4 kN
PASS - Compression resistance exceeds design compression force
Design compression force;
Fc = 575 kN;
Compression resistance;
Pcy = 2497.9 kN
PASS - Compression resistance exceeds design compression force
Compression members with moments - Section 4.8.3
Comp.and bending check;
Fc / (A  py) + M / Mc = 0.623
PASS - Combined bending and compression check is satisfied
Member buckling resistance - cl.4.8.3.3.2
Buckling resistance checks;
Fc / Pcx + mx  M / Mc  (1 + 0.5  Fc / Pcx) = 0.673
Fc / Pcy + mLT  MLT / Mb = 0.663
PASS - Member buckling resistance checks are satisfied
Page 49 of 72
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
COLUMN C2
COLUMN C3
STEEL MEMBER DESIGN (BS5950)
STEEL MEMBER DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Section details
Section type;
Section details
UKC 254x254x89 (Corus Advance);
Steel grade;
S275
Section type;
Classification of cross sections - Section 3.5
Tensile strain coefficient;
UKC 254x254x89 (Corus Advance);
 = 1.02;
Section classification;
Plastic
Tensile strain coefficient;
 = 1.02;
Fv = 60 kN;
Design shear resistance;
Py,v = 426.3 kN
Design shear force;
Fv = 60 kN;
Shear capacity - Section 4.2.3
Moment capacity - Section 4.2.5
Moment capacity - Section 4.2.5
M = 145 kNm;
Moment capacity low shear;
Mc = 324.3 kNm
Design bending moment;
Buckling resistance moment - Section 4.3.6.4
M = 170 kNm;
Mb / mLT = 323.7 kNm
Buckling resistance moment;
Compression members - Section 4.7
Mc = 324.3 kNm
Mb / mLT = 323.7 kNm
Compression members - Section 4.7
Fc = 575 kN;
Compression resistance;
Pcx = 2899.4 kN
Design compression force;
Fc = 575 kN;
Compression resistance;
Pcy = 2497.9 kN
Fc = 255 kN;
Compression resistance;
Pcx = 2899.4 kN
PASS - Compression resistance exceeds design compression force
Design compression force;
Compression members with moments - Section 4.8.3
Fc = 255 kN;
Compression resistance;
Pcy = 2497.9 kN
PASS - Compression resistance exceeds design compression force
Compression members with moments - Section 4.8.3
Fc / (A  py) + M / Mc = 0.639
Comp.and bending check;
Fc / (A  py) + M / Mc = 0.609
PASS - Combined bending and compression check is satisfied
Member buckling resistance - cl.4.8.3.3.2
Buckling resistance checks;
Py,v = 426.3 kN
PASS - Buckling resistance moment exceeds design bending moment
PASS - Compression resistance exceeds design compression force
Comp.and bending check;
Design shear resistance;
Moment capacity low shear;
Mb = 323.7 kNm;
PASS - Compression resistance exceeds design compression force
Design compression force;
Plastic
Buckling resistance moment - Section 4.3.6.4
Mb = 323.7 kNm;
PASS - Buckling resistance moment exceeds design bending moment
Design compression force;
Section classification;
PASS - Design shear resistance exceeds design shear force
Shear capacity - Section 4.2.3
Buckling resistance moment;
S275
Shear capacity - Section 4.2.3
PASS - Design shear resistance exceeds design shear force
Design bending moment;
Steel grade;
Classification of cross sections - Section 3.5
Shear capacity - Section 4.2.3
Design shear force;
TEDDS calculation version 3.0.04
PASS - Combined bending and compression check is satisfied
Member buckling resistance - cl.4.8.3.3.2
Fc / Pcx + mx  M / Mc  (1 + 0.5  Fc / Pcx) = 0.690
Buckling resistance checks;
Fc / Pcy + mLT  MLT / Mb = 0.678
Fc / Pcx + mx  M / Mc  (1 + 0.5  Fc / Pcx) = 0.635
Fc / Pcy + mLT  MLT / Mb = 0.627
PASS - Member buckling resistance checks are satisfied
Page 50 of 72
PASS - Member buckling resistance checks are satisfied
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
GROUND BEAM 1
STEEL BEAM ANALYSIS & DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Load Envelope - Com bination 1
574.000
0.0
mm
A
800
1
4500
2
B
C
Load Envelope - Com bination 2
517.200
0.0
mm
A
800
1
4500
2
B
C
Support conditions
Support A
Vertically free
Rotationally free
Support B
Vertically restrained
Rotationally free
Page 51 of 72
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
Support C
Vertically restrained
Moment capacity at span 2 - Section 4.2.5
Rotationally free
Design bending moment;
M = 460.1 kNm;
Moment capacity low shear;
Mc = 911.6 kNm
PASS - Moment capacity exceeds design bending moment
Applied loading
Beam loads
Dead self weight of beam  1
Check vertical deflection - Section 2.5.2
Dead point load 330 kN at 0 mm
Consider deflection due to dead and imposed loads
Imposed point load 70 kN at 0 mm
Limiting deflection;
lim = 4.444 mm;
Maximum deflection;
 = 4.286 mm
PASS - Maximum deflection does not exceed deflection limit
Wind point load 31 kN at 0 mm
Dead point load 330 kN at 5300 mm
Imposed point load 70 kN at 5300 mm
Wind point load -31 kN at 5300 mm
Analysis results
Maximum moment;
Mmax = 0 kNm;
Mmin = -460.1 kNm
Maximum moment span 1;
Ms1_max = 0 kNm;
Ms1_min = -460.1 kNm
Maximum moment span 2;
Ms2_max = 0 kNm;
Ms2_min = -460.1 kNm
Maximum shear;
Vmax = 108.4 kN;
Vmin = -576.2 kN
Maximum shear span 1;
Vs1_max = -517.2 kN;
Vs1_min = -576.2 kN
Maximum shear span 2;
Vs2_max = 108.4 kN;
Vs2_min = 86.9 kN
Deflection;
max = 4.3 mm;
min = 3.9 mm
Deflection span 1;
s1_max = 4.3 mm;
s1_min = 0 mm
Deflection span 2;
s2_max = 0 mm;
s2_min = 3.9 mm
Maximum reaction at support A;
RA_max = 0 kN;
RA_min = 0 kN
Maximum reaction at support B;
RB_max = 684.5 kN;
RB_min = 616.4 kN
Unfactored dead load reaction at support B;
RB_Dead = 394.7 kN
Unfactored imposed load reaction at support B;
RB_Imposed = 82.4 kN
Unfactored wind load reaction at support B;
RB_Wind = 36.5 kN
Maximum reaction at support C;
RC_max = 477.9 kN;
Unfactored dead load reaction at support C;
RC_Dead = 275.6 kN
Unfactored imposed load reaction at support C;
RC_Imposed = 57.6 kN
Unfactored wind load reaction at support C;
RC_Wind = -36.5 kN
RC_min = 355.9 kN
Section details
Section type;
UC 305x305x198 (BS4-1);
Steel grade;
S275
 = 1.02;
Section classification;
Plastic
Fv = 576.2 kN;
Design shear resistance;
Pv = 1032.2 kN
Classification of cross sections - Section 3.5
Tensile strain coefficient;
Shear capacity - Section 4.2.3
Design shear force;
PASS - Design shear resistance exceeds design shear force
Page 52 of 72
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
PAD FOUNDATION P1 DESIGN
PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997)
TEDDS calculation version 2.0.05.06
178.8 kN/m
178.8 kN/m
2
178.8 kN/m
2
178.8 kN/m
2
2
Partial safety factors for loads
Pad footing details
Length of pad footing;
L = 2000 mm;
Width of pad footing;
Depth of pad footing;
h = 400 mm;
Depth of soil over pad footing; hsoil = 0 mm
B = 1500 mm
Density of concrete;
conc = 23.6 kN/m
Dead loads;
fG = 1.40;
Wind loads;
fW = 0.00
HyuA = 0.0 kN
Ultimate foundation loads
Fu = 39.6 kN
Column base length;
lA = 350 mm;
Column base width;
bA = 350 mm
Ultimate horizontal loading on column
Column eccentricity in x;
ePxA = 0 mm;
Column eccentricity in y;
ePyA = 0 mm
Ult.horizontal load in x dir;
Density of soil;
soil = 20.0 kN/m
Soil details
Depth of soil over pad footing; hsoil = 0 mm;
Pbearing = 200 kN/m
HxuA = 0.0 kN;
Ultimate moment on column
3
Ult.moment on column in x dir; MxuA = 0.000 kNm;
2
Ult.moment on column in y dir; MyuA = 0.000 kNm
Ultimate pad base reaction
Axial loading on column
Dead axial load;
PGA = 395.0 kN;
Imposed axial load;
PQA = 82.0 kN
Wind axial load;
PWA = 31.0 kN;
Total axial load;
PA = 508.0 kN
Ultimate base reaction;
Tu = 723.8 kN
Ecc.of ult.base reaction in x;
eTxu = 0 mm;
2
2
Dead surcharge load;
FGsur = 0.000 kN/m ;
Pad footing self weight;
Fswt = 9.440 kN/m
Soil self weight;
Fsoil = 0.000 kN/m ;
Imposed surcharge load;
FQsur = 0.000 kN/m
Total foundation load;
F = 28.3 kN
2
2
2
Ecc.of ult.base reaction in y;
eTyu = 0 mm
Calculate ultimate pad base pressures
Foundation loads
2
2
q1u = 241.283 kN/m ;
q2u = 241.283 kN/m ;
Minimum ult.base pressure;
qminu = 241.283 kN/m ;
2
2
q3u = 241.283 kN/m ;
q4u = 241.283 kN/m
Maximum ult.base pressure;
qmaxu = 241.283 kN/m
2
Library item: Ultimate pressures summaryUltimate moments
Ultimate moment in x dir;
Calculate pad base reaction
Mx = 171.050 kNm;
Ultimate moment in y dir;
My = 128.288 kNm
Char.strength of reinft;
fy = 500 N/mm
Material details
T = 536.3 kN
Base reaction eccentricity in x; eTx = 0 mm;
Base reaction eccentricity in y; eTy = 0 mm
Base reaction acts within middle third of base
2
Char.strength of concrete;
fcu = 35 N/mm ;
Char.strength of shear reinft;
fyv = 500 N/mm ;
Nom.cover to reinforcement;
cnom = 40 mm
xB = 12 mm;
Tens.reinforcement depth;
dx = 354 mm
Kx = 0.026;
Kx’ = 0.156
2
2
Moment design in x direction
Calculate pad base pressures
2
Ult. horizontal load in y dir;
Ultimate axial load on column; PuA = 684.2 kN
3
Ultimate foundation load;
Total base reaction;
fQ = 1.60
Ultimate axial loading on column
Column details
Allowable bearing pressure;
Imposed loads;
2
2
q1 = 178.773 kN/m ;
q2 = 178.773 kN/m ;
Minimum base pressure;
qmin = 178.773 kN/m ;
2
Tens.reinforcement diameter;
2
q3 = 178.773 kN/m ;
q4 = 178.773 kN/m
Maximum base pressure;
qmax = 178.773 kN/m
2
Design formula for rectangular beams (cl 3.4.4.4);
Kx < Kx' compression reinforcement is not required
PASS - Maximum base pressure is less than allowable bearing pressure
2
Tens.reinforcement required;
As_x_req = 1169 mm ;
Minimum tens.reinforcement;
Tens.reinforcement provided;
11 No. 12 dia. bars btm;
As_xB_prov = 1244 mm
2
As_x_min = 780 mm
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Moment design in y direction
Tens.reinforcement diameter;
Page 53 of 72
yB = 12 mm;
Tens.reinforcement depth;
dy = 342 mm
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
Design formula for rectangular beams (cl 3.4.4.4);
Ky = 0.016;
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Tens.reinforcement required;
Tens.reinforcement provided;
2
As_y_req = 908 mm ;
Minimum tens.reinforcement;
13 No. 12 dia. bars btm;
As_yB_prov = 1470 mm
As_y_min = 1040 mm
2
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Calculate ultimate shear force at d from right face of column
Ult.pressure for shear;
Area loaded for shear;
qsu = 241.283 kN/m
2
2
As = 0.706 m ;
Ult.shear force;
Vsu = 161.129 kN
Shear stresses at d from right face of column (cl 3.5.5.2)
Design shear stress;
vsu = 0.303 N/mm
2
2
Design concrete shear stress; vc = 0.445 N/mm ;
Allowable design shear stress; vmax = 4.733 N/mm
2
PASS - vsu < vc - No shear reinforcement required
Calculate ultimate punching shear force at face of column
Ult.press.for punching shear;
2
qpuA = 241.283 kN/m ;
Avg.effective reinf.depth;
d = 348 mm
Area loaded;
ApA = 0.123 m ;
Length of shear perimeter;
upA = 1400 mm
Ult.punching shear force;
VpuA = 656.262 kN;
Eff.punching shear force;
VpuAeff = 656.262 kN
2
Punching shear stresses at face of column (cl 3.7.7.2)
Design shear stress;
vpuA = 1.347 N/mm
2
PASS - Design shear stress is less than allowable design shear stress
Calculate ultimate punching shear force at perimeter of 1.5 d from face of column
Ult.press.for punching shear;
2
qpuA1.5d = 241.283 kN/m ;
Avg.effective reinf.depth;
d = 348 mm
Area loaded;
ApA1.5d = 2.091 m ;
Length of shear perimeter;
upA1.5d = 3000 mm
Ult.punching shear force;
VpuA1.5d = 207.313 kN;
Eff.punching shear force;
VpuA1.5deff = 259.141 kN
2
Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2)
Design shear stress;
vpuA1.5d = 0.248 N/mm
2
PASS - vpuA1.5d < vc - No shear reinforcement required
Page 54 of 72
Pringuer-James Consulting Engineers Ltd
L1675 – 32 Lansdowne Road
PAD FOUNDATION P2 DESIGN
PAD FOOTING ANALYSIS AND DESIGN (BS8110-1:1997)
TEDDS calculation version 2.0.05.06
173.9 kN/m
173.9 kN/m
2
173.9 kN/m
2
173.9 kN/m
2
2
Partial safety factors for loads
Dead loads;
fG = 1.40;
Wind loads;
fW = 0.00
Imposed loads;
fQ = 1.60
Ult. horizontal load in y dir;
HyuA = 0.0 kN
Ultimate axial loading on column
Pad footing details
Ultimate axial load on column; PuA = 479.2 kN
Length of pad footing;
L = 1500 mm;
Width of pad footing;
Depth of pad footing;
h = 400 mm;
Depth of soil over pad footing; hsoil = 0 mm
B = 1500 mm
Density of concrete;
conc = 23.6 kN/m
Ultimate foundation loads
Ultimate foundation load;
3
Ultimate horizontal loading on column
Column details
Ult.horizontal load in x dir;
Column base length;
lA = 350 mm;
Column base width;
bA = 350 mm
Column eccentricity in x;
ePxA = 0 mm;
Column eccentricity in y;
ePyA = 0 mm
Density of soil;
soil = 20.0 kN/m
Depth of soil over pad footing; hsoil = 0 mm;
Pbearing = 200 kN/m
Ult.moment on column in x dir; MxuA = 0.000 kNm;
Axial loading on column
PGA = 276.0 kN;
Imposed axial load;
PQA = 58.0 kN
Wind axial load;
PWA = 36.0 kN;
Total axial load;
PA = 370.0 kN
Imposed surcharge load;
FQsur = 0.000 kN/m
Total foundation load;
F = 21.2 kN
Pad footing self weight;
Fswt = 9.440 kN/m
Soil self weight;
Fsoil = 0.000 kN/m ;
Minimum base pressure;
2
Minimum ult.base pressure;
qminu = 226.194 kN/m ;
2
Mx = 89.850 kNm;
2
2
q2 = 173.884 kN/m ;
q3 = 173.884 kN/m ;
2
qmin = 173.884 kN/m ;
Maximum base pressure;
q4 = 173.884 kN/m
2
q3u = 226.194 kN/m ;
q4u = 226.194 kN/m
Maximum ult.base pressure;
qmaxu = 226.194 kN/m
Ultimate moment in y dir;
My = 89.850 kNm
2
Char.strength of reinft;
fy = 500 N/mm
Char.strength of concrete;
fcu = 35 N/mm ;
Char.strength of shear reinft;
fyv = 500 N/mm ;
Nom.cover to reinforcement;
cnom = 40 mm
xB = 12 mm;
Tens.reinforcement depth;
dx = 354 mm
Kx = 0.014;
Kx’ = 0.156
2
Tens.reinforcement diameter;
Design formula for rectangular beams (cl 3.4.4.4);
Calculate pad base pressures
q1 = 173.884 kN/m ;
2
q2u = 226.194 kN/m ;
2
2
Moment design in x direction
Base reaction eccentricity in y; eTy = 0 mm
Base reaction acts within middle third of base
2
eTyu = 0 mm
Material details
T = 391.2 kN
Base reaction eccentricity in x; eTx = 0 mm;
Ecc.of ult.base reaction in y;
q1u = 226.194 kN/m ;
Ultimate moment in x dir;
Calculate pad base reaction
Total base reaction;
eTxu = 0 mm;
Library item: Ultimate pressures summaryUltimate moments
2
2
2
Tu = 508.9 kN
Ecc.of ult.base reaction in x;
2
Foundation loads
2
Ultimate base reaction;
Calculate ultimate pad base pressures
Dead axial load;
FGsur = 0.000 kN/m ;
Ult.moment on column in y dir; MyuA = 0.000 kNm
Ultimate pad base reaction
3
2
Dead surcharge load;
HxuA = 0.0 kN;
Ultimate moment on column
Soil details
Allowable bearing pressure;
Fu = 29.7 kN
2
Kx < Kx' compression reinforcement is not required
2
qmax = 173.884 kN/m
PASS - Maximum base pressure is less than allowable bearing pressure
2
Tens.reinforcement required;
As_x_req = 614 mm ;
Minimum tens.reinforcement;
Tens.reinforcement provided;
11 No. 12 dia. bars btm;
As_xB_prov = 1244 mm
2
As_x_min = 780 mm
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Moment design in y direction
Tens.reinforcement diameter;
Page 55 of 72
yB = 12 mm;
Tens.reinforcement depth;
dy = 342 mm
Pringuer-James Consulting Engineers Ltd
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Design formula for rectangular beams (cl 3.4.4.4);
Ky = 0.015;
Ky’ = 0.156
Ky < Ky' compression reinforcement is not required
Tens.reinforcement required;
Tens.reinforcement provided;
2
As_y_req = 636 mm ;
Minimum tens.reinforcement;
11 No. 12 dia. bars btm;
As_yB_prov = 1244 mm
2
As_y_min = 780 mm
2
PASS - Tension reinforcement provided exceeds tension reinforcement required
Calculate ultimate shear force at d from top face of column
Ult.pressure for shear;
Area loaded for shear;
qsu = 226.194 kN/m
2
2
As = 0.349 m ;
Ult.shear force;
Vsu = 74.436 kN
Shear stresses at d from top face of column (cl 3.5.5.2)
Design shear stress;
vsu = 0.145 N/mm
2
2
Design concrete shear stress; vc = 0.454 N/mm ;
Allowable design shear stress; vmax = 4.733 N/mm
2
PASS - vsu < vc - No shear reinforcement required
Calculate ultimate punching shear force at face of column
Ult.press.for punching shear;
2
qpuA = 226.194 kN/m ;
Avg.effective reinf.depth;
d = 348 mm
Area loaded;
ApA = 0.123 m ;
Length of shear perimeter;
upA = 1400 mm
Ult.punching shear force;
VpuA = 453.110 kN;
Eff.punching shear force;
VpuAeff = 453.110 kN
2
Punching shear stresses at face of column (cl 3.7.7.2)
Design shear stress;
vpuA = 0.930 N/mm
2
PASS - Design shear stress is less than allowable design shear stress
Calculate ultimate punching shear force at perimeter of 1.5 d from face of column
Ult.press.for punching shear;
2
qpuA1.5d = 226.194 kN/m ;
Avg.effective reinf.depth;
d = 348 mm
Area loaded;
ApA1.5d = 2.091 m ;
Length of shear perimeter;
upA1.5d = 3000 mm
Ult.punching shear force;
VpuA1.5d = 33.863 kN;
Eff.punching shear force;
VpuA1.5deff = 42.329 kN
2
Punching shear stresses at perimeter of 1.5 d from face of column (cl 3.7.7.2)
Design shear stress;
vpuA1.5d = 0.041 N/mm
2
PASS - vpuA1.5d < vc - No shear reinforcement required
11 No. 12 dia. bars btm (150 c/c)
11 No. 12 dia. bars btm (150 c/c)
Shear at d from column face
Punching shear perimeter at 1.5 × d from column face
Page 56 of 72
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Page 57 of 72
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RAFT ANALYSIS & DESIGN
CONTINUOUS BEAM ANALYSIS - RESULTS
RC SLAB DESIGN (BS8110:PART1:1997)
TEDDS calculation version 1.0.04
CONTINUOUS ONE WAY SPANNING SLAB DEFINITION
;
Overall depth of slab; h = 400 mm
Sagging steel
;
Cover to tension reinforcement resisting sagging; c sag = 30 mm
;
Trial bar diameter; Dtryx = 25 mm
Depth to tension steel (resisting sagging)
dx = h - csag - Dtryx/2 = 358 mm
Hogging steel
;
Cover to tension reinforcement resisting hogging; c hog = 40 mm
;
Trial bar diameter; Dtryxhog = 25 mm
Depth to tension steel (resisting hogging)
dxhog = h - chog - Dtryxhog/2 = 348 mm
Materials
Page 58 of 72
2
;
Characteristic strength of reinforcement; fy = 500 N/mm
;
Characteristic strength of concrete; fcu = 40 N/mm
2
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Asyhog
dx
h
Nominal 1 m width
h
Tension steel
Asxhog
;;Provide 25 dia bars @ 150 centres; outer tension steel resisting hogging
2
Asxhog_prov = Asxhog = 3270 mm /m
dxhog
Area of outer tension steel provided sufficient to resist hogging
Asy
Asx
Nominal 1 m width
(hogging)
(sagging)
TRANSVERSE TOP STEEL - INNER
One-way spanning slab
;;Inner layer of transverse steel;
(continuous)
Provide 25 dia bars @ 150 centres
2
Asyhog_prov = Asyhog = 3270 mm /m
ONE WAY SPANNING SLAB (CL 3.5.4)
Check min and max areas of steel resisting sagging
MAXIMUM DESIGN MOMENTS IN SPAN
;
2
;Total area of concrete; Ac = h = 400000 mm /m
Design sagging moment (per m width of slab); m sx = 310.0 kNm/m
;
Minimum % reinforcement; k = 0.13 %
CONCRETE SLAB DESIGN – SAGGING – OUTER LAYER OF STEEL (CL 3.5.4)
Ast_min = k  Ac = 520 mm /m
;
Design sagging moment (per m width of slab); m sx = 310.0 kNm/m
Ast_max = 4 %  Ac = 16000 mm /m
;
Moment Redistribution Factor; bx = 1.0
2
2
Steel defined:
;
Area of reinforcement required
;;
2
Outer steel resisting sagging; Asx_prov = 3270 mm /m
Kx = abs(msx) / ( dx  fcu ) = 0.061
2
Area of outer steel provided (sagging) OK
K'x = min (0.156 , (0.402  (bx - 0.4)) - (0.18  (bx - 0.4) )) = 0.156
2
;
2
Inner steel resisting sagging; Asy_prov = 3270 mm /m
Area of inner steel provided (sagging) OK
Outer compression steel not required to resist sagging
Slab requiring outer tension steel only - bars (sagging)
;;
Check min and max areas of steel resisting hogging
zx = min (( 0.95  dx),(dx(0.5+0.25-Kx/0.9)))) = 332 mm
;Total area of concrete; Ac = h = 400000 mm /m
Neutral axis depth; xx = (dx - zx) / 0.45 = 58 mm
;
2
Ast_min = k  Ac = 520 mm /m
2
Area of tension steel required
;;;
Minimum % reinforcement; k = 0.13 %
Asx_req = abs(msx) / (1/ms fy  zx) = 2151 mm /m
2
Ast_max = 4 %  Ac = 16000 mm /m
2
Tension steel
Steel defined:
;;Provide 25 dia bars @ 150 centres; outer tension steel resisting sagging
2
Asx_prov = Asx = 3270 mm /m
;
2
Outer steel resisting hogging; Asxhog_prov = 3270 mm /m
Area of outer steel provided (hogging) OK
Area of outer tension steel provided sufficient to resist sagging
;
2
Inner steel resisting hogging ; Asyhog_prov = 3270 mm /m
Area of inner steel provided (hogging) OK
TRANSVERSE BOTTOM STEEL - INNER
;;Inner layer of transverse steel;
SHEAR RESISTANCE OF CONCRETE SLABS (CL 3.5.5)
Provide 25 dia bars @ 150 centres
2
Asy_prov = Asy = 3270 mm /m
Outer tension steel resisting sagging moments
;
Depth to tension steel from compression face; dx = 358 mm
MAXIMUM DESIGN MOMENTS OVER SUPPORT
;
Area of tension reinforcement provided (per m width of slab); A sx_prov = 3270 mm /m
;
;
Design ultimate shear force (per m width of slab); V x = 465 kN/m
;
Characteristic strength of concrete; fcu = 40 N/mm
Design hogging moment (per m width of slab); m sxhog = 345.0 kNm/m
CONCRETE SLAB DESIGN – HOGGING – OUTER LAYER OF STEEL (CL 3.5.4)
;
Design hogging moment (per m width of slab); m sxhog = 345.0 kNm/m
;
Moment Redistribution Factor; bx = 1.0
2
Applied shear stress
2
vx = Vx / dx = 1.30 N/mm
Check shear stress to clause 3.5.5.2
Area of reinforcement required
;;
2
vallowable = min ((0.8 N /mm)  (fcu ), 5 N/mm ) = 5.00 N/mm
1/2
Kxhog = abs(msxhog) / ( dxhog  fcu ) = 0.071
2
2
2
Shear stress - OK
K'x = min (0.156 , (0.402  (bx - 0.4)) - (0.18  (bx - 0.4) )) = 0.156
2
Outer compression steel not required to resist hogging
Design shear stress
Slab requiring outer tension steel only - bars (hogging)
;;
2
2
zxhog = min (( 0.95  dxhog),(dxhog(0.5+0.25-Kxhog/0.9)))) = 317 mm
fcu_ratio = if (fcu > 40 N/mm , 40/25 , fcu/(25 N/mm )) = 1.600
Neutral axis depth; xxhog = (dxhog - zxhog) / 0.45 = 67 mm
vcx = 0.79 N/mm  min(3,100  Asx_prov / dx)
2
Area of tension steel required
;;;
Shear stresses to clause 3.5.5.3
vcx = 0.74 N/mm
Asxhog_req = abs(msxhog) / (1/ms fy  zxhog) = 2501 mm /m
2
1/3
 max(0.67,(400 mm / dx) ) / 1.25  fcu_ratio
1/4
1/3
2
Applied shear stress
Page 59 of 72
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vx = 1.30 N/mm
2
Shear reinforcement required
If shear reinforcement is required (outer tension steel resisting sagging)
Table 3.16 - Note 1, advises that shear reinforcement is not used in slabs less than 200mm thick.
It is assumed that reinforcement provided by bent up bars is not considered.
The shear stress component resisted by shear reinforcement
;;
2
vsvx = max (abs(vx) - vcx , 0.4 N/mm ) = 0.56 N/mm
2
Reinforcement Requirements
;
Characteristic strength of shear reinforcement; fyv = 500 N/mm
2
;Asvx_to_sv_reqd = (1 m)  vsvx / (1/ms fyv ) = 1.294 mm
Define Links Provided
;;
Link diameter; Ldiax = 10 mm
;
Link spacing ( parallel to span ); svx = 200 mm
;
Link leg spacing ( perp to span); svx_perp = 200 mm
Asvx_to_sv_provided = 1m    Ldiax / (4  svx_perp  svx) = 1.963 mm
2
Area of links sufficient
CONCRETE SLAB DEFLECTION CHECK (CL 3.5.7)
;
Slab span length; lx = 4.700 m
;
Design ultimate moment in shorter span per m width; m sx = 310 kNm/m
;
Depth to outer tension steel; dx = 358 mm
Tension steel
2
;
Area of outer tension reinforcement provided; Asx_prov = 3270 mm /m
;
Area of tension reinforcement required; Asx_req = 2151 mm /m
;
Moment Redistribution Factor; bx = 1.00
2
Modification Factors
;Basic span / effective depth ratio (Table 3.9); ratio span_depth = 20
The modification factor for spans in excess of 10m (ref. cl 3.4.6.4) has not been included.
;fs = 2  fy  Asx_req / (3  Asx_prov  bx ) = 219.2 N/mm
2
factortens = min ( 2 , 0.55 + ( 477 N/mm - fs ) / ( 120  ( 0.9 N/mm + msx / dx ))) = 1.196
2
2
2
Calculate Maximum Span
This is a simplified approach and further attention should be given where special circumstances exist. Refer to clauses 3.4.6.4
and 3.4.6.7.
Maximum span; lmax = ratiospan_depth  factortens  dx = 8.55 m
Check the actual beam span
Actual span/depth ratio; lx / dx = 13.15
Span depth limit; ratiospan_depth  factortens = 23.92
Span/Depth ratio check satisfied
;
Page 60 of 72
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PROPPING LOADING
CONTINUOUS BEAM ANALYSIS - RESULTS
Unfactored support reactions
Page 61 of 72
Dead
(kN)
Imposed
(kN)
Support A;
-3.4
-3.3
0.0
0.0
0.0
0.0
0.0
0.0
Support B;
-34.7
-8.2
0.0
0.0
0.0
0.0
0.0
0.0
Support C;
-88.6
-6.0
0.0
0.0
0.0
0.0
0.0
0.0
Support D;
-41.3
-2.5
0.0
0.0
0.0
0.0
0.0
0.0
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Imposed full UDL 6 kN/m
WALING BEAM DESIGN
Analysis results
STEEL BEAM ANALYSIS & DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Load Envelope - Com bination 1
134.463
0.0
mm
A
1400
1
B
4800
2
C
3100
3
D
2100
4
E
1500
5
F
Mmax = 165.8 kNm;
Mmin = -221.6 kNm
Maximum moment span 1;
Ms1_max = 0 kNm;
Ms1_min = -221.6 kNm
Maximum moment span 2;
Ms2_max = 165.8 kNm;
Ms2_min = -221.6 kNm
Maximum moment span 3;
Ms3_max = 36.4 kNm;
Ms3_min = -221.3 kNm
Maximum moment span 4;
Ms4_max = 26.5 kNm;
Ms4_min = -51.4 kNm
Maximum moment span 5;
Ms5_max = 19 kNm;
Ms5_min = -44 kNm
Maximum shear;
Vmax = 322.8 kN;
Vmin = -322.6 kN
Maximum shear span 1;
Vs1_max = -64.2 kN;
Vs1_min = -252.4 kN
Maximum shear span 2;
Vs2_max = 322.8 kN;
Vs2_min = -322.6 kN
Maximum shear span 3;
Vs3_max = 263.2 kN;
Vs3_min = -153.6 kN
Maximum shear span 4;
Vs4_max = 144.7 kN;
Vs4_min = -137.7 kN
Maximum shear span 5;
Vs5_max = 130.2 kN;
Vs5_min = -71.5 kN
Deflection;
max = 16.4 mm;
min = 1.2 mm
Deflection span 1;
s1_max = 0 mm;
s1_min = 1.2 mm
Deflection span 2;
s2_max = 16.4 mm;
s2_min = 0 mm
Deflection span 3;
s3_max = 0.3 mm;
s3_min = 1.2 mm
Deflection span 4;
s4_max = 0.4 mm;
s4_min = 0 mm
Deflection span 5;
s5_max = 0.2 mm;
s5_min = 0 mm
Maximum reaction at support A;
RA_max = -64.2 kN;
RA_min = -64.2 kN
Unfactored dead load reaction at support A;
RA_Dead = -42.6 kN
Unfactored imposed load reaction at support A;
RA_Imposed = -2.9 kN
Maximum reaction at support B;
RB_max = 575.2 kN;
Unfactored dead load reaction at support B;
RB_Dead = 381.5 kN
Unfactored imposed load reaction at support B;
RB_Imposed = 25.7 kN
Maximum reaction at support C;
RC_max = 585.9 kN;
Unfactored dead load reaction at support C;
RC_Dead = 388.6 kN
Unfactored imposed load reaction at support C;
RC_Imposed = 26.1 kN
Maximum reaction at support D;
RD_max = 298.3 kN;
Unfactored dead load reaction at support D;
RD_Dead = 197.9 kN
Unfactored imposed load reaction at support D;
RD_Imposed = 13.3 kN
Maximum reaction at support E;
RE_max = 267.9 kN;
Unfactored dead load reaction at support E;
RE_Dead = 177.7 kN
Unfactored imposed load reaction at support E;
RE_Imposed = 12 kN
Maximum reaction at support F;
RF_max = 71.5 kN;
Unfactored dead load reaction at support F;
RF_Dead = 47.4 kN
Unfactored imposed load reaction at support F;
RF_Imposed = 3.2 kN
RB_min = 575.2 kN
RC_min = 585.9 kN
RD_min = 298.3 kN
RE_min = 267.9 kN
RF_min = 71.5 kN
Section details
Support conditions
Support A
Maximum moment;
Section type;
Vertically restrained
UKC 203x203x60 (Corus Advance);
Steel grade;
S355
Rotationally free
Support B
Vertically restrained
Rotationally free
Support C
Vertically restrained
Rotationally free
Support D
Vertically restrained
Rotationally free
Support E
Vertically restrained
Rotationally free
Support F
Vertically restrained
Rotationally free
Applied loading
Beam loads
Dead self weight of beam  1
Dead full UDL 88.6 kN/m
Page 62 of 72
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Classification of cross sections - Section 3.5
Tensile strain coefficient;
 = 0.88;
Section classification;
Plastic
Design shear resistance;
Pv = 419.7 kN
Shear capacity - Section 4.2.3
Design shear force;
Fv = 322.8 kN;
PASS - Design shear resistance exceeds design shear force
Moment capacity at span 2 - Section 4.2.5
Design bending moment;
M = 221.6 kNm;
Moment capacity high shear;
Mc = 222.3 kNm
PASS - Moment capacity exceeds design bending moment
Check vertical deflection - Section 2.5.2
Consider deflection due to dead and imposed loads
Limiting deflection;
lim = 19.2 mm;
Maximum deflection;
 = 16.446 mm
PASS - Maximum deflection does not exceed deflection limit
Page 63 of 72
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Compressive strength - Section 4.7.5
PROP DESIGN
STEEL MEMBER DESIGN (BS5950)
In accordance with BS5950-1:2000 incorporating Corrigendum No.1
TEDDS calculation version 3.0.04
Limiting slenderness;
0 = 0.2  (  E / py)
Strut curve - Table 23;
b
Robertson constant;
x = 3.5
Perry factor;
x = x  (x - 0) / 1000 = 0.182
Euler stress;
pEx =   E / x = 450.7 N/mm
Section details
UKC 203x203x60 (Corus Advance)
Steel grade;
S355
Compressive strength - Annex C.1;
Modulus of elasticity;
= 15.099
2
2
2
pcx = pEx  py / (x + (x - pEx  py) ) = 251.6 N/mm
2
0.5
2
Compression resistance - Section 4.7.4
From table 9: Design strength py
Design strength;
2
0.5
x = (py + (x + 1)  pEx) / 2 = 443.8 N/mm
Section type;
Thickness of element;
2
Compression resistance - cl.4.7.4;
max(T, t) = 14.2 mm
py = 355 N/mm
Pcx = A  pcx = 1921.3 kN
PASS - Compression resistance exceeds design compression force
2
2
E = 205000 N/mm
Effective length for minor (y-y) axis buckling - Section 4.7.3
Effective length for buckling;
LEy = Ly  Ky = 6000 mm
Slenderness ratio - cl.4.7.2;
y = LEy / ryy = 115.399
Compressive strength - Section 4.7.5
Limiting slenderness;
0 = 0.2  (  E / py)
Strut curve - Table 23;
c
Robertson constant;
y = 5.5
Perry factor;
y = y  (y - 0) / 1000 = 0.552
Euler stress;
pEy =   E / y = 151.9 N/mm
2
2
0.5
2
= 15.099
2
y = (py + (y + 1)  pEy) / 2 = 295.4 N/mm
Compressive strength - Annex C.1;
2
pcy = pEy  py / (y + (y - pEy  py) ) = 112.9 N/mm
2
0.5
2
Compression resistance - Section 4.7.4
Compression resistance - cl.4.7.4;
Pcy = A  pcy = 862 kN
PASS - Compression resistance exceeds design compression force
Lateral restraint
Distance between major axis restraints;
Lx = 6000 mm
Distance between minor axis restraints;
Ly = 6000 mm
Effective length factors
Effective length factor in major axis;
Kx = 1.00
Effective length factor in minor axis;
Ky = 1.00
Effective length factor for lateral-torsional buckling; KLT = 1.00;
Classification of cross sections - Section 3.5
 = [275 N/mm / py] = 0.88
2
Internal compression parts - Table 11
Depth of section;
d = 160.8 mm
Stress ratios;
r1 = min(Fc / (d  t  pyw), 1) = 1
r2 = Fc / (A  pyw) = 0.216
d / t = 19.4   <= max(80   / (1 + r1), 40  );
Class 1 plastic
Outstand flanges - Table 11
Width of section;
b = B / 2 = 102.9 mm
b / T = 8.2   <= 9  ;
Class 1 plastic
Section is class 1 plastic
Compression members - Section 4.7
Design compression force;
Fc = 585 kN
Effective length for major (x-x) axis buckling - Section 4.7.3
Effective length for buckling;
LEx = Lx  Kx = 6000 mm
Slenderness ratio - cl.4.7.2;
x = LEx / rxx = 67.002
Page 64 of 72
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FACING WALL ANALYSIS & DESIGN (TWO STOREY)
CONTINUOUS BEAM ANALYSIS - RESULTS
Unfactored support reactions
Dead
(kN)
Imposed
(kN)
Support A;
4.4
-2.7
0.0
0.0
0.0
0.0
0.0
0.0
Support B;
-76.1
-11.0
0.0
0.0
0.0
0.0
0.0
0.0
Support C;
-96.3
-6.2
0.0
0.0
0.0
0.0
0.0
0.0
RC SLAB DESIGN (BS8110:PART1:1997)
TEDDS calculation version 1.0.04
CONTINUOUS ONE WAY SPANNING SLAB DEFINITION
;
Overall depth of slab; h = 200 mm
Sagging steel
;
Cover to tension reinforcement resisting sagging; csag = 30 mm
;
Trial bar diameter; Dtryx = 20 mm
Depth to tension steel (resisting sagging)
dx = h - csag - Dtryx/2 = 160 mm
Hogging steel
;
Page 65 of 72
Cover to tension reinforcement resisting hogging; c hog = 40 mm
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Area of reinforcement required
;;
;
Kxhog = abs(msxhog) / ( dxhog  fcu ) = 0.091
2
K'x = min (0.156 , (0.402  (bx - 0.4)) - (0.18  (bx - 0.4) )) = 0.156
2
Trial bar diameter; Dtryxhog = 20 mm
Depth to tension steel (resisting hogging)
Outer compression steel not required to resist hogging
Slab requiring outer tension steel only - bars (hogging)
dxhog = h - chog - Dtryxhog/2 = 150 mm
;;
Materials
Neutral axis depth; xxhog = (dxhog - zxhog) / 0.45 = 38 mm
2
;
Characteristic strength of reinforcement; fy = 500 N/mm
;
Characteristic strength of concrete; fcu = 35 N/mm
Area of tension steel required
2
;;;
Asyhog
dx
h
zxhog = min (( 0.95  dxhog),(dxhog(0.5+0.25-Kxhog/0.9)))) = 133 mm
Nominal 1 m width
h
Asxhog_req = abs(msxhog) / (1/ms fy  zxhog) = 1247 mm /m
2
Tension steel
Asxhog
;;Provide 20 dia bars @ 150 centres; outer tension steel resisting hogging
2
Asxhog_prov = Asxhog = 2090 mm /m
dxhog
Area of outer tension steel provided sufficient to resist hogging
Asy
Asx
Nominal 1 m width
(hogging)
(sagging)
TRANSVERSE TOP STEEL - INNER
One-way spanning slab
;;Inner layer of transverse steel;
(continuous)
Provide 12 dia bars @ 150 centres
2
Asyhog_prov = Asyhog = 754 mm /m
ONE WAY SPANNING SLAB (CL 3.5.4)
Check min and max areas of steel resisting sagging
2
;Total area of concrete; Ac = h = 200000 mm /m
MAXIMUM DESIGN MOMENTS IN SPAN
;
Design sagging moment (per m width of slab); m sx = 33.0 kNm/m
;
Minimum % reinforcement; k = 0.13 %
CONCRETE SLAB DESIGN – SAGGING – OUTER LAYER OF STEEL (CL 3.5.4)
Ast_min = k  Ac = 260 mm /m
;
Design sagging moment (per m width of slab); m sx = 33.0 kNm/m
Ast_max = 4 %  Ac = 8000 mm /m
;
Moment Redistribution Factor; bx = 1.0
2
2
Steel defined:
;
Area of reinforcement required
;;
2
Outer steel resisting sagging; Asx_prov = 754 mm /m
Kx = abs(msx) / ( dx  fcu ) = 0.037
2
Area of outer steel provided (sagging) OK
K'x = min (0.156 , (0.402  (bx - 0.4)) - (0.18  (bx - 0.4) )) = 0.156
2
;
2
Inner steel resisting sagging; Asy_prov = 754 mm /m
Area of inner steel provided (sagging) OK
Outer compression steel not required to resist sagging
Slab requiring outer tension steel only - bars (sagging)
;;
zx = min (( 0.95  dx),(dx(0.5+0.25-Kx/0.9)))) = 152 mm
Check min and max areas of steel resisting hogging
2
;Total area of concrete; Ac = h = 200000 mm /m
Neutral axis depth; xx = (dx - zx) / 0.45 = 18 mm
;
Ast_min = k  Ac = 260 mm /m
2
Area of tension steel required
;;;
Minimum % reinforcement; k = 0.13 %
Asx_req = abs(msx) / (1/ms fy  zx) = 499 mm /m
2
Ast_max = 4 %  Ac = 8000 mm /m
2
Tension steel
Steel defined:
;;Provide 12 dia bars @ 150 centres; outer tension steel resisting sagging
2
Asx_prov = Asx = 754 mm /m
;
2
Outer steel resisting hogging; Asxhog_prov = 2090 mm /m
Area of outer steel provided (hogging) OK
Area of outer tension steel provided sufficient to resist sagging
;
2
Inner steel resisting hogging ; Asyhog_prov = 754 mm /m
Area of inner steel provided (hogging) OK
TRANSVERSE BOTTOM STEEL - INNER
;;Inner layer of transverse steel;
SHEAR RESISTANCE OF CONCRETE SLABS (CL 3.5.5)
Provide 12 dia bars @ 150 centres
2
Asy_prov = Asy = 754 mm /m
Outer tension steel resisting hogging moments
;
Depth to tension steel from compression face; dxhog = 150 mm
MAXIMUM DESIGN MOMENTS OVER SUPPORT
;
Area of tension reinforcement provided (per m width of slab); A sxhog_prov = 2090 mm /m
;
Design hogging moment (per m width of slab); m sxhog = 72.0 kNm/m
;
Design ultimate shear force (per m width of slab); V xhog = 145 kN/m
CONCRETE SLAB DESIGN – HOGGING – OUTER LAYER OF STEEL (CL 3.5.4)
;
Characteristic strength of concrete; fcu = 35 N/mm
;
Design hogging moment (per m width of slab); msxhog = 72.0 kNm/m
;
Moment Redistribution Factor; bx = 1.0
2
2
Applied shear stress
vxhog = Vxhog / dxhog = 0.97 N/mm
Page 66 of 72
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Pringuer-James Consulting Engineers Ltd