Job Number Sheet

Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Column Base Design : F1
C15
Input Data

Base length A (m) 1.40

Base width B (m) 1.70
Column(s) Col 1 Col 2
C (m) 0.2
D (m) 0.4
E (m)
F (m)
Stub column height X (m) 0.65
Base depth Y (m) 0.35
Soil cover Z (m) 0.60
Concrete density (kN/m 3) 25
Soil density (kN/m 3) 18
Soil friction angle (°) 22
Base friction constant 0.5
Rebar depth top X (mm) 50
Rebar depth top Y (mm) 60
Rebar depth bottom X (mm) 50
Rebar depth bottom Y (mm) 60
ULS ovt. LF: Self weight 0.9
ULS LF: Self weight 1.4
Max. SLS bearing pr. (kN/m²) 120
S.F. Overturning (ULS) 1
S.F. Slip (ULS) 1
fcu base (MPa) 35
fcu columns (MPa) 35
fy (MPa) 460
 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Loading Data

Unfactored Loads

Load Col LF LF P Hx Hy Mx My
Case no. ULS ULS (kN) (kN) (kN) (kNm) (kNm)
ovt
1 1 1.6 1.6 230
Sketch of Base

BS8110 - 1997 Y

C1
X
B

D1

A
Y

Mx
P

Hx

X
Z

Y
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Output for Load Case 1
Note: The calculated flexural reinforcement was used in the calculations.

Output for Load Case 1

Soil pressure (ULS)(kN/m²) 182.25
Soil pressure (SLS)(kN/m²) 116.37
SF overturning (SLS) >100
SF overturning (ULS) >100
Safety Factor slip (ULS) >100
Safety Factor uplift (ULS) >100
Bottom
Design moment X (kNm/m) 27.88
Reinforcement X (mm²/m) 223
Design moment Y (kNm/m) 32.72
Reinforcement Y (mm²/m) 271
Top
Design moment X (kNm/m) 0.00
Reinforcement X (mm²/m) 0
Design moment Y (kNm/m) 0.00
Reinforcement Y (mm²/m) 0
Linear Shear X (MPa) 0.155
vc (MPa) 0.404
Linear Shear Y (MPa) 0.191
vc (MPa) 0.407
Linear Shear Other (MPa) 0.000
Punching Shear (MPa) 0.107
vc (MPa) 0.405
vcu (MPa) 4.73
v col face (MPa) 1.18
Cost 370.14

Load Case:1 Legend

Max M (+)
BS8110 - 1997 Y Y Max M (-)
Max Shear

0.20 0.20
X X
1.70

1.70

0.40 0.40

1.40 1.40
Y Y

Mx Mx
P P

Hx Hx
0.60

0.60

0.65 0.65
0.35 0.35

116.37 kN/m² 182.25 kN/m²

Soil Pressures at SLS Soil Pressures at ULS
 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Reinforcement Calculation for X-direction:Positive Moments

Load Case 1

Design moment:27.88 kNm/m

Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
300
=
.00219
1+
.0035
= 184.534 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 184.519 mm

Depth of rectangular stress block:

a = b. x
= .9 ×150
= 135.000 mm

Concrete moment capacity: 3.4.4.4

 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
135
= 135 × 300 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 488.255 kNm

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
2788×104
= 300 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×300 2 ×35
= 296.983 mm

Lever arm limitation 3.4.4.4

z = min(296.983 , 0.95*300.000) = 285.000 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
2788×104
=
438.1 ×285
= 223.293 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(300 - 285 )
=
.9
= 33.333 mm

Reinforcement Calculation for Y-direction:Positive Moments

Load Case 1

Design moment:32.72 kNm/m

 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
290
=
.00219
1+
.0035
= 178.383 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 178.368 mm

Depth of rectangular stress block:

a = b. x
= .9 ×145
= 130.500 mm

Concrete moment capacity: 3.4.4.4

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
130.5
= 130.5 × 290 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 456.247 kNm
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 a f b . d 2 . f cu
. . .

3272×104
= 290 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×290 2 ×35
= 286.327 mm

Lever arm limitation 3.4.4.4

z = min(286.327 , 0.95*290.000) = 275.500 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
3272×104
=
438.1 ×275.5
= 271.093 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(290 - 275.5 )
=
.9
= 32.222 mm

Punching Shear Stress/Reinforcement Calculation

Load Case 1

Column number:1

The shear force outside the critical perimeter amounts to:44.23 kN

Calculation of required punching shear reinforcement:

Area used for shear

 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Ash = ucrit. d
= 1400 ×295
= 413.0×103 mm²

% Rebar in X - direction tab 3.11

100 . Asx
F1x =
xb. d
100 ×312.59
=
1400 ×295
= 0.0757

% Rebar in Y - direction tab 3.11

100 . Asy
F1y =
yb. d
100 ×404.57
=
1 492.5 ×295
= 0.0919

tab 3.11

F1x = min(0.15 , F1x ) = 0.150

tab 3.11

F1y = min(0.15 , F1y ) = 0.150

tab 3.11

400
F2 =
d
400
=
295
= 1.356

tab 3.11
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

0.79 .
v cx = F1x0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

tab 3.11

0.79 .
v cy = F1y0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

Weighted average of vc

v cx. nx. xb + v cy. ny. yb

vc =
nx. xb + ny. yb
.36236 ×2 ×1400 + .36236 ×0 ×1 492.5
=
2 ×1400 + 0 ×1 492.5
= 0.3624 MPa

Actual shear stress

Veff
vv =
Ash
44233
=
413000
= 0.1071 MPa

tab 3.11

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

tab 3.11
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

v c = v c_. F3
= .36236 ×1.1187
= 0.4054 MPa

Shear capacity without reinforcement

v c. Ash
Vcap =
1000
.40537 ×413000
=
1000
= 167.418 kN

The concrete shear capacity for this perimeter is sufficient: no shear reinforcement required

Shear Stress/Reinforcement Calculation for Y-direction

Load Case 1

Design shear force:55.29 kN

Based on flexural reinforcement of:271mm²/m

Shear Calculations:
Area used for shear calculations:
Ash = bw. de - (bw - bf ). bf
= 1000 ×290 - (1000 - 0 )×0
= 290.0×103 mm²

Reinforcement ratio 3.4.5.4-6

Ast
F1 =
Ash
271.07
=
290000
= 0.0009

f 1 = f 1i. 100
= .00093 ×100
= 0.0930
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
F1 = min(3 , F1 ) = 0.093

F1 = max(0.15 , F1 ) = 0.150 3.4.5.4-6

400
f2 =
de
400
=
290
= 1.379

0.79 . 0.33333. 0.25

vc = F1 F2
1.25
0.79
= ×.15 0.33333×1.3793 0.25
1.25
= 0.3639 MPa

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

v c = v c1. F3
= .36391 ×1.1187
= 0.4071 MPa

Maximum allowable shear stress

v cu = 0.8 . f cu
= 0.8 × 35
= 4.733 MPa

vcu = min(5 , vcu ) = 4.733

Actual Shear stress:

V . 1×103
v=
Ash
55.292 ×1×103
=
290000
= 0.1907 MPa
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Shear force capacity

Vco = v c. Ash. 1×10-3

= .40711 ×290000 ×1×10-3
= 118.062 kN

Shear reinforcement Table 3.16

(v - v c). b
Asv/sv =
0.95 . f yv
(.19066 - .40711 )×1000
=
0.95 ×460
= -0.4953

Minimum links Table 3.16

0.40 . b
Asv/svn=
0.95 . f yv
0.40 ×1000
=
0.95 ×460
= 0.9153
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Column Base Design : F2
C15
Input Data

Base length A (m) 1.25

Base width B (m) 1.5
Column(s) Col 1 Col 2
C (m) 0.2
D (m) 0.4
E (m)
F (m)
Stub column height X (m) 0.65
Base depth Y (m) 0.35
Soil cover Z (m) 0.60
Concrete density (kN/m 3) 25
Soil density (kN/m 3) 18
Soil friction angle (°) 22
Base friction constant 0.5
Rebar depth top X (mm) 50
Rebar depth top Y (mm) 60
Rebar depth bottom X (mm) 50
Rebar depth bottom Y (mm) 60
ULS ovt. LF: Self weight 0.9
ULS LF: Self weight 1.4
Max. SLS bearing pr. (kN/m²) 120
S.F. Overturning (ULS) 1
S.F. Slip (ULS) 1
fcu base (MPa) 35
fcu columns (MPa) 35
fy (MPa) 460
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Loading Data

Unfactored Loads

Load Col LF LF P Hx Hy Mx My
Case no. ULS ULS (kN) (kN) (kN) (kNm) (kNm)
ovt
1 1 1.6 1.6 185

Sketch of Base

BS8110 - 1997 Y

C1
X
B

D1

A
Y
Mx
P

Hx

X
Z

Y
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Output for Load Case 1
Note: The calculated flexural reinforcement was used in the calculations.

Output for Load Case 1

Soil pressure (ULS)(kN/m²) 185.56
Soil pressure (SLS)(kN/m²) 118.45
SF overturning (SLS) >100
SF overturning (ULS) >100
Safety Factor slip (ULS) >100
Safety Factor uplift (ULS) >100
Bottom
Design moment X (kNm/m) 21.80
Reinforcement X (mm²/m) 175
Design moment Y (kNm/m) 23.93
Reinforcement Y (mm²/m) 198
Top
Design moment X (kNm/m) 0.00
Reinforcement X (mm²/m) 0
Design moment Y (kNm/m) 0.00
Reinforcement Y (mm²/m) 0
Linear Shear X (MPa) 0.119
vc (MPa) 0.404
Linear Shear Y (MPa) 0.143
vc (MPa) 0.407
Linear Shear Other (MPa) 0.000
Punching Shear (MPa) 0.052
vc (MPa) 0.405
vcu (MPa) 4.73
v col face (MPa) 0.94
Cost 284.45

Load Case:1 Legend

Max M (+)
BS8110 - 1997 Y Y Max M (-)
Max Shear

0.20 0.20
X X
1.50

1.50

0.40 0.40

1.25 1.25
Y Y

Mx Mx
P P

Hx Hx
0.60

0.60

0.65 0.65

0.35 0.35

118.45 kN/m² 185.56 kN/m²

Soil Pressures at SLS Soil Pressures at ULS
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Reinforcement Calculation for X-direction:Positive Moments

Load Case 1

Design moment:21.80 kNm/m

Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
300
=
.00219
1+
.0035
= 184.534 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 184.519 mm

Depth of rectangular stress block:

a = b. x
= .9 ×150
= 135.000 mm

Concrete moment capacity: 3.4.4.4

 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
135
= 135 × 300 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 488.255 kNm

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
2180×104
= 300 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×300 2 ×35
= 297.646 mm

Lever arm limitation 3.4.4.4

z = min(297.646 , 0.95*300.000) = 285.000 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
2180×104
=
438.1 ×285
= 174.598 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(300 - 285 )
=
.9
= 33.333 mm

Reinforcement Calculation for Y-direction:Positive Moments

Load Case 1

Design moment:23.93 kNm/m

 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
290
=
.00219
1+
.0035
= 178.383 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 178.368 mm

Depth of rectangular stress block:

a = b. x
= .9 ×145
= 130.500 mm

Concrete moment capacity: 3.4.4.4

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
130.5
= 130.5 × 290 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 456.247 kNm
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 a f b . d 2 . f cu
. . .

2393×104
= 290 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×290 2 ×35
= 287.323 mm

Lever arm limitation 3.4.4.4

z = min(287.323 , 0.95*290.000) = 275.500 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
2393×104
=
438.1 ×275.5
= 198.266 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(290 - 275.5 )
=
.9
= 32.222 mm

Punching Shear Stress/Reinforcement Calculation

Load Case 1

Column number:1

The shear force outside the critical perimeter amounts to:19.28 kN

Calculation of required punching shear reinforcement:

Area used for shear

 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

Ash = ucrit. d
= 1250 ×295
= 368.8×103 mm²

% Rebar in X - direction tab 3.11

100 . Asx
F1x =
xb. d
100 ×218.26
=
1250 ×295
= 0.0592

% Rebar in Y - direction tab 3.11

100 . Asy
F1y =
yb. d
100 ×276.04
=
1 392.5 ×295
= 0.0672

tab 3.11

F1x = min(0.15 , F1x ) = 0.150

tab 3.11

F1y = min(0.15 , F1y ) = 0.150

tab 3.11

400
F2 =
d
400
=
295
= 1.356

tab 3.11
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

0.79 .
v cx = F1x0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

tab 3.11

0.79 .
v cy = F1y0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

Weighted average of vc

v cx. nx. xb + v cy. ny. yb

vc =
nx. xb + ny. yb
.36236 ×2 ×1250 + .36236 ×0 ×1 392.5
=
2 ×1250 + 0 ×1 392.5
= 0.3624 MPa

Actual shear stress

Veff
vv =
Ash
19280
=
368750
= 0.0523 MPa

tab 3.11

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

tab 3.11
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

v c = v c_. F3
= .36236 ×1.1187
= 0.4054 MPa

Shear capacity without reinforcement

v c. Ash
Vcap =
1000
.40537 ×368750
=
1000
= 149.480 kN

The concrete shear capacity for this perimeter is sufficient: no shear reinforcement required

Shear Stress/Reinforcement Calculation for Y-direction

Load Case 1

Design shear force:41.53 kN

Based on flexural reinforcement of:198mm²/m

Shear Calculations:
Area used for shear calculations:
Ash = bw. de - (bw - bf ). bf
= 1000 ×290 - (1000 - 0 )×0
= 290.0×103 mm²

Reinforcement ratio 3.4.5.4-6

Ast
F1 =
Ash
198.24
=
290000
= 0.0007

f 1 = f 1i. 100
= .00068 ×100
= 0.0680
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
F1 = min(3 , F1 ) = 0.068

F1 = max(0.15 , F1 ) = 0.150 3.4.5.4-6

400
f2 =
de
400
=
290
= 1.379

0.79 . 0.33333. 0.25

vc = F1 F2
1.25
0.79
= ×.15 0.33333×1.3793 0.25
1.25
= 0.3639 MPa

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

v c = v c1. F3
= .36391 ×1.1187
= 0.4071 MPa

Maximum allowable shear stress

v cu = 0.8 . f cu
= 0.8 × 35
= 4.733 MPa

vcu = min(5 , vcu ) = 4.733

Actual Shear stress:

V . 1×103
v=
Ash
41.525 ×1×103
=
290000
= 0.1432 MPa
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Shear force capacity

Vco = v c. Ash. 1×10-3

= .40711 ×290000 ×1×10-3
= 118.062 kN

Shear reinforcement Table 3.16

(v - v c). b
Asv/sv =
0.95 . f yv
(.14319 - .40711 )×1000
=
0.95 ×460
= -0.6039

Minimum links Table 3.16

0.40 . b
Asv/svn=
0.95 . f yv
0.40 ×1000
=
0.95 ×460
= 0.9153
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Column Base Design : F3
C15
Input Data

Base length A (m) 1.20

Base width B (m) 1.30
Column(s) Col 1 Col 2
C (m) 0.2
D (m) 0.4
E (m)
F (m)
Stub column height X (m) 0.65
Base depth Y (m) 0.35
Soil cover Z (m) 0.60
Concrete density (kN/m 3) 25
Soil density (kN/m 3) 18
Soil friction angle (°) 22
Base friction constant 0.5
Rebar depth top X (mm) 50
Rebar depth top Y (mm) 60
Rebar depth bottom X (mm) 50
Rebar depth bottom Y (mm) 60
ULS ovt. LF: Self weight 0.9
ULS LF: Self weight 1.4
Max. SLS bearing pr. (kN/m²) 120
S.F. Overturning (ULS) 1
S.F. Slip (ULS) 1
fcu base (MPa) 35
fcu columns (MPa) 35
fy (MPa) 460
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Loading Data

Unfactored Loads

Load Col LF LF P Hx Hy Mx My
Case no. ULS ULS (kN) (kN) (kN) (kNm) (kNm)
ovt
1 1 1.6 1.6 155
Sketch of Base

BS8110 - 1997 Y

C1
X
B

D1

A
Y
Mx
P

Hx

X
Z

Y
 Job Number Sheet
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Output for Load Case 1
Note: The calculated flexural reinforcement was used in the calculations.

Output for Load Case 1

Soil pressure (ULS)(kN/m²) 186.74
Soil pressure (SLS)(kN/m²) 119.19
SF overturning (SLS) >100
SF overturning (ULS) >100
Safety Factor slip (ULS) >100
Safety Factor uplift (ULS) >100
Bottom
Design moment X (kNm/m) 19.92
Reinforcement X (mm²/m) 160
Design moment Y (kNm/m) 16.14
Reinforcement Y (mm²/m) 134
Top
Design moment X (kNm/m) 0.00
Reinforcement X (mm²/m) 0
Design moment Y (kNm/m) 0.00
Reinforcement Y (mm²/m) 0
Linear Shear X (MPa) 0.105
vc (MPa) 0.404
Linear Shear Y (MPa) 0.086
vc (MPa) 0.407
Linear Shear Other (MPa) 0.000
Punching Shear (MPa) 0.029
vc (MPa) 0.405
vcu (MPa) 4.73
v col face (MPa) 0.78
Cost 232.76

Load Case:1 Legend

Max M (+)
BS8110 - 1997 Y Y Max M (-)
Max Shear

0.20 0.20
X X
1.30

1.30

0.40 0.40

1.20 1.20
Y Y

Mx Mx
P P

Hx Hx
0.60

0.60

0.65 0.65

0.35 0.35

119.19 kN/m² 186.74 kN/m²

Soil Pressures at SLS Soil Pressures at ULS
 Job Number Sheet
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Reinforcement Calculation for X-direction:Positive Moments

Load Case 1

Design moment:19.92 kNm/m

Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
300
=
.00219
1+
.0035
= 184.534 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 184.519 mm

Depth of rectangular stress block:

a = b. x
= .9 ×150
= 135.000 mm

Concrete moment capacity: 3.4.4.4

 Job Number Sheet
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a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
135
= 135 × 300 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 488.255 kNm

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 a f b . d 2 . f cu
. . .

1992×104
= 300 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×300 2 ×35
= 297.850 mm

Lever arm limitation 3.4.4.4

z = min(297.850 , 0.95*300.000) = 285.000 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
1992×104
=
438.1 ×285
= 159.541 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(300 - 285 )
=
.9
= 33.333 mm

Reinforcement Calculation for Y-direction:Positive Moments

Load Case 1

Design moment:16.14 kNm/m

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
290
=
.00219
1+
.0035
= 178.383 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 178.368 mm

Depth of rectangular stress block:

a = b. x
= .9 ×145
= 130.500 mm

Concrete moment capacity: 3.4.4.4

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
130.5
= 130.5 × 290 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 456.247 kNm
 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
1614×104
= 290 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×290 2 ×35
= 288.200 mm

Lever arm limitation 3.4.4.4

z = min(288.200 , 0.95*290.000) = 275.500 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
1614×104
=
438.1 ×275.5
= 133.724 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(290 - 275.5 )
=
.9
= 32.222 mm

Punching Shear Stress/Reinforcement Calculation

Load Case 1

Column number:1

The shear force outside the critical perimeter amounts to:22.37 kN

Calculation of required punching shear reinforcement:

Area used for shear

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Ash = ucrit. d
= 2600 ×295
= 767.0×103 mm²

% Rebar in X - direction tab 3.11

100 . Asx
F1x =
xb. d
100 ×173.11
=
1085 ×295
= 0.0541

% Rebar in Y - direction tab 3.11

100 . Asy
F1y =
yb. d
100 ×173.8
=
1300 ×295
= 0.0453

tab 3.11

F1x = min(0.15 , F1x ) = 0.150

tab 3.11

F1y = min(0.15 , F1y ) = 0.150

tab 3.11

400
F2 =
d
400
=
295
= 1.356

tab 3.11
 Job Number Sheet
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0.79 .
v cx = F1x0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

tab 3.11

0.79 .
v cy = F1y0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

Weighted average of vc

v cx. nx. xb + v cy. ny. yb

vc =
nx. xb + ny. yb
.36236 ×2 ×1085 + .36236 ×2 ×1300
=
2 ×1085 + 2 ×1300
= 0.3624 MPa

Actual shear stress

Veff
vv =
Ash
22375
=
767000
= 0.0292 MPa

tab 3.11

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

tab 3.11
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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v c = v c_. F3
= .36236 ×1.1187
= 0.4054 MPa

Shear capacity without reinforcement

v c. Ash
Vcap =
1000
.40537 ×767000
=
1000
= 310.919 kN

The concrete shear capacity for this perimeter is sufficient: no shear reinforcement required

Shear Stress/Reinforcement Calculation for Y-direction

Load Case 1

Design shear force:24.86 kN

Based on flexural reinforcement of:134mm²/m

Shear Calculations:
Area used for shear calculations:
Ash = bw. de - (bw - bf ). bf
= 1000 ×290 - (1000 - 0 )×0
= 290.0×103 mm²

Reinforcement ratio 3.4.5.4-6

Ast
F1 =
Ash
133.69
=
290000
= 0.0005

f 1 = f 1i. 100
= .00046 ×100
= 0.0460
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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F1 = min(3 , F1 ) = 0.046

F1 = max(0.15 , F1 ) = 0.150 3.4.5.4-6

400
f2 =
de
400
=
290
= 1.379

0.79 . 0.33333. 0.25

vc = F1 F2
1.25
0.79
= ×.15 0.33333×1.3793 0.25
1.25
= 0.3639 MPa

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

v c = v c1. F3
= .36391 ×1.1187
= 0.4071 MPa

Maximum allowable shear stress

v cu = 0.8 . f cu
= 0.8 × 35
= 4.733 MPa

vcu = min(5 , vcu ) = 4.733

Actual Shear stress:

V . 1×103
v=
Ash
24.861 ×1×103
=
290000
= 0.0857 MPa
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Shear force capacity

Vco = v c. Ash. 1×10-3

= .40711 ×290000 ×1×10-3
= 118.062 kN

Shear reinforcement Table 3.16

(v - v c). b
Asv/sv =
0.95 . f yv
(.08573 - .40711 )×1000
=
0.95 ×460
= -0.7354

Minimum links Table 3.16

0.40 . b
Asv/svn=
0.95 . f yv
0.40 ×1000
=
0.95 ×460
= 0.9153
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
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Column Base Design : F4
C15
Input Data

Base length A (m) 1.20

Base width B (m) 1.20
Column(s) Col 1 Col 2
C (m) 0.2
D (m) 0.4
E (m)
F (m)
Stub column height X (m) 0.65
Base depth Y (m) 0.35
Soil cover Z (m) 0.60
Concrete density (kN/m 3) 25
Soil density (kN/m 3) 18
Soil friction angle (°) 22
Base friction constant 0.5
Rebar depth top X (mm) 50
Rebar depth top Y (mm) 60
Rebar depth bottom X (mm) 50
Rebar depth bottom Y (mm) 60
ULS ovt. LF: Self weight 0.9
ULS LF: Self weight 1.4
Max. SLS bearing pr. (kN/m²) 120
S.F. Overturning (ULS) 1
S.F. Slip (ULS) 1
fcu base (MPa) 35
fcu columns (MPa) 35
fy (MPa) 460
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Loading Data

Unfactored Loads

Load Col LF LF P Hx Hy Mx My
Case no. ULS ULS (kN) (kN) (kN) (kNm) (kNm)
ovt
1 1 1.6 1.6 140
Sketch of Base

BS8110 - 1997 Y

C1
X
B

D1

A
Mx Y
P

Hx

X
Z

Y
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Output for Load Case 1
Note: The calculated flexural reinforcement was used in the calculations.

Output for Load Case 1

Soil pressure (ULS)(kN/m²) 183.35
Soil pressure (SLS)(kN/m²) 117.08
SF overturning (SLS) >100
SF overturning (ULS) >100
Safety Factor slip (ULS) >100
Safety Factor uplift (ULS) >100
Bottom
Design moment X (kNm/m) 19.50
Reinforcement X (mm²/m) 156
Design moment Y (kNm/m) 12.48
Reinforcement Y (mm²/m) 103
Top
Design moment X (kNm/m) 0.00
Reinforcement X (mm²/m) 0
Design moment Y (kNm/m) 0.00
Reinforcement Y (mm²/m) 0
Linear Shear X (MPa) 0.103
vc (MPa) 0.404
Linear Shear Y (MPa) 0.058
vc (MPa) 0.407
Linear Shear Other (MPa) 0.000
Punching Shear (MPa) 0.029
vc (MPa) 0.405
vcu (MPa) 4.73
v col face (MPa) 0.70
Cost 213.34

Load Case:1 Legend

Max M (+)
BS8110 - 1997 Y Y Max M (-)
Max Shear

0.20 0.20
X X
1.20

1.20

0.40 0.40

1.20 1.20
Y Y

Mx Mx
P P

Hx Hx
0.60

0.60

0.65 0.65

0.35 0.35

117.08 kN/m² 183.35 kN/m²

Soil Pressures at SLS Soil Pressures at ULS
 Job Number Sheet
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Reinforcement Calculation for X-direction:Positive Moments

Load Case 1

Design moment:19.50 kNm/m

Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
300
=
.00219
1+
.0035
= 184.534 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 184.519 mm

Depth of rectangular stress block:

a = b. x
= .9 ×150
= 135.000 mm

Concrete moment capacity: 3.4.4.4

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
135
= 135 × 300 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 488.255 kNm

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 a f b . d 2 . f cu
. . .

1950×104
= 300 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×300 2 ×35
= 297.896 mm

Lever arm limitation 3.4.4.4

z = min(297.896 , 0.95*300.000) = 285.000 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
1950×104
=
438.1 ×285
= 156.177 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(300 - 285 )
=
.9
= 33.333 mm

Reinforcement Calculation for Y-direction:Positive Moments

Load Case 1

Design moment:12.48 kNm/m

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
290
=
.00219
1+
.0035
= 178.383 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 178.368 mm

Depth of rectangular stress block:

a = b. x
= .9 ×145
= 130.500 mm

Concrete moment capacity: 3.4.4.4

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
130.5
= 130.5 × 290 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 456.247 kNm
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
1248×104
= 290 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×290 2 ×35
= 288.610 mm

Lever arm limitation 3.4.4.4

z = min(288.610 , 0.95*290.000) = 275.500 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
1248×104
=
438.1 ×275.5
= 103.400 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(290 - 275.5 )
=
.9
= 32.222 mm

Punching Shear Stress/Reinforcement Calculation

Load Case 1

Column number:1

The shear force outside the critical perimeter amounts to:10.11 kN

Calculation of required punching shear reinforcement:

Area used for shear

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Ash = ucrit. d
= 1200 ×295
= 354.0×103 mm²

% Rebar in X - direction tab 3.11

100 . Asx
F1x =
xb. d
100 ×178.41
=
1 142.5 ×295
= 0.0529

% Rebar in Y - direction tab 3.11

100 . Asy
F1y =
yb. d
100 ×124.06
=
1200 ×295
= 0.0350

tab 3.11

F1x = min(0.15 , F1x ) = 0.150

tab 3.11

F1y = min(0.15 , F1y ) = 0.150

tab 3.11

400
F2 =
d
400
=
295
= 1.356

tab 3.11
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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0.79 .
v cx = F1x0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

tab 3.11

0.79 .
v cy = F1y0.33333. F20.25
1.25
0.79
= ×.15 0.33333×1.3559 0.25
1.25
= 0.3624 MPa

Weighted average of vc

v cx. nx. xb + v cy. ny. yb

vc =
nx. xb + ny. yb
.36236 ×2 ×1 142.5 + .36236 ×1 ×1200
=
2 ×1 142.5 + 1 ×1200
= 0.3624 MPa

Actual shear stress

Veff
vv =
Ash
10107
=
354000
= 0.0286 MPa

tab 3.11

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

tab 3.11
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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v c = v c_. F3
= .36236 ×1.1187
= 0.4054 MPa

Shear capacity without reinforcement

v c. Ash
Vcap =
1000
.40537 ×354000
=
1000
= 143.501 kN

The concrete shear capacity for this perimeter is sufficient: no shear reinforcement required

Shear Stress/Reinforcement Calculation for Y-direction

Load Case 1

Design shear force:16.85 kN

Based on flexural reinforcement of:103mm²/m

Shear Calculations:
Area used for shear calculations:
Ash = bw. de - (bw - bf ). bf
= 1000 ×290 - (1000 - 0 )×0
= 290.0×103 mm²

Reinforcement ratio 3.4.5.4-6

Ast
F1 =
Ash
103.38
=
290000
= 0.0004

f 1 = f 1i. 100
= .00036 ×100
= 0.0360
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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F1 = min(3 , F1 ) = 0.036

F1 = max(0.15 , F1 ) = 0.150 3.4.5.4-6

400
f2 =
de
400
=
290
= 1.379

0.79 . 0.33333. 0.25

vc = F1 F2
1.25
0.79
= ×.15 0.33333×1.3793 0.25
1.25
= 0.3639 MPa

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

v c = v c1. F3
= .36391 ×1.1187
= 0.4071 MPa

Maximum allowable shear stress

v cu = 0.8 . f cu
= 0.8 × 35
= 4.733 MPa

vcu = min(5 , vcu ) = 4.733

Actual Shear stress:

V . 1×103
v=
Ash
16.846 ×1×103
=
290000
= 0.0581 MPa
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Shear force capacity

Vco = v c. Ash. 1×10-3

= .40711 ×290000 ×1×10-3
= 118.062 kN

Shear reinforcement Table 3.16

(v - v c). b
Asv/sv =
0.95 . f yv
(.05809 - .40711 )×1000
=
0.95 ×460
= -0.7987

Minimum links Table 3.16

0.40 . b
Asv/svn=
0.95 . f yv
0.40 ×1000
=
0.95 ×460
= 0.9153
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019
Column Base Design : F5
C15
Input Data

Base length A (m) 0.75

Base width B (m) 1.1
Column(s) Col 1 Col 2
C (m) 0.2
D (m) 0.4
E (m)
F (m)
Stub column height X (m) 0.65
Base depth Y (m) 0.35
Soil cover Z (m) 0.60
Concrete density (kN/m 3) 25
Soil density (kN/m 3) 18
Soil friction angle (°) 22
Base friction constant 0.5
Rebar depth top X (mm) 50
Rebar depth top Y (mm) 60
Rebar depth bottom X (mm) 50
Rebar depth bottom Y (mm) 60
ULS ovt. LF: Self weight 0.9
ULS LF: Self weight 1.4
Max. SLS bearing pr. (kN/m²) 120
S.F. Overturning (ULS) 1
S.F. Slip (ULS) 1
fcu base (MPa) 35
fcu columns (MPa) 35
fy (MPa) 460
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Loading Data

Unfactored Loads

Load Col LF LF P Hx Hy Mx My
Case no. ULS ULS (kN) (kN) (kN) (kNm) (kNm)
ovt
1 1 1.6 1.6 41
Sketch of Base

BS8110 - 1997 Y

C1

X
B

D1

A
Mx Y
P

Hx

X
Z

Y
 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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Output for Load Case 1
Note: The calculated flexural reinforcement was used in the calculations.

Output for Load Case 1

Soil pressure (ULS)(kN/m²) 107.63
Soil pressure (SLS)(kN/m²) 69.78
SF overturning (SLS) >100
SF overturning (ULS) >100
Safety Factor slip (ULS) >100
Safety Factor uplift (ULS) >100
Bottom
Design moment X (kNm/m) 3.03
Reinforcement X (mm²/m) 24
Design moment Y (kNm/m) 4.92
Reinforcement Y (mm²/m) 41
Top
Design moment X (kNm/m) 0.00
Reinforcement X (mm²/m) 0
Design moment Y (kNm/m) 0.00
Reinforcement Y (mm²/m) 0
Linear Shear X (MPa) 0.000
vc (MPa) 0.404
Linear Shear Y (MPa) 0.017
vc (MPa) 0.407
Linear Shear Other (MPa) 0.000
Punching Shear (MPa) N.A.
vc (MPa) N.A.
vcu (MPa) 4.73
v col face (MPa) 0.23
Cost 117.18

Load Case:1 Legend

Max M (+)
BS8110 - 1997Y Y Max M (-)
Max Shear

0.20 0.20
X X
1.10

1.10

0.40 0.40

0.75 0.75
Y Y

Mx Mx
P P

Hx Hx
0.60

0.60

0.65 0.65

0.35 0.35

69.78 kN/m² 107.63 kN/m²

Soil Pressures at SLS Soil Pressures at ULS
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Reinforcement Calculation for X-direction:Positive Moments

Load Case 1

Design moment:3.03 kNm/m

Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
300
=
.00219
1+
.0035
= 184.534 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 184.519 mm

Depth of rectangular stress block:

a = b. x
= .9 ×150
= 135.000 mm

Concrete moment capacity: 3.4.4.4

 Job Number Sheet
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GR FLOOR SERV BLOCK-FOUNDATION DESIGN
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a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
135
= 135 × 300 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 488.255 kNm

Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
3035×103
= 300 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×300 2 ×35
= 299.674 mm

Lever arm limitation 3.4.4.4

z = min(299.675 , 0.95*300.000) = 285.000 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
3035×103
=
438.1 ×285
= 24.308 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(300 - 285 )
=
.9
= 33.333 mm

Reinforcement Calculation for Y-direction:Positive Moments

Load Case 1

Design moment:4.92 kNm/m

 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
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Calculation of the maximum steel strain:

f st
est =
Es
438.1
=
200000
= 0.0022

Balanced neutral axis

d
xb =
est
1+
ec
290
=
.00219
1+
.0035
= 178.383 mm

%redis
bb = 1 -
100
0
=1 -
100
= 1.0000

Maximum allowed neutral axis depth: 3.2.2.1

x = min(d/(1+ e st/ e c) , ( b b-0.4)*d) = 178.368 mm

Depth of rectangular stress block:

a = b. x
= .9 ×145
= 130.500 mm

Concrete moment capacity: 3.4.4.4

a . . . . .
Mu = a . d - b a f f cu 1×10-6
2
130.5
= 130.5 × 290 - ×1000 ×.66667 ×.66667 ×35 ×1×10-6
2
= 456.247 kNm
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
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Lever arm 3.4.4.4

M
z = d . 0.5 + 0.25 -
2 . a . f. b . d 2 . f cu
4915×103
= 290 × 0.5 + 0.25 -
2 ×.66667 ×.66667 ×1000 ×290 2 ×35
= 289.454 mm

Lever arm limitation 3.4.4.4

z = min(289.454 , 0.95*290.000) = 275.500 mm

Tension steel 3.4.4.4

M
Ast =
f st. z
4915×103
=
438.1 ×275.5
= 40.722 mm²

Neutral axis depth 3.4.4.4

2 . (d - z )
Neutralaxis
=
b
2 ×(290 - 275.5 )
=
.9
= 32.222 mm

Punching Shear Stress/Reinforcement Calculation

Load Case 1

Column number:1

Shear Stress/Reinforcement Calculation for Y-direction

Load Case 1

Design shear force:4.86 kN

 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
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Based on flexural reinforcement of:41mm²/m

Shear Calculations:
Area used for shear calculations:

Ash = bw. de - (bw - bf ). bf

= 1000 ×290 - (1000 - 0 )×0
= 290.0×103 mm²

Reinforcement ratio 3.4.5.4-6

Ast
F1 =
Ash
40.723
=
290000
= 0.0001

f 1 = f 1i. 100
= .00014 ×100
= 0.0140

F1 = min(3 , F1 ) = 0.014

F1 = max(0.15 , F1 ) = 0.150 3.4.5.4-6

400
f2 =
de
400
=
290
= 1.379

0.79 . 0.33333. 0.25

vc = F1 F2
1.25
0.79
= ×.15 0.33333×1.3793 0.25
1.25
= 0.3639 MPa
 Job Number Sheet
Job Title
GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
E-Mail : mail@prokon.com Calcs by Checked by Date
2/26/2019

f cu 0.33333
F3 =
25
35 0.33333
=
25
= 1.119

v c = v c1. F3
= .36391 ×1.1187
= 0.4071 MPa

Maximum allowable shear stress

v cu = 0.8 . f cu
= 0.8 × 35
= 4.733 MPa

vcu = min(5 , vcu ) = 4.733

Actual Shear stress:

V . 1×103
v=
Ash
4.8554 ×1×103
=
290000
= 0.0167 MPa

Shear force capacity

Vco = v c. Ash. 1×10-3

= .40711 ×290000 ×1×10-3
= 118.062 kN

Shear reinforcement Table 3.16

(v - v c). b
Asv/sv =
0.95 . f yv
(.01674 - .40711 )×1000
=
0.95 ×460
= -0.8933

Minimum links Table 3.16

 Job Number Sheet
Job Title GR FLOOR SERV BLOCK-FOUNDATION DESIGN
Software Consultants (Pty) Ltd Client
Internet: http://www.prokon.com ABBAS AL BALOSHI
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0.40 . b
Asv/svn=
0.95 . f yv
0.40 ×1000
=
0.95 ×460
= 0.9153