### Document

```Check Beam Shear
 V C  0 . 75 
25
 530  2500 / 1000  828 . 13 kN
6
V U at d from column
VU
face 
 223  245 

 * 1 . 5  0 . 53  * 2 . 5
2


 567 . 5 kN
VU   V C
Bending moment design
2.5m
Short direction
M U at d from column
  245  166
MU  
2


face 

1  3 . 5   0 . 5  719 . 25 kN .m
245

b  3500 , d  5 30 mm
6
25 
2  10 * 719.25
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 530 * 3500

A S  0 . 002  530  1000  1060 mm
245
166
2
 10 . 1cm
2

  0.002

short direction
Central band ratio =
 
3.5
 1.4
2.5
2
 1

2
 0.83
2.4
Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2
2 14
7  16/m
7  14/m
2 14
Footing Design
Part II
Combined footing
Example 1
Design a combined footing As shown
q all ( net )  20 t / m  200 kPa
2
f c  25 N mm
2
f y  420 N mm
2
Dimension calculation
The base dimension to get uniform distributed load
2000kN
800 kN
1200 kN
A
x1=0.2m
x2=6.2m
x
800(0.2)+1200(6.2)=2000(x)
x = 3.8m
800 kN
1200 kN
Try thickness
=80cm
2x =7.6 m
Area required
q all ( net )  20 t / m  200 kPa ,
2
Pu  1 . 3 ( Ps )  1 . 3 ( 2000 )  2600 kN
Ag 
qu 
Ps

q all ( net )
Pu
A

2000  10
200  10
 2600   10 3
7 .6 * 1 .8
3
3
 10 m  7 . 6 * 1 . 8
2
 190  10 Pa  190 kPa
3
Check for punching Shear
d = 730 mm
1.13m
0.765
A
b o  2 ( 765 )  1130  2260 mm
fc '
 VC  
3

 VC    2 

b o d  0 . 75 
 sd 
25
 730  2260 / 1000  2062 . 3 kN
3
30  730 
25

b
d

0
.
75
2


 730  2260 / 1000  6027 kN



o
b  12
2260 
12

fc '
V U  800 (1 . 3 )  1 . 13 * 0 . 765 * 190  875 . 8 kN   V c
oK
B
b o  4 ( 730  400 )   4520 mm
 VC  
fc '
3
b o d  0 . 75 
25
 730  4520 / 1000  4124 . 4 kN
3
 sd  fc '
40  730 
25


 VC    2 
b o d  0 . 75  2 
 730  4520 / 1000  13322 . 5 kN


b
12
4520
12




V U  1200 (1 . 3 )  1 . 13 * 1 . 13 * 190  1317 . 4 kN   V c
oK
Draw S.F.D & B.M.D
Stress under footing
= 190 *1.8 = 342 kN/m
Check for beam shear
b = 1800mm, d = 730mm
 V C  0 . 75 
25
 730  1800 / 1000  821 . 25 kN
6
Max .  V U at d from column
VU   V C
face  762 . 34 kN
Bending moment Long
direction
 ve M  1366 kN .m
b  1 800 mm , d  7 30 mm
6
25 
2  10 * 1366
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 730 * 1800

A S  0 . 0039  730  1000  2847 mm
2
 28 . 5 cm

  0.0039

use 9 20 / m
2
Top
 ve M  246 . 7 kN .m
b  1 800 mm , d  7 30 mm
6
25 
2  10 * 246.7
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 730 * 1800

A S min  0 . 0018  800  1000  1440 mm
2

  0.0007   min

 14 . 4 cm
2
use 7  16 / m
Bottom
Bending moment Short
direction
Under Column A
2
0 . 765  1 . 8  0 . 4 
M 


  141 . 6
(1 . 8 * 0 . 765 )
2 
2

1040
b  765 mm , d  7 30 mm
6
25 
2  10 * 141.6
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 730 * 765

A S min  0 . 0018  800  765  1101 . 6 mm
2

   min

 11 cm
2
use 7  14 / m
Under Column B
2
1 . 13  1 . 8  0 . 4 
M 


  212 . 33
(1 . 8 * 1 . 13 )
2 
2

1560
b  1130 mm , d  7 30 mm
6
25 
2  10 * 212.33
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 730 * 1130


   min

A S min  0 . 0018  800  765  1101 . 6 mm
2
2
 11 cm
use 7  14 / m
Shrinkage Reinforcement in short direction
A S min  0 . 0018  800  765  1101 . 6 mm
2
 11 cm
2
use 7  14 / m
Footing Design
Part III
Combined footing, strip footing, & Mat foundation
Example 2
Design a combined footing As shown
q all ( net )  18 t / m  180 kPa
2
f c  25 N mm
2
f y  420 N mm
2
Dimension calculation
The base dimension to get uniform distributed load
1950kN
1200 kN
750 kN
750(4.2)+1200(0.2)=1950 (x)
x = 1.75m
A
x1=0.2m
x2=4.2
m
x
 B1  2 B 2
x  
 B1  B 2
L

3

Area required
q all ( net )  20 t / m  200 kPa ,
2
Ag 
Ps

q all ( net )
1950  10
180  10
3
3
 10 . 8 m
2
 B1  B 2 

 L  10 . 8
2


 B1  B 2 

 4 . 35  10 . 8
2


 B1  B 2 

  2 .5
2


B1  B 2  5
 B  2 B2
x   1
 B1  B 2
 L  5  B 2  4 . 35
 

3
5

 3

1 . 75  1 . 45  0 . 29 B 2
B 2  1m
B1 4m
qu 
Pu
A

1 . 3 1950   10
10 . 8
3
 235  10 Pa  235 kPa
3
Check for punching Shear
h= 750mm
d = 732 mm
A
B2=1m
b o  2 ( 732 )  1065  2590 mm
fc '
 VC  
3

 VC    2 

b o d  0 . 75 
 sd 
25
B1=4m
 665  2590 / 1000  2160 . 4 kN
3
30  665 
25

b
d

0
.
75
2


 665  2590 / 1000  5222 kN



o
b  12
2590 
12

fc '
V U  1200 (1 . 3 )  1 . 065 * 0 . 733 * 235  1376 . 6 kN   V c
oK
B
b o  2 ( 633 )  965  2231 mm
 VC  
fc '
3
b o d  0 . 75 
25
 665  2231 / 1000  1854 . 5 kN
3
 sd  fc '
30  665 
25


 VC    2 
b o d  0 . 75  2 
 665  2231 / 1000  5273 kN


b
12
2231
12




V U  800 (1 . 3 )  0 . 965 * 0 . 633 * 235  896 . 5 kN   V c
oK
Draw S.F.D & B.M.D
Empirical S.F.D & B.M.D
975  0 . 70  682
m
w ave  235 
2
3
wl
2
 M max 
8
( 940  235 )  705

705 3 . 65 
2
 1174 kN .m
8
Clear distance between column
B in moment design = ave. width = 2.5m
1560  ( 0 . 70 )  1092
Mmax
Check for beam shear
d = 665mm
b  1  2 ( Lx )  y
at x  0.815  0.15
1  2 ( 04.965
)  1 . 5  1 . 7 m  1700 mm
. 35
 V C  0 . 75 
25
 665  1700 / 1000  696 kN
6
Max .  V U at d from column
B face ( the most critical)
 668 kN
VU   V C
b
Y=1.5m
x
4m
1m
Bending moment Long
Top
 ve M  1260 kN .m
direction
b  1  2 ( 24 ..25
)  1 . 5  2 . 60 m  2600
35
d  7 30 mm
6
25 
2  10 * 1260
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 665 * 2600

A S  0 . 003  665  1000  1995 mm
2
 20 cm
2

  0.003

use 1 0 16 / m
Top
Bottom
A S min  0 . 0018  750  1000  1350 mm
2
 13 . 5 cm
2
use 9 14 / m
Bottom
Bending moment Short
direction
Under Column A
b '  1  2 ( 34..62
)  1 . 5  3 . 5 m  3500 mm
35
b '  1  2 ( 04.633
)  1 . 5  1 . 44 m  1440 mm
. 35
b
3 .5  4
m  3 . 75 m  3750 mm
2
2
0 . 733  3 . 75  0 . 4 
M 


  583 . 6
( 3 . 75 * 0 . 733 )
2 
2

1560
d  665 mm
6
25 
2  10 * 583.57
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 665 * 733

A S  0 . 005  665  733  3325 mm
2
 33 cm
2

  0.005

use 1 0 20
Under Column B
b
1 . 44  1
m  1 . 22 m  1220 mm
2
2
0 . 633  1 . 22  0 . 3 
M 


  84 . 6
(1 . 22 * 0 . 633 )
2 
2

975
b  633 mm , d  665 mm
6
25 
2  10 * 84.6
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 665 * 633

A S min  0 . 0018  750  633  854 . 6 mm
2

   min

 8 . 6 cm
use 6 14
2
Shrinkage Reinforcement in short direction
A S min  0 . 0018  1000  750  1350 mm
2
 13 . 5 cm
2
use 9 14 / m
Reinforcement details
Example 3 (Strip footing)
Design a combined footing As shown
q all ( net )  20 t / m  200 kPa
2
f c  25 N mm
2
f y  420 N mm
2
Dimension calculation
The base dimension to get uniform distributed load
3040kN
800 kN
1280 kN
960 kN
A
Assume
L1=0.6
x1=5.2m
x
x2=10.7m
800(0.6)+1280(5.1)+960(10.6)=
3040 (x)
x = 5.65m,
2(x)=11.3m
L2=11.3 - (10.6)=0.7
L2
q all ( net )  18 t / m  180 kPa ,
2
Ag 
qu 
Ps

q all ( net )
Pu
A

3040  10
180  10
3
3
1 . 3 3040   10
11 . 3  1 . 8
 16 . 9 m  11 . 3 m  1 . 8 m
2
3
 195  10 Pa  195 kPa
3
Check for punching Shear
h = 700 mm
d=630mm
Example
B
b o  4 ( 630  400 ))  4120 mm
 VC  
fc '
3
b o d  0 . 75 
25
 630  4120 / 1000  3244 . 5 kN
3
 sd  fc '
40  630 
25


 VC    2 
b o d  0 . 75  2 
 630  4120 / 1000  6584 kN


b
12
4120
12




V U  1280 (1 . 3 )  1 . 03 * 195  1457 . 1kN   V c
2
You can check other columns
oK
Draw S.F.D & B.M.D
Stress under footing
= 195 *1.8 = 351 kN/m
Check for beam shear
b = 1800mm, d = 630mm
 V C  0 . 75 
25
 630  1800 / 1000  708 . 75 kN
6
Max .  V U at d from column
VU   V C
face  0 . 7 (1009 )  706 . 3 kN
Bending moment Long
direction
 ve M  1366 kN .m
b  1 800 mm , d  7 30 mm
6
25 
2  10 * 1365
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 630 * 1800

A S  0 . 0053  630  1000  3362 mm
2
 33 . 6 cm

  0.0053

use 9 22 / m
2
Top
 ve M  246 . 7 kN .m
b  1 800 mm , d  7 30 mm
6
25 
2  10 * 81
  0 . 85 *
1 - 1 2
420 
0 . 9 0.85 25 * 730 * 1800

A S min  0 . 0018  700  1000  1260 mm
Design Short
2

   min

 12 . 6 cm
2
use 8 14 / m
direction as example 1 (lecture 11)
Bottom
Reinforcement details
Mat Foundation
Check for punching Shear
General Example, Ref. 2