presentation for 3 types of shallow foundations : - combined footing - strap beam footing (neighbor's footing) - raft foundation

1. SHALLOW FOUNDATIONS 2
DESIGN OF COMBINED, STRAP & RAFT FOOTING

2. INTRODUCTION

3. Contents :
Combined Footing
Strap Footing
Raft Foundation
1
2
3

4. 1. Combined Footing

5. Combined Footing
 A combined Footing is a long footing supporting two or more
columns in (typically two) one row.
 A combined Footing is a rectangular or Trapezoidal shaped
footing.

6. Using of Combined Footing
Construction Practice may dictate using only one footing
for two or more columns due to:
a) Closeness of Columns
b) Due to property line constraint, which may limit the size of footing at
boundary.

8. The Design of Combined Footing requires that the centroid of the
area be as close as possible to the resultant of the 2 column loads
for uniform pressure and settlement.
Which Means :

9. Eccentricity = ZERO

19. STEPS OF DESIGN

20. DimensioningforP.C
Find Area
“L” & “B”
DesignofR.CFooting
(longitudinal)
Find “Max
Moment”
Find “d”
Check Shear
Check
Punching
DesignofR.CFooting
(Short)
Find Moment @
Hidden beam 1
Find Moment @
hidden beam 2
Ensure That
C1 > 2.3
RFT
RFT in Long
Direction
RFT in Short
Direction

21. C1 C2
L

22. RC
Max
S.F.D
B.M.D
Zero shear

23. c1 c2

24. c1 c2
H.B.1 H.B.2

26. c1 c2

27. EXAMPLE
C1 (30*70) C2 (30*90)
4 m
Design a combined footing to support a working load of p1=160 ton & p2=220 ton.
Bearing capacity: 12.5 ton/m2
Thickness of plain concrete 15 cm

28. Solution

29. 1- Dimensions of footing :
R = p1+p2 = 160 + 220 = 380 ton  R . X = p2 . S
380 x X = 220 x 4  X= 2.32 m
Lpc / 2 = X + b1 / 2 + 0.5 + tpc
Lpc / 2 = 2.32 + 0.7 / 2 + 0.5 + 0.3 = 3.47m  Lpc = 6.94 m
LRc = LPc - 2 x tPC = 6.94 - 2 X 0.3 = 6.34m
APC = 380 / 12.5 = 30.4
LPc X BPc = 30.4
6.34 x BPc = 30.4
BPc = 4.8m
BRc = 4.8 - 2 x 0.3 = 4.2m

30. 2Design of R.c footing
P1u = 160 x 1.5 = 240 ton
P2u = 220 x 1.5 = 330 ton
Ru = 380 x 1.5 = 570 ton
Wu = 570 / 6.34 = 89.9 ton
qu = 570 / 6.34 x 4.2 = 21.4 t/㎡

31. Design of Footing in
Longitudinal Direction

32. At point of zero shear :
P1u = wu . X
240 = 89.9 . X
X = 2.67m
Mmax = 89.9 x 7.1289 / 2 - 240 x 1.82 = 116.3
D =5 × √116.35×10^7 / 25 × 4200 = 526.3㎜
D = 530㎜

33. 2- check shear
Qsu = 0.16√25/1.5 = 0.653 N/m㎡
Qsumax = 155.56 - 89.99 × 0.53/2 = 131.7365
Qs = 131.73×10^4 / 530×4200 = 0.59 ≤ qsu
 Safe

34. c1 c2
1.23 1.43
0.83
0.83

35. For c1 (30x70)
Qpcu1 = 0.316 ( 0.5 + 300 / 700) √25/1.5 = 1.197
Qpu1 = 240 - 21.4 x (1.23x0.83)=218.15 ton
Qpu1 = 218.15 x 10^4 / 530x2 (1230+830) =0.99 ≤ qpcu1  Safe
For c2 (30*90)
Qpcu2 = 0.316 x (0.5 + 300/900) √25/1.5 = 1.07
Qpu2 = 220 -21.4 (1.43x0.83) = 194.6
Qpu2 = 194.6 x 10^4 / 530 (1430+830)x2 = 0.81 ≤ 1.07  safe

36. Design of Footing in
Short Direction

37. c1 c2
1.73 m 1.96 m

38. For hidden Beam1
Qu1 = 240 / 4.2 x 1.73 = 33.03t/㎡
M1 = 33.03 * 1.95^2 /2 = 62.9 t.m
For hidden Beam 2
Qu2 = 330 / 4.2 x 1.96 = 40.1t/㎡
M2 = 40.1 x 1.95^2 / 2 = 76.2 t.m
d = 530 = c1 √76.2x10^7 / 25 x 1000
C1 = 3.1≥ 2.3

39. Reinforcement

40. Asmin =1.5 X d = 1.5 x 530 = 795 mm2
Asmin = 4 1 16

41. RFT in long direction
Astop = 116.35 x 107 / 360 x 0.826 x 530
= 72382mm2/4.2m =1757.7 mm2  7 ϕ18
Asbottom = 48.61 x 107 / 360x 0.826 x 530
= 3084mm2 /4.2 = 734.3 mm2  4ϕ16

42. RFT in short direction
As1 = 62.79 X 107 / 360 X 0.826 X 530
= 3984 mm2  11ϕ22
As2 = 76.2 X 107 / 360 X 0.826 X 530
= 4835 mm2  10ϕ25

43. 7ϕ18
4ϕ1611ϕ22 10ϕ25
4ϕ16

44. c1 c2
7ϕ18
4ϕ16
4ϕ16
11ϕ22
4ϕ16
4ϕ16
10ϕ25
4ϕ16

45. 2. Strap Footing

46. Design a strap footing to support an exterior column (30*50cm) and an
interior column (30*90cm). The un factored Loads C1 =685 KN, C2 =1270
kN . Assume the allowable Bearing capacity is 150 kN/m2, fcu=25 N/mm2
and Fy = 360 N/mm2 , P.C. thickness = 40 cm
EXAMPLE

47. C2C1
0.5 0.9
4.9 m

48. Calculation of reaction:
Assume e=0.1 to 0.2 (L)
e = 0.5 to 1m (take it 1m)
R1=
685∗4.9
3.9
= 860.6 KN
R2=685 + 1270 - 860.6 = 1094.4 KN
C2C1
0.5 0.9
e

49. Area of plain concrete footing:
A1 =
860.6
150
= 5.73 𝑚2
Dimension ((1+0.25)*2) = 2.5 (2.5*2.3)
A2 =
1094
150
= 7.29 𝑚2
Dimension = (2.7*2.7) = 7.29
R.C dimensions:
F1 = (2.1*1.5) m
F2 = (1.9*1.9) m

50. C2C1
0.5 0.9
0.50.90.50.5 2.11.6
M = 602720
695
432
262
B.M.D
S.F.D

51. Design of strap beam:
qu1=
1.5∗860.6
2.1
= 614 kN/m’
qu2=
1.5∗1094
1.9
= 864 kN/m’
Point of zero shear:
614X-1027=0 , X=1.67m , Mmax=602kN.m
d = 5 *
602∗106
25∗400
=1227m
t=130cm d=123cm

52. Reinforcement:
Astop =
602∗106
0.826∗360∗1230
=1646𝑚𝑚2 = 7∅18/m’
Asbot =
108∗106
0.826∗360∗1230
= 295𝑚𝑚2
Asmin = 0.15% b*d = 0.15%*400*1230 =738𝑚𝑚2
= 4∅16 /𝑚′

53. CHECK SHEAR:
Qs = (720*0.55)/1.17 = 341.53kN
qs =
341.53∗1000
1230∗400
= 0.694N/𝑚𝑚2
qcu =0.24
25
15
= 0.98N/𝑚𝑚2
qs< 𝑞𝑐𝑢  use min stirrups 5∅10/𝑚′
0.550.62
720

54. Design of Footings

55. Critical Sec
Footing 1:
qu =
860.6∗1.5
2.1∗1.5
= 409.5 kN/𝑚2
L1 = (1.5-0.4)/2 = 0.55 m
Mu = 409.5*(0.552
/2) = 61.9kN.m
d = 5
61.9∗106
25∗1000
= 249mm
t = 35cm d = 28cm

56. Check shear:
Qs = 409.5*(0.55-(0.28/2)) = 168kN
qs =
168∗1000
1000∗280
= 0.6 MPa
qcu = 0.16
25
1.5
= 0.65 MPa
qs < 𝑞𝑐𝑢  safe
As =
61.9∗106
0.826∗360∗280
=743.44 𝑚𝑚2
= 5∅14/𝑚′

57. Footing 2:
qu=
1094∗1.5
1.9∗1.9
=455kN/𝑚2
L2=
1.9−0.4
2
=0.75m
Mu=455*
0.75
2
=128kN.m
d=5
128∗106
25∗1000
=358mm
t=45cm d=38cm
Critical Sec

58. Check shear:
Qs = 455*(0.75-
0.38
2
) = 255KN
qs =
255∗1000
1000∗380
=0.67N/𝑚𝑚2
qcu =0.65N/𝑚𝑚2
 qs > 𝑞𝑐𝑢  unsafe
0.65=
255∗1000
1000𝑑
d=392mm
t=50cm d=43cm
As =
128∗106
0.826∗360∗430
=1001𝑚𝑚2=5∅16/𝑚′
Asmin = 1.5d = 1.5*430 = 645𝑚𝑚2

59. 3ϕ18 + 4ϕ18 = 7ϕ18
4ϕ16
5ϕ10
2ϕ12
2ϕ12

60. 0.4m
2.3m
1.5m
2.1 m
2.5 m 2.7 m
1.9 m
5ϕ12 5ϕ12
5ϕ14
5ϕ16

61. 3. Raft Foundation