Isolated Footing Design

Footing No.
1
Footin
g No.
1

MARK
P1F1

Length
2832.000 mm

Foundation Geometry
Width
2832.000 mm

Footing Reinforcement

Thickness
300.000 mm
Pedestal Reinforcement

Bottom
Top
Top
Bottom
Main
Reinforcement(M Reinforcement(M Reinforcement(M
Trans Steel
Reinforcement(Mz)
Steel
)
)
)
x
z
x
#16 @ 228 mm #16 @ 228 mm
#10 @ 304
N/A
N/A
12 - #20
c/c
c/c
mm
Isolated Footing

Input Values
Footing Geomtery
Design Type : Calculate Dimension
Footing Thickness (Ft) : 300.000 mm
Footing Length - X (Fl) : 800.000 mm
Footing Width - Z (Fw) : 800.000 mm
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in

Pedestal
Include Yes
Pedest
al?
Pedest Rectangular
al
Shape :
Pedest 2700.0
al 00 mm
Height
(Ph) :
Pedest 400.00
al 0 mm
Length
-X
(Pl) :
Pedest 400.00
al 0 mm
Width Z (Pw) :

Design Parameters
Concrete and Rebar Properties
Unit Weight of Concrete :

24.000 kN/m3

Strength of Concrete : 4.000 ksi

Yield Strength of Steel : 60.000 ksi

Minimum Bar Size : #16
Maximum Bar Size : #25
Minimum Bar Spacing : 100.000 mm
Maximum Bar Spacing : 250.000 mm
Pedestal Clear Cover (P, CL) : 75.000 mm
Footing Clear Cover (F, CL) : 75.000 mm

Soil Properties
Soil Type : UnDrained
Unit Weight : 15.000 kN/m3
Soil Bearing Capacity : 90.000 kN/m2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 1500.000 mm
Undrained Shear Strength : 0.000 kip/in2
Sliding and Overturning
Coefficient of Friction :0.500
Factor of Safety Against Sliding :1.500
Factor of Safety Against Overturning :1.500

Design Calculations
Footing Size
Initial Length (Lo) = 800.000 mm
Initial Width (Wo) = 800.000 mm

Load Combination/s- Service Stress Level

Load
Combinatio
n Number

Load Combination Title

1.4DL

1.2DL+1.6LL

1.2DL+1.6W+LL

1.2DL+E+LL
Load Combination/s- Strength Level

Load
Combination
Number

Load Combination Title

1.4DL

1.2DL+1.6LL

1.2DL+1.6W+LL

1.2DL+E+LL
Applied Loads - Service Stress Level

LC

Axial
(kN)

Shear X
(kN)

Shear Z
(kN)

Moment X
(kNm)

Moment Z
(kNm)

12.600

9.800

0.000

0.000

0.000

33.200

26.000

0.000

0.000

0.000

77.600

59.400

0.000

0.000

0.000

25.800

22.400

0.000

0.000

0.000

Applied Loads - Strength Level

LC

Axial
(kN)

Shear X
(kN)

Shear Z
(kN)

Moment X
(kNm)

Moment Z
(kNm)

12.600

9.800

0.000

0.000

0.000

33.200

26.000

0.000

0.000

0.000

77.600

59.400

0.000

0.000

0.000

25.800

22.400

0.000

0.000

0.000

Reduction of force due to buoyancy =

0.000 kN

Effect due to adhesion =

0.000 kN

Area from initial length and width, A o =Lo X Wo = 640000.000 mm2

Min. area required from bearing pressure,
P / qmax = 1148612.260 mm2
Amin =
Note: Amin is an initial estimation.
P = Critical Factored Axial Load (without self
weight/buoyancy/soil).
qmax = Respective Factored Bearing Capacity.
Final Footing Size
Length (L2) =

2832.000 mm

Governing Load Case :

#7

Width (W2) =

2832.000 mm

Governing Load Case :

#7

Depth (D2) =

300.000 mm

Governing Load Case :

#7

Area (A2) =

8020223.6 mm
32 2
Pressures at Four Corners

Pressure Pressure at Pressure

Pressure
at corner 1
corner 2
at corner 3 at corner 4
Load Case
(q1)
(q2)
(q3)
(q4)
(kN/mm2)
(kN/mm2)
(kN/mm2)
(kN/mm2)

Area of
footing in
uplift (Au)
(mm2)

0.0000

0.0000

0.0000

0.0000

0.000

-0.0000

0.0001

0.0001

-0.0000

641617.891

-0.0000

0.0001

0.0001

-0.0000

641617.891

0.0000

0.0000

0.0000

0.0000

0.000

If Au is zero, there is no uplift and no pressure adjustment is necessary. Otherwise,

to account for uplift, areas of negative pressure will be set to zero and the pressure
will be redistributed to remaining corners.

Summary of Adjusted Pressures at 4 corners Four Corners

Pressure at
corner 1 (q1)

Pressure at
corner 2 (q2)

Pressure at
corner 3 (q3)

Pressure at
corner 4 (q4)

Load Case

(kN/mm2)

(kN/mm2)

(kN/mm2)

(kN/mm2)

0.0000

0.0000

0.0000

0.0000

0.0000

0.0001

0.0001

0.0000

0.0000

0.0001

0.0001

0.0000

0.0000

0.0000

0.0000

0.0000

Check for stability against overturning and sliding

Factor of safety against

sliding

Factor of safety against

overturning

Load
Case
No.

Along XDirection

Along ZDirection

About XDirection

About ZDirection

13.141

N/A

N/A

12.405

5.349

N/A

N/A

5.050

2.715

N/A

N/A

2.563

6.044

N/A

N/A

5.705

Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding X Direction
Critical Load Case for Sliding along X- 7
Direction :
Governing Disturbing Force : 59.400 kN
Governing Restoring Force : 161.280 kN
Minimum Sliding Ratio for the Critical 2.715
Load Case :
Critical Load Case for Overturning about 5
X-Direction :
Governing Overturning Moment : 0.000 kNm
Governing Resisting Moment : 364.698 kNm
Minimum Overturning Ratio for the N/A
Critical Load Case :

Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding Z Direction
Critical Load Case for Sliding along Z- 5
Direction :
Governing Disturbing Force : 0.000 kN
Governing Restoring Force : 128.780 kN
Minimum Sliding Ratio for the Critical Load N/A
Case :
Critical Load Case for Overturning about Z- 7

Direction :
Governing Overturning Moment : -178.197 kNm
Governing Resisting Moment : 456.737 kNm
Minimum Overturning Ratio for the Critical 2.563
Load Case :

Shear Calculation
Punching Shear Check

Total Footing Depth, D 300.000mm

=
Calculated Effective
Depth, deff =
For rectangular
column,

D - Ccover - 199.60 1 inch is deducted from total depth to cater

1.0 = 0 mm bar diameter(US Convention).
Bcol / Dcol = 1.000

Effective depth, deff, increased until 0.75XVc

Punching Shear Force

Punching Shear Force, Vu = 74.121 kN, Load Case # 7

From ACI Cl.11.12.2.1, bo for
column=
Equation 11-33, Vc1 =

2398.400
mm
1252.514
kN

Equation 11-34, Vc2 =

1112.417
kN

Equation 11-35, Vc3 =

835.009 kN

Punching shear strength, Vc =

0.75 X minimum of (Vc1, Vc2, Vc3) =

626.257 kN

0.75 X Vc > Vu hence, OK

Along X Direction
(Shear Plane Parallel to Global X Axis)

From ACI Cl.11.3.1.1, Vc =

492.984 kN

Distance along X to design for

shear, Dx =

1016.400
mm

Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a
distance deff from the face of the column caused by bending about the X axis.
From above calculations,
Critical load case for Vux is # 7

0.75 X Vc =

369.73 kN
8
49.230 kN

0.75 X Vc > Vux hence, OK

One-Way Shear Check

Along Z Direction
(Shear Plane Parallel to Global Z Axis)

From ACI Cl.11.3.1.1, Vc =

492.984 kN
1815.60
mm
0

Distance along X to design for shear, Dz =

Check that 0.75 X Vc > Vuz where Vuz is the shear force for the critical load cases at
a distance deff from the face of the column caused by bending about the Z axis.
From above calculations,

0.75 X Vc =

Critical load case for Vuz is # 7

369.73 kN
8
55.206 kN

0.75 X Vc > Vuz hence, OK

Design for Flexure about Z Axis

(For Reinforcement Parallel to X Axis)

Calculate the flexural reinforcement along the X direction of the footing. Find the
area of steel required, A, as per Section 3.8 of Reinforced Concrete Design (5th ed.)
by Salmon and Wang (Ref. 1)
Critical Load Case # 7
The strength values of steel and concrete used in the formulae are in ksi
Factor

from ACI Cl.10.2.7.3 =

0.850

From ACI Cl. 10.3.2,

0.0285
1

From ACI Cl. 10.3.3,

0.0213
8

From ACI Cl. 7.12.2,

0.0018
0

From Ref. 1, Eq. 3.8.4a, constant m =

17.647

Calculate reinforcement ratio for critical load case

Design for flexure about Z axis is
performed at the face of the
column at a distance, Dx =

1216.0
mm
00

Ultimate moment,

164.53 kN
6 m

Nominal moment capacity, Mn =

182.81 kN
7 m

0.0040
6

Required =
Since

OK
3.559 in2

Area of Steel Required, As =

Selected bar Size = #5

Minimum spacing allowed (Smin) = = 100.000 mm
Selected spacing (S) = 242.375 mm
Smin <= S <= Smax and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 193.500 mm

Based on spacing reinforcement increment; provided reinforcement is

#16 @ 228.600 mm o.c.

Required development length for bars

=

=304.8
mm
00

Available development length for

bars, DL =

1141.00
mm
0

Try bar size

#5

Area of one bar =

Number of bars required, Nbar =

0.310 in2
12

Because the number of bars is rounded up, make sure new reinforcement
ratio < max
Total reinforcement area, As_total =
deff =

Nbar X (Area of one bar) =

3.720 in2

D - Ccover - 0.5 X (dia. of one

bar) =

217.06 mm
3

Reinforcement ratio, =
From ACI Cl.7.6.1, minimum req'd max (Diameter of one bar, 1.0,

0.0039
0
242.37 mm

clear distance between bars, Cd =

Min. User Spacing) =

Check to see if width is sufficient to accommodate bars

Design for Flexure about X axis

(For Reinforcement Parallel to Z Axis)

Calculate the flexural reinforcement along the Z direction of the footing. Find the
area of steel required, A, as per Section 3.8 of Reinforced Concrete Design (5th ed.)
by Salmon and Wang (Ref. 1)
Critical Load Case # 7
The strength values of steel and concrete used in the formulae are in ksi
Factor

from ACI Cl.10.2.7.3 =

0.850

From ACI Cl. 10.3.2,

0.0285
1

From ACI Cl. 10.3.3,

0.0213
8

From ACI Cl.7.12.2,

0.0018
0

From Ref. 1, Eq. 3.8.4a, constant m

=

17.647

Calculate reinforcement ratio for critical load case

Design for flexure about X axis is
performed at the face of the
column at a distance, Dz =

1216.0
mm
00

Ultimate moment,

47.723

kN
m

Nominal moment capacity, Mn =

53.026

kN
m

Required =

0.0013
6

Since
Area of Steel Required, As =

OK
1.452 in2

Selected Bar Size = #5

Minimum spacing allowed (Smin) = 100.000 mm
Selected spacing (S) = 250.000 mm
Smin <= S <= Smax and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 193.500 mm

Based on spacing reinforcement increment; provided reinforcement is

#16 @ 228.600 mm o.c.

Required development length for bars

=

=304.8
mm
00

Available development length for

bars, DL =

1141.00
mm
0

Try bar size

#5

Area of one bar =

0.310 in2

Number of bars required, Nbar =

11

Because the number of bars is rounded up, make sure new reinforcement
ratio < max
Total reinforcement area, As_total =
deff =

Nbar X (Area of one bar) =

3.410 in2

D - Ccover - 0.5 X (dia. of one bar) 192.99 mm

=
6
0.0040
3

Reinforcement ratio, =
From ACI Cl.7.6.1, minimum req'd max (Diameter of one bar, 1.0,
clear distance between bars, Cd =
Min. User Spacing) =

250.00 mm
0

Check to see if width is sufficient to accommodate bars

Pedestal Design Calculations

Critical Load
Case:

Strength and Moment Along Reinforcement in X direction

Bar size :

# 20

Number of Bars :

12

Steel Area :
Neutral Axis Depth (Xb):

3319.99 sq.m
98 m
87.1250 mm

Strength and Moment from Concrete

Cc =

694.41
kN
6

Mc =

113.16
kNm
8

Calculate strength and moment from one bar.

Distance between extreme

fiber and bar,

db

91.129 mm

Strain in bar,

0.0001

Maximum Strain,

0.0021

as
-0.028

kN/mm
2

0.0019
as
0.000

kN/mm
2

kN
22.587
-2.459 kNm
Total Bar Capacity,

Capacity of Column =

Cs =

Cc + C s =

- kN
723.08
3
- kN
28.667

Total Bar Moment,

Ms =

68.885 kNm

Total Moment =

Mc + M s =

182.05 kNm
3