unit 15 ( design of foundations )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT15/

UNIT 15

DESIGN OF FOUNDATIONS

GENERAL OBJECTIVES

To establish the basic principles underlying the design of isolated spread

foundations.

At the end of this unit, you will be able to: -

1. calculate the area of base required in the design of foundations.

2. determine whether punching shears is a criterion.

3. check sections for bending and local bond stresses.

4. check for transverse shears.

5. check for cracks.

6. calculate the area of starter bar required.

7. sketch the reinforcements.

1

OBJECTIVES

SPECIFIC OBJECTIVES


8.2 Introduction .

The function of foundation is to transfer the load from a vertical member to a

horizontal slab. The thickness and area will facilitate the load to a lower

concentration level at the underside of the slab. However, this depends on the

capacity of the ground on which the foundation rests.

To avoid the failure of the foundation, a suitable size and thickness should be

used to ensure that the stress encountered does not exceed the permissible stress.

Failure of the foundation can cause a severe effect on the structure as a whole.

Failure of the foundation may also cause instability to the whole structure or it

even collapse.

In the design of foundation, the soil characteristics are important as it may vary.

For example from soft clay to solid rocks. There is always an element of the

unknown in work below ground. Therefore, it is imperative to carry site

investigation prior to erecting the foundation so that the safe bearing capacity of

the soil can be determined.

2

INPUT 1


The design of foundation or bases is covered in section 3.11, BS 8110. The

pressure of the supporting soil is assumed to be uniformly distributed throughout

the area of the base (Clause 3.11.2.1). This assumption is based on the fact that

soil act as an elastic material and that the base possesses a significant degree of

stiffness. When there is moment, the distribution of the underside of the

foundation will be variable in a straight line. This is shown in Figure 15.1a and

Figure 15.1b

3

N

Figure 15.1a: Uniform soil pressure

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT15/ 4

N

M

ACTIVITY 15a

Figure 15.1b: Soil pressure varies in a straight line.


Complete the following statements.

15.1 The function of foundation or base is to

____________________________________________________________

____________________________________________________________

15.2 The commonest type of foundation is ____________________________.

15.3 A suitable thickness and _________________________________ should

be used.

15.4 Site investigation is carried out to determine_______________________.

15.5 For foundation to be safe, the stress due to imposed load must not exceed

the ___________________________________ of the supporting soil.

15.6 When there is only axial load, the pressure distribution underneath the

base is ________________________________________________.

5

`FEEDBACK 15a


Check your answers. Are you happy with your answers?

15.1transfer load from column or wall to the ground.

15.2isolated spread foundation (pad foundation)

15.3area

15.4soil bearing capacity

15.5bearing capacity

15.6uniform

6

INPUT 2


15.2 Considerations affecting design details

Since concrete foundation is exposed to a severe environment, a higher grade of

concrete is used. A greater nominal cover should also be provided. With reference

to Table 3.2 and Table 3.4, BS 8110, the lowest grade of concrete to be used is

C35 and the corresponding nominal cover is not less than 40 mm as stated in

3.3.1.4.

A thickness nominal cover to reinforcement than those given in Table 3.4, BS

8110 is provided when fresh concrete is poured onto an uneven surface. This is to

ensure that minimum thickness is provided.

If fresh concrete is poured directly onto the ground, it is recommended that a

nominal cover of less than 75 mm is provided. A nominal cover of less than 40

mm is not recommended even though concrete blinding is provided. Concrete

blinding is used to protect the soil after excavation and to provide an even surface

for the foundation. In order to protect steel and concrete from the soil during

construction a blinding layer of concrete, 50 to 100 mm thick is usually laid at the

bottom of the excavation.

7

ACTIVITY 15b


Fill in all the boxes provided with the corresponding values.

8

Lowest concrete grade

Blinding LayerMinimum cement content

Exposure condition

Maximum free water/ cement ratio

Nominal cover

Foundation

`FEEDBACK 15b


Here are the correct answers.

15.3 Design of pad foundation

9

Lowest concrete grade

C40

Blinding Layer

50 to 100 mm thick

Minimum cement content

325 kg/m3

Exposure condition

SEVERE

Maximum free water/ cement ratio

0.55

Nominal cover

40 mm

Foundation

INPUT 3


Pad or slab foundation is most commonly used in framed buildings. This is a

square or rectangular slab of concrete carrying a single column. Reinforcement is

placed at the bottom in both directions to resist the bending stresses set up by the

double-cantilever action of the slab on the column base. Shear reinforcement is

normally not provided. The thickness of the slab may be reduced towards the

edges to economise concrete either by stepping or tapering the top face. Figure

15.2 below shows a typical pad foundation.

15.4 Service Load

The bearing capacity for a soil in the design of pad foundation is in terms of

service units, while the loads are in ultimate units. The sizing of the foundation is

based on service loads and the design of section (ultimate load). This is shown in

the following example.

10

Slab reinforcement

Column reinforcement

Figure 15.2: Cross-Section of a Typical Pad Foundation


15.4.1 Example (Isolated Pad Foundation Carrying Axial Load Only)

A reinforced concrete column of 400mm x 400mm in section and supports an

ultimate load of 2200 kN. Of this total load 1400 kN is dead load and 800 kN is

imposed load. The maximum permissible increase of ground pressure against

service loads is 150 kN/m2 with fcu = 40N/mm2 and fy = 460 N/mm2. Design an

isolated square foundation to meet the design criteria given.

Solution: Service Load

Dead Load = 1400 = 1000 kN 1.4

Imposed Load = 800 = 500 kN 1.6

Total service load = 1500 kN

15.5 Area of base

Self-weight must be allowed for in advance of any consideration of the support of

the load. If soil is removed and replaced with concrete, then the self-weight of

base is the difference in density between soil and concrete. This is shown below.

Density of concrete = 2400 kg/m3

Density of soil = 1500 kg/m3

Net difference = 900 kg/m3

This is equivalent to 9 kN/m3

11


Assuming that the base thickness is 650 mm,

The net self-weight per unit area = 0.65 x 9

= 5.8 kN/m 2 , (say 6.0 kN/m2.)

The net allowable bearing pressure becomes

150 – 6 = 144 kN/m 2

The area of base required for this load is

1500 = 10.4 m 2 144

If the base is made 3.25m x 3.25 m,

The total area provided = 10.6 m2.

15.6 Ground Pressure

Reverting to ultimate loads, the net ultimate pressure on the underside of the base

is,

2200 = 207.5 kN/m 2 10.6

15.7 Punching shear

The critical section for punching shear is shown in figure 15.3. The shaded areas

represent the load to be considered.

12

h

1.5d1.5d

c

c

1.5d

1.5d 1.5d

1.5d

3d+c

Figure 15.3: The critical section for punching shear


The critical perimeter for punching shear is 1.5d around the column. For punching

shear to be under control, the nominal design shear stress () should not exceed

the allowable value.

is calculated using equation 28, BS 8110 as follows: -

= N …………equation 28 ud

Where u = 4(3d+c) and the allowable design shear stress is given by equation 21,

BS 8110 as follows: -

= V …………..equation 21 bd

13


If 1.5d exceeds the outer areas of the base, punching shear will not occur and

hence it need not be checked.

For the given example, the value of d, is established first. This is calculated as

follows: -

With a base thickness of 650 mm and allowing for two layers of reinforcement

on each direction, the average effective depth is

d = 650 – 40 –20 (assume T20 is used)

= 590 mm

Then, the perimeter the punching shear is

U = 4 (3d + c)

= 4 (3 x 590 + 400) mm

= 8680 mm

The nominal design is,

= V bd

= 207.5 x 3.25 x 3.25 x 10 3 N 3250 x 590 mm2

= 1.14 N/mm 2

Therefore, the nominal design shear does not exceed the allowable value; thus

punching shear value is under control.

14


15.8 Bending

According to Clause 3.11.2.2, BS 8110, the critical section for moment is at the

column face. This is calculated as follows: -

Given: column size 400mm x 400mm

Gk =1400 kN

Qk = 800 kN

Allowable bearing pressure = 207.5 kN/m2

fcu = 40 N/mm2

fy = 460 N/mm2

Cover = 40 mm

h = 650 mm

d = 590 mm

Figure 15.4:Critical sections for bending

15

= 1.425m

3.25 m


= 210.7 kNm

= 0.005 < 0.156

= 0.995d > 0.95d

Use Z = 0.95d

= 0.95 x 590

=560.5 mm

= 939 mm 2

Asmin=

= 2746 mm 2

Provide 6T25 (As=2946 mm2) for both directions.

15.9 Bar anchorage (Bond)

16


Now check the requirement for anchorage bond lengths, which is given in Table

3.29, BS 8110. This is shown as follows: -

Anchorage bond length (extending beyond the face of column)

= 34 x bar diameter

= 34 x 25 mm

= 850 mm

Available space = 1425 mm – 400 mm

= 1385 mm > 850 mm

Therefore the anchorage length provided is satisfactory and the bar need not be

bent upwards.

15.10 Bar distribution of reinforcement

It is good practice to concentrate more of the transverse reinforcement

immediately under the column than elsewhere because the outer limits of the base

will be principally concerned with longitudinal bending.

In accordance with Clause 3.11.3.2, BS 8110, it states that,

where exceeds ( ), two thirds of the required reinforcement

should be concentrated within a zone from the centre line of the

column to a distance 1.5d from the face of the column; otherwise the

reinforcement should be uniformly distributed over .

17


is half the spacing between column centre and c is the column width.

For this particular example,

Therefore,

= 1627.5 mm

= 1.6 m

Since exceeds ( ), of As will be banded in the centre and the

remaining will be uniformly distributed in the other region.

18

c=400

Figure 15.5: Clearification of c


Amount of reinforcement in the central band is

= x 2746 mm2

= 1831 mm 2 (Use 4T25)

The remaining amount is 915 mm2 (This is 2T25)

This is shown below:-

19

1.5d=1.5 x 590=885mmC

Figure 15.6a: Reinforcement Banding

L


15.11 Transverse shear (Vertical shear)

The next item to be checked is transverse shear and the shear stress to be

considered is related to that load in the critical section. The critical section is

shown in the Figure 15.8 below, which is 1.5d from the column face.

20

1.5d

Figure 15.6b: Reinforcement Banding

Figure 15.7: The critical section of traverse shear

CL


* The shaded areas show the load to be considered for transverse shear.

For this particular example,

1.5d = 1.5 x 590 mm

= 885 mm

Remaining distance = 1425 – 885 = 540 mm

Load causing transverse shear = 207.5 x 0.54 x 3.25

= 364 kN

Now calculate the shear stress using equation 21, BS 8110 as follows: -

Check whether transverse shear is satisfactory if does not exceed

(from Table 3.9)

= (0.632)(0.5389)(0.8969)(1.1678)

= 0.36 N/mm 2

21


Therefore is not greater than . Transverse shear is under controlled.

15.12 Maximum Shear at column face

The next item to be checked is the maximum shear around the column face. The

critical section is shown below: -

The maximum shear should not exceed the lesser of or 5N/mm2.

The maximum shear is calculated as follows: -

22

ud

Nmax

15.9: The critical section at column face.


N = design axial load

u = perimeter

d = effective depth

where, u is the column perimeter.

For this particular example

N = 1.4Gk + 1.6 Qk

= 2200 kN

= 2.33 N/mm 2

= 0.840 = 5.06 N/mm 2

This shows that <

Therefore check is all right.

15.13 Check for cracks

This is done similar to the checking of slabs. This was already covered in Unit 11.

With reference to Clause 3.12.11.2.7, BS 8110,

23


But

= 01.5 < 0.3

The permissible clear distance of bars

= 750 mm (because 3d > 750 mm)

Actual clear distance between bars (6T25)

=

= 604 mm < 750 mm

Therefore, the solutions above are suitable to use.

15.14 Column Starter Bar

The remaining reinforcement to be calculated is the requirement for the

requirement of starter bars and for this equation 39, BS 8110 is used. From Table

3.27, the maximum percentage of reinforcement required in the column is given

by;

24


= 640 mm 2

Use 4T16 bars, one is in each corner

The arrangement of reinforcement is shown below: -

25

4T16

Figure 15.10: Starter Bars Arrangement


15.7 Answer the following questions

a) Given that Gk = 1150 kN and Qk = 350 kN. What is the service

load on the base?

b) Given the dimensions of a base are 2.8 m x 2.8 m x 0.6 m,

calculate the self-weight of the base in kN.

c) If the service load of a base is 1500 kN and the permissible

pressure is 200kN/mm2, what is the area of the base required?

d) What is the dimension suitable for a square base in question c?

e) For the same data in question 1, calculate the ultimate load from

the column to the base.

26

ACTIVITY 15c


f) If the ultimate load, N=1960 kN and the area of base is 7.84 m2,

what is the ground pressure?

g) A square pad foundation is reinforced with T20 bars in both

directions. A nominal cover of 40 mm is provided. If the starter bar

used is T25 (deformed Type 2), what is the thickness of the base

required? Take fcu = 35N/mm2. The starter bar is to bend 200 mm

in length.

15.8 Answer the following questions.

Given the following data: -

i) Base dimensions: 2.8 m x 2.8 m x 0.6 m

ii) Column dimension: 400 mm square

a) Calculate the critical perimeter, U for punching shear if d = 520

mm.

b) If N = 1960 kN and d = 520 mm, calculate the shear stress, at

the column face.

c) If the punching shear, = 1000 kN, calculate the punching shear

stress, .

d) Draw the base on plan then shade and calculate the dimension of

the critical region for bending.

e) If fy=460 mm and Z=494 mm, calculate the area of the

reinforcement required. (Ground pressure = 250 kN/m2).

27


f) Propose a suitable bar size and its numbers and distance. Use data

from question e. The cover to be provided is 50mm.

Answers:

15.7 a) Service load = 1.0 Gk +1.0 Qk

=1.0(1150) + 1.0(350)

=1500 kN

b) Self-weight = volume of base x concrete density

=2.8 m x 2.8 m x 0.6 m x 24 kN/m3

= 113 kN

28

FEEDBACK 15c


c) Area of base =

=

= 7.5 m 2

d) Base width =

=2.74 m

Dimension of base is 2.74 m square.

e) Ultimate load = 1.4 Gk +1.6 Qk

= 1.4(1150) + 1.6(350)

= 2170 kN

f) Ground pressure = kN/m2

= 250 kN/m 2

g) From Table 3.29, BS 8110;

Anchorage length of starter bar = 27

= 27 x 25

= 675 mm

d = anchorage length + 2(20) + cover

= 675 + 40 + 40

= 755 mm

29


Starter Bar is bending 200 mm

Therefore, thickness = h = 755 – 200

= 555 mm

550 mm

This is shown below:

15.8 The answers are as below:

30

675 mm

200

Starter Bars Arrangement


a) Critical perimeter = (column perimeter) + (8 x 1.5d)

= (4 x 4000) + (8 x 1.5 x 520)

= 7840 mm

b)

c)

= 0.25 N/mm 2

d)

d)

31

1.2m

2.8 m

*critical section for bending


= 2550 mm 2

e) As,required = 2550 mm 2

Use 9T20 (Asprov = 2830 mm2)

Centre to centre distance

= 2800 – 2(cover) – bar diameter) 8

= 2800 – 2(50) – 20 8

= 335 mm c/c

32


The steps in the design of reinforced concrete isolated pad foundation are

summarized as follows;

1. Calculate the size of the base using allowable bearing pressure at

serviceability limit state.

2. Calculate the ground pressure at ultimate limit state.

3. Take a suitable value for the thickness of the base (h) and the effective

depth (d). Check and ensure that the shear stress around the peripheral of

the column does not exceed the lesser of 0.8 or 5 N/mm2.

4. Check the base thickness for punching shear stress. The ultimate concrete

shear stress is obtained from Table 3.9 of BS 8110.

5. Determine the reinforcement to resist bending moment.

6. When is determined accurately, do final check for punching shear.

33

SUMMARY


7. Check shear stress at the critical section.

8. Determine the size and number of starter bars required.

Answer all the questions given by circling the alphabets corresponding to the

correct answer of your choice. Award 1 point for every correct answer. You may

start now when you are ready. GOOD LUCK!

SECTION A (10 MARKS)

1. The base area of pad foundation is calculated using _____________.

A. ultimate load permissible pressure

B. service load permissible pressure

C. ultimate load x permissible pressure

D. service load x permissible pressure

34

SELF-ASSESSMENT


2. The main reinforcement in reinforced concrete pad foundation is placed

………………..

A. parallel to the shorter side

B. parallel to the longer side

C. both ways at bottom section

D. both ways at top section

3. The minimum percentage of mild steel reinforcement in reinforced

concrete pad foundation is …

A. 0.13%

B. 0.18%

C. 0.24%

D. 0.26%

4. A higher concrete grade is required for pad foundation because it is

exposed to _________________________ environment.

A. mild

B. moderate

C. severe

D. extreme

5. The reinforcement is uniformly distributed over the pad foundation width

when the distance of x spacing between columns is..…

A. less than

B. exceeds

35


C. exceeds

D. less than

6. The anchorage length of starter bar is equal to …

A. 25

B. 27

C. 29

D. 31

7. The critical section for punching shear is at _____________ from the

column face.

A. 0.5d

B. 1.0d

C. 1.5d

D. 2.0d

8. The maximum horizontal clear spacing of nominal reinforcement in

reinforced concrete pad foundation is …

A. 3d

B. 24

C. 450 mm or 5d

36


D. 200mm or 3d

9. The minimum cover to the main reinforcement in R.C pad foundation

when fresh concrete is poured directly on to the ground is …

A. 40 mm

B. 55 mm

C. 60 mm

D. 75 mm

10. Total service load in the design of R.C pad foundation is …

A. 1.4 Gk + 1.6 Qk

B. 1.0Gk + 1.0Qk

C.

D. 1.4 Gk +

SECTION B (30 marks)

A column 400 mm x 400 mm carries a dead load of 800kN and an imposed load

of 300kN. Assuming that the weight of the base is 60kN, safe bearing pressure on

the ground is 200 kN/m2; design the base to resist the loads. The materials to be

37


used are concrete of grade 30 and high yield steel reinforcement. Cover to

reinforcement is to be taken as 40mm. (For preliminary trial, h = 420mm and T25

bars are used). Sketch details of the reinforcement.

Now, check your answers with those answers given below. Calculate the

percentage of your answer you have scored too.

SECTION A

1. B

2. A

3. C

4. C

5. B

6. B

7. C

8. A

9. D

10. C

38

(1 x 10 = 10 marks)

FEEDBACK ON SELF-ASSESSMENT


SECTION B

Size of base

Working loads = 800 + 300 + 60 = 1160 kN………………

Area =

= 5.8 m 2 …………………………………………..

Try to use the base as 2.5 m x 2.5 m square base…………..

Moment steel

Ultimate load = (1.4 x 800) + (1.6 x 300) = 1600 kN..…

Ultimate pressure = = 256 kN/m 2 ……………….

Critical section is shown below;

39

2500

1050y

1

1

1

1

1


At column face,

Shear,

= 672 kN…………………………………….

Moment,

=352.8 kNm……………………………..

Base thickness

Try h = 420 mm and assuming T25 bars, the effective depth is calculated

as follows;

= 355 mm…………………………………………….

Lever arm

K=

=

40

y

1

1

1

1


= 0.032…………………………………………………..

= 0.96d > 0.95d…………………………………………

Hence, z = 0.95d = 0.95 x 355 = 337.25 mm……………

=

= 2614 mm 2 …………………………………………..

Use 6T25 (As = 2946 mm 2 )

Nominal steel =

= 1365 mm2…………………………………

As, provided: o.k.

41

1

1

1

1


Shear

The critical section for shear is shown below;

Critical section Y1Y1 is at 1.5d = 1.5 x 355 = 532.5 mm ………..

(from the face of the column).

Shear, VY1Y1 = 0.533 x 2.5 x 256 = 342 kN………………………

= 0.384 N/mm 2 …………………………………………………

The bars extend a distance (532.5 – 40 = 492.5 mm) more than d beyond the

critical section for shear and so the tension reinforcement is effective in increasing

the shear stress.

42

532.5

Y1

Y1

2500

1

1

1


= 0.632

= 0.51 N/mm 2 ………………………………………….

Therefore, vertical shear stress is satisfactory……………………

Punching shear

Overall depth = h = 420 mm……………………………

Critical perimeter is the shaded area shown below;

43

1.5d =532.5 mm from the column

1

1

1


Critical perimeter

= column perimeter + (8 x 1.5d)

= 4 x 400 + 12 x 355

= 5860 mm…………………………………………………

Area within this perimeter = (400 + 3d)2

= (400 + 3 x 355)2

= 2.15 x 10 6 mm 2 …………………………………………

Punching shear force, V = ground pressure (base area – perimeter area)

V = 256(2.52-2.15)

= 1049.6 kN…………………….

Punching shear stress,

= 0.50 N/mm 2 < = 0.51 N/mm 2 ……………….

The punching shear stress is safe. The bars have extended 478 mm beyond the

critical section.

44

1

1

1

1


Bond

Anchorage length = 34

= 34 x 25

= 850 mm (from column face) ……….

Available space =

= 1010 mm ………………………………

Anchorage length is satisfactory……………………….

Cracking

The bar spacing = 2500 – 2(40) – 25 5

= 479 mm centre to centre……………..

The clear distance between bars is not

to exceed 3d = 3(355)

= 1065 mm…………………………………………

45

Clear distance between bars

1

1

1

1

1


or 750 mm

Actual clear distance = 479 – 25

=454 mm < 750 mm………………

Therefore it complies with requirements for crack control…..

Reinforcement’s details: -

46

6T25-480

6T25-480

T25 T25 T25 T25 T25 T25

1

1

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT15/ 47

unit 15 ( design of foundations )

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