Lecture notes for Stage 2 Structural Design A, Civil Engineering Department, Newcastle University (CIV231)


First, consider the stress/strain behaviour of the two materials used in reinforced concrete, concrete and steel.

a) Concrete


The graph shows the behaviour of concrete when compressed such that the final strain is 0.0035, after which it fails. It is important to understand the different stresses used in design. Concrete strength is measured by crushing cubes when they are 28 days old and the term fcu is the characteristic concrete cube strength. Characteristic strength is obtained by calculating the average strength of a series of cubes and subtracting 1.64 times the standard deviation. Characteristic strength can be thought of as that strength below which no more than 5% of the concrete will fall. In the graph above, fcu is multiplied by 0.67 to account for the difference between cube crushing strength and bending strength. The term g m is the partial safety factor for material and has a value of 1.5 for concrete. This means that the designer is allowed to use a concrete strength of 0.67fcu/1.5 or 0.45fcu. Concrete is referred to by its grade as in the table below. The grade is simply the value of fcu. Therefore, the designer is allowed to use almost half of the grade number as a stress.

Grade Typical use Allowable Stress (N/mm2)

C7 Unreinforced concrete 3.1

C20 Reinforced concrete foundation 9

C30 or C40 Most reinforced concrete 13.5

C40 Pre-stressed concrete 18

Note that this is not the only safety factor. Loads are increased by a partial safety factor as below


Load Serviceability

Combination Dead Imposed Wind


Dead + 1.4 or 1.6 or 1.4 1.0

Imposed 1.0 0.0


Dead + 1.4 or - 1.4 1.0

Wind 1.0


Dead + 1.2 1.2 1.2 1.0

Imposed +


The full range of material partial safety factors g m is

Limit State Concrete Steel

Ultimate flexure 1.5 1.15

Ultimate shear 1.25 1.15

Ultimate bond 1.4 -

Serviceability 1.0 1.0

This means that as well as the designer using a stress which is only 0.45 of the characteristic concrete strength, he may be using a load which is between 1.2 and 1.6 times greater than the true load so that the concrete is stressed to about a third of its characteristic strength, even in the most severe adverse combination of loading.

b) Steel

Two grades of steel are used in reinforced concrete,

i) Mild steel fy = 250 N/mm2

ii) High tensile steel fy = 460 N/mm2

fy = characteristic strength of reinforcement.





Analysis of Bending

From the concrete and steel stress-strain graphs, we can draw the following sections showing stress and strain in a beam subjected to bending. We will assume that the concrete has reached its ultimate strain capability of 0.0035 strain.


When the ultimate design moment M is applied to the beam, a tensile force Fst forms in the steel and a compressive force Fcc forms in the concrete, acting through the centroid of the effective area of the compressive part of the concrete. For equilibrium between the externally applied moment M and the moment within the concrete we must have :

M = Fcc . Z = Fst . Z ............ (1)

where Z = lever arm between Fcc and Fst

We can write :

Fcc = stress x area of concrete in compression

= 0.45fcu . b. s

i.e. M = 0.45 fcu . b. s. z

Now, let Z = d - s/2 or S = 2 (d-z) and substitute in the above :

M = 0.9fcu . b (d - z) z ............ (2)

Now, we introduce the term K = M


which we substitute into the above equation to give :

(z/d)2 - (z/d) + K/0.9 = 0

which can be solved to give :

z = d[0.5 + Ö (0.25 - K/0.9)] ............ (3)

We can now write

Fst = fy As = 0.87 fy As

g m

So from (1) above :

As = M

0.87 fy . z ............ (4)


Equations (3) and (4) are used to calculate the area of steel required As in a beam to resist a bending moment M.

So far, we have assumed that the concrete takes all of the bending compression and the steel takes all of the bending tension. Sometimes (often, actually), this would result in very large beams, so steel is also placed in the compressive zone to help the concrete to take the compression. Going back to equation (2), the maximum moment which can be taken before the concrete fails in compression is :

M = 0.9 fcu . b (d-z) z

BS8110, the Code of Practice for concrete design prohibits reinforced concrete from being designed such that the concrete fails before the steel. This is achieved by not allowing z to fall below 0.775d. Therefore, we can substitute this value of z into the above equation to find the value of M which is the cut off point between compression steel being required. When z is equal to 0.775d, then the depth of the neutral axis, x, is d/2 which is the maximum value of x allowed by BS8110 for a section with tensile steel only.

Equation (2) then becomes :

M = 0.9 fcub (d-0.775d). 0.775d


M = 0.156 fcu . b.d2

Therefore, compression reinforcement is required to assist the concrete when :

M = K > 0.156

bd2 fcu

Example of beam design with tension reinforcement only


The ultimate design moment to be resisted by this section is 185 kN-m. Find the area of tension reinforcement required when :

Characteristic steel strength fy = 460 N/mm2

Characteristic concrete strength fcu = 30N/mm2


K = M = 185000000

bd2fcu 260x440x440x30


= 0.122

This is less than 0.156 so compression reinforcement is not required.

Fom Equation (3) :

Z = d[0.5 + Ö (0.25-K/0.9)]

= 440[0.5 + Ö (0.25-0.122/0.9)]

= 369mm

From Equation (4) :

As = M

0.87. fy . z

= 185000000


= 1253 mm2

The following table shows commonly used bars and can be used to select the number and size of bars

Bar Number of bars


(mm) 1 2 3 4 5 6


6 28 56 85 113 142 170

10 79 157 236 314 393 471

12 113 226 339 452 566 679

20 314 628 943 1260 1570 1890

25 491 982 1470 1960 2450 2950

Areas of various bar arrangements (mm2)


Section analysis with compression reinforcement


We already know that compression reinforcement is required when M > 0.156 fcubd2

For the above to be true x > 0.5d

So, z = d - s/2 = d - 0.9x/2

= d - 0.9. 0.5. d/2

= 0.775d

From the above figure, for horizontal equilibrium,

Fst = Fcc + Fsc

Assume both compressive and tensile steel have achieved their yield stress fy, then :

0.87 fy . As = 0.45 fcu. b.s + 0.87 fy . A's


s = 0.9 x d/2 = 0.45d

Therefore :

0.87 fy . As = 0.201 fcu . b.d + 0.87 fy . A's ............(5)

Now take moments about the line of action of the tension steel and equate the externally applied moment M to the sum of the forces Fcc and Fsc multiplied by their lever arms :

M = Fcc . z + Fsc (d - d')

= 0.201 fcu . b.d. 0.775d + 0.87. fy .A's(d-d')

= 0.156 fcu . b.d2 + 0.87 fy . A's (d-d')


A's = M - 0.156 fcu.b.d2

0.87 fy(d-d') ............ (6)

Multiplying both sides of equation (5) by z = 0.775d gives :

As = 0.156 fcu.b.d2 + A's

0.87 fy . z ............ (7)

We can use equations (6) and (7) to determine the area of the compression steel A's and the area of the tension steel As. Equation (6) shows that we do not need compression steel if the applied ultimate moment M is less than 0.156 fcu. b.d2

In BS8110, equations (6) and (7) are written in a slightly different way and if we let :

K' = 0.156

K = M/bd2 fcu

we can write them in as BS8110 :

A's = (K-K')fcu bd2

0.87 fy (d-d')


As = K' fcu .b.d2 + A's

0.87 fy. z





Example of beam design with tension and compression reinforcement

Ultimate design moment = 285 kN-m

Characteristic steel strength = 460 N/mm2

Characteristic concrete strength 30N/mm2

Calculate As and A's

K = M

b.d2. fcu

= 285000000 = 0.189


This is greater than 0.156 so compression reinforcement is required

A's = (K-K') fcu. b.d2

0.87. fy . (d-d')


= (0.189 - 0.156).30.260.440.440

0.87.460 (440-50)

= 319mm2


As = K'. fcu . b. d2 + A's

0.87 . fy . Z


= 0.156 x 30 x 260 x 440 x 440 + 319

0.87 x 460 (0.775 x 440)

= 1726 + 319

= 2045 mm2



Imposed Loads Load (kN/m2)

Domestic 1.5

Flat roofs, no access 0.75

Flat roofs with access 1.5

Hospital wards/hotel bedrooms 2.0

General use offices 2.5

Classrooms 3.0

Dance halls 5.0

Garages, passenger cars 2.5

Corridors 4.0

Storage areas 7.5


Weight (kN/m3)

Reinforced concrete 24

Water 9.81

Steel 77

Asphalt 23

Brickwork 22

Glass 27

Lead 112

Timber 10

Sand 20

Clay 23

Wind loads should be calculated using the procedure in CP3 : Ch V : Part 2 "Wind Loading". However, conservatively you can take a horizontal pressure of 1.0kN/m2 and a suction of 0.8kN/m2 in populous regions of the UK. Note the following relationships :

1 lb = 0.454kg = 4.448 N

1 lb/ft2 = 4.88 kg/m2 = 47.9 N/m2

1 lb ft3 = 16.02 kg/m3 = 157N/m3








Consider a column subject to an applied moment M and a compressive force N. There are two conditions to consider :

1) bending moment M is large in relation to

compressive force N so one side of column

is in tension

2) bending moment M is small in relation to

compressive force N so both sides of

column are in compression, one side more

so than the other

Fcc = compressive force in concrete acting through centroid of stress block

Fsc = compressive force in reinforcement at bending compressive side of column

Fs = tensile or compressive force in reinforcement at bending tension side of column

Consider vertical equilibrium :

Applied compressive force N = Fcc + Fsc + Fs

(Fs is negative when the reinforcement on the bending tension side of the column is in tension)

Therefore :

N = 0.45 fcu.b.s + fsc A's + fs As ..........(8)


fsc = compressive stress in reinforcement A's

fs = tensile or compressive stress in reinforcement As

Now take moments about the mid-depth of the section and equate the externally applied moment M to the components of the internal moment of resistance

M = Fcc h - s + Fsc h - d' + Fs h - d

2 2 2 2


M = 0.45fcu. b. s h - s + Fsc A's h - d' - fs As d -h ............ (9)

  2 2 2 2

Note that when the whole of the concrete is subject to uniform compressive stress (as in the second case), the concrete provides no contribution to the moment of resistance and the term containing fcu disappears. In practice, we usually ensure that the area of reinforcement at each side of the column is similar, i.e. A's = As and d' = h - d

Equations (8) and (9) can be rewritten :

N = 0.45 fcu .s + fsc As + fs As

bh h bh bh

............ (8a)


M = 0.45 fcu .s (0.5 - s)

bh2 h 2h


+ fsc As (d - 0.5) - fs As d - 0.5

bh h bh h

............ (9a)

hese equations have been used to draw column design charts in BS 8110.

Example of column section design

Note that much of the effort normally needed to design a column is in determining the maximum design ultimate moment and direct force at critical positions (usually just below and above floors). In this case, assume that a column is to withstand an applied ultimate moment, M of 169kN-m together with a compressive load of 1550kN. Assumptions :

Depth of column = 400mm

Width of column = 300mm

Characteristic strength of steel, fy = 460N/mm2

Characteristic strength of concrete, fcu = 30 N/mm2

The area of steel is required. Using the design chart on P. 83 of "Manual for the design of reinforced concrete building structures", we need to calculate

N and M

b.h.fcu b.h2.fcu


N = 1550000 = 0.43

b.h.fcu 300x400x30


M = 169000000 = 0.12

bh2fcu 300x400x400x30

Projecting each of these values into the chart gives

p. fy = 0.42 or p = 0.027



Asc = p.b.h = 3286 mm2

Half of this steel is needed at each side of the column (1643 mm2) which could be provided by 4 No. 25mm diameter bars at each side. Note that the chart has a dashed eccentricity line. The reason for this is that a column is never designed for zero bending. If there is no applied bending moment, then a moment which would be generated if the vertical load were applied with an eccentricity of h/20 is included in the analysis. The eccentricity line shows the effect of this - never work to the left of this line.

Analysis of Shear

The figure shows the pattern of principal stresses in a simply supported unreinforced beam. At mid span, the stresses are purely tensile near the bottom and compressive near the top. At the supports, the principal stresses are inclined at 45o to the horizontal. This means that the horizontal tensile reinforcement will not resist the tensile stress and cracks could develop even if a rigorous bending analysis has been carried out.

We prevent such cracks from developing by providing additional reinforcement in the form of stirrups and this additional reinforcement is known as the shear reinforcement. Note that some shear resistance is taken by the shear strength of the concrete but usually we need to provide stirrups. The simplest way of understanding shear reinforcement is to construct the following truss analogy in which the top and bottom chords are the compressive and tensile reinforcement, the vertical members are stirrups acting in tension and the diagonal members are the concrete itself

For the time being, assume that the stirrups are spaced at a distance d equal to the effective depth of the section so that the diagonal concrete members are inclined at 45o.

Let Asv = cross-sectional area of two legs of a stirrup

fyv = characteristic strength of stirrup reinforcement

V = shear force resulting from ultimate loads

By considering vertical equilibrium on section XX,

V = 0.87 fyv. Asv at the ultimate limit state

or v.b.d = 0.87. fyv. Asv

where v = V/bd is the average shear stress over XX.

Often, the stirrup spacing is less than the effective depth. We could consider such a situation by analysing a series of superimposed equivalent trusses. If the stirrup spacing is Sv, then the above equation is :

v.b.d (Sv) = 0.87. fyv. Asv



Asv = v . b

sv 0.87. fyv

The concrete can take some shear and we will let let Vc = concrete ultimate shear strength. We can take this value into account and rewrite the above equation

Asv = b(V - Vc)

sv 0.87.fyv

The following table shows values of Vc for 30 N/mm2 characteristic strength concrete. The value of Vc increases as the amount of steel increases because the steel contributes shear strength.

Values of ultimate shear stress Vc (N/mm2) for concrete strength fcu = 30 N/mm2

Effective depth (mm)

100 As

bd 150 200 250 300

0.25 0.54 0.50 0.48 0.46

0.5 0.68 0.64 0.59 0.57

1.0 0.86 0.80 0.75 0.72

2.0 1.08 1.01 0.95 0.91


We can rearrange the previous equation to form the following one which allows us to check that the shear resistance of a given stirrup is greater than the shear force to be withstood :

shear resistance of stirrup = vbd =

(Asv . 0.87 fyv + b.vc ) . d


Example of checking shear resistance of stirrups

  2 No. T25

As = 982mm2


fyv = 250 N/mm2 for the stirrups

fcu = 30 N/mm2 for the concrete

Reaction at support = 400 kN

First, calculate 100 As = 100x982 =0.43

bd 350x650

From the previous table, vc = 0.5N/mm2

The cross-sectional area of a 12mm diameter stirrup is 113mm2

Therefore Asv = 2 x 113/100 = 2.26


Shear resistance of stirrup

= 2.26 x 0.87 x 250 x 650 + 350 x 0.5 x 650

= 433000N

= 433 kN

The maximum shear force is at the support and is equal to the support reaction (400 kN in this case). Therefore, this arrangement of stirrups (which is a common one) is satisfactory.







In order that the reinforcement and the concrete act compositely, there must be bond between the two materials. Consider a bar cast into concrete as shown. Bond is achieved by ensuring that sufficient length of bar is embedded in the concrete. We can calculate the bond length as follows.



Let L = minimum length needed to avoid debonding

Æ = bar diameter

fbu = ultimate anchorage bond stress

fs = tensile (or compressive) stress in bar

The tensile force in the bar is :

p Æ 2 fs


The bond, or anchorage force is the product of the surface area of the embedded part of the bar multiplied by the ultimate bond stress fbu i.e.

bond or anchorage force = L.p .Æ .fbu

Equating tensile force to anchorage force gives :

L = fs Æ


Let fs = fy/g m = 0.87fy

where fy = characteristic strength of reinforcement (250 N/mm2 or 460 N/mm2)

Then, the bond length required can be checked from the following equation :

L = 0.87 fy Æ

4 fbu

The value of fbu, the ultimate anchorage bond stress is related to fcu the characteristic concrete cube strength by :

fbu = b Ö fcu

Values of b are shown below

Bar type b for bars in tension b for bars in compression


Plain bars 0.28 0.35

Deformed bars 0.40 0.50

Fabric 0.65 0.81


Anchorage can also be provided by hooks or bends in the reinforcement. Tests have confirmed the following contribution

Anchorage value = 4r but not greater than 12 Æ

Anchorage value = 8r but not greater than 24 Æ

For mild steel bars, minimum r = 2 Æ

For high yield bars, minimum r = 3 Æ

Example of bond length check

How far does the tensile reinforcement need to be embedded in the case of this cantilever where :

fcu = 30 N/mm2

fy = 250 N/mm2

25mm Æ plain bars

The ultimate anchorage bond stress, fbu = b Ö fcu

From the previous table, b = 0.28


fbu = 0.28 Ö 30 = 1.5 N/mm2

Anchorage length L = 0.87 fy Æ

4 fbu

= 0.87 x 250 x 25

4 x 1.5

= 910 mm

Note that a 180o hook could take 8Æ anchorage length i.e. 200mm so L would be reduced to

710 mm









The figure shows a frame of a heavily loaded industrial structure for which the centre columns along line PQ are to be designed in this example. The frames at 4m centres, are braced against lateral forces, and support the following floor loads :

dead load gk = 10 kN/m2

live load qk = 15 kN/m2

Characteristic material strengths are fcu = 30 N/mm2 for the concrete and fy = 460 N/mm2 for the steel.

Maximum ultimate load at each floor

= 4.0(1.4gk + 1.6qk) per metre length of beam

= 4(1.4 x 10 + 1.6 x 15)

= 152 kN/m

Minimum ultimate load at each floor

= 4.0 x 1.0 gk

= 4.0 x 10 = 40 kN per metre length of beam

Consider first the design of the centre column at the underside (u.s.) of the first floor. The critical arrangement of load which will cause the maximum moment in the column is shown in the figure.

Column loads

second and third floors = 2 x 152 x 10/2 = 1520kN

first floor = 152 x 6/2 + 40 x 4/2 = 536kN

Column self-weight, say 2 x 14 = 28kN

N = 2084kN

Similar arrangements of load will give the axial load in the column at the underside (u.s. and top side (t.s.) of each floor level and these values of N are shown in table 1 (coming soon).

The moments on the column are not large. Try a 300 x 400 column

N = 0.35fcubh + 0.67 fy Asc

2084 x 103 = 0.35 x 30 x 300 x 400 + 0.67 x 460 x Asc

from which Asc = 2674 mm2 and 100 Asc/bh = 2.23

This provides an adequate cross-section and a 300 x 400 column is to be used.

Column Moments

The loading arrangement and the substitute frame for determining the column moments at the first and second floors are shown in the figure. Member stiffnesses are


kAB = bh3 = x 0.3 x 0.73 = 0.71 x 10-3

2 12LAB 12x6


kBC = x 0.3 x 0.73 = 1.07 x 10-3

2 12 x 4


kcol = 0.3 x 0.43 = 0.46 x 10-3

12 x 3.5


S k = (0.71 + 1.07 + 2 x 0.46) 10-3

= 2.70 x 10-3

and distribution factor for the column

= kcol = 0.46 = 0.17

S k 2.70

Fixed end moments at B are

F.E.MBA = 152 x 62 = 456 kNm



F.E.MBC = 40 x 42 = 53kN m


Column moment M = 0.17(456 - 53) = 69kNm

At the 3rd floor

S k = (0.71 + 1.07 + 0.46) 10-3

= 2.24 x 10-3

Column moment M = 0.46 (456 - 53)


= 83 kNm

The areas of reinforcement in table 1 are determined by using the column design chart. Sections through the column are shown in the figure.


Cover for the reinforcement is taken as 50mm and d/h = 320/400 = 0.8. The minimum area of reinforcement allowed in the section is given by

Asc = 0.004 bh = 0.004 x 300 x 400 = 480 mm2

and the maximum area is

Asc = 0.06 x 300 x 400 = 7200 mm2

or at laps

Asc = 0.1 x 300 x 400 = 12000 mm2

and the reinforcement provided is within these limits.

Table 1


Floor N M N M 100Asc Asc

(kN) (kNm) bh bh2 bh mm2


3rd u.s. 536 83.0 4.47 1.73 0.4 480

2nd t.s. 774 69.0 6.45 1.44 0.4 480


2nd u.s. 1310 69.0 10.92 1.44 0.4 480

1st t.s. 1548 69.0 12.9 1.44 0.9 1080


1st u.s. 2084 69.0 17.37 1.44 2.1 2520

Foundation 2098 34.5 17.48 0.72 1.6 1920


A smaller column section could have been used above the first floor but this would have involved changes in formwork and also increased areas of reinforcement. For simplicity in this example no reduction was taken in the total live load although this is permitted with some structures.


Moment Redistribution

An important consideration in reinforced concrete design is moment redistribution. Usually, we do not design reinforced concrete redundant structures to withstand the bending moments which we calculate by area moments, slope deflexion or virtual work. We first of all calculate bending moments by those methods, then we redistribute them. We do this because reinforced concrete does not behave elastically. The graph shows the relationship between moment and curvature for a typical reinforced concrete section.

Even when the ultimate moment Mu has been reached in reinforced concrete, usually the beam or other element can carry on rotating without breaking. During this further rotation, in a redundant structure, subjected to increasing load, some other part of the structure will eventually reach its ultimate moment and failure finally takes place when two or more "hinges" have formed.

This can be illustrated in a fixed ended beam loaded by a UDL.


We could use area moments to prove that the elastic bending moment diagrams has the shape above. This means that Mu would be reached at each end first. When that happens, the reinforced concrete reaches the horizontal part of the previous graph i.e. each end begins to rotate as Mu remains constant. During this time, the bending moment at the centre of the beam increases and eventually reaches the same value as the ends. Only when the central section reaches Mu does the beam fail. This means that a redundant structure has an extra reserve of strength beyond that which elastic analysis would predict. In this case, the ultimate moment is when the central moment and the support moments have an equal value i.e. wL2/16.

By this means, we can see that the bending moment to cause failure is a third greater than elastic analysis would have predicted - we therefore redistribute moments by 33%. In fact, only 30% redistribution is permitted by BS8110 and for multi-storey structures, a maximum of 10% is permitted. In order to undertake a rigorous moment redistribution, a plastic analysis of the structure has to be carried out, i.e. the "hinges" have to be identified and the ultimate bending moment calculated there. This is beyond the scope of this course. Therefore, be conservative and redistribute moments by no more than 10%, and only with redundant structures.

Serviceability of Reinforced Concrete

In designing reinforced concrete, it is not enough just to ensure that the steel and concrete are not overstressed. The structure must remain serviceable and this is achieved by the following :

a) Cover to reinforcement to avoid corrosion

If the reinforcement is too close to the surface of the concrete, it may corrode. Its likelihood of corroding depends upon the exposure conditions and upon the quality of the concrete covering it. The following table is taken from BS8110 and shows values of cover for various conditions.

Environment Cover (mm)


Mild - protected 25 20 20 20

against weather


Moderate - - 35 30 25

sheltered from severe

rain and freezing when

wet. Subject to



Severe - exposed - - 40 30

to severe rain,

occasional freezing


Very severe - - - 50 40

exposed to sea water or de-icing salts


Extreme - exposed to abrasion - - - 60


Minimum cement content 275 300 325 350


Lowest permissible grade C30 C35 C40 C45




b) Fire resistance

The cover values in the previous table may have to be increased to ensure fire resistance. The Building Regulations state that various types of structure require different fire resistance times and the following table gives the corresponding cover values

Fire Resistance Cover for fire resistance

(hr) Beams Slabs Columns


0.5 20 20 20

1.0 20 20 20

2.0 40 35 25

4.0 70 55 25


BS8110 also gives minimum member sizes to ensure fire resistance as below

Fire Beam Floor Column Wall

Resistance Width Thickness Width Thickness

(hr) (mm) (mm) (mm) (mm)


0.5 200 75 150 150

1.0 200 95 200 150

2.0 200 125 300 150

4.0 280 170 500 180


c) Maximum spacing of reinforcement

It might seem economical in labour to have relatively few very large diameter bars at large spacings rather than many smaller bars. Doing so leads to cracking of the concrete. Generally speaking, all reinforced concrete cracks. Good reinforcement detailing keeps the cracks small enough to ensure that they do not reach the reinforcements. The following table ensures that cracks no greater than 0.3mm wide form and this is usually sufficient to prevent corrosion. Note that in water retaining concrete, the crack width limit is 0.2mm generally and 0.1mm when staining of the concrete surface would be unacceptable.






fy Maximum clear reinforcement spacing (mm) for tension bars

% Moment Redistribution


-30 -20 -10 0 10 20 30

250 210 240 270 300 300 300 300

460 115 130 145 160 180 195 210


d) Minimum spacing of reinforcement

In order to allow the concrete to be placed successfully, the reinforcement bars must not be too close together. The minimum spacings should be

i) hagg + 5mm horizontally

ii) 2 hagg/3 vertically

where hagg = maximum aggregate size

e) Minimum reinforcement areas

When concrete is cast, some of the mixing water evaporates with the result that the concrete shrinks. Also, cement gives off heat when it is hydrating - an exothermic reaction takes place - so concrete is usually warm when it sets. For both of these reasons, concrete which is restrained tries to crack, usually in its early life. This cracking is inhibited by providing minimum quantities of reinforcement, which may be greater than the quantities needed for primary structural purposes. The following table gives values from BS 8110.


Mild Steel High Yield Steel

fy=250N/mm2 fy=460N/mm2


Pure tension 100 As/Ac = . . .8% 0.45%

Bending tension100 As/Ac = . 0.24% 0.13%

Compression 100 Asc/Acc = . 0.4% 0.4%


f) Side Face Beam Reinforcement

Beams deeper than 750mm require reinforcement in their sides as well as in the top and bottom. The bars should run horizontally along the beams and should be at not greater than 250mm, centre. The diameter of these bars must be not less than _(sb.b/fy) where :

sb = bar spacing (mm)

b = breadth of section (mm)

fy = yield strength of steel

(250 N/mm2 or 460 N/mm2)

g) Span-effective depth ratio

The following span-effective depth ratios will ensure that deflexions are kept to reasonable values which are usually taken to be :

1) Span/250 resulting from self-weight plus imposed load

2) Span/500 resulting from imposed load

If the following are adhered to, deflexions need not be calculated. For beams of span greater than 10m, multiply the ratio by 10/span. This avoids damage to finishes (plaster etc.)


Type of Element Span/effective depth ratio


Cantilever 7

Simply supported beam 20

Continuous beam 26


The above three ratios should be multiplied by a factor in one of the next two tables according to the maximum stress in the reinforcement when serviceability partial safety factor are applied and and according to the ratio M/bd2 for tension reinforcement or 100 A's/bd for compression reinforcement.

Tension Reinforcement :

Reinforcement M/bd2

Service Stress

(N/mm2) 0.5 0.75 1.0 1.5 2.0 3.0 4.0


100 2.0 2.0 2.0 1.86 1.63 1.36 1.19


156 2.0 2.0 1.96 1.66 1.47 1.24 1.10

200 2.0 1.95 1.76 1.51 1.35 1.14 1.02


288 1.68 1.50 1.38 1.21 1.09 0.95 0.87


Compression Reinforcement :

100 A's Factor



0 1

0.25 1.08

0.5 1.14

1 1.25

2 1.4

3 1.5


Example of deflexion check

A rectangular continuous beam spans 12m with a mid-span ultimate moment of 400 kN-m. If its breadth is 300mm, check that an effective depth of 600mm is acceptable in terms of deflexion. High yield reinforcement is used (fy = 460 N/mm2) and two 16mm diameter bars are used in the compression zone (area of compression steel = 402mm2). Basic span - effective depth ratio = 26

Because the span is greater than 10m, the above figure has to multiplied by 10/span

i.e. 26 x 10/12 = 21.7

Tensile reinforcement modification factor :

M/bd2 = 400000000 = 3.7

300 x 600 x 600

Modification factor = 0.89. Compression reinforcement modification factor :

100A's = 100 x 402 = 0.22

bd 300 x 600

Modification factor = 1.07

Modified span-effective depth ratio = 21.7 x 0.89 x 1.07 = 20.7

Actual span-effective depth ratio =

12000 = 20


The proposed section does not deflect unduly.

h) Stability

An important consideration in reinforced concrete design is that should an accident occur to part of the structure, the resulting damage should not be disproportionate to the collapse. A relatively minor mishap must not be allowed to cause structural collapse, as happened when a gas explosion caused the whole corner of a multi-storey block of flats to collapse at Ronan Point. Several precautions have to be taken. Any structure must be able to resist 1.5% of the total characteristic load acting as a horizontal load. Extremely vulnerable members, e.g. columns beside a road should be protected by barriers. The most important stability measure is to provide reinforcement which acts as ties. This reinforcement can be exactly the same material as that provided in element design but it has to be detailed to hold the structure together in the event of an accident as follows.


Vertical ties

For buildings five storeys high or greater, the reinforcement must be kept continuous from roof to ground floor by load bearing laps. The reinforcement in the columns must be able to resist a tensile force equal to the maximum ultimate compressive load in the column. In reality, this only affects joint detailing.

Horizontal ties

In all buildings, three types of tie should be provided :

1) Peripheral ties

2) Internal ties

3) Column ties

Each of these ties should resist a force Ft of either 60kN or 20 + 4 x number of storeys in structure kN, whichever is less.

Peripheral Ties

A peripheral tie is provided by reinforcement which is effectively continuous around the perimeter of the building at each floor level and at roof level. It must take a force Ft.

Internal ties 

Internal ties should be provided at each floor in two orthogonal directions and should be anchored at each end to the peripheral tie. Either continuous ties should be used i.e. each individual bar in a slab linked to the peripheral tie or alternatively, the reinforcement in beams can be used. The tie force required is Ft kN/metre or

Ft (dead load + live load) . L

7.5 5

where L = greatest horizontal distance along the tie between centres of columns.

Column ties

Column ties are horizontal ties which must resist a force of 3% of the total vertical ultimate load for which the column has been designed. Also they must resist a force of 2Ft or Ft. Lo/2.5kN where Lo is the floor to ceiling height in metres.

Example of Stability Design

Calculate the stability tie requirements for an eight storey building comprising in-situ concrete beams and columns with a precast concrete floor as shown

Clear storey height under beam =2.9m

Floor to ceiling height (Lo) = 3.4m

Characteristic dead load= 6kN/m2 on each floor

Characteristic live load = 3kN/m2 on each floor

Characteristic steel strength (fy) =460N/mm2

Ft = (20 + 4 x number of storeys) = 20 + 4 x 8 = 52kN

(a) Peripheral ties

Force to be resisted = Ft = 52kN

Area of steel required = 52000 = 113mm2


This could be provided by one T12 bar

(b) Internal ties

Force to be resisted = Ft (live load + dead load) x L kN/m

7.5 5

(1) Transverse direction :

force = 52(6+3) x 7 = 87.4kN/m

7.5 5

force per bay = 87.4 x 6.5 = 568 kN

Bar area required in each transverse interior beam is

568000 = 1235mm2


This could be provided by 4 T20 bars

(2) Longitudinal direction

force = 52 (6 + 3) x 6.5 = 81.1kN/m

7.5 5

Force along length of building

= 81.1 x 7 = 567kN

Bar area required each side of building =

567000 = 617mm2

2 x 460

This could be provided by 2 T20 bars

(3) Column ties

force to be designed for is

Lo Ft = 3.4 . 52 = 70.7kN

2.5 2.5

Alternatively 3% of ultimate floor load on a column is

8 [ 3/100(1.4 x 6 + 1.6 x 3) x 6.5 x 7/2 ] = 72 kN at ground level

We must also add 3% of the column self-weight so increase the 72kN to 75kN

Area of ties required = 75000 = 163mm2


This could be provided by a 1 T20 bar

(c) Vertical ties

Maximum column load corresponding with one storey is :

(1.6 x 3 + 1.4 x 6) x 3.5 x 6.5 = 300kN

Bar area required throughout each column is :


300000 = 653mm2


This could be provided by 4 T16 bars.