Basic Concepts , Rectangular and T Beams.pdf

5/20/2018 Basic Concepts , Rectangular and T Beams.pdf

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1.1 INTRODUCTION

The design of different reinforced concrete sections of beams will be considered in this chapter.

1.2 DESIGNANDANALYSIS

The main task of a structural engineer is the analysis and design of structures. The two approaches of design an

will be used in this chapter:

Design of a section.This implies that the external ultimate moment is known, and it is required to comdimensions of an adequate concrete section and the amount of steel reinforcement. Concrete strength and yiel

used are given.

Analysis of a section. This implies that the dimensions and steel used in the section (in addition to concrete

yield strengths) are given, and it is required to calculate the internal ultimate moment capacity of the section so t

be compared with the applied external ultimate moment.

1.3 BASICASSUMPTIONSINFLEXURETHEORY

Five basic assumptions are made:

1. Plane sections before bending remain plane after bending.

2. Strain in concrete is the same as in reinforcing bars at the same level, provided that the bond between the

concrete is sufficient to keep them acting together under the different load stages i.e., no slip can occur bet

two materials.

3. The stress-strain curves for the steel and concrete are known.

4. The tensile strength of concrete may be neglected.

5. At ultimate strength, the maximum strain at the extreme compression fiber is assumed equal to 0.00

Egyptian Code.

The assumption of plane sections remaining plane (Bernoulli's principle) means that strains above and below th

axisNA are proportional to the distance from the neutral axis, Fig. 1.1. Tests on reinforced concrete memb

indicated that this assumption is very nearly correct at all stages of loading up to flexural failure, provided gexists between the concrete and steel. This assumption, however, does not hold for deep beams or in region

shear.


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FIGURE 1.1. Single reinforced beam section with strain distribution.

1.4 BEHAVIOROFAREINFORCEDCONCRETEBEAMSECTIONLOADEDTOFAILURE

To study the behavior of a reinforced concrete beam section under increasing moment, let us examine how s

and stresses progress at different stages of loading:

1.4.1 Noncracked, Linear Stage

As illustrated in Fig. 1.2, where moments are small, compressive stresses are very low and the maximum tensileconcrete is less than its rupture strength,f . In this stage the entire concrete section is effective, with the steel b

tension side sustaining a strain equal to that of the surrounding concrete ( ) but the stress in the steel bar

to that in the adjacent concrete multiplied by the modular ratio n. Utilizing the Transformed Area Concept, in

steel is transformed into an equivalent concrete area , the conventional elastic theory may be used to analy

concrete" area in Fig. 1.2.

FIGURE 1.2. Transformed section for flexure before cracking.

This stage should be considered as the basis for calculating the cracking momentM , which produces tensil

at the bottom fibers equal to the modulus of rupture of concrete, Fig. 1.3. The Egyptian Code recommends th

formulaM/Zto compute the flexural strength of the section:

(1.1a)

where is the moment of inertia of gross concrete section about the centroidal axis, neglecting the reinforcem

the distance from the centroidal axis of cross section, neglecting steel, to extreme fiber tension and f is the m

rupture of concrete. The Egyptian code (ECCS) suggests an imperical formula relates the modulus of rupture of

to its compressive strength:

N/mm (1.1b)

ctr

cr

ctr

2


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FIGURE 1.3. Transformed section for flexure just prior to cracking.

1.4.2 Cracked, Linear Stage

When the moment is increased beyondM , the tensile stresses in concrete at the tension zone increased until t

greater than the modulus of rupturef , and cracks will develop. The neutral axis shifts upward, and cracks ext

to the level of the shifted neutral axis. Cracked concrete below the neutral axis is assumed to be not effective

steel bars resist the entire tensile force. The stress-strain curve for concrete is approximately linear up to 0.40 f

the concrete stress does not exceed this value, the elastic (straight line) theory formulaM/Zmay be used to an

"all concrete" area in Fig. 1.4.

FIGURE 1.4. Transformed section for flexure somewhat after cracking.

1.4.3 Cracked, Nonlinear Stage

For moments greater than these producing stage 2, the maximum compressive stress in concrete exceeds

However, concrete in compression has not crushed. Although strains are assumed to remain proportional to the

from the neutral axis, stresses are not and, therefore, the flexural formula M/Z of the conventional elastic theo

be used to compute the flexural strength of the section. The Internal Couple Approach, instead, will be used to

the section strength. This approach allows two equations for equilibrium, for the analysis and design of

members, that are valid for any load and any section. As Fig. 1.5 indicates, the compressive force C should be

the tensile force T, otherwise the section will have a linear displacement plus rotation. Thus,

C = T (1.2a)

cr

ctr

cu


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The internal moment is equal to either the tensile force T multiplied by its arm yct or the compressive

multiplied by the same lever arm. Thus,

(1.2b)

FIGURE 1.5. Transformed section for flexure after cracking.

The resultant internal tensile force Tis given by

(1.3)

where is the area of steel and is the steel stress. The resultant internal compressive force is obtained by i

the stress block over the area bc. Taking an infinitesimal strip dyof area dAequals bbydy, located at a distan

the neutral axis and subject to an assumed uniform compressive stress fand strain Xthe compressive force C

by

(1.4)

This stage may be considered as the basis for calculating the flexural strength of the section at first yield of th

steel (known as the yield moment ). When the tension steel first reaches the yield strain ( ), the str

extreme fiber of the concrete may be appreciably less than 0.003. If the steel reaches the yield strain and the

reaches the extreme fiber compression strain of 0.003, simultaneously, the yield moment occurs and equals the

momentM. Otherwise, if the concrete crushed before the steel yields, the yield moment will never take place.

1.4.4 Ultimate Strength Stage

For the given section, when the moment is further increased, strains increased rapidly until the maximum

capacity of the beam was reached at ultimate momentM. The section will reach its ultimate flexural strength

concrete reaches an extreme fiber compression strain X of 0.003 and the tensile steel strain X cloud have

higher or lower than the yield strain .

As Fig. 1.6 indicates, the compressive forces C and C are obtained by integrating the parabolic and recta

stress blocks over the rectangular areasA andA of and , respectively.

u

u

cu s

1 2

1 2


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FIGURE 1.6. Single reinforced beam section with flexure at ultimate.

The corresponding lever armsy andy are given by

The resultant force Cis, then, computed from

(1.5)

The position of Cis at a distanceyfrom the top fiber whereyis computed from

The distance between the resultant internal forces, known as the internal lever arm, is

y =d - 0.4c (1.6)

where d, the distance from the extreme compression fiber to the centroid of the steel area, is known as thedepth. The ultimate strengthMis therefore

(1.7)

1.5 EQUIVALENTRECTANGULARCOMPRESSIONSTRESSBLOCK

As a means of simplification, the Egyptian Code has suggested the replacement of the actual shape of the

compressive stress block (a second-degree parabola up to 0.002 and a horizontal branch up to 0.003) by an e

rectangular stress block, Fig. 1.7.

1 2

ct

u


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FIGURE 1.7. Actual and equivalent stress distribution at failure.

A concrete stress of is assumed uniformly distributed over an equivalent compression zone bo

the edges of the cross section and a line parallel to the neutral axis at a distance from the fiber of

compressive strain, where cis the distance between the top of the compressive section and the neutral axisNA.

For the resultant compressive forces of the actual and equivalent stress blocks of Fig. 1.7, to have

magnitude and line of action, the average stress of the equivalent rectangular stress block and its depth are

and where and . These values are as already derived when calculating th

strengthMin Section 1.4.4.

The equivalent rectangular stress block applies, as the Egyptian Code permits, to rectangular, Tand tr

sections, Fig. 1.8.

FIGURE 1.8. Applicability of equivalent rectangular stress block to some sections.

For sections as shown in Fig. 1.9, stress distribution should be based on the actual stress-strain diagram. Tprocedure, however, can be implemented to obtain the parameters and that correspond to these section

u


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FIGURE 1.9. Inapplicability of equivalent rectangular stress block to some sections.

1.6 TYPESOFFLEXURALFAILURE

The types of flexural failure possible (tension, compression and balanced) and the nominal (ideal) strengthM

beam section (a singly reinforced rectangular section) are discussed next.

1.6.1 Tension Failure

If the steel content of the section is small (an under-reinforced concrete section), the steel will reach its yieldbefore the concrete reaches its maximum capacity. The flexural strength of the section is reached when the str

extreme compression fiber of the concrete is approximately 0.003, Fig. 1.10. With further increase in strain, the

of resistance reduces, and crushing commences in the compressed region of the concrete. This type of failure, b

is initiated by yielding of the tension steel, could be referred to as a "primary tension failure,"or simply "tensio

The section then fails in a "ductile"fashion with adequate visible warning before failure.

FIGURE 1.10. Single reinforced section when the tension failure is reached.

For a tension failure, ; for equilibrium, C = T. Hence from from

and

we have which results in

(1.8)

The nominal strengthM(which obtained from theory predicting the failure of the section on assumed section

geom

specified materials strengths i.e., = = 1.0), is

(1.9)

u


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1.6.2 Compression Failure

If the steel content of the section is large (an over-reinforced concrete section), the concrete may reach its

capacity before the steel yields. Again the flexural strength of the section is reached when the strain in the

compression fiber of the concrete is approximately 0.003, Fig. 1.11. The section then fails suddenly in a "brittle

if the concrete is not confined and there may be little visible warning of failure.

FIGURE 1.11. Single reinforced section when the compression failure is reached.

For a compression failure, as the steel remains in the elastic range. The steel stress may be dete

terms of the neutral axis depth considering the similar triangles of the strain diagram of Fig. 1.11.

\ (1.10)

The steel stress is

(1.11a)

or, sinceE= 200 kN/mm ,

(1.11b)

For equilibrium, , hence

(1.12)

The above quadratic equation may be solved to find cand, on substituting a= 0.8c, the nominal strength is

(1.13)

1.6.3 Balanced Failure

At a particular steel content, the steel reaches the yield strength and the concrete reaches its extreme fiber com

strain of 0.003, simultaneously, Fig. 1.12. Then, and from the similar triangles of the strain diagra

s2


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1.12 we can write

(1.14)

where = neutral axis depth for a balanced failure. Then

(1.15)

or, on substituting = 0.80 , Eq. 1.15 becomes

(1.16)

FIGURE 1.12. Single reinforced section when the balanced failure is reached.

For equilibrium, ; hence we have

(1.17)

which results in

(1.18)

where is the balanced steel ratio.

The type of failure that occurs will depend on whether the steel ratio m (where m= ) is less than or gre

. Figure 1.13 shows the strain profiles at a section at the flexural strength for three different steel contents

1.13 indicates, if for the section mis less than , then c< c and ; hence a tension failure occurs. Sim

is greater than , then c> c and , and a compression failure occurs.

b

b


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FIGURE 1.13. Strain profiles at the flexural strength of a section.

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1.7 FLEXURALDESIGNOFREINFORCEDCONCRETERECTANGULARSECTIONS

The use of design equations, as well as design aids and tables, in designing reinforced concrete rectangular sections is

next.

1.7.1 Design of Singly Reinforced Sections

Compression failures are dangerous in practice because they occur suddenly, giving little visible warning and are brittl

failures, however, are preceded by wide cracking of the concrete and have a ductile character. To ensure that all be

adequate visible warning if failure is imminent, as well as reasonable ductility at failure, it is necessary to limit the area of te

in singly reinforced sections to a proportion of the balanced area because, as Eq. 1.18 indicates, if the yield strength of t

higher or the concrete strength is lower, a compression failure may occur in a beam that is loaded to the flexural strength.

In design, a dependable (design) strength of characteristic (nominal or ideal) strength is used, where g is th

reduction factor. Therefore, the balanced steel ratio in Eq.1.18 becomes

(1.19)

and, since g = 1.15 and g = 1.5, we can write

(1.20)

The Egyptian Code (as compared to the ACI Code which limits the maximum allowable steel ratio m to 0.75 ) recomm

the neutral axis depth cin singly reinforced beams not exceed two-thirds that for a balanced failure:

(1.21)

The maximum allowable steel ratio is thus

(1.22)

As Fig. 1.14 indicates, for a tension failure, , for equilibrium, C= T, we have

s c

max

max


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(1.23)

which results in a. The ultimate strengthM is then given by

(1.24)

where

(1.25)

and

(1.26)

FIGURE 1.14. Single reinforced section when the flexural strength is reached.

For a given concrete strength and steel yield strength,R is a function of . This means thatR cannot be increased b

valueR that correspond to . Therefore, using ,

(1.27)

It should be mentioned that the design strength of a cross section is limited to the value that correspond to R or

Therefore, usingR ,

(1.28)

u

u u

max max max

max

max


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The values of , , and for all grades of steel are given in Table 1.1. Table 1.2 is used if a fraction of th

10% is redistributed

TABLE 1.1. Values of , , and - No moment redistribution.

Type of steel

240/350 0.742 0.50 8.56 10 0.21

280/450 0.711 0.48 7.00 10 0.20

360/520 0.657 0.44 5.00 10 0.19

400/600 0.633 0.42 4.31 10 0.18

450/520 0.605 0.40 3.65 10 0.18

Apply only to rectangular sections with only and in N/mm .

TABLE 1.2. Values of , , and - Moment redistribution = 10%.

Type of steel

240/350 0.597 0.40 6.85 10 0.18

280/450 0.567 0.38 5.58 10 0.17

360/5200.507 0.34 10 0.15

400/600 0.477 0.32 3.29 10 0.15

450/520 0.447 0.30 2.74 10 0.14

Apply only to rectangular sections with only and in N/mm .

In design, the variables in Eq. 1.24 can be b, dandA . It is evident that there is a range of satisfactory sections having

strength, and before a solution can be obtained the designer must assume the value for one or more of these variables.

a) Minimum Effective Depth with Maximum Steel

In this case, when b is given or assumed and the effective depth dis unknown, the section may be designed to have a

depth by putting . Such a design requires a very high steel content. Unless a very shallow depth is ess

of is not economical and it is better to use a deep section with less steel. Also, the deflections of a beam with th

possible depth may be excessive and may need to be checked.

-4fcu

-4fcu

-4fcu

-4fcu

-4fcu

* 2

-4fcu

-4fcu

-4

fcu

-4fcu

-4fcu

* 2

s


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FIGURE 1.15. Single reinforced section with minimum effective depth and maximum steel.

Design Equations:

As Fig. 1.15 indicates, the depth will be a minimum, d = , if v is the maximum allowed, . First calculate

(1.29)

and from

(1.30)

Design Aids:

The ultimate design momentMis given by

(1.31a)

Or

(1.31b)

and on substituting

and

Eqs. 1.31aand 1.31bbecome

u


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(1.32)

and

(1.33)

Table A.1 gives values for and for a range of commonly used steel yield and concrete strengths. Enter

with the known values of and . Traverse vertically to the value, then horizontally to the value and finally

values of and to be used.

Example 1.1:

A 250 mm wide, single reinforced rectangular section is to carry an Mu of 200 kNm. Using = 25 N/mm2 and stee

design the section for minimum depth.

Solution:

Design Equations:

From Table 1.1, we have 0.208 and 7 10 f

Calculate from

\ = 480 mm

Calculate from

2100 mm (Use 7 f20)

Design Aids:

1[1]

-4cu

2
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Enter Table A.1 with = 25 N/mm and = 280 N/mm and obtain

= 0.534 and = 196.7

Then, calculate and as follows:

mm

cm (Use 7 f20)

The Egyptian Code specifies that the compression steel be not less than 0.10 . Therefore, 2.17 cm (Us

b) Great Effective Depth with less Steel

Here, both the width band the effective depth d are known. If d is assumed to be greater than d ; therefore, the

adequate without compression reinforcement. This implies also that will be less than .

FIGURE 1.16. Single reinforced section with great effective depth and less steel.

Design Equations:

First calculate from

If dis greater than , therefore, the section is adequate without compression steel. As Fig. 1.16 indicates, for equ=T, hence from

and

2 2

2

2

min


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we have (1.34)

The ultimate design moment is thus

(1.35)

The above quadratic equation may be solved to find .

Design Aids:

The ultimate design moment is given by

Or

and on substituting

and

Eqs. 1.36aand 1.36bbecome

(1.37)

and

(1.38)


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Tables B.1 through B.5 give values for and for all grades of steel and a range of commonly used concrete strengt

calculateK from Eq. 1.37. Then, with the known value of , determine the design table that corresponds (Tables B.1

B.5). Traverse vertically to theK value, then horizontally to thef value, and finally obtain the value ofKto be used. C

from Eq. 1.38.

Note:

It is also necessary to provide a minimum reinforcement ratio that should always be exceeded. This is recommended bec

reinforcement ratio is very small, the computed flexural strength as a reinforced concrete section becomes less than th

momentM , and on cracking, failure is sudden and brittle. To prevent this, the Egyptian Code specifies that the area

beams be not less than the minimum area of steelA that should be provided. The minimum area of steel, accord

Egyptian Code, equals the least of:

(units are inN andmm) (1.39)

and

1.30 ( = area of steel required by the analysis) (1.40)

Provided thatA should be not less than

0.25% bd (for normal mild steel) (1.41)

and

0.15% bd (for high grade steel) (1.42)

In T-shaped andL-shaped sections where the web is in tension, the minimum steel ratio is computed using the web width

Example 1.2:

A 250 mm wide, singly reinforced rectangular section is to carry anMof 200 kNm. Using = 25 N/mm and steel280/450, design the section for an effective depthdof 550 mm.

Solution:

Design Equations:


\ = 480 mm

because dis greater than , therefore, the section is adequate without compression steel. Calculate from

1

1 cu 2

cr

smi n

smi n

u2


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= 1730 mm (Use 2 f22 + 2 f25)

Design Aids:


which results inK= 0.6166.

Assume that dis greater than and enter Table B.2 withK= 0.6166, then traverse horizontally to = 25, and fin

K= 210. Calculate from

mm (Use 2 f22 + 2 f25)

A = the least of:

540 mm 2406 mm

1.3 = 1.3 x 1731.6 = 2251 mm

But not less than

344 mm

A = 540 mm A = 2406 mm

Here, = 1731.6 mm which is greater thanA and less than .

= 173.0 mm (Use 2f12)

2

1

1

2

2

smi n

2

2

2

2

smin2

smax2

2smi n

2


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1.7.2 Design of Doubly Reinforced Sections

When a beam of shallow depth is used, the flexural design strength obtained that is allowed for the section as singly

M ( if and 0) may be insufficient. The design moment capacity may be increased by placing compre

and additional tension steel. In addition, to increasing the section strength when its depth is limited, compression ste

required in design for the following reasons:

1. Compression steel may be used in design to increase the ductility of the section at the flexural strength.

2. Compression steel may be used to reduce deflection of beams at the service load. Compression steel also reduce

term deflections of beams due to creep. Curvatures due to shrinkage of concrete are also reduced by compression stee

3. For the beams of continuous frames under gravity and lateral loading, consideration of possible combinations

loading reveal that the bending moment can change sign. Such members require reinforcement near both faces to carry th

tensile forces and therefore act as doubly reinforced members.

4. Compression steel provides hangers for stirrups.

When designing double reinforced concrete sections, the Egyptian Code specifies that the maximum spacings betwe

should not be greater than 15 times the diameter of the compression steel. This helps to prevent buckling.

Design Equations:


(1.43)

If dis less than , therefore, compression steel is required. As Fig. 1.17 indicates,

(1.44)

and

(1.45)

umax


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FIGURE 1.17. Double reinforced section when the flexural strength is reached.

The difference in momentM is given by

(1.46)

which results in and . Of course,

(1.47)

If , we can write

and (1.48)

Otherwise; if the compression steel is not yielding, the stress in it may be found in terms of c , using the strain diagram of

(1.49)

and thus,

(1.50)

FIGURE 1.18. Double reinforced section when the flexural strength is reached.

Design Aids:

With reference to Fig. 1.18, taking the moments of forces about T, Cand Ceach a time:

(1.51a)

(1.51b)

u2

max

c s


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(1.51c)

and on substituting

, ,

Equations 1.51a, 1.51band 1.51cgive

(1.52)

(1.53)

and

(1.54)

First, calculateKfrom Eq. 1.52. Then, with the known value of , determine the design table that correspon

C.1 through C.3). Traverse vertically to the andf values, then horizontally to theKvalue, and finally obtain the val

and a to be used. In so doing, calculate A from Eq. 1.53 and from Eq. 1.54. In addition, the Egyptian Code

if:

Table 1.4

Steel

240/350 0.20

360/520 0.15

450/520 0.10

Example 1.3:

A 250 mm wide, single reinforced rectangular section is to carry anMof 200 kNm. Using = 25 N/mm and steel280/450, design the section for aneffective depthof 450 mm.

Solution:

Design Equations:


and

1

cu 1

s

u 2


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\ = 480 mm

because dis less than , compression steel is required.

175.5 kNm

and

1968.75 mm

The difference in momentM is given by

kNm

which results in

= 251.6 mm ( = 0.11 which is less than 0.20; \ )

= 1968.75 + 251.6 = 2220.35 mm

Design Aids:

First, calculateK from

which results inK= 0.503. Enter Table B.2, the first value ofK(that corresponds tof =

25 N/mm ) is 0.534 which is greater than 0.503. This implies that compression steel is required.

Enter Table C.3 (where 0.10) and obtainK= 199 and a= 0.10.

2233.4 mm (Use 6 f22)

2

u2

2

2

1

1 1 cu

2

2

2


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223.34 mm (Use 2 f12)

2010 mm

A = the least of:

442 mm

1.3 = 1.3 x 1968.75 = 2559.4 mm

But not less than

mm

A = 442 mm

Here, 1968.75 mm which is greater thanA .

251.6 mm which is greater than 0.10A .

"Design Aids For Limit States Design," Third Edition, 2002, Dr.Mohamed E. El-Zoughiby.

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2

2

smi n

2

2

2

smin2

2smi n

2

s

[1] 1
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1.8 SPACINGOFREINFORCEMENTANDCONCRETECOVER

The spacing of reinforcement and the concrete cover should be sufficient to make concreting more easier; cons

the concrete surrounding the reinforcement can be efficiently vibrated, resulting in a dense concrete cover which

suitable protection of the reinforcement against corrosion.

1.8.1 Spacing of Reinforcement

Figure 1.19 shows two reinforced concrete sections. The bars are placed such that the clear spacingsshall b

equal to the maximum diameter of the bars, or 25 mm, or 1.50 times maximum size of aggregate, whichever is

according to the Egyptian Code. Vertical clear spacing between bars, in more than one layer, shall not be les

mm.

FIGURE 1.19 Spacing of steel bars (a) in one row or (b) in two rows.

1.8.2 Concrete Cover

The specified minimum concrete cover for different structural members, according to their degree of exposure,

the Egyptian Code, Table 4-13. Concrete cover for beams is equal to 25 mm for main bars and 20 mm for sti

that for slabs is equal to 15 mm, when concrete is not exposed to weather or in contact with ground.

1.8.3 Number of Stee l Layers and Overall depth of Concrete Section

The general equation for the required width of a concrete section is as follows:

(1.55)

The total depth tis equal to the effective depth dplus the distance from the centroid of the tension reinforc

the extreme tension concrete fibers, which depends on the number of layers of the steel bars. In application to th

shown in Fig. 1.19a,

(1.56)


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for one row of steel bars and

(1.57)

for two layers of steel bars, Fig. 1.19b. The overall depth tshall be increased to the nearest 5 cm. If No. 8 (2

smaller bars are used, a practical estimate of the overall depth can be made as follows:

t= d+ 50 mm, for one layer of steel bars

t= d+ 75 mm, for two layers of steel bars

Example 1.4:

For the cantilever beam shown in Fig. 1.20, ifDL= 13.5 kN/m' (including own weight) andLL= 35 kN, it isto:

a. Design the beam section for a minimum depth when b= 250 mm.

b. Design the beam section for a minimum depth when b= 120 mm.

c. Design the beam section for an effective depth d= 450 m when b= 250 mm.

d. Design the beam section for an overall deptht= 700 m when b= 120 mm.

Given: = 25 N/mm and = 280 N/mm .

FIGURE 1.20 Example 1.4.

Solution:

The ultimate moment as specified by the Egyptian Code (where and are the dead and liv

moments, respectively) is to be:

2 2


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150 kNm.

Part a:

Enter Table A.1 withf = 25 N/mm and = 280 N/mm and obtain

= 0.534 and = 196.7

Then, calculate and as follows:

414 mm

mm

For = 1842 mm , different choices of steel bars may be selected as follows:

Steel Bars Area of steel, mm

6 f 20 1884

4 f 25 1960

5 f 22 1960

9 f16 1800

2 f25 + 2 f22 1740

The area of steel bars must be closest to the required steel area. If 2 f25 plus 2 f22 are chosen, A = 17

which is 102 mm less than the required area of 1842 mm . But since the overall depth tmay be increased a f

50 mm, the actual effective depth will be a little greater than the calculated d , consequently reducing the requi

The 2 f25 plus 2 f22 would have to be placed in one row as 250 mm width is sufficient. Calcu

required width to place 2 f25 plus 2 f22 in one layer:

cu2 2

2

2

2

s

2 2

min


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= 2(f + 22f )+ 3s + 2f + 2c

= 2 (25 + 22) + 3 25 + 2 8 + 2 25 = 235 mm

which is less than b = 250 mm. The overall depth t, is then computed from:

t= d+ 0.5f +f + c

= 414 + 0.5 25 + 8 + 25 = 461.5 mm; say 500 mm

The actual effective depth d= 500 - 50 = 450 mm

which is greater than the calculated dof 414 mm. Because of the small variation, reduction in the required steel

be approximated by the ratio of the calculated dto the actual d. A actually needed is as follows

1693 mm

which is less than 1740 mm (2 f25 plus 2 f22) provided, Fig. 1.21.

FIGURE 1.21 Example 1.4, part a.

Part b:

The minimum effective depth that correspond to b= 120 mm equals 597.5 mm. The area of steelA requir

A or 1276 mm . If 4 f20 is chosen,A = 1256 mm , which is 20 mm less than 1276 mm . If the stee

placed in one row:

= 4 20 + 3 25 + 2 8 + 2 25 = 211 mm

25 22 str

25 str

s

2

2

s

sma x2

s2 2 2


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which is greater than b= 120 mm, therefore, the steel bars have to be placed in two rows as 120 mm wid

sufficient. The overall depth tis thus,

t= 597.5 + 25 + 8 + 20 + 0.5 25 = 663 cm ; say 700 mm

The actual d= 700 - 75 = 625 mm

FIGURE 1.22 Example 1.4, part b.

Part c:

First calculateK from:

which results inK = 0.581.

Enter Table B.2 withK = 0.581, then traverse horizontally to = 25, and finally obtain

K = 208. Then, 1630 mm (Use 3 f22 + 3 f19)

A = the least of :

442 mm

1

1

1

22

smi n

2


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mm

1.3 = 1.3 1630 = 2119 mm

But not less than

mm

A = 442 mm A = 1968.75 mm

Here, 1630 mm which is greater thanA and less than .

163 mm (Use 2 f12)

Part d:

If steel is assumed to be placed in two layers as 120 mm width is not sufficient.

d= 600 - 75 = 525 mm

First calculateK from:

which results inK = 0.468.

Enter Table B.2 withK = 0.468, the first value ofK (that correspond to = 25) is 0.534 which is gr

0.468. This implies that is required. Enter Table C.3 (where 0.10) and obtainK = 201.8 and a= 0.

1416 mm

319 mm

Since is less than 0.20,

2

2

2

smin2

smax2

2smi n

2

1

1

1 1

2

2

2


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.

1.9 FLANGEDSECTIONS

Concrete floor slabs and beams are normally tied together by means of stirrups and bent-up bars if any and the

form one mass of concrete. Such a monolithic system will act integrally i.e., it is allowed to assume that part o

acts with the beam and they form what is known as a flanged beam, Fig. 1.23.

FIGURE 1.23 Slab-beam floor system.

The part of the slab acting with the beam is called the flange, and it is indicated in Fig. 1.24a by the areaBt. T

the section confining the area (t-t)b is called the stem or web. As Fig. 1.24b indicates, in an I-section there

flanges, a compression flange, which is actually effective, and a tension flange, which is ineffective as it lies bneutral axis and is thus neglected completely. Therefore, the design of anI-section is similar to that of a T-sectio

FIGURE 1.24 (a) T-section and (b)I-section.

1.9.1 Effective Flange Width,B

As Fig. 1.25 indicates, the compressive stresses, in a T-section, are at a maximum value at points adjacent to

and decrease approximately in a parabolic form to zero at a distancexfrom the face of the beam. Stresses

vertically from a maximum at the top fibers of the flange to a minimum at the lower fibers of the flange.

\

s

s


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FIGURE 1.25 Effective flange width,B.

As a means of simplification, rather than varying with distance from the web, an effective widthBof uniform s

be assumed. The effective widthBis a function of span length of the beam and depends on:

1. Spacing of beams

2. Width of web of beam

3. The ratio of the slab thickness to the total beam depth

4. End conditions of the beam (simply supported or continuous)

5. The way in which the load is applied (distributed load or point load)

6. The ratio of the length of beam between points of zero moment to the width of the web and the distancewebs.

1. Spacing of beams

2. Width of web of beam

3. The ratio of the slab thickness to the total beam depth

4. End conditions of the beam (simply supported or continuous)

5. The way in which the load is applied (distributed load or point load)

6. The ratio of the length of beam between points of zero moment to the width of the web and the distancewebs.

T-andI -Shaped Sections

The Egyptian Code prescribes that the effective flange widthBof a T-section, as in Fig. 1.26, shall be taken a

width bplus the effective overhanging flange sides x andx . Thus,

B= b+ (x +x ) (1.58)

wherex +x equal the least of:

(1.59a)

orwhent = t (1.59b)

(1.59c)

1 2

1 2

1 2

s1 s2


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whereL is the distance between the points of zero moments. For a simply supported beam, the distanceL re

above is just the span distance between centers of supports. For beams continuos from one end and simply support

the other end, the distanceL may be taken as 0.80 times the span distance between centers of supports. For beam

continuos from both ends, the distanceL may be taken as 0.70 times the span distance between centers of suppor

t and t are the thicknesses of the right and left slabs and S and S are the clear distances to the next right and le

beams.

FIGURE 1.26 Effective flange width of T-beams.

IsolatedT- Shaped Sections

To increase the compression force capacity of isolated rectangular beams, concrete overhanging flange sides a

Fig. 1.27. This isolated T-shaped section is most commonly used as prefabricated units. The Egyptian Code

the size of isolated T-sections as:

and (1.60)

FIGURE 1.27 Isolated T-shaped sections.

InvertedL-Shaped Sections

2 2

2

2

s1 s2 1 2


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The end beam of a slab-beam girder floor is called a spandrel beam. The beam joins the slab from only one side

FIGURE 1.28 Effective flange width ofL-beams.

The Egyptian Code specifies that the effective flange widthB shall be taken as the web width b plus the

overhanging flange widthx . Thus,

B= b+x (1.61)

wherex equals the least of:

(1.62a)

(1.62b)

(1.62c)

The design of invertedL-shaped sections may approximately follow the same procedure of T- and

sections but with employing the respective effective widthB.

1.10 FLEXURALDESIGNOFREINFORCEDCONCRETEFLANGEDSECTIONS

In flanged sections, it can be seen that a large area of the compression flange, forming a part of the slab, is e

resisting a great part or all of the compressive force due to bending. If the section is designed on this basis, the

the web will be small; consequently the moment army is small, resulting in a large amount of tension steel wh

favorable.

Because of the large area of the compression flange, the design of a T-section does not need, in most practi

to consider a doubly reinforced section. But, in case of precast units, when the width of the flange is smal

effective depth is limited, compression steel may be added.

1.10.1 Effective Depth d

In many cases, the effective depth d can be known based on the flexural design of the section at the sup

1

1

1

ct


35/49

continuous beam, e.g. section 2-2 in Fig. 1.29a. The section at the support is subjected to a negative momen

being under tension and ignored, and the beam width is that of the web b.

FIGURE 1.29 Slab and beam systems

If the effective depth dof section 1-1 in Fig. 1.29b is not known, an approximate effective depth can be ob

considering a rectangular section with a reduced width , Fig. 1.30. The reduced width is greater than the

the web b and less than the effective flange width B. A reasonable choice of ratio varies between

depending on the applied moment and shear requirements. If shear is high or a small amount of is required,

depth is needed; i.e. approaches . For shallow sections, a higher ratio is used; i.e. the ratio may app

After determining the ratio , the next step is to estimate the effective depth using equation

(1.63)

FIGURE 1.30 Reduced width of T-section.

It is also possible to estimate the effective depth dusing


36/49

(1.64)

Table D.1 gives values forK for all grades of steel and a range of commonly used concrete strengths.

1.10.2 Design of T- andI-Sections

As already stated in Section 1.9, the design of an I-section is similar to that of a T-section. When the dep

equivalent stress block alies within the flange; i.e. at, the section behaves as a rectangular section with

width equal to the flange width. Otherwise, if ais greater than t, a T-section design is a must.

T-Section Behaves as a Rectangular Section

If at , the section may be designed as a rectangular section of widthB, Fig. 1.31.

FIGURE 1.31. Rectangular section behavior.

The design may be commenced by assuming that a t. Taking moments of forces about the tension steel, we h

(1.65)

solution of the quadratic equation yields a. If atas assumed, the tension steel can be found using

(1.66)

T-Section Behavior

When the depth of the equivalent stress block is greater than the flange thickness, i.e. a> t, the section may be

using the equations for a doubly reinforced beam, as follows. As Fig. 1.32 indicates, the tension steel A

considered to be divided into an areaA , which resists the compression in the concrete over the web, and an

orA , which resists the compression in the concrete in the overhanging of the flange.

1min

s

s

s

s

s

s

s1

sf


37/49

FIGURE 1.32. Design of a T-Section when aa .

Assuming that the tension steel is yielding, considering equation T = C , then

(1.67)

or

(4.68)

The ultimate moment of the section is the sum of the two momentsM andM :

(1.69)

where

(1.70)

and

(1.71)

solving the quadratic equation yields a.

max

2 2

u1 u2


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If aa

This implies that the section is adequate without . Considering equation, T = C , then

(1.72)

or

(1.73)

The total steel used in the T-section is

(1.74)

If a> a

This implies that is necessary, Fig. 1.33. Here also,

(1.73)

The ultimate moment of the section is the sum of the three momentsM ,M andM :

(1.75)

where

(1.70)

(1.76)

max

1 1

max

u1 u2 u3


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and

(1.77)

and

(1.7

The total steel used in the T-section is

(1.79)

If , then

(1.80)

FIGURE 1.33. Design of a T-Section, a> a .

Egyptian Code Solution

The Egyptian Code allows another approach to determineA when a> t. Ignoring the compression in the

below the flange as shown in Fig 1.34, the tension steel can be obtained from:

(1.81)

max

s s


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giving

(1.82)

FIGURE 1.34. Design of a T-section.

Example 1.5

A T-beam section withB= 1000 mm, b= 250 mm and t= 100 mm is to have a design flexural strength M

kNm. If f = 25 N/mm and steel 360/520, calculate the required steel area when:

a. d= 550 mm

b. d= 440 mm

c. d= 400 mm

FIGURE 1.35. Example 1.5.

Solution:

Assume at. Then,

giving :

s

cu2

s


41/49

Part a: d= 550 mm

solution of the quadratic equation gives a = 79 mm which is less than t. Therefore, the section will beh

rectangular section. For equilibrium, C= T, we have

A = 2818 mm ; use 6 25.

mm ; use 3 12.

Part b: d= 440 mm

solution of the quadratic equation gives a= 104 mm which is greater than t. Therefore, a T-section design is

With reference to Fig. 1.32, for equilibrium, , hence from

we have

kNm

giving

kNm

s

s2

2

s


42/49

Solving the quadratic equation yields a= 116 mm and c= 145 mm.

But c = 0.44 d= 0.44 x 440 = 193.6 mm which is greater than c. This implies that the section is

without .

For equilibrium, T= C + C , hence from

kN

we have

givingA = 3710 mm

Another Solution

For equilibrium, C = T , we can put

which results inA = 1034 mm

Also, for equilibrium, C = T , we can put

A =A = 2675 mm

A =A +A =A +A = 1034 + 2675 = 3709 mm

A =A -A = 1034 mm which is less than

A = b d= 5 x 10 x 25 x 250 x 440 = 1375 mm

and greater thanA = b d= x 250 x 440 = 336.11 mm


Upon neglecting the compression in the web part below the neutral axis, we have

max

1 2

s 2

1 1

s12

2 2

sf s22

s s1 s2 s1 sf2

s1 s sf2

smax m max-4 2

smin mmin2


43/49

mm

Part c: d= 400 mm

solution of the quadratic equation gives a= 118.5 mm which is greater than t. Therefore, a T-section design is

For equilibrium, , hence from

kN

we have

kNm

giving

kNm

Solving the quadratic equation yields a= 182 mm and c= 227.5 mm.

But c = 0.44 d= 0.44 x 400 = 176 mm which is less than c. This implies that compression steel is requ

1.33.

Here, also kNm

2

s

max


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A =A = 2675 mm

a = 0.80 c = 140 mm

kNm

A =A = m b d= 5 x 10 x 25 x 250 x 400 = 1250 mm

kNm

Since which is less than 0.15 for steel 360/520, this implies that .

= 255.56 mm

A =A +A +A =A +A + = 1250 + 2675 + 255.56 = 4180.6 mm


Upon neglecting the compression in the web part below the neutral axis, we have

mm

= 0.10A = 410.7 mm or more.

Example 1.6

In a slab-beam floor system, the smallest effective flange widthBwas found to be 1450 mm, the web width b

sf s22

max max

s1 smax max-4 2

2

s s1 s2 s3 smax sf2

2

s2


45/49

mm and the slab thickness was 120 mm, Fig. 1.36a. Design a T-section to resist an ultimate external momentM

kNm. Given:f = 20 N/mm and steel 240/350.


Solution:

Since the effective depth is not given, a reduced flange width is assumed; say

mm.

That is, an equivalent rectangular section, Fig. 1.36b, can be chosen withB = 580 mm and

which results in d= 380 mm. Assume two rows of steel bars (to be checked later).

t= 380.8 + 75 = 455.8 mm; say t= 500 mm

actual d= 500 - 75 = 425 mm

Proceed as in the previous example to calculateA .

Assume at

cu2

r

s

s


46/49

a= 46 mm which is less than t

For equilibrium, T= C, we have

mm choose 6 f25 (2950 mm )

should not be less than 0.10A , use 3 f12.


1.11 DESIGNOFTANDISECTIONSUSINGDESIGNAIDS

Once band dare known, the design of a T-section simulates that of a rectangular section when a , with

B, Fig. 1.38a. Otherwise, if a> as in Fig. 1.38b, the code allows the neglecting of compression in the web p

the flange as shown in Fig 1.38c.

First calculate the ratio andK from

(1.67)

Then, with the known value of , determine the design table that corresponds (Tables E.1 through E.5).

vertically to the value and also to value, then horizontally to theK value, and finally obtain the vabe used. Then, calculate A from

s

2 2

s

1

1

s


47/49

(1.68)

If a> take the valueK that correspond to a= .

FIGURE 1.38. Design of TandIsections

Example 1.7:

In a slab-beam floor system, the smallest effective flange widthBwas found to be 1450 mm, the web width b

mm and the slab thickness was 120 mm. Design a T-section to resist an ultimate external moment Mof 2

Given:f = 20 N/mm and steel 240/350.

Solution:

= 580mm

= 380 mm

Assume two rows of steel bars (to be checked later)

t= 380 + 75 = 455 mm; say t= 500 mm and therefore, actual d= 500 -75 = 425 mm

2

u

cu2


48/49

= and which results inK = 1.0446

Enter Table E.1 and obtainK = 197.3 and a= 0.40 t= 48 mm. Then,

2862 mm (Use 6 f25) and = 286.2 mm (Use 3 f12)

Example 1.8:

A T-beam section withB= 1000 mm, b= 250 mm and t= 100 mm is to have a design flexural strengthM

kNm. Usingf = 25 N/mm and steel 360/520, calculate the required steel area when d= 550, 440 and 400

Solution:

a. d= 550 mm

= and which results inK = 0.8198 Enter Table E.3 and obta

K = 291.2 and a= 0.80 t= 80 mm. Then,

= 2810 mm (Use 6 25) and = 281 mm (Use 3 12)

b. d= 440 mm


Enter Table E.3, since a> t, takeK = 277.8 at a= t. Then,

1

2 s

2 2

s

cu2

1

2 s

2 2

1

s 2 s


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= 3682 mm and = 368.2 mm

c. d= 400 mm


Enter Table E.3, since a> t, takeK = 273.9 at a= t. Then,

= 4107 mm and = 410.7 mm

Back Up

2 2

1

s 2 s

2 2
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