Principles of RCC

8/18/2019 Principles of RCC

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UNIT 2: PRINCIPLES OF LIMIT STATE DESIGN AND ULTIMATE

STRENGTH OF R.C. SECTION

2.1 DESIGN STRESS STRAIN CURVE FOR CONCRETE AND STEEL

2.1.1 Concrete

The idealised stress-strain curve for concrete is shown in Fig. 2.1. The compressive

strength of concrete in the actual structure is assumed to be 0.67f ck (curve 2). The factor

0.67 is introduced to account in for the size effect based on the assumption that the

concrete in the structure develops strength of 0.67 times the strength of the cube. The

partial safety factor m =1.5 shall be applied in addition to this. Thus, the design curve

(curve 3) is obtained by using a partial safety factor of m =1.5. Thus the maximum

compressive stress in concrete for design purpose is equal to 0.67f ck /m = 0.67f ck /1.5 =

0.446f ck ≈ 0.45f ck . It is to be noted that each curve is parabolic in the initial portion upto astrain of 0.002. At a strain of 0.002 (0.2% strain), the stress remains constant with the

increasing load, until a strain of 0.0035 (0.35%) is reached, when the concrete is said to

have failed.

Fig. 2.1: Characteristic and design stress-strain curves for concrete in flexural

compression

For the purpose of limit states design, the appropriate partial safety factor γ c has to be

applied, and γc is equal to 1.5 for the consideration of ultimate limit states. Thus, the

„design curve‟ is obtained by simply scaling down the ordinates of the characteristic

curve, i.e., dividing by γc (Fig. 2.1). Accordingly, the maximum design stress becomes

equal to 0.447f ck , and the formula for the design compressive stress f c corresponding to

any strain ε ≤ 0.0035 is given by:


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= 0.447 2 0.002 − 0.0022 < 0.0020.447 0.002 ≤ ≥ 0.0035

When concrete is subjected to uniform compression, as in the case of a concentricallyloaded short column, the ultimate strain is limited to 0.002, and the corresponding

maximum design stress is 0.447f ck . The stress−strain curve has no relevance in the limit

state of collapse by (pure) compression of concrete, and hence is not given by the Code.

2.1.2 Steel

The stress-strain curve for steel according to IS 456: clause 37.1 is assumed to depend on

the type of steel. Mild steel bar (f y= 250) is assumed to have a stress-strain curve as

shown in Fig. 2.2 and cold worked deformed bar (Fe 415) a stress-strain curve as shown

in Fig. 2.3 (Fig. 22 of IS 456).

Fig. 2.2: Characteristic and design stress-strain curves for Fe 250 grade mild steel

Fig. 2.3: Characteristic and design stress-strain curves for Fe 415 grade cold-worked

steel


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The stress-strain curves for steel, both in tension and compression in the structure, are

assumed to be the same as obtained in the tension test. As the yield strength of IS grade

steel has a minimum guaranteed yield strength, the partial safety factor to be used for

steel strength need not be as large as that for concrete. The partial safety factor

recommended for steel is 1.15, and this is to be applied to the stress-strain curve as shownin Figs. 2.2 and 2.3 (IS: 456-2000, Fig. 22). It should be noted that for cold worked

deformed bars the factor 1.15 is applied to points on the stress-strain curve from 0.87f y to

f y only. The strain corresponding to 0.87f y stress is,

= 0.87 + 0.002 2.2 LIMIT STATES

A limit state is a state of impending failure, beyond which a structure ceases to perform

its intended function satisfactorily, in terms of either safety or serviceability; i.e., it either

collapses or becomes unserviceable.

There are two types of limit states:

1. Ultimate limit states (or „limit states of collapse‟), which deal with strength,

overturning, sliding, buckling, fatigue fracture, etc.

2. Serviceability limit states, which deal with discomfort to occupancy and/or

malfunction, caused by excessive deflection, crack-width, vibration, leakage, etc., and

also loss of durability, etc.

2.2.1 Limit state of collapse (Safety requirements)The limit state of collapse of the structure or part of the structure could be assessed from

rupture of one or more critical sections and from buckling due to elastic or plastic

instability or overturning. The resistance to bending, shear, torsion and axial loads at

every section shall not be less than the appropriate value at that section produced by the

probable most unfavourable combinations of loads on the structure using the appropriate

partial factors.

The following limit states of collapse are considered in design:

a. Limit state of collapse in compression.

b.

Limit state of collapse in shear.c.

Limit state of collapse in torsion.

2.2.2 Limit state of serviceability

The limit state of serviceability consists of

a.

Excessive deflection.

b.

Premature or excessive cracking.

c.

Corrosion.

d. Excessive vibrations.


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Out of these, the important limit states of serviceability are excessive deflection and

cracking. The deflection of a structure or part thereof shall not adversely affect the

appearance or efficiency of the structure or finishes or partitions. Cracking of concrete

should not adversely affect the appearance or durability of the structure. Generally crack

widths at surface shall not exceed 0.3 mm. The limit state of excessive deflection andcrack width is applicable at service loads and is estimated on the basis of elastic analysis

(working stress method). The limit state of collapse (or failure), however, depends upon

ultimate strength.

2.3 ASSUMPTIONS MADE IN LIMIT STATE OF COLLAPSE IN FLEXURE

The following assumptions are relevant in the computations of ultimate flexural strength

of reinforced concrete sections as specified in IS: 456-2000 clause 38.1.

1. Plane sections normal to the axis remain plane after bending. It means the strain

diagram across the depth of the cross section is linear as shown in Fig. 2.4.

Fig. 2.4: Strain and Stress blocks

2. The maximum strain in concrete at the extreme compression fibre is assumed as

0.0035 in flexure.

3. The stress strain curve for concrete is having parabolic shape up to 0.002 strain and

then constant up to limit state of 0.0035. However, IS code do not prevent using othershapes like rectangle, trapezoidal which result in prediction of strength in substantial

agreement with the result of the tests.

For design purpose, the compressive strength in the structure (size effect) may be

assumed to be 0.67 times the characteristic strength. In addition to this the partial

safety factor m may be taken as 1.5. It means the maximum compressive strength in

the extreme fibre of the section will be (0.67/1.5)f ck = 0.446f ck or it may be taken as

0.45f ck also. The stress diagram assumed in the beam is as shown in Fig. 2.5.

4. The tensile strength of concrete is ignored.

5.

The stresses in the reinforcement are obtained from the stress-strain curves shown inFigs. 2.3 and 2.4. For design purposes the partial safety factor m, equal to 1.15 is


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applied to compute the design strength. Hence, the maximum stress in steel is limited

to f y/1.15 = 0.87f y.

6. The maximum strain in the tension reinforcement in the section at the failure shall not

be less than

Where, f y is characteristic strength of steel.

Es is modulus of elasticity of steel.

Actually, this value corresponds to the beginning of the flat portion in design stress strain

diagram for steel as shown in Fig. 2.5.

Fig. 2.5: Design stress train curve for steel

2.4 STRESS BLOCK PARAMETERS

The diagram showing the distribution of compressive stress in concrete across the depth

xu of the section is termed as “stress block”. Since the strain diagram is linear over this

depth xu, the shape of stress block is the same as the idealized stress-strain curve of

concrete. It has zero stress at neutral axis. It varies parabolically upto a height of 47 xu i.e., 0.0020.0035

xu and has constant value equal to design stress of 0.446f ck i.e., 0.671.5 f ck forthe 3

7xu i.e., 0.0035−0.0020.0035 xu. The shape of stress block is shown in Fig. 2.6 below.

Area of stress block: It may be found as explained below,

Area A of stress block = Area of rectangular portion + area of parabolic portion

= 0.446

∗3

7xu +

2

3

∗0.446f ck

∗4

7xu

= 0.361xuf ck ≅ 0.36xuf ck


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Fig. 2.6: Stress block parameters

Now we look at 0.36f ck as the average stress over the depth xu as shown in figure above.

Area of stress block = f av*xu

Then the compressive force on the section, C = b*xu*f av = 0.36f ck *xu*b=k 1f ck xu b

Where k 1 is called as stress block parameter and is equal to 0.36.

The distance of compressive force from the extreme compression fibre can be calculated

as 0.42xu = k 2xu

Where k 2 is another important parameter of the stress block and is equal to 0.42.

2.5 BALANCED, UNDER REINFORCED AND OVER REINFORCED SECTIONS

In bending, strain varies linearly across the depth of cross section of the member. Once

edge of the beam is in maximum compression and the other edge is in maximum tension.

Hence, somewhere across the depth, there is an axis where strain is zero. This axis is

called neutral axis. Depth of the axis from the maximum compression fibre is called depth

of neutral axis and is denoted by xu (Fig. 2.7). In limiting case maximum compressive

strain in concrete 0.0035 corresponding strain in steel is,

= − 0.0035 From the stress strain curve we find that when this value exceeds

0.87

+ 0.002 the stress

in steel is yield stress f y.

A section is called balanced section if for the same applied moment the strain in concrete

and the strain in steel reach their limiting values simultaneously. In other words, in

balanced sections maximum compressive strain c in concrete is reached 0.0035 when the

tensile strain in steel is reached its limiting value of

=

0.87

+ 0.002


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Fig. 2.7 Different beam sections

Sections in which tensile strain reaches yield strain of0.87 + 0.002 earlier to

compressive strain in concrete reaching the limiting value of 0.0035, are called under

reinforced section. In these cases as moment increases, first steel reaches yield strain.The stress in steel remains same (f y) but strain goes on increasing. When the moment

corresponding to 0.0035 strain in concrete is reached, concrete is crushed and failure

takes place. The excess strain in steel beyond0.87 + 0.002 amounts to considerable

cracks in concrete. The deflection will increase. They serve as a warning to the user and

one can take precautions to avoid disaster. Hence, IS code specifies that the maximum

strain in tension reinforced shall not be less than0.87 + 0.002. In other words, IS code

prefers design of under reinforced sections and at the most it can be balanced section.

This type of failure in under reinforced section is called primary tensile failure.

RC sections in which the limiting strain in concrete is reached earlier than the yield strain

of steel are called over reinforced sections. At failure steel is not yet yielded and

concrete bursts out. As there are no warning of failure in such sections, IS code

recommends avoiding such designs. Hence, a designer should not provide extra steel to

get the feeling of making design safer. No doubt, providing extra steel increases the load

carrying capacity of the section, but in case of over loading, it results into sudden

collapse.

If xulim is the value of depth of neutral axis in balanced section, it may be noted thatxuxulim in over reinforced sections. These

sections are shown in Fig. 2.7.

2.6 DEPTH OF NEUTRAL AXIS

Beams are assumed to fail in bending when the strain in concrete reaches limiting

compression strain of cu = 0.0035. But in all cases of design tensile strain in steel need

not be equal to limiting strain

=

0.87

+ 0.002. It can be less or more than it.


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Fig. 2.8: Neutral axis

However, designs with s < su (over reinforced sections) are to be avoided. Hence, for all

cases

Total compression, C = 0.36*f ck *b*xu

Total tension, T = f s*Ast

Where f s is the stress in steel corresponding to a strain of 0.0035 in concrete [Note:

Maximum design value, f s = 0.87f y]

Equilibrium requirement in horizontal direction gives, C = T

i.e., 0.36f ck bxu = f sAst = 0.36 Where, Ast = area of tension steel

xu = depth of neutral axis from top compression fibre

b, d = width and effective depth of the beam.

Case 1: Balanced section:

From strain diagram of Fig. 2.8 (d), − = 0.0035 Rearranging, − = 0.0035

=

0.0035

+ 0.0035


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To avoid compression failure, IS: 456-2000 recommends minimum strain in steel

corresponding to 0.0035 strain in concrete as,

= 0.87

+ 0.002

Therefore, limiting value of xu is given by, = 0.00350.87 + 0.002 + 0.0035 = 0.00350.87 + 0.0055

Since, E = 2*10

5

N/mm

2

for all types of steels, values for various types of steel areas shown in Table 2.1.Table 2.1: Limiting values of depth of neutral axis

Type of steel f in N/mm2 xulim/d

Mild steel (Fe250) 250 0.53

Fe 415 415 0.48

Fe500 500 0.46

Case 2: Under reinforced section:

For under reinforced sections, strain in steel is greater than its limiting value (s > slim).

Hence, from idealized stress strain curve, we get f s = 0.87f y. Substituting this we get,

= 0.87 0.36 Case 3: Over reinforced section:

For over reinforced section, s < slim. Hence, the actual strain in steel at failure es is to be

found from the equation, − = 0.0035 Then using stress strain diagram of steel, corresponding stress is to be found. Since, s

and xu are independent; it is not possible to get one value from the other. Trial and error

method is to be used.


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2.7 STRENGTH OF RECTANGULAR SECTIONS IN FLEXURE

The flexural strength of RC section is also known as moment carrying capacity of the

section. The compressive force C in concrete and tensile force T in steel are equal and

opposite and are separated by distance (d-0.42xu) (Fig. 2.8), which is called as lever armLA. Hence, they form a couple. The couple moment is the moment of resistance and it is

called moment carrying capacity when c = 0.0035. Thus moment carrying capacity is

given by, = × = 0.36 ( − 0.42)

= 0.36

1

−0.42

2

Mu is called as the strength of section in flexure.

For limiting case of balanced section, = 0.36 1 − 0.42 2 = 2

Where,

= 0.36 1 − 0.42 Substituting the values of (xulim/d) for different grades of steel, we get the values of „k‟ for

finding Mulim for different grade of steel as shown in Table 2.2 below.

Table 2.2: Limiting moment carrying capacity for different grade of steels

Type of steel xulim/d Mulim

Fe250 0.53 0.148f ck bd2

Fe415 0.48 0.138f ck bd

Fe500 0.46 0.133f ck bd

Approximate expression for moment of resistance:

In case of under reinforced and balanced sections, x u < xulim, the stress in steel reaches the

limiting value of 0.87f y earlier. Hence, the equilibrium equation for horizontal forces

gives,

=

0.36 = 0.87


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= 0.87 0.36 Moment equilibrium equation gives,

=

×

= 0.87

−0.42

= 0.87 − 0.42 0.87 0.36

= 0.87 1 − 1.015 Approximating 1.015 ≈ 1.0, we get,

= 0.87

1

−

In case of over reinforced sections (xu> xulim), the actual moment of resistance of the

section may be obtained by usual formula C x LA or T x LA. However, to avoid

compression failure, the strength of such sections to be considered as that of balanced

sections only, i.e., = = 0.36 1 − 0.42 2 2.8 FLANGED SECTIONS IN FLEXURE

In case of flanged sections (Tee and L-sections), the ultimate flexural strength is

influenced by the position of neutral axis which may lie in the flange or outside the flange

depending upon the area of the reinforcement on the tension face. The IS code specifies

equations for computing the moment of resistance of flanged sections by assuming the

stress block and the following parameters.

Let, bw = width of the rib or web. xu = neutral axis depth

bf = width of the flange. Ast = area of tension reinforcement

D = effective depth.

Df = depth of flange.

Fig. 2.9: Stress diagram for Tee beam.


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The effective width of flange may be taken as the following but in no case greater than

the breadth of web plus half the sum of the clear distance to the adjacent beams on either

side.

(a) For T-beams

= 6 + + 6 (b) For L-beams =

12 + + 3

For continuous beams and frames lo may be assumed as 0.7 times the effective span.

The following cases are considered.

(a) Neutral axis lies within the flangeIn this case since, xu < Df , the section can be considered as rectangular with the width of

compression flange as the width of the section.

i.e., b = bf

The moment of resistance of under reinforced section is computed by the relation:

= 0.87 1 − In the case of over reinforced sections, xu > xulim, the moment of resistance is computed

by, = 0.36 1 − 0.42 2 (b) Neutral axis lies outside the flange and (Df /d) 0.2

When the neutral axis falls outside the flange (xu > Df ) and the ration (Df /d) ≯ 0.2 and(Df /xu) < 0.43, the flexural strength can be computed by using the stress block parameters

shown in figure above. The stress blocks are considered separately for the web portion

and the flanges. Considering the tensile and compressive forces shown in figure above,

the moment of resistance of the flanged section is expresses as, = 1 − 0.42 + 2( − 0.5 ) Where, 1 = 0.36 2 = 0.45 ( − ) The final expression being,

= 0.36

2

1

−0.42

+ 0.45

−

−0.5

The above equation is valid for the case (Df /d) < 0.2 and (Df /xu) < 0.43


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(c) Neutral axis lies outside the flange and (Df /d) > 0.2

When the ratio (Df /d) > 0.2, the moment equation is modified by substituting yf for Df in

above equation where,

= 0.15 + 0.65 The modified equation for moment of resistance is expressed as, = 0.36 2 1 − 0.42 + 0.45 − − 0.5

The Indian standard further states that for xumax > xu > Df , the moment of resistance may

be calculated by using (b) condition when (D f /xu) does not exceed 0.43 and when (Df /xu)

exceeds 0.43, the moment of resistance is computed by the condition (c) by substituting

xumax by xu.

(d) Computation of tension steel in tee beam

(i) When xu < Df , the tee beam is considered as rectangular and the area of reinforcement

is computed by the following equation for a known value of Mu.

= 0.87 1 −

(ii) When xu > Df , (Df /d) ≯ 0.2 and (Df /xu) ≯ 0.43, the area of tension reinforcement iscomputed by the following equation,

Force equilibrium, T1 = C1

0.87 f y Astw = 0.36 f ck bw xu

∴ = 0.36 0.87

Also, T2 = C2

0.87 f y Astf = 0.45 f ck (bf -bw) Df

∴ = 0.45 ( − ) 0.87

Therefore,

= 0.36 0.87

+ 0.45 − 0.87


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(iii) When xu > Df , (Df /d) ≯ 0.2 and (Df /xu) > 0.43, the area of tension reinforcement iscomputed by the following equation,

Replace Df by yf ,

= 0.36

0.87 + 0.45

− 0.87 (iv) When xu > Df , (Df /d) 0.43, this is similar to case iii.

2.9 DOUBLY REINFORCED SECTIONS IN FLEXURE

Doubly reinforced sections are required in beams of restricted depth due to head room

requirements. When the singly reinforced section is insufficient to resist the bending

moment on the section, additional tension and compression reinforcements are designed

based on steel beam theory.

There are several reasons to add compression steel. Keep in mind, supported steel

(meaning it can't buckle) resists compression as well.

Compression steel helps reduce long term deflections. Concrete creeps under

sustained loads. Steel lessens the compression, meaning less sustained compressive

stress to cause creep deflection.

It makes members more ductile. Since the steel takes some of the compressivestress, the compression block depth is reduced, increasing the strain in the tension

steel at failure, resulting in more ductile behaviour (the moment at first yield

remains largely the same with compression steel added, but the increase in capacity

after yield is significant).

Compression steel insures that the tension steel yields before the concrete crushes,

meaning it helps change the failure mode to tension controlled.

It makes beams easier to construct. With bars in the top and bottom, you have

longitudinal reinforcement in all 4 corners of the shear stirrups to keep them in

place when pouring the concrete. Also, for continuous members, its often easier to

run your negative moment steel the full length of the beam rather than trying to cut

it off in the positive moment regions.

Serviceability concerns. You're going to end up putting steel in that region anyway

to for temperature and shrinkage.


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Fig. 2.10: Doubly reinforced section

The design equation is obtained as follows,

Let, Mu = ultimate flexural strength of the doubly reinforced section

Mu1 = Mulim = Limiting or maximum moment of resistance of the singly reinforced

section.

Mu2 = moment of resistance of the steel beam neglecting the effect of concrete

= f sc Asc (d-d`)

f sc = the stress in compression steel corresponding to the strain reached by it

when the extreme concrete fibre reaches a strain of 0.0035.

Asc = area of compressive reinforcement

d = effective depth of tension steel

d` = depth of compression reinforcement from compressive face

Ast1 = area of tensile reinforcement for a singly reinforced section

Ast2 = area of tensile reinforcement required to balance the compression

reinforcement

Ast = Ast1 + Ast2

The following procedure is used to compute the ultimate flexural strength of doubly

reinforced section.

1. Computing the limiting moment Mulim of the singly reinforced section as,

= 0.36 1 − 0.42 2 2. Calculate

y

limuck 1st

f 87.0

bxf 36.0A

3. Compute 1stst2st AAA


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4. Calculate

sc

2sty

scf

Af 87.0A

Where, sscsc Ef

5. The ultimate moment capacity of the section is given by,

= + − ` The strain,

maxu

maxusc

x

`dx0035.0

For Fe 250 grade steel, Es = 2*105 N/mm2

For Fe 415 and Fe 500 grade steel refer the Table 2.3 (Table A of SP-16) to get the stress

value for the calculated strain value.

Table 2.3: Salient points on the design stress strain curve

2.10 ULTIMATE SHEAR STRENGTH OF REINFORCED CONCRETE

SECTIONS

Reinforced concrete members are generally subjected to maximum shear forces normally

near the support sections of simply supported flexural members. The shear stress

developed is accompanied by diagonal tension as shown in Fig. 2.11. In continuous

beams the support sections are subjected to shear coupled with moments. In the case of

corbels and brackets, large shear forces develop at the junction of the corbel and column.

Fig. 2.11: Diagonal tension in beams


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The prominent types of shear failures observed in reinforced concrete members can be

categorized under the following types:

(a) Shear-tension or diagonal tension failure.

(b) Flexure-shear failure.

(c) Shear-compression failure.(d) Shear-bond failure.

(a) Diagonal tension failure (b) Flexure-shear failure

(c) Shear-compression failure (d) Shear-bond failure

The shear stress distribution in a reinforced concrete beam is influenced by the shearforce acting on the section and the shape of cross section in the elastic stage. At the

ultimate stage, concrete below the neutral axis is ineffective due to cracking. Hence, for

simplicity the nominal shear stress across the section is computed as average shear stress

and is expressed as,

= Where, Vu = ultimate shear force at the section.

v = nominal shear stress

b = breadth (width of rib in flanged sections)

d = effective depth

The design shear strength in beams without shear reinforcement depends upon the grade

of concrete and the percentage of tension reinforcement in the section. The permissible

design shear strength (c) of concrete in beams without shear reinforcement is complied in

Table 2.4 (Table 19 of IS: 456-2000). The values given in the table are applicable for

beams. In the case of slabs having an overall thickness less than 300mm, the shear

strength being higher, the IS code suggests an enhanced shear strength computed as k c


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where „k‟ is a multiplying factor depending upon the overall depth of slab as shown in

clause 40.2.1.1 (Table 2.5).

The code also specifies an upper limit for the design of shear strength of concrete

strengthened by shear reinforcements. Accordingly the maximum shear stress in concrete(cmax) should not exceed the values specified in Table 2.6 (Table 20 of IS: 456-2000). If

the values of nominal shear stress (v) exceeds the value of cmax, the section should be

redesigned with increased cross sectional dimensions.

Table 2.4: Design shear strength of concrete ( c) N/mm

2

Table 2.5: shear strength factor ‘k’ for slabs

Table 2.6: Maximum shear stress ( cmax) N/mm

2


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Design of shear reinforcements

In case of reinforced concrete sections where the nominal shear stress (v) exceeds the

design shear strength of concrete (c), shear reinforcements are to be designed comprising

of,

1. Vertical stirrups.2. Tension reinforcement bent up near supports to resist the shear forces.

The typical arrangement of these types is shown in Fig. 2.12. At the limit state of collapse

in shear, the forces are resisted by the combined action of concrete and steel.

Let, Vu = total design shear force

Vc = shear resisted by concrete

Vus = shear resisted by reinforcements in the form of links or bent up bars.

Then, Vus = Vu – Vc

= (v – c) b d

If, Sv = spacing of stirrups

Asv = total area of the legs of shear reinforcements

d = effective depth of section

Then,

us

svy

vV

dAf 87.0S

The shear resisted by the bent up bars inclined at an angle „‟ to the horizontal is given

by,

sinAf 87.0V svyus

These equations are recommended for the design of shear reinforcements in IS: 456-2000,

clause 40.4.

(a) Vertical stirrups

(b) Bent up bars

Fig. 2.12: Types of shear reinforcements


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In designing reinforced concrete beams, the IS code stipulates that minimum shear

reinforcements are to be designed even if the design shear strength of concrete (c)

exceeds the nominal shear stress (v) to safeguard against local cracking and nominal

safety requirements.

The minimum shear reinforcements to be designed using the relation,

≥ 0.40.87 Provision of nominal shear reinforcement is equivalent to designing the shear

reinforcement for a shear stress of (v – c) = 0.4 N/mm2 and it safeguards against spalling

of concrete cover and bond failures.

2.11 ULTIMATE TORSIONAL STRENGTH OF REINFORCED CONCRETE

SECTIONS

Design of reinforced concrete structures subjected to torsion requires a proper

understanding of the torsional strength of reinforced concrete sections. Pure torsion is

exceptional in reinforced concrete. Normally torsion associated with flexure and shear

develops in reinforced concrete structures such as L-beams, circular girders, corner lintels

where the loading is eccentric to the line of reaction at supports. Primary torsion is

generally induced by eccentric loading and equilibrium conditions are sufficient to

evaluate the torsional moments acting at critical sections.

The effect of torsion is to induce shear stresses and causes warping of non-circular

sections. The failure of plain concrete members in torsion is due to diagonal tensile cracks

since concrete is weak in tension. Hence, the IS code provides a method of designing

suitable reinforcements in concrete sections subjected to combined effects of torsion,

flexure and shear by introducing the concept of enhanced equivalent bending moment,

shear and torsion.

The design rules specified in IS code applies to beams of solid rectangular section andflanged sections in which the width of rib is considered for computations. Sections

subjected to torsion and shear are to be designed for an equivalent shear force computed

as,

Ve = Vu + 1.6 (Tu/b)

Where, Ve = equivalent shear

Vu = transverse shear

Tu = torsional moment

b = breadth of beam


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The equivalent nominal shear stress is expresses as,

= + 1.6

The values of ve should lie between c, the permissible shear stress compiled in Table 19

of IS code and the maximum shear stress cmax given in Table 20 of IS code.

In cases where ve > cmax, the section has to be suitably redesigned by increasing the

cross-sectional area and/or increasing the grade of concrete.

In ve < c, minimum shear reinforcements are designed. Longitudinal reinforcements are

designed to resist an equivalent bending moment expressed as,

Me = Mu + Mt

Where, Me = equivalent bending moment

Mu = design bending moment

Mt = bending moment developed due to torsion and expressed as,

= 1 +

1.7 Where, Tu = torsional moment

D = overall depth

b = breadth of section

In cases where the numerical value of Mt exceeds the numerical value of Mu, longitudinal

reinforcement should be provided on flexural compression face such that the beam can

also withstand an equivalent moment Me computed as Me2 = (Mt – Mu), the moment Me2 being taken as acting in the opposite sense to the moment Mu.

Transverse reinforcements comprising of two legged closed hoops enclosing the corner

longitudinal bars should have the area given by,

= 11(0.87 ) + 2.51(0.87 )


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However, the total transverse reinforcement should not be less than the values computed

as,

− 0.87

Where, Tu = torsional moment

Vu = transverse shear force

sv = spacing of vertical links

b1 = centre to centre distance between corner bars in the direction of width

d1 = centre to centre distance between corner bars in the direction of depth

f y = characteristic strength of stirrup reinforcement

ve = equivalent shear stress

c = shear strength of concrete

2.12 CONCEPT OF DEVELOPMENT LENGTH AND ANCHORAGE

The bond between steel and concrete is very important and essential so that they can act

together without any slip in a loaded structure. With the perfect bond between them, the

plane section of a beam remains plane even after bending. The length of a member

required to develop the full bond is called the anchorage length. The bond is measured by

bond stress. The local bond stress varies along a member with the variation of bending

moment. The average value throughout its anchorage length is designated as the average

bond stress. In our calculation, the average bond stress will be used.

Thus, a tensile member has to be anchored properly by providing additional length on

either side of the point of maximum tension, which is known as „Development length in

tension‟. Similarly, for compression members also, we have „Development length Ld in

compression‟.

It is worth mentioning that the deformed bars are known to be superior to the smooth mild

steel bars due to the presence of ribs. In such a case, it is needed to check for the

sufficient development length Ld only rather than checking both for the local bond stressand development length as required for the smooth mild steel bars. Accordingly, IS 456,

clause 26.2 stipulates the requirements of proper anchorage of reinforcement in terms of

development length Ld only employing design bond stress τ bd.

Design Bond Stress (τbd)

The design bond stress τ bd is defined as the shear force per unit nominal surface area of

reinforcing bar. The stress is acting on the interface between bars and surrounding

concrete and along the direction parallel to the bars. This concept of design bond stress

finally results in additional length of a bar of specified diameter to be provided beyond a


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given critical section. Though, the overall bond failure may be avoided by this provision

of additional development length Ld, slippage of a bar may not always result in overall

failure of a beam. It is, thus, desirable to provide end anchorages also to maintain the

integrity of the structure and thereby, to enable it carrying the loads. Clause 26.2 of IS

456 stipulates, “The calculated tension or compression in any bar at any section shall bedeveloped on each side of the section by an appropriate development length or end

anchorage or by a combination thereof.”

The local bond stress varies along the length of the reinforcement while the average bond

stress gives the average value throughout its development length. This average bond

stress is still used in the working stress method and IS 456 has mentioned about it in cl.

B-2.1.2. However, in the limit state method of design, the average bond stress has been

designated as design bond stress τ bd and the values are given in cl. 26.2.1.1. The same is

given below as Table 2.7.

Table 2.7: Design bond stress in limit state method for plain bars in tension

For deformed bars conforming to IS 1786, these values shall be increased by 60 percent.

For bars in compression, the values of bond stress in tension shall be increased by 25

percent.

2.12.1 Development Length

Fig. 2.13: Development length of bar


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(a) A single bar

Fig. 2.13(a) shows a simply supported beam subjected to uniformly distributed load.

Because of the maximum moment, the Ast required is the maximum at x = L/2. For any

section 1-1 at a distance x < L/2, some of the tensile bars can be curtailed. Let us thenassume that section 1-1 is the theoretical cut-off point of one bar. However, it is necessary

to extend the bar for a length Ld as explained earlier. Let us derive the expression to

determine Ld of this bar.

Fig. 2.13(b) shows the free body diagram of the segment AB of the bar. At B, the tensile

force T trying to pull out the bar is of the value T = (π υ2σs /4), where υ is the nominal

diameter of the bar and σs is the tensile stress in bar at the section considered at design

loads. It is necessary to have the resistance force to be developed by τ bd for the length Ld

to overcome the tensile force. The resistance force = π υ (L d) (τ bd). Equating the two, we

get

∅ = ∅24

= ∅4

The above equation is given in cl. 26.2.1 of IS 456 to determine the development length

of bars.

The example taken above considers round bar in tension. Similarly, other sections of the

bar should have the required Ld as determined for such sections. For bars in compression,

the development length is reduced by 25 percent as the design bond stress in compression

τ bd is 25% more than that in tension (see the last lines below Table 2.7). Following the

same logic, the development length of deformed bars is reduced by 60 percent of that

needed for the plain round bars. Tables 64 to 66 of SP-16 present the development lengths

of fully stressed plain and deformed bars (when f s = 0.87f y) both under tension and

compression. It is to be noted that the consequence of stress concentration at the lugs of

deformed bars has not been taken into consideration.

(b) Bars bundled in contact

The respective development lengths of each of the bars for two, three or four bars in

contact are determined following the same principle. However, cl. 26.2.1.2 of IS 456

stipulates a simpler approach to determine the development length directly under such

cases and the same is given below:

“The development length of each bar of bundled bars shall be that for the individual bar,

increased by 10 percent for two bars in contact, 20 percent for three bars in contact and 33 per cent for four bars in contact.”


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However, while using bundled bars the provision of cl. 26.1.1 of IS 456 must be satisfied.

According to this clause:

• In addition to single bar, bars may be arranged in pairs in contact or in groups of

three or four bars bundled in contact.

•

Bundled bars shall be enclosed within stirrups or ties to ensure the bars remainingtogether.

• Bars larger than 32 mm diameter shall not be bundled, except in columns.

Curtailment of bundled bars should be done by terminating at different points spaced

apart by not less than 40 times the bar diameter except for bundles stopping at support (cl.

26.2.3.5 of IS 456).

Checking of Development Lengths of Bars in Tension

The following are the stipulation of cl. 26.2.3.3 of IS 456.

(i) At least one-third of the positive moment reinforcement in simple members and

one-fourth of the positive moment reinforcement in continuous members shall be

extended along the same face of the member into the support, to a length equal to

Ld/3.

(ii)

Such reinforcements of (i) above shall also be anchored to develop its design stress

in tension at the face of the support, when such member is part of the primary lateral

load resisting system.

(iii)The diameter of the positive moment reinforcement shall be limited to a diameter

such that the Ld computed for f s = f d does not exceed the following:

= ≤ 1 + where M1 = moment of resistance of the section assuming all reinforcement at the

section to be stressed to f d,

f d = 0.87f y,

V = shear force at the section due to design loads,

Lo = sum of the anchorage beyond the centre of the support and the equivalent

anchorage value of any hook or mechanical anchorage at simple support.

At a point of inflection, Lo is limited to the effective depth of the member

or 12υ, whichever is greater, and

υ = diameter of bar.

It has been further stipulated that M1/V in the above expression may be increased by 30

percent when the ends of the reinforcement are confined by a compressive reaction.


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2.12.2. Anchoring Reinforcing Bars

The bars may be anchored in combination of providing development length to maintain

the integrity of the structure. Such anchoring is discussed below under three sub-sections

for bars in tension, compression and shear respectively, as stipulated in cl. 26.2.2 of IS456.

Bars in tension (Clause 26.2.2.1 of IS 456)

The salient points are:

Derformed bars may not need end anchorages if the development length

requirement is satisfied.

Hooks should normally be provided for plain bars in tension.

Standard hooks and bends should be as per IS 2502 or as given in Table 67 of SP-

16, which are shown in Figs. 2.14 (a) and (b).

The anchorage value of standard bend shall be considered as 4 times the diameter

of the bar for each 45o bend subject to a maximum value of 16 times the diameter

of the bar.

The anchorage value of standard U-type hook shall be 16 times the diameter of the

bar.

Fig. 2.14: Standard hook and bend

Bars in compression (cl. 26.2.2.2 of IS 456)

Here, the salient points are:

•

The anchorage length of straight compression bars shall be equal to its development

length as mentioned in pervious section.

• The development length shall include the projected length of hooks, bends and

straight lengths beyond bends, if provided.


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Bars in shear (cl. 26.2.2.4 of IS 456)

The salient points are:

• Inclined bars in tension zone will have the development length equal to that of bars

in tension and this length shall be measured from the end of sloping or inclined

portion of the bar.

• Inclined bars in compression zone will have the development length equal to that of

bars in tension and this length shall be measured from the mid-depth of the beam.

•

For stirrups, transverse ties and other secondary reinforcement, complete

development length and anchorage are considered to be satisfied if prepared as

shown in Figs. 2.15.

Fig. 2.15: Anchorage of stirrups

Bearing Stresses at Bends (cl. 26.2.2.5 of IS 456)

The bearing stress inside a bend is to be calculated from the expression:

= ∅ where F bt = tensile force due to design loads in a bar or group of bars,

r = internal radius of the bend, and

= size of the bar or bar of equivalent area in bundled bars


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The calculated bearing stress shall not exceed the following:

≯ 1.5 1 + 2

∅

where f ck = characteristic cube strength of concrete

a = center to center distance between bars or groups of bars perpendicular to the

plane of the bend. For bars adjacent to the face of the member, „a‟ shall be

taken as cover plus size of the bar υ.

Splicing of reinforcement: Splices are required when bars placed short of their required

length (due to non-availability of longer bars) need to be extended. Splices are also

required when the bar diameter has to be changed along the length (as is sometimes done

in columns). The purpose of „splicing‟ is to transfer effectively the axial force from the

terminating bar to the connecting (continuing) bar with the same line of action at the

junction. This invariably introduces stress concentrations in the surrounding concrete.

These effects should be minimised by:

• using proper splicing techniques;

• keeping the splice locations away from sections with high flexural/shear stresses; and

• staggering the locations of splicing in the individual bars of a group (as, typically in a

column).

The Code recommends that “splices in flexural members should not be at sections where

the bending moment is more than 50 percent of the moment of resistance; and not more

than half the bars shall be spliced at a section” (Cl. 26.2.5).

Principles of RCC

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