507 33 Powerpoint-slides Ch6 DRCS

5/24/2018 507 33 Powerpoint-slides Ch6 DRCS

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Oxford University Press 2013. All rights reserved.

Design of Reinforced

Concrete Structures

N. Subramanian


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Chapter 6

Design For Shear


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According to traditional design philosophy, bending moment and

shear force are treated separately, even though they coexist.

It is important to realize that shear analysis and design are not really

concerned with shear as such. The shear stresses in most beams may be

below the direct shear strength of concrete.

Shear failure is often termed as diagonal tension failure.

Introduction


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In the sectional design model, the flexural longitudinal reinforcement

is designed for the effects of flexure and any additional axial force, andthe transverse reinforcement is designed for shear and torsion.

In the case of slabs, this type of shear is called one-way shear, which is

different from the two-way or punching shear, which normally occurs in

flat slabs near the slab-column junctions.

The main objective of an RC designer is to produce ductile behaviour

in the members such that ample warning is provided before failure. For

this, RC beams are often provided with shear reinforcement.

Introduction


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Behaviour of Reinforced Concrete Beams

under Shear

The behaviour of RC beams under shear may be categorized into the

following three types:

1. Behaviour when the beam is not cracked

2. Cracked beam behaviour when no shear reinforcements are

provided

3. Cracked beam behaviour when shear reinforcements are provided


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Behaviour of Uncracked Beams

The loads acting on a structural element is in equilibrium with the

reactions, and the bending moment and shear force diagrams can be

drawn for the entire span as shown in Fig. 6.1 (in the following slide).

Before cracking, the RC beam may be assumed to behave like a

homogenous beam.


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Fig. 6.1 Shear force (S.F.) and bending moment (B.M.) diagrams for typical beam elements (a) Simply

supported beam with concentrated load (b) Cantilever beam (c) Simply supported beam with

uniformly distributed load (d) Continuous beam




The bending and shear stress distributions across the cross section of

rectangular beam are shown in Fig. 6.2 (in the following slide).

It should be noted that the shear stress variation is parabolic, with the

maximum value at the neutral axis and zero values at the top and

bottom of the section. Thus, the maximum shear stress is 50 per cent

more than the average shear stress.




Fig. 6.2 Flexural and shear stress variation across the cross section

of a rectangular beam



Neglecting any vertical normal stress caused by the surface loads, the

combined flexural and shear stresses can be resolved into equivalent

principal stresses acting on orthogonal planes and inclined at an angle a

to the beam axis, as shown in Figs 6.3(a)(f).

The direction of the principal compressive stresses is in the shape of

an arch, whereas that of the principal tensile stresses is in the shape of a

catenary or suspended chain.




The maximum bending stresses occur at mid-span and the direction ofstresses tends to be parallel to the axis of the beam. Near the supports,

the shear forces have the greatest value and hence the principal

stresses become inclined; greater the shear force, greater the angle of

inclination.

Since concrete is weak in tension, tension cracks as shown in Fig.

6.3(c) will develop in a direction perpendicular to the principal tensilestresses. Thus, the compressive stress trajectories indicate the potential

crack pattern (depending on the magnitude of tensile stresses

developed).




Fig. 6.3 Stress distribution in RC beams (a) Beam with loading (b)(e) Stresses in

elements 1 and 2 (f) Principal stress distribution




Cracking of Beams

Fig. 6.4 Cracking of beams due to tensile stresses (a) Typical cracking

(b) Theoretical reinforcement required to resist such cracking



Types of Cracks1. Near the mid-span, where the bending moment predominates, the

tensile stress trajectories are crowded and are horizontal in directionas shown in Fig. 6.3(f). Hence, flexural cracks perpendicular to the

horizontal stress trajectories will appear even at small loads. These

flexural cracks are controlled by the longitudinal tension bars.

2. In the zones where shear and bending effects combine together, that

is, in zones midway between the support and mid-span, the cracks

may start vertically at the bottom, but will become inclined as theyapproach the neutral axis due to shear stress (see Fig. 6.5). These

cracks are called flexure shear cracks.



Types of Cracks

3. Near the supports that contain concentrated compressive forces, the

stress trajectories have a complicated pattern. As shear forces are

predominant in this section, the stress trajectories are inclined (see

Fig. 6.3f) and cracks inclined at about 45 appear in the mid-depth of

the beam. These cracks are termed as web-shear cracks or diagonaltension cracks.

4. Sometimes, inclined cracks propagate along the longitudinal tensionreinforcement towards the support. Such cracks are termed as tensile

splitting cracks or secondary cracks.



Types of Cracks

Fig. 6.5 Typical crack pattern in an RC beam

B h i f B i h Sh



Behaviour of Beams without Shear

Reinforcement

The behaviour of beams failing in shear may vary widely, depending

on the av/dratio (shear span to effective depth ratio) and the amount of

web reinforcement (see Fig. 6.6 in the following slide).

Very short shear spans, with av/d ranging from zero to one, develop

inclined cracks joining the load and the support. These cracks, in effect,

change the behaviour from beam action to arch action (see Fig. 6.7).

Such beams with the a/dratio of zero to one are termed as deep

beams. These beams normally fail due to the anchorage failure at the

ends of the tension tie.

B h i f B i h Sh




Reinforcement

Fig. 6.6 Effect of a/d ratio on shear strength of beams without stirrups (a) Beam, shear force, and

moment diagrams (b) Variation in shear capacity with a/dfor rectangular beams



Modes of Failure in Deep Beams

Fig. 6.7 Modes of failure of deep beams (a) Arch action (b) Types of failures

B h i f B ith t Sh




ReinforcementBeams with a/dranging from 1 to 2.5 develop inclined cracks and

carry some additional loads due to arch action. These beams may fail by

splitting failure, bond failure, shear tension, or shear compression

failure.

For slender shear spans, having av/d ratio in the range of 2.5 to 6,

When the load is applied and gradually increased, flexural cracks appear

in the mid-span of the beams. With further increase of load, inclined

shear cracks develop in the beams which are sometimes called primary

shear cracks.

Very slender beams, with a/dratio greater than 6.0, will fail in flexure

prior to the formation of inclined cracks.

B h i f B ith t Sh




Reinforcement

Fig. 6.8 Behaviour of beam without shear reinforcement (a) Typical crack pattern (b) Typical failure of

beam without shear reinforcement (c) Shear compression failure


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Types of Shear or Web Reinforcements

Shear or web reinforcements, called stirrups, links, or studs, may be

provided to resist shear in several different ways such as the following:

1. Stirrups perpendicular to the longitudinal flexural (tension)

reinforcement of the member, normally vertical (Fig. 6.9a in the

following slide).

2. Inclined stirrups making an angle of 45 or more with the

longitudinal flexural reinforcement of the member (Fig. 6.9b)

3. Bent-up longitudinal reinforcement, making an angle of 30 or

more with the longitudinal flexural reinforcement (Fig. 6.9c)


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4. Welded wire mesh, which should not be used in potential plastichinge locations (Fig. 6.9d). They are used in small, lightly loaded

members with thin webs and in some precast beams

5. Spirals (Fig. 6.9e)

6. Combination of stirrups and bent-up longitudinal reinforcement (Fig.

6.9f)


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7. Mechanically anchored bars (head studs) with end bearing plates ora head having an area of at least 10 times the cross-sectional area of

bars

8. Diagonally reinforced members

9. Steel fibres in potential plastic hinge locations of members


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Fig. 6.9 Types and arrangement of stirrups (a) Vertical stirrups (b) Inclined stirrups

(c) Longitudinal bent bars (d) Welded wire fabric (e) Spirals (f) Combined bent bars and vertical

stirrups


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The presence of stirrups contributes to the strength of shear

mechanisms in the following ways:

1. They carry part of the shear.

2. They improve the contribution of the dowel action. The stirrup

can effectively support a longitudinal bar that is being crossed by a

flexural shear crack close to a stirrup.

3. They limit the opening of diagonal cracks within the elastic range,

thus enhancing and aiding the shear transfer by aggregate interlock.

Stirrups in Shear Mechanisms


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4. When stirrups are closely spaced, they provide confinement to the core

concrete, thus increasing the compression strength of concrete, which will behelpful in the locations affected by the arch action.

5. They prevent the breakdown of bond when splitting cracks develop in theanchorage zones because of the dowel and anchorage forces.

6. The strength of the concrete tooth between two adjacent shear cracks of

the beam and located below the neutral axis is important for developing shear

strength. The stirrups suppress the flexural tensile stresses in the cantilever

blocks by means of the diagonal compressive force resulting from the truss

action.

Stirrups in Shear Mechanisms


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Vertical Stirrups

The transverse reinforcement in the form of shear stirrups will usuallybe vertical and taken around the outermost tension and compression

longitudinal reinforcements along the faces of the beam, as shown in

Fig. 6.10. In T- and I-beams, they should pass around the longitudinal

bars located close to the outer face of the web.

The most common types are shown in Figs 6.10(a)(e). The stirrup

arrangements shown in Figs 6.10(a)(e) are not closed at the top andhence their placement at site is relatively easy compared to the closed

stirrups. However, they should be used in beams with negligible

torsional moment.


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Vertical StirrupsClosed stirrups, which are suitable for beams with significant torsion

and in earthquake zones, are shown in Figs 6.10(f)(k).

The vertical hoop is a closed stirrup having a 135 hook with a 610

diameter extension at each end that is embedded in the confined core

(see Figs 6.10f and j).

It can also be made of two pieces of reinforcement as shown in Fig.

6.10(g) with a U-stirrup having a 135 hook and a 10 diameter extension

at each end, embedded in concrete core and a cross-tie. It is alsopossible to have the cross-tie with a 135 hook at one end and 90 hook

at the other end for easy fabrication, as shown in Figs 6.10(h) and (k).

The hooks engage peripheral longitudinal bars.


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Vertical Stirrups

Fig 6.10 Types of vertical stirrups (a)(e) Open stirrups for beams with negligible torsion

(f)(i) Closed stirrups with significant torsion (j)(k) Detail of 135hook


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Bent-up Bars

The performance of bent-up bars in shear is

illustrated in Fig. 6.11 (in the following slide).

As seen in this figure, large stresses

concentrate in the region of such bars,

leading to the splitting of concrete when

spaced far apart or when placed

asymmetrically with reference to the verticalaxis of cross section.


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Bent-up Bars

Fig. 6.11 Performance of bent-up bars in shear


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Disadvantages of Bent-up Bars

1. They are widely spaced and are few in number. Hence, a crack maynot be intercepted by more than one bar, thus resulting in wider

cracks than those in beams with stirrups.

2. When some of the bars at a section are bent up, the remainingflexural bars are subjected to higher stresses, resulting in wider

flexural cracks.

3. Concrete at the bends may be subjected to splitting forces, resultingin possible web cracking.


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Disadvantages of Bent-up Bars

4. They do not confine the concrete in the shear region.

5. Reduction of flexural steel due to bent-up bars may result in the

shifting of the neutral axis upwards, causing wider cracks in the

tension zone.

6. They are less efficient in tying the compression flange and web

together.


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Inclined Stirrups

Inclined stirrups are similar to vertical stirrups, except that they are

placed at an angle of about 45 to the longitudinal axis of the beam.

Their behaviour is similar to the bent-up bars.

As they are nearly perpendicular to the cracks, they are more efficient

than all other shear reinforcements.


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Advantages of Inclined Stirrups

1. They can be closely spaced, and hence the cracks may beintercepted by more than one bar, resulting in less wider cracks than

those in beams with bent-up bars.

2. They confine the concrete in the shear region.

3. They are efficient as vertical stirrups in tying the compression flange

and web together.


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Disadvantages of Inclined Stirrups

1. They are difficult to fabricate and construct.

2. When there is a reversal of shear force (due to earthquakes), they

may be inefficient.


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Spirals

Helical reinforcement in RC columns results in increase in strength of

core concrete and ductility due to confinement reinforcement.

If the correct pitch is utilized for effective confinement, helical

reinforcement will provide an economical solution for enhancing the

strength of flexural members.


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Spirals

Fig 6.12 Ways of providing spirals (a) Spirals (tensile zone) (b) Spirals (compression zone)

(c) Double spirals (d) Interlocking spirals


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Headed Studs

Conventional stirrups are being increasingly replaced by headed studs,which are smooth or deformed bars provided with forged or welded

heads for anchorage at one or both the ends (see next slide).

The two common types are the single-headed studs welded to a

continuous base rail and the double-headed studs welded to spacer

rails. The base rail is used to position the studs at the required spacing,which is determined by readymade software or calculation.

d d d


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Headed Studs

Fig. 6.13 Headed stud with deformed stem

and heads at both ends

d d d


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Headed Studs

To be fully effective, the size of the heads should be

capable of developing the specified yield strength of the

studs.

When the studs are used, it is not essential to place

longitudinal bars behind the heads. Without the

longitudinal bars, the heads can produce sufficient

anchorage to develop yield force in the studs.

Headed studs reduce congestion in beam-column

joints and in zones of lap splices.

l ib


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Steel Fibres

Fibre-reinforced concrete (FRC) with a minimum volume fraction of0.5 per cent fibres can be used to replace minimum shrinkage and

temperature reinforcement.

Steel fibres (crimped or hooked) increase the post-cracking resistance

across an inclined crack, which in turn increases the aggregate interlock

and shear resistance of concrete. The use of fibres also results in

multiple inclined cracks and gradual shear failure.

Ad f S l Fib


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Advantages of Steel Fibres

The replacement of stirrups by fibres in FRC has the followingadvantages:

1. The random distribution of fibres throughout the volume of

concrete at much closer spacing than is practical for the stirrups

can lead to distributed cracking with reduced crack width.

2. The first-crack tensile strength and the ultimate tensile strength

of the concrete are increased by the fibres.

3. The shear friction strength is increased by resistance to pull-out

and by fibres bridging cracks.

Behaviour of Beams with


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Behaviour of Beams withShear / Web Reinforcements

When a beam with transverse shear reinforcement is loaded, most of

the shear force is initially carried by the concrete. Between flexural and

inclined cracking, the external shear is resisted by the concrete the

interface shear transfer, and the dowel action.

The first branch of shear cracking of the beams with transverse

reinforcement is typically the same in nature as that of beams without

transverse reinforcement. The shear crack in this case also involves two

branches.



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The formation of the second crack and the corresponding load may be

assumed to be the same, as shown in Fig. 6.11 (in the following slide).

The presence of shear reinforcements restricts the growth of diagonal

cracks and reduces their penetration into the compression zone.

Cracked RC transmits shear in a relatively complex manner. The

highest reinforcement stresses and the lowest concrete tensile stresses

occur at the cracks.



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Fig. 6.14 Distribution of internal shears in beam with web reinforcement

Factors Affecting Shear Strength of


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Factors Affecting Shear Strength ofConcrete

1. Tensile strength of concrete: The inclined cracking load in shear is a

function of the tensile strength of concrete.

2. Longitudinal reinforcement ratio: The shear strength of the RC

beams is found to drop significantly if the longitudinal reinforcement

ratio is decreased below 1.21.5 per cent.

3. Shear span to effective depth ratio: Its effect is pronounced when av/dis less than two and has no effect when it is greater than six.

Factors Affecting Shear Strength of


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Factors Affecting Shear Strength ofConcrete

4. Lightweight aggregate concrete: They reduce tensile strength than

concrete with normal aggregates.

5. Size of beam: As the depth of the beam increases, the shear stress

at failure decreases.

6. Axial forces : Axial tension decreases the inclined cracking load and

the shear strength of concrete, whereas axial compression does just

the opposite.

7. Size of coarse aggregate: Increasing the size of coarse aggregates

increases the roughness of the crack surfaces, thus allowing higher

shear stresses to be transferred across the cracks.

Si Eff t


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Size Effect

With increasing beam depth, the crack spacing and crack width tendto increase.

Fig. 6.15 shows that the shear stress at failure decreases when the

member depth increases or the maximum aggregate size decreases.

Failure shear stress does not significantly change when the width of

beams is increased.

Si Eff t


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Size Effect

Crack widths increase nearly linearly with the tensile strain inreinforcement and the spacing between cracks.

The size effects are mitigated considerably if the depth of the beam is

less than 1 m. However, in beams having depths greater than 1 m, the

size effect cannot be neglected.

Size effect is not felt in beams with web reinforcement.

Si Eff t


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Size Effect

Fig 6.15 Influence of member depth and maximum aggregate size on shear stress at failure

C St d


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Case Study

Design Shear Strength of Concrete in


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Design Shear Strength of Concrete inBeams

The design shear strength of concrete depends on factors such as the

following:

1. Grade of concrete

2. Longitudinal reinforcement ratio

3. Shear span to depth ratio

4. Type of aggregate used

5. Size of beam6. Axial force

7. Size of coarse aggregate used

Design Shear Strength of Concrete in


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Design Shear Strength of Concrete inBeams

Maximum Shear Stress: The shear strength of beams cannot be

increased beyond a certain limit, even with the addition of closelyspaced shear reinforcement because large shear forces in the beam will

result in compressive stresses that may cause crushing of web concrete.

Effects Due to Loading Condition: The shear strength of beams, either

slender or deep, under the uniform load is much higher than that of

beams under a loading arrangement of two concentrated loads at

quarter points or one concentrated load at mid-span. In this case, thesplitting failure along the line of the second branch of the critical

diagonal crack occurs near the support reaction and not near a

concentrated load.

Critical Section for Shear


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Before designing the beam for shear, the

critical section for shear should first be

located.

The maximum shear force in a beam usually

occurs at the face of the support and reducesprogressively away from the support.

When there are concentrated loads, shear

force remains high in the span between thesupport and the first concentrated load (Figs

6.16af).



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Critical Section for ShearClause 22.6.2 of IS 456 allows a section located at a distance d(effective

depth) from the face of the support to be treated as a critical section inthe following cases (see Figs 6.16ac):

1. Support reaction, in the direction of applied shear force,

introduces compression into the end regions of the member.

2. Loads are applied at or near the top of the member.

3. No concentrated load occurs between the face of the support and

the location of the critical section, which is at a distance dfrom

the face of the support.



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Critical Section for ShearThis clause cannot be applied in the following situations:

1.Beams framing into a supporting member in tension (see Fig.

6.16d)

2.Beams loaded near the bottom, as in the case of inverted beam

(see Fig. 6.16e)

3.Concentrated load introduced within a distance 2dfrom the face

of the support, as in the beam on the left side of Fig. 6.16(b). In

this case, closely spaced stirrups should be designed and provided

in the region between the support and the concentrated load.


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Fig. 6.16 Critical sections for shear near support (a)(c) Critical section at a distance d from

the face of the support (d)(f) Critical section at the face of the support

Enhanced Shear Strength near Supports


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It has been observed from tests that shear failure at sections of beamsand cantilevers without shear reinforcement occurs at a plane inclined

at an angle 30 as shown in Fig. 6.17a (in the following slide).

When the failure plane is inclined more steeply than this, the shear

force required to produce the failure is increased (see Fig. 6.17b in the

following slide).



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Fig. 6.17 Enhanced shear strength (a) Steep failure plane (b) Influence of shear span to depth ratio



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A good design is one in which shear failure iseliminated & the flexure governs the design .

Hence, reducing the shear reinforcement near thesupports and increasing the vulnerability to shear

failure, is not advisable, especially in seismic zones.

The reduction in the quantity of shear

reinforcement achieved through clause 40.5 of IS

456 is marginal and hence it is better to ignore it.

Minimum and Maximum Shear


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When the principal tensile stress within the shear span exceeds the

tensile strength of concrete, diagonal tension cracks are initiated in theweb of concrete beams. These cracks later propagate through the beam

web, resulting in brittle and sudden collapse. When shear

reinforcements are provided, they restrain the growth of inclined

cracking.

Minimum shear reinforcement should be provided in all the beams

when the calculated nominal shear stress is less than half of designshear strength of concrete. The minimum shear reinforcement is also a

function of concrete strength.

Reinforcement

Minimum and Maximum Shear


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In seismic regions, web reinforcement is required in most beams

because the shear strength of the concrete is taken as equal to zero ifthe earthquake-induced shear exceeds half the total shear.

Maximum Spacing: The maximum yield strength of web

reinforcement is limited to avoid the difficulties encountered in bendinghigh-strength stirrups (they may be brittle near sharp bends) and also to

prevent excessively wide inclined cracks.

Upper Limit on Area of Shear Reinforcement: Maximum shearreinforcement are proportional to the square root of the concrete

compressive strength as per the Indian code.

Reinforcement

Minimum Shear Reinforcement


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Minimum Shear Reinforcement

Fig. 6.18 Minimum shear reinforcement as a function of fcas per different codes

The Ritter Mrsch Truss Model


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The RitterMrsch Truss Model

The truss model was originally introduced by Ritter, who proposed a45 truss model for computing the shear strength of the RC members;

this model was refined by Mrsch.

Ritter assumed that after the cracking of concrete, the behaviour of an

RC member is similar to that of a truss with a top longitudinal concrete

chord, a bottom longitudinal steel chord (consisting of longitudinalreinforcement), vertical steel ties (stirrups), and diagonal concrete struts

inclined at 45, as shown in Fig. 6.19(a).



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The RitterMrsch Truss ModelIt was further assumed that the diagonally cracked concrete cannot

resist tension and the shear force is resisted by transverse steel,commonly referred to as the steel contribution and the uncracked

concrete contribution.

When a shear force is applied to this truss, the concrete struts are

subjected to compression whereas tension is produced in the transverse

ties and in longitudinal chords.

The design of stirrups is usually based on the vertical component of

diagonal tension, whereas the horizontal component is resisted by the

longitudinal tensile steel of the beam.



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Fig. 6.19 Truss models for beams with web reinforcement (a) RitterMrsch truss model

(b) Variable angle truss model



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The concrete contribution is generally considered to be a combinationof force transfer by the dowel action of the main flexural steel,

aggregate interlock along a diagonal crack, and uncracked concrete

beyond the end of the crack.

It is also difficult to calculate the exact proportion of each of these

forces. Hence, it was vaguely rationalized to adopt the diagonal cracking

load of the beam without web reinforcement as the concretecontribution to the shear strength of an identical beam with web

reinforcement.




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Experience with the 45 truss analogy revealed that the results of thismodel were quite conservative, particularly for beams with small

amounts of web reinforcement.

The truss model does not consider the size effects.

The truss model has been modified by several others in the past 30

years.

Based on that, it was realized that the angle of inclination of theconcrete struts, q, may be in the range 2565, instead of the constant

45 assumed in the RitterMrsch model. These developments lead to

the variable angle truss model.


Design of Vertical and Inclined


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gStirrups

Fig. 6.20 Design of stirrups (a) Vertical stirrups (b) Inclined stirrups

Modified Compression Field Theory


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The modified compression field theory (MCFT) developed by Collins,Vecchio, andBentz uses the strain conditions in the web to determine

the inclination qof the diagonal compressive stresses (see Fig. 6.21).

The equilibrium conditions, compatibility conditions, and stressstrain

relationships (constitutive relationships) are formulated in terms of

average stresses and average strains.




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Figure 6.21 gives the 15 equations used in MCFT. The MCFT assumed

that the directions of the inclined compression field (i.e., the strut angle

and the crack angle) and the principal compressive stress coincide.

Solving the equations of the MCFT is tedious, if attempted by hand,

and hence software programs called Membrane-2000and Response-

2000 were developed.

Over the last 20 years, the MCFT has been applied to the analysis of

numerous RC structures and found to provide accurate simulations of

behaviour.



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Fig. 6.21 Equations of modified compression field theory

Design Procedure for Shear


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gReinforcement

The design of an RC beam for shear using vertical stirrups involves the

following steps:

1.Determine the maximum factored shear force Vuat the critical

sections of the member (see Fig. 6.13).

2.Check the adequacy of the section for shear. Compute the nominal

shear stress and check whether it is less than the maximum

permissible shear stress. If it is greater than the maximumpermissible shear stress, increase the size of the section or the

grade of concrete and recalculate steps 1 and 2.



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gReinforcement

3. Determine the shear strength provided by the concrete (for thepercentage of tensile reinforcement available at the critical section)

Vc.

4. If Vu> Vc, shear reinforcements have to be provided for Vus= Vu Vc.

5. Compute the distance from the support beyond which only minimum

shear reinforcement is required.



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gReinforcement

6. Design of stirrups: Where stirrups are required, it is usuallyadvantageous to select a bar size and type and determine the

required spacing. The spacing for vertical stirrups is calculated as:

In regions where only minimum stirrups are required, it is:

6. Check anchorage requirements and details.

Transverse Spacing of Stirrups in


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In wide beams with large number of longitudinal rods and carryingheavy shear forces (such as those encountered in raft foundations), it is

advisable to provide multi-legged stirrups, so that the longitudinal

forces are evenly distributed among the longitudinal rods of the beam.

The effectiveness of the shear reinforcement decreases as the spacing

of the web reinforcement legs across the width of the memberincreases.

p g pWide Beams

Anchoring of Bent-up Bars


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The bent-up bars should be anchored adequately. The development

length should be provided in the compression zone, measuring from the

mid-depth of the beam. If the bent-up bars are anchored in the tension

zone, the development length can be measured from the end of the

sloping or inclined portion of the bar.

Fig. 6.22 Anchoring of bent-up bars

Anchoring of Bent up Bars

Anchoring of Shear Stirrups


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Anchoring of Shear StirrupsThe stirrups should be well anchored to develop the yield stress in the

vertical legs, as follows:

1. The stirrups should be bent close to the compression and

tension surfaces, satisfying the minimum cover.

2. The ends of the stirrups should be anchored by standard hooks.

3. Each bend of the stirrups should be around a longitudinal bar. The

diameter of the longitudinal bar should not be less than thediameter of the stirrups.



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In addition to providing anchorage, these specifications are provided forother reasons too, including the following:

1. Constructability purposes

2. Prevention of presumed concrete crushing at the corner of the

stirrup, resulting from the high stress concentrations that develop in

this region when the member is loaded

Shear Design of Flanged Beams


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Shear Design of Flanged Beams

The behaviour and cracking pattern of T-beams under two-pointloading or one-point loading in the middle of the beam are similar to

that of rectangular beams.

An increase in the shear capacity results from an increase of the cross-

sectional area of the compressive zone of a beam. It has been found

that the shear capacity of T-beams is 3040 per cent higher than the

shear strength of their web. This increased strength is due to the size ofthe flanges, an increase in the tensile strength of concrete, and the

neutral axis depth.

Concept of Shear Funnel


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Concept of Shear FunnelTo account for the effect of flange thickness on the shear area of the T-

beams, the concept of shear funnel was developed where the area ofconcrete bounded by the neutral axis and the two angled lines is

defined as the effective shear area.

Fig. 6.23 Concept of shear funnel (a) Neutral axis within flange

(b) Neutral axis outside flange

Shear Design of Beams with Varying


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Beams of varying depth are encountered in haunched beams. In suchmembers, it is necessary to account for the contribution of the vertical

component of the flexural tensile force Tu, which is inclined at an angle b

to the longitudinal direction, in the nominal shear stress, tv.

The following two cases may arise in practice:

1. The bending moment increases numerically in the same direction

in which the effective depth increases.

2. The bending moment decreases numerically in the direction in

which the effective depth increases.

Depth



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A similar situation arises in tapered base slabs or footings, where

flexural compression is inclined to the longitudinal axis of the beam,

since the compression face may be sloping.

In the case of cantilever beams, the depth increases in the same

direction as the bending moment.

Depth



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Depth

Fig. 6.24 Beams of variable depth (a) Bending moment increases with increasing depth(b) Bending moment decreases with increasing depth

Shear Design of Beams Located in


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Earthquake Zones

When there is a reversal of stresses, due to earthquakes or reversedwind directions, the shear strength of concrete cannot be relied upon,

as the cracks will criss-cross the cross section and hence cracked

concrete will be present in the tension and compression zones.

Hence, the stirrups should be designed to take the entire shear with

zero contribution from concrete.

In earthquake zones, only vertical closed stirrups or those placedperpendicular to the member axis are to be used, with 135 hooks.



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Earthquake Zones

According to Clause 6.3.3 of IS 13920:1993, the shear capacity of thebeam shall be more than the following:

1. Calculated factored shear force as per analysis

2. Shear force due to the formation of plastic hinges at both ends ofthe beam plus the factored gravity load on the span (see Fig. 6.21 in

the following slide).

Clause 6.3.3 of IS 13920:1993 ensures that a brittle shear failure does

not precede the actual yielding of the beam in flexure.



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Earthquake Zones

Fig. 6.25 Calculation of design shear force in case of earthquake loading

Stirrup Arrangement for Beams


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Located in Earthquake Zones

Fig. 6.26 Stirrup arrangement for beams located in earthquake zones

Shear in Beams with High-strengthC d i h h S l


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Concrete and High-strength SteelHSC beams exhibited increased capacity and improved hysteretic

performance compared to NSC beams. Flexure deformation-dominatedductile responses were achieved by designing the beam shear strength

based on the seismic provision of the current ACI 318 code.

Beams made of HSC were found to exhibit more significant size effectthan NSC beams.

The width of the diagonal cracks is directly related to the strain in the

stirrups. Hence, the Indian and US codes do not permit the design yieldstress of stirrups to exceed 415 MPa. This requirement limits the width

of cracks that can develop. When the width of the crack is limited, the

aggregate interlock is enhanced.

Shear in Beams with High-strengthC d Hi h h S l


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Concrete and High-strength Steel

A further advantage of a limited yield stress is that the requiredanchorage length at the top of the stirrups is not as stringent as it would

be for stirrups with higher yield strength. Pairing high-strength steel

(HSS) with HSC is more beneficial.

The limitation of 420 MPa for design yield stress of stirrups is relaxed

for deformed welded wire fabric because research has shown the use of

higher strength wires to be quite satisfactory.

The width of inclined shear cracks at service loads is found to be less

for beams with higher strength wire fabric than for beams with stirrups

having yield strength of 415 MPa.

Shear in Beams with High-strengthC d Hi h h S l


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Concrete and High-strength Steel

Fig. 6.27 Comparison for HSC beams with different transverse reinforcement details (a) HSC beam

with 10 mm diameter stirrups (b) HSC beam with 6 mm diameter stirrups

Shear Design Beams with Web Opening


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g p g

Transverse openings are provided in concrete beams for

accommodating utility services, which will result in compact design andoverall saving in terms of total building height. The provision of

openings changes the behaviour of the beam from a simple one to a

more complex one.

Although numerous shapes of openings are possible, circular (to

accommodate service pipes) and rectangular (to accommodate air-

conditioning ducts) openings are most common.

The openings must be located in such a way that no potential failure

planes passing through several openings could develop.



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g p g

Entire shear resistance may be assigned to the compression chord.Longitudinal and transverse reinforcements should be placed in both

sides of the opening to resist 1.5 times the shear force and bending

moment generated by the shear across the opening (see Fig. 6.28 in the

following slide).

To control the horizontal splitting and diagonal tension cracks at the

corners of the opening, transverse reinforcements should be designedfor two times the design shear force and provided over a distance not

less than 0.5don both sides of the opening (see Fig. 6.28 in the

following slide).



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g p g

Fig. 6.28 Beams with large web openings

Shear Strength of Members withA i l F


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Axial Force

The beams in moment-resistant frames are often subjected to axialforces in addition to the bending moments and shears. Columns are also

subjected to axial loads, bending moments, and shear forces.

Axial tensile forces tend to decrease the shear strength of concrete,

whereas axial compression tends to increase it.

The compressive force acts like prestressing and delays the onset of

flexural cracking; also, flexural cracks do not penetrate to a greater

extent into the beam.

Shear Strength of Members withA i l F


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Axial Force

Tensile forces directly increase the stress

and hence the strain in the longitudinal

reinforcement.

Axial tension increases the inclined crack

width and reduces the aggregate interlock,

and hence, the shear strength provided by

the concrete is reduced.

Design of Stirrups at SteelC t off Points


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Cut-off PointsLongitudinal tension reinforcement is often curtailed in order to

provide the required reduced area of steel in locations where thebending moment is less than the maximum value.

The termination of flexural tensile reinforcement gives rise to sharp

discontinuity in the steel, causing early appearance of flexural cracks,which in turn may turn into diagonal shear cracks.

Clause 26.2.3.1 of IS 456 insists that the bars should extend beyond

the theoretical cut-off point to reduce stress concentration.

Design of Stirrups at SteelCut off Points


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Cut-off PointsClause 26.2.3.2 of IS 456 stipulates that flexural reinforcement in beams

may be terminated in the tension zone, only if any one of the followingconditions is satisfied:

1. The shear at the cut-off point does not exceed 2/3 of Vu(i.e., cut-

off is allowed in low shear zones).

2. Extra shear reinforcements are provided over a distance equal to

0.75d from the cut-off point.

3. When 36 mm diameter or smaller bars are used, excess flexuralsteel is available (continuing bars provide double the area

required for flexure) along with excess shear capacity (shear

capacity is greater than 1.33Vu).

Shear Friction


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Theconcept of Shear friction is used where direct shear is transferred

across a given plane.

The situations where this concept will be useful include the interface

between concretes cast at different times, interface between concreteand steel, connections of precast constructions, and corbels (see Fig.

6.29).

The correct application of this concept depends on the properselection of the assumed location of crack or slip. The reinforcement

must be provided crossing the potential or actual crack or shear plane to

prevent direct shear failure.

Shear Friction


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Fig. 6.29 Locations of potential cracks where shear friction concept is applied

Shear Friction Design Method


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The shear friction design method is quite simple and the behaviour can

be easily visualized as follows (see Figs 6.30a-c):

1. A cracked block of concrete with the intercepted reinforcement is

assumed. The shear force acts parallel to the crack, and the

tendency for the upper block to slip relative to the lower blockhas to be resisted by the friction along the interface of the crack,

by the resistance to the shearing off of protrusions on the crack

faces, and by the dowel action of the reinforcement crossing the

crack.



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2. The dowel effect is usually neglected for simplicity, and to

compensate for this factor a high value of friction coefficient isassumed. The irregular surface may separate the two blocks slightly,

as shown in Fig. 6.30(b). If the crack surface is rough, the coefficient

of friction may be high. The reinforcement provides a clamping

force across the crack faces.

While using the shear friction method of design, reinforcement should

be well anchored to develop the yield strength of steel, by fulldevelopment length, hooks, or bends in the case of reinforcement bars

and by proper heads or welding in the case of studs joining the

concrete to structural steel.



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The shear friction reinforcement anchorage should engage the

primary reinforcement; otherwise, a potential crack may pass between

the shear friction reinforcement and the body of the concrete.

Care must be exercised to consider all possible failure planes and to

provide sufficient well-anchored reinforcement across the planes.

Basis of Shear Friction Design


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Fig. 6.30 Basis of shear friction design (a) Applied shear (b) Enlarged crack surface

(c) Free body of concrete above crack

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