507 33 Powerpoint-slides Ch8 DRCS

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Oxford University Press 2013. All rights reserved.

Chapter 8

Design For Torsion


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The diagonal tension stresses produced by

torsion are very similar to those caused by

shear. But they occur on all the faces of the

member; hence, they have to be added to

the stresses caused by the shear on one

face whereas subtracted from the stresseson the other face.

As the torsional cracks spiral around the

beams, it is necessary to provide closedstirrups as well as additional longitudinal

reinforcement, especially at the corners of

the faces of the beams.

Introduction

Torsional failure of columns ofMianyang Airport Viaduct duringthe M7.9 Wenchuan Earthquake,May 12, 2008 (source: FHWA-HRT-11-029)


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Equilibrium and Compatibility Torsion

Primary torsion, also called equilibrium torsion or statically

determinate torsion, exists when the external load has no alternativeload path but must be supported by torsion (see Figs 8.1a and b). For

such cases, the torsion required to maintain static equilibrium can be

uniquely determined from statics alone.

Secondary torsion, also called compatibility torsion or statically

indeterminate torsion, occurs due to the requirements of continuity,

that is, due to compatibility of deformation between the adjacentelements of a structure (see Fig. 8.1d). Torsional moments cannot be

found based on static equilibrium alone. The beams in a grid structure

also have compatibility torsion.


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Equilibrium and Compatibility Torsion

Disregard to compatibility torsion in the design will often lead to

extensive cracking, but generally will not cause collapse. An internal

readjustment of forces will take place and an alternative equilibrium of

forces will be found.

The amount of torsion in a member depends on its torsional stiffness

in relation to the torsional stiffness of the interconnecting members.


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Fig. 8.1 Structural elements subjected to torsion (a) Beams supporting cantilevered canopy slabs (b)

Cantilever beam supporting eccentric load (c) Box-girder bridges (d) Edge beams in framed structures

(e) Circular ring beams



Structures Subjected to Torsion

Fig. 8.2 Structures subjected to torsion (a) Curved continuous beams or box girders in

BandraWorli sea link bridge (b) Helicoidal girders



Torsion in Curved Beams

Curved beams (e.g., ring beams under circular water tanks supported

by columns) are subjected to bending and torsion. The magnitude and

distribution of the bending and torsional moments along the

circumference are influenced by the number of supports and the radius

of the curved beam.

A typical curved beam circular in plan and supported by eight columns

is shown in Fig. 8.3(a) (in the following slide). By considering Fig. 8.3(b),the maximum positive and negative bending moments and the torsional

moments can also be determined.




The critical sections for design are the support sections subjected to

maximum negative and positive bending moments and the sections

subjected to maximum torsion associated with some shear force; at this

section, the bending moment will be zero. Hence, it has to be designed

for combined torsion and shear.

The magnitude and position of maximum positive and negative

bending moments and torsional moments in a semicircular beam

supported on three equally spaced supports can be similarly

determined.




Fig. 8.3 Beams curved in plan (a) Ring beam supported on eight columns (b) Position of

maximum moments



Torsional Analysis

Elastic Analysis

This theory is applicable to homogeneous material such as steel of

prismatic circular, non-circular, and thin-walled cross sections. From this

theory, it may be observed that torsion causes shear stresses. In non-

circular sections, there is considerable warping of the cross section andthe plane sections do not remain plane, as shown in Fig. 8.4.

Because of the advantageous distribution of shear stresses, thin-walled

tubular sections are more efficient in resisting torsion. The concept of

shear flow around the thin-walled tube is useful when the role of

reinforcement in torsion is considered.



Elastic Analysis

Fig. 8.4 Elastic torsional behaviour of rectangular beams (a) Beam subjected to torsion

(b) Warping of the cross section (c) Torsional stress (d) Crack pattern



In the case of compatibility torsion, if the spandrel beam as shown in

Fig. 8.1(d) is uncracked, the torsional moment carried by it may be verylarge. As the beam cracks, the torsional stiffness reduces considerably

and the beam will rotate, reducing the torsional moment carried by it.

Cracked section stiffness requires the knowledge of the steelreinforcement. To solve this problem, Lampert (1973) and Collins and

Lampert (1973) proposed expressions for torsional rigidity of cracked

sections based on their studies.

As per Collins and Lampert (1973) the analysis can also be based on

zero torsional stiffness; such an analysis and the subsequent design

based on flexure and shear, neglecting torsion is satisfactory. However

torsional reinforcement increase ductility & distribute cracks.

Elastic Analysis



Torsion in a Thin-walled Rectangular Tube

Fig. 8.5 Torsion in a thin-walled rectangular tube (a) Thin-walled tube (b) Area enclosed

by shear flow path



Torsional Analysis

Plastic Analysis

The value of stress to be used in the limit

states design should be based on plastic

analysis, even though the assumption of fully

plasticized section is not justifiable for

materials like concrete.

In plastic analysis, a uniform shear stress over

the cross section is assumed, whereas the

elastic analysis shows a non-linear stress

distribution, as shown in Fig. 8.4(c).


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Plastic Analysis

The ultimate torque can now be easily obtained by using the sand

heap analogy, which is based on the

following principles:

1. Ultimate torque = Twice the volume of sand heap

2. Slope of sand heap = 2 constant plastic shear stress

The ultimate torque of T-, L-, or I-sections can be obtained in a similar

manner by dividing them into component rectangles (see Fig. 8.6).


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Plastic Analysis

Fig. 8.6 Sand heap analogy for different sections

(a) Rectangle (b) T-section (b) L-section (d) I-section


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Behaviour of Plain Concrete Members

When a rectangular concrete beam is subjected to pure torsion, a

state of pure shear develops at the top and side faces of the beam, with

direct tensile and compressive stresses along the diagonal directions,

similar to the beam subjected to shear.

The principal tensile and compressive stress trajectories form in

orthogonal directions at 45 to the axis of the beam. When the principal

tensile stress reaches the value of tensile strength of concrete, cracks

form at the maximum stressed location centre of the beam (at the

middle of wider face).


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Behaviour of Plain Concrete Members

These inclined cracks tend to extend around the member in a spiral

fashion, as shown in Figs 8.7(b) and 8.4(d).

Once the crack is formed, the crack will penetrate inwards from the

outer surface of the beam, due to the brittle nature of the concrete and

will lead to a sudden failure of the beam unless torsional reinforcements

are provided.


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Stresses Caused by Torsion

Fig. 8.7 Stresses caused by torsion (a) Shear and principal stresses (b) Crack pattern

Behaviour of Beams with Torsional


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Reinforcement

The torsional reinforcements come into play only after the cracks form

due to diagonal tensile stresses. As the cracks spiral around the beam,

the best way to provide reinforcement is to have them in the form of

spirals to resist the tensile stresses.

However, it is impractical to provide such reinforcement. Hence,

usually torsional reinforcement is provided in the form of a combination

of longitudinal bars at the corners of the beam and stirrups placedperpendicular to the beam axis. Since the cracks spiral around the

beam, four-sided closed stirrups are required.



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ReinforcementOnce the crack is formed, the angle of twist increases without any

increase in the external torque, as the forces are redistributed to thetorsional reinforcement. Then, the cracking extends to the central core

of the member, rendering the central core ineffective. After this, the

failure may take several forms, such as the following:

1. The yielding of longitudinal reinforcement or the stirrups oryielding of both at the same time

2. The crushing of concrete between the inclined cracks due to

principal compression before the yielding of steel

Ductile behaviour is achieved only when both the longitudinal and

transverse reinforcements yield prior to the crushing of concrete.


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Plastic Space Truss Model

The design theory called the thin-walled tube

or plastic space truss model combines the

thin-walled tube analogy with the plastic

truss analogy for shear and leads to simpler

calculations than the skew bending theory.


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Design Strength in Torsion

In the case of solid and hollow beams, once cracking has occurred, the

concrete in the centre of the member has little effect on the torsional

strength of the cross section and can be ignored.

The beams can be considered to be equivalent tubular members.

Hence, solid members can be considered as equivalent tubes. The solid

rectangular or square beams may be idealized as a thin-walled tube as

shown in Fig. 8.8(a).


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This hollow trussed tube consists of closed stirrups forming transverse

tension tie members, longitudinal bars in the corners of the stirrups that

act as tension chords, and concrete compression diagonals, which spiral

around the member between the torsional cracks at an angle (which can

take load parallel to but not perpendicular to the torsional cracks), asshown in Fig. 8.8(b).


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After torsional cracking develops, the torsional resistance is provided

mainly by a space truss consisting of closed stirrups, longitudinal bars,

and compression diagonals, as shown in Fig. 8.8(c).

The thickness of the walls of the imaginary tube representing a solid

member is large and is in the range of one-sixth to one-fourth of the

minimum width of the rectangular beam.


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Fig. 8.8 Thin-walled tube or plastic space truss analogy (a) Thin-walled

tube analogy (b) Space truss analogy (c) Idealized section of the truss


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In a tube wall, the concrete will crack when the tensile stress exceeds

the tensile strength of concrete. in this situation, concrete is underbiaxial tension and compression.

In the case of compatibility torsion (See Fig. 8.9), the design torsional

moment can be reduced, because there will be redistribution of internal

forces to other adjoining members after cracking. Here the designer canreduce the design by an amount equal to the cracking torsion, given

by(ACI 318):

Cracking Torque


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Threshold Torsion

The Threshold Torsion, below which torsion

can be ignored in solid cross-section is (ACI

318):

For thin walled hollow sections, it is given by:


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Cracking Torque

Fig. 8.9 Example of an indeterminate structure where design torque can be

reduced


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For beams cast monolithically with a floor slab, the valuesAcpandpcpshould be calculated by including the parts of adjacent slabs of the

resulting T- or L-shaped beams (whereAcpis the area of the full concrete

cross section andpcpis the perimeter of the full concrete cross section).

The width of the slab that should be included is shown shaded in Fig.

8.10 and should not exceed the projection of beam above or below theslab or four times the thickness of slab whichever is smaller.

Consideration of Flanged Beams


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Consideration of Flanged Beams

Fig. 8.10 Consideration in the case of flanged beams


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To find out the area of stirrups that is necessary to resist torsion, let us

consider Figs 8.8(b) and 8.11(a). The angle of the cracks is initially takenat about 45 but may become flatter at higher torques. The ACI code

Clause 11.5.3.6 suggests taking the angle as 45, as this corresponds to

the assumed angle in the derivation of the equation for designing

stirrups for shear.

With reference to Fig. 8.8(b), the torsional resistance provided by the

member with a rectangular cross section can be found to be the sum ofthe contributions of the shears in each of the four walls of the

equivalent hollow tube.

Area of Stirrups for Torsion


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To find the torsional resistance, the shear flow or shear force per unit

length of the perimeter of the tube are obtained. Then the shear forces

acting in the right- and left-hand vertical walls of the tube and in the top

and bottom walls of the tube are determined. Assuming that the

stirrups crossing the crack are yielding, the shear in vertical walls aredetermined.



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Tests have shown that the concrete outside the stirrups is relatively

ineffective. Hence, the gross area enclosed by the shear flow path

around the perimeter of the tube after cracking may be defined in terms

of the area enclosed by the centre line of the outermost closed

transverse torsional reinforcement. Required area of stirrup steel:

When significant torsion is present, it is economical to select a larger

beam than a smaller one with closely spaced stirrups and longitudinal

steel required for the torsion design.



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Stirrups for Torsion

Fig. 8.11 Stirrups for torsion (a) Closed stirrup in rectangular beam (b) Closed stirrup in T-beam section


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Transverse Torsional Reinforcement

Fig. 8.12 Area enclosed by centre line of the outermost closed transverse torsional

reinforcement for rectangular, I, L, and box section beams

Area of Longitudinal Reinforcement for


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The longitudinal reinforcement must be proportioned to resist the

longitudinal tensile forces that occur due to torsion. It is required todistribute the longitudinal torsional steel around the perimeter of the

cross section.

Area of Longitudinal Reinforcement for

Torsion

Fig. 8.13 Free body of horizontal equilibrium


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Longitudinal Steel & Capacity

The required longitudinal steel is:

The capacity of rectangular cross-section is:

Limiting Crack Width for Combined Shear


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The space truss analogy assumes that all the torsion is carried by thereinforcements, without any torsion being carried by the concrete. The

codes often limit the maximum shear stresses (approximately c,max

= 0.631 fck) carried by stirrups in order to control the crack width

(where fckis characteristic cube compressive strength of concrete).

This concept is extended in the case of torsion too and an upper limit

of 0.6 fckplus the stress causing shear cracking is specified; this limit is

intended to control the crack width due to shear and torsion.

However, a better correlation is achieved when the square root of the

sum of the squares of nominal shear stresses is used.


and Torsion


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Limiting Shear Stress

The following equation is suggested for solid

sections with specified limit for crack control:



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and Torsion

Fig. 8.14 Addition of torsional and shear stresses (a) Hollow sections (b) Solid sections

Sk B di Th


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Skew Bending Theory

The skew bending theory assumes that some shear and torsion isresisted by the concrete and the rest by the shear or torsion

reinforcement. In this theory, the behaviour is studied on the basis of

the mechanism of failure, rather than on the basis of stresses.

Under the action of bending, the failure is vertical, with the primary

yielding of tension steel in under-reinforced beams and secondarycompression crushing of concrete.

Sk B di Th


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Skew Bending TheoryThe effect of adding even a little torque skews the failure surface. The

skewing is in the direction of the resultant momenttorque vector. Thecompression face is at an angle to the vertical face of the beam cross

section.

This compression failure can occur at the top, sides, or bottom of thebeam as shown in Fig. 8.15. Such a failure surface intersects some of the

stirrups, which essentially provide torsional resistance.

The tension steel may yield first followed by the stirrups. If both yieldbefore the crushing of concrete, the beam is under-reinforced. If the

concrete crushes before both types of steel yield, it is over-reinforced.

Sk B di Th


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Beams with large bending moment and small torsion fail with the

compression fibres crushing at the top; this type of failure is termed asMode 1 or modified bending failure(Fig. 8.15a). Mode 1 is the most

common type of failure and likely to occur in wide beams, even if the

torsion is relatively high.

If the beam is narrow (D >> b) and deep with equal amounts of top

and bottom steel, the failure may be by crushing at the sides. This

failure is termed as Mode 2 or lateral bending failure(Fig. 8.15b).

If the top longitudinal steel is much less than the bottom steel, the

failure may occur by crushing at the bottom fibre. This type of failure is

termed as Mode 3 or negative bending failure (Fig. 8.15c).

Skew Bending Theory

Sk B di Th


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Large torsion and low flexure may result in Mode 2 and Mode 3

failures. Large moment may force the Mode 1 failure. High shear andlow torsion sometimes result in Mode 4 failure. It is necessary to

investigate these several modes systematically and choose the lowest

capacity for a given beam.

In a square beam with symmetrical longitudinal reinforcement

subjected to pure torsion, the three modes will become identical.

The presence of shear in addition to the bending and torsion willcause the beam to fail at a lower strength. The Indian code attempts to

prevent such a possibility and suggests to design the beam using the

concept of equivalent shear.

Skew Bending Theory

Sk B di Th


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Skew Bending Theory

Fig. 8.15 Failure modes as per skew bending theory (a) Mode 1 (bending

and torsion) (b) Mode 2 (low shearhigh torsion) (c) Mode 3 (low bending

high torsion; weaker top steel) (d) Mode 4 (high shearlow torsion)

Interaction Curves for Combined Flexure


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and TorsionTorsion is normally accompanied by bending and shear. In general,

flexural and torsional shears are of significance in those regions wherethe bending moment is low.

Tests on rectangular, L-shaped, and T-shaped beams have indicated

that a quarter circle interaction relationship is acceptable for memberswithout web reinforcement.

For members with web reinforcement, the interaction curve is found

to be flatter than the quarter circle. The behaviour of asymmetricallyreinforced beams may differ significantly from that of symmetrically

reinforced beams.


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IS 456:2000 Provisions



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and Torsion

In pure torsion, the additional bottom longitudinal steel available inasymmetrically reinforced sections does not increase the ultimate

capacity because the weaker top steel is critical. The presence of

bending moment introduces compression in the weaker steel and

increases its resistance to the torsional shear stresses.

However, the presence of bending moment reduces the torsional

ductility of beams with symmetrical or asymmetric longitudinal steel. Ithas to be noted that the presence of torsion invariably reduces the

flexural strength of RC members.

T i Fl I t ti C


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TorsionFlexure Interaction Curves

Fig. 8.16 TorsionFlexure interaction curves for asymmetrically

reinforced members with transverse reinforcement

Interaction Curves for Combined Shear


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Interaction Curves for Combined Shear

and Torsion

Fig. 8.17 TorsionShear interaction (a) Experimental results (b) Curves in the literature

Indian Code Provisions for Design of


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g

Longitudinal and Transverse ReinforcementsThe Indian code provisions are based on the simplified skew bending

theory.

In this approach, the longitudinal and torsional reinforcements are not

calculated separately. Instead, the total longitudinal reinforcement iscalculated based on a fictitious, equivalent bending moment, which is a

function of the actual bending moment and torsion.

Similarly, the transverse reinforcement is determined from a fictitious,

equivalent shear, which is a function of the actual shear and torsion.

Indian Code Provisions for Design of


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g

Longitudinal and Transverse Reinforcements

In T-beams, the flanges are neglected and the beam is designed by

considering the rectangular web alone.

Clause 41.2 of the code also states that the sections located at a

distance less than the effective depth, d, from the face of the support

may be designed for the same torsion as computed at a distanced fromthe support.

Equivalent Shear and Moment


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For the case of pure torsion equal longitudinal

reinforcement is required at the top and

bottom of the rectangular beam.

The equivalent B.M. and equivalent Shear are

given by IS 456:2000 as (Clause 41.4.2):

Equivalent Shear and Moment

f f


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In order to avoid a brittle torsional failure, a

minimum amount of torsional reinforcement

(including both transverse and longitudinal

steel) is required in a member subjected to

torsion.

Minimum Reinforcement for Torsion

Design of Transverse Reinforcements


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The code assumes that both the longitudinal

and transverse steel reach design strength

before failure occurs.

The area of two-legged closed stirrups are

calculated as (Clause 41.4.3):

Design of Transverse Reinforcements

The code also specifies the following minimum limit:

Distribution of Torsional Reinforcement


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The longitudinal reinforcement for torsionshould be placed as close as possible to the

corners of the cross section.

At least one longitudinal bar should beplaced at the corners of the stirrups.

The hooks of the closed stirrup should be

developed into the core with 135 bendsotherwise the corners of the beam may spall

off.

Distribution of Torsional Reinforcement

Recommended Closed Stirrups for Torsion


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Recommended Closed Stirrups for Torsion

Fig. 8.18 Recommended closed stirrups for torsion

Ineffective Closed Stirrups for Members


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p

under High Torsion

Fig. 8.19 Ineffective closed stirrups for members under high torsion

Design and Detailing for Torsion as


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g g

per IS 456 Code

The following design steps are required for the design of flexural and

shear reinforcement as per IS 456:

1. Determine the equivalent bending moment, Me1and equivalent

shear.

2. Calculate the required longitudinal steel for Me1

.

Design and Detailing for Torsion as


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g g

per IS 456 Code

3. Check for shear. Calculate the equivalent shear stress, ve. The valueof veshould not exceed the value of c,maxas given in Table 20 of

the code; if it exceeds, revise the section or increase the grade of

concrete.

4. Calculate the transverse reinforcement.

5. Check the spacing: It should not exceedx1,

(x1 + y1)/4, and 300 mm

6. Check if side face reinforcement is required. If the size is greater than

450 mm, provide 0.05 per cent side face reinforcement at each face.

Graphical Methods


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Graphical Methods

Two graphical methods have also been developed as follows:

1. Rahal developed a simplified method for combined stress

resultants based on the MCFT (Rahal 2007).

2. Leu and Lee (2000) proposed a graphical solution to the softened

truss model developed by Hsu (1988).

Rahals Method


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Rahalsgraphical method is applied to beams

subjected to torsion by idealizing the section as a

hollow tube and by adopting simplified assumptions

regarding the thickness of the hollow tube and thesize of the shear flow zone.

Figure 8.20 gives the relationship between the

reinforcing indices and the normalized shear strengthobtained using the results of the MCFT.

Rahal sMethod

Rahals Method


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Rahal sMethod

Figure 8.20 shows a curve passing through those points beyond whichconcrete crushes before the transverse steel yields (over-reinforced

case). The figure also shows a similar curve for the over-reinforced case

in the longitudinal direction.

The two balanced yield curves divide the graph into four regions. The

relative position of a point of coordinates (reinforcing indices) withrespect to these curves or regions indicates the expected mode of

failure of an element with these reinforcement ratios.

Graphical Methods


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Graphical Methods

Fig. 8.20 Normalized shear strength curve for RC members

Rahals Method


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Rahal sMethodAs shown in the figure, 4 modes of failure are possible:

1. Partially over-reinforced (only longitudinal steel yields, Mode 3)

2. Partially over-reinforced (only transverse reinforcement yields,

Mode 2)

3. Completely over-reinforced (concrete crushing before steel

yielding, Mode 4)

4. Completely under-reinforced (longitudinal and transverse steel

yield, Mode 1)

Other Considerations


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Other ConsiderationsThe following are the other considerations that should be taken into

account:1. Maximum yield strength of torsional reinforcement

2. High-strength concrete

3. Lightweight concrete

4. Size effect

5. Precast L-shaped spandrel beams


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