II. MARCO METODOLÓGICO
2.3. Metodología
7.4.1 Design Torsional Strength without Torsional Reinforcement As already indicated, the strength of a torsionally reinforced member at torsional cracking is practically the same as the failure strength of a plain concrete
under pure torsion. Although several methods have been developed to compute , the plastic theory approach is described here, as the Code can be explained on its basis.
Cracking Torque
As explained earlier, the assumption that the unreinforced section is fully at the point of failure implies that the shear stress is constant throughout the section, having a magnitude [Fig. The resultant shears are obtainable from the shear flow diagram, as shown in 7.5.
The resultant horizontal shear is simply obtained multiplying the tributary area, (equal to b 14) by Similarly, the resultant vertical shear is obtained by multiplying the trapezoidal area by
.
The two equal and opposite forces a couple which has a lever D-
Similarly, the two equal and opposite force form another couple, with a lever which can be shown to be:The of the two gives the desired value of
DESIGN FOR 275
Fig. 7.5 Plastic theory to determine
The above relation also be derived using the so-called 'sand heap' analogy.
Evidently, the assumption of full plastification of the section is not justified for a material like concrete. Hence, n correction factor has to be applied, by either modifying the expression in Eq. 7.5, or by using a reduced value of (which is equal to the of concrete)
-
to with experimental Test results indicate an ultimate value of units) of about to be used with Eq.Torsional Shear Stress
As the hehaviour cracking is linear
[Fig. 7.41, torsional shear Z, corresponding to any factored may be obtained 7.5 for a concrete rectangular section
Extending this expression for a reinforced concrete with effccuve 7.6
=
-
X --- constantwhere the constant, equal to takes values in range 0.8 1.15 for most rectangular sections practice. Considering an average value for this constant and further a factor for the assumption of full
276 REINFORCED CONCRETE DESIGN
olastification of the section, the above expression reduces to the following simplified form, given in the Code: exceeds the shear strength of the plain concrete Where flexural shear occurs in combination with torsional shear (as is commonly the case), the combined shear stress (flexural plus torsional) has to bc considered. For this purpose, the term equivalent shear is used by the Code (CI. 41.3.1) to express the combined shear effects on a reinforced concrete beam, subject to flexural shear and shear:
=
+
1.6-b
It may be noted that the shear' due to and are additive only on one side of the beam; they act in opposite directions on other side.
The equivalent nominal shear stress, is given by
If exceeds [refer Section 6.61, the section has to be suitably redesigned increasing the cross-sectional area (especially width)
the grade of concrete. If is less than the design shear strength of concrete [refer Section 6.61, minimum reinforcement has to he provided, as explained in Section If the value of lies between 7, and , suitable torsional reinforcement (both transverse and longitudinal) has to bc
for the combined of shear and
,
must be to express V,, and T,, in consistent in N and in b in and d i n
FOR TORSION 277 7.4.2 Design Torsional S t r e n g t h with R e i n f o r c e m e n t
Several theories have been proposed for computation of the torsional strength of reinforced concrete members with torsional reinforcement notably the space-truss
and the skew bending theory [Ref. 7.9 - 7.121.
Space Truss A n a l o g y
The space truss analogy is essentially an extension of the plane truss analogy [Fig. 6.91 used to explain flexural shear resistance. The 'space-truss model' (illustrated in Fig. 7.6) is an idealisation of the effective portion of the beam, comprising the longitudinal and transverse torsional reinforcement and the surrounding layer of concrete. It is this 'thin-walled tube' which becomes fully effective at the post-torsional cracking phase. The truss is made up of the corner longitudinal bars as stringers, the closed stirmp legs as transverse ties, the concrete between diagonal cracks as compression diagonals.
For a closed thin-walled tube, the shear flow q (force per unit length) the thickness of the tube [Ref. is given by:
where A, is the area enclosed by the centreline of the thickness. Tho proof for 7.10 is indicated in Fig. For the box section under consideration,
where and denote the centre-to-centre distances between the comer bars in the directions of the width and the depth respectively. Accordingly, substituting Eq. 7.11 in Eq. 7.10,
I;,
Assuming torsional cracks (under pure torsion) at 45" to the longitudinal axis of the beam, and considering equilibrium of forces to section AB [Fig.
the total force in each is given by tan where is the spacing of the (vertical) stirrups. Further, assuming that the stirmp has yielded in tension at the ultimate limit state (with a design stress of it follows from force equilibrium that
where A, is the cross-sectional of the (equal to for two legged stirrups). Substituting Eq. 7.12 in the above equation, the following expression is obtained for the ultimate strength = in torsion:
=
Further, assuming that the longitudinal steel (symmetrically placed with respect to the beam axis) has also yielded at the ultimate limit state, it follows from longitudinal force equilibrium [Fig. that:
.
CONCRETE DESIGN
(a) space-truss model
(b) detail 'X' (c) thin-walled tube
(box section) 7.6 The idealised space-truss model
= ---- tan 45"
is the total of the longitudinal steel and its yield strength.
Eq. 7.12 in the above equation, the is obtained for the ultimate strength in torsion:
= (0.87
+
(7.16)FOR TORSION 279 The two alternative expressions for Eq. 7.14 and Eq. 7.16, will give identical results only if the following the areas of steel and transverse steel (as torsional is
If given Eq. 7.17 is not satisfied, then may he by combining. 7.14 and Eq. 7.16 [Ref. taking into account the areas of both transverse and longitudinal
To that the member does not in a brittle after the development of torsional torsional of cracked reinforced section he at equal to cracking torque (computed without considering any safety
Skew Bending Theory
The behaviour of concrete may be alternatively
studied on the basis of the of rather on basis of [Ref. 7.10, In the consideration of the failure action of torsion flexure shear to be taken into account.
Three modes of failure have been identified for subjected to
flexure and torsion [Fig. action of torsion is to the failure surface (which is otherwise vertical under action of flexure alone); the skewing is in the direction of the resultant moment-torsion vector. type of failure is as shown in Mode 1 [Fig. - with bending over torsion and the compression zone (shown shaded) on top, albeit skewed 45'). This type of failure called 'modified failure') will occur in wide beams, even if torsion is relatively high. However, if a beam with a section is subject .to predominant torsion, a Mode 2 type of failure [Fig. is likely, with the compression zone skewed to a side of the section; type of failure is sometimes called a failure. A third of failure Mode 3 [Fig.
-
is possible when compression zone at the bottom and the area of the longitudinal top is much less than that of the bottom steel; this typeof is sometimes called a failure.
280 REINFORCED CONCRETE DESIGN
moment vector
(a) rectangular beam section (under combined flexure-torsion)
in tension
(c) MODE 2
(lateral bending failure)
bars in tension (b) MODE
(modified bending
(d) MODE 3
(negative' bending failure)
Fig. 7.7 Failure Modes for combined flexure and torsion
In a beam with a square cross-section, with longitudinal reinforcement, subiected to torsion, the three modes become identical.
Expressions the strength in torsion have been derived for each of the three modes of failure. The interested reader is advised to refer to 7.13 (or Ref. 7.14) for the derivation of these expressions. It is customary to check for all the three modes and to choose the lowest value of the torsional strength.
It may be noted that the presence of shear may a beam to fail at a lower strength. The Code attempts to prevent the possibility of suchshear type nf failure by the concept of designing for shear [refer Section
7.4.3 Design S t r e n g t h in Torsion C o m b i n e d Flexure
DESIGN FOR TORSION 281
respectively the strengths of the member under pure torsion and pure respectively. The following parabolic interaction formulas [Ref. have been proposed, based on experimental studies on rectangular reinforced
Mode failure
Mode 3 failure
where A, and A,' denote, respectively, the areas of longitudinal steel provided in the 'flexural tension zone' and 'flexural compression zone' of the rectangular beam section
7.8 Torsion-Flexure Interaction
The torsion-flexure interaction based on Eq. 7.19, are depicted in Fig. 7.8 for in the range 0.3 to 1.0. Each curve represents a 'failure envelope', in the sense that any combination of and that falls outside the area bounded by the curve and the coordinate axes is 'unsafe'. In general, it is seen that the torsional strength increases beyond the 'pure torsion' strength in the presence of bending moment -provided is low and is also low.
In such cases, failure may occur in Mode 3 initiated by the yielding of the compression steel) at very low values of [Fig. In general, however, a Mode failure is likely to occur initiated by the yielding of the 'tension' steel);
this becomes inevitable when equal to A,.
REINFORCED CONCRETE DESIGN
It may also be noted from 7.8 that the presence of brings down the flexural strength of the reinforced concrete member.
IS Code for of Longitudinal Reinforcement
The Code (C1.41.4.2) recommends a simplified skew-bending based' formulation for the design of longitudinal reinforcement to resist torsion combined with flcnure in beams with rectangular sections. The torsional moment is convertedinto an effective bending moment defined
r
as follows:M,
=+
(7.20)where D is the overall depth and b the width beam.
so calculated, is combined with the actual bending moment at the section, to give 'equivalent bending moments', and
The longitudinal area A,, is designed to the equivalent moment and this steel is to be located in the 'flexural tension hi addition, if M, then a reinforcement area A,,' is to be designed to resist this equivalent moment, and this steel is to be located in the 'flexural compression zone'.
It follows from the above that in limiting case of 'pure torsion'
equal longitudinal reinforcement is required at the top and bottom of the rectangular beam, each capable of resisting an equivalent bending moment equal to
7.4.4 Design Strength in Torsion Combined with Shear
Tonion-shear interaction curves have been proposed [Ref. similar to interaction curves. In general, the interaction between and takes the following form:
and are the given twisting moment and factored force and are the ultimate in 'pure torsion' and 'flexural shear' (without torsion) respectively;
a
is a constant, for which values in the range 1 to 2 have been proposed [Fig. A value ofa
equal to unity results in a linear interaction and generally provides a conservative estimate.This formula can alternatively be generated from the space analogy [Fig. by the longitudinal tensile forces in the bars (located either at top or at bottom) as those required to resist an effective bending which can be shown to be equal to
the Code has simplified this to the form given in 7.20.
DESIGN FOR
Fig. 7.9 Torsion-shear interaction
Code Provisions for Design of Transverse Reinforcement
The Code provisions 41.4.3) for the design of transverse reinforcement (2-legged, closed) are on ihc and are at resisting a Mode 2 failure [Fig. by a large combined with a small shear:
where A,, is the total nrca of two legs of the stirrup: is
of the stirrups; and are the centre-to-centre between the along the width depth respectively; and and are the factored
and Eq. 7.23 becomes exactly equivalent to Eq. 7.14, which was derived using the space-truss
addition to Eq. 7.23, Code 41.4.3) specifies a limit to the area of transverse reinforcement:
where is the 'equivalent shear stress' by Eq. 7.7. The purpose of Eq. 7.24 is to provide adequate resistance against flexural shear failure, which is indicated in situations where is negligible in Indeed, for the extreme case of = 0, 7.24 becomes exactly equivalent to Eq. 6.25, which was derived for flexural shear. It that the contribution of inclined and bent up bars can be included in the calculation of A,, in Eq. 7.24, but not Eq. 7.23.
Distribution of Torsional Reinforcement
The Code specifics maximum limits to the of the stirrups provided as torsional reinforcement - ensure the of post-cracking
284 REINFORCED CONCRETE DESIGN
torsional resistance, to control crack-widths and to control the fall in torsional stiffness on account of torsional cracks:
+ (7.25)
mm
where and are, respectively, short and long centre-to-centre dimensions of the closed The spacing . should satisfy all the limits given in Eq. 7.25.
The Code also recommends that "longitudinal shall be placed a s close a s is practicable to the comers of cross-section,
cases, there shall be least longitudinal bar corner of ties".
Further, if the torsional member has a cross-sectional dimension (usually, overall depth rather than width) that exceeds additional longitudinal bars are required to be provided as side face reinforcement, with an area not less than 0.1 percent of the web area. These bars are to be distributed equally on the two faces at a spacing not exceeding 300 or web thickness, whichever is less.
7.5 ANALYSIS AND DESIGN EXAMPLES