The principal stress (tensile strength) concept would suggest that failure cracks should develop at each face of the beam along a spiral running at 450 to the beam axis. However, this is not possible because the boundary of the failure surface must form a closed loop. Hsu has suggested that bending occurs about an axis parallel to the planes that is at approximately 450 to the beam axis and of the long faces of a rectangular beam. This bending causes compression beam. The latter tension cracking eventually and tensile stresses in the 450 plane across the initiates a surface crack. As soon as flexural occurs the flexural strength of the section is reduced, the crack rapidly propagates, and sudden failure follows. Hsu observed this sequence of failure with the aid of high-speed motion pictures. For most structures little use
can be made of the torsional (tensile) strength of unreinforced concrete members.
8.2.3 Tubular Sections
Because of the advantageous efficient in resisting distribution of shear stresses, tubular sections are most resisting torsion. They are widely used in bridge construction .Figure8.7 illustrates the basic forms used for bridge girders. The torsional properties of the girders improve in progressing from Figs. 8.7a to 8.7g.
When the wall thickness h is small relative to the overall dimensions of the section, uniform shear stress across the thickness can be assumed. By considering the moments exerted about a suitable point by the shear stresses,acting over infinitesimal elements of the tube section, as in Fig. 8.8a, the torque of resistance can be expressed.as
The product hvt = vo is termed the shear flow,.and this is constant; thus
where Ao = the area enclosed by the center Jine of the tube wall (shaded area in Fig. 8.8). The concept of shear flow around the thin wall tube is useful when the role of reinforcement in torsion is considered.
The ACI code 8.2 suggests that the equation relevant to solid sections. 8.8, be used also for hollow sections, with the following modification when the wall thickness is not less than x/l0 (see Fig. 8.8c):
where x ≤y.
Equation 8.9b follows from first principles and has the advantage of being applicable to both the elastic and fully plastic state of stress.
The torque-twist relationship for hollow sections may be readily derived from strain energy considerations. By equating the work done by the applied torque (external work) to that of the shear stresses (internal work), the torsion constant CO for tubular sections can be found thus:
Hence by equating the two expressions and using Eq. 8.9b, the relationship between torque and angle of twist is found to be
and the torsional stiffness of such member is therefore
where C0 is the equivalent polar moment of inertia of the tubular section and is given by
where s is measured around the wall centerline. The same expression for the more common form of box section (Fig. 8.8b) becomes
For uniform wall thickness Eq. 8.11 reduces further to
where p is the perimeter measured along the tube centerline.
It is emphasized that the preceding discussion on elastic and plastic behavior relates to plain concert .and the propositions are applicable only at low load intensities before cracking. They may be used for predicting the one of diagonal cracking.
8.3 BEAMS WITHOUT WEB REINFORCEMENT SUBJECT TO
FLEXURE AND TORSION
The failure mechanism of beams subjected to torsion and bending depends on the predominance of one or the other. The ratio of ultimate torque to moment, TJ/MU,is a suitable parameter to measure the relative magnitude of these actions. The flexural resistance depends primarily on the amount of flexural reinforcement. The -torsional behavior of a concrete beam without web reinforcement is more difficult to assess in the presence of flexure.
Flexural stresses initiate diagonal cracks in the case of torsion, much as they do in the case of shear. In the presence of flexure these cracks are arrested in the compression zone. For this reason a diagonally cracked beam is capable of carrying a certain amount or torsion. The manner in which this torsion is resisted is, at present, a matter, of speculation. Clearly the compression zone of the beam is capable of resisting a limited amount of torsion,.and horizontal reinforcement can also contribute to torsional resistance by means of dowel action.
It has been found (e.g., by Mattock\approximately one-half the ultimate torsional strength of the uncracked section, provided a certain -amount of bending is present.
Thus one half the torque causing cracking can be sustained after the formation of cracks. The torque thus carried is so small that its influence on flexure on can be ignored.
The nominal torsional shear stress, corresponding to this limited torsion is conservatively assumed by ACI 318-71. to be 40 % of a cracking stress of
and the torque supplied by the .concrete section only, after the onset of cracking, is revealed by Eq. 8.8 to be
Similarly, for compound sections, Eq. 8.8a gives
with the limitations on overhanging parts as indicated in Fig. 8.4.
When T./M > 0.5 (i.e., when torsion is significant), brittle failure has been observed. When the bending moment is more pronounced, (i.e., when T/Mu < 0.5), a more ductile failure can be expected. The torsional strength of abeam can be increased only with the addition of web reinforcement. The amount of flexural reinforcement appears to have no influence on the torsional capacity of the concrete section, T .
In T or L beams the overhanging part of the flanges contribute to torsional. strength. This has been verified on isolated beams. The effective width of flanges, when these are part of a floor slab, is difficult to assess.When a yield line can develop along an edge beam because of negative bending moment in the slab, as illustrated in Fig. 8.9 it is unlikely that much of the flange can contribute toward torsional strength.