Torsion. Design of Reinforced Concrete Structures I

Transcription

1 Design of Reinforced Concrete Structures I Torsion Table of Contents 6.1 Introducon Torsional Stresses When Torsional Reinforcing Is Required by the ACI Torsional Moment Strength Torsional reinforcement... 9 Msc. Wissam Nadir [Company name] [Date]

2 Chapter six Torsion Page 1 of Introducon The average designer probably does not worry about torsion very much. They thinks almost exclusively of axial forces, shears, and bending moments, and yet most reinforced concrete structures are subject to some degree of torsion. Appreciable torsion does occur in many structures, such as in the main girders of bridges, which are twisted by transverse beams or slabs. Earthquakes can cause dangerous torsional forces in all buildings. This is particularly true in asymmetrical structures, where the centers of mass and rigidity do not coincide. Other cases where torsion may be significant are in curved bridge girders, spiral stairways.

3 Page 2 of Torsional Stresses The torsional stresses add to the shear stresses on one side of a member and subtract from them on the other. This situation is illustrated for a hollow beam in Figure. Torsional stresses are quite low near the center of a solid beam. Because of this, hollow beams (assuming the wall thicknesses meet certain ACI requirements) are assumed to have almost exactly the same torsional strengths as solid beams with the same outside dimensions. In solid sections, the shear stresses due to the torsion, Tu, are concentrated in an outside tube of the member, as shown in Figure above (a), while the shear stresses resulting from Vu are spread across the width of the solid section, as shown in part (b) of the figure. As a result, the two types of shear stresses (those from shear and torsion) are combined, using a square root expression shown in the next section of this chapter.

4 Page 3 of 14 After cracking, the resistance of concrete to torsion is assumed negligible. The torsion cracks tend to spiral around members (hollow or solid) located at approximately 45 angles with the longitudinal edges of those members. Torsion is assumed to be resisted by an imaginary space truss located in the outer tube of concrete of the member. Such a truss is shown in Figure below. The longitudinal steel in the corners of the member and the closed transverse stirrups act as tension members in the truss, while the diagonal concrete between the stirrups acts as struts of compression members. The cracked concrete is still capable of taking compression stresses.

5 Page 4 of When Torsional Reinforcing Is Required by the ACI The design of reinforced concrete members for torsion is based on a thin-walled tube space truss analogy in which the inside or core concrete of the members is neglected. After torsion has caused a member resistance the torsion by almost entirely closed stirrups and the longitudinal reinforcing located near the member surface. Once cracking occurs, the concrete is assumed to have negligible torsional strength left. In ACI , it is stated that torsion effects may be neglected for nonprestressed members if < 1 4 Where: = = 3! ="h $ =0.75 = Equals the area of the entire cross sections (including the area of any voids in hollow members). =) h + " =Represents the perimeters of the entire cross sections. =2 -)+h/ + "

6 Page 5 of 14 If the beam cast monolithically with a slab, the values of and may be assumed to include part of the adjacent slabs of the resulting T- or L-shaped sections. The widths of the slabs that may be included as parts of the beams may not exceed the projections of the beams above or below the slab (h w ) or four times the slab thickness, whichever is smaller. When appreciable torsion is present, it may be more economical to select a larger beam than would normally be selected so that torsion reinforcing does not have to be used. Such a beam may very well be more economical than a smaller one with the closed stirrups and additional longitudinal steel required for torsion design. On other occasions, such a practice may not be economical, and sometimes-architectural considerations may dictate the use of smaller sections.

7 6.4 Torsional Moment Strength Page 6 of " 1 " $ The ACI Code limits the sizes of members subject to shear and torsion so that unsightly cracking is reduced and crushing of the surface concrete caused by inclined compression stresses is prevented. This objective is accomplished with the condition that follow. To ensure that solid sections fail in a ductile manner, the following condition should satisfied: 23 4 ) 5 $ ! ) 5 $ To ensure that hollow sections fail in a ductile manner, the following condition should satisfied: 3 4 ) 5 $ ! ) 5 $ Where: = > =Is the perimeter of the centerline of the outermost closed torsional reinforcing. 7 =2-? +A / B C> =is the cross-sectional area of the member that is enclosed within this centerline. 87 =-? A /

8 Page 7 of 14 Note: 1- Should the wall thickness of a hollow section be less than A oh /p h, the second term in ACI Equaon is to be taken not as D EF G H.I J L but as KG D E H.I J KG, Where (t) is the thickness of the wall where stresses are, being checked (ACI ). 2- For the hollow sections is that the distance from the centerline of the transverse torsion reinforcing to the inside face of the wall must not be less than 0.5A oh /p h

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10 Page 9 of Torsional reinforcement The torsional strength of reinforced concrete beams can be greatly increased by adding torsional reinforcing consisting of closed stirrups and longitudinal bars. If the factored torsional moment for a particular member is larger than the value given in ACI Secon H M the code provides an expression to compute the absolute minimum area of transverse closed stirrups that may be used. Minimum torsion reinforcement - N +2 O /=0.062 ) 5Q RO ) 5Q RO ACI Where: N = is the area of reinforcing required for shear in a distance(s). O =which represents the area of the stirrups needed for torsion, is for only one leg of the stirrup. Therefore, the value - N +2 O / is the total area of both legs of the stirrup (for two legged stirrups) needed for shear plus torsion. It is considered desirable to use equal volumes of steel in the stirrups and the added longitudinal steel so that they will participate equally in resisting torsional moments. V = 2 8 O RO Q cotz ACI This equation is usually written in the following form: O Q = Where: V = D E V 2 8 RO cotz Is the ultimate applied torque O Area of one leg of stirrups RO Yield stress of stirrups must not exceed 420 [ Also 8 = and Z =45 $

11 Page 10 of 14 As given in the ACI Commentary (R ), the required srrup areas for shear and torsion are added together as follows for a two-legged stirrup: + 3 N\O Q 6= N Q +2 O Q The spacing of transverse torsional reinforcing may not be larger than " ]^_ =1. ` 7/8 300 Remember also the maximum spacing of srrups for shear d/2 and d/4 given in ACI Secons and

12 Longitudinal reinforcement Page 11 of 14 Furthermore, extra longitudinal reinforcement is also added as it capable of resisting torsion as follows: b = O " 7 RO R!cot Z ch R =Is the yield stress of longitudinal reinforcement However, the amount of longitudinal torsional reinforcement b must not be less than the minimum longitudinal torsional reinforcement b,]ev, which is given as follows: b,]ev = 0.42 R 3 O " 6 RO 7 ACI R Note that O /" shall not be taken less than 0.175) 5 / RO The longitudinal reinforcing must be distributed around the inside perimeter of the closed stirrups and must be spaced no farther apart than 300 mm. At least one bar must be placed in each corner of the srrups to provide anchorage for the stirrup legs. These bars have to be 10 mm or larger in size, and they must have diameters no less than mes the srrup spacings (ACI ).

13 Page 12 of 14 Ex: Design the torsional reinforcing for the beam shown in Figure, for which =28 [g,a 420 [g,4=190 h $ =30 h.1. Assume Ø13 srrups and a clear cover equal to 40 mm. Solution 1- Is the torsional reinforcement necessary? < 12! = = = /= ! 10jk =8.55 h.1 < =30 h.1 Torsional reinforcement is required 2- Is the concrete section sufficient large to support? 23 4 ) 5 $ ! 87 4 =190h,) 5 =35011 $ =30h.1 $ = = ? H = =25811 A H = = = = ) 5 $ =? H A H 8 = = = =2 -? H +A H /= /=

14 4 = 6 ) 5$= jl = h Page 13 of k ! 1.67 [g <3.3[g Section is sufficiently large 3- Determine the torsional reinforcement required V = = 30 =40 h O Q = V 2 8 RO cotz = = k "g 4- Determine the shear reinforcement required 4 =190h> 4 = h 4 n = 4 4 = = h 0.75 N Q = 4 n A $ = = o Select stirrups spacing Shear reinforcement is required + 3 N\O Q 6= N Q +2 O Q = = o 11 c +" "g

15 6- Check the minimum area of stirrup - N +2 O /=0.062 ) 5Q ) 5Q RO - N +2 O / " - N +2 O / " " -J p\ J q / n =0.062 ) 5 RO ) 5 RO RO = = =0.29 = o 11 < + r J psq u=1.075 t Page 14 of 14 " "= Q= Check the allowable spacing of stirrups = n =72.6 h< 1 3 ) 5 $=361.45h " ]^_ =1. y 7 w8,=1632 = x w v $ 2 =585.5 = S>Smax. Use Ø12 mm