Minimum steel ratios in reinforced concrete beams made of

Minimum Steel Ratios In Reinforced Concrete Beams Made Of-Free PDF

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Materials and Structures Mat riaux et Constructions Vol 36 January February 2003. Generally minimum steel ratios are provided in rein. forced concrete elements for one or more reasons duc. tility and residual strength at collapse minimum strength. at ULS minimum strength at PCLS crack distribution. and durability, In this work a theoretical approach for the minimum. longitudinal and transversal steel ratios in beams sub. jected to flexure shear and torsion associated only with. the ductility and minimum strength at ULS is pre, sented The ductility through out this work is under. stood as the capacity of a section a structure member or. a structure as whole to undergo a reasonable amount of. plastic deformation without significant loss of strength. during its collapse ULS Fig 1 Stress distribution in a rectangular concrete section. under cracking moment,2 MINIMUM FLEXURAL STEEL RATIO IN. FLEXURE a minimum moment of resistance M and a certain. amount of ductility at failure of a beam should satisfy. From the previous definition of ductility the mini both Equations 1 and 2. mum flexural steel reinforcement can be any tensile longi The previous analysis is valid only if the beam is pre. tudinal steel ratio that is smaller than the balance steel ratio cracked in f lexure If on the other hand the beam is. provided that the steel deformation is less than the ultimate uncracked the following analysis should be considered. steel deformation deformation at steel rupture i e taking into consideration the cracking moment Mcr. A M M In this case it is important to guarantee that if the. s min s min s bal 1, b d b d z fy b d d 0 5 x fy cracking moment Mcr is reached due to eventual over. loading the forces resisted by concrete in tension is. with transmitted to tensile longitudinal steel capable of resist. cu fc ing Mcr i e,cu y fy s min b d fy z M cr, where are the parameters of the rectangular stress M cr.
block 0 85 and 0 8 for normal strength concrete s min 3. Equation 1 is valid for any concrete strength pro, vided that the rectangular stress block parameters The value of the cracking moment M cr can be. and the ultimate concrete strain cu are adjusted to the obtained from the linear elastic stress distribution along. strength used The CSA A23 3 6 that covers concrete the section and the tensile strength of the concrete. with strength up to 80 MPa for example propose Fig 1b In the case of rectangular cross section without. 0 85 0 0015fc 0 67 0 97 0 0025fc 0 67 normal forces ignoring shrinkage and temperature. and a constant cu 3 5 effects the neutral axis depth is x 0 5 h and therefore. In order to avoid the sudden rupture of the reinforce the cracking moment can be written as. ment when the concrete reaches its ultimate strain cu. fct b h 2 4a, the maximum steel deformation in the section should be M cr 0 167 fct b h 2. limited to ultimate strain su 6,or in the non dimensional form. s max cu su M cr f,x cr 0 167 ct 4b,b h 2 fc fc, or s min Considering the flexural tensile strength of concrete. d fc su as given by the CEB FIP MC90 9,and consequently h.
fct f 0 2 fck 0 7,cu the non dimensional cracking moment cr can be. Then the flexural minimum steel ratio that provides expressed as a function of fck. Shehata Shehata Garcia, plastic moment can be evaluated from the plastic tensile. strength of concrete given as fctp fct where 1 0 is the. effectiveness factor The value of this factor is obtained by. equating the areas underneath the stress strain diagrams for. the actual and idealised elasto plastic diagrams see Fig 2. for a given relation between the ultimate deformation tu. and the deformation of the concrete o at peak stress fct. Fig 3 gives the values of obtained as function of the. ratio tu o for a range of variation of this ratio between. 1 5 and 2 0 and the stress strain curves given in Fig 2. Considering that the concrete modulus of elasticity is. the same for tension and compression from the equilib. rium of the section the plastic moment at failure is. Fig 2 Assumed actual and plastic stress strain curves for con. crete under uniaxial tension 1 x x ycg,Mp fct f b h 2 1 2 3. ycg 1 1 tu h x for tu 2,c fct f tu,Assuming 1 85 which corresponds to 2 3. Fig 3 Variation of the effectiveness factor with the relationship o. between the maximum deformation and the deformation at peak. stress for concrete in tension one can obtain see Fig 3. ycg 0 302 h position of the resultant of the tension. 0 7 forces with respect to neutral axis,h c 1 36 fct f.
0 33 Mp 0 168fct f b h2,cr 0 0333 fck 0 7, 6 This moment practically the same as the cracking. moment M cr given by expression 4a leads to s min, 100 similar to the one obtained from expression 7b. Fig 4 shows comparisons made between s min fy as, In the case of prestressed beams Mcr is evaluated tak. ing into consideration the normal force,Admitting z 0 8h and d 0 85h from expressions. 3 and 4a one can get to,or as a function of fck,s min 0 05 0 7.
Fig 4 Comparison between Equation 7b and the proposals of. On the other hand if the elasto plastic tensile stress some codes for the longitudinal minimum steel in concrete. distribution is considered in the ULS Fig 1c the collapse beams for h 500 mm. Materials and Structures Mat riaux et Constructions Vol 36 January February 2003. Fig 7 Shear stress distribution in a rectangular concrete. Fig 5 Comparison between Equation 7b and available test cross section at the instant of diagonal crack formation. results 4 5, The influence of the beam height h and the concrete. strength fc on the cracking moment and on the mini. mum steel ratio to insure ductile behaviour of beams at. failure have been investigated experimentally 4 5 and. analytically 11 but there are divergences between the. obtained results The analysis of the results of Bosco et al. 4 and 5 made by Queir z 12 have indicated that the. inf luence of h on the non dimensional cracking, moment Mcr bh2fct is restricted to a small range of vari. ation of h For practical values of h h 300 mm this. influence can be ignored when compared to that of fc. The expression for s min based on the fracture, mechanics approach suggested by Bosco et al 4 and 5. Fig 6 Comparison between Equation 2 su su 20 and 30. indicates a decrease in the minimum steel ratio as the. and Equation 7b h 200 mm and 1000 mm beam height h increases in the proportion of h 0 5. According to Oz bolt and Bruckner 11 this only hap. pens up to a certain value of h if the beams are not pro. given by expression 7b for h varying between 200 mm vided with distributed reinforcement along their heights. and 1000 mm and by the proposals of five codes of prac In addition to that in high beams without distributed. tice 1 3 6 9 and 10 To obtain the values of s min fy for steel higher minimum steel ratios are required in order. CEB FIP MC 90 9 in Fig 4 fy 500 MPa was used It to avoid the sudden loss of resistance at ultimate load. can be seen from this figure that expression 7b leads to Ruiz et al 14 on the other hand concluded that. values of s min fy less than those suggested by the codes besides the beam height and the concrete strength the. except for the CEB FIP MC 90 and the proposed values minimum f lexural steel ratio depends also on the steel. for the minimum steel ratios given by these codes are type and the concrete cover which in part substantiate. greatly different from one another Equation 2,Fig 5 shows a comparison between expression 7b. and the results of the beams tested by Bosco et al 4 and. 5 with height h varying between 100 mm and 800 mm 3 MINIMUM SHEAR STEEL RATIO. and yielding moments M y close to the cracking, moments In Fig 6 Equation 7b is also compared with 3 1 Analysis based on the diagonal cracking.
Equation 2 Two values of the ratio su cu were used load. in Equation 2 20 and 30 in order to represent two lev. els of steel ductility medium and high The adopted val Considering a concrete beam with rectangular cross. ues of and were as proposed by the CSA 23 3 6 section subjected to shear and moment from the elastic. From Figs 5 and 6 it can be concluded that expres shear stress distribution second degree parabola. sion 7b can be used to determine the minimum longi Fig 7b the shear crack appears when max fct and the. tudinal steel ratio for beams reinforced with high ductil associated shear force is calculated as. ity steel su cu 30 In the case of beams reinforced. with medium and low ductility steel Equation 2 2 2. becomes dominant and should be used to define the Vcr b h max b h fct 8. minimum reinforcement It is worth noting that the val. ues of s min fy given by Equation 2 with su cu 20 When an axial force is applied to the beam resulting. lie in between those given by the ACI 318 99 1 and the in an average normal stress equal to cp positive when. NS 3474E92 10 shown in Fig 4 compressive the maximum shear stress that causes the. Shehata Shehata Garcia, first diagonal shear crack according to the modified nal crack occurs in a region with no flexural cracks for. Mohr Coulomb criteria is example in the case of fully prestressed beams In beams. with a rectangular cross section and without shear rein. ct fct cp 9 forcement and normal forces test results have shown. that the diagonal crack in a pre cracked beam in flexure. and the associated shear force is starts from or joins an existing f lexural one and at a. shear force lesser than the one given by expression 8. The analysis of pre cracked beams in f lexure is dealt. fct cp 10 with in a later section, Given that in both cases when the shear force Vcr is. reached a sudden rupture of the beam will occur with a 3 2 Analysis based on the truss model. total loss of strength it is important that the beam con. tains a minimum shear reinforcement to provide it with From the truss analogy the stress field in the web of. certain ductility at failure Considering the failure plane an uncracked concrete beam subject to flexure and shear. to be inclined at an angle u to the beam axis and the can be considered as two orthogonal principal stress. stirrups at right angle to the beam axis from the equilib state uniform stress fields one in tension and the other. rium of forces in the vertical direction along a projected in compression Considering and the angles of incli. length of the failure plane h cot u the minimum shear nation of the compression and tension fields comple. reinforcement is mentary angles respectively the stress in the tension. field can be obtained from the truss equilibrium at the. Asw min fyw Vcr diagonal crack formation as,Asw min sw min b h cot u Vcr Vcr. or b z cot cot sin 2 b z cot 16, 2 f cp From expressions 11 and 16 the cracking shear. sw min ct 1 tg u force can be found,Vcr fct b z cot Asw min fyw sw min fyw b h cot.
The angle of the failure plane u normally varies, between 20 and 45 tending to the smaller value for the Making 2 3 and z 0 8h in Equation 17 the. minimum transversal steel ratio Considering then that minimum transversal steel ratio becomes. sw min 0 5 18,sw min 0 24 1, fct 13 Note that Equation 18 differently from Equations. 13 and 15 does not take into account the influence of. In the case of prestressed beams with inclined cables the normal force at the section This occurs because in the. the minimum transversal steel ratio can be reduced by truss analysis the top and bottom cords resist the normal. deducting the vertical component of the prestressing forces and the stress fields of the web remains unaltered. force Vp so as, Asw min fyw Vcr V p 3 3 Minimum shear steel ratio in beams. and therefore cracked in flexure, Vcr V p The diagonal crack formation process is very complex. sw min 0 14 and depends on the development of the flexural cracks in. b h fyw cot u, length and width and on the dowel effect of the flexural.
If the plastic shear stress distribution uniform Fig 7c steel Earlier studies have shown that the section dimen. is used instead of the elastic distribution and the plastic sions b and d the concrete strength and the longitudinal. tensile strength of the concrete is taken as fctp fct with steel ratio inf luence the cracking shear force Analyses. 2 3 the following expression for sw min can be found made by Castro 8 and Queir z 12 among others have. shown that the non dimensional nominal cracking shear. ct fct cp 2 f 3 cp stress Vcr bdfct decreases with the increase of fc and d and. 1 2 f tg u 15 increases with the increase of It was concluded also. fyw cot u 3 fyw ct that for beams with practical heights h 300 mm the. Expression 15 gives the same values for sw min as influence of the effective depth of the beam d is not signif. Equation 12 when the normal stress cp 0 and icant and according to Regan 13 the scale effect tends. slightly different values when cp 0 to vanish in beams with some shear reinforcement. This analysis is valid only for beams where the diago Queir z 12 has also shown that in beams without. Materials and Structures Mat riaux et Constructions Vol 36 January February 2003. Fig 8 Comparisons between Equations 20 and 21 and code Fig 9 Comparison between Equations 20 and 21 and avail. equations for the minimum transversal steel values able test results. shear reinforcement and with small flexural steel ratios. cases of beams with minimum reinforcement the, cracking shear force Vcr is about 40 of the one obtained. Minimum reinforcement is provided in concrete beams in order to improve their behaviour towards crack ing and ductility at failure Generally codes of practice equations for the mini mum steel ratios longitudinal and transversal are mainly empirical and do not include all the influential parameters in them For this reason and due to the

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