Architectural design and practice Phần 8

Design for axial load=800kN and moment=50+31.3=81.3kNm Assume that dc=300 mm and As1=As2=905 mm2 (two T24 bars). Since dc is between t/2 and (t-d2), fs2 can be determined fromTake fs1=0.83fy. Then10.6 10.6.1REINFORCED MASONRY COLUMNS, USING ENV 1996–1–1 IntroductionThe Eurocode does not refer separately to specific design procedures for reinforced masonry columns although in section 4.7.1.6 of the code reference is made to reinforced masonry members subjected to bending and/or axial load. In the section a diagram showing a range of strain distributions, in the ultimate state, for all the possible load combinations is given and these are based on three limiting...
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Architectural design and practice Phần 8Design for axial load=800kN and moment=50+31.3=81.3kNmAssume that dc=300 mm and As1=As2=905 mm2 (two T24 bars). Since dc isbetween t/2 and (t-d2), fs2 can be determined fromTake fs1=0.83fy. Then10.6 REINFORCED MASONRY COLUMNS, USING ENV 1996–1–110.6.1 IntroductionThe Eurocode does not refer separately to specific design procedures forreinforced masonry columns although in section 4.7.1.6 of the codereference is made to reinforced masonry members subjected to bendingand/or axial load. In the section a diagram showing a range of straindistributions, in the ultimate state, for all the possible load combinationsis given and these are based on three limiting strain conditions for thematerials.1. The tensile strain of the reinforcement is limited to 0.01.2. The compressive strain in the masonry due to bending is limited to - 0.0035.©2004 Taylor & Francis3. The compressive strain in the masonry due to pure compression is limited to -0.002.Using these conditions a number of strain profiles can be drawn. For example if it is decided that at the ultimate state the strain in thereinforcement has reached its limiting value then the range of straindiagrams take the form shown in Fig. 10.12. In Fig. 10.12 the straindiagrams all pivot about the point A, the ultimate strain in thereinforcement. Line 2 would represent the strain distribution if theultimate compressive strain was attained in the masonry at the sametime as the ultimate strain was reached in the reinforcement and line 1an intermediate stage. In the Eurocode additional strain lines, such asline 3, are included in the diagram but since no tension is allowed inthe masonry these strain distributions would require upperreinforcement. If the limiting condition is assumed to be that the strain in themasonry has reached its limiting value then the strain distributiondiagrams would be as shown in Fig. 10.13. In Fig. 10.13 the straindiagrams all pivot about the point B, the ultimate compressive strain inthe masonry. Line 3 would represent the strain distribution if theultimate tensile strain was attained in the reinforcement at the same timeas the ultimate compressive strain was reached in the masonry and line 2an intermediate stage. Line 1, representing the limiting line for thisrange, occurs when the depth of the compression block equals the depthof the section. Compare section 10.5.2. To allow for pure compression, with a limiting strain value of -0.002,the Eurocode allows for a third type of strain distribution as shown inFig. 10.14. In Fig. 10.14 the strain diagrams all pivot about the point C at Fig. 10.12 Strain diagrams with reinforcement at ultimate.©2004 Taylor & Francis10.6.2 Comparison between the methods of BS 5628 and ENV 1996–1–1(a) Strain diagramsThe strain diagrams shown in Fig. 10.14 differ from those used in BS 5628in the selection of the pivotal point; the Eurocode uses the pivot C whilstBS 5628 uses the pivot B. As a result of this, Eurocode calculations in thisrange might result in the maximum compressive stress in the masonrybeing less than the allowable and also the stress in the reinforcementbeing slightly larger than that calculated by BS 5628; compare line 2 ofFig. 10.14 with Fig. 10.10(c). To determine the strain in the lowerreinforcement, using the Eurocode, it would be necessary to know thevalue of the maximum compressive strain ( 0.0035) and then use thegeometry of the figure to calculate the strain at the level of thereinforcement. The calculation can be expressed in the form: (10.17)where ε2=strain in the reinforcement at depth d and e=strain in the upperface of the masonry.(b) Stress-strain diagram for the reinforcing steelIn the Eurocode the stress-strain relationship for steel is taken as bilinearas shown in Fig. 10.15 rather than the trilinear relationship used in BS5628 (see Fig. 10.3.).(c) ConclusionThe main difference between the two codes occurs when the straindistribution is such that the section is in compression throughout. (This is Fig. 10.15 Stress-strain diagram for reinforcement (ENV 1966–1–1).©2004 Taylor & Francisillustrated in Fig. 10.10(d) for BS 5628 and Fig. 10.14 (line 2) for ENV1996–1–1.) Additionally, the method of obtaining the stress, for thesecases, will differ because of the different representations of thestressstrain relationship. For other distributions the design approach for BS 5628 would satisfythe requirements of ENV 1996–1–1 and it is suggested that the methodsdescribed in section 10.5 could be used for all cases. No guidance is givenin the Eurocode with regard to biaxial bending or slender columns andfor these cases the methods described in sections 10.5.3 (b) and 10.5 ...