Manual for Design and Detailings of Reinforced Concrete to Code of ...

Version 2.3 May 2008can be worked out as per the strain distribution profile of concrete and steel asindicated in Figure 3.8.d’εult= 0.0035xdneutral axisStress DiagramStrain DiagramFigure 3.8 – Stress Strain diagram for BeamThe solution for the neutral axis depth ratio dxfor singly reinforced beam isthe positive root of the following quadratic equation where εult= 0. 0035 forconcrete grades not exceeding 60 (Re Appendix C for detailed derivation) :0.67 fγmcu⎡⎢−⎢⎣12+13ε0εult−112⎛ ε0⎜⎝ εult⎞⎟⎠2⎤⎛⎥⎜⎥⎦⎝xd⎞ ⎟⎠20.67 f +γmcu⎛ 1 ε0⎞⎜1−3⎟⎝ εult ⎠xdM−bd2= 0(Eqn 3-1)With neutral axis depth ratio determined, the steel ratio can be determined byAstbd1 0.67 fcu⎛ 1 ε=f⎜ −010.87yγm ⎝ 3 εult⎞⎟⎠xd(Eqn 3-2)As dx is limited to 0.5 for singly reinforcing sections for grades up to 45under moment redistribution not greater than 10% (Clause 6.1.2.4 of the Code),Mby (Eqn 3-1), will be limited to K ' values as in2bdf cuK ' = 0.154 for grade 30; K ' = 0. 152 for grade 35;K ' = 0.151 for grade 40; K ' = 0. 150 for grade 45which are all smaller than 0.156 under the simplified stress block.However, for grades exceeding 45 and below 70 where neutral axis depth ratiois limited to 0.4 for singly reinforced sections under moment redistribution not17