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

Version 2.3 May 2008AndAstb dw0.67 fcu=γ 0.87 fmy⎡⎛b⎢⎜⎣⎝beffw⎞ hf−1⎟⎠ dx ⎤+ η ⎥d ⎦(Eqn 3-10)(iii) Doubly reinforcing section :By following the procedure in (ii), if dxobtained by (Eqn 3-8)exceeds ϕ where ϕ = 0. 5 for fcu> 45 ; 0.4 for fcu> 70 and 0.33for f > 100 , then double reinforcements will be required withcurequiredAscandA stasA 1 ⎡ 0.67 ⎡⎛b ⎞ h ⎛ 1 h ⎞ ⎛ ⎞⎤⎤scM fcueff ff 1= ⎢ − ⎢⎜−1⎟⎜1−⎟ + ηϕ⎜1− ϕ ⎟⎥b d 0.87 fwy( ) ⎥ ⎥ 21−d'/ d⎟ ⎜⎢⎣b⎜wdγm ⎣⎝bw⎠ d ⎝ 2 d ⎠ ⎝ 2 ⎠⎦⎦A 0.67 f ⎡⎛beff⎞ hf⎤cu= ⎢⎥bwdmf⎜ −1+y ⎣ b⎟ ηϕγ 0.87 ⎝ w ⎠ d ⎦st+25Ascb dw(Eqn 3-11)(Eqn 3-12)3.5.2 Worked Examples for Flanged Beam, grade 35 ( η = 0. 9 )x hf(i) Worked Example 3.7 : Singly reinforced section where 0 .9 ≤d dConsider the previous example done for a rectangular beam 500 (h) ×400 (w), fcu= 35 MPa, under a moment 486 kNm, with a flangedsection of width = 1200 mm and depth = 150 mm :b = 400 , d = 500 − 40 − 20 = 440 , b = 1200 h = 150wFirst check ifK =fcuMbeffdx hf0 .9 ≤ based on beam width of 1200,d d6486×10=0.0598235×1200×4402=x ⎛K ⎞ 1By (Eqn 3-5), = ⎜0 .5 0.25 ⎟− − = 0. 1590.9;d ⎝⎠ 0.45x hf 150z x∴0 .9 = 0.143 < = = 0.341. = 1 − 0.45 = 0. 928 ; Thusd d 440 d d6AstM486×10=== 0.005632 2b d b d × 0.87 f z / d 1200×440 × 0.87×460×0.928effeffy( )bw400> 0.18% (minimum for = = 0.33 < 0. 4 in accordance withbeff1200Table 9.1 of the Code)∴ A st= 2974mm 2 . Use 2T40 + 1T25As in comparison with the previous example based on rectangularefff