DESIGN & ANALYSIS OF DOUBLY REINFORCED BEAM | IS 456 | Limit State Design | Mumbai University

C = Cc + Cs Cc = 0.36 fck b xu – fcc Asc = 0.36 fck b xu . Cs = fsc Asc and T = 0.87 fy Ast . C = T gives 0.36 fck b xu + fsc Asc = 087 fy Ast . From it xu may be found. * The strain at the level of compression steel esc = 0.0035 The corresponding stress in compression steel may be obtained from stress-strain curve. Then fsc = esc Es = 0.0045 × 2 × 10 5 = 700 , subject to maximum of 0.87 fy . Mu = Cc (d – 0.42 xu ) + Cs (d – d¢) = 0.36 fck b xu (d – 0.42 xu ) + fck Asc (d – d¢). Strength of R.C. Section in Shear, Torsion and Bond * Design procedure is given by IS: 456–2000 based on average shear stress across the section. Average shear stress Tv = where Vu = factored (design) shear force b = bredth d = effective depth In case of varying depth, tv = where Mu = factored bending moment b = angle between the top and bottom edges. * Table 19 in IS 456–2000 gives design shear strength (tc ) for various mixes for different percentage of reinforcements. * Shear strength Vu can be enhanced by p...