Limit state design
1. Design of beam
           a. Reinforcement calculation
                                  0.0035
                     k
                                                fy
                           (0.0055  0.87          )
                                                E
                                          M
                     d lim 
                                  k (1  0.42k ) f ck b
                                                 f y Ast 
                     Msteel  0.87 f y Ast  d          
                                                  f ck b 
                     Mconcrete  0.36 f ck bd d  0.42 x 
            b. Check minimum reinforcement
                                                                    0.85
                                                        A min           bd
                                                                     fy
            c. Check span depth ratio L/d=R*F1*F2*F3
               F1,f2,f3 from graphs or following formula
                      Modification factor for
                      tension steel
                      F1                               1/(0.225+0.00322fs+0.625logpt)<=2.0
                                                       fs=0.58*fy*As,req/As,provided*β
                                                       pt=100*As provided/b/D
                                                       redistribution factor, β
                      Modification factor for
                      compression steel
                                                       1+(pc/(3+pc))<
                      F2                               =1.5
                                                       pc=100*As
                                                       provided/b/D
                      Correction of flanged
                      beams
                                                       0.8+2/7*(bw/bf-
                      F3                               0.3)
             d. Shear calculation
Maximum permissible shear stress for this section   τmax=0.62*sqrt(fck) <= 4
Allowable design shear stress                       τc=0.85 sqrt(0.8fck)sqrt(1+5β-1)/6β
                                                    β=0.8fck/6.89pt
                                                    τc=0.79pt^(1/3)*(400/d)^(1/4)*(1/SF)*(fck/25)^(1/3)