Case I : Balanced Section
Case II a) : Under-Reinforced Section (Xu < Df)
Position of NA,
C=T
0.36 fck b Xu = 0.87 fy Ast
MoR = C x LA = 0.36 fck b Xu (d-0.42 Xu)
or,
MoR = T x LA = 0.87 fy Ast (d-0.42 Xu)
Case II b) : Under-Reinforced Section (Xu > Df)
Case I ) Df < 3/7 Xu
Consider under reinforced section, Xu>Df and Df < 3/7 Xu
This case can be analysed by follows :
For position of NA
C=T
C1+C2 = T1 +T2
Xu > Df, Df < 3/7Xu
For Moment of Resistance,
MR = MR1+MR2 = C1 x LA1+C2 x LA2
or, MoR = T x LA (Not preferrable )
