Calculate the strength of section is torsion when it is also subjected to an ultimate shear force of $12kN$ and ultimate bending moment of $40kNm.$ Assume M20 concrete $F_{e415}$ steel. Adopt LSM.

$$b= 230mm $$

$D=400mm, dc = 35mm \\ D= 365mm \\ Asc = 3-16mm \phi \\ Ast = 4-20 mm ∅ \\ Md = 40 kNm \\ Vu= 12kN, M20 \space and \space Fe415 $

Tu based on compression Reinforcement

$Asc = \dfrac{(M_t-M_d)}{(f_{sc}-f_{cc} )×(d-d_c )} \\ Now, \dfrac {d_c}d=\dfrac {35}{365}=0.095$

0.05 355

$0.095 f_{sc}=353.2 N/mm^2$

0.1 353

Also $f_{cc}=0.446 f_{ck}=0.446×20=8.92 N/mm^2 \\ A_{sc}= \dfrac {Mt-Md}{(f_{sc}-f_{cc} )×(d-d_c ) }\\ 3×\dfrac\pi4× 16^2=\dfrac {M_t-40×10^6}{[(353.2-8.92)×(365-35)]}\\ M_t=108.51 kNm\\ Now, M_t=\dfrac{T_u×(1+D⁄b)}{1.7} \\ 108.51×10^6 =\dfrac{T_u×(1+400/230)}{1.7} \\ T_u=67.34 kNm $

$T_u$ based on Tension Reinforcement

$$A_{st}=\dfrac{0.5 f_{ck} bd}{f_{ck}} \times [1-\sqrt{\dfrac{1-4.6M_u}{f_{ck} bd^2 }}]$$ $$A_{st}=\dfrac{0.5 \times 20\times 230\times 365}{415} \times [1-\sqrt{\dfrac{1-4.6M_u}{20\times 230\times 365^2 }}]$$ $$ M_u=114.07 kNm $$ $$ M_u= M_t+ M_d \ 114.07= M_t+40 $$ $$ M_t=74.07 kNm $$ $$ Now, M_t = T_u\times \dfrac{(1+D/b)}{1.7} $$ $$74.07\times 10^6= T_u\times \dfrac{(1+400/230)}{1.7}$$ $$ T_u=T_{u2}=45.97kN/m $$

The strength /Capacity of section in torsion $= 45.97kNm$

[Take minimum of $2 T_u's$]