Correct Answer - Option 1 :
110 N/mm2
d = 400 mm
xu = 180 mm
C = 5 N/mm2
m = 18
\(\frac{{\rm{c}}}{{\frac{{\rm{t}}}{{\rm{m}}}}}{\rm{ = }}\frac{{{{\rm{x}}_{\rm{u}}}}}{{{\rm{d - }}{{\rm{x}}_{\rm{u}}}}}\)
\({\rm{t = }}\frac{{{\rm{18 \times 5}}}}{{{\rm{180}}}}{\rm{ \times }}\left( {{\rm{400 - 180}}} \right)\)
t = 220/2
t = 110 N/mm2
Hence the tensile stress developed in the steel is 110 N/mm2