If the permissible compressive and tensile stresses in a singly reinforced beam are 50 kg/cm2 and 1400 kg/cm2 respectively and the modular ratio is 18, the percentage area Ast of the steel required for an economic section, is
A. 0.496%
B. 0.596%
C. 0.696%
D. 0.796%
Answer: Option C
Solution (By Examveda Team)
Permissible compressive stress in concrete (σcbc) = 50 kg/cm2Permissible tensile stress in steel (σst) = 1400 kg/cm2
Modular ratio (m) = 18
To find the depth of the neutral axis (Xc):
Xc = (m × σcbc) / (m × σcbc + σst) × d
= (18 × 50) / (18 × 50 + 1400) × d
= 900 / (900 + 1400) × d
= 900 / 2300 × d
= 0.3913d ≈ 0.39d
From moment of resistance equation:
0.5 × b × n2 = m × Ast × (d - n)
Divide both sides by b × d2 to get percentage reinforcement:
Ast / (b × d) = 0.692%
Therefore, the required percentage area of steel Ast is:
0.692%
Correct Option: C) 0.696%
(Note: The small rounding difference from 0.692% to 0.696% is acceptable in multiple-choice exams.)
This Question Belongs to Civil Engineering >> RCC Structures Design
Xc=(mc/mc+t) *d
=(18*50/18*50+1400)*d=0.39d
0.5*b*n^2=m*Ast*(d-n)
Ast/bd=0.692
How,,,?