Q. reinforced concrete beam 230 mm wide by 500 mm overall depth is reinforced with 4 bars of 12 mm diameter at an effective cover of 50 mm. Using M 20 grade concrete and Fe 415 steel, estimate the moment of resistance of the section. If the beam is simply supported over an effective span of 5m, find the maximum uniformly distributed load the beam can carry, inclusive of its own weight.
Q. reinforced concrete beam 230 mm wide by 500 mm overall depth is reinforced with 4 bars of 12 mm diameter at an effective cover of 50 mm. Using M 20 grade concrete and Fe 415 steel, estimate the moment of resistance of the section. If the beam is simply supported over an effective span of 5m, find the maximum uniformly distributed load the beam can carry, inclusive of its own weight.
SOLUTION:
WIDTH b= 230mm
EFFECTIVE DEPTH d = 500 – 50 = 450mm
AREA OF TENSION STEEL Ast = 4× π /4 ×12² = 452.4mm²
1. PERMISSIBLE STRESSES.
σcbc = 7 N/mm²
σst = 230 N/mm²
2. MODULAR RATIO
•m = 280/3σcbc
= 280/ 3×7
= 13.33
•Critical depth of neutral axis NC = 0.289d
3. DEPTH OF NEUTRAL AXIS
Equating the moments of areas of compression and tension zone about the neutral axis
bn²/2= m.Ast(d – n)
1/2×230 n² = 13.33 x 452.4 (450 – n)
115 n² = 2.714×10⁶ – 6030.5 n
n² +52.44 n – 23600= O
n= (- 52.44 + √52.44² +4 × 23600)/2
n = 129.6 mm
critical depth of neutral axis Nc= 0.289×450 = 130mm
since n < nc, the section’s is under Reinforced
4. MOMENT OF RESISTANCE
MR = σst . Ast (d-na/3)
= 230×452.5 (450-129.6/3)
=42.34×10⁶ N-mm = 42.34kNm
MOMENT OF RESISTANCE = 42.34 kNm
5. MAXIMUM LOAD:
let w KN/m be the uniformly distributed load the beam can carry maximum Bending moment = wI²/8
=w×5²/8
=3.125
Equating maximum bending moment to the moment of Resistance Of the section
