Q.A reinforced concrete beam 300 mm x 600 mm overall depth is reinforced with 4 bars of 20mm diameter at an effective 600 over what uniformly distributed load this beam can carry excluding self weight, over a simply supported span of 5 m. Assume M 25 grade concrete and Fe 415 steel. Use working stress method
SOLUTION:
WIDTH b= 300mm
EFFECTIVE DEPTH d = 600 – 50 = 550mm
AREA OF TENSION STEEL Ast = 4× π /4 ×20² = 1256.6mm²
1. PERMISSIBLE STRESSES.
σcbc = 8.5 N/mm²
σst = 230 N/mm²
2. MODULAR RATIO
•m = 280/3σcbc
= 280/ 3×8.5
= 10.89
•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)
1/2×300 n² = 10.89 x 1256.6 (550 – n)
150 n² = 7.526×10⁶ – 13684.4 n
n² +91.23 n – 50173= O
n= (- 91.23 + √91.23² +4 × 50173)/2
n = 182.98 mm
critical depth of neutral axis Nc= 0.289×550 = 158.98mm
since n > nc, the section’s is over Reinforced
4. MOMENT OF RESISTANCE
MR = σcbc/2 . bn (d-na/3)
=8.5/2 × 300×182.98 (550-182.98/3)
=114.09×10⁶ N-mm = 114.09kNm
MOMENT OF RESISTANCE = 114.09 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
=3.125w
Equating maximum bending moment to the moment of Resistance Of the section
3.125 w = 114.09
w = 114.09/3.125
= 36.51kN/m ( including self weight)
self weight of the beam = 0.3×0.6×1×25 = 4.5kN/m
Uniformly Distributed load excluding Self weight the beam carry
= 36.51 – 4.5= 32.01 kN m
