Evaluate the maximum bending stresses:
A beam of I-section illustrated in Figure is simply supported over a span of 10 m. It carries an uniform load of 4 kN/m over the whole span. Evaluate the maximum bending stresses.
Cross-section of Beam Bending Stress Distribution
Figure
Solution
Moment of inertia, I =1/12 (BD^{3} - bd^{3})
= (1/12) (300 × 660^{3} - 280 × 600^{3})
= 21.474 × 10^{8} mm^{4}
Span of beam, l = 10 m
Uniformly distributed load, w = 4 kN/m
Maximum bending moment at the centre of beam, M = wl ^{2}/ 8
Therefore, M =4 × 10 ^{2} / 8
= 50 kN m
= 5 × 10^{7} N mm
Neutral axis passes by the centroid of I-section.
The distance of top and bottom layer from neutral axis, y = 330 mm
Thus, Bending stress, σ = (M / I )× y
= (5 × 10 ^{7} / 21.474 × 10^{8} )× 330 = 7.68 N/mm^{2}
So the bending stress at top and bottom layers = 7.68 N/mm^{2}.