Determine Maximum fibre stress:
A beam of rectangular section of 80 mm to 120 mm carries a uniformly distributed load of 40 kN/m over a span of 2 m an axial compressive force of 10 kN. Determine
1. Maximum fibre stress,
2. Fibre stress at a point 0.50 m from the left end of the beam & 40 mm below the neutral axis.
(1) Loading (2) Cross-section
Figure
Solution
Bending moment, M = (w × e ^{2})^{ }/8= (40 × 2^{2})/8 = 20 kN-m = 20 × 10^{6} N-mm
Section modulus, Z = (1/6) × 80 × (120)^{2} = 1.92 × 10^{5} mm^{3}
Moment of inertial, I = (1/12) × (80) × (120)^{3} = 11.52 × 10^{6} mm^{4}
Axial load, P = 10 kN = 10 × 10^{3} N
Direct stress, f_{0} = P/A =10 × 10^{3}/ (80 × 120) = 1.04 N/mm^{2}
Bending stress, f b =± M/ Z =( 20 × 10^{6} )/(1.92 × 10^{5}) = ± 104.16 N/mm2
∴ Maximum fibre stress = 1.04 + 104.16 = 105.20 N/mm^{2 }(compressive)
∴ Bending moment at 0.50 m from left end will be,
M = ( - 40 × 0.50 + 40 × 0.50^{2} / 2)
= - 15 kN-m
= 15 × 10^{6} N-mm (sagging)
∴ Bending stress at 40 mm below the neutral axis will be,
= (M/I) . y
= (15 × 10^{6 })/(11.52 × 106 ) × (- 40)
=- 52.08 N/mm^{2} (tensile)
∴ Resultant fibre stress = 1.04 - 52.08
=- 51.04 N/mm^{2} (tensile)