Determine the maximum stress developed in the beam:
An I section in Figure is utilized as a beam. The beam is subjected to a bending moment of 2.5 kN m at its neutral axis. Determine the maximum stress developed in the beam.
Beam Section Bending Stress Distribution
Figure
Solution
Assume be the distance of centroid from the bottom face.
Then, =∑ ay / ∑ a
= ((100 × 20) ×10 + (20 ×100) × 70 + (60 × 20) ×130) /((100 × 20) + (20 ×100) + (60 × 20))
= 60.8 mm
Moment of inertia of the section around the horizontal axis passing through the centriod,
I = [(1/12) ×100 × (20)^{3} + (100 × 20)(60.8 - 10)^{2} ]
+[ (1/12) × 20 (100)^{3} + (20 ×100)(70 - 60.8)^{2} ]
+ [(1/12) × 60 (20)^{3} + (60 × 20)(130 - 60.8)^{2} ]
= 1285.04 × 10^{4} mm4
Topmost layer distance, y_{t} = 140 - 60.8 = 79.2 mm
Bottommost layer distance, y_{b} = 60.8 mm
∴ y_{max} = y_{t} = 79.2 mm
Through the relation, M/ I = σ/ y
We obtain, σ_{max} = ( M/ I) × y_{max}
Bending moment, M = 2.5 kN m
= 2.5 × 10^{6} mm
∴ σ_{max} = (2.5 × 10 ^{6}/1285.04 × 10^{4}) × 79.2 = 15.41 N/mm
∴ The maximum bending stress in the beam = 15.41 N/mm^{2}.