I and T Sections:

In these sections, first we have to find the centroid of the section. After that the moment of inertia of the section around the centroidal axis is determined. Since the neutral axis coincides along the centroidal axis, the moment of inertia around the neutral axis is the same as that around the centroidal axis.

In unsymmetrical sections, the distance of the extreme layer, y_max, is not the same for the topmost & bottommost layers. Thus, the topmost layer distance, y_t, & the bottommost layer distance, y_b, are discovered. The section modulus is found utilizing the higher value of y_t or y_b viz. the maximum value of y. After that, the moment of resistance of the section is equivalent to the product of the maximum bending stress & the section modulus.

Example

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 × 10⁴ 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 × 10⁶ mm

∴          σ_max =  (2.5 × 10 ⁶/1285.04 × 10⁴)  × 79.2 = 15.41 N/mm

∴  The maximum bending stress in the beam = 15.41 N/mm².