Analysis of stress in pure bending
Normal Stresses in beams
As per Hooke’s law, εx = σ/E
Here, σ is the bending stress in longitudinal direction and E is the Young’s modulus of elasticity.
σ/E = y/R
σ/y = E/R
In simply supported beam, when subjected to downward loading bending stresses are compressive above neutral axis while these are tensile below the neutral axis.
Due to these stress, a moment is produced which is bending moment for the section and it acts along z-axis.
As the resultant force in x-direction is zero.
∫ 0xdA = O [∵ There is no other horizontal force]
When dA is small area element on cross-section on section.
σ/y = E/R
σ= Ey/R …(v)
Putting eq. (v) in (iv), ∫ Ey/R dA =0
As E and R are constant, ∫ydA=0
Eq. (vi) implies that neutral axis which is in z-direction passes through centroid of section.
Also, resultant moment acting on section is equal to bending moment say M, then
M= ∫ σ, ydA
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