Analysis of strain in pure bending
Normal strain in beams
Consider a simply supported beam of length L subjected to moments at ends A and B as shown in figure.
Take two section x1-x1 and x2-x2 which are dx length apart and section x1-x1 is at distance x from end A, as shown. Due to applied moments M0, the beam will sag in a manner as shown.
As it is case of pure bending, the elastic curve will be circular in shape.
Point O is the centre of curvature. Cross-section x1-x1 and x2-x2 will remain plane and normal to longitudinal fibres of beam so as to keep radius of curvature constant.
As it is evident , longitudinal fibres EF are shortened while fibres AB are elongated due to bending.
Thus, top fibres are in compression while bottom fibres are in tension. Somewhere between top and bottom fibres, a surface of layer exist at which there is no change in length known as neutral layer as represented by CD. The intersection of neutral surface with any cross-sectional plane is called neutral axis.
Radius of curvature R is taken as distance from centre of curvature O to neutral layer.
As derived earlier,
Radius of curvature, R = dx/dθ
Curvature, C = dθ/dx
To calculate longitudinal strain, take a fibre GH at distance y from neutral layer CD as shown.
The length L1 of fibre GH =(R + y) dθ
= Rdθ+ dx/R
= dx+ y dx/R
Longitudinal strain in fibre GH, εx = L1-dx/dx=y/R
Hence, longitudinal strain in any fibre,
εx = y/R
Here, y is the distance of fibre from neutral axis. Strains are positive below neutral axis and negative above neutral axis.
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