This question was previously asked in

UP Jal Nigam E&M 2016 Official Paper

Option 3 : Maximum bending stress remains constant

__Explanation:__

Equation of pure bending:

Pure bending or bending is that in which bending moment *M* is constant along the length i.e. \(\frac{{dM}}{{dx}} = 0\), or shear force is zero.

Its empirical relationship is given by –

\(\frac{{\rm{M}}}{{\rm{I}}} = \frac{{\rm{\sigma }}}{{\rm{y}}} = \frac{{\rm{E}}}{{\rm{R}}}\)

where, M = Bending moment, I = MOI of cross-section about NA, E = Young’s Modulus of Elasticity, σ = Bending stress at a distance y from NA

Section Modulus (Z):

The ratio of Moment of Inertia I of beam cross-section about NA to the distance of extreme fiber ymax from the neutral axis.

It also represents the strength of the section. It is given by -

\({\rm{Z}} = \frac{{\rm{I}}}{{{{\rm{y}}_{{\rm{max}}}}}}\)

\(\therefore {\rm{\sigma }} = \frac{{\rm{M}}}{{\rm{Z}}}\)

For pure bending, when M = Constant

∴ σ = constant.

- In presence of transverse shear load, a prismatic beam is said to be a non-uniform strength because bending stress varies as bending moment vary.
- To obtain a beam of uniform strength under the transverse shear load, the non-prismatic beam should be used. It can be done by two methods
- Uniform width varying depth
- Uniform depth varying width.