Why is it said that the flange of an I beam carries most of the moment?

What is really proportional to the distance from the neutral axis (let's denote it $z$) is the strain.

For pure bending and a symmetric cross-section the equation for strain is given by $$epsilon(z) = frac{z}{rho}$$

where:

• $rho$ or $R$ is the radius of curvature of the beam

Because in the linear region the constitutive equation is $sigma(z) = Ecdot epsilon(z)$, the development of stress at distance z is proportional to the strain.

Because stress in a small area $Delta A$ is defined as $sigma(z) =frac{Delta F}{Delta A}$, in that small area $Delta A$ which is at distance z you are getting a Force :

$$Delta F(z) = Ecdot Delta Acdot epsilon(z)$$

Those forces produce bending moments the further away they are from the neutral axis (about which the cross-section rotates), the contribution of the bending moment is greater.

$$Delta M(z) = Delta F(z) cdot z= Ecdot Delta Acdot epsilon(z)cdot z$$

So at a distance (z) any small area $Delta A$ will produce a bending moment equal to

$$Delta M(z) = frac{Ecdot}{rho} Delta Acdot z^2$$

However, as you can see at the flanges you get more area at the same (more or less distance) (e.g. $z=pmfrac{A}{2}$ for the image below).

So, although the stresses and strain (in the linear region) follow a linear distribution wrt to the distance from the neutral axis, you can see that the actual bending moment eventually is proportional

• to the square of the distance z ,
• to the area at distance z from the neutral axis.

Therefore, because the flanges have a significant area, further away from the neutral axis they have greater contribution to the bending moment which resists bending.