enter image description here

It is often said for I beams that the flange carries most of the moment subjected to the beam, while the web carries most of the shear force. The latter is obvious from pictures like above, showing the distribution of shear stress.

But why does the flange carry most of the moment? The normal stress on a beam is linearly proportional to distance from the neutral axis. So why is it assumed that the flange carries the bending moment?

  • 1
    $\begingroup$ Because, as you rightly note, it's proportional to the distance from the neutral axis. So, stuff further from the neutral axis will take more of the moment. The flange is all quite far from the neutral axis, but the web is closer, or at zero distance even for the middle. Therefore, the flange (being furthest from the neutral axis) takes most of the bending moment. $\endgroup$ Mar 14, 2021 at 18:00
  • $\begingroup$ But the stress is only linearly proportional to the distance. Does the thin flange really carry so much of the moment even though its further? $\endgroup$
    – S. Rotos
    Mar 14, 2021 at 18:04
  • $\begingroup$ @S Rotos On half of the cross section, simply sum the flexural stress on the flange and the web, represent by Ff and Fw, then get moments Mf = Efdf and Mw = Fwdw, in which df and dw are distances measured from the neutral axis to the respective forces. Now you can easily see which one is larger, and affects the moment capacity of a wide flange, or I beam, the most. $\endgroup$
    – r13
    Mar 14, 2021 at 19:01

3 Answers 3


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

enter image description here

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


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

enter image description here

Because in the linear region the constitutive equation is $\sigma(z) = E\cdot \epsilon(z)$, 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 :

$$\Delta F(z) = E\cdot \Delta A\cdot \epsilon(z)$$

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.

$$\Delta M(z) = \Delta F(z) \cdot z= E\cdot \Delta A\cdot \epsilon(z)\cdot z$$

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

$$\Delta M(z) = \frac{E\cdot}{\rho} \Delta A\cdot z^2$$

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

enter image description here

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.


See the bending stress diagram below (the rightmost). The compression and tension force is the sum of the respective normal stress times the area the stress is acting on. Usually, the flange has an area much larger than the web, as a result, the flanges take the majority of stresses induced by moment. Also, geometrically, the flanges are further out from the neutral axis, which, the longer moment arm, contributes to the fact that the flanges resist the majority of the moment.

enter image description here


Imagine a beam with a very thin flange, with an area A and height H and a web that has the same area A to make it easy to compare the effect. so the web thickness is A/H.

The flange moment resistance is

$ 2AY^2= 2 * A*(H/2)^2= AH^2/2$

the web moment resistance is

$ bH^3/12 = A/H*H^3/12= AH^2/12$

So the contribution of our example beam's web is 1/6 of the flange.

And of course in I beams the flanges are much thicker than the web, so the contribution of the flanges are much more.


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