Consider a rectangular cross sectional beam for instance with dimensions a = 1 mm and b = 1 mm, fixed at one end and transverse force at the other. I will get the maximum stress at the farthest away point from the neutral axis. Now, if I decrease the dimensions to a = 0.5 mm and b = 0.5 mm, then we know that the maximum stress at the farthest away point from neutral axis will increase.

But my question is does this increase in stress happens because the tensile force (or the compressive force) at the farthest away point has increased, or is it solely due to the fact that I have decreased the cross sectional area of the beam?


2 Answers 2


It is actually more complicated than that.

  • the magnitude of the stress increases 8 fold
  • the tensile forces increase so that they generate the counteracting bending moment.

When you decrease the cross-section from 1mm x 1mm to 0.5mm x 0.5m, the cross-sectional area decreases to 1/4 of the original but

  • the second moment of area decreases by $ \left(\frac{1}{2}\right)^4= \frac{1}{16} $(i.e. $I_{new} = \frac{I_0}{16}$ and
  • the section modulus decreases by $\frac{1}{8}$ (ie. $W_{new} = \frac{W_0}{8}$)

The maximum stress due to bending is given from:

$$\sigma_b = \frac{M}{W}$$

So the ratio of the original (0) and the (new) will be

$$\frac{\sigma_{b,new}}{\sigma_{b,0}} = \frac{\frac{M}{W_{new}}}{\frac{M}{W_{0}}}$$

$$\frac{\sigma_{b,new}}{\sigma_{b,0}} = \frac{W_{0}}{W_{new}}=8$$

So by reducing by half both dimensions, the stresses due bending increase 8 fold.

why do the tensile forces increase

The tensile (and compressive) forces during bending increase so that their combined effect generates the required bending moment to counteract the bending moment from the external loads.

enter image description here

Figure (source a builders engineer

The above illustration shows in (1) a section with a bending moment M is presented. Notice that the resulting normal forces are such that they counteract M.

The idea is that:

  • the displacement changes linearly with the distance from the neutral center.
  • A force at a distance from the neutral axis ($F_i$) develops that is proportional to the displacement
  • Each force ($F_i$) produces a moment $M_i$
  • The integral of all the moment over the cross-section should be equal to M $$M= \sum M_i$$

Figures 2 and 3 show what happens to the tensile forces when the cross-section is reoriented.

  • $\begingroup$ I appreciate your reply. It was helpful. I would like to ask one more thing. Assume a T section beam, and I calculate (using integrals) the total compressive force and tensile force acting on above and below the neutral axis (Ofcourse, above or below depends how bending moment is applied). I know that these two forces must be equal (for equilibirum), but how can I find the exact location of these net compressive and tensile forces on above and below the neutral axis? I want to know this because I wanna know the distance between these forces. $\endgroup$ Sep 11, 2021 at 12:44
  • $\begingroup$ In the case of pure bending, for an isotropic and homogeneous material the location of the neutral axis coincides with the geometric center of the cross-section. $\endgroup$
    – NMech
    Sep 12, 2021 at 8:36

Section properties in a fluxural member play a dual role as opposed to members in simple tension or compression.

in a tension member stress increases jist because there are less fibers to share the load.

But in a beam's reduction in section has dual effect of rediucing the second area moment which weakens the beam's bending resistance and also the fact that now there is less fibers to share the load.

reducing the height by half reduces the I, 2nd area moment by 1/4 and reducing the width by half reduces the the srea by half. so 8 times load on th ultimate fiber.


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