Shear stress is introduced in statics as $\frac{P}{A}$.

However, the maximum shear stress with neutral axis is different. It is $\frac{3P}{2bh}$ (Where $b$ is width and $h$ is height)

What is the difference between these concepts?

  • $\begingroup$ Whose concept? source? $\endgroup$
    – Solar Mike
    Feb 20, 2018 at 9:35
  • $\begingroup$ not enough info given to provide a meaningful answer $\endgroup$
    – agentp
    Feb 20, 2018 at 11:10

1 Answer 1


The difference is in the assumptions. The first "shear stress" is simply assuming the stress is uniform over the cross section. As a result, we see a shearing force $P$ shearing a rod of uniform cross section $A$, and due to the assumption or uniform shear stress across the cross section, we have the resulting $\frac{P}{A}$ shear stress.

When we abandon this assumption of uniform shear stress, and we use the assumptions for beam shear, we find "maximum shear stress". Here, we find the shear stress develops parabolically across the cross section, and develops a maximum sheer stress at the center line of the cross section, for symmetric cross sections. For a rectangular cross section, this maximum sheer stress happens to be $\frac{3P}{2bh}$, but for other cross sections the value differs.

In general, the value is $$\frac{PQ}{Ib}$$

where $b$ is the thickness of the cross section at the cut, $Q$ is the first moment of the area above the cut referenced to the centroid of the area: $$\int^{h_t-c}_{h_c-c} x dA$$

Where $c$ is the height of centroid from the bottom of the cross section $$\frac{\int^{h_t}_{0} x dA}{\int^{h_t}_{0} dA}$$, $h_t$ is the height of the top of the area from the bottom of the cross section, $h_c$ is the height of the cut from the bottom of the cross section. $I$ is the second moment of the area above the cut referenced to the centroid:


(Note different uses of the reference point other than the bottom of the cross section will result in easier formulas for Q and I). A related post may be able to help you with more visuals.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.