# What does negative shear stress mean $\tau \:=\:\frac{VQ}{It\:}$?

$$\tau \:=\:\frac{VQ}{It\:}$$ My confusion lies within $$Q$$ term. $$Q = yA$$ where $$A$$ is the cross-sectional area of the segment that is connected to the beam at the juncture where the shear flow is calculated, and $$y$$ is the distance from the neutral axis to the centroid of $$A$$.

So $$y$$ (orange) can be negative if the area is below the neutral axis right? What does that mean?

$$\tau = \dfrac{V}{Ib}\int ydA = \dfrac{VQ}{Ib}$$

The static/first moment of area, $$Q = \int y dA$$, is a property of a shape, which is always positive.

However, depending on the direction of shear force, $$V$$, the resulting shear stress can be negative, for which the negative sign indicates the location of a section cut along the span of the beam, and on which face of the section cut the shear stress is acting on/calculated. The sketch below shows where negative shear force occurs, and what negative shear stress means.

• The blue areas you have drawn are arbitrary area right? and $Y$ is always positive? Mar 24, 2022 at 6:14
• Yes and yes. Note, for calc Q, "y" is defined as the "distance" between two points projected on the 'Y-axis". It is sign-convention neutral, and always positive.
– r13
Mar 24, 2022 at 12:25

Just the direction you are measuring ie above the line is positive, below is negative.

If you put line A at the bottom of the item then all values would be positive.

The area moment,q, quation is

$$q=\int{\bar{y}}{da}$$

And is always positive because the distance, $$\bar{y} \$$ is always positive.

If "y" was a signed value all beams sections would end up having zero shears because the lower half of the section with supposedly a negative shear would cancel the half with positive shear.