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$\tau \:=\:\frac{VQ}{It\:}$ 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? enter image description here

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3 Answers 3

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

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

enter image description here

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.

enter image description here

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  • $\begingroup$ The blue areas you have drawn are arbitrary area right? and $Y$ is always positive? $\endgroup$
    – CountDOOKU
    Mar 24, 2022 at 6:14
  • $\begingroup$ 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. $\endgroup$
    – r13
    Mar 24, 2022 at 12:25
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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.

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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.

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