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I understand that for a rectangular c-s the shear stress distribution is parabolic and the max shear stress occurs at the neutral axis and has a value of 1.5V/A. Where V is the 'applied shear force' and A is the cross-sectional area.

But this in turn then means that the shear force at this point is equal to 1.5V ( 1.5 times larger than the applied shear force) - which seems a tad strange physically.

Is this due to the fact the the average shear force (average shear stress x cs area) is equal to the applied shear force? This is the only way it makes sense to me.

Thanks!

enter image description here

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You've got your terms confused.

The maximum shear stress at the midpoint is equal to

$$\tau_{max} = 1.5\frac{V}{A} = 1.5\overline\tau$$

where $\dfrac{V}{A}=\overline\tau$, which is the average shear stress along the entire section.

That is the only viable comparison to be made, stress to stress. And having a maximum stress greater than the average stress is totally reasonable.

Your doubt, however, is that "the shear force at this point is equal to $1.5V$". That is not the case. There is no shear force at any point in the section. There is only a shear stress. The entirety of the shear stress must then be integrated over the area to obtain the shear force.

You may be thinking "stress is just equal to force divided by area, so can't I just do $$\begin{align} \tau &= \frac{V}{A} \\ \therefore \tau_{max} &= \frac{V_{max}}{A} \\ \tau_{max} &= 1.5\frac{V}{A} \\ \frac{V_{max}}{A} &= 1.5\frac{V}{A} \\ V_{max} &= 1.5V \end{align}$$

and prove that the shear force at the midpoint is greater than the applied shear force?" But I already beat you to it. After all, as I mentioned at the start, $\dfrac{V}{A}$ gives you the average stress along the section. So $\dfrac{V_{max}}{A}$ is equivalent to the following stress profile, which clearly isn't the one you're expecting:

enter image description here

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  • $\begingroup$ Great answer! Thanks! That was exactly what I was thinking and actually did a similar calculation! This has helped a lot. To follow on can I ask: so does the average shear stress multiplied by the area equal the applied shear force? As V / A = Tavg. Additionally, could you please elaborate on "there is no shear force at any point in the section" - is this just a fundamental rule? $\endgroup$
    – massey95
    Aug 3, 2016 at 18:00
  • $\begingroup$ @massey95: Yes, the average shear stress times the cross-sectional area equals the applied force. This can be understood by Newton's Third Law: the internal stress state in the beam is created to withstand the external force, so the total internal force must be equal to the external force. If it were different (greater or lesser), then the system wouldn't be in equilibrium. Since the internal force is found by $V_I = \int_A \tau\text{d}A \equiv \overline\tau A = V_E$. $\endgroup$
    – Wasabi ♦
    Aug 3, 2016 at 19:05
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    $\begingroup$ @massey95: This also answers your second question. External loads generate internal stresses along the beam. These stresses can be integrated along the entire area to obtain equivalent internal forces. As the integral states, $V_{int} = \int_A \tau\text{d}A$. For any given point $\text{d}A=0 \therefore V_{int}=0$ for that point. The entire concept of internal force requires one to look at a larger scale than any single point. If you're looking at the section as a whole (or large part), then forces make sense. If you're looking at a point, then only stresses make sense. $\endgroup$
    – Wasabi ♦
    Aug 3, 2016 at 19:22
  • $\begingroup$ Fantastic, thanks very much for your help! Definitely helped my understanding! Slowly making my way through static mechanics! $\endgroup$
    – massey95
    Aug 3, 2016 at 19:26
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V/A represents the average shear stress, that is the total shear force resisted by the whole section, which has an area of A.

As we understand, if the stress distribution is uniform, the maximum shear stress will be equal to the average stress.

However, if we have a parabolic distribution of shear stress, some areas will be stressed less and others more than the average stress V/A, as shown in the figure posted.

What the expression tau(max)=(3/2)(V/A) shows is that the worst case of the stress (not force) is 50% higher than the average.

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  • $\begingroup$ Yes, thank you! I think I understand it better now! I was incorrectly assuming that as the max shear stress was 1.5 times larger than the average shear stress, that on an infinitesimal area, the maximum shear force would be 1.5 larger than the applied shear force. But am I correct in saying that although theoretically that may be true, it does not hold (or matter) as no shear forces, only shear stresses act on the cross section? Or am I still missing something? $\endgroup$
    – massey95
    Aug 3, 2016 at 18:08
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If we need to calculate how much shear a rectangular beam can take this is the formula.
V= 2/3 [A x tau(allowable)].

We find the allowable tau in charts readily available but for lumber under normal humidity it is around 80-90 psi.

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  • $\begingroup$ Not strictly what I was looking for, Wasabi and Mathmate have helped with the theory, but this is actually very intersting as from my unexperience assumption that formula is simply derived for the Max shear stress = 1.5 x applied shear load / area. Re-arranged so that the shear force the beam can withstand. Very simple but always nice to see the practical use! $\endgroup$
    – massey95
    Aug 3, 2016 at 18:23
  • $\begingroup$ Yes, exactly: it basically is what you had but re-arranged. Of course it would not be the same for I beams or pipes, etc. You'd have to figure the distribution of tau stress on the surface and integrate over the entire surface of the member. $\endgroup$
    – kamran
    Aug 3, 2016 at 19:28

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