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Say the cross section carries a vertical shearing force of 40 kN. And say I want to find the value of the shear stress just below the section a-a.

I am very confused about which area $(A)$ to take When calculating $Q = y \times A$ in the $\tau = \frac{VQ}{It}$, shear force equation. I have drawn 2 sketches and can someone please explain which one is correct?

Also where is the maximum shear in this case, I assume in a location where the width is small? ie the upper rectangle?

enter image description here

enter image description here

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2 Answers 2

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enter image description here

You shall first identify the location of the NA, then calculate the Q by either considering the shaded area above "line-a", or the unshaded area below "line-a". If you are doing correctly, Q for both areas should be identical, so the stress over the neck area (b = 50 mm).

Discussion:

In order to resolve your confusion on this matter, we need to explore the significance of indicating the direction of shear/shear stress.

Shear force/stress occurs on a plane within a rigid body, on that plane, the segment above moves in one direction and below moves in another direction. This phenomenon can be observed in the graph below, which shows a deformed/deflected beam in the longitudinal direction.

enter image description here

Conclusion:

So, does indicating the direction of shear stress have significance?

  • Yes, when there is the need to show the direction of the stress graphically as in the sketch above, and in 3D cases.

  • No, there is no significance for the general purpose of calculating shear stress for checking stress level or for design. For which $Q = \sum A|d|$, and $\tau = \dfrac {V|Q|}{Ib}$ are routinely used without the directional indicators (+/-signs) attached to the $V, Q, d$ terms.

Hope this helps.

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  • $\begingroup$ Thank you so much for the detailed explanation. I am practising these problems a lot, hoping to reach your level of understanding. $\endgroup$
    – CountDOOKU
    Mar 26 at 14:21
  • $\begingroup$ You'll be fine in due time. After a while of practice, and paying attention to details, everything learned in school will start to make sense, which in turn will promote better understanding. Good luck. $\endgroup$
    – r13
    Mar 26 at 15:06
  • $\begingroup$ Thanks for the encouragement! And Many thanks for the help, saved me so much time! $\endgroup$
    – CountDOOKU
    Mar 26 at 15:07
  • $\begingroup$ Hello r13, I have a new question regarding fluids. Do you mind taking a look? Thanks. engineering.stackexchange.com/questions/50337/… $\endgroup$
    – CountDOOKU
    Mar 26 at 15:16
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By definition shear flow on a beam's cross-section is the same on the two sides of any horizontal plane.

eg, the top surface is trying to move to the right, the bottom surface is reacting by trying to move to the left with an equal amount of shear to maintain equilibrium. Otherwise, the beam will start to move to the left or right.

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  • $\begingroup$ Thanks for the quick answer! $\endgroup$
    – CountDOOKU
    Mar 26 at 15:13

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