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r13
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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.

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).

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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r13
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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 = 50050 mm).

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 = 500 mm).

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).

Source Link
r13
  • 8.2k
  • 3
  • 9
  • 28

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 = 500 mm).