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In the figure , we can notice that the moment due to Pe is counterclokwise , but moment due to shear flow is also counterclockwise , How can the Pe prevent the twisting of the beam ?

Or is the direction of the shear flow wrong ? Should it flow from the below to the top ?1 22

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Any time you cut a beam to look at the shear flow distribution, the equilibrium condition will give you the correct orientation of the distribution. Additionally, the shear flow distribution is always equal and opposite when looking at the two faces of the cut. From Fig. 7-23a the loading from $P$ is causing a clockwise moment and rotation of the beam. Which is indicated by the equilibrium shear flow in the non-cut image of Fig. 7-23a. When the beam is cut in Fig. 7.23-b, we are looking at the shear flow in the cross section in an equal and opposite manner as indicated by the equilibrium condition.

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  • $\begingroup$ By saying When the beam is cut in Fig. 7.23-b, we are looking at the shear flow in the cross section in an equal and opposite manner as indicated by the equilibrium condition , do you mean the moment generated by Fh is clockwise ? But , it's clear that the moment generated by the Fh is anticlockwise , right ? How could it be clockwise ? I didnt get it $\endgroup$ Jan 2, 2017 at 22:51
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    $\begingroup$ What force do you mean by $F_h$? $\endgroup$
    – TRF
    Jan 2, 2017 at 22:54
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    $\begingroup$ If you were looking at the shear flow distribution on the face where $P_e$ is applied in Fig. 7.23-a, you would see the distribution presented in Fig. 7.23-b, which is counterclockwise. That part is clear as the shear flow in the vertical web must be in the same direction as the applied force. However, if you were to cut the beam, your equilibrium equation would provide a shear flow distribution in the clockwise direction at the opposite end. This ensures equilibrium and also provides the mechanism for twist in the beam. This is shown in the non-cut image of Fig. 7.23-a. Hope this helps. $\endgroup$
    – TRF
    Jan 2, 2017 at 23:09
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    $\begingroup$ I am not sure what else to say. You are aware that when you cut the beam (a hypothetical cut), that the forces at the cut are equal in magnitude and opposite in direction right? Additionally, depending on which piece of the beam you are analyzing after the cut, the shear flow can be either clockwise or counterclockwise? I honestly don't see how you are studying shear center loading, if you aren't aware of this aspect of mechanics of materials at this point. $\endgroup$
    – TRF
    Jan 2, 2017 at 23:28
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    $\begingroup$ Yes, the forces are equal in opposite as a consequence of equilibrium. The shear flow in Fig. 7.23-a is the equilibrium shear flow distribution provided one where to cut where the distribution is drawn. Also, if you actually cut there, that is the shear flow distribution on the piece that you remove from the wall. However, on the other face, where $P$ is actually applied, the shear flow distribution is shown in FIg. 7.23-b. Obviously the two shear flow distributions are equal and opposite as consequence of equilibrium, but also provide the mechanism for twisting in the beam. $\endgroup$
    – TRF
    Jan 3, 2017 at 0:22

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