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In an I-beam, the shear flow is usually shown drawn as on the left, not as drawn on the right. I do not understand why. I do understand:

  1. The total horizontal shear force must be zero, consistent with statics.This is satisfied in both images.
  2. The vertical direction matches the direction of the external load.

My fundamental question is, Why is the image on the left correct and the image on the right incorrect?

It matters because when the direct shear due to a horizontal force (pushing to the right) we will subtract the shears on the upper right hand side of the left I-beam to get the net. If we were to use the right I-beam, we'd have to add the shears to get the net.

1. How can I understand why a vertical force produces horizontal shear and 2. Why the direction is chosen as it is?

enter image description here

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Imagine we take an element of this beam between two cross sections of thickness dx as shown in (a).

enter image description here

The bending moment on one face is M and on the other is M+dM. Because the bending moments are different, at any distance y from the NA the bending stress would be different on the faces of this element.

Now I take a sub-element in this element (as shown in (a) by shaded part). This sub-element (see (b)) will have normal stress distributions on either side of it. However, since the stress on one of the faces is greater than the other there will be a net force on the element (see (c)). Since this sub-element should be in equilibrium there will be an internal force developed on the side face (the one which has one of the dimensions as dx). This internal force gives us a shear stress and because of the complimentary property of the shear stress we get a shear stress on the face that you are interested in, in the direction as shown in (d). You can repeat this analysis by taking an element in the lower region of the element, to get the results that you want. (Note in the lower region the stresses would be tensile).

As per my understanding, in the flange there will be some vertical shear stresses as well. However, in thin walled members, this vertical shear is small (in the flange) and we can assume that the vertical shear force is completely carried by the web.

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  • $\begingroup$ I like this. When we add in the analysis for the longitudinal and shear transverse, wouldn't we have to say there are two components of shear stress on every non free face? And these all have to force and moment balance for the element to be in equilibrium. $\endgroup$
    – Karlton
    Apr 16, 2022 at 22:59
  • $\begingroup$ If, that is, (d) were inside the meat of the flange, then the shear state would be complex. $\endgroup$
    – Karlton
    Apr 16, 2022 at 23:15
  • $\begingroup$ If we take a differential element in the flange, yes there will be vertical and horizontal shear stresses acting together. If you want to verify why vertical stresses would be present in the flange, try slicing the element (the first element in my ans) in the flange region from a horizontal plane. You would find that in order for that subelemt to remain in equilibrium there must be vertical shear stresses in the flange. $\endgroup$
    – Harshit Rajput
    Apr 17, 2022 at 6:04
  • $\begingroup$ Let me ask you one last thing. Would you agree that in your (d) the shear on the right face, tau_xz must equal the shear on the far, hidden face and that the sum of these will equal the shear on the front face, tau_zx? There are only three faces that can have shear stress, since the far face is a free surface and so are the top and bottom. When I perform a moment balance about a vertical axis through the centroid of the cut in (d), this is my result. But it seems strange to say tau_zx = 2tau_xz. Yet a moment balance with three shear forces compels it. What do you think? $\endgroup$
    – Karlton
    Apr 17, 2022 at 20:09
  • $\begingroup$ There are no stresses on the far hidden face. You're right that only three faces can carry shear stress in (d). You're forgetting about the normal stress though. The ckw moment (as seen from the top) of the forces due to the shear on right face and the left hidden face would be balanced by the antickw moment of the forces due to normal stresses. The moment of the force due to shear on the front face will be zero because it passes through the axis about which we take the moment. $\endgroup$
    – Harshit Rajput
    Apr 18, 2022 at 8:39
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I think it is drawn based on the tendency of deformation of the beam, and the corresponding deformed shape.

enter image description here

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