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I have calculated the shear flows and center of an L shaped profile. I assume they are correctly calculated (I followed the same steps as during the lectures for an I shaped profile). However, I have some problems with the interpretation of the results. The results are shown in my simple sketch.

The height of the profile is 1.5 times the base (but this actually doesn't matter) and thickness is very small compared to other dimensions. Given this, I calculated that the shear center is about 0.49xH to the left of the center of gravity (c.g.) to counter act the torsion of the shear flow. enter image description here

My confusion regarding the results:

The net result of the vertical shear flow is equal to the vertical force V. The net result of torsion of the shear flows is equal to the torsion moment of the vertical force (0.49xHxV). But there is no horizontal force to counter act the horizontal shear flows. How is this possible?

I am confused because in the lectures we have only seen shear flows in symmetric profiles (like C an I).

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Firstly, the shear center is the point at which an applied load produces no torsion on the section. For a singly or doubly symmetric section, the shear center will lie on the axis of symmetry. For the unequal leg angle shown here, we can say by inspection that the shear center must be at the intersection of the two legs because the shear flow will produce no moment about this point.

Shear Center of Unequal Leg Angle

The general equation for shear flow in an unsymmetric section is:

$$q=\frac{I_xV_x - I_{xy}V_y}{I_xI_y - I_{xy}^2}\sum xA - \frac{I_yV_y - I_{xy}V_x}{I_xI_y - I_{xy}^2}\sum yA$$

If the section is symmetric, the $I_{xy}$ term is equal to zero. For a symmetric section subjected only to vertical shear $V_y$ this equation reduces to:

$$q = \frac{V_y}{I_x}\sum yA = \frac{VQ}{I}$$

But for an unsymmetric section we need the full generalized equation. The result is something like this:

Shear Flow Diagrams for Unequal Leg Angle

Now you can see that if we integrate the shear force over the length of the horizontal leg, the result will be zero. Integrating over the vertical leg will equal the applied shear force $V_y$

There is a fairly good discussion/example of this here and here (page 474)

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You are right that for vertical applied load, the shear flow in the horizontal direction must add up to zero to satisfy equilibrium between the internal stress and the externally applied load. Therefore the shear flow diagram which is drawn cannot be right.

The key thing is that shear flow (like bending stress) must be calculated with respect to the principal axis directions of the cross section. Unlike I beams, the principal axes of an angle section do not line up wit the horizontal and vertical directions. The principal axes are rotated slightly.

It is necessary to calculate the section properties about the principal (rotated) axes. Then consider your applied load as a vector sum of two forces, one acting in each of the principal directions. And then calculate the shear flow separately for the forces acting about the two axes, and add together the results. It's a bit long winded unfortunately!

There is a simplified method of doing the calculation as well, which uses only the x and y coordinates of the section (not the rotated coordinates of the principal axes). This saves on a lot of trig! You can find an explanation of this here: bending stress and shear flow in thin walled sections

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