Logic behind location of shear centre

Question

When we apply a vertical shear force Sy and the structure is symmetric about x axis, then why is it logical to have the position of shear center at the location of intersection of line of action of shear force and x axis?

If we go by basic definition then the moment about the shear center due to shear forces should be 0, so the line of action of external shear force is logical, but when we consider moment about x- axis, rather than getting cancelled, I think they get added.

eg. This is the situation.

Now look at the final solution depicting the shear flows

So as I suggested, instead of moments being equal and opposite, I think they exactly equal in magnitude and direction. So where am I getting a wrong interpretation?

EDIT:- I tried to keep the question general and explained accordingly, to be specific the question is as follows:-

The thin-walled single cell beam shown in Fig. 20.11 has been idealized into a combination of direct stress carrying booms and shear stress only carrying walls. If the section supports a vertical shear load of 10 kN acting in a vertical plane through booms 3 and 6, calculate the distribution of shear flow around the section.

Boom areas: B1 =B8 =200mm2, B2 =B7 =250mm2, B3 =B6 =400mm2, B4 = B5 =100mm2.

x is horizontal and y is vertical

It is a closed section beam. An idealized version of an airplane wing

Answer 1

why is it logical to have the position of shear center at the location of intersection of line of action of shear force and x axis?

That statement isn't logical. I think you have misunderstood how the shear centre is defined.

The shear centre is the point such that an applied force passing through the S.C. does not cause any rotation of the section. In other words, if you apply a shear force through the shear centre, it does not cause any torsion in the beam but only bending.

The position of the S.C. depends only on the geometry of the beam section, not on the applied loads.

You can apply a shear force at any point on the beam. If the force does not pass through the S.C, you need to replace it by an equal shear force through the S.C, plus a moment about the S.C. You can then find the deflections and stresses due to bending (caused by the force through the S.C.) and torsion (caused by the moment about the S.C.) separately, and add them together to get the total deflections and stresses.

In your aircraft wing example, the shear forces caused by the aerodynamic lift and drag will act through the centre of pressure of the wing, and the centre of pressure is usually not the same point as the shear centre. The aerodynamic forces will therefore cause a combination of bending and twisting in the wing.