Find the location of the shear center of this shape

[Dell]

the open cross section shown is thin walled with a constant thickness.
find the location of the shear center

i have solved this sort of question but never for a circular cross section, at first i thought it should be the same just using polar coordinates instead of Cartesian coordinates

i know that when the force is applied at the shear center, the moment e*V is equal to the moment of the stresses

since i have one axis of symmetry (y) i know that the shear center must be on this line, now all that's left is to find the distance "e" from the center of the shape to the shear center

computing the moment about the center of the cross section

ΣMo=Vz*e

Qy=\intzdA=\int\int(r*sinθ*r*dr*dθ)=-cosθ*R²t (since t²<<R)

Iyy=πR³t (since t²<<R)



ΣMo=\int\int\frac{Vz*Qy}{Iyy*t}*da*R=\frac{Vz}{π*t}*\int\int-cosθ*r*dr*dθ

this integration gives me 0

 

[pongo38]

There's more than one axis of symmetry. How many are there? What is the implication for the position of the shear centre?

 

[Dell]

had there been more than one axis of symmetry the shear center would be the point where they meet, in this case since the shape is open at the right hand side, i do not see more that the one axis of symmetry

 

[pongo38]

Sorry. I missed the gap.

 

[pongo38]

You are probably familiar with cases such as a channel section with linearly varying shear stresses. How do you think the shear stresses might vary with theta?

 

[Dell]

i would say 0 at θ=0, θ=2pi, and maximum at θ=pi,