the open cross section shown is thin walled with a constant thickness.(adsbygoogle = window.adsbygoogle || []).push({});

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 thats 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=[tex]\int[/tex]zdA=[tex]\int[/tex][tex]\int[/tex](r*sinθ*r*dr*dθ)=-cosθ*R^{2}t (since t^{2}<<R)

Iyy=πR^{3}t (since t^{2}<<R)

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

this integration gives me 0

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# Homework Help: Find the location of the shear center of this shape

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