5003 Lectures - Faculty of Engineering and Applied Science

E5003 - Ship Structures I 210
© C.G. Daley
The lower flange is symmetrical with the upper
and will have a shear flow of the same magnitude
but opposite in direction.
The shear flow as drawn shows the directions of
shear in the direction of the applied force. If we
think instead of the reaction to the applied force,
we have the sketch at left.
In this case the applied force is shown pushing
directly down on the web. In this case the vertical
forces oppose each other and produce no moment.
However, the horizontal forces, while equal in
magnitude, are separated by 190mm and produce
a couple of 1879 x 190 = 355300 N-mm. This couple
is a torsion acting on the section.
In order to eliminate the torsion, we would need to
apply the load Q at the shear center ‘e’ to the left
of the web. We can find the location of ‘e’ as
follows;