Reading about shearing stresses in beams, it is commonly stated that it is necessary to load a section through its shear center, so no twisting is applied to the beam. This is specially important in thin-walled open section, that have very weak torsion strength.

But how do you desing a beam in such a way? For example: if I'm designing a floor beam with a C-shape section, how do I enforce that the shearing forces act through the shear center?



For a C channel with the following dimensions:

b = width of the flange measured to the center of the web, inches or mm

c = Distance to the shear center, inches or mm

h = Height of section measured to the center of flanges, inches or mm

I = The second moment of area, in^4 or mm^4

t = flange thickness, inches or mm

$ c=\frac{b^2h^2t}{4I_x}$

If we apply load on this offset falling outside of the channel opposing the flanges there will be no twisting torque.

But usually in the construction C channels are rarely used as beams, the favorite section fo steel is I beams, or similar shapes because of their symmetry, a large relative I and the fact that they have been designed as beam and already include many requirements to be used as a beam by codes. However, C channels in most cases (if ever) are used in situations where they are laterally braced by the floor slabs or roof sheathing so the torque created by loading, not through the shear center is resisted by those members.

  • $\begingroup$ Thanks for the explanation! It is now more clear. $\endgroup$ – Msegade Apr 22 '20 at 9:44

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