twisting at shear center / centroid

In the notes , the author stated that when the force is applied through the centorid of cross section , the channel will bend and twist. but , on the second page , the author stated that the shear center lies on an axis of symmetry of member's cross sectional area... So, i am confused whether the member will twist or not when the force P is applied thru the centroid or shear center ....

Why the shear center is located at O , which is located outside of the C channel ?

Is there anything wrong with the notes ?

Nothing wrong with the book or notes. This is just a little unintuitive at first until you take some time to internalize it.

The shear center lies on an axis of symmetry. If two axis of symmetry exist, it will lie on both of them at the centroid. Otherwise, it is only constrained to lie on one of them. The location of the shear center on the unsymmetric axis is determined by the equation for e.

If the centroid is located at the shear center, then it will not twist. However, if they are in different locations, twisting will happen at the centroid if a force is applied there.

According to McGraw Hill Companies inc,

Shear Center: Of any cross section of a beam, that point in the plane of the cross section through which a transverse load must be applied in order that there will be only bending of the section and no twisting.

This picture shows how this works:

If the downwards force is applied along the screw at any point other than the shear center, twisting will occur.

• if it is only constrained to lie on one of them. then , it is considered to be the location of the shear center on the unsymmetric axis ? just like the case in 7-23a ? Nov 28 '16 at 7:10
• Your grammar makes your question a bit confusing. I understand if you aren't a native speaker, but please throw me a bone here. Dec 4 '16 at 19:57
• Can you explain what is the force is applied along the once vertical , unsymmetrical axis that passes thru centroid ??? Since the applied force already passing thru centroid , it's alreday the symmertical axis , right ? Dec 28 '16 at 11:19