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Shear centre

The Shear centre is a point where a shear force can act without producing any twist in the section. At Shear centre of a section the applied force is balanced by the set of shear forces obtained by summing the shear stresses over the section.
Shear centre is also known as the centre of twist.
When the load does not act through the shear center, in addition to bending, a twisting moment will develop  in the section. 
The location of the shear center is independent of the direction and magnitude of the transverse forces.
In unsymmetrical sections and in particular angle and channel sections, the summation of the Shear Stresses in each leg gives a set of Forces which must be in equilibrium with the applied Shearing Force.
(a) Consider the angle section which is bending about a principal axis and with a Shearing Force F at right angle to this axis. The sum of the Shear Stresses produces a force in the direction of each leg as shown above. It is clear that their resultant passes through the corner of the angle and unless F is applied through this point there will be a twisting of the angle as well as Bending. This point is known as The Shear Centre or Centre of Twist
(b) For a channel section with loading parallel to the Web, the total Shearing Force carried by the web must equal F and that in the flanges produces two equal and opposite horizontal forces. It can be seen that for equilibrium the applied load causing F must lie in a plane outside the channel as indicated. 
Note:
1. In case of a beam having two axes of symmetry, the shear centre coincides with the centroid.
2. In case of sections having one axis of symmetry, the shear centre does not coincide with the centroid but lies on the axis of symmetry.
3. When the load passes through the shear centre then there will be only bending in the cross section and no twisting.

Example:

1. A H-section with unequal flanges as shown in fig. If the smaller flange is 16mm x 100mm the larger flange is 16mm by 200mm, and the web is 9mm by 284mm. Find the location of the shear center from the center of the bigger flange.

Solution:
The shear centre lies on the axis of symmetry and at a distance of x form the center of large flange.
Taking the moments about shear center
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Comments

  1. Praveen30 November 2011 at 05:47

    thank u sir....I got the concept...

    ReplyDelete
  2. shrikant M23 January 2013 at 22:56

    can this concept used for unsymmetrical lipped c section ?

    ReplyDelete
  3. Prof. Kodali Srinivas11 November 2015 at 21:00

    Sir,
    I have some conceptual problems about shear centre of angular sections.How can i contact you?i have attached the problem and just waiting for the answer of e=? with all the procedures that will clarify my doubt.my doubt is how to calculate ý at angular sections.

    With regards

    Attachments area
    Preview attachment 20151110_185806.jpg
    Image
    20151110_185806.jpg

    ReplyDelete
    Replies
    1. Prof. Kodali Srinivas11 November 2015 at 21:02

      Dear Abisek,
      For a channel section with loading parallel to the Web, the total Shearing Force carried by the web must equal F and that in the flanges produces two equal and opposite horizontal forces. It can be seen that for equilibrium the applied load causing F must lie in a plane outside the channel. Your problem is also have shear center out side of the section. But the total shear force carried by the web and vertical projections, vertical components of inclined portion must equal to external force. The horizontal component of inclined portions will produce anti clock wise moment and that will be balanced clock wise moment produced by the force witch will act at shear center. Using this concept you can find 'e"

      Delete
    2. Unknown9 March 2018 at 21:31

      The flange of channel section are situated same side. How can they carry equal and opposite force.please explain..

      Delete
  4. Unknown3 November 2017 at 01:48

    sir i need the formula of shear center for lipped channel section

    ReplyDelete

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