If a beam is subjected to bending moments and shear force in a plane, other than the plane of geometry, which passes through the centroid of the section, then bending moment will be accompanied by twisting. In order to avoid twisting and cause bending only, the transverse forces must act through a point which may not coincide with the centroid, but will depend upon the shape of the section and such a point is termed as shear centre.

Figure 1

Consider a channel section as shown in figure 1. Now we shall find the position of the plane through which the vertical loads must act so as to produce simple bending, with the x-axis as neutral axis.

It may be assumed that the vertical shearing force, F at the section is taken up by the web alone. In the flanges, there will be horizontal shear stresses which will be denoted by q.

Let us consider an element ‘abcd’ cut from the lower flange by two adjacent cross-sections apart and by a vertical plane parallel to the web and at distance ’u’ (which is variable) from the free end of the lower flange. The difference in tensile forces T and must be equal to the shear force on the side ‘ad’ of the element. Assuming a uniform distribution of shear stress (since the thickness is small) over the thickness, we have,

The integration being carried out over the portion ‘ab’ of the flange.

The stress per unit length of the centre line of the section,

Therefore, it is seen that q is proportional to u.

The maximum value of .

At the junction of the flange and web, the distribution of the shear stress is complicated, so we may assume that the equation holds good for u = 0 and u = b.

The average shear stress

The longitudinal shear force in the top and bottom of the flange

The couple about the z-axis of these shear forces

Let us assume that the vertical shear force F acts through point ‘o’, the shear centre at a distance c from O on the centre line of the web.

The twisting of this section is avoided if

which gives the position of the shear centre.

**Note:** the shear centre for cross-sectional areas having one axis of symmetry, is always located on the axis of symmetry. In the case of the I-beam which is symmetrical about both the x-axis and y-axis, the shear centre coincides with the centroid of the section. The exact location of the shear centre for unsymmetrical sections are complicated and can be located by inspection.

**EXAMPLE – 1**

#### To locate the shear centre of the unsymmetrical I-beam cross section as shown in figure below:

Here

Taking moment about the point D

**EXAMPLE – 2**

**TO DETERMINE THE SHEAR CENTRE FOR THE SECTION SHOWN IN FIGURE:**

Resolving at A and B and equating moments

**Example – 3**

**TO DETERMINE THE SHEAR CENTRE OF THE CHANNEL SECTION SHOWN IN FIGURE**

shashank

thanks for the above cases …they helped a lot.

Obeka Ogiri Samuel

this is wonderful.

Robert Ouko

it's worth it for us engineers.

Rahul Dornal

Thanks…..

Anonymous

thank you……. sir can u help me to find both x and y ordinate of shear center of unequal lipped c section (having lips inward)?

Anonymous

thank you for above explanation.

Sir please help me for y ordinate(vertical) of shear center of unequal c section.

Enamul Haque

Excelent.Thanks.

Ibrahim Yusuf

omg

Aditya Adi

thanks

Kalle Musalilwa Silwimba

thanks alot

Khat Sam Ath

Thanks so much

Abhijith Abhi

thank u

Prathikshen Nambiar Selvadorai

Can someone help me figure out how to calculate the shear center of a completely unsymmetrical box shape cross-section. Where all its four sides have different thickness. Any help is much appreciated.