Concept of Shear centre with example

1. SHEAR CENTRE
PRESENTED BY:
1. Aiswarya Ray-B120100ME
2. Anirudh Ashok-B120325ME
3. Mahesh M.S.-
4. Harikumar
5. Ashish Ranjan
6. Deepesh

2. NEED FOR FINDING LOCATION
OF SHEAR CENTRE
 In unsymmetrical sections, if the external applied forces
act through the centroid of the section, then in
addition to bending, twisting is also produced.
 To avoid twisting, and cause only bending, it is
necessary for the forces to act through the particular
point, which may not coincide with the centroid.
 The position of the this point is a function only of the
geometry of the beam section. It is termed as shear
center.

3. WHAT IS SHEAR CENTRE??
 Shear center is defined as the point on the
beam section where load is applied and no
twisting is produced.
- At shear center, resultant of internal
forces passes.
- On symmetrical sections, shear center is
the center of gravity of that section.
- Such sections in which there is a sliding
problem, we place loads at the shear
center.

4. 4
PROPERTIES OF SHEAR CENTRE
1) The shear center lays on the axis of symmetry.
2) Thus for twice symmetrical section the shear centre is
the point of symmetry axes intersection.
3) If the cross section is composed of segments
converging in a single point, this point is the shear
centre.
4) The transverse force applied at the shear centre does
not lead to the torsion in thin walled-beam.
5) The shear centre is the centre of rotation for a section
of thin walled beam subjected to pure shear.
6) The shear center is a position of shear flows resultant
force, if the thin-walled beam is subjected to pure
shear.

5. DETERMINING LOCATION OF
SHEAR CENTRE

7. SHEAR STRESSES IN THIN-WALLED
OPEN SECTIONS
Consider a beam having a thin-walled open section as shown
above.
Now consider an element of length Δx at section x as shown.
Now for Force equilibrium in x-direction,
휏푠푥 푡푠Δx - 0∫s 휎푥tds + 0∫s (σx +
휕휎푥
휕x
Δx) tds = 0

8. ie. 휏푠푥= –
1
푡푠
0∫s 휕σx
휕x
tds ts → wall thickness at s
Observing My = 0, normal stress, σx is given by
σx =
y퐼푦−푧퐼푦푧
퐼푦푧
2−퐼푦퐼푧
Mz
Hence,
휕σx
휕x
=
y퐼푦−푧퐼푦푧
퐼푦푧
2−퐼푦퐼푧
휕Mz
휕푥
Recalling from strength of materials that
휕Mz
휕푥
= – Vy, and substituting
We get, 휏푠푥 =
Vy
푡푠
1
퐼푦푧
2−퐼푦퐼푧
푠
y퐼푦 − 푧퐼푦푧 푡푑푠
0
휏푠푥 =
Vy
푡푠
1
퐼푦푧
2−퐼푦퐼푧
푠
푦푡 푑푠 − 퐼푦푧 0
[퐼푦 0
푠
푧푡 푑푠]
휏푠푥 = 휏푥푠 =
Vy
푡푠
1
퐼푦푧
2−퐼푦퐼푧
[퐼푦푄푧 − 퐼푦푧푄푦]

9. 푄푧 , 푄푦 → 1st moments of area about the z and y-axis respectively
Since for shear centre twisting caused is zero,
The moment due to shear stress = The moment due to the load applied
푉푦푒푧 = Moment of 휏푠푥 about centroid

10. Shear Centre in Real
Life Situations

11. Purlins
 Construction of Purlins
 A purlin is any longitudinal, horizontal, structural
member in a roof.

12. The point of application of load is important
, depending on the cross-section of purlin.

13.  If an unsupported channel section is loaded closer to its
shear centre, it'll take more load before buckling than if
you put the load over the centre of the channel, the
application being that you can get more load out of the
same member.
 Useful in design of thin walled open steel sections as
they are weak in resisting torsion.

14. SHEAR CENTRE PROBLEM
 A beam has the cross section composed of
thin rectangles as shown in fig. The loads on
the beam lie in a plane perpendicular to the
axis of symmetry of cross section and so
located that the beam does not twist .Bending
load cause for any section a vertical shear V.
determine the location of shear center.

17. Q. (a) Locate the shear center S of
the hat section by determining the
eccentricity, e. (b) If a vertical shear
V =10 kips acts through the shear
center of this hat what are the values
of the shear stresses τA at the location
and direction indicated in Figure.

18. Solution:
The centroidal principal moment of inertia of the beam section is

19. The first moment areas of the locations shown in figure are

20. The shear stresses at these
locations are
τA = 2618 psi

21. The moment about point A is
the resultant forces are