SHEAR CENTRE

PRESENTED BY:

1. Aiswarya Ray-B120100ME

2. Anirudh Ashok-B120325ME

3. Mahesh M.S.-

4. Harikumar

5. Ashish Ranjan

6. Deepesh

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.

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

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.

DETERMINING LOCATION OF

SHEAR CENTRE

Now for Force equilibrium in x-direction,

𝜏𝑠𝑥𝑡𝑠∆x - 0∫s 𝜎𝑥tds + 0∫

s (σx + 𝜕𝜎𝑥

𝜕x∆x) tds = 0

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 ∆x at section x as shown.

ie. 𝜏𝑠𝑥= –1

𝑡𝑠0∫

s 𝜕σx

𝜕xtds ts → wall thickness at s

Observing My = 0, normal stress, σx is given by

σx = y𝐼𝑦−𝑧𝐼𝑦𝑧

𝐼𝑦𝑧2−𝐼𝑦𝐼𝑧

Mz

Hence,

𝜕σx

𝜕x=

y𝐼𝑦−𝑧𝐼𝑦𝑧

𝐼𝑦𝑧2−𝐼𝑦𝐼𝑧

𝜕Mz

𝜕𝑥

Recalling from strength of materials that 𝜕M

z

𝜕𝑥= – Vy, and substituting

We get, 𝜏𝑠𝑥 =Vy

𝑡𝑠

1

𝐼𝑦𝑧2−𝐼𝑦𝐼𝑧

0𝑠y𝐼𝑦 − 𝑧𝐼𝑦𝑧 𝑡𝑑𝑠

𝜏𝑠𝑥 =Vy

𝑡𝑠

1

𝐼𝑦𝑧2−𝐼𝑦𝐼𝑧

[𝐼𝑦 0𝑠𝑦𝑡 𝑑𝑠 − 𝐼𝑦𝑧 0

𝑠𝑧𝑡 𝑑𝑠]

𝜏𝑠𝑥 = 𝜏𝑥𝑠 =Vy

𝑡𝑠

1

𝐼𝑦𝑧2−𝐼𝑦𝐼𝑧

[𝐼𝑦𝑄𝑧 − 𝐼𝑦𝑧𝑄𝑦]

𝑄𝑧 , 𝑄𝑦 → 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

𝑉𝑦𝑒𝑧 = Moment of 𝜏𝑠𝑥 about centroid

Shear Centre in Real

Life Situations

Purlins

Construction of Purlins

A purlin is any longitudinal, horizontal, structural

member in a roof.

The point of application of load is important

, depending on the cross-section of purlin.

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.

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.

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.

Solution:

The centroidal principal moment of inertia of the beam section is

The first moment areas of the locations shown in figure are

The shear stresses at these

locations are

τA = 2618 psi

The moment about point A is

the resultant forces are