Shear Centre definition:

To avoid twisting and cause only bending, the transverse forces must act through a point which may not necessarily coincide with the centroid, but will depend upon the shape of the section. Such a point is known as the ‘Shear Centre’.

Shear centre

The point of intersection of the bending axis with the cross section of the beam is defined as ‘shear centre’.

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Shear Centre definition:

The shear center always falls on a cross-sectional axis of symmetry.

If the cross section contains two axes of symmetry, then the shear center is located at their intersection. Notice that this is the only case where shear center and centroid coincide.

Shear centre is the point through which the resultant shear forces acts.

Shear flows from tension side to compression side when looking on section in the direction of increasing bending moment.

Shear centre

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Shear Centre of thin walled open sections:Examples

Channel section:Shear stress at a distance ‘𝑥 ′ in the bottom horizontal

member AB as shown in

Figure is given by ,

1𝑞 =𝐹𝐴𝑦

𝐼 𝑡

=𝐹

× 𝑥𝑡 ×ℎ

𝐼 𝑡 2𝑏

𝐹1 = 𝑞1𝑡 𝑑𝑥

0𝑏𝐹 ℎ

∴ 𝐹1 = 𝐼 𝑡

× 𝑥𝑡 ×2

× 𝑡 𝑑𝑥

0

Shear centre

1

𝑥 AB

C

b

𝑡𝑓

𝐹1

𝑥𝑧

G

𝐹2

h 𝑡𝑤

a b𝛿𝑥𝐹

cd

S

e

𝑡

T

T+ 𝛿𝑇

𝑦u

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1∴ 𝐹 =𝐹𝑏

ℎ

𝐼 𝑡× 𝑥𝑡 ×

2 × 𝑡 𝑑𝑥

=𝐹𝑡ℎ 𝑏2

0

× =𝐹𝑡ℎ𝑏2

2𝐼 2 4𝐼

The shear force 𝐹1is same in the top

member also.

Shearing stress in the vertical member BC

at a distance ′𝑦′ from B (as shown in Fig.)

𝑞2 =𝐹𝐴𝑦

𝐼 𝑡= 0 (since 𝑦 = 0)

∴ 𝐹2 = 0

Shear centre

𝐹1

𝑥 AB

C

b

𝑡𝑓

𝐹1

𝑥𝑧

G

𝐹2

h 𝑡𝑤

a b𝛿𝑥

cd

S

e

𝑡

T

T+ 𝛿𝑇

𝑦u

O

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Taking moments about‘O’,𝐹1 × ℎ = 𝐹 ×𝑒

𝐹𝑡ℎ𝑏2

4𝐼× ℎ = 𝐹 × 𝑒

∴ 𝑒 =𝑏2ℎ2𝑡

4𝐼

Shear centre

B

C

b

𝑡𝑓

𝐹1

𝑥𝑧G

𝐹2

h 𝑡𝑤

a b𝛿𝑥𝐹1

cd

S

e

𝑡

T

T+ 𝛿𝑇

𝑦

𝑥 Au

O

𝐹

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Shear centre

G X

Y

Fig. Shear centre of an equal –leg angle section

S

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Fig. Shear centres S of sections consisting of two intersecting narrowrectangles

Shear centre

S S

S

SS

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Shear centre of a semi circularsection:

𝑞 =𝐹𝐴𝑦𝐼 𝑡

𝜃

𝐴𝑦 = 𝑅 𝑑𝛼 . 𝑡 × 𝑅 cos 𝛼 = 𝑅2t sin𝜃0

𝜋 𝜋

𝐼 = 𝑦2 dA = 𝑅 cos 𝜃 2 𝑅𝑑𝜃. 𝑡

=

0 0

𝜋𝑅3𝑡

2

𝜋𝑅𝑡∴ 𝑞 =

2𝐹sin𝜃

Force acting on the area 𝑅𝑑𝜃 × 𝑡 = 𝑞 × 𝑅𝑑𝜃× 𝑡

Shear centre

eO

R𝛼𝑞 𝑅

cos𝛼

𝑡

𝑑𝛼

𝑑𝜃 𝜃

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Shear centre of a semi circular section:

Moment of this force about O =𝑞 × 𝑅𝑑𝜃 × 𝑡× 𝑅

Total moment of the force due to

shearing stress on the cross section𝜋

0

𝜋

= 𝑞𝑅2𝑡 𝑑𝜃 = 𝑞𝑅2𝑡 𝑑𝜃

0𝜋

0

=2𝐹

𝑅2𝑡sin 𝜃 𝑑𝜃 = 𝐹 ×4𝑅

𝜋𝑅𝑡 𝜋

This must be equal to 𝐹 × 𝑒

𝐹 × 𝑒 = 𝐹 ×4𝑅

𝜋

𝜋∴ 𝑒 =

4𝑅≅ 1.27𝑅

Shear centre

eO

R𝛼𝑞 𝑅

cos𝛼

𝑡

𝑑𝛼

𝑑𝜃 𝜃

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9/11

Problem:

A channel section has flanges 12 cm x 2 cm and web 16 cm x 1cm. determine the shear centre of the channel.

Shear centre - Problem

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Solution:

Flanges dimensions=12 cm x

2 cm web dimension=16

cm x 1 cm.

Shear centre, 𝑒 =𝑏2ℎ 2

4𝐼

b= 12 cm, h=16 cm, t= 2 cm

𝐼 =1 × 163 12 × 23

12 12+ 2 × +

2 × 12 × 2 × 92 = 4245.33 𝑐𝑚2

2 2

4 × 4245.33∴ 𝑒 =

12 × 16 × 2 = 4.3413 cm

Shear centre - Problem

𝐹1

B

C

b

𝑡𝑓

𝐹1

𝑥𝑧G

𝐹2

h 𝑡𝑤

a b𝛿𝑥

cd

S

e

𝑡

T

T+ 𝛿𝑇

𝑦

𝑥 Au

O

𝐹

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