The Shear-Center concept - NTNU · Shear Center A lateral load acting on a beam will produce...

Georgia Tech - Aerospace Eng 7/7/2009 1

The Shear-Center concept

Recall restrictions on shear stress associated

with bending

– Symmetric section

– Lateral acting in plane of symmetry

What happens when loads act in a plane that is

not a plane of symmetry?

– Loads must be applied at particular point in the

cross section, called shear center, if the beam is to

bend without twisting


2/hy

yy11

1

dAF

2/hy

yy22

1

dAF

zz1

I

yM

zz2

I

ydxx

MM


1F 2F

2/hy

yy 1

ydAQ

dAFFdxb2/hy

yy1212

1

zzbI

QV

0


Shear Stress distribution

Constrained by the shape of the cross

section

– Its resultant acts at the shear center

Not necessarily the centroid


Shear Center A lateral load acting on a beam will produce

bending without twisting only if it acts through the shear center

The shear center

– Is a property of the cross section like the centroid

– It lies on an axis of symmetry For a doubly symmetric section S and C coincide


yV

Why a property of the cross

section?

Resultant shear yV

Twist moment changes


Shear Center

Doubly or singly symmetric section

yP

zP

zz

zy

bI

QP

yy

yz

bI

QP

ydAQz

zdAQy

yP

zP


Shear Center

How to locate Shear Center?

– Doubly symmetric cross sections

Coincides with centroid

– Singly symmetric cross sections

Lies on the axis of symmetry

Unsymmetric Cross sections

– Thin-walled open sections

Opposite side of open part


Thin-walled Open Cross

sections Shear stress distribution

– Derived from equilibrium similar to rectangular

sections and I beams

– Assumed constant over thickness

Restrictions

– Wall thickness << cross section dimensions (width

and height)

– Open section


Restrictions

– Wall thickness << cross section dimensions

(width and height)t

w

a

x x

2centxx aAreaII

23

xx wta12

wtI

1

a

t

12

1wtaI

22

xx

0

wt


Thin-walled open Cross

Sections

diagramforceshearpositive



Sections s

011 dAF

s

022 dAF t

zIII

IMy

III

IM

2yzzzyy

yzz

2yzzzyy

yyz1

zIII

Idxx

MM

yIII

Idxx

MM

2yzzzyy

yzz

z

2yzzzyy

yyz

z

2



Sections s

011 dAF

s

022 dAF t

dAFFdxts

02121



Sections

zIII

IMy

III

IM

2yzzzyy

yzz

2yzzzyy

yyz1

zIII

Idxx

MM

yIII

Idxx

MM

2yzzzyy

yzz

z

2yzzzyy

yyz

z

2

dAFFdxts

02121

yyzzyy2yzzzyy

yQIQI

III

Vft

s

0z dAyQ

s

0y dAzQ



Sections

yyzzyy2yzzzyy

yQIQI

III

Vft

For Bending by loads parallel to the y axis

For Bending by loads parallel to the z axis

zyzyzz2yzzzyy

z QIQIIII

Vft

Principal axes

zz

zy

I

QVft

yy

yz

I

QVft


Summary of Equations for

Thin-walled Open Cross

Sections

zyzyzz2yzzzyy

zyyzzyy2

yzzzyy

yQIQI

III

VQIQI

III

Vft

Principal axes

yy

yz

zz

zy

I

QV

I

QVft

tf:flowShear

Shear flow due to combined loads Vy and Vz

s

0z dAyQ

s

0y dAzQ


Shear stress distribution

strategy

1. Determine location of centroid and Iyy, Izz and Iyz as needed

- (symmetric sections subject to Vy needs only Izz)

2. Divide section into elements according to geometry (change in slope)

3. Start with a vector s following element center line from a free end

4. Calculate first moment of area(s). This determines the shear flow distribution

- Negative shear value indicate direction of shear flow opposite to assumed vector s


Shear stress distribution

strategy 4. Calculate first moment of area(s). This

determines the shear flow distribution

- For symmetric sections subject to bending about one axis

- Elements parallel to bending axis

- Linear distribution

- Elements normal to bending axis

- Parabolic distribution

- For unsymmetric sections shear flow in all elements is parabolic

5. When moving from one element to another the end value of shear in one element equals the initial value for the subsequent element (from equilibrium)


Ex. A cantilevered beam is

subject to tip load P. The

beam has a channel section

shown in figure.

Determine the shear flow

distribution, the maximum

shear stress and the shear

center

P

a

a

4a

t

y

z


Solution:

Max. shear Vy= +P

Section is symmetric

y and z are principal axes

Section subject to Vy only

P

a

a

4a

t

y

z

P P

+

zz

z

zz

zy

I

QP

I

QVft

3

32

3

zz ta3

40

12

a4ta2at

12

at2I

12

34

Element 1-2

zz

z21

I

QPf 21


z

Element 1-2

zz

z21

I

QPf 21

12

s

2a

a2tsQ21z

2321a20

sP3

ta40

a2tsP3sf

a20

P32f 21

a20

P3


Element 2-3

zz

z

zz

z2132

I

QP

a20

P3

I

QP2ff 3232

2

sa2tsQ

32z

z

12

sa20

P3

3

s

2

sa2

0a4sQ0sQ3232 zz

2z ta2a2sQ

32

332ta40

2

sa2tsP3

a20

P3)s(f

a20

P3)a4(f)0(f 3232

a20

P3

a20

P3)a2(f 32


Element 2-3

12

sa20

P3

3

s332

ta40

2

sa2tsP3

a20

P3)s(f

a20

P3)a4(f)0(f 3232

a20

P3

a20

P3)a2(f 32

a20

P3

a20

P3


Element 3-4

343

ta40

a2tsP3

a20

P3)s(f

a20

P3)0(f 43

0)a(f 43

12

sa20

P3

3

s

a20

P3

a20

P3

4s

-2a


Resultant force in each element

40

P3a

a20

P3

2

1

dssfFFas

0s214321

21F 12

sa20

P3

3

s

a20

P3

a20

P3

4s

Pa4a20

P3

3

2a4

a20

P3

dssfFa4s

0s3232

32F

43F


Shear Center

Is on an axis of symmetry

Take moment about any point

Points 2 or 3 are the easiest

( no contribution of element 2-3)40

P3F 21

12

sa20

P3

3

s

a20

P3

a20

P3

4s

PF 32

40

P3F 43

e

PVy

a4FeP3@M 21

a10

3ea4

40

P3eP

The Shear-Center concept - NTNU · Shear Center A lateral load acting on a beam will produce...

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