unsymmetric beams.pdf

R&DE (Engineers), DRDO

Unsymmetrical

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Ramadas Chennamsetti

Bending of Beams


Introduction

� Beam – structural member – takes transverse loads

� Cross-sectional dimensions much smaller than length

� Beam width same range of thickness/depth

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Ramadas Chennamsetti2

� Beam width same range of thickness/depth� Thin & thick beams� If l ≥≥≥≥ 15 t – thin beam� Thin beam – Euler – Bernoulli’s beam� Thick beam – Timoshenko beam


Introduction

� In thin beams – deformation due to shear negligible

� Thick beams – shear deformation considered

� Beam – one dimensional structural member

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� Beam – one dimensional structural member – length is very high than lateral dimensions

� Various parameters => function of single independent variable


Sign convention

� Following sign convention is followed

y

z

oy

z

My

Mz

+ ve moments

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x

yo

Orientation of beam

y

x

Mx

x

z

ψψψψx

ψψψψy

ψψψψz

y+ve rotations

Beam bending in xy and xzplanes


Sign convention

� Shear force and bending moments

y

z

My

Mz + +z

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y

x

Mx

M M F F

y

z

+

M M

+

F F

x

z

Positive shear force and bending moments

Convex upward +ve direction


Unsymmetrical bending

P

z

P

zθ

Load applied in the plane of symmetry

Load applied at some orientation

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yG

Symmetrical cross-sectional – load applied in the plane of symmetry - xz

Bending takes place in xz plane

yG

Bending takes place in both planes – xz and xy


Unsymmetrical bending

� Cross-section is unsymmetrical

z P

z

P

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y y

Bending takes place in both planes


Euler – Bernoulli’s beam theory

� Basic assumptions � Length is much higher than lateral dimensions

– l ≥ 15 t� Plane cross section remains plane before and

after bending AA’

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after bending� Stresses in lateral directions

negligible� Thin beam

strain variation is linear across cross-section� Hookean material – linear elastic

A

B

z

x

A

B

A’

B’


Bending of arbitrary cross section beam

� An arbitrary cross-section beam oriented along x-direction

( )cbzayEE

cbzay

xxxx

xx

++==++=

εσε

M

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y

z

x

dAσxx

Mz

MyG

z

y

Nx

Equilibrium of the section requires

( )dAcbzayEdAdN

MM

NF

xxx

zy

xx

++==

=>

=

∑∑∑

σmoments External ,

Integrate this for total force in x - direction

xyz csys coincides with centroid



� This equals to external force in ‘x’ direction = Nx

( )dAcbzayEdANAA

xxx ++==

∫∫∫

∫∫ σ

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EA

NcEcAN

dAEcdAzEbdAyEaN

xx

AAA

x

==>=

++= ∫∫∫

Origin of xyz co-ordinate system is selected at centroid

- Coefficients of ‘Ea’ and ‘Eb’ constants vanish



� Moment wrt ‘y’ axis ( )

dAEzdM

zdAdM

xxy

xxy

εσ

=

=

Integrating over cross-sectional

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y

z

x

dAσxx

Mz

MyG

z

y

Nx

Integrating over cross-sectional area ‘A’

( )

dAzEcdAzEb

dAyzEaM

dAcbzayEzM

AA

A

y

A

y

2∫∫

∫

∫

+

+==>

++=



� Moment about ‘y’ axis

(1) -

2

yyyzy

AAA

y

IEbIEaM

dAzEcdAzEbdAyzEaM

+==>

++= ∫∫∫

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� Moment about ‘z’ axis

(1) - yyyzy IEbIEaM +==>

( )

(2) -

2

yzzzz

AAAA

z

IEbIEaM

dAyEcdAyzEbdAyEadAycbzayEM

+==>

++=++= ∫∫∫∫

‘a’ and ‘b’ unknowns – solve simultaneous algebraic equations (1) and (2)



� Solving (1) and (2) ‘a’ and ‘b’ values are as following

( ) ( )zzyyyz

yzzzzy

zzyyyz

yyzyzy

IIIE

IMIMb

IIIE

IMIMa

−+−

=−−

=22

,

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( ) ( )zzyyyzzzyyyz IIIEIIIE −−

Strain is

( ) ( ) EA

Nz

IIIE

IMIMy

IIIE

IMIMx

zzyyyz

yzzzzy

zzyyyz

yyzyzyxx +

−+−

+−−

= 22

ε

This represents a straight line



� Bending stress – multiply axial strain (bending strain) with Young’s modulus (E)

� Neutral axis /plane – no stress/strain => ‘0’

=++−

+− yzzzzyyyzyzy NIMIMIMIM

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( ) ( )

( )

−−

==>

−−

−−

==>

=+−

+−+

−−

yzzzzy

yyzyzy

zzyyyzx

yzzzzy

yyzyzy

x

zzyyyz

yzzzzy

zzyyyz

yyzyzy

IMIM

IMIM m

IIIA

Ny

IMIM

IMIMz

A

Nz

III

IMIMy

III

IMIM

slope

0

2

22



� Equation of neutral axis

( )zzyyyzx

yzzzzy

yyzyzy IIIA

Ny

IMIM

IMIMz −−

−−

= 2

Neutral axis doesn’t pass through

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y

z

G

N

A

α

Neutral axis doesn’t pass through origin – centroid of cross-section

If no axial load => Nx = 0

N.A equation z = m y Passes through centroid

yzzzzy

yyzyzy

IMIM

IMIMα m

−−

== tan slope,



� Neutral axis (N. A)� Equation of neutral axis completely depends

on geometry and loading – moments and axial load

� Axial load offset N. A. – does not pass through centroid of cross-section

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centroid of cross-section� Orientation (slope) purely depends on

moments and geometry� Orientation not decided by axial load� Independent of material properties - isotropic� Simplifying equation if co-ordinate system

coincides with principal axes



� Bending in a single plane x-z – no axial load=> Nx = 0 ( )

yIMIM

z

IIIA

Ny

IMIM

IMIMz

yyzyzy

zzyyyzx

yzzzzy

yyzyzy

−

=

−−

−−

= 2

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yI

Izy

IM

IMz

yIMIM

IMIMz

zz

yz

zzy

yzy

yzzzzy

yyzyzy

==>==>

−−

=

y

z

x

dAσxx

MyG

z

Mz = 0Equation of N.A depends on area moment of inertia


Principal axes

� Area moment of inertia – geometric property

zdA

p (y, z)

y’

z’

θ

dAyIdAzI zzyy ∫∫ == 22 ,

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o yθAA

∫∫

y’

y

zz’

oθ

p (y, z)

Product moment of inertia

∫=A

yz yzdAI

Rotate yozcsys to a new csys y’oz’ at an angle ‘θ’


Principal axes

� Transformation

θθθθ

cossin

sincos''

''

zyz

zyy

+=−= y’

y

zz’

oθ

p (y, z)y

z

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( )

(1) - 2sinsincos

2sinsincos

sincos

''''''22

''2'22'2

2''2

θθθ

θθθ

θθ

zyyyzzzz

AAA

zz

AA

zz

IIII

dAzydAzdAyI

dAzydAyI

−+=

−+=

−==

∫∫∫

∫∫

yo


Principal axes

� Transformation

( )2sincossin

cossin

''2'22'2

2''2

θθθ

θθA A

yy

dAzydAzdAyI

dAzydAzI

++=

+==

∫∫∫

∫ ∫

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(2) - 2sincossin

2sincossin

''''''22

''2'22'2

θθθ

θθθ

yzyyzzyy

AAA

yy

IIII

dAzydAzdAyI

++=

++= ∫∫∫


Principal axes

� Product moment of inertia – a property defined wrt a set of perpendicular axes lying in the same plane of area

∫=yz yzdAI

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( )( )

( )∫∫∫

∫

∫

−+

−==>

+−==>

=

AAA

yz

A

yz

A

yz

dAzydAzdAyI

dAzyzyI

yzdAI

''222'2'

''''

sincossincos

cossin sincos

θθθθ

θθθθ


Principal axes

� Product M. I

� Product M. I can be +ve or –ve – depends on selection of co-ordinate system

( ) (3) - 2cos2sin2

1'''''' zyyyzzyz IIII θθ +−=

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selection of co-ordinate system

� Variation of Iyy, Izz and Iyz wrt ‘θ’ – harmonic

� When product M. I changes sign – it passes through zero also

� Principal axes is a csys, where product M.I is zero


Maximum value of IyyI

Principal axes

� Variation of Iyy, Izz and Iyz

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Ramadas Chennamsetti

θ

Minimum value of Izz

Iyz = 0

23


Principal axes

� Product M. I = 0, area M. I – maximum and minimum

� Orientation corresponding to max. & min. – principal axes => θ, θ + 900

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� From (3), Iyz = 0

( )

( )''''

''

''''''

2tan2

0 2cos2sin2

1

zzyy

zy

zyyyzzyz

II

I

IIII

−==>

=+−=

θ

θθ

Area moment of inertia – a second order tensor


Principal axes

� Symmetric sections

y

z

o

z

+ y-y

zProduct moment of inertia

∫=A

yz yzdAI

Co-ordinate system is selected symmetrically, I is positive

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Co-ordinate system is selected symmetrically, I yz is positive and negative.

Summation over the area vanishes zero. YoZ– principal co-ordinate system.

y

z

O’

Yo’Z – another co-ordinate system. Product moment of inertia doesn’t vanish.

One axis of symmetry – principal axis



� Co-ordinate system is aligned with principal csys – product M. I => Iyz = 0

( ) ( ) EA

Nz

IIIE

IMIMy

IIIE

IMIMx

zzyyyz

yzzzzy

zzyyyz

yyzyzyxx +

−+−

+−−

= 22

εMz

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y

z

x

dAσxx

MyG

z

y

Nx

EA

Nz

EI

My

EI

M x

yy

y

zz

zxx ++= ε

Simplified expression

Again, N. A equation can be obtained by making strain zero

z

zzx

yy

zz

z

y

M

I

A

Nz

I

I

M

My −−=



� When no axial load acts –Nx = 0 –equation of N. A

M

I

A

Nz

I

I

M

My

z

zzx

yy

zz

z

y −−=zN Mz

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zI

I

M

My

MAIM

yy

zz

z

y

zyyz

−==> yG

A

My

Straight line passing through centroid

If, Mz = 0 => Equation of N. A => z = 0

This is ‘y’ axis


Axial deformation

� Axial deformation in bending

A

B

z

x

A

B

A’

B’

z

ψz

Mz

w

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View - V

ψy

vo

x

yMyψy

V


Axial deformation

� Axial deformation due to axial load, Nx => uo

� Axial deformation due to bending� Bending in xz plane => -zψy

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� Bending in xz plane => -zψy

� Bending in xy plane => -yψz

� Axial deformation due to all loads

( ) ( ) ( )xyxzuxu zyo ψψ −−=-ve signs – rotations are opposite to moments


Axial deformation

� In Euler-Bernoulli’s theory – shears strains negligible

zyo yzuu ψψ −−=

xy x

v

y

uγ =∂∂+

∂∂= 0 xz x

w

z

uγ =∂∂+

∂∂= 0

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z

z

xy

x

vx

v

xy

ψ

ψ

γ

=∂∂=>

=∂∂+−=>

=∂

+∂

=

0

0

y

y

xz

x

wx

wxz

ψ

ψ

γ

=∂∂=>

=∂∂+−=>

=∂

+∂

=

0

0


Axial deformation

� Axial deformation in terms of deflection gradient

22 vwuu

x

vy

x

wzuu o

∂∂∂∂∂∂−

∂∂−=

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2

2

2

2

2

2

2

2

, ,

strain,

x

va

x

wb

x

uc

aybzcx

vy

x

wz

x

u

x

u

o

xx

oxx

∂∂−=

∂∂−=

∂∂=

++=∂∂−

∂∂−

∂∂=

∂∂=

ε

ε


Curvatures

� Curvature – gradient of slope

zz

z

zz

z

MwMw

I

M

x

vE

EI

M

x

va

∂∂

=∂∂−=>=

∂∂−=

22

2

2

2

2

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xoxo

yy

y

yy

y

Nx

uEA

EA

N

x

uc

I

M

x

wE

EI

M

x

wb

=∂∂=>=

∂∂=

=∂∂−=>=

∂∂−=

2

2

2

2

Moment – curvature relationships


Shear stress distribution

� Shear stress in a solid section

x

y

z

τ

dx

FbArea ‘A’

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Solid section

Fb – force due to bending stress

τ - shear acting on plane b dx

b

dx

Fb*

‘b’ – local width on which shear to be estimated



� Force due to bending stress on area ‘A’

dAFA

xxb ∫= σ

Bending force at x+dx

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dAx

dAdxx

FFF

A

xx

A

xxb

bb ∫∫ ∂∂+=

∂∂+= σσ*

Shear force acting on plane b dx

bdxFs τ=



� Equilibrium

τ

dx

FbArea ‘A’

F

bdxFdxx

FF

FFF

b

bb

b

sbb

=∂=>

=−−∂

∂+

=−−

τ

τ 0

0*

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b

Fb*

dAx

bbdAx

dAF

bx

F

A

xx

A

xx

A

xxb

b

∫∫

∫

∂∂==>=

∂∂=>

=

=∂

∂=>

σττσ

σ

τ

A = area, which is above the plane on which shear to be found



� Bending stress

MM

A

Nz

I

My

I

ME

y

x

yy

y

zz

zxxxx

∂+∂=∂

++==

11

σ

εσ

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zx

M

Iy

x

M

Ixy

yy

z

zz

xx

∂∂

+∂

∂=∂

∂ 11σ

If a beam is subjected to pure bending – no variation of bending moment along length

00 , 0 ,0 ==>=∂

∂=∂

∂=

∂∂ τσ

xx

M

x

M xxyz



� Shear stress exists when transverse loads are applied q

V* Vq

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x

zM* M

dx

Force equilibrium of infinitesimal element

(1) - 0

0*

qx

VVqdxdx

x

VV

VqdxV

−=∂∂=>=−+

∂∂+=>

=−+

y

z

x

dAσxx

G

y

Nx

Vy

Vz



� Moment equilibrium

z

q

V* V

M* M( )

0

02

2

**

dxqdxdx

VVMdx

MM

dxqdxdxVMM

=−

∂+−−∂+

=−−−

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x

zM* M

dx

( )

(2) -

02

Vx

M

qdxdxx

VMdxx

M

=∂

∂

=−

∂+−−

∂+

Higher order terms are neglected

From (1) and (2)

qx

Mq

x

V

x

M

x−=

∂∂=>−=

∂∂=

∂∂

∂∂

2

2



� When transverse loads are acting

Using equation (2) ,x

MV

x

MV z

yy

z ∂∂=

∂∂

=

dAzV

yV

b zy

∫

+= τ

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zI

Vy

I

V

x

zx

M

Iy

x

M

Ix

dAx

b

yy

z

zz

yxx

y

yy

z

zz

xx

A

xx

+=∂

∂

∂∂

+∂

∂=∂

∂

∂∂= ∫

σ

σ

στ

11

dAz

I

Vy

I

Vb

A yy

z

zz

y

∫

+= τ

∫∫ ==AA

dAzzAdAyyA , --

--

zAI

VyA

I

Vb

yy

z

zz

y +=τ



� Transverse loading in xzplane - Vz

y

z

G

Vz

-

--

zAV

b

zAI

VyA

I

Vb

z

yy

z

zz

y

=

+=

τ

τVy = 0

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y-

zAI

Vb

yy

z=τ

Resembles shear flow, q

Shear flow at a given horizontal section depends on moment of area above that section wrt csys passing through centroid

This shear flow is not constant. Varies from zero to maximum



� Thin walled beam subjected to transverse loads z

ττττ

Fb

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y

x

t

dx

ττττ

F*b

Equilibrium of infinitesimal element gives stress distribution in the wall



� Shear stress in thin walled beam

y

z

qzAI

VyA

I

Vt

yy

z

zz

y =+=--

τ

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Shear flow in the wall of thin beam

Sum of the shear forces in the cross-section = externally applied shear force => vertical equilibrium of the section

Shear flow is not constant – varies from point to point along the wall

Shear is purely because of transverse loads, which cause bending => Bending – shear

Shear due to pure torsion => Torsion - shear


Shear center

� Equilibrium of beam cross-section requires – Forces and moment equilibrium

zTwisting will not take place if moment due to shear developed in the wall is equal to moment due to external transverse load

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y

Vz

qmoment due to external transverse load

Variable – location of application of external load

Shear flow is a function of external transversal load.

Moments of forces taken wrt any point in space – gives point of application of external load – shear center

Shear center – independent of external load – depends only on geometry


Shear center

� Shear center (S. C) may or may not pass through centroid of beam cross-section

� When there is no axial load, N. A passes through centroid

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� Loads acting in both planes

y

z

Qz

q

eyy

z

Qyq ez

y

z

ez

eyS.C


Shear center

� Some specific cross-sectionsz

e h

F1

F3

o

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yGey

Qz

h

F2

o

Moment of forces wrt ‘o’ hQ

Fe

hF

hFeQ

zyyz

111 0

22==>=−−

Equilibrium in ‘z’ direction => Qz = F3

Equilibrium in ‘y’ direction => F1 - F2 = 0


Shear center

� When load acts at S.C, it produces only bending – no twisting takes place

zQz

ey

Taking moments wrt ‘G’

yzeQbFaF 21 0=−−

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yG

ey

F1F2

a b zy

yz

Q

bFaFe

eQbFaF

21

21 0

−==>

=−−

Vertical equilibrium => F1 + F2 = Qz

Horizontal equilibrium – sum of forces in horizontal web = 0

Express F1 and F2 in terms of transverse load, Qz

ey will be a function of geometric parameters


Shear center

� Symmetric cross-sectionSince cross-section is symmetric, shear distribution in each leg is symmetric wrt xz plane

Taking moments about ‘O’

z

F1 F1

o

Qz

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yG

F2 F2

F3hmoment no - 022 =− hFhF

External load should pass through axis of symmetry – shear center lies in the plane symmetry

In symmetric cross-sections shear center lies in the plane of symmetry


Shear center

� Load acting at some angle

Q = Q sinαSC

z Qα

SC

z Qz = Q cosα

SC

z

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Qy = Q sinαSC

G y

SC

G y

SC

G y

= +

Point of application of load has to through S.C – no twisting takes place

Singly symmetric


Shear center

� Doubly symmetric beam

z

Qz

z

In a doubly

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yG

Symmetric in xz plane – S.C lies on z - axis

yGQy

Symmetric in xy plane – S.C lies on y - axis

In a doubly symmetric beam shear center lies at centroid


Shear center

Location of shear center for some cross-sections

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unsymmetric beams.pdf

Documents

Transcript of unsymmetric beams.pdf