R&DE (Engineers), DRDO
Unsymmetrical
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Ramadas Chennamsetti
Bending of Beams
R&DE (Engineers), DRDO
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
R&DE (Engineers), DRDO
Introduction
� In thin beams – deformation due to shear negligible
� Thick beams – shear deformation considered
� Beam – one dimensional structural member
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Ramadas Chennamsetti3
� Beam – one dimensional structural member – length is very high than lateral dimensions
� Various parameters => function of single independent variable
R&DE (Engineers), DRDO
Sign convention
� Following sign convention is followed
y
z
oy
z
My
Mz
+ ve moments
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Ramadas Chennamsetti4
x
yo
Orientation of beam
y
x
Mx
x
z
ψψψψx
ψψψψy
ψψψψz
y+ve rotations
Beam bending in xy and xzplanes
R&DE (Engineers), DRDO
Sign convention
� Shear force and bending moments
y
z
My
Mz + +z
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Ramadas Chennamsetti5
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
R&DE (Engineers), DRDO
Unsymmetrical bending
P
z
P
zθ
Load applied in the plane of symmetry
Load applied at some orientation
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Ramadas Chennamsetti6
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
R&DE (Engineers), DRDO
Unsymmetrical bending
� Cross-section is unsymmetrical
z P
z
P
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Ramadas Chennamsetti7
y y
Bending takes place in both planes
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti8
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’
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti9
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
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� This equals to external force in ‘x’ direction = Nx
( )dAcbzayEdANAA
xxx ++==
∫∫∫
∫∫ σ
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Ramadas Chennamsetti10
EA
NcEcAN
dAEcdAzEbdAyEaN
xx
AAA
x
==>=
++= ∫∫∫
Origin of xyz co-ordinate system is selected at centroid
- Coefficients of ‘Ea’ and ‘Eb’ constants vanish
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� Moment wrt ‘y’ axis ( )
dAEzdM
zdAdM
xxy
xxy
εσ
=
=
Integrating over cross-sectional
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Ramadas Chennamsetti11
y
z
x
dAσxx
Mz
MyG
z
y
Nx
Integrating over cross-sectional area ‘A’
( )
dAzEcdAzEb
dAyzEaM
dAcbzayEzM
AA
A
y
A
y
2∫∫
∫
∫
+
+==>
++=
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� Moment about ‘y’ axis
(1) -
2
yyyzy
AAA
y
IEbIEaM
dAzEcdAzEbdAyzEaM
+==>
++= ∫∫∫
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Ramadas Chennamsetti12
� 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)
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� Solving (1) and (2) ‘a’ and ‘b’ values are as following
( ) ( )zzyyyz
yzzzzy
zzyyyz
yyzyzy
IIIE
IMIMb
IIIE
IMIMa
−+−
=−−
=22
,
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Ramadas Chennamsetti13
( ) ( )zzyyyzzzyyyz IIIEIIIE −−
Strain is
( ) ( ) EA
Nz
IIIE
IMIMy
IIIE
IMIMx
zzyyyz
yzzzzy
zzyyyz
yyzyzyxx +
−+−
+−−
= 22
ε
This represents a straight line
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� 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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Ramadas Chennamsetti14
( ) ( )
( )
−−
==>
−−
−−
==>
=+−
+−+
−−
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
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� Equation of neutral axis
( )zzyyyzx
yzzzzy
yyzyzy IIIA
Ny
IMIM
IMIMz −−
−−
= 2
Neutral axis doesn’t pass through
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Ramadas Chennamsetti15
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,
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� 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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Ramadas Chennamsetti16
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
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� 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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Ramadas Chennamsetti17
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
R&DE (Engineers), DRDO
Principal axes
� Area moment of inertia – geometric property
zdA
p (y, z)
y’
z’
θ
dAyIdAzI zzyy ∫∫ == 22 ,
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Ramadas Chennamsetti18
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’
R&DE (Engineers), DRDO
Principal axes
� Transformation
θθθθ
cossin
sincos''
''
zyz
zyy
+=−= y’
y
zz’
oθ
p (y, z)y
z
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Ramadas Chennamsetti19
( )
(1) - 2sinsincos
2sinsincos
sincos
''''''22
''2'22'2
2''2
θθθ
θθθ
θθ
zyyyzzzz
AAA
zz
AA
zz
IIII
dAzydAzdAyI
dAzydAyI
−+=
−+=
−==
∫∫∫
∫∫
yo
R&DE (Engineers), DRDO
Principal axes
� Transformation
( )2sincossin
cossin
''2'22'2
2''2
θθθ
θθA A
yy
dAzydAzdAyI
dAzydAzI
++=
+==
∫∫∫
∫ ∫
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Ramadas Chennamsetti20
(2) - 2sincossin
2sincossin
''''''22
''2'22'2
θθθ
θθθ
yzyyzzyy
AAA
yy
IIII
dAzydAzdAyI
++=
++= ∫∫∫
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti21
( )( )
( )∫∫∫
∫
∫
−+
−==>
+−==>
=
AAA
yz
A
yz
A
yz
dAzydAzdAyI
dAzyzyI
yzdAI
''222'2'
''''
sincossincos
cossin sincos
θθθθ
θθθθ
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti22
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
R&DE (Engineers), DRDO
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti24
� From (3), Iyz = 0
( )
( )''''
''
''''''
2tan2
0 2cos2sin2
1
zzyy
zy
zyyyzzyz
II
I
IIII
−==>
=+−=
θ
θθ
Area moment of inertia – a second order tensor
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti25
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
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� 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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Ramadas Chennamsetti26
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 −−=
R&DE (Engineers), DRDO
Bending of arbitrary cross section beam
� 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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Ramadas Chennamsetti27
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
R&DE (Engineers), DRDO
Axial deformation
� Axial deformation in bending
A
B
z
x
A
B
A’
B’
z
ψz
Mz
w
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Ramadas Chennamsetti28
View - V
ψy
vo
x
yMyψy
V
R&DE (Engineers), DRDO
Axial deformation
� Axial deformation due to axial load, Nx => uo
� Axial deformation due to bending� Bending in xz plane => -zψy
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Ramadas Chennamsetti29
� 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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti30
z
z
xy
x
vx
v
xy
ψ
ψ
γ
=∂∂=>
=∂∂+−=>
=∂
+∂
=
0
0
y
y
xz
x
wx
wxz
ψ
ψ
γ
=∂∂=>
=∂∂+−=>
=∂
+∂
=
0
0
R&DE (Engineers), DRDO
Axial deformation
� Axial deformation in terms of deflection gradient
22 vwuu
x
vy
x
wzuu o
∂∂∂∂∂∂−
∂∂−=
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Ramadas Chennamsetti31
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti32
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
R&DE (Engineers), DRDO
Shear stress distribution
� Shear stress in a solid section
x
y
z
τ
dx
FbArea ‘A’
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Ramadas Chennamsetti33
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
R&DE (Engineers), DRDO
Shear stress distribution
� Force due to bending stress on area ‘A’
dAFA
xxb ∫= σ
Bending force at x+dx
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Ramadas Chennamsetti34
dAx
dAdxx
FFF
A
xx
A
xxb
bb ∫∫ ∂∂+=
∂∂+= σσ*
Shear force acting on plane b dx
bdxFs τ=
R&DE (Engineers), DRDO
Shear stress distribution
� Equilibrium
τ
dx
FbArea ‘A’
F
bdxFdxx
FF
FFF
b
bb
b
sbb
=∂=>
=−−∂
∂+
=−−
τ
τ 0
0*
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Ramadas Chennamsetti35
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
R&DE (Engineers), DRDO
Shear stress distribution
� Bending stress
MM
A
Nz
I
My
I
ME
y
x
yy
y
zz
zxxxx
∂+∂=∂
++==
11
σ
εσ
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Ramadas Chennamsetti36
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
R&DE (Engineers), DRDO
Shear stress distribution
� Shear stress exists when transverse loads are applied q
V* Vq
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Ramadas Chennamsetti37
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
R&DE (Engineers), DRDO
Shear stress distribution
� Moment equilibrium
z
q
V* V
M* M( )
0
02
2
**
dxqdxdx
VVMdx
MM
dxqdxdxVMM
=−
∂+−−∂+
=−−−
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Ramadas Chennamsetti38
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
R&DE (Engineers), DRDO
Shear stress distribution
� When transverse loads are acting
Using equation (2) ,x
MV
x
MV z
yy
z ∂∂=
∂∂
=
dAzV
yV
b zy
∫
+= τ
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Ramadas Chennamsetti39
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
R&DE (Engineers), DRDO
Shear stress distribution
� 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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Ramadas Chennamsetti40
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
R&DE (Engineers), DRDO
Shear stress distribution
� Thin walled beam subjected to transverse loads z
ττττ
Fb
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Ramadas Chennamsetti41
y
x
t
dx
ττττ
F*b
Equilibrium of infinitesimal element gives stress distribution in the wall
R&DE (Engineers), DRDO
Shear stress distribution
� Shear stress in thin walled beam
y
z
qzAI
VyA
I
Vt
yy
z
zz
y =+=--
τ
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Ramadas Chennamsetti42
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti43
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti44
� Loads acting in both planes
y
z
Qz
q
eyy
z
Qyq ez
y
z
ez
eyS.C
R&DE (Engineers), DRDO
Shear center
� Some specific cross-sectionsz
e h
F1
F3
o
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Ramadas Chennamsetti45
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti46
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
R&DE (Engineers), DRDO
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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Ramadas Chennamsetti47
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
R&DE (Engineers), DRDO
Shear center
� Load acting at some angle
Q = Q sinαSC
z Qα
SC
z Qz = Q cosα
SC
z
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Ramadas Chennamsetti48
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
R&DE (Engineers), DRDO
Shear center
� Doubly symmetric beam
z
Qz
z
In a doubly
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Ramadas Chennamsetti49
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
R&DE (Engineers), DRDO
Shear center
Location of shear center for some cross-sections
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R&DE (Engineers), DRDO
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Ramadas Chennamsetti51