R&DE (Engineers), DRDO
rd_mech@yahoo.co.in
Unsymmetrical
Bending of Beams
Ramadas Chennamsetti
Introduction
Beam – structural member – takes
transverse loads
Cross-sectional dimensions much smaller
than length
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
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R&DE (Engineers), DRDO
Introduction
In thin beams – deformation due to shear
negligible
Thick beams – shear deformation
considered
Beam – one dimensional structural member
– length is very high than lateral
dimensions
Various parameters => function of single
independent variable
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Sign convention
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Following sign convention is followed
z
z
+ ve moments
Mz
o
My
y
x
y
Mx
z
ψz
Orientation of beam
+ve rotations
Beam bending in xy and xz planes
Ramadas Chennamsetti
ψy
y
ψx
x
4
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x
Sign convention
R&DE (Engineers), DRDO
Shear force and bending moments
z
Mz
+
y
z
M
M
Mx
y
x
Positive shear force and
bending moments
Convex upward +ve
direction
F
F
+
+
z
M
x
Ramadas Chennamsetti
M
F
F
5
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My
+
Unsymmetrical bending
R&DE (Engineers), DRDO
Load applied in the plane of
symmetry
z
Load applied at some
orientation
θ
P
z
P
G
y
Symmetrical cross-sectional – load
applied in the plane of symmetry - xz
Bending takes place in xz plane
y
Bending takes place in both
planes – xz and xy
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G
Unsymmetrical bending
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Cross-section is unsymmetrical
P
z
y
y
Bending takes place in both planes
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z
P
Euler – Bernoulli’s beam theory
R&DE (Engineers), DRDO
Basic assumptions
z
’
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x
8
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Length is much higher than lateral dimensions
– l ≥ 15 t
Plane cross section remains plane before and
after bending
A
A’
Stresses in lateral directions
BB
A
negligible
Thin beam
B
strain variation is linear across cross-section
Hookean material – linear elastic
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
An arbitrary cross-section beam oriented
along x-direction
ε xx = ay + bz + c
σ xx = Eε xx = E (ay + bz + c )
Mz
Equilibrium of the section requires
z
σxx
G
Nx
dA
z
My
x
∑F
x
y
∑M , ∑M
y
xyz csys coincides
with centroid
z
= Nx
=> External moments
dN x = σ xx dA = E (ay + bz + c )dA
Integrate
thisChennamsetti
for total force in x - direction
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9
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y
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
This equals to external force in ‘x’
direction = Nx
N x = ∫ σ xx dA = ∫ E (ay + bz + c ) dA
A
A
N x = Ea ∫ y dA + Eb ∫ z dA + Ec ∫ dA
A
A
Nx
N x = EcA => c =
EA
Origin of xyz co-ordinate system is selected at centroid
- Coefficients of ‘Ea’ and ‘Eb’ constants vanish
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A
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Moment wrt ‘y’ axis
= (σ
dM
y
= Ez ε xx dA
xx
dA )z
M y = ∫ Ez (ay + bz + c ) dA
y
σxx
G
dA
z
My
y
A
=> M y = Ea ∫ yz dA +
A
Eb ∫ z 2 dA + Ec ∫ z dA
A
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A
11
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z
x
y
Integrating over cross-sectional
area ‘A’
Mz
Nx
dM
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Moment about ‘y’ axis
M y = Ea ∫ yz dA + Eb ∫ z 2 dA + Ec ∫ z dA
A
A
=> M y = Ea I yz + Eb I yy
A
- (1)
M z = ∫ E (ay + bz + c ) y dA = Ea ∫ y 2 dA + Eb ∫ yz dA + Ec ∫ y dA
A
A
=> M z = Ea I zz + Eb I yz
A
- (2)
‘a’ and ‘b’ unknowns – solve simultaneous algebraic
equations (1) and (2)
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A
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Moment about ‘z’ axis
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Solving (1) and (2) ‘a’ and ‘b’ values are as
following
a=
M y I yz − M z I yy
(
E I − I yy I zz
2
yz
)
, b=
− M y I zz + M z I yz
(
E I yz2 − I yy I zz
)
ε xx =
M y I yz − M z I yy
(
E I − I yy I zz
2
yz
)
y+
− M y I zz + M z I yz
(
E I yz2 − I yy I zz
)
Nx
z+
EA
This represents a straight line
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Strain is
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Bending stress – multiply axial strain
(bending strain) with Young’s modulus (E)
Neutral axis /plane – no stress/strain => ‘0’
(I
2
yz
− I yy I zz
)
y+
− M y I zz + M z I yz
(I
2
yz
− I yy I zz
)
Nx
z+
=0
A
 M y I yz − M z I yy 
Nx 2


=> z =
y−
I yz − I yy I zz
M I −M I 
A
z yz 
 y zz
 M y I yz − M z I yy 

slope => m = 
M I −M I 
y zz
z yz 

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(
)
14
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M y I yz − M z I yy
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Equation of neutral axis
 M y I yz − M z I yy 
Nx 2


z=
y−
I yz − I yy I zz
 M I −M I 
A
z yz 
 y zz
(
z
)
Neutral axis doesn’t pass through
origin – centroid of cross-section
A
α
G
N
y
N.A equation z =
through centroid
slope, m = tan α =
Ramadas Chennamsetti
m y Passes
M y I yz − M z I yy
M y I zz − M z I yz
15
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If no axial load => Nx = 0
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
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
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
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Neutral axis (N. A)
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Bending in a single plane x-z – no axial load
z
σxx
G
dA
z
My
y
x
Mz = 0
 M y I yz − M z I yy 
Nx 2


z=
y−
I yz − I yy I zz
 M I −M I 
A
z yz 
 y zz
 M y I yz − M z I yy 
y
z =
 M I −M I 
z yz 
 y zz
M y I yz
I yz
=> z =
y => z =
y
M y I zz
I zz
(
Equation of N.A depends on area
moment of inertia
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17
)
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=> Nx = 0
Principal axes
R&DE (Engineers), DRDO
Area moment of inertia – geometric
property
z’ z
I yy = ∫ z 2 dA, I zz = ∫ y 2 dA
A
dA
p (y, z)
A
o
θ
y’
y
Product moment of inertia
I
yz
=
∫
z
p (y, z)
yzdA
A
y’
Rotate yoz csys to a new csys
y’oz’ at an angle ‘θ’
Ramadas Chennamsetti
θ
o
y 18
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z’
Principal axes
R&DE (Engineers), DRDO
z’
Transformation
z
p (y, z)
y
y = y cos θ − z sin θ
'
'
z = y ' sin θ + z ' cos θ
θ
o
(
y’
z
y
)
2
I zz = ∫ y dA = ∫ y cos θ − z sin θ dA
2
'
A
I zz = cos 2 θ ∫ y ' 2 dA + sin 2 θ ∫ z ' 2 dA − sin 2θ ∫ y ' z ' dA
A
A
A
I zz = I z ' z ' cos 2 θ + I y ' y ' sin 2 θ − I y ' z ' sin 2θ - (1)
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A
'
Principal axes
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Transformation
(
)
2
I yy = ∫ z dA = ∫ y sinθ + z cosθ dA
2
A
'
'
A
I yy = sin θ ∫ y dA + cos θ ∫ z dA + sin 2θ ∫ y z dA
'2
A
2
'2
A
' '
A
I yy = I z' z ' sin θ + I y' y' cos θ + I z ' y' sin 2θ - (2)
2
2
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2
Principal axes
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Product moment of inertia – a property
defined wrt a set of perpendicular axes
lying in the same plane of area
I yz = ∫ yzdA
(
A
)(
)
=> I yz = ∫ y ' cosθ − z ' sin θ y ' sin θ + z ' cosθ dA
 '2

'2
=> I yz = cosθ sin θ  ∫ y dA − ∫ z dA + cos2 θ − sin 2 θ
A
A

(
Ramadas Chennamsetti
)∫ y z dA
' '
A
21
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A
Principal axes
R&DE (Engineers), DRDO
Product M. I
(
)
Product M. I can be +ve or –ve – depends on
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
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1
I yz = sin 2θ I z ' z ' − I y ' y ' + cos 2θ I y ' z ' - (3)
2
Principal axes
R&DE (Engineers), DRDO
Variation of Iyy, Izz and Iyz
Maximum value of Iyy
Minimum value of Izz
Iyz = 0
θ
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I
Principal axes
R&DE (Engineers), DRDO
I yz
(
)
1
= sin 2θ I z ' z ' − I y ' y ' + cos 2θ I y ' z ' = 0
2
2 I y'z'
=> tan2 θ =
I y' y' − I z'z'
(
)
Area moment of inertia – a second order tensor
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Product M. I = 0, area M. I – maximum and
minimum
Orientation corresponding to max. & min.
– principal axes => θ, θ + 900
From (3), Iyz = 0
Principal axes
R&DE (Engineers), DRDO
z
Symmetric sections
-y + y
Product moment of inertia
I yz =
z
z
∫ yzdA
o
y
A
Summation over the area vanishes zero. YoZ – principal coordinate system.
z
Yo’Z
– another co-ordinate system.
Product moment of inertia doesn’t
vanish.
One axis of symmetry – principal axis
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y
O’
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Co-ordinate system is selected symmetrically, Iyz is positive
and negative.
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
Co-ordinate system is aligned with
principal csys – product M. I => Iyz = 0
ε xx =
Mz
G
Nx
dA
z
My
)
(
E I yz2 − I yy I zz
)
z+
Nx
EA
My
Nx
Mz
ε xx =
y+
z+
EI zz
EI yy
EA
y
σxx
(
E I yz2 − I yy I zz
y+
− M y I zz + M z I yz
y
Simplified expression
x
Again, N. A equation can be obtained by making strain zero
y = −
M
y
M
z
I zz
N x I zz
z−
I yy
A Mz
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z
M y I yz − M z I yy
Bending of arbitrary cross section beam
R&DE (Engineers), DRDO
When no axial load acts – Nx = 0 –
equation of N. A
M y I zz
=> y = −
z
M z I yy
Straight line passing through centroid
N
z Mz
My
y
G
A
If, Mz = 0 => Equation of N. A => z = 0
This is ‘y’ axis
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M y I zz
N x I zz
y=−
z−
M z I yy
A Mz
Axial deformation
R&DE (Engineers), DRDO
Axial deformation in bending
z w
Mz
A’
A
z
’
BB
A
ψz
x
B
o
yv
ψy
My
View - V
x
V
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ψy
Axial deformation
R&DE (Engineers), DRDO
Axial deformation due to axial load,
Nx => uo
Axial deformation due to bending
Bending in xz plane => -zψy
Bending in xy plane => -yψz
u(x) = uo − zψ y (x) − yψ z (x)
-ve signs – rotations are opposite to moments
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Axial deformation due to all loads
Axial deformation
R&DE (Engineers), DRDO
∂u ∂v
γ xy =
+
=0
∂y ∂x
∂v
=> −ψ z +
=0
∂x
∂v
=>
=ψ z
∂x
∂u ∂w
γ xz = +
=0
∂z ∂x
∂w
=> −ψ y +
=0
∂x
∂w
=>
=ψ y
∂x
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In Euler-Bernoulli’s theory – shears strains
negligible
u = uo − zψ y − yψ z
Axial deformation
Axial deformation in terms of deflection
gradient
∂w
∂v
u = uo − z
− y
∂x
∂x
∂u o
∂u
∂ 2w
∂ 2v
strain, ε xx =
=
−z
− y
2
2
∂x
∂x
∂x
∂x
ε xx = c + bz + ay
∂u o
∂ w
∂ v
c=
, b=−
, a=−
2
2
∂x
∂x
∂x
2
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2
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Curvatures
R&DE (Engineers), DRDO
Curvature – gradient of slope
Mz
∂ 2v
∂ 2v M z
a=− 2 =
=> − E 2 =
EI zz
I zz
∂x
∂x
∂u o N x
∂u o
c=
=
=> EA
= Nx
∂x
EA
∂x
Moment – curvature relationships
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∂2w M y
∂2w M y
b=− 2 =
=> − E 2 =
I yy
∂x
EI yy
∂x
Shear stress distribution
R&DE (Engineers), DRDO
Shear stress in a solid section
Fb
z
Area ‘A’
τ
y
x
dx
Fb*
Fb – force due to bending stress
Solid section
τ - shear acting on plane b dx
‘b’ – local width on which shear
to be estimated
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b
Shear stress distribution
R&DE (Engineers), DRDO
Force due to bending stress on area ‘A’
Fb = ∫ σ xx dA
A
Bending force at x+dx
∂Fb
∂σ xx
F = Fb +
dx = ∫ σ xx dA + ∫
dA
∂x
∂x
A
A
Shear force acting on plane b dx
Fs = τ bdx
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34
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*
b
Shear stress distribution
R&DE (Engineers), DRDO
Equilibrium
Fb
Fb* − Fb − Fs = 0
Area ‘A’
∂ Fb
Fb +
dx − Fb − τ bdx = 0
∂x
∂ Fb
=>
= τb
∂x
Fb = ∫ σ xx dA
τ
dx
b
A
∂
σ xx dA = τ b => τ b =
=>
∫
∂x A
∫
A
∂ σ xx
dA
∂x
A = area, which is above the plane on which shear to be found
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Fb*
Shear stress distribution
R&DE (Engineers), DRDO
Bending stress
σ xx = E ε xx
My
Nx
Mz
=
y+
z+
I zz
I yy
A
If a beam is subjected to pure bending – no variation of bending
moment along length
∂M y
∂σ xx
∂M z
= 0,
= 0 => τ = 0
= 0,
∂x
∂x
∂x
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∂ σ xx
1 ∂M z
1 ∂M y
y+
=
z
∂x
I zz ∂ x
I yy ∂ x
Shear stress distribution
R&DE (Engineers), DRDO
Shear stress exists when transverse loads
q
are applied
M
M*
z
y
σxx
G
z
dx
dA
x
y
Nx
Vy
Force equilibrium of infinitesimal element
V * + qdx − V = 0
∂V
∂V
=> V +
dx + qdx − V = 0 =>
= − q - (1)
∂x
∂x
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37
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Vz
x
V
V*
q
Shear stress distribution
R&DE (Engineers), DRDO
Moment equilibrium
dx
dx − qdx
=0
2
2
(
∂V 
dx )
+
dx  dx − q
=0
∂x 
2
*
∂M

M+
dx − M −  V
∂x

∂M
= V - (2)
∂x
V
V*
z
M
M*
dx
x
Higher order terms are neglected
From (1) and (2)
∂  ∂M  ∂V
∂ 2M
= − q =>
= −q

=
2
∂x  ∂x  ∂x
∂x
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M − M −V
*
q
Shear stress distribution
R&DE (Engineers), DRDO
When transverse loads are acting
∂M z
Vz =
, Vy =
∂x
∂x
y
 Vy

V
z
τ b = ∫ 
y+
z dA
I
I yy 
A  zz
∂ σ xx
τb=∫
dA
∂x
A
∂ σ xx
1 ∂M z
1 ∂M y
=
y+
z Ay=
∂x
I zz ∂ x
I yy ∂ x
∂ σ xx V y
Vz
=
y+
z
∂x
I zz
I yy
-
∫ y dA,
A z = ∫ z dA
A
τb=
Ramadas Chennamsetti
A
Vy
I zz
Vz
Ay+
Az
I yy
-
39
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Using equation (2)
∂M
Shear stress distribution
R&DE (Engineers), DRDO
Transverse loading in xz plane - Vz
Vz τ b=
Ay+
Az
I zz
I yy
-
z
Vy = 0
Vz τ b=
Az
I yy
G
y
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
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Vy
Vz
Shear stress distribution
R&DE (Engineers), DRDO
Thin walled beam subjected to transverse
loads
z
Fb
τ
dx
y
F*b
x
Equilibrium of infinitesimal element
gives stress distribution in the wall
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41
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t
Shear stress distribution
R&DE (Engineers), DRDO
Shear stress in thin walled beam
Vy
z
Vz τt=
Ay+
Az=q
I zz
I yy
-
y
Shear flow in the wall of thin beam
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
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Sum of the shear forces in the cross-section = externally
applied shear force => vertical equilibrium of the section
Shear center
R&DE (Engineers), DRDO
Equilibrium of beam cross-section requires
– Forces and moment equilibrium
Variable – location of application of external
load
z
q
Shear flow is a function of external
transversal load.
y
Vz
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
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Twisting will not take place if moment due
to shear developed in the wall is equal to
moment due to external transverse load
Shear center
R&DE (Engineers), DRDO
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
Loads acting in both planes
z
q
q
y
ey
Qy
ez
y
ey
S.C
ez
y
Qz
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z
z
Shear center
R&DE (Engineers), DRDO
Some specific cross-sections
z
F1
F3
ey
h
o
y
G
F2
Equilibrium in ‘z’ direction => Qz = F3
Equilibrium in ‘y’ direction => F1 - F2 = 0
Moment of forces wrt ‘o’
Qz e y − F1
h
h
F
− F1 = 0 => e y = 1 h
2
2
Qz
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Qz
Shear center
R&DE (Engineers), DRDO
When load acts at S.C, it produces only
bending – no twisting takes place
Taking moments wrt ‘G’
z
F1a − F2 b − Q z e y = 0
ey
y
G
F2
F1
a
b
Vertical equilibrium => F1 + F2 = Qz
Horizontal equilibrium – sum of
forces in horizontal web = 0
F1a − F2 b
=> e y =
Qz
Express F1 and F2 in terms of
transverse load, Qz
ey will be a function of
geometric parameters
Ramadas Chennamsetti
46
rd_mech@yahoo.co.in
Qz
Shear center
R&DE (Engineers), DRDO
Symmetric cross-section
z
Qz
F1
o
Taking moments about ‘O’
y
G
F3
h
F2 h − F2 h = 0
- no moment
External load should pass through
axis of symmetry – shear center lies
in the plane symmetry
F2
F2
In symmetric cross-sections shear
center lies in the plane of symmetry
Ramadas Chennamsetti
47
rd_mech@yahoo.co.in
F1
Since cross-section is symmetric,
shear distribution in each leg is
symmetric wrt xz plane
Shear center
R&DE (Engineers), DRDO
Load acting at some angle
Q
α
z Qz = Q cosα
SC
G
SC
y
G
=
z
SC
y
G
Qy = Q sinα
y
+
Singly symmetric
Point of application of load has to through S.C
– no twisting takes
place
Ramadas
Chennamsetti
48
rd_mech@yahoo.co.in
z
Shear center
R&DE (Engineers), DRDO
Doubly symmetric beam
Qz
G
G
y
Symmetric in xz plane – S.C
lies on z - axis
y
Qy
In a doubly
symmetric
beam shear
center lies
at centroid
Symmetric in xy plane – S.C
lies on y - axis
Ramadas Chennamsetti
49
rd_mech@yahoo.co.in
z
z
Shear center
R&DE (Engineers), DRDO
Ramadas Chennamsetti
50
rd_mech@yahoo.co.in
Location of shear center for some cross-sections
Ramadas Chennamsetti
51
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO