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AOE2024: Thin Walled Structures
Module 6: Flexure due to Bidirectional Bending in
Thin-Walled Cross Sections
Prof. Gilbert
Virginia Polytechnic Institute and State University
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Food for Thought
• Consider the frame structure inside a wing as a beam and consider the load due to lift
and drag.
Bishay, P.L.; Burg, E.; Akinwunmi, A.; Phan, R.;
Sepulveda, K. Development of a New Span-Morphing
Wing Core Design. Designs 2019, 3, 12.
https://doi.org/10.3390/designs3010012
Jan R. Wright and Jonathan E. Cooper,
Introduction to Aircraft Aeroelasticity and Loads, Second
Edition, 2015 John Wiley & Sons, Ltd.
Jan R. Wright and Jonathan E. Cooper,
Introduction to Aircraft Aeroelasticity and Loads,
Second Edition, 2015 John Wiley & Sons, Ltd.
A. Kabir, M. S. Chowdhury, M. J. Islam and M. Islam,
”Numerical Assessment of the Backward Facing Step for
NACA 0015 Airfoil using Computational Fluid
Dynamics,” 2019 1st International Conference on
Advances in Science, Engineering and Robotics
Technology (ICASERT), Dhaka, Bangladesh, 2019, pp.
1-6, doi: 10.1109/ICASERT.2019.8934501.
• Due to bending about the z-axis (from lift), and bending
about the y-axis (from drag), the wing (modeled as a
beam) is subjected to bidirectional bending.
• Furthermore, due to the potential of the beam cross
section being asymmetric, this problem becomes even
more complex.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 1 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Overview
20 Introduction to Flexure
Deriving the General Flexure Formula
Principal Moment of Inertia Orientation
Plane of Loading Orientation
Plane of Bending Orientation
Example Problems
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 2 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
Bidirectional Beam Bending
• Consider a Beam with arbitrarily shaped cross section with the x-axis passing through
the bending axis at point O of the cross section (i.e. the neutral axis for bending)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 3 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• To derive the flexure formula, we start by assuming an x-displacement field consistent
with axial extension and pure bending about the z- and y- axis:
ux(x, y, z) = ux0(x)︸ ︷︷ ︸
Axial Extension
−y duy0
dx
+ c1(y, z)︸ ︷︷ ︸
Bending about z-axis
−z duz0
dx
+ c2(y, z)︸ ︷︷ ︸
Bending about y-axis
,
where ux0, uy0, and uz0 are the x-displacement, y-deflection, and z-deflection of the
neutral axis, respectively, and are functions of x only.
• We can therefore determine the strain and stress as:
xx(x, y, z) =
∂ux
∂x
=
dux0
dx
− y d
2uy0
dx2
− z d
2uz0
dx2
σxx(x, y, z) = Exx(x, y, z) = E
[
dux0
dx
− y d
2uy0
dx2
− z d
2uz0
dx2
]
Note:
Provided σyy and σzz are zero.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 4 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, we can take the expressions for resultant force and moment and plug in the
stress:
P (x) =
∫ ∫
σxxdydz =
∫ ∫
E
dux0
dx
dydz −
∫ ∫
Ey
d2uy0
dx2
dydz −
∫ ∫
Ez
d2uz0
dx2
dydz
My(x) =
∫ ∫
zσxxdydz =
∫ ∫
Ez
dux0
dx
dydz −
∫ ∫
Eyz
d2uy0
dx2
dydz −
∫ ∫
Ez2
d2uz0
dx2
dydz
Mz(x) =
∫ ∫
−yσxxdydz =
∫ ∫
−Eydux0
dx
dydz−
∫ ∫
−Ey2 d
2uy0
dx2
dydz−
∫ ∫
−Eyz d
2uz0
dx2
dydz
• We can pull out all the terms that are functions of x only.
• In this case we assume E = E(x).
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 5 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
P (x) = E
dux0
dx
∫ ∫
dydz︸ ︷︷ ︸
A
−Ed
2uy0
dx2
∫ ∫
ydydz︸ ︷︷ ︸
Qz
−Ed
2uz0
dx2
∫ ∫
zdydz︸ ︷︷ ︸
Qy
My(x) = E
dux0
dx
∫ ∫
zdydz︸ ︷︷ ︸
Qy
−Ed
2uy0
dx2
∫ ∫
yzdydz︸ ︷︷ ︸
??
−Ed
2uz0
dx2
∫ ∫
z2dydz︸ ︷︷ ︸
Iyy
Mz(x) = −Edux0
dx
∫ ∫
ydydz︸ ︷︷ ︸
Qz
+E
d2uy0
dx2
∫ ∫
y2dydz︸ ︷︷ ︸
Izz
+E
d2uz0
dx2
∫ ∫
yzdydz︸ ︷︷ ︸
??
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 6 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, we recognize the integral expressions as cross-section parameters corresponding
to the area (A), the first area moment about the z-axis, Qz, the first area moment
about the y-axis, Qy, the area moment of inertia (second area moment) about the
z-axis, Izz, the area moment of inertia about the y-axis, Iyy, and the product of
inertia, Iyz =
∫ ∫
yzdydz:
P (x) = E
dux0
dx
A− Ed
2uy0
dx2
Qz − Ed
2uz0
dx2
Qy
My(x) = E
dux0
dx
Qy − Ed
2uy0
dx2
Iyz − Ed
2uz0
dx2
Iyy
Mz(x) = −Edux0
dx
Qz + E
d2uy0
dx2
Izz + E
d2uz0
dx2
Iyz
So these are the general formulas
for coupled extension and bending!
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 7 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• If the x-axis corresponds to the centroid of the cross-section, then: Qy = Qz = 0
P (x) = E
dux0
dx
A
My(x) = −Ed
2uy0
dx2
Iyz − Ed
2uz0
dx2
Iyy
Mz(x) = E
d2uy0
dx2
Izz + E
d2uz0
dx2
Iyz
So these formulas are specific
to the centroid, and therefore
decouple extension and
bending
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 8 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, simultaneously solve the two moment equations for −E d2uy0dx2 and −E d
2uz0
dx2 :
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
• Identifying the axial strain and stress due to flexure:
xx,flex(x, y, z) = −y d
2uy0
dx2
− z d
2uz0
dx2
→ σxx,flex(x, y, z) = E
[
−y d
2uy0
dx2
− z d
2uz0
dx2
]
• And substituting from above we obtain the flexure formula for moments relative to the
centroid:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 9 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
• Thus pure bending is coupled when Iyz 6= 0
• Thus for arbitrary (asymmetrical) cross section, you must use flexure formula (even if
one of the bending moments is zero!)!!
• If Iyz = 0, then
σxx,flex(x, y, z) =
(My(x)Izz) z − (Mz(x)Iyy) y
IyyIzz
=
My(x)z
Iyy
− Mz(x)y
Izz
,
• which is the superposition of normal stress for symmetric cross sections.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 10 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
Determine the stresses and deflections in each of the cross
sections of the beam:
Note: Mz(x) 6= 0 and My(x) 6= 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 11 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The governing differential equations for the shear and moment diagrams are the same
whether the cross-section is symmetric (Module 5) or non-symmetric (Module 7):
Bending about z-axis
dVy
dx
= −py(x) =
dMz
dx
= −Vy(x) =
Bending about y-axis
dVz
dx
= −pz(x) =
dMy
dx
= Vz(x) =
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 12 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The governing differential equations for the shear and moment diagrams are the same
whether the cross-section is symmetric (Module 5) or non-symmetric (Module 7):
Bending about z-axis B.C.s Bending about y-axis B.C.s
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 13 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
So, the flexure stress is obtained as:
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 14 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The deflection of the neutral axis in the y-direction is obtained as:
−Ed
2uy0
dx2
=−
(
Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 15 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The deflection of the neutral axis in the z-direction is obtained as:
−Ed
2uz0
dx2
=
(
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 16 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Moment of Inertia Orientation
• In order to simplify the flexure formula, one can ask if there is an orientation of y − z
axis (i.e. y′ − z′) passing through the centroid such that Iy′z′ = 0.
For this coordinate system, the flexure formula would reduce to:
σxx,flex(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
,
where
My′(x) = My(x) cos θ +Mz(x) sin θ
Mz′(x) = −My(x) sin θ +Mz(x) cos θ
uy′0(x) = uy0(x) cos θ + uz0(x) sin θ
uz′0(x) = −uy0(x) sin θ + uz0(x) cos θ
• Due to this coordinate transformation, it turns out we can use a Mohr’s circle analogy
here.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 17 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Moment of Inertia Orientation
• Expressing the moment of inertia components as a 2×2 matrix, we can determine the
principal moments of inertia, Iy′y′ and Iz′z′ , as well as the angle theta relating the y-z
and y′-z′ coordinate systems using the Mohr circle equations:
II =
[
Iyy Iyz
Iyz Izz
]
,
where the y-face has coordinates: (Iyy, Iyz) and the z-face has coordinates (Izz,−Iyz).
The Mohr’s Circle Equations are:
c =
Iyy + Izz
2
r =
√(
Iyy − Izz
2
)2
+ I2yz
Ip1,2 = c± r
Angle to horizontal axis:
φ = sin−1
(
Iyz
r
)
,
where φ is always measured CCW from the
y-face.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 18 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
b
h
b/2
t
t t
y
z c
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 19 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the flexure stress in the principal orientation:
My′(x) = My(x) cos θ +Mz(x) sin θ
Mz′(x) = −My(x) sin θ +Mz(x) cos θ
σxx(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 20 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the deflection in the y′-direction:
−Ed
2uy′0
dx2
= −Mz′(x)
Iz′z′
d2uy′0
dx2
=
Mz′(x)
EIz′z′
=
1
EIz′z′
[(
p0x
2
2
− p0Lx+ p0L
2
2
)
cos θ −M0 sin θ
]
duy′0
dx
=
1
EIz′z′
[(
p0x
3
6
− p0Lx
2
2
+
p0L
2
2
x
)
cos θ −M0x sin θ
]
+ c3
uy′0(x) =
1
EIz′z′
[(
p0x
4
24
− p0Lx
3
6
+
p0L
2
4
x2
)
cos θ − M0x
2
2
sin θ
]
+ c3x+ c4
Rigid Wall, so
BC’s are:
uy′0(0) = 0→ c4 = 0
duy′0
dx
∣∣∣∣
x=0
= 0→ c3 = 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 21 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the deflection in the z′-direction:
−Ed
2uz′0
dx2
=
My′(x)
Iy′y′
d2uz′0
dx2
= −My′(x)
EIy′y′
= − 1
EIy′y′
[
M0 cos θ +
(
p0x
2
2
− p0Lx+ p0L
2
2
)
sin θ
]
duz′0
dx
= − 1
EIy′y′
[
M0x cos θ +
(
p0x
3
6
− p0Lx
2
2
+
p0L
2
2
x
)
sin θ
]
+ d3
uz′0(x) = − 1
EIy′y′
[
M0x
2
2
cos θ +
(
p0x
4
24
− p0Lx
3
6
+
p0L
2
4
x2
)
sin θ
]
+ d3x+ d4
Rigid Wall, so
BC’s are:
uz′0(0) = 0→ d4 = 0
duz′0
dx
∣∣∣∣
x=0
= 0→ d3 = 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 22 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Loading Orientation
• Alternatively, one can ask if there is an orientation of y-z axis (i.e. y′′′-z′′′) passing
through the centroid such that Mz′′′ = 0 to simplify the flexure formula.
For this coordinate system, the flexure formula would reduce to:
σxx,flex(x, y
′′′, z′′′) =
(My′′′(x)Iz′′′z′′′) z
′′′ − (My′′′(x)Iy′′′z′′′) y′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
,
where
My′′′(x) = My(x) cos γ +Mz(x) sin γ 6= 0
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
uy′′′0(x) = uy0(x) cos γ + uz0(x) sin γ
uz′′′0(x) = −uy0(x) sin γ + uz0(x) cos γ
• Note: γ only depends on the loads (relative bending moments), and NOT geometry
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 23 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Loading Axis Example
b
h
b/2
t
t t
y
z c
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 24 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• Alternatively, one can ask if there is an orientation of y-z axis (i.e. y′′-z′′) passing
through the centroid such that uy′′0 = 0 so that deflection is all in the uz′′0
direction, so the flexure formula becomes:
σxx,flex(x, y
′′, z′′) =
(My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′) z
′′ − (Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′) y′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
,
where
My′′(x) = My(x) cosα+Mz(x) sinα
Mz′′(x) = −My(x) sinα+Mz(x) cosα
uy′′0(x) = uy0(x) cosα+ uz0(x) sinα = 0
uz′′0(x) = −uy0(x) sinα+ uz0(x) cosα 6= 0
So the question is: how does one determine the orientation of the
plane of bending?
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 25 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• We choose a line where there is no axial stress when bending, which is perpendicular
to the direction of bending.
σxx(x, y, z) =
My(x)z
Iyy
− Mz(x)y
Izz
= 0→ z
y
=
Mz(x)Iyy
My(x)Izz
= tanα
• Note that α depends on the geometry of the cross section and the loads.
• If the moments are equal to Mo i.e. the plane of loading is γ = 45◦ and the
cross-section is square, then: α = 45◦ (Plane of Bending and Plane of Loading
coincide)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 26 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• More generally, the plane of bending is determined by:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
= 0,
z
y
=
Mz(x)Iyy +My(x)Iyz
My(x)Izz +Mz(x)Iyz
= tanα
• Plane of bending depends on both geometry and loads!
• In general, the principal plane of the cross-section, the
plane of loading and the plane of bending are not equal to
one another. We can show the definition of plane of
loading in the plane of bending formula:
tanα =
My(x) tan γIyy +My(x)Iyz
My(x)Izz +My(x) tan γIyz
=
tan γIyy + Iyz
Izz + tan γIyz
• If the cross section is symmetric, Iyz = 0, then
tanα = tan γ
Iyy
Izz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 27 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Comparison of Orientations
Comparing the flexure formula, bending moments and deflections in the three coordinate
systems (all of which pass through the centroid):
For the given y-z coordinate system:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
For the y′-z′ Principal Orientation coordinate system: Iy′z′ = 0
σxx,flex(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
For the y′′-z′′ Plane of Bending coordinate system: uy′′0(x) = 0
σxx,flex(x, y
′′, z′′) =
(My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′) z
′′ − (Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′) y′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
For the y′′′-z′′′ Plane of Loading coordinate system: Mz′′′(x) = 0
σxx,flex(x, y
′′′, z′′′) =
(My′′′(x)Iz′′′z′′′) z
′′′ − (My′′′(x)Iy′′′z′′′) y′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 28 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Z-beam with y-z axis located at centroid with y parallel to the midsection:
Let h1 = b2 = h and b1 = h2 = h3 = t << 1
So, Iyy =
2h3t
3 , Izz =
7h3t
12 , and Iyz =
h3t
2 .
Let: L = 10 m, M0 = 5 N-m, E = 1 GPa,
h = 0.2 m, and t = 0.003 m.
dVy
dx
= −py(x) = 0
dMz
dx
= −Vy(x)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 29 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
We can therefore identify the flexure stress as:
σxx =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
We can also identify the plane of loading orientation as:
γ = tan−1
(
Mz(x)
My(x)
)
=
We can also identify the planing of bending orientation as:
α = tan−1
(
MzIyy +MyIyz
MyIzz +MzIyz
)
=
We need to express numeric dimensions to be able to determine moments of inertia and therefore numerically determine orientation angles to the principal directions and to
plane of bending orientation.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 30 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Now, we can determine the deflection in the y-direction:
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 31 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Now determine the deflection in the z-direction:
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
So bending about z produced not only deflection in the y-direction, but also deflection in the unloaded z-direction because of the non-symmetric cross-section.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 32 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Z-beam with y-z axis located at centroid with y parallel to the midsection:
Let h1 = b2 = h and b1 = h2 = h3 = t << 1
So, Iyy =
2h3t
3 , Izz =
7h3t
12 , and Iyz =
h3t
2 .
Let: L = 10 m, M0 = 5 N-m, E = 1 GPa,
h = 0.2 m, and t = 0.003 m.
dVy
dx
= −py(x) = 0
Vy(x) = 0
Vy(x) = c1 but Vy(L) = 0
similarly Vz(x) = 0
dMz
dx
= −Vy(x) = 0
Mz(x) = M0
Mz(x) = c2 but Mz(L) = M0
similarly My(x) = M0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 33 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Now determine the deflection in the y-direction:
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
= −M0Iyy +M0Iyz
IyyIzz − I2yz
= −A
d2uy0
dx2
=
A
E
duy0
dx
=
A
E
x+ c3
uy0(x) =
A
E
x2
2
+ c4
But...
duy0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
But... uy0(0) = 0,∴ c4 = 0
uy0(x) =
A
E
x2
2
= 0.000875x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 34 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Now determine the deflection in the z-direction:
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
=
M0Izz +M0Iyz
IyyIzz − I2yz
= B
duz0
dx
=
−B
E
x+ c5
uz0(x) = −B
E
x2
2
+ c5x+ c6
But...
duz0
dx
∣∣∣∣
x=0
= 0,∴ c5 = 0
But... uz0(0) = 0,∴ c6 = 0
uz0(x) = −B
E
x2
2
= −0.0008125x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 35 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can therefore identify the flexure stress as:
σxx(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
= Bz −Ay
or
σxx(x, y, z) = 1.25E10 (0.00013z − 0.00014y) Pa
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 36 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can now determine the orientation to the principal directions using Mohr’s Circle:
II =
[
Iyy Iyz
Iyz Izz
]
=
[
1.6 1.2
1.2 1.4
]
× 10−5 y-face: (Iyy, Iyz) = (0.000016, 0.000012)z-face: (Izz,−Iyz) = (0.000014, -0.000012)
c =
Iyy + Izz
2
= 0.000015 r =
√(
Iyy − Izz
2
)2
+ I2yz = 1.20416× 10−5
Ip1 = c+ r = 2.70416× 10−5 → Iy′y′
Ip2 = c− r = 2.95841× 10−5 → Iz′z′
φ = sin−1
(
Iyz
r
)
2θ = 2pi − φ = 2pi − sin−1
(
Iyz
r
)
θ = 137.382◦
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 37 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′-deflection as:
−Ed
2uy′0
dx2
= −Mz′(x)
Iz′z′
= −A′
My′(x) = My(x) cos θ +Mz(x) sin θ = −0.293864 N-m
Mz′(x) = −My(x) sin θ +Mz(x) cos θ = −7.06496 N/m
Can check against the formula:
uy′0(x) = uy0(x) cos θ + uz0(x) sin θ
d2uy′0
dx2
=
A′
E
duy′0
dx
=
A′
E
x+ c3
But
duy′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′0 =
A′
E
x2
2
+ c4
But uy′0(0) = 0,∴ c4 = 0
uy′0(x) =
A′
E
x2
2
= −0.00119405x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 38 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′-deflection as:
−Ed
2uz′0
dx2
=
My′(x)
Iy′y′
= B′
My′(x) = My(x) cos θ +Mz(x) sin θ = −0.293864 N-m
Mz′(x) = −My(x) sin θ +Mz(x) cos θ = −7.06496 N/m
d2uz′0
dx2
=
−B′
E
duz′0
dx
=
−B′
E
x+ c3
But
duz′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′0 =
−B′
E
x2
2
+ c4
But uz′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′0(x) = −uy0(x) sin θ + uz0(x) cos θ uz′0(x) =
−B′
E
x2
2
= 5.43356E − 6x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 39 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the orientation to the plane of loading and using Mohr’s Circle,
evaluate the moments of inertia in the plane of loading orientation:
tan γ =
Mz(x)
My(x)
=
M0
M0
γ = 45◦
φ = pi − 2γ − Φ
Iy′′′y′′′ = c− r cos Φ = 3E − 6
Iz′′′z′′′ = c+ r cos Φ = 2.7E − 5
Iy′′′z′′′ = r sin Φ = 1E − 6
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 40 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′′′-deflection as:
−Ed
2uy′′′0
dx2
= −My′′′(x)Iy′′′y′′′ +My′′′(x)Iy′′′z′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
= −A′′′
My′′′(x) = My(x) cos γ +Mz(x) sin γ = 5
√
2 N-m
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
d2uy′′′0
dx2
=
A′′′
E
duy′′′0
dx
=
A′′′
E
x+ c3
But
duy′′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′′′0 =
A′′′
E
x2
2
+ c4
But uy′′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uy′′′0(x) = uy0(x) cos γ + uz0(x) sin γ
uy′′′0(x) =
A′′′
E
x2
2
= 4.41942E − 5x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 41 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′′′-deflection as:
−Ed
2uz′′′0
dx2
=
My′′′(x)Iz′′′z′′′ +My′′′(x)Iy′′′z′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
= B′′′
My′′′(x) = My(x) cos γ +Mz(x) sin γ = 5
√
2 N-m
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
d2uz′′′0
dx2
= −B
′′′
E
duz′′′0
dx
= −B
′′′
E
x+ c3
But
duz′′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uz′′′0 =
−B′′′
E
x2
2
+ c4
But uz′′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′′′0(x) = −uy0(x) sin γ + uz0(x) cos γ
uz′′′0(x) =
−B′′′
E
x2
2
= −1.19324E − 3x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 42 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the orientation of the plane of bending and using Mohr’s circle,
evaluate the moments of inertia in the plane of bending orientation:
tanα =
(Mz(x)Iyy +My(x)Iyz)
(My(x)Izz +Mz(x)Iyz)
α = 47.1211◦
Φ = pi − 2α− φ
Iy′′y′′ = c− r cos Φ = 2.9589E − 6
Iz′′z′′ = c+ r cos Φ = 2.70411E − 5
Iy′′z′′ = r sin Φ = 1.09589E − 7
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 43 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′′-deflection as:
−Ed
2uy′′0
dx2
= −Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
= −A′′
My′′(x) = My(x) cosα+Mz(x) sinα = 7.06622 N-m
Mz′′(x) = −My(x) sinα+Mz(x) cosα = −0.261712 N-m
d2uy′′0
dx2
=
A′′
E
duy′′0
dx
=
A′′
E
x+ c3
But
duy′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′′0 =
A′′
E
x2
2
+ c4
But uy′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uy′′0(x) = uy0(x) cosα+ uz0(x) sinα
uy′′0(x) =
A′′
E
x2
2
= 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 44 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′′-deflection as:
−Ed
2uz′′0
dx2
=
My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
= B′′
My′′(x) = My(x) cosα+Mz(x) sinα = 7.06622 N-m
Mz′′(x) = −My(x) sinα+Mz(x) cosα = −0.261712 N-m
d2uz′′0
dx2
= −B
′′
E
duz′′0
dx
= −B
′′
E
x+ c3
But
duz′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uz′′0 =
−B′′
E
x2
2
+ c4
But uz′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′′0(x) = −uy0(x) sinα+ uz0(x) cosα
uz′′0(x) =
−B′′
E
x2
2
= −0.00119406x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 45 / 45
学霸联盟
Module 6: Flexure due to Bidirectional Bending in
Thin-Walled Cross Sections
Prof. Gilbert
Virginia Polytechnic Institute and State University
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Food for Thought
• Consider the frame structure inside a wing as a beam and consider the load due to lift
and drag.
Bishay, P.L.; Burg, E.; Akinwunmi, A.; Phan, R.;
Sepulveda, K. Development of a New Span-Morphing
Wing Core Design. Designs 2019, 3, 12.
https://doi.org/10.3390/designs3010012
Jan R. Wright and Jonathan E. Cooper,
Introduction to Aircraft Aeroelasticity and Loads, Second
Edition, 2015 John Wiley & Sons, Ltd.
Jan R. Wright and Jonathan E. Cooper,
Introduction to Aircraft Aeroelasticity and Loads,
Second Edition, 2015 John Wiley & Sons, Ltd.
A. Kabir, M. S. Chowdhury, M. J. Islam and M. Islam,
”Numerical Assessment of the Backward Facing Step for
NACA 0015 Airfoil using Computational Fluid
Dynamics,” 2019 1st International Conference on
Advances in Science, Engineering and Robotics
Technology (ICASERT), Dhaka, Bangladesh, 2019, pp.
1-6, doi: 10.1109/ICASERT.2019.8934501.
• Due to bending about the z-axis (from lift), and bending
about the y-axis (from drag), the wing (modeled as a
beam) is subjected to bidirectional bending.
• Furthermore, due to the potential of the beam cross
section being asymmetric, this problem becomes even
more complex.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 1 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Overview
20 Introduction to Flexure
Deriving the General Flexure Formula
Principal Moment of Inertia Orientation
Plane of Loading Orientation
Plane of Bending Orientation
Example Problems
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 2 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
Bidirectional Beam Bending
• Consider a Beam with arbitrarily shaped cross section with the x-axis passing through
the bending axis at point O of the cross section (i.e. the neutral axis for bending)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 3 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• To derive the flexure formula, we start by assuming an x-displacement field consistent
with axial extension and pure bending about the z- and y- axis:
ux(x, y, z) = ux0(x)︸ ︷︷ ︸
Axial Extension
−y duy0
dx
+ c1(y, z)︸ ︷︷ ︸
Bending about z-axis
−z duz0
dx
+ c2(y, z)︸ ︷︷ ︸
Bending about y-axis
,
where ux0, uy0, and uz0 are the x-displacement, y-deflection, and z-deflection of the
neutral axis, respectively, and are functions of x only.
• We can therefore determine the strain and stress as:
xx(x, y, z) =
∂ux
∂x
=
dux0
dx
− y d
2uy0
dx2
− z d
2uz0
dx2
σxx(x, y, z) = Exx(x, y, z) = E
[
dux0
dx
− y d
2uy0
dx2
− z d
2uz0
dx2
]
Note:
Provided σyy and σzz are zero.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 4 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, we can take the expressions for resultant force and moment and plug in the
stress:
P (x) =
∫ ∫
σxxdydz =
∫ ∫
E
dux0
dx
dydz −
∫ ∫
Ey
d2uy0
dx2
dydz −
∫ ∫
Ez
d2uz0
dx2
dydz
My(x) =
∫ ∫
zσxxdydz =
∫ ∫
Ez
dux0
dx
dydz −
∫ ∫
Eyz
d2uy0
dx2
dydz −
∫ ∫
Ez2
d2uz0
dx2
dydz
Mz(x) =
∫ ∫
−yσxxdydz =
∫ ∫
−Eydux0
dx
dydz−
∫ ∫
−Ey2 d
2uy0
dx2
dydz−
∫ ∫
−Eyz d
2uz0
dx2
dydz
• We can pull out all the terms that are functions of x only.
• In this case we assume E = E(x).
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 5 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
P (x) = E
dux0
dx
∫ ∫
dydz︸ ︷︷ ︸
A
−Ed
2uy0
dx2
∫ ∫
ydydz︸ ︷︷ ︸
Qz
−Ed
2uz0
dx2
∫ ∫
zdydz︸ ︷︷ ︸
Qy
My(x) = E
dux0
dx
∫ ∫
zdydz︸ ︷︷ ︸
Qy
−Ed
2uy0
dx2
∫ ∫
yzdydz︸ ︷︷ ︸
??
−Ed
2uz0
dx2
∫ ∫
z2dydz︸ ︷︷ ︸
Iyy
Mz(x) = −Edux0
dx
∫ ∫
ydydz︸ ︷︷ ︸
Qz
+E
d2uy0
dx2
∫ ∫
y2dydz︸ ︷︷ ︸
Izz
+E
d2uz0
dx2
∫ ∫
yzdydz︸ ︷︷ ︸
??
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 6 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, we recognize the integral expressions as cross-section parameters corresponding
to the area (A), the first area moment about the z-axis, Qz, the first area moment
about the y-axis, Qy, the area moment of inertia (second area moment) about the
z-axis, Izz, the area moment of inertia about the y-axis, Iyy, and the product of
inertia, Iyz =
∫ ∫
yzdydz:
P (x) = E
dux0
dx
A− Ed
2uy0
dx2
Qz − Ed
2uz0
dx2
Qy
My(x) = E
dux0
dx
Qy − Ed
2uy0
dx2
Iyz − Ed
2uz0
dx2
Iyy
Mz(x) = −Edux0
dx
Qz + E
d2uy0
dx2
Izz + E
d2uz0
dx2
Iyz
So these are the general formulas
for coupled extension and bending!
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 7 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• If the x-axis corresponds to the centroid of the cross-section, then: Qy = Qz = 0
P (x) = E
dux0
dx
A
My(x) = −Ed
2uy0
dx2
Iyz − Ed
2uz0
dx2
Iyy
Mz(x) = E
d2uy0
dx2
Izz + E
d2uz0
dx2
Iyz
So these formulas are specific
to the centroid, and therefore
decouple extension and
bending
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 8 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
• Now, simultaneously solve the two moment equations for −E d2uy0dx2 and −E d
2uz0
dx2 :
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
• Identifying the axial strain and stress due to flexure:
xx,flex(x, y, z) = −y d
2uy0
dx2
− z d
2uz0
dx2
→ σxx,flex(x, y, z) = E
[
−y d
2uy0
dx2
− z d
2uz0
dx2
]
• And substituting from above we obtain the flexure formula for moments relative to the
centroid:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 9 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Deriving the General Flexure Formula
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
• Thus pure bending is coupled when Iyz 6= 0
• Thus for arbitrary (asymmetrical) cross section, you must use flexure formula (even if
one of the bending moments is zero!)!!
• If Iyz = 0, then
σxx,flex(x, y, z) =
(My(x)Izz) z − (Mz(x)Iyy) y
IyyIzz
=
My(x)z
Iyy
− Mz(x)y
Izz
,
• which is the superposition of normal stress for symmetric cross sections.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 10 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
Determine the stresses and deflections in each of the cross
sections of the beam:
Note: Mz(x) 6= 0 and My(x) 6= 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 11 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The governing differential equations for the shear and moment diagrams are the same
whether the cross-section is symmetric (Module 5) or non-symmetric (Module 7):
Bending about z-axis
dVy
dx
= −py(x) =
dMz
dx
= −Vy(x) =
Bending about y-axis
dVz
dx
= −pz(x) =
dMy
dx
= Vz(x) =
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 12 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The governing differential equations for the shear and moment diagrams are the same
whether the cross-section is symmetric (Module 5) or non-symmetric (Module 7):
Bending about z-axis B.C.s Bending about y-axis B.C.s
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 13 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
So, the flexure stress is obtained as:
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 14 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The deflection of the neutral axis in the y-direction is obtained as:
−Ed
2uy0
dx2
=−
(
Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 15 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Flexure Initial Comparison Example
The deflection of the neutral axis in the z-direction is obtained as:
−Ed
2uz0
dx2
=
(
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 16 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Moment of Inertia Orientation
• In order to simplify the flexure formula, one can ask if there is an orientation of y − z
axis (i.e. y′ − z′) passing through the centroid such that Iy′z′ = 0.
For this coordinate system, the flexure formula would reduce to:
σxx,flex(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
,
where
My′(x) = My(x) cos θ +Mz(x) sin θ
Mz′(x) = −My(x) sin θ +Mz(x) cos θ
uy′0(x) = uy0(x) cos θ + uz0(x) sin θ
uz′0(x) = −uy0(x) sin θ + uz0(x) cos θ
• Due to this coordinate transformation, it turns out we can use a Mohr’s circle analogy
here.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 17 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Moment of Inertia Orientation
• Expressing the moment of inertia components as a 2×2 matrix, we can determine the
principal moments of inertia, Iy′y′ and Iz′z′ , as well as the angle theta relating the y-z
and y′-z′ coordinate systems using the Mohr circle equations:
II =
[
Iyy Iyz
Iyz Izz
]
,
where the y-face has coordinates: (Iyy, Iyz) and the z-face has coordinates (Izz,−Iyz).
The Mohr’s Circle Equations are:
c =
Iyy + Izz
2
r =
√(
Iyy − Izz
2
)2
+ I2yz
Ip1,2 = c± r
Angle to horizontal axis:
φ = sin−1
(
Iyz
r
)
,
where φ is always measured CCW from the
y-face.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 18 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
b
h
b/2
t
t t
y
z c
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 19 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the flexure stress in the principal orientation:
My′(x) = My(x) cos θ +Mz(x) sin θ
Mz′(x) = −My(x) sin θ +Mz(x) cos θ
σxx(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 20 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the deflection in the y′-direction:
−Ed
2uy′0
dx2
= −Mz′(x)
Iz′z′
d2uy′0
dx2
=
Mz′(x)
EIz′z′
=
1
EIz′z′
[(
p0x
2
2
− p0Lx+ p0L
2
2
)
cos θ −M0 sin θ
]
duy′0
dx
=
1
EIz′z′
[(
p0x
3
6
− p0Lx
2
2
+
p0L
2
2
x
)
cos θ −M0x sin θ
]
+ c3
uy′0(x) =
1
EIz′z′
[(
p0x
4
24
− p0Lx
3
6
+
p0L
2
4
x2
)
cos θ − M0x
2
2
sin θ
]
+ c3x+ c4
Rigid Wall, so
BC’s are:
uy′0(0) = 0→ c4 = 0
duy′0
dx
∣∣∣∣
x=0
= 0→ c3 = 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 21 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Principal Axis Example
• Determining the deflection in the z′-direction:
−Ed
2uz′0
dx2
=
My′(x)
Iy′y′
d2uz′0
dx2
= −My′(x)
EIy′y′
= − 1
EIy′y′
[
M0 cos θ +
(
p0x
2
2
− p0Lx+ p0L
2
2
)
sin θ
]
duz′0
dx
= − 1
EIy′y′
[
M0x cos θ +
(
p0x
3
6
− p0Lx
2
2
+
p0L
2
2
x
)
sin θ
]
+ d3
uz′0(x) = − 1
EIy′y′
[
M0x
2
2
cos θ +
(
p0x
4
24
− p0Lx
3
6
+
p0L
2
4
x2
)
sin θ
]
+ d3x+ d4
Rigid Wall, so
BC’s are:
uz′0(0) = 0→ d4 = 0
duz′0
dx
∣∣∣∣
x=0
= 0→ d3 = 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 22 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Loading Orientation
• Alternatively, one can ask if there is an orientation of y-z axis (i.e. y′′′-z′′′) passing
through the centroid such that Mz′′′ = 0 to simplify the flexure formula.
For this coordinate system, the flexure formula would reduce to:
σxx,flex(x, y
′′′, z′′′) =
(My′′′(x)Iz′′′z′′′) z
′′′ − (My′′′(x)Iy′′′z′′′) y′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
,
where
My′′′(x) = My(x) cos γ +Mz(x) sin γ 6= 0
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
uy′′′0(x) = uy0(x) cos γ + uz0(x) sin γ
uz′′′0(x) = −uy0(x) sin γ + uz0(x) cos γ
• Note: γ only depends on the loads (relative bending moments), and NOT geometry
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 23 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Loading Axis Example
b
h
b/2
t
t t
y
z c
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 24 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• Alternatively, one can ask if there is an orientation of y-z axis (i.e. y′′-z′′) passing
through the centroid such that uy′′0 = 0 so that deflection is all in the uz′′0
direction, so the flexure formula becomes:
σxx,flex(x, y
′′, z′′) =
(My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′) z
′′ − (Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′) y′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
,
where
My′′(x) = My(x) cosα+Mz(x) sinα
Mz′′(x) = −My(x) sinα+Mz(x) cosα
uy′′0(x) = uy0(x) cosα+ uz0(x) sinα = 0
uz′′0(x) = −uy0(x) sinα+ uz0(x) cosα 6= 0
So the question is: how does one determine the orientation of the
plane of bending?
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 25 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• We choose a line where there is no axial stress when bending, which is perpendicular
to the direction of bending.
σxx(x, y, z) =
My(x)z
Iyy
− Mz(x)y
Izz
= 0→ z
y
=
Mz(x)Iyy
My(x)Izz
= tanα
• Note that α depends on the geometry of the cross section and the loads.
• If the moments are equal to Mo i.e. the plane of loading is γ = 45◦ and the
cross-section is square, then: α = 45◦ (Plane of Bending and Plane of Loading
coincide)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 26 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Plane of Bending Orientation
• More generally, the plane of bending is determined by:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
= 0,
z
y
=
Mz(x)Iyy +My(x)Iyz
My(x)Izz +Mz(x)Iyz
= tanα
• Plane of bending depends on both geometry and loads!
• In general, the principal plane of the cross-section, the
plane of loading and the plane of bending are not equal to
one another. We can show the definition of plane of
loading in the plane of bending formula:
tanα =
My(x) tan γIyy +My(x)Iyz
My(x)Izz +My(x) tan γIyz
=
tan γIyy + Iyz
Izz + tan γIyz
• If the cross section is symmetric, Iyz = 0, then
tanα = tan γ
Iyy
Izz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 27 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Comparison of Orientations
Comparing the flexure formula, bending moments and deflections in the three coordinate
systems (all of which pass through the centroid):
For the given y-z coordinate system:
σxx,flex(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
For the y′-z′ Principal Orientation coordinate system: Iy′z′ = 0
σxx,flex(x, y
′, z′) =
My′(x)z
′
Iy′y′
− Mz′(x)y
′
Iz′z′
For the y′′-z′′ Plane of Bending coordinate system: uy′′0(x) = 0
σxx,flex(x, y
′′, z′′) =
(My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′) z
′′ − (Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′) y′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
For the y′′′-z′′′ Plane of Loading coordinate system: Mz′′′(x) = 0
σxx,flex(x, y
′′′, z′′′) =
(My′′′(x)Iz′′′z′′′) z
′′′ − (My′′′(x)Iy′′′z′′′) y′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 28 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Z-beam with y-z axis located at centroid with y parallel to the midsection:
Let h1 = b2 = h and b1 = h2 = h3 = t << 1
So, Iyy =
2h3t
3 , Izz =
7h3t
12 , and Iyz =
h3t
2 .
Let: L = 10 m, M0 = 5 N-m, E = 1 GPa,
h = 0.2 m, and t = 0.003 m.
dVy
dx
= −py(x) = 0
dMz
dx
= −Vy(x)
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 29 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
We can therefore identify the flexure stress as:
σxx =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
We can also identify the plane of loading orientation as:
γ = tan−1
(
Mz(x)
My(x)
)
=
We can also identify the planing of bending orientation as:
α = tan−1
(
MzIyy +MyIyz
MyIzz +MzIyz
)
=
We need to express numeric dimensions to be able to determine moments of inertia and therefore numerically determine orientation angles to the principal directions and to
plane of bending orientation.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 30 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Now, we can determine the deflection in the y-direction:
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 31 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 1
Now determine the deflection in the z-direction:
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
So bending about z produced not only deflection in the y-direction, but also deflection in the unloaded z-direction because of the non-symmetric cross-section.
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 32 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Z-beam with y-z axis located at centroid with y parallel to the midsection:
Let h1 = b2 = h and b1 = h2 = h3 = t << 1
So, Iyy =
2h3t
3 , Izz =
7h3t
12 , and Iyz =
h3t
2 .
Let: L = 10 m, M0 = 5 N-m, E = 1 GPa,
h = 0.2 m, and t = 0.003 m.
dVy
dx
= −py(x) = 0
Vy(x) = 0
Vy(x) = c1 but Vy(L) = 0
similarly Vz(x) = 0
dMz
dx
= −Vy(x) = 0
Mz(x) = M0
Mz(x) = c2 but Mz(L) = M0
similarly My(x) = M0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 33 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Now determine the deflection in the y-direction:
−Ed
2uy0
dx2
= −Mz(x)Iyy +My(x)Iyz
IyyIzz − I2yz
= −M0Iyy +M0Iyz
IyyIzz − I2yz
= −A
d2uy0
dx2
=
A
E
duy0
dx
=
A
E
x+ c3
uy0(x) =
A
E
x2
2
+ c4
But...
duy0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
But... uy0(0) = 0,∴ c4 = 0
uy0(x) =
A
E
x2
2
= 0.000875x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 34 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
Now determine the deflection in the z-direction:
−Ed
2uz0
dx2
=
My(x)Izz +Mz(x)Iyz
IyyIzz − I2yz
=
M0Izz +M0Iyz
IyyIzz − I2yz
= B
duz0
dx
=
−B
E
x+ c5
uz0(x) = −B
E
x2
2
+ c5x+ c6
But...
duz0
dx
∣∣∣∣
x=0
= 0,∴ c5 = 0
But... uz0(0) = 0,∴ c6 = 0
uz0(x) = −B
E
x2
2
= −0.0008125x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 35 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can therefore identify the flexure stress as:
σxx(x, y, z) =
(My(x)Izz +Mz(x)Iyz) z − (Mz(x)Iyy +My(x)Iyz) y
IyyIzz − I2yz
= Bz −Ay
or
σxx(x, y, z) = 1.25E10 (0.00013z − 0.00014y) Pa
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 36 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can now determine the orientation to the principal directions using Mohr’s Circle:
II =
[
Iyy Iyz
Iyz Izz
]
=
[
1.6 1.2
1.2 1.4
]
× 10−5 y-face: (Iyy, Iyz) = (0.000016, 0.000012)z-face: (Izz,−Iyz) = (0.000014, -0.000012)
c =
Iyy + Izz
2
= 0.000015 r =
√(
Iyy − Izz
2
)2
+ I2yz = 1.20416× 10−5
Ip1 = c+ r = 2.70416× 10−5 → Iy′y′
Ip2 = c− r = 2.95841× 10−5 → Iz′z′
φ = sin−1
(
Iyz
r
)
2θ = 2pi − φ = 2pi − sin−1
(
Iyz
r
)
θ = 137.382◦
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 37 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′-deflection as:
−Ed
2uy′0
dx2
= −Mz′(x)
Iz′z′
= −A′
My′(x) = My(x) cos θ +Mz(x) sin θ = −0.293864 N-m
Mz′(x) = −My(x) sin θ +Mz(x) cos θ = −7.06496 N/m
Can check against the formula:
uy′0(x) = uy0(x) cos θ + uz0(x) sin θ
d2uy′0
dx2
=
A′
E
duy′0
dx
=
A′
E
x+ c3
But
duy′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′0 =
A′
E
x2
2
+ c4
But uy′0(0) = 0,∴ c4 = 0
uy′0(x) =
A′
E
x2
2
= −0.00119405x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 38 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′-deflection as:
−Ed
2uz′0
dx2
=
My′(x)
Iy′y′
= B′
My′(x) = My(x) cos θ +Mz(x) sin θ = −0.293864 N-m
Mz′(x) = −My(x) sin θ +Mz(x) cos θ = −7.06496 N/m
d2uz′0
dx2
=
−B′
E
duz′0
dx
=
−B′
E
x+ c3
But
duz′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′0 =
−B′
E
x2
2
+ c4
But uz′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′0(x) = −uy0(x) sin θ + uz0(x) cos θ uz′0(x) =
−B′
E
x2
2
= 5.43356E − 6x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 39 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the orientation to the plane of loading and using Mohr’s Circle,
evaluate the moments of inertia in the plane of loading orientation:
tan γ =
Mz(x)
My(x)
=
M0
M0
γ = 45◦
φ = pi − 2γ − Φ
Iy′′′y′′′ = c− r cos Φ = 3E − 6
Iz′′′z′′′ = c+ r cos Φ = 2.7E − 5
Iy′′′z′′′ = r sin Φ = 1E − 6
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 40 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′′′-deflection as:
−Ed
2uy′′′0
dx2
= −My′′′(x)Iy′′′y′′′ +My′′′(x)Iy′′′z′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
= −A′′′
My′′′(x) = My(x) cos γ +Mz(x) sin γ = 5
√
2 N-m
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
d2uy′′′0
dx2
=
A′′′
E
duy′′′0
dx
=
A′′′
E
x+ c3
But
duy′′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′′′0 =
A′′′
E
x2
2
+ c4
But uy′′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uy′′′0(x) = uy0(x) cos γ + uz0(x) sin γ
uy′′′0(x) =
A′′′
E
x2
2
= 4.41942E − 5x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 41 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′′′-deflection as:
−Ed
2uz′′′0
dx2
=
My′′′(x)Iz′′′z′′′ +My′′′(x)Iy′′′z′′′
Iy′′′y′′′Iz′′′z′′′ − I2y′′′z′′′
= B′′′
My′′′(x) = My(x) cos γ +Mz(x) sin γ = 5
√
2 N-m
Mz′′′(x) = −My(x) sin γ +Mz(x) cos γ = 0
d2uz′′′0
dx2
= −B
′′′
E
duz′′′0
dx
= −B
′′′
E
x+ c3
But
duz′′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uz′′′0 =
−B′′′
E
x2
2
+ c4
But uz′′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′′′0(x) = −uy0(x) sin γ + uz0(x) cos γ
uz′′′0(x) =
−B′′′
E
x2
2
= −1.19324E − 3x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 42 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the orientation of the plane of bending and using Mohr’s circle,
evaluate the moments of inertia in the plane of bending orientation:
tanα =
(Mz(x)Iyy +My(x)Iyz)
(My(x)Izz +Mz(x)Iyz)
α = 47.1211◦
Φ = pi − 2α− φ
Iy′′y′′ = c− r cos Φ = 2.9589E − 6
Iz′′z′′ = c+ r cos Φ = 2.70411E − 5
Iy′′z′′ = r sin Φ = 1.09589E − 7
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 43 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the y′′-deflection as:
−Ed
2uy′′0
dx2
= −Mz′′(x)Iy′′y′′ +My′′(x)Iy′′z′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
= −A′′
My′′(x) = My(x) cosα+Mz(x) sinα = 7.06622 N-m
Mz′′(x) = −My(x) sinα+Mz(x) cosα = −0.261712 N-m
d2uy′′0
dx2
=
A′′
E
duy′′0
dx
=
A′′
E
x+ c3
But
duy′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uy′′0 =
A′′
E
x2
2
+ c4
But uy′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uy′′0(x) = uy0(x) cosα+ uz0(x) sinα
uy′′0(x) =
A′′
E
x2
2
= 0
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 44 / 45
AOE 2024
Prof. Gilbert
Introduction to
Flexure
Deriving the
General Flexure
Formula
Principal
Moment of
Inertia
Orientation
Plane of Loading
Orientation
Plane of Bending
Orientation
Example
Problems
Example 2
We can also determine the z′′-deflection as:
−Ed
2uz′′0
dx2
=
My′′(x)Iz′′z′′ +Mz′′(x)Iy′′z′′
Iy′′y′′Iz′′z′′ − I2y′′z′′
= B′′
My′′(x) = My(x) cosα+Mz(x) sinα = 7.06622 N-m
Mz′′(x) = −My(x) sinα+Mz(x) cosα = −0.261712 N-m
d2uz′′0
dx2
= −B
′′
E
duz′′0
dx
= −B
′′
E
x+ c3
But
duz′′0
dx
∣∣∣∣
x=0
= 0,∴ c3 = 0
uz′′0 =
−B′′
E
x2
2
+ c4
But uz′′0(0) = 0,∴ c4 = 0
Can check against the formula:
uz′′0(x) = −uy0(x) sinα+ uz0(x) cosα
uz′′0(x) =
−B′′
E
x2
2
= −0.00119406x2
AOE 2024: Thin Walled Structures Module 6: Flexure due to Bidirectional Bending Gilbert 45 / 45
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