Summary Structural Analysis and Design
This summary covers all material for the second year course of Aerospace Engineering Structural
Analysis and Design (AE-) at the TU Delft. During this course, the book ‘Aircraft Structures for
Engineering Students’ from T. H. G. Megson is used. The following chapters (in the order as they
were treated during the course) are included in this summary:
- Chapter 16: Bending of open and closed, thin-walled beams
- Chapter 18: Torsion of beams
- Chapter 17: Shear of beams
- Chapter 19: Combined open and closed section beams
- Chapter 20: Structural idealization
- Chapter 8: Columns
- Chapter 4: Virtual work and energy methods
- Chapter 5: Energy methods
- Chapter 23: Wings
A last note is that the equation numbers refer to the numbers in the book.
,Chapter 16: Bending of open and closed, thin-walled beams
§16.1: Symmetrical bending
Assumptions
- Plane cross-sections remain plane and normal to the longitudinal fibers of the beam after
bending.
- Linear elastic materials (obeying Hooke’s law).
- Homogeneous materials.
Direct stress distribution
The strain within a beam subjected to bending is equal to:
𝑦
𝜖𝑧 = − 𝑅 (16.1)
Therefore, Hooke’s law says that the stress distribution is:
𝑦
𝜎𝑧 = −𝐸 𝑅 (16.2)
Integrating the direct stress over the cross-section should equal zero, as the beam is in
equilibrium. Therefore:
∫𝐴 𝑦𝑑𝐴 = 0 (16.5)
This equation states that the position of the neutral axis (𝑦 = 0) should be such that the direct
stress distribution is zero over the cross-section and should therefore pass through the centroid of
the cross-section.
Considering a moment equilibrium gives the following relationship:
𝑦2 𝐸𝐼
𝑀 = − ∫𝐴 𝐸 𝑅
𝑑𝐴 =−𝑅 (16.7)
Combining equations (16.2) and (16.7) gives a relation between the direct stress and the bending
moment:
𝑀𝑦
𝜎𝑧 = 𝐼 (16.9)
In addition, the curvature (1/𝑅) of the bending is given by:
1 𝑀
𝑅
= − 𝐸𝐼 (16.10)
Biaxial bending
In the case where the moment is not along the x- or y-axis, the bending is biaxial and should be
decomposed into an x- and y-component:
𝑀𝑥 = 𝑀 cos 𝜃
𝑀𝑦 = 𝑀 sin 𝜃
The combined stress distribution due to these two moments is equal to (the Normal Flexure
Formula):
𝑀𝑥 𝑦 𝑀𝑦 𝑥
𝜎𝑧 = 𝐼𝑥𝑥
+ 𝐼𝑦𝑦
In this equation, the moment should be taken positive if it produces a tensile normal stress in the
positive quadrant of the coordinate system.
The neutral axis will still pass through the centroid, however, the angle of this axis is not equal to
the angle of the moment. The angle of neutral axis is equal to:
𝑀 𝐼
tan 𝛼 = 𝑀𝑦𝐼𝑥𝑥
𝑥 𝑦𝑦
, §16.2: Unsymmetrical bending
If the Normal Flexure Formula is applied to an unsymmetrical problem, force equilibrium gives
that ∫𝐴 𝑦𝑑𝐴 = 0, which shows that the neutral axis still has to pass through the centroid of the
cross-section. In addition, moment equilibrium gives that ∫𝐴 𝑥𝑦𝑑𝐴 = 𝐼𝑥𝑦 = 0.
Therefore, the Normal Flexure Formula can be applied as long as 𝐼𝑥𝑦 = 0.
In the case that 𝐼𝑥𝑦 ≠ 0, the following formula for the normal stress is valid:
(𝑀𝑥 𝐼𝑦𝑦 −𝑀𝑦 𝐼𝑥𝑦 )𝑦+(𝑀𝑦 𝐼𝑥𝑥 −𝑀𝑥 𝐼𝑥𝑦 )𝑥
𝜎𝑧 = 2 (16.18)
𝐼𝑥𝑥 𝐼𝑦𝑦 −𝐼𝑥𝑦
The angle of the neutral axis will then be equal to:
𝑀𝑦 𝐼𝑥𝑥 −𝑀𝑥 𝐼𝑥𝑦
tan 𝛼 = (16.21)
𝑀𝑥 𝐼𝑦𝑦 −𝑀𝑦 𝐼𝑥𝑦
Equation (16.18) indeed shows that the normal stress equation reduces to the Normal Flexure
Formula whenever 𝐼𝑥𝑦 is zero. This is the case when the coordinate axes are the principal axes of
the cross-section
Identifying principal axes
Using the principal axes can simplify the stress equation a lot. Therefore, it is sometimes more
convenient to transfer the loads to the principal axes of the cross section. The principal axes can
be found using the following rules:
- A plane of symmetry will always be a principal axes.
- The axes will always be orthogonal and will both pass through the centroid.
§16.3: Deflections due to bending
When 𝑢 is the deflection in x-direction and 𝑣 the deflection in y-direction, the general formula for
deflection due to bending is:
𝑀𝑥 𝐼𝑥𝑦 𝐼𝑥𝑥 𝑢′′
[𝑀 ] = −𝐸 [ ] [ ′′ ] (16.29)
𝑦 𝐼𝑦𝑦 𝐼𝑥𝑦 𝑣
§16.4: Calculation of section properties
Parallel axes theorem
Consider an area which is a distance 𝑏 away from the neutral axis. The second moment of inertia
is then equal to:
𝐼𝑁 = 𝐼𝐶 + 𝐴𝑏 2 (16.32)
Where 𝐼𝐶 is the second moment of inertia around the centroid of the area. For the product
second moment of inertia, this theorem gives the following equation (with 𝑎 the horizontal
distance from the centroid of the area to the neutral axis and 𝑏 the vertical distance):
𝐼𝑁 = 𝐼𝐶 + 𝐴𝑎𝑏
Theorem of perpendicular axes
The polar moment of inertia (𝐽 or 𝐼𝑜 ) is equal to:
𝐽 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦 (16.33)
Second moments of inertia
The second moment of inertia around the x-axis can be calculated as follows:
𝐼𝑥𝑥 = ∫𝐴 𝑦 2 𝑑𝐴
This summary covers all material for the second year course of Aerospace Engineering Structural
Analysis and Design (AE-) at the TU Delft. During this course, the book ‘Aircraft Structures for
Engineering Students’ from T. H. G. Megson is used. The following chapters (in the order as they
were treated during the course) are included in this summary:
- Chapter 16: Bending of open and closed, thin-walled beams
- Chapter 18: Torsion of beams
- Chapter 17: Shear of beams
- Chapter 19: Combined open and closed section beams
- Chapter 20: Structural idealization
- Chapter 8: Columns
- Chapter 4: Virtual work and energy methods
- Chapter 5: Energy methods
- Chapter 23: Wings
A last note is that the equation numbers refer to the numbers in the book.
,Chapter 16: Bending of open and closed, thin-walled beams
§16.1: Symmetrical bending
Assumptions
- Plane cross-sections remain plane and normal to the longitudinal fibers of the beam after
bending.
- Linear elastic materials (obeying Hooke’s law).
- Homogeneous materials.
Direct stress distribution
The strain within a beam subjected to bending is equal to:
𝑦
𝜖𝑧 = − 𝑅 (16.1)
Therefore, Hooke’s law says that the stress distribution is:
𝑦
𝜎𝑧 = −𝐸 𝑅 (16.2)
Integrating the direct stress over the cross-section should equal zero, as the beam is in
equilibrium. Therefore:
∫𝐴 𝑦𝑑𝐴 = 0 (16.5)
This equation states that the position of the neutral axis (𝑦 = 0) should be such that the direct
stress distribution is zero over the cross-section and should therefore pass through the centroid of
the cross-section.
Considering a moment equilibrium gives the following relationship:
𝑦2 𝐸𝐼
𝑀 = − ∫𝐴 𝐸 𝑅
𝑑𝐴 =−𝑅 (16.7)
Combining equations (16.2) and (16.7) gives a relation between the direct stress and the bending
moment:
𝑀𝑦
𝜎𝑧 = 𝐼 (16.9)
In addition, the curvature (1/𝑅) of the bending is given by:
1 𝑀
𝑅
= − 𝐸𝐼 (16.10)
Biaxial bending
In the case where the moment is not along the x- or y-axis, the bending is biaxial and should be
decomposed into an x- and y-component:
𝑀𝑥 = 𝑀 cos 𝜃
𝑀𝑦 = 𝑀 sin 𝜃
The combined stress distribution due to these two moments is equal to (the Normal Flexure
Formula):
𝑀𝑥 𝑦 𝑀𝑦 𝑥
𝜎𝑧 = 𝐼𝑥𝑥
+ 𝐼𝑦𝑦
In this equation, the moment should be taken positive if it produces a tensile normal stress in the
positive quadrant of the coordinate system.
The neutral axis will still pass through the centroid, however, the angle of this axis is not equal to
the angle of the moment. The angle of neutral axis is equal to:
𝑀 𝐼
tan 𝛼 = 𝑀𝑦𝐼𝑥𝑥
𝑥 𝑦𝑦
, §16.2: Unsymmetrical bending
If the Normal Flexure Formula is applied to an unsymmetrical problem, force equilibrium gives
that ∫𝐴 𝑦𝑑𝐴 = 0, which shows that the neutral axis still has to pass through the centroid of the
cross-section. In addition, moment equilibrium gives that ∫𝐴 𝑥𝑦𝑑𝐴 = 𝐼𝑥𝑦 = 0.
Therefore, the Normal Flexure Formula can be applied as long as 𝐼𝑥𝑦 = 0.
In the case that 𝐼𝑥𝑦 ≠ 0, the following formula for the normal stress is valid:
(𝑀𝑥 𝐼𝑦𝑦 −𝑀𝑦 𝐼𝑥𝑦 )𝑦+(𝑀𝑦 𝐼𝑥𝑥 −𝑀𝑥 𝐼𝑥𝑦 )𝑥
𝜎𝑧 = 2 (16.18)
𝐼𝑥𝑥 𝐼𝑦𝑦 −𝐼𝑥𝑦
The angle of the neutral axis will then be equal to:
𝑀𝑦 𝐼𝑥𝑥 −𝑀𝑥 𝐼𝑥𝑦
tan 𝛼 = (16.21)
𝑀𝑥 𝐼𝑦𝑦 −𝑀𝑦 𝐼𝑥𝑦
Equation (16.18) indeed shows that the normal stress equation reduces to the Normal Flexure
Formula whenever 𝐼𝑥𝑦 is zero. This is the case when the coordinate axes are the principal axes of
the cross-section
Identifying principal axes
Using the principal axes can simplify the stress equation a lot. Therefore, it is sometimes more
convenient to transfer the loads to the principal axes of the cross section. The principal axes can
be found using the following rules:
- A plane of symmetry will always be a principal axes.
- The axes will always be orthogonal and will both pass through the centroid.
§16.3: Deflections due to bending
When 𝑢 is the deflection in x-direction and 𝑣 the deflection in y-direction, the general formula for
deflection due to bending is:
𝑀𝑥 𝐼𝑥𝑦 𝐼𝑥𝑥 𝑢′′
[𝑀 ] = −𝐸 [ ] [ ′′ ] (16.29)
𝑦 𝐼𝑦𝑦 𝐼𝑥𝑦 𝑣
§16.4: Calculation of section properties
Parallel axes theorem
Consider an area which is a distance 𝑏 away from the neutral axis. The second moment of inertia
is then equal to:
𝐼𝑁 = 𝐼𝐶 + 𝐴𝑏 2 (16.32)
Where 𝐼𝐶 is the second moment of inertia around the centroid of the area. For the product
second moment of inertia, this theorem gives the following equation (with 𝑎 the horizontal
distance from the centroid of the area to the neutral axis and 𝑏 the vertical distance):
𝐼𝑁 = 𝐼𝐶 + 𝐴𝑎𝑏
Theorem of perpendicular axes
The polar moment of inertia (𝐽 or 𝐼𝑜 ) is equal to:
𝐽 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦 (16.33)
Second moments of inertia
The second moment of inertia around the x-axis can be calculated as follows:
𝐼𝑥𝑥 = ∫𝐴 𝑦 2 𝑑𝐴
