University of the West of England

1. Aero Structures-Basic Elasticity
By
Dr. Mahdi Damghani
2017-2018
1

2. Suggested Readings
Chapter 1
of
Aircraft Structural Analysis
2

3. Questions
• Please make your comments in;
• https://padlet.com/mahdi_damghani/AeroLec1
3

4. Topics
• Familiarisation with the following concepts;
4
• Stress in deformable bodies
• Equations of equilibrium
• Plane stress condition
• Principal stresses
• Mohr’s circle of stress
• Von Mises stress and failure
criterion
• Strain in deformable bodies
• Equations of compatibility
• Plane strain condition
• Principal strains
• stress-strain relationship
• Mohr’s circle of strain

5. Introduction
• As stress engineers, you are required to obtain stresses
inside a structure due to external loading
• Before learning structural analysis methods, it is
necessary to have a better understanding of stress and
strain in a 3D body
5

6. Stress
6
Arbitrary shaped 3D body
Free body diagram
(cut through plane nn)
In equilibrium
Continuous
deformable
material
Resultant of external
forces at point O is δP

7. Notation of stresses
• The resultant stress is
resolved into the following
components;
• Perpendicular to δA, i.e. .
It is also called “direct stress”
• Parallel to δA, i.e. .
They are also called “shear
stresses”
7
z
zyzx  ,
The first subscript denotes
the plane it is acting on
The second subscript denotes
the direction it is acting on

8. Sign convention of stresses
• Direct stresses are
positive if they point out
of the surface, i.e.
tensile stresses
• Shear stresses are
positive if they act in the
positive direction of the
relevant axis in a plane
on which the direct
tensile stress is in the
positive direction of the
axis
8

9. Equations of equilibrium
9
Generally, except in cases of uniform stress, the direct
and shear stresses on opposite faces of an element
are not equal but differ by small amounts

10. Equations of equilibrium
• Imposing equations
of equilibrium on
the body
about the red line
gives;
• Following similar
manner we can get;
10
0M

11. Equations of equilibrium
• Imposing equations
of equilibrium on
the body
,
gives;
11
0 xF
0 yF 0 zF
Body forces in x, y
and z directions
Body forces in x, y
and z directions
Body forces in x, y
and z directions

12. Plane stress
• Due to thin sections in most aircraft structures,
stresses in one of the dimensions could be regarded
as negligible or zero.
• Let’s assume z axis is the thin direction, i.e.
• Our problem simplifies to a 2D problem as below;
12
0 yzxzz 
0
0
0

13. Boundary conditions
• Equilibrium must
also be satisfied at
all positions on the
boundary of the
body where the
components of the
surface force per
unit area are
13
.,, ZYX
where l, m, and n become the direction cosines of the
angles that a normal to the surface of the body
makes with the x, y, and z axes, respectively

14. Stresses on inclined planes
• Stresses on planes
blue, orange and
green might not be
the critical ones
• What if the yellow
plane is critical?
• How do we
calculate stresses
on the yellow plane
knowing stresses
on the others?
14

15. Stresses on inclined planes
15
Stresses on
2D body
Stresses on an
inclined plane

16. Principal stresses
• Direct and shear stresses values change on each plane as θ changes
• There are two planes, Principal planes, on which shear stresses are
zero and direct stresses are either maximum or minimum
• Direct stresses on Principal planes are called Principal stresses
• To obtain the max and min we equate derivative of stress to zero
leading to identification of that particular plane;
16

17. Principal stresses
• There are two answers for this, i.e. 𝜃 and 𝜋
2 + 𝜃
• tan 2𝜃 = tan 2
𝜋
2
+ 𝜃
• It means there are two mutually perpendicular planes.
Stress on one of them is max and on the other one is
min
17

18. Principal stresses
18

19. Principal stresses
• From trigonometry we know that;
• Replacing α with 2𝜃 and 2
𝜋
2
+ 𝜃 and considering
produces;
19
1tan
tan
sin
2





1tan
1
cos
2





20. Principal stresses
20
Major (max)
principal stress
Minor (min)
principal stress
• Students are required to find the plane with
maximum shear stresses as an extra activity.
III  

21. Mohr’s circle of stress
• The state of stress in a deformable body can be
illustrated by Mohr’s circle
• Previously we established that;
21
Re-writing
Square it and add it to
the shear term below

22. Reminder from maths
• Equation of circle;
• Compare the following
with the equation of a
circle;
22

23. Mohr’s circle
23
Major (max)
principal stress
Minor (min)
principal stress
Angles on the
circle become
twice as much

24. Example
• If stress state on two mutually perpendicular planes,
i.e. A and B, at a point in the structure is given,
determine the following;
• Draw Mohr’s circle
• Obtain principal stresses
• Obtain maximum shear stress
• The angle of principal planes with
planes A and B
24

25. Solution


)0,( t)0,( c
)25,80( A
)25,50( B
c15
2
5080


 avec 
R
  6.69251550 22
R
MPaRct 6.546.6915 
MPaRcc 6.846.6915 
),( maxave
),( max ave
MPaR 6.69max 

26. Solution


)0,6.54()0,6.84(
)25,80( A
)25,50( B
c
R

6.69
2525
2sin
R

),( maxave
),( max ave
2
deg52.10
deg05.212





27. von-Mises stress
• von Mises theory of elastic failure suggests that, in ductile
materials (metals such as aluminium, steel, etc), material fails if
the von Mises stress (𝜎𝑣) exceeds the yield stress (𝜎 𝑦) of
material
• For general 3D stress state;
• For plane stress state;
• Any point on or outside the ellipse
represents failure of material
27
        222222
65.0 zxyzxyxzzyyxv  
222
3 xyyyxxv  

28. Example
• Stress analysis of a spacecraft structural member
gives the state of stress as shown below. If the part is
made from an alloy with 𝜎 𝑦 = 500 MPa, check yielding
according to von Mises criterion.
28
-50MPa
100MPa
200MPa
-30MPa
-30MPa

29. Solution
29
        222222
65.0 zxyzxyxzzyyxv  
MPa
MPa
MPa
z
y
x
200
50
100






MPa
MPa
MPa
zx
yz
xy
0
0
30






MPaMPav 50005.224 

30. Tutorial 1
• Direct stresses of 160N/mm2 (tension) and 120N/mm2
(compression) are applied at a particular point in an
elastic material on two mutually perpendicular planes.
The principal stress in the material is limited to
200N/mm2 (tension). Using Mohr’s circle calculate the
allowable value of shear stress at the point on the
given planes. Determine also the value of the other
principal stress and the maximum value of shear
stress at the point.
(see Example 1.3 of Ref [1] on page 18 for solution)
30

31. Tutorial 2
• A structural member supports loads that produce, at a
particular point, a direct tensile stress of 80N/mm2
and a shear stress of 45N/mm2 on the same plane.
Calculate the values and directions of the principal
stresses at the point and also the maximum shear
stress, stating on which planes this will act.
(see P1.1 of Ref [1] on page 41 for the answer. You may also
refer to “Stress-Strain_Tutorial_Solution.pdf” uploaded on BB for
detailed solution)
31

32. Tutorial 3
• Listed here are varying combinations of stresses
acting at a point and referred to axes x and y in an
elastic material. Using Mohr’s circle of stress,
determine the principal stresses at the point and their
directions for each combination.
(see P1.3 of Ref [1] on page 41 for the answer. You may also refer
to “Stress-Strain_Tutorial_Solution.pdf” uploaded on BB for detailed
solution)
32

33. Strain
• stresses described in previous slides bring about
linear and angular strains within the structure
• Direct strains (𝜀) relate to change in length as the
result of direct stresses
• Shear strains (𝛾) relate to change in the angle due to
shear stresses
33

34. Strain
• Direct (normal) strain (+)
34
• Shear strain (+)
y
y
yx
yx
xy
y
x
yy E






xyxyyx G 

35. Strain
35
Three mutually perpendicular
line elements OA, OB & OC with
original length 𝛿𝑥, 𝛿𝑦, 𝛿𝑧
Coordinates of O in unstrained body; (𝑥, 𝑦, 𝑧)
Coordinates of A, B and C;
(𝑥 + 𝛿𝑥, 𝑦, 𝑧), (𝑥, 𝑦 + 𝛿𝑦, 𝑧) & (𝑥, 𝑦, 𝑧 + 𝛿𝑧),
respectively.

36. Strain
36

37. Compatibility equations
• So far we have defined components of strain in a
deformable body in terms of components of
displacement, i.e.
37

38. Compatibility equations
• Continuity or Compatibility equations suggest that
displacements within the body are continuous and
single valued, i.e. no voids in the structure would
form.
• In other words displacement can be defined as;
38

39. Compatibility equations
• The result of this is six extra equations relating strain
components s to each other. This is because strains
cannot have independent arbitrary values.
• Strain compatibility equations in 3D space are;
39

40. Physical meaning of compatibility of
strains
• Look at three line elements in a deformable body;
• Material particles need to meet up like pieces of a
jigsaw puzzle, i.e. they stay together after
deformation
40
After
deformation
compatible
Not
compatible

41. Example
• Can strains in a deformable elastic body be defined
as the following?
41
22
3
3
3
1
3
1
yx
yx
xy
xy
y
x







42. Solution
42
22
3
3
3
1
3
1
yx
yx
xy
xy
y
x






xy
xxx
yy
22
2












 
xy
yyy
xx
22
2












 
xy
yxyx
xyxy
4
2












 Yes

43. Plane strain
• The state of strain in which it is assumed that particles of
the body suffer displacement in one plane only is known as
plane strain
• Let’s assume that plane is xy, then we end up having the
following equations, i.e. three strains and one compatibility;
43
0
0
0

44. Strain on inclined planes
44
An element in 2D body subject
to complex stress
Deformed shape of the element
after application of stresses
We know strains at planes, i.e.
𝜀 𝑥, 𝜀 𝑦, 𝛾𝑥𝑦
We want strains on the CF
plane, i.e. 𝜀 𝑛, 𝛾
See Ref [1] as how
these are derived

45. Principal strains
• Comparison of stress terms with strain terms;
45

46. Mohr’s circle of strain
• Similar to that of stress with the difference that shear
stress will be replaced by half of shear strain
• We will see this in the tutorial
46

47. Stress-strain relationship
• Experiments show that (Hooke’s Law);
• This means application of direct stress in x dir
produces only direct strain without any shear
deformation (change in angles)
• The direct stress also causes strain in other
directions, i.e.
47
E is Young’s modulus
(modulus of elasticity)
Poisson’s ratio
0 ≤ 𝜗 ≤ 0.5 for isotropic material

48. Stress-strain relationship
• For a body subject to direct 3D stresses 𝜎 𝑥, 𝜎 𝑦, 𝜎𝑧 and
using the principle of superposition we have the
following;
48

49. Stress-strain relationship for plane
stress
• If we put 𝜎𝑧 = 0 then we have;
49
0
0
0
0

50. Example
• The steel block shown is subjected to a uniform
pressure on all its faces. Knowing that the change in
length of edge AB is -1.2x10-3 inches, determine (a)
the change in length of the other two edges, (b) the
pressure p applied to the faces of the block. Assume
E=29x106 psi and 𝜗=0.29.
50

51. Solution
51
pzyx  
  21


E
p
zyx
6
3
10300
4
102.1 




AB
x
x



52. Example
• A rectangular element in a linearly elastic isotropic
material is subjected to tensile stresses of 83 and
65N/mm2 on mutually perpendicular planes.
Determine the strain in the direction of each stress
and in the direction perpendicular to both stresses.
Find also the principal strains, the maximum shear
stress, the maximum shear strain, and their directions
at the point. Take E=200,000N/mm2 and 𝜗=0.3.
52

53. Solution
53

54. Solution
• There are no
shear stresses
on given planes
hence they are
principal planes
54
5.0

)0,10175.3( 4

c
R
)5.0,( max
)0,10005.2( 4

44
maxmax
1017.110
2
005.2175.3
2
25.0






 

 RR 
24
maxmax
/9769231017.1 mmNG  


55. Solution
• What if I had used
Mohr’s circle for
stress?
5555


)0,83(
c
R
),( max
)0,65(
2
max
/9
2
6583
mmN
R





 


56. Example
• At a particular point in a structural member, a two-
dimensional stress system exists where σx=60N/mm2,
σy=−40N/mm2, and 𝜏xy=50N/mm2. If Young’s modulus
E=200,000N/mm2 and Poisson’s ratio 𝜗=0.3,
calculate the direct strain in the x and y directions and
the shear strain at the point. Also calculate the
principal strains at the point and their inclination to the
plane on which σx acts (use Mohr’s circle)
56

57. Solution
57

58. Solution
58
6
105.0 

6
10

)6505.0,360( 
)6505.0,290( 
c
R
2 111
  66
1035102903605.0 
c
Rc
Rc


11
1


 
o
OC
452
35360
6505.0
360
6505.0
2tan









  61.459
7.0
3256505.0
2sin 

 R
R

 
  66
11
66
1
1061.4241061.45935
1061.4941061.45935






o
5.22
O

59. Student centred activity
• You are strongly advised to read section 1.16 of Ref.
[1] on “Experimental Measurement of Surface
Strains”.
• Please do so prior to the tutorial session.
59
     22
2
1
5.0 bcbacaI  
     22
2
1
5.0 bcbacaII  
ca
cab






2
2tan

60. Tutorial 4
60
(see P1.10 of Ref [1] on page 43 for the answer. You may also refer to “Stress-
Strain_Tutorial_Solution.pdf” uploaded on BB for detailed solution)

61. Tutorial 5
61
• The simply supported rectangular beam shown in Figure is
subjected to two symmetrically placed transverse loads each of
magnitude Q. A rectangular strain gauge rosette located at a
point P on the centroidal axis on one vertical face of the beam
gave strain readings as follows: εa=−222×10−6, εb=−213×10−6,
and εc=+45×10−6. determine principal strains at point P.
(see P1.11 of Ref [1] on page
44 for the answer. You may
also refer to “Stress-
Strain_Tutorial_Solution.pdf”
uploaded on BB for detailed
solution). Next slide shows
how the lecturer would solve
it.

62. Solution of Tutorial 5
62
6
105.0 

6
10

)5.0,222( a
C
R
111 O
)'5.0,213( b
)5.0,45( c
 
  6
105.8845452225.0
5.0


 OGOC ca 
N
  66
105.133105.88222 

 OCOCONCN a
2
2
Q
  66
105.124105.88213 

 OCCQaN b
6226
22
1054.1825.1335.12410 

 CNaNR
 
6
6
1
1004.94
1054.1825.88



 RC
G
 
6
6
11
10271
1054.1825.88



 RC