HASAN AFTAB SAEED, PhD, PE
STRESS
TRANSFORMATION
Mechanics of Materials II
HASAN AFTAB SAEED, PhD, PE
REPRESENTATION OF STRESS AT A POINT
Infinite possibilities: consider 2 of those possibilities for stress at point 𝐶:
Which one is correct?
Which one is more useful?
HASAN AFTAB SAEED, PhD, PE
…REPRESENTATION OF STRESS AT A POINT
Same point
∴ Same state of stress
∴ Both are equivalent
∴ Both are correct
The normal and shear stress components at a point clearly depend on the orientation of
the stress element in addition to its position in the object
However, the state of stress (being a physical reality) at that point is one and the same
and must be independent of our choice of the orientation of the stress element. (Stress at
a point only depends on the geometry and loading details)
Some orientations are more useful than others though. The orientation on the left
(without the shear stress), for example, is of great significance and utility in design and
analysis, which will become clear shortly
Although demonstrated using a simple uniaxial loading problem, the above is true of
stress in general
HASAN AFTAB SAEED, PhD, PE
GENERAL STATE OF STRESS
In the general three-dimensional case, we expect to find at most one normal
stress component and two shear stress components on each interface of the threedimensional stress element
There are three such interfaces, and therefore there
are nine stress components in all
In cartesian coordinate system, taking these interfaces parallel to the x, y, z axes,
these components (called rectangular components) are (in array form):
𝜎𝑥
𝜎𝑖𝑗 = 𝜏𝑦𝑥
𝜏𝑧𝑥
𝜏𝑥𝑦
𝜎𝑦
𝜏𝑧𝑦
𝜏𝑥𝑧
𝜏𝑦𝑧
𝜎𝑧
Owing to the complementary property of shear, only six of these are independent
quantities (the above array is symmetric)
HASAN AFTAB SAEED, PhD, PE
THE STRESS TENSOR
Quantities that need this sort of information for their complete specification, in
this case the magnitude as well as orientation of the area (the first subscript) and
the direction of stress component (the second subscript) are called second order
tensors
In contrast, a vector (e.g., force) only needs three components along three
coordinate axes directions for its complete specification
𝐹𝑥
𝐅 = 𝐹𝑦
𝐹𝑧
A vector is a first order tensor
A scalar (e.g., temperature), which is completely specified by one number is a
zeroth order tensor
HASAN AFTAB SAEED, PhD, PE
TENSOR TRANSFORMATION
We often need to transform second order tensors from one set of
coordinates 𝑥𝑦𝑧 to another 𝑥ƴ 𝑦ƴ 𝑧.ƴ We will learn the procedure
now
You are familiar with the procedure to transform a first order
tensor (vector) from one set of coordinates 𝑥𝑦𝑧 to another 𝑥ƴ 𝑦ƴ 𝑧,ƴ
resulting in different components but preserving the physical
reality of the vector
cos 𝜃 sin 𝜃
−sin 𝜃
cos 𝜃
Of course, being just a number, a scalar does not require
transformation from one set of coordinates 𝑥𝑦𝑧 to another 𝑥ƴ 𝑦ƴ 𝑧ƴ
For simplicity, the instances shown are all 2D transformations
HASAN AFTAB SAEED, PhD, PE
…TENSOR TRANSFORMATION
Once the second order tensor transformation rules are mastered, they can be
used to transform any second order tensor, whether stress, strain, or inertia
tensor (although physically each is very different from the others)
𝜎𝑥
𝜎𝑖𝑗 = 𝜏𝑦𝑥
𝜏𝑧𝑥
𝜏𝑥𝑦
𝜎𝑦
𝜏𝑧𝑦
𝜏𝑥𝑧
𝜏𝑦𝑧
𝜎𝑧
𝜖𝑥
𝛾𝑦𝑥
𝜖𝑖𝑗 =
2
𝛾𝑧𝑥
2
𝛾𝑥𝑦
2
𝜖𝑦
𝛾𝑧𝑦
2
𝛾𝑥𝑧
2
𝛾𝑦𝑧
2
𝜖𝑧
𝐼𝑥
𝐼𝑖𝑗 = −𝐼𝑦𝑥
−𝐼𝑥𝑧
−𝐼𝑥𝑦
𝐼𝑦
−𝐼𝑧𝑦
−𝐼𝑥𝑧
−𝐼𝑦𝑧
𝐼𝑧
HASAN AFTAB SAEED, PhD, PE
STRESS TRANSFORMATION: UTILITY
Stress transformation is indispensable in mechanical engineering design and
analysis because the normal and shear stress components given or calculated
with respect to an arbitrary set of coordinates 𝑥𝑦𝑧 are usually not the extreme
values that are critical to decide whether an object can be expected to withstand
them without failing
Stress transformation allows the designer/analyst to view the stress components
in any set of coordinates 𝑥ƴ 𝑦ƴ 𝑧ƴ including those for which the normal and shear
stress components reach their extreme values
In this course, instead of the general three-dimensional stress transformation, we
will focus on plane stress transformation only
HASAN AFTAB SAEED, PhD, PE
PLANE STRESS
Also called a two-dimensional state of stress
Consider a thin plate on which all loads and constraints are in its midplane
Since the upper and lower faces are subject only to atmospheric pressure (which
can be ignored), therefore 𝜎𝑧 ≈ 0 on the two faces. Also, since the plate is thin, it
is concluded that 𝜎𝑧 is zero inside too
Since there is no shear traction on the upper and lower faces, it is similarly
concluded that 𝜏𝑧𝑥 = 𝜏𝑧𝑦 = 0 on the two faces as well as throughout the entire
thickness of the plate
As non-zero stresses, this leaves only 𝜎𝑥 , 𝜎𝑦 and 𝜏𝑥𝑦 , which are considered not to
vary over the thickness of the plate (not functions of 𝑧)
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS
𝜎𝑥
𝜎𝑖𝑗 = 𝜏
𝑦𝑥
𝜏𝑥𝑦
𝜎𝑦
Very useful approximation as it occurs extensively in engineering practice
It is not found exclusively in thin plates but can also exist in more complex
geometries such as surfaces of thin-walled shells, beams, and rods under torsion
HASAN AFTAB SAEED, PhD, PE
PLANE STRESS TRANSFORMATION
The element shown is of finite
dimensions so all stress values are
average values, indicated by the *
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS TRANSFORMATION
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS TRANSFORMATION
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS TRANSFORMATION
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS TRANSFORMATION
HASAN AFTAB SAEED, PhD, PE
…PLANE STRESS TRANSFORMATION
To avoid confusion in signs of the terms, some people find it
simpler to use the expression for 𝜎𝑥ሶ and adding 90° to the
angle while solving a particular problem
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 1
Determine the normal stress and shear stress acting on the inclined plane 𝐴𝐵.
Show the result on a stress element.
𝑦′
85 MPa
40 MPa
5 MPa
𝑥′
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 2
The steel beam is subjected to a load of 12 kN. Show the stress components at
point 𝐴 on the 25° inclined line.
1.7787 MPa
𝑦′
0.958 MPa
0.507 MPa
𝑥′
HASAN AFTAB SAEED, PhD, PE
EXTREME VALUES OF STRESS COMPONENTS
Depending on the material, failure of structures occurs owing to the inability of
the material to resist normal or shear stresses
The extreme values of normal stress are found on planes on which shear stresses
are zero
These planes are called principal planes and the (normal) stresses on them are
called the principal stresses
We will now evaluate principal stresses from the stress transformation equations
just derived
This will be followed by evaluating the extreme values of shear stresses
HASAN AFTAB SAEED, PhD, PE
PRINCIPAL PLANES
HASAN AFTAB SAEED, PhD, PE
…PRINCIPAL PLANES
HASAN AFTAB SAEED, PhD, PE
…PRINCIPAL PLANES
There are two principal stresses in the plane of interest (𝑥𝑦), one being the
maximum normal stress and the other being the minimum normal stress on this
plane
The two principal stresses are 90° apart
On the principal planes, there is no shear stress
Even in the plane stress situation, 𝜎𝑧 = 0 is by definition (no shear stress) the third
principal stress
The three principal axes are therefore always mutually perpendicular; the only
difference from the general 3D case is that in the latter, the third principal stress is
non-zero
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 3
The steel beam is subjected to a load of 12 kN. Determine, and show on a stress
element, the principal stresses at 𝐴.
𝑦′
7.21 kPa
𝑥
3.21°
2.29 MPa
𝑥′
HASAN AFTAB SAEED, PhD, PE
PLANES OF MAXIMUM IN-PLANE SHEAR STRESS
HASAN AFTAB SAEED, PhD, PE
…PLANES OF MAXIMUM IN-PLANE SHEAR STRESS
HASAN AFTAB SAEED, PhD, PE
…PLANES OF MAXIMUM IN-PLANE SHEAR STRESS
There are two planes of maximum shear stress, 90° apart from one another
From the complementary property of shear, the two stresses are equal
On both these planes, the normal stress is equal to the average normal stress
The planes of maximum shear stress are oriented at 45° to the principal planes
Unlike the principal stresses, where the sign of the stress (tension or
compression) has a marked effect on how the material would behave, the sign of
shear stress has no relevance to material behavior
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 4
The steel beam is subjected to a load of 12 kN. Determine, and show on a stress
element, the maximum in-plane shear stresses at 𝐴.
1.1428 MPa
1.1428 MPa
HASAN AFTAB SAEED, PhD, PE
PLANE STRESS: MOHR’S CIRCLE
An ingenious graphical method for representing stress at a point and its
transformation
Developed by Christian Otto Mohr, a German structural engineer and a professor
of mechanics
Advantages:
▪ Extremely convenient, easy to remember and time saving
▪ Gives a visual insight into variation of stress at a point, which equations do not
▪ One circle gives all the information (transformation to any plane, principal
stresses, maximum shear stress and the associated average stress, for each of
which separate, lengthy analyses are required otherwise (also, tasks such as
ascertaining which principle direction corresponds to which principal stress
are rendered unnecessary with this tool)
HASAN AFTAB SAEED, PhD, PE
MOHR’S CIRCLE: BASIS
HASAN AFTAB SAEED, PhD, PE
…MOHR’S CIRCLE: BASIS
2
The axes could also
be the other way
around but by
convention 𝜎 and 𝜏
are taken on x and y
axes respectively
2
2
HASAN AFTAB SAEED, PhD, PE
MOHR’S CIRCLE: HOW TO DRAW (5 STEPS)
1. Take the positive 𝜎 axis to the right of the origin and
the positive 𝜏 axis downwards (the reason to be
explained presently)
2. Mark the center 𝐶 of the circle at a horizontal
distance 𝜎avg from the origin 𝑂
3. Mark 𝜎𝑥 and 𝜏𝑥𝑦 and label the point 𝐴 (this represents the right-hand vertical
side of the element for which 𝜃 = 0 and therefore 𝜎𝑥ƴ = 𝜎𝑥 , 𝜏𝑥𝑦ƴ = 𝜏𝑥𝑦 )
4. Join 𝐶 to 𝐴, which is the radius 𝑅
5. With 𝐶 and 𝑅 known, draw the circle
The circle is usually drawn in free-hand and the various quantities are marked
and read analytically, instead of drawing it graphically to some scale
HASAN AFTAB SAEED, PhD, PE
MOHR’S CIRCLE: PROPERTIES
With no shear stress, points 𝐵 and 𝐸 represent the principal
planes, stresses here being 𝜎1 and 𝜎2 respectively
Points 𝐷 and 𝐹 represent the planes of maximum shear
stress ±𝜏maxin−plane . They are in-line with C, so the normal
stress on each is 𝜎avg
On the circle, the angle between the two principal planes and between the two
maximum shear planes is 180° each, while between the principal planes and the
maximum shear planes it is 90°. On the stress elements these angles are 90° and
45° respectively. So, all angles need to be doubled when bringing them to the
stress circle and halved when taking them back to the stress element
For example, a rotation 𝜃 for orientation 𝑃 on the stress element corresponds to
a rotation of 2𝜃 on the stress circle
HASAN AFTAB SAEED, PhD, PE
DIRECTION OF POSITIVE 𝜏 AXIS
With our convention of positive shear stress, for the stress configuration shown,
the maximum principal stress on the stress element occurs by rotating it by an
angle 𝜃𝑝 counterclockwise (as shown)
𝐴
Taking the positive 𝜏 axis downwards, the principal plane is reached by going
an angle 2𝜃𝑝 again counterclockwise. If the positive 𝜏 axis is taken upwards, a
counterclockwise rotation on the stress element will correspond to a clockwise
rotation on the Mohr’s circle (and vice versa), which is cumbersome and
aesthetically undesirable
HASAN AFTAB SAEED, PhD, PE
EXAMPLES 2, 3 AND 4 REVISITED
Now, by Mohr’s circle
Example 2
The steel beam is subjected to a load of 12 kN. Show the stress components at
point 𝐴 on the 25° inclined line.
Example 3
The steel beam is subjected to a load of 12 kN. Determine, and show on a stress
element, the principal stresses at 𝐴.
Example 4
The steel beam is subjected to a load of 12 kN. Determine, and show on a stress
element, the maximum in-plane shear stresses at 𝐴.
Answers same as earlier (done using equations)
HASAN AFTAB SAEED, PhD, PE
UNI-, BI- AND TRIAXIAL STRESS STATES
At any arbitrary point in any object subjected to any kind of loading, there are
ALWAYS three mutually perpendicular principal stresses
In 1D axial problems (uniaxial state of stress) two of these principal stresses are
zero
Plane stress is a biaxial state of stress, where one principal stress (𝜎𝑧 ) happens
to be zero
In the general 3D or triaxial state of stress, all three principal stresses are nonzero
HASAN AFTAB SAEED, PhD, PE
MOHR’S CIRCLES FOR TRIAXIAL STRESS
In pane stress analysis, we have already constructed a Mohr’s circle for the 𝑥𝑦
plane while looking down from the positive 𝑧-axis (the axis which coincides with
𝜎𝑧 = 0). The maximum ‘in-plane’ shear stress in this case is equal to the radius of
the Mohr’s circle
It so happens that we would have the exact same Mohr’s circle for plane 𝑥𝑦 if 𝜎𝑧 ≠
0 (provided 𝑧 is a principal axis), with the same maximum ‘in-plane’ shear stress
HASAN AFTAB SAEED, PhD, PE
AN ASIDE: TRIAXIAL STATE WHERE 𝜎𝑧 ≠ 0
Although 3D stress transformation is beyond the scope of this course, the stress
transformation equations are valid for stress transformation, although if 𝑧 is not a
principal direction, the two extreme values of stress (although the extreme values
in the plane) will not be the principal stresses
HASAN AFTAB SAEED, PhD, PE
…MOHR’S CIRCLES FOR TRIAXIAL STRESS
Based on the above, two more Mohr’s circles can be drawn, one for the 𝑦𝑧 plane
while looking down from the positive 𝑥-axis, and the other for the 𝑧𝑥 plane
while looking down from the positive 𝑦-axis. Each has the maximum in-plane
shear stress equal to the radius of the Mohr’s circle
The absolute maximum shear stress is the greatest of the three in-plane shear
stresses, i.e., equal to the radius of the biggest Mohr’s circle of the three
Design and analysis are governed by the absolute maximum shear stress
IMPLICATION FOR PLANE STRESS
In plane stress, by definition, one principal stress 𝜎𝑧 = 0
HASAN AFTAB SAEED, PhD, PE
1-1 corres. of 𝛾𝑖𝑗 and 𝜏𝑖𝑗
𝜏𝑥𝑧 in plane stress because
of 𝛾𝑥𝑧
𝜏𝑦𝑧 is similarly there
𝑥
Three scenarios are possible:
𝑧
𝜎
−𝜎
1. The non-zero principal stresses are of opposite signs: 𝜏ABS MAX = max min
2.
3.
𝜎
The non-zero principal stresses are both positive: 𝜏ABS MAX = max
2
𝜎min
The non-zero principal stresses are both negative: 𝜏ABS MAX =
2
2
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 5
A cylindrical pressure vessel is constructed from a long, narrow
steel plate by wrapping the plate around a mandrel and then
welding along the edges of the plate to make a helical joint. The
helical weld makes an angle of 𝛼 = 55° with the longitudinal axis.
The vessel has inner radius of 1.8 m and wall thickness of 20 mm.
The internal pressure is 800 kPa.
Calculate:
▪ the circumferential and longitudinal stresses
▪ the maximum in-plane and the absolute maximum shear stresses
▪ the normal stress and shear stress acting perpendicular and
parallel to the welded seam
𝜎1 = 72 MPa, 𝜎2 = 36 MPa
𝜏maxin−plane = 18 MPa, 𝜏maxabs = 36 MPa
𝜎𝑥 ′ = 47.84 MPa, 𝜏𝑥 ′ 𝑦′ = 16.91 MPa
55°
35°