6

Principal Stresses
and Planes
(Complex state of stresses)
Stress Transformation
Learning Outcomes

• Define stresses on inclined planes

• Define the sign convention for stresses on inclined planes

• Determine the stresses on any given plane at a point

using Mohr’s Circle

3
 Plane Stresses

y-axis

x-axis

4
z-axis
Stress Analysis

P P

V
N
P

N
P
P V
5
 y-axis

𝜏𝑥 𝑦
x-axis
𝜎𝑦
z-axis plane
direction
𝜏 𝑦𝑥
y-axis 𝜏 𝑦𝑧 𝜏 𝑥𝑦 If the body is
𝜏 𝑧𝑦
𝜎𝑥
in equilibrium
y-plane
𝜏 𝑧𝑥 𝜏 𝑥𝑧 then all the
𝜎𝑧 stresses will be
ne
x - pla balanced by
z-plane x-axis corresponding
opposite stress.
z-axis

3-D Stress Element 6

 Plane Stresses
???
Two dimensions

y-axis
In engineering practices, to simplify the
complexity the stresses on one axis is
considered as minimum or neglected.

x-axis

7
z-axis
 y-axis

xy-plane

x-axis
𝜎𝑦

𝜏 𝑦𝑥

𝜏 𝑥𝑦

𝜎𝑥 𝜎𝑥

Plane Stresses 𝜏 𝑥𝑦
𝜏 𝑦𝑥

8
𝜎𝑦
Sign Conventions Tension Compression

𝜎𝑥 𝜎𝑥 𝜎𝑥 𝜎𝑥

𝜎𝑦 𝜎𝑦

Tension

𝜎𝑦 𝜎𝑦
Compression
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Sign Conventions
𝜏 𝑥𝑦 𝜏 𝑥𝑦

𝜏 𝑥𝑦 𝜏 𝑥𝑦

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Principal Planes
At any point in a strained material, there are three planes,
mutually perpendicular to each other, which carry direct stresses
only and no shear stress.

Out of these direct stresses one will be maximum, one minimum

and one intermediate between the two. These particular planes
having no shear stress are known as principal planes.

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Why Principal Stress are important to
determine ??
The magnitude of direct stress across principal plane is known as principal
stress

Determination of principal planes and then principal stress is an important

factor in design of various structures:

1. Understanding Material Behavior:

•Principal stresses represent the maximum and minimum normal stresses acting
on a material at a specific point. Knowing these values helps engineers
understand how a material will respond to different loading conditions.
•Stress Orientation: Principal stresses also indicate the orientation of these
maximum and minimum stresses. This information is crucial for understanding
how cracks might initiate and propagate in a material.
18
Why Principal Stress are important to
determine ??
2. Predicting Failure:
• Many failure theories, such as the Maximum Principal Stress Theory, are
based on the concept of principal stresses. These theories help engineers
predict when a material will fail under various loading conditions.
• Weak Points: By identifying the maximum principal stress, engineers can
pinpoint the locations within a structure that are most susceptible to
failure.

3. Analyzing Complex Stress States:

• Multi-directional Loads: In many real-world scenarios, materials are
subjected to complex, multi-directional loads. Principal stresses provide a
way to simplify and analyze these complex stress states.
19
Why Principal Stress are important to
determine ??

SUMMARY:

Principal Stresses are essential for understanding material behavior,

predicting failure, designing safe structures, and analyzing complex stress
states. They provide critical information that engineers need to ensure
the safety and reliability of their designs.

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Stress Analysis

P P

V
N
P

N
P
P V
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Stresses on an Inclined section/plane

N
P
P V

Force per unit area perpendicular to the

cut surface

Assume Uniformly Distributed

(+)
Tension
(-)
Compress
ion

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Stresses on an Inclined section/plane

N
P
P V

Force per unit area parallel to the cut

surface

Assume Uniformly
Distributed

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Stresses on an Inclined section/plane

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Stresses on an Inclined section/plane

Transformation Equations

For stresses in 𝝈′ 𝒇𝒂𝒄𝒆,

𝒚 𝒋𝒖𝒔𝒕 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒆 𝜽 + 𝟗𝟎° 𝒇𝒐𝒓 𝜽

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Stresses on an Inclined section/plane

Transformation Equations

For stresses in 𝝈′ 𝒇𝒂𝒄𝒆,

𝒚 𝒋𝒖𝒔𝒕 𝒔𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒆 𝜽 + 𝟗𝟎° 𝒇𝒐𝒓 𝜽

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 Transformation of Plane Stresses

𝜎𝑦 y'
a' 𝜎 𝑦′ 𝜏𝑥 ′ 𝑦 ′
x'
𝜎𝑥′
𝜏 𝑥𝑦
is measured counter
𝜃 clockwise, and taken
𝜎𝑥 𝜎𝑥
positive.

𝜎𝑥′
𝜏 𝑥𝑦 𝜃
𝜎 𝑦′
𝜏𝑥 ′ 𝑦 ′
𝜎𝑦 a' 27
 𝜏𝑥 ′ 𝑦 ′
𝜎𝑥′
Apply the conditions of equilibrium.
𝜎𝑥
𝜃
First, convert stresses to forces.

𝜏 𝑥𝑦 𝐹𝑜𝑟𝑐𝑒
𝑆𝑡𝑟𝑒𝑠𝑠=
𝐴𝑟𝑒𝑎
𝐹𝑜𝑟𝑐𝑒=𝑆𝑡𝑟𝑒𝑠𝑠 × 𝐴𝑟𝑒𝑎
𝜎𝑦 𝜎𝑦
a'
Let the area of inclined face is dA.
𝜏 𝑥𝑦

𝜃 𝑑𝐴
𝜎𝑥 𝜃 𝜎𝑥 𝑑𝐴cos 𝜃

𝜏 𝑥𝑦

𝜎𝑦 a' 28
 (x component)

𝜏 𝑥 ′ 𝑦 ′ 𝑑𝐴
(y component)
∑ 𝐹 𝑥=0
𝜃
∑ 𝐹 𝑦 =0
𝜎 𝑥 ′ 𝑑𝐴
𝜎 𝑥 𝑑𝐴 cos 𝜃
(y component)

𝜏 𝑥𝑦 𝑑𝐴 cos 𝜃
(x component)
𝜏 𝑥𝑦 𝑑𝐴 sin 𝜃

𝜎 𝑦 𝑑𝐴 sin 𝜃

∑ 𝐹 𝑥=0 𝜎 𝑥 ′ 𝑑𝐴 cos 𝜃− 𝜏 𝑥 𝑦 𝑑𝐴sin 𝜃 −𝜏 𝑥𝑦 𝑑𝐴 sin 𝜃− 𝜎 𝑥 𝑑𝐴 cos 𝜃=0

′ ′

∑ 𝐹 𝑦 =0 𝜎 𝑥 ′ 𝑑𝐴 sin 𝜃+ 𝜏 𝑥 ′ 𝑦 ′ 𝑑𝐴 cos 𝜃 − 𝜎 𝑦 𝑑𝐴 sin 𝜃 − 𝜏 𝑥𝑦 𝑑𝐴 cos 𝜃=0

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 𝜎 𝑥 ′ 𝑑𝐴 cos 𝜃− 𝜏 𝑥 𝑦 𝑑𝐴sin 𝜃 −𝜏 𝑥𝑦 𝑑𝐴 sin 𝜃− 𝜎 𝑥 𝑑𝐴 cos 𝜃=0
′ ′

𝜎 𝑥 ′ 𝑑𝐴 sin 𝜃+ 𝜏 𝑥 ′ 𝑦 ′ 𝑑𝐴 cos 𝜃 − 𝜎 𝑦 𝑑𝐴 sin 𝜃 − 𝜏 𝑥𝑦 𝑑𝐴 cos 𝜃=0

dA is common in both the equations, hence dividing both the equations with dA,
we have.

𝜎 𝑥 ′ cos 𝜃 − 𝜏 𝑥 𝑦 sin 𝜃− 𝜏 𝑥𝑦 sin 𝜃 − 𝜎 𝑥 cos 𝜃=0

′ ′

𝜎 𝑥 ′ sin 𝜃+𝜏 𝑥′ 𝑦 ′ cos 𝜃− 𝜎 𝑦 sin 𝜃 −𝜏 𝑥𝑦 cos 𝜃=0

Now we have two equations, and we want to find out stresses on inclined plane
i.e., and .
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𝜎 𝑥 ′ cos 𝜃 − 𝜏 𝑥 𝑦 sin 𝜃− 𝜏 𝑥𝑦 sin 𝜃 − 𝜎 𝑥 cos 𝜃=0
′ ′ Eq. (A)
Two equations and
two variables.
𝜎 𝑥 ′ sin 𝜃+𝜏 𝑥′ 𝑦 ′ cos 𝜃− 𝜎 𝑦 sin 𝜃 −𝜏 𝑥𝑦 cos 𝜃=0 Eq. (B)

Multiply first equation with and second equation with . We have

2 2
𝜎 𝑥 ′ cos 𝜃 −𝜏 𝑥 𝑦 sin 𝜃 cos 𝜃 − 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃 − 𝜎 𝑥 cos 𝜃=0
′ ′

2 2
𝜎 𝑥 ′ sin 𝜃 +𝜏 𝑥 ′ 𝑦 ′ cos 𝜃 sin 𝜃 − 𝜎 𝑦 sin 𝜃 −𝜏 𝑥𝑦 cos 𝜃 sin 𝜃=0

2 2 2 2
𝜎 𝑥 ′ cos 𝜃+ 𝜎 𝑥 ′ sin 𝜃− 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃 − 𝜎 𝑥 cos 𝜃 − 𝜎 𝑦 sin 𝜃 −𝜏 𝑥𝑦 cos 𝜃 sin 𝜃=0
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 2 2
𝜎 𝑥 ′ − 2𝜏 𝑥𝑦 sin 𝜃 cos 𝜃− 𝜎 𝑥 cos 𝜃 − 𝜎 𝑦 sin 𝜃=0

2 2
𝜎 𝑥 ′ =2 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃+ 𝜎 𝑥 cos 𝜃+ 𝜎 𝑦 sin 𝜃

Now to determine , again take both the equations (A) and (B).

𝜎 𝑥 ′ cos 𝜃 − 𝜏 𝑥 𝑦 sin 𝜃− 𝜏 𝑥𝑦 sin 𝜃 − 𝜎 𝑥 cos 𝜃=0 Eq. (A)

′ ′

𝜎 𝑥 ′ sin 𝜃+𝜏 𝑥′ 𝑦 ′ cos 𝜃− 𝜎 𝑦 sin 𝜃 −𝜏 𝑥𝑦 cos 𝜃=0 Eq. (B)

Multiply first equation with and second equation with . We have

32
 2 2
𝜎 𝑥 ′ sin 𝜃 cos 𝜃 −𝜏 𝑥 𝑦 sin 𝜃 −𝜏 𝑥𝑦 sin 𝜃− 𝜎 𝑥 sin 𝜃 cos 𝜃=0
′ ′

c ti ng
btra
Su
2 2
𝜎 𝑥 ′ sin 𝜃 cos 𝜃+𝜏 𝑥′ 𝑦 ′ cos 𝜃− 𝜎 𝑦 sin 𝜃 cos 𝜃 −𝜏 𝑥𝑦 cos 𝜃=0
− − +¿ +¿
−𝜏 𝑥 𝑦 sin2 𝜃 −𝜏 𝑥𝑦 sin 2 𝜃 −𝜎 𝑥 sin 𝜃 cos 𝜃 −𝜏 𝑥 𝑦 cos 2 𝜃+ 𝜎 𝑦 sin 𝜃 cos 𝜃+ 𝜏 𝑥𝑦 cos 2 𝜃=0
′ ′ ′ ′

−𝜏 𝑥 𝑦 ( sin 2 𝜃 +cos 2 𝜃 ) =𝜏 𝑥𝑦 sin2 𝜃 − 𝜏 𝑥𝑦 cos2 𝜃 +𝜎 𝑥 sin 𝜃 cos 𝜃 − 𝜎 𝑦 sin 𝜃 cos 𝜃

′ ′

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 2 2
𝜎 𝑥 ′ =2 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃+ 𝜎 𝑥 cos 𝜃+ 𝜎 𝑦 sin 𝜃

??
a'
𝜎 𝑦 =100 𝑀𝑝𝑎 y'
𝜎 𝑦′ 
𝜏𝑥 ′ 𝑦 ′
 x'
𝜎𝑥′
𝜏 𝑥𝑦 =25 𝑀𝑃𝑎

𝜃 𝜎 𝑥 =50 𝑀𝑃𝑎

Orientation of is 𝜎𝑥′
𝜏 𝑥𝑦 𝜃
𝜎 𝑦′
𝜏𝑥 ′ 𝑦 ′
𝜎𝑦 a' 34
 2 2
𝜎 𝑥 ′ =2 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃+ 𝜎 𝑥 cos 𝜃+ 𝜎 𝑦 sin 𝜃

𝜎 𝑦 ′ =2 𝜏 𝑥𝑦 sin ( 𝜃 + 90 ) cos ( 𝜃 + 90 ) + 𝜎 𝑥 cos ( 𝜃 +90 ) cos ( 𝜃 +90 )+ 𝜎 𝑦 sin ( 𝜃 +90 ) sin ( 𝜃 + 90 )

cos ( 𝜃+ 90 ) =−sin 𝜃

2 2
𝜎 𝑦 ′ =−2 𝜏 𝑥𝑦 cos 𝜃 sin 𝜃 +𝜎 𝑥 sin 𝜃+ 𝜎 𝑦 cos 𝜃

35
 2 2
𝜎 𝑥 ′ =2 𝜏 𝑥𝑦 sin 𝜃 cos 𝜃+ 𝜎 𝑥 cos 𝜃+ 𝜎 𝑦 sin 𝜃

2 2
𝜎 𝑦 ′ =−2 𝜏 𝑥𝑦 cos 𝜃 sin 𝜃 +𝜎 𝑥 sin 𝜃+ 𝜎 𝑦 cos 𝜃

1+ cos 2 𝜃 y'
2
cos 𝜃 = 𝜎 𝑦′ 𝜏𝑥 ′ 𝑦 ′
2 x'
𝜎𝑥′

2 1 −cos 2 𝜃
sin 𝜃=
2

1 𝜎𝑥′
sin 𝜃 cos 𝜃= sin 2 𝜃 𝜃
2 𝜎 𝑦′
𝜏𝑥 ′ 𝑦 ′
36
 𝜎 𝑥 +𝜎 𝑦 𝜎 𝑥 − 𝜎 𝑦
𝜎 𝑥 ′= + cos 2 𝜃 +𝜏 𝑥𝑦 sin 2 𝜃
2 2

𝜎 𝑥+ 𝜎 𝑦 𝜎 𝑥 − 𝜎 𝑦
𝜎 𝑦 ′= − cos 2 𝜃 −𝜏 𝑥𝑦 sin 2 𝜃
2 2

y'
𝜎𝑥− 𝜎𝑦 𝜎 𝑦′ 𝜏𝑥 ′ 𝑦 ′
𝜏 𝑥 ′ 𝑦 ′ =− sin 2 𝜃 + 𝜏 𝑥𝑦 cos 2 𝜃 x'
2 𝜎𝑥′

𝜎𝑥′
𝜃
𝜎 𝑦′
𝜏𝑥 ′ 𝑦 ′
37
 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠,
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
𝜎𝑃1 = + + 𝜏𝑥𝑦
2
2 2

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠,

𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
𝜎𝑃2 = − + 𝜏𝑥𝑦
2
2 2

41
Maximum in-plane shear stress
𝑑
sin 𝜃=cos 𝜃
𝑑𝜃
𝜎𝑥− 𝜎𝑦
𝜏 𝑥 ′ 𝑦 ′ =− sin 2 𝜃 + 𝜏 𝑥𝑦 cos 2 𝜃
2 𝑑
cos 𝜃=− sin 𝜃
𝑑𝜃
𝑑𝜏 𝑥′𝑦′
=0
𝑑𝜃

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