Transformation of Stress

CE 213: Mechanics of Solids-II

Md. Hossain Nadim
Department of Civil Engineering
Ahsanullah University of Science and Technology
1
Transformation of Stress

Stress at a General Point in an Arbitrarily Loaded Body

y
P2

Vxy
ΔA
P1
P3 Fx
Vxz Q

P4

x x
z
z

Δ𝐹 Δ𝑉𝑥𝑦 Δ𝑉𝑥𝑧
𝜎𝑥 = lim 𝜏𝑥𝑦 = lim 𝜏𝑥𝑧 = lim
Δ𝐴→0 Δ𝐴 Δ𝐴→0 Δ𝐴 Δ𝐴→0 Δ𝐴

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Transformation of Stress
y

y y

σy
τxy τzy

τzx
σx τyz τyx
τxz Q σz

x
x
x
z
z 𝒚 − 𝒛 𝑷𝒍𝒂𝒏𝒆 z 𝒙 − 𝒚 𝑷𝒍𝒂𝒏𝒆
𝒛 − 𝒙 𝑷𝒍𝒂𝒏𝒆

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 Transformation of Stress

𝝉𝒚𝒛 𝝉𝒚𝒙
Q
y 𝝉𝒙𝒚
𝝉𝒛𝒚

x 𝝉𝒛𝒙 𝝉𝒙𝒛
𝝈𝒚
z
y
x
𝝈𝒙

z
𝝈𝒛

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Transformation of Stress

The x-face of the element is perpendicular to the x-axis.

𝟐. denotes the direction 𝝈𝒙 Denotes the face

𝝉𝟏𝟐 Thus, 𝜎𝑥 denotes the normal stress acting on an
𝑥‐face; 𝜏𝑥𝑦 is the shear stress on the 𝑥‐face
𝟏. denotes the face
acting in the 𝑦‐direction; and so on.

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Transformation of Stress

y 𝒚′ The transformation of stress components, however, is more

complicated than vector addition. In considering stresses, the
transformation must account for not only the magnitude and
direction of each stress component, but also the orientation of
the area upon which the stress component acts.
x

𝒙′

The components are different in the two

coordinate systems, but both sets of
components represent the same resultant
force. Failure of the material will occur in response to the largest
stresses that are developed in the object, regardless of the
orientation at which those critical stresses are acting. To find the
critical stresses at a point in a material object, methods must be
developed so that stresses acting at all possible orientations can
be investigated.

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 Transformation of Stress

Stress Sign Conventions

• The normal stress acting on a face of the stress element is
positive if it points in the outward normal direction. In other
words, normal stresses are positive if they cause tension in the
material. Compressive normal stresses are negative.
• Positive Shear stresses act in positive coordinate directions on
positive faces of the element and acts in the negative coordinate
direction on a negative face of the stress element.
Equilibrium of the Stress Element

If a shear stress exists on any plane, there

must also be a shear stress of the same
magnitude acting on an orthogonal plane
𝜏𝑥𝑦 Δ𝐴 (i.e., a plane perpendicular to the original
So the result concludes that: plane).
𝜏𝑥𝑦 Δ𝐴

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Transformation of Stress

Plane Stress
• Plane Stress - state of stress in which two faces of
the cubic element are free of stress. For the
illustrated example, the state of stress is defined by
 x ,  y ,  xy and  z   zx   zy  0.

• State of plane stress occurs in a thin plate subjected

to forces acting in the mid-plane of the plate.

• State of plane stress also occurs on the free surface

of a structural element or machine component, i.e.,
at any point of the surface not subjected to an
external force.

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 Transformation of Stress

General Equations of Plane Stress Transformation

Consider a state of stress represented by a plane stress element
subjected to stresses 𝜎𝑥 , 𝜎𝑦 , and 𝜏xy = 𝜏𝑦𝑥 , as shown in Fig.(𝑎). To A
derive equations applicable to any orientation, we begin by θ
defining a plane surface 𝐴 − 𝐴 oriented at some angle 𝜃 with 𝝈𝒙
respect to a reference axis 𝑥. The normal to surface 𝐴 − 𝐴 is A
termed the 𝑥′ axis. The axis parallel to surface 𝐴 − 𝐴 is termed the 𝝉𝒙𝒚
𝑦′ axis. The 𝑧 axis extends out of the plane of the stress element.
y' y
Both the 𝑥 − 𝑦 − 𝑧 and the 𝑥′ − 𝑦′ − 𝑧′ axes are arranged as x'
right‐hand coordinate systems. This process of changing stresses
𝑦′ y σx'
from one set of coordinate axes (𝑖. 𝑒. , 𝑥 − 𝑦 − 𝑧) to another set of σy'
axes (𝑖. 𝑒. , 𝑥′ − 𝑦′ − 𝑧′) is termed stress transformation. x'

θ θ
x x

τx'y'
Fig.(𝑎).
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Transformation of Stress

Fig.(𝑏) is a free‐body diagram of a wedge‐shaped element in which

the areas of the faces are 𝐴 for the inclined face (plane 𝐴 − 𝐴), ΔA
A cos θ θ
𝐴 cos 𝜃 for the vertical face (𝑖. 𝑒. , 𝑡ℎ𝑒 𝑥 face), and 𝐴 sin 𝜃for the
horizontal face (𝑖. 𝑒. , 𝑡ℎ𝑒 𝑦face). The equilibrium equation for the sum
of forces in the 𝑥′ direction gives A sin θ

y
y'
x'
 x  A
θ  x  A
x
θ  x A cos θ
θ
 x A cos θ θ
 xyA
 xyA  xy A cos 
 xy A cos 
 xy A sin 
 xy A sin  θ

 y A sin 
 y A sin 
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Transformation of Stress

෍ 𝐹𝑥 ′ = 0

≫ 𝜎𝑥 ′ 𝐴 − (𝜎𝑥 𝐴 cos 𝜃) cos 𝜃 − (𝜏𝑥𝑦 𝐴 cos 𝜃) 𝑠𝑖𝑛 𝜃 − (𝜎𝑦 𝐴 si𝑛 𝜃) sin 𝜃 − (𝜏𝑥𝑦 𝐴 sin 𝜃) 𝑐𝑜𝑠 𝜃 = 0
≫ 𝜎𝑥 ′ − 𝜎𝑥 cos2 𝜃 − 𝜏𝑥𝑦 cos 𝜃𝑠𝑖𝑛 𝜃 − 𝜎𝑦 si𝑛2 𝜃 − 𝜏𝑥𝑦 sin 𝜃𝑐𝑜𝑠 𝜃 = 0
 x  A
2 2
≫ 𝜎𝑥 ′ = 𝜎𝑥 cos 𝜃 + 𝜎𝑦 si𝑛 𝜃 + 2𝜏𝑥𝑦 sin 𝜃𝑐𝑜𝑠 𝜃
𝜎𝑦
 x A cos θ
θ
𝜎𝑥
≫ 𝜎𝑥 ′ = (2 cos2 𝜃) + (2si𝑛2 𝜃) + 𝜏𝑥𝑦 (2 sin 𝜃𝑐𝑜𝑠 𝜃) θ
2 2
 xyA
≫ 𝜎𝑥 ′ =
𝜎𝑥 𝜎𝑦
1 + 𝑐𝑜𝑠2𝜃 + (1 − 𝑐𝑜𝑠2𝜃) + 𝜏𝑥𝑦 (𝑠𝑖𝑛2𝜃)
 xy A cos 
2 2
𝜎𝑥 𝜎𝑦 𝜎𝑥 𝜎𝑦
θ
 xy A sin 
≫ 𝜎𝑥 ′ = + + − 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2 2 2
𝝈𝒙 + 𝝈𝒚 𝝈𝒙 − 𝝈𝒚
≫ 𝜎𝑥 ′ = + 𝒄𝒐𝒔𝟐𝜽 + 𝝉𝒙𝒚 𝒔𝒊𝒏𝟐𝜽  y A sin 
𝟐 𝟐
(1)

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 Transformation of Stress

the equilibrium equation for the sum of forces in the 𝑦’ direction gives

෍ 𝐹𝑦 ′ = 0

≫ 𝜏𝑥 ′ 𝑦 ′ 𝐴 + (𝜎𝑥 𝐴 cos 𝜃) sin 𝜃 − (𝜏𝑥𝑦 𝐴 cos 𝜃) 𝑐𝑜𝑠 𝜃 − (𝜎𝑦 𝐴 si𝑛 𝜃) cos 𝜃 + (𝜏𝑥𝑦 𝐴 sin 𝜃) 𝑠𝑖𝑛 𝜃 = 0
≫ 𝜏𝑥 ′ 𝑦 ′ = −𝜎𝑥 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 + 𝜏𝑥𝑦 cos 2 𝜃 + 𝜎𝑦 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 − 𝜏𝑥𝑦 𝑠𝑖𝑛2 𝜃

≫ 𝜏𝑥 ′ 𝑦 ′ = 𝜎𝑦 − 𝜎𝑥 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛2 𝜃

𝜎𝑦 − 𝜎𝑥
 x  A
θ
≫ 𝜏𝑥 ′ 𝑦 ′ = 2 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃
2  x A cos θ
θ
𝝈𝒙 − 𝝈𝒚 y
𝑦′
≫ 𝝉𝒙′ 𝒚′ = −
𝟐
𝒔𝒊𝒏𝟐𝜽 + 𝝉𝒙𝒚 𝒄𝒐𝒔𝟐𝜽
θ  xyA
x'  xy A cos 
(2)
θ θ
x
 xy A sin 

 y A sin 

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 Transformation of Stress

The normal stress acting on the 𝑦’ face can also be obtained from Equation (1) by substituting θ + 90° in place
of, θ giving the following equation:
y' y
x'
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑦 ′ = + 𝑐𝑜𝑠2(900 + 𝜃) + 𝜏𝑥𝑦 𝑠𝑖𝑛2(900 + 𝜃)
2 2 σy' σx'
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
≫ 𝜎𝑦 ′ = + 𝑐𝑜𝑠 1800 + 2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛 1800 + 2𝜃 τx'y'
2 2 τx'y' θ
𝝈𝒙 + 𝝈𝒚 𝝈𝒙 − 𝝈𝒚
θ
≫ 𝝈 𝒚′ = − 𝒄𝒐𝒔𝟐𝜽 − 𝝉𝒙𝒚 𝒔𝒊𝒏𝟐𝜽
x
𝟐 𝟐 τx'y'
(3) τx'y'

Stress Invariance σx' σy'

If the expressions from equations (1) and (3) are added, the following relationship is obtained:

𝜎𝑥′ + 𝜎𝑦′ = 𝜎𝑥 + 𝜎𝑦
Which, shows that the sum of the normal stresses acting on any two orthogonal faces of a plane stress element is a
constant value, independent of the angle, θ . This mathematical characteristic of stress is termed stress invariance.

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Transformation of Stress

Correct angle for stress transformations

Top-Drop-Sweep the clock method
Steps:
1. Start at the topmost corner of the inclined plane.
2. Drop a vertical line from the topmost corner and make a right angle
triangle with the inclined plane as the hypotenuse.
3. Sweep the angle from the vertical line to the hypotenuse.
4. If the angle sweeps in a counterclockwise direction, θ is positive. If the
angle turns in a clockwise direction, θ is negative
y y y
y

σy σy σy σy
12
τxy τxy τxy τxy

x x θ x x
σx σx σx 9
θ 3
σx

Step 1 Step 2 Step 3 Step 4

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 Transformation of Stress

Determine the angle for stress transformations

13.6 ksi 13.6 ksi

70
7 ksi 7 ksi

40

𝜃 = +300 𝜃 = +600 𝜃 = −200 𝜃 = −500

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 Transformation of Stress

Principal Stresses and Maximum Shear Stress

It can be seen that the magnitudes of 𝜎𝑥′ and 𝜏𝑥′ 𝑦′ depend on the angle of inclination 𝜃 of the planes on
which these stresses act. In engineering practice, it is often important to determine the orientation of the
element that causes the normal stress to be a maximum and a minimum and the orientation that causes
the shear stress to be a maximum. In this section each of these problems will be considered.
Principal stresses and principal planes
The maximum and minimum normal stresses at a point are called the principal stresses at that point. The
planes on which the principal stresses act are referred to as the principal planes. To determine the
maximum and minimum normal stress, we must differentiate Eq. 1 with respect to and set the result
equal to zero. This gives
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 ≫ 𝜎𝑥 − 𝜎𝑦 𝑠𝑖𝑛2𝜃 = 2𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃
𝜎𝑥 ′ = + 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2
𝑠𝑖𝑛2𝜃 2𝜏𝑥𝑦
𝑑𝜎𝑥 ′ 𝜎𝑥 − 𝜎𝑦 ≫ =
=0+ 2 −𝑠𝑖𝑛2𝜃 + 2𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃 = 0 cos 2𝜃 𝜎𝑥 − 𝜎𝑦
𝑑𝜃 2
𝛕𝐱𝐲
≫ 0 = − 𝜎𝑥 − 𝜎𝑦 𝑠𝑖𝑛2𝜃 + 2𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃 ≫ 𝐭𝐚𝐧 𝟐𝛉 = 𝛔 − 𝛔
𝐱 𝐲
(4)
𝟐
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 Transformation of Stress

For a given set of stress components 𝜎𝑥 , 𝜎𝑦 , and 𝜏𝑥𝑦 , Eq. (4) has two solutions 2𝜃𝑝1 and 2𝜃𝑝2 , and these two values will
be separated by 180° . Accordingly, the values of 𝜃𝑝 will differ by 90° . From this result, we can conclude that
(a) there will be only two planes where either a maximum or a minimum normal stress occurs, and
(b) these two planes will be 90𝑜 apart (i.e., orthogonal to each other).
Magnitude of Principal Stresses
The maximum normal stress (i.e., the most positive value algebraically) acting at a point is denoted as 𝜎𝑝1 , and
the minimum normal stress (i.e., the most negative value algebraically) is denoted as 𝜎𝑝2 . To obtain these values
by substituting 2𝜃𝑝1 and 2𝜃𝑝2 in Eq.(1). The construction of these triangles is based on Eq.(4) assuming that 𝜏𝑥𝑦
and (𝜎𝑥 − 𝜎𝑦 ) are both positive or both negative quantities. τ xy
𝑡𝑎𝑛 2𝜃𝑝1 = σ − σ
x y
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
2 2
∴ 𝜎𝑝1 = + + 𝜏𝑥𝑦
2 2 σx − σy
𝑐𝑜𝑠 2𝜃𝑝1 = 2
Similarly, for 2𝜃𝑝2
σx − σy 2 2
+ 𝜏𝑥𝑦
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2 2
2
∴ 𝜎𝑝2 = − + 𝜏𝑥𝑦 τxy
2 2 𝑠𝑖𝑛 2𝜃𝑝1 =
σx − σy 2 2
+ 𝜏𝑥𝑦
2

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Transformation of Stress

These two equations can then be combined into a single equation for the two in‐plane principal stresses 𝜎𝑝1 and
𝜎𝑝2 :
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
2
𝜎𝑝1 , 𝜎𝑝2 = ± + 𝜏𝑥𝑦
2 2

Furthermore, if the trigonometric relations for 𝜃𝑝1 or 𝜃𝑝2 are substituted into Eq. 2, it can be seen that
𝜏𝑥′𝑦′ = 0; in other words, no shear stress acts on the principal planes.

The principal stresses determined may both be

positive, may both be negative, or may be of
opposite signs. In naming the principal stresses,
𝜎𝑝1 is the more positive value algebraically (i.e.,
the algebraically larger value). If one or both of
the principal stresses are negative, 𝜎𝑝1 can have a
smaller absolute value than 𝜎𝑝2 .

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Transformation of Stress

Maximum In-Plane Shear Stress

To determine the planes where the maximum in-plane shear stress 𝜏 max occurs, Eq.2 is differentiated with respect
to θ and set equal to zero, yielding

𝜎𝑥 − 𝜎𝑦 Eq. (5) has two solutions 2𝜃𝑠1 and 2𝜃𝑠2 , and these two
𝜏𝑥 ′ 𝑦 ′ =− 𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃 values will be separated by 180° . Accordingly, the values of
2
𝑑 𝜏𝑥 ′ 𝑦 ′ 𝜎𝑥 − 𝜎𝑦 𝜽𝒔 will differ by 𝟗𝟎° . We also note that Eq. (5) is the
=− 2 𝑐𝑜𝑠2𝜃 − 2𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃 negative reciprocal of Eq. (4), meaning that the angles 2𝜃
𝑑𝜃 2
≫ 0 = −(𝜎𝑥 −𝜎𝑦 ) 𝑐𝑜𝑠2𝜃 − 2𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
defined by these two equations differ by 90.
𝜎𝑥 − 𝜎𝑦 We thus conclude that the planes of maximum in-plane
≫ 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃 = − 𝑐𝑜𝑠2𝜃
2 shear stress are inclined at 45 to the principal planes.
𝜎𝑥 − 𝜎𝑦
𝑠𝑖𝑛2𝜃 2 Maximum In-plane Shear Stress magnitude
≫ =−
cos 2𝜃 𝜏𝑥𝑦 A general equation can be derived to give the magnitude of
𝜎𝑥 − 𝜎𝑦 𝜏max by substituting angle functions obtained from Eq. (5)
≫ 𝑡𝑎𝑛 2𝜃 = − 2 into Eq.(2). The result is
𝜏𝑥𝑦

(5)

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Transformation of Stress

𝜎𝑥 − 𝜎𝑦 2
2
𝜏𝑚𝑎𝑥 = + 𝜏𝑥𝑦
2
Similarly, for 2𝜃𝑝2

𝜎𝑥 − 𝜎𝑦 2
2
𝜏𝑚𝑖𝑛 = − + 𝜏𝑥𝑦
2

𝝈𝒙 − 𝝈𝒚 𝟐
𝟐
𝐒𝐨, 𝝉𝒎𝒂𝒙/𝒎𝒊𝒏 = ± + 𝝉𝒙𝒚 𝜎𝑥 − 𝜎𝑦
𝟐
𝑡𝑎𝑛 2𝜃𝑠1 =− 2
The sign of 𝜏 max/min is ambiguous. The maximum shear stress differs 𝜏𝑥𝑦
from the minimum shear stress only in sign. Unlike normal stress, which 𝜎𝑥 − 𝜎𝑦
−
can be either tension or compression, the sign of the maximum in-plane sin 2𝜃𝑠1 = 2
shear stress has no physical significance for the material behavior of a σx − σy 2 2
+ 𝜏𝑥𝑦
stressed body. The sign simply indicates the direction in which the shear 2
stress acts on a particular plane surface. 𝜏𝑥𝑦
𝑐𝑜𝑠 2𝜃𝑠1 =
σx − σy 2 2
+ 𝜏𝑥𝑦
2

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 Transformation of Stress

Normal Stresses on Maximum In-plane Shear Stress Surfaces

After substituting angle functions obtained from Eq. (5) into Eq. (1) and Eq. (2) and simplifying, we find that the
normal stress acting on a plane of maximum in-plane shear stress is
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑥 ′ = + 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2
𝜎𝑥 − 𝜎𝑦
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜏𝑥𝑦 −
≫ 𝜎𝑥 ′ = + + 𝜏𝑥𝑦 2
2 2 σx − σy 2 2 σx − σy 2 2
+ 𝜏𝑥𝑦 + 𝜏𝑥𝑦
2 2
𝝈𝒙 + 𝝈𝒚
∴ 𝝈 𝒙′ =
𝟐
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝒚′ = − 𝑐𝑜𝑠2𝜃 − 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2
𝜎𝑥 − 𝜎𝑦
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜏𝑥𝑦 −
≫ 𝜎𝑦′ = − − 𝜏𝑥𝑦 2
2 2 σx − σy 2 2 σx − σy 2 2
+ 𝜏𝑥𝑦 + 𝜏𝑥𝑦
2 2
𝝈𝒙 + 𝝈𝒚
∴ 𝝈𝒚′ =
𝟐
𝝈𝒙 + 𝝈𝒚
𝝈=
𝟐

Md. Hossain Nadim Dept. of CE, AUST Page 21 of 37

Transformation of Stress

Relation between Principal Stress (𝝈𝟏 and 𝝈𝟐 ) and Maximum/Minimum Shear Stress (𝝉𝒎𝒂𝒙/𝒎𝒊𝒏 )
𝜎1 − 𝜎2
= 𝜏𝑚𝑎𝑥/𝑚𝑖𝑛
2

𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2 𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
2 2
𝜎1 = + + 𝜏𝑥𝑦 𝜎1 = − + 𝜏𝑥𝑦
2 2 2 2

𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2 𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
2 2
𝜎2 = − + 𝜏𝑥𝑦 𝜎2 = + + 𝜏𝑥𝑦
2 2 2 2

𝜎𝑥 − 𝜎𝑦 2 𝜎𝑥 − 𝜎𝑦 2
2 2
𝜏𝑚𝑎𝑥 = + 𝜏𝑥𝑦 𝜏𝑚𝑖𝑛 = − + 𝜏𝑥𝑦
2 2

𝜎𝑥 − 𝜎𝑦 2 𝜎𝑥 − 𝜎𝑦 2
2 2
𝜎1 − 𝜎2 = 2 + 𝜏𝑥𝑦 𝜎1 − 𝜎2 = −2 + 𝜏𝑥𝑦
2 2
𝜎1 − 𝜎2 𝜎1 − 𝜎2
≫ = 𝜏𝑚𝑎𝑥 ≫ = 𝜏𝑚𝑖𝑛
2 2
𝜎1 − 𝜎2
= 𝜏𝑚𝑎𝑥/𝑚𝑖𝑛
2

Md. Hossain Nadim Dept. of CE, AUST Page 22 of 37

Transformation of Stress

Problem:
For the following element as shown in Figure, find the followings :
1) The principal stress and orientation.
2) The maximum & minimum shear stress, orientation and associated normal stress.
3) The transformed stress on the given inclined plane.

8 ksi

𝟐𝟎𝟎
10 ksi

6 ksi

Md. Hossain Nadim Dept. of CE, AUST Page 23 of 37

Transformation of Stress

1) The Principal Stress and Orientation:

𝜎𝑥 = −10 𝑘𝑠𝑖, 𝜎𝑦 = −8 𝑘𝑠𝑖,
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 2
2 𝜏𝑥𝑦 = −6 𝑘𝑠𝑖, 𝜃 = −200
𝜎𝑝1 𝑜𝑟 𝜎𝑝2 = ± + 𝜏𝑥𝑦
2 2 τxy
tan 2𝜃𝑝 = σ − σ
x y
2
−10 − 8 −10 + 8 2
2
= ± + −6 = −9 ± 37
2 2 −6
= =6
−10 + 8
= −𝟐. 𝟗𝟐 𝒌𝒔𝒊, −𝟏𝟓. 𝟎𝟖 𝒌𝒔𝒊 2

Now 𝜃𝑝 = 𝟒𝟎. 𝟐𝟕𝟎 , 𝟏𝟑𝟎. 𝟐𝟕𝟎

𝝈𝒙 + 𝝈𝒚 𝝈𝒙 − 𝝈𝒚
𝝈𝒙′ = + 𝒄𝒐𝒔𝟐𝜽 + 𝝉𝒙𝒚 𝒔𝒊𝒏𝟐𝜽
𝟐 𝟐
𝜽𝒑 = 𝟒𝟎. 𝟐𝟕𝟎 𝝈𝒙′ = −𝟏𝟓. 𝟎𝟖 𝒌𝒔𝒊

Md. Hossain Nadim Dept. of CE, AUST Page 24 of 37

Transformation of Stress

2) Maximum & minimum shear stress, orientation and associated normal stress.

2 𝜎𝑥 − 𝜎𝑦 −10 + 8
𝜎𝑥 − 𝜎𝑦 2
𝜏𝑚𝑎𝑥/𝑚𝑖𝑛 = ± + 𝜏𝑥𝑦 𝑡𝑎𝑛 2𝜃𝑠 = − 2 =− 2
2 𝜏𝑥𝑦 −6
2 = −0.167
−10 + 8 2
=± + −6
2 𝜃𝑠 = −𝟒. 𝟕𝟒𝟎 , 𝟖𝟓. 𝟐𝟔𝟎
−10 + 8
2 Now
=± + −6 2 𝜎𝑥 + 𝜎𝑦 −10 − 8
2 𝜎= = = −𝟗 𝒌𝒔𝒊
2 2
= +𝟔. 𝟎𝟖𝟑 𝒌𝒔𝒊, −𝟔. 𝟎𝟖𝟑 𝒌𝒔𝒊

𝝈𝒙 − 𝝈𝒚
𝝉𝒙′ 𝒚′ =− 𝒔𝒊𝒏𝟐𝜽 + 𝝉𝒙𝒚 𝒄𝒐𝒔𝟐𝜽
𝟐
4.74°
𝒊𝒇 𝜽𝒔 = −𝟒. 𝟕𝟒𝟎 𝒕𝒉𝒆𝒏 𝝉𝒎𝒊𝒏 = −𝟔. 𝟎𝟖𝟑 𝒌𝒔𝒊

Md. Hossain Nadim Dept. of CE, AUST Page 25 of 37

Transformation of Stress

3) The transformed stress on the given inclined plane.

𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑥 ′ = + 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2
−10 − 8 −10 + 8
= + cos 2 × − 200 − 6𝑠𝑖𝑛 2 × − 200
2 2
= −9 − cos 400 + 6𝑠𝑖𝑛 400 = −𝟓. 𝟗𝟏 𝒌𝒔𝒊
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑦 ′ = − 𝑐𝑜𝑠2𝜃 − 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃
2 2
−10 − 8 −10 + 8
= − cos 2 × − 200 + 6𝑠𝑖𝑛 2 × − 200
2 2
= −9 + cos 400 − 6𝑠𝑖𝑛 400 = −𝟏𝟐. 𝟎𝟗 𝒌𝒔𝒊
𝜎𝑥 − 𝜎𝑦
𝜏𝑥 ′ 𝑦 ′ = − 𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃
2
−10 + 8
=− sin −400 − 6 cos −400 = −𝟓. 𝟐𝟒 𝒌𝒔𝒊
2

Md. Hossain Nadim Dept. of CE, AUST Page 26 of 37

Transformation of Stress

Practice Problem 01
A plane-stress condition exists at a point on the surface of a loaded structure, where the
stresses have the magnitudes and directions shown on the stress element of Figure.
Determine the stresses acting on an element that is oriented at a clockwise angle of 15°
with respect to the original element.
Answer

Ans: 𝝈𝒙′ = −𝟑𝟐. 𝟔 𝑴𝒑𝒂, 𝝈𝒚′ = −𝟏. 𝟒 𝑴𝒑𝒂 & 𝝉𝒙′ 𝒚′ = −𝟑𝟏 𝑴𝒑𝒂

Md. Hossain Nadim Dept. of CE, AUST Page 27 of 37

Transformation of Stress

Practice Problem 02
The state of plane stress at a point on a body is represented on the element shown in
Figure. Represent this stress state in terms of the maximum in-plane shear stress and
associated normal stress.

Answer

Md. Hossain Nadim Dept. of CE, AUST Page 28 of 37

Transformation of Stress

Mohr’s Circle for Plane Stress

the transformation equations represent a circle
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑥 ′ = + 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃 (1)
2 2
𝜎𝑥 − 𝜎𝑦
𝜏𝑥 ′ 𝑦 ′ = − 𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃 (2)
2
by squaring both sides of Eq. (1) and (2) and then adding the equations,
2 2
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎𝑥 ′ − + 𝜏𝑥2′ 𝑦 ′ = 2
+ 𝜏𝑥𝑦 (6)
2 2

ത 2 + 𝜏𝑥2′ 𝑦 ′ = 𝑅2
Which can be written more compactly as, (𝜎𝑥 ′ − 𝜎)
𝜎𝑥 + 𝜎𝑦
Where, 𝜎=
2
𝜎𝑥 − 𝜎𝑦 2
2
𝑅= + 𝜏𝑥𝑦
2
We recognize Eq. (6) as the equation of a circle in the 𝜎𝑥′ 𝜏𝑥′𝑦′ −plane. The radius of the circle is 𝑅,
and its center has the coordinates (𝜎,
ത 0) .
Md. Hossain Nadim Dept. of CE, AUST Page 29 of 37
 Transformation of Stress

y τ

r
a R
σ

b
y
𝑥 −𝑎 2 + 𝑦−𝑏 2 = 𝑟2

Md. Hossain Nadim Dept. of CE, AUST Page 30 of 37

Transformation of Stress

Sign convention used in Mohr’s Circle

Normal Stress: Tensile normal stresses are plotted on the right side of the 𝜏 axis, and compressive
normal stresses are plotted on the left side of the axis. In other words, tensile normal stress is plotted
as a positive value (algebraically) and compressive normal stress is plotted as a negative value.

Shear Stress:

• If the shear stress acting on a face of the stress element tends to

rotate the stress element in a clockwise direction, then the shear
stress is plotted above the 𝜎 axis.
• If the shear stress tends to rotate the stress element in a
counterclockwise direction, then the shear stress is plotted below
the 𝜎 axis.

Md. Hossain Nadim Dept. of CE, AUST Page 31 of 37

 Transformation of Stress

Construction of Mohr’s circle 𝝈𝒙 > 𝝈𝒚 > 𝟎

1. Draw a set of axes, with the abscissa representing the
𝝉
normal stress 𝜎 and the ordinate representing shear stress
𝜏.
2. Plot the point labeled ⓧ with coordinates (𝜎𝑥 , −𝜏𝑥𝑦 ) and 𝜎𝑦
𝜎𝑦 ′ ⓨ
the point labeled ⓨ with coordinates (𝜎𝑦 , 𝜏𝑥𝑦 ).
3. Joint the point ⓧ and ⓨ with a straight line, and draw a 𝜏𝑥𝑦
circle with this line as its diameter. 𝜏𝑥 ′ 𝑦 ′

′ 𝒚 ′ axes
𝝈
Transformation to 𝒙 𝜏𝑥 ′ 𝑦 ′
𝜏𝑥𝑦
4. Rotate the diameter ⓧ−ⓨ of Mohr’s circle through the 2𝜃
angle 2𝜃 in the same sense as 𝜃. Label the endpoint of this
new diameter 𝑥 ′ as 𝑦 ′ and as shown in Fig with ⓧ
𝜎
coordinates (𝜎𝑥 ′ , −𝜏𝑥 ′ 𝑦 ′ ), and (𝜎𝑦 ′ ,𝜏𝑥 ′ 𝑦 ′ ) respectively.
𝜎𝑥
𝜎𝑥 ′

Md. Hossain Nadim Dept. of CE, AUST Page 32 of 37

 Transformation of Stress

The Principal Stresses and Maximum In-Plane Shear Stress 𝝈𝒙 > 𝝈𝒚 > 𝟎
𝝉
5. The principal planes (planes of maximum and minimum
normal stress) are labeled ① and ② respectively. Where
𝜎𝑝1 = 𝜎 + 𝑅 and 𝜎𝑝2 = 𝜎 − 𝑅 and their shear stress 𝜎𝑝1
coordinates are zero. The principal directions differ from 𝜎𝑦 ⓐ
𝜎𝑦 ′ ⓨ
the xy- coordinates by the angle 𝜃𝑝 which is represented
by 2𝜃𝑝 on the Mohr’s circle.
𝜏𝑚𝑎𝑥
6. The maximum in plane shear stress 𝜏𝑚𝑎𝑥 equals the ①
𝝈
radius R of the circle. The planes on which 𝜏𝑚𝑎𝑥 acts are 2𝜃𝑝 𝜏𝑥 ′ 𝑦 ′
②
represented by points ⓐ and ⓑ. The 𝝈-coordinates of 𝜏𝑚𝑎𝑥 𝜏𝑥𝑦
𝜎𝑝2 2𝜃
both ⓐ and ⓑ are 𝜎. These points on the circle differ by 2𝜃𝑠
90 from the points representing principal planes.
ⓧ
Therefor, The difference between the planes of maximum 𝜎 ⓑ
shear and principal planes is 45. 𝜎𝑥
𝜎𝑥 ′

Md. Hossain Nadim Dept. of CE, AUST Page 33 of 37

Transformation of Stress

Checking the Mohr’s Circle 𝜎𝑥 − 𝜎𝑦

𝜏𝑥𝑦
From Fig 𝜎 = 𝜎 + 𝑅 cos 2𝜃𝑝 − 2𝜃 𝝉 𝑡𝑎𝑛 2𝜃𝑠 = − 2 𝑡𝑎𝑛 2𝜃𝑝 = 𝜎 − 𝜎
𝑥′ 𝑥 𝑦
𝜏𝑥𝑦
Using, cos 2𝜃𝑝 − 2𝜃 = cos 2𝜃𝑝 cos 2𝜃 + sin 2𝜃𝑝 sin 2𝜃 2
𝜎𝑥 +𝜎𝑦
and substituting 𝜎 = 2 , we obtain 𝜎𝑝1
σ𝑥 +σ𝑦 𝜎𝑦 ⓐ
𝜎𝑥 ′ = + 𝑅 cos 2𝜃𝑝 cos 2𝜃 + sin 2𝜃𝑝 sin 2𝜃 ⓨ
2 𝜎𝑦 ′
𝜏𝑥𝑦 𝜎𝑥 −𝜎𝑦
Again, sin 2𝜃𝑝 = and, cos 2𝜃𝑝 = 𝜏𝑚𝑎𝑥
𝑅 2𝑅
①
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝝈
𝜎𝑥 ′ = + cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃 ② 2𝜃𝑝 𝜏𝑥 ′ 𝑦 ′
2 2
𝜎𝑝2 𝜏𝑚𝑎𝑥 𝜏𝑥𝑦
Similarly, from Fig. 2𝜃
2𝜃𝑠
𝜏𝑥 ′ 𝑦′ = 𝑅 sin 2𝜃𝑝 − 2𝜃
ⓧ
Using, sin 2𝜃1 − 2𝜃 = sin 2𝜃𝑝 cos 2𝜃 − cos 2𝜃𝑝 sin 2𝜃 𝜎 ⓑ
𝜎𝑥 − 𝜎𝑦 𝜎𝑥
𝜏𝑥 ′ 𝑦 ′ = − sin 2𝜃 + 𝜏𝑥𝑦 cos 2𝜃
2 𝜎𝑥 ′

Md. Hossain Nadim Dept. of CE, AUST Page 34 of 37

Transformation of Stress

Problem 01
The state of plane stress at a point with respect to the 𝑥𝑦-axes is shown in Fig. (a). Using Mohr’s circle,
determine (1) the principal stresses and principal planes; (2) the maximum in-plane shear stress; and (3) the
equivalent state of stress with respect to the inclined plane. Show all results on sketches of properly oriented
elements.
y Solution
Given data,
𝜎𝑥 = 40 𝑀𝑃𝑎
20 MPa
𝜎𝑦 = 20 𝑀𝑃𝑎
𝜏𝑥𝑦 = 16 𝑀𝑃𝑎
𝜃 = +50° (C.C.W) 20 MPa

40 MPa x

40 𝜃 40 MPa
16 MPa
40
16 MPa

Md. Hossain Nadim Dept. of CE, AUST Page 35 of 37

 Transformation of Stress

From Fig
𝝉
(𝑀𝑃𝑎)
40+20
𝜎= 2
= 30 MPa 𝝈𝒑𝟏
𝑅 = (10)2 + (16)2
= 18.87 MPa
ⓐ
ⓨ
Part 1 (20, 16)
The principal stresses are 𝜏𝑚𝑎𝑥

𝜎𝑝1 = 𝜎 + 𝑅 = 30 + 18.87 = 48.9 MPa

𝟏𝟎𝟎° ①
𝜎𝑝2 = 𝜎 − 𝑅 = 30 − 18.87 = 11.13 MPa
② 𝝈
The principal directions 2𝜃𝑝 (𝑀𝑃𝑎)
𝝈𝒑𝟐
16 2𝜃𝑠
tan 2𝜃𝑝 =
10
2𝜃𝑝 = 58. 0∘ ⓧ (40, −16)
ⓑ
𝜃𝑝 = 29.0∘

Md. Hossain Nadim Dept. of CE, AUST Page 36 of 37

 Transformation of Stress

Part 2 𝝉
Maximum In−Plane Shear Stress (𝑀𝑃𝑎)
𝜏𝑚𝑎𝑥 = 𝑅 = 18.87 MPa
𝝈𝒑𝟏
Normal stresses, 𝜎 = 30 MPa
Planes of Maximum In-Plane Shear
Stress ⓐ
ⓨ
𝜃𝑠 = 29.0∘ − 45° = −16° (C.W) (20, 16)
Part 3 𝜏𝑚𝑎𝑥
From the geometry of the Mohr’s circle
𝜎𝑥 ′ = 30 + 18.87 cos 100° − 58. 0° 𝟏𝟎𝟎° ①
② 𝝈
= 44 MPa 2𝜃𝑝
𝝈𝒑𝟐 (𝑀𝑃𝑎)
𝜎𝑦 ′ = 30 − 18.87 cos 100° − 58.0° 2𝜃𝑠
= 15.98 MPa
ⓧ (40, −16)
𝜏𝑥 ′ 𝑦 ′ = 18.87 sin 100° − 58.0° ⓑ
= 12.6 MPa

Md. Hossain Nadim Dept. of CE, AUST Page 37 of 37

Transformation of Stress

29.0

40  16 .0

Part 1
Part 2
y'

x'

50 .0

12.63 MPa Part 3

Md. Hossain Nadim Dept. of CE, AUST Page 38 of 37