Mohr’s Circle
ETM 2201
Dr. Farhana Abedin
Mohr’s circle
• Mohr’s circle is named after Christian Otto Mohr
• Mohr’s circle visualizes the change in stress components with rotation of
the coordinate axes
• The transformation equations can be written as an equation of a circle
(𝜎𝑥′ − 𝜎)2 +𝜏𝑥2′ 𝑦′ = 𝑅2
Where, 𝑅 =
𝜎𝑥 −𝜎𝑦 2
2
2
+ 𝜏𝑥𝑦
& 𝜎=
𝜎𝑥 +𝜎𝑦
2
• The center of the Mohr’s circle is given by (𝜎, 0)
• The y-axis of Mohr’s circle represents shear stress, 𝜏 and the x-axis normal
stress, 𝜎
Properties of Mohr’s circle
• Stress on the y face is drawn as
(𝜎𝑦 , 𝜏𝑥𝑦 )
• Stress on the x face is drawn as
(𝜎𝑥 , −𝜏𝑥𝑦 )
• Stress on x face is 180o from the y-face
• The sign of shear stress on the x face
of Mohr’s circle is reversed compared
to the actual sign on the x-face
𝜏
0
𝜎
𝜎
Properties of Mohr’s circle
• The angle between two
diameters on the Mohr’s circle is
twice the transformation angle,
𝜃
 The stress at any x’ face at an angle, 𝜃
from the x face can be obtained by going
2𝜃 around the circle
𝜏
0
𝜎
𝜎
y
Properties of Mohr’s circle
• When the shear stress on the y-face is
clockwise, it is considered positive and
plotted on the Mohr’s circle above the 𝜎axis
𝜏
0
x
𝝉 plotted above 𝝈-axis
y
𝜎
𝜎
x
𝝉 plotted below 𝝈-axis
Properties of Mohr’s circle
• The maximum and minimum stresses on
the 𝜎-axis are the principal stresses
• Principal stresses are 180o apart on the
Mohr’s circle
• The two faces on which the principal
stresses act are actually oriented 90o to
each other
• These planes are called principal planes
and the axes are called principal axes
• Shear stress is zero on principal planes
𝜏
0
𝜎2
𝜎
𝜎
𝜎1
𝜎1 and 𝜎2 are the principal stresses
Construction of the Mohr’s circle
1. Draw the 𝜎 and 𝜏 axes where 𝜎 is the abscissa and 𝜏 as the ordinate
2. Plot the point representing x face with coordinates (𝜎𝑥 , −𝜏𝑥𝑦 ) and the
point representing y face with coordinates (𝜎𝑦 , 𝜏𝑥𝑦 )
3. Join the two points with a straight line and draw a circle with this line as
the diameter
4. You can calculate the center and radius of the circle using the following
equations
𝜎𝑥 − 𝜎𝑦 2
2
𝑅=
+ 𝜏𝑥𝑦
2
𝜎𝑥 + 𝜎𝑦
𝜎=
2
List the given stresses:
𝜎𝑥 = 20 MPa
𝜎𝑦 = -60 MPa
𝜏𝑥𝑦 = -30MPa
Problem 8.49
𝜎 = −20 𝑀𝑃𝑎
• For the state of stress shown (a) draw
the Mohr’s circle (b) determine the
radius R and the coordinate 𝜎 of its
center
𝑅=
402 + 302 = 50 𝑀𝑃𝑎
𝜏 (MPa)
60 MPa
x
30
20
30
10
𝜎
y
-60
x
20 MPa
-20
-10
30
-20
y
30 MPa
-40
0
-30
20
(MPa)
60 MPa
𝜏 (MPa)
−𝜏𝑥 ′ 𝑦′
x’
y
x
30
2𝜃
𝜎𝑥′
-60
20
10
-40
-20
x
30
𝜎𝑦′
0
30
𝜎
20
-10
20 MPa
(MPa)
30 MPa
-20
y
-30
𝜏𝑥 ′ 𝑦′
y’
𝜏𝑥 ′ 𝑦′
y
𝜃
x
𝜏𝑥 ′ 𝑦′
Maximum in-plane shear stress
• Maximum in-plane shear stress is
denoted as 𝜏𝑚𝑎𝑥
• Maximum in-plane shear stress is
equal to the radius of the Mohr’s
circle
𝜎1 − 𝜎2
𝜏𝑚𝑎𝑥 = 𝑅 =
2
• 𝜏𝑚𝑎𝑥 may not be necessarily the
maximum shear stress at a point
• The largest shear stress is called the
absolute maximum shear stress, 𝜏𝑎𝑏𝑠
𝜏
R
0
𝜎
𝜎
y
𝜎2
Absolute maximum shear stress
• Let us consider a state of stress where x
and y axes coincide with the principal
directions
• Three Mohr’s circle in the xy, xz and yz
planes can be drawn
• Maximum in-plane shear stress is the
radius of the Mohr’s circle in the xy plane
• Absolute maximum shear stress is the
radius of the largest Mohr’s circle
𝜎1 − 𝜎2 𝜎1 𝜎2
𝜏𝑎𝑏𝑠 = max(
,
,
)
2
2
2
𝜎1
x
z
y
y
𝜎2
𝜎2
𝜎1
𝜏
𝜎1
2
𝜎1 − 𝜎2
2
𝜎2
𝜎1
x
x
z
𝜎2
2
𝜏
𝜏
𝜎1
z
𝜎
𝜎1
𝜎
𝜎2
𝜎
y
𝜎2
𝜎2
𝜏
𝜎1
𝜎1
x
z
0
𝜎