LECTURE
Third Edition
FAILURE CRITERIA: MOHR’S
CIRCLE AND PRINCIPAL STRESSES
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
22
Chapter
7.4
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
The Stress Transformation
Equations for Plane Stress
„
Slide No. 1
ENES 220 ©Assakkaf
Example 2
The stresses shown in Figure 12a act at a
point on the free surface of a stressed
body. Determine the normal stresses σn
and σt and the shearing stress τnt at this
point if they act on the rotated stress
element shown in Figure 12b.
1
Slide No. 2
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
The Stress Transformation
Equations for Plane Stress
„
Example 2 (cont’d)
Figure 12
n
t
70 MPa
σt
τ nt
40 MPa
10 MPa
σn
150
(a)
(b)
Slide No. 3
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
The Stress Transformation
Equations for Plane Stress
„
Example 2 (cont’d)
The given values are as follows:
σ x = −10 MPa, σ y = −70 MPa, τ xy = +40 MPa
n
θ n = 150 , θ t = 150 + 900 = 1050 t
σt
t
τ nt
n
σn
900
150
150
2
Slide No. 4
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
The Stress Transformation
Equations for Plane Stress
„
Example 2 (cont’d)
Applying Eq. 12 for the given values
σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ
= −10 cos 2 (15) − 70 sin 2 (15) + 2(40) sin (15) cos(15)
σ n = 5.981 MPa = 5.981 MPa (Tension)
σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ
= −10 cos 2 (105) − 70 sin 2 (105) + 2(40) sin (105) cos(105)
σ t = −85.98 MPa = 86 MPa (comprssion)
Slide No. 5
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
The Stress Transformation
Equations for Plane Stress
„
Example 2 (cont’d)
τ = −(σ − σ )sin θ cosθ + τ (cos θ − sin θ )
= −(− 10 − ( −70) )sin(15) cos(15) + 40(cos (15) − sin (15))
2
nt
x
y
2
xy
2
τ nt = 19.64 MPa
2
n
t
σ t = 86 MPa
τ nt = 19.64 MPa
70 MPa
40 MPa
10 MPa
σ n = 5.98 MPa
150
3
Slide No. 6
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses
– The transformation equations (Eq. 12 or
13) provides a means for determining the
normal stress σn and the shearing stress τnt
on different planes through a point O in
stressed body.
– Consider, for example, the state of stress
at a point O of the free surface of a
structure or machine component (Fig. 13).
Slide No. 7
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
y
Principal Stresses
Surfaces perpendicular to z-axis
are stress-free.
x
z
7 ksi
12 ksi
⋅⋅ o
(a)
25 ksi
Figure 13
(b)
4
Slide No. 8
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses
– As the element is rotated through an angle
θ about an axis perpendicular to the stressfree surface, the normal stress σn and the
shearing stressτnt on different planes vary
continuously as shown in Figure 14.
– For design purposes, critical stress at the
point are usually the maximum tensile (or
compressive) and shearing stresses.
Slide No. 9
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses
– The principal stresses are the maximum
normal stress σmax and minimum normal
stress σmin.
– In general, these maximum and minimum
or principal stresses can be determined by
plotting curves similar to those of Fig. 14.
– But this process is time-consuming, and
therefore, general methods are needed.
5
Slide No. 10
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Variation of Stresses as Functions of θ
σ
40
n sn
Stress
Stress
τ tnt
Stress (Ksi)
7 ksi
nt
30
20
t
10
y
θ
0
0
60
120
180
240
300
12 ksi
n
25 ksi
x
360
-10
-20
Angle θ (Degrees)
Figure 14
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 11
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses
– Principal Stresses for Special Loading
Conditions:
• Bar under axial load
P
and
σ max=
A
τ max =
P
2A
(14)
• Shaft under Pure Bending
σ max = τ max =
Tmax c
J
(15)
6
Slide No. 12
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
P
Principal Stresses for Axially Loaded
Bar
Figure 15
Inclined Area, An
Original Area, A
θ
P
F
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 13
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
P
Principal Stresses for Axially Loaded
Bar
Figure 16a
N
θ
θ
N = P cosθ
V = P sin θ
A
An =
cosθ
θ
σn =
V
N P cosθ P
P
(1 + cos 2θ )
=
= cos 2 θ =
A
An
A
2A
cosθ
7
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 14
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
P
Principal Stresses for Axially Loaded
Bar
Figure 16b
N
θ
θ
N = P cosθ
V = P sin θ
A
An =
cosθ
θ
τn =
V
V
P sin θ P
P
=
= sin θ cosθ =
sin 2θ
A
An
A
2A
cosθ
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 15
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses for Axially Loaded
Bar
σn is maximum when θ = 00 or 1800
τn is maximum when θ = 450 or 1350
– Also
τ max =
σ max
(16)
2
Therefore
σ max=
P
A
and
τ max =
P
2A
(17)
8
Slide No. 16
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
Principal Stresses for Shaft under Pure
Fig.17 Torsion
y
„
t
y
A
(c)
n
τx y dA cos α
τn t dA
α
α
σn dA
x
y
τ xy
τyx dA sin α
x
τ yx
(a)
τ xy
α
τ yx
x
(b)
Slide No. 17
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses for Shaft under Pure
Torsion
σ n = 2τ xy sin α cos α = τ xy sin 2α
τ nt = τ xy (cos 2 α − sin 2 α ) = τ xy cos 2α
σ max = τ max =
Tmax c
J
(18)
(19)
(20)
τx y dA cos α
t
y
τn t dA
α
α
n
σn dA
τyx dA sin α
9
Slide No. 18
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Development of Principal Stresses
Equations
Recall Eq 13
σn =
σ x +σ y
τ nt = −
2
+
σ x −σ y
σ x −σ y
2
2
(13a)
cos 2θ + τ xy sin 2θ
sin 2θ + τ xy cos 2θ
(13b)
Slide No. 19
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Development of Principal Stresses
Equations
Differentiating the first equation with
respect to θ and equate the result to zero,
gives
or

dσ n
d σ x + σ y σ x − σ y
=
+
cos 2θ + τ xy sin 2θ 

dθ dθ  2
2

set
= −(σ x − σ y )sin 2θ + 2τ xy cos 2θ = 0
tan 2θ p =
2τ xy
σ x −σ y
 2τ xy 

or 2θ p = tan −1 
 σ −σ 
x
y


(21)
10
Slide No. 20
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Development of Principal Stresses
Equations
Substituting the expression for 2θp into Eq.
13a, yields
σ p1, p 2 =
σ x +σ y
2
2
 σ −σ y 
 + τ xy2
±  x
 2 
(22)
Eq. 21 gives the two principal stresses in the
xy-plane, and the third stress σp3 = σz = 0.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 21
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Principal Stresses
Principal stresses σmax and σmin can be
computed from
σ p1, p 2 =
σ x +σ y
2
σ x −σ y
± 
 2
2

 + τ xy2

(22a)
where subscript p refers to the planes of
maximum and minimum values of σn.
11
Slide No. 22
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Location of the Plane of Principal
Stresses
 2τ xy 
1

θ p = tan −1 

2
σ x −σ y 
(22b)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 23
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Notes on Principal Stresses Equation
1. Eq. 22 gives the angle θp and θp + 900
between x-plane (or y-plane) and the
mutually perpendicular planes on which
the principal stresses act.
2. When tan 2θp is positive, θp is positive,
and the rotation is counterclockwise from
the x- and y-planes to the planes on
which the two principal stresses act.
12
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 24
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
Notes on Principal Stresses Equation
„
3. When tan 2θp is negative, θp is negative,
and the rotation is clockwise.
4. The shearing stress is zero on planes
experiencing maximum and minimum
values of normal stresses.
5. If one or both of the principal stresses
from Eq.22 is negative, the algebraic
maximum stress can have a smaller
absolute value than the minimum stress.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 25
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Development of Maximum Shearing
Stress Equation
Recall Eq 13b: τ nt = −
σ x −σ y
2
sin 2θ + τ xy cos 2θ
 σ x −σ y

sin 2θ + τ xy cos 2θ 
−
2


set
= −(σ x − σ y )cos 2θ − 2τ xy sin 2θ = 0
dτ nt
d
=
dθ
dθ
or
tan 2θ τ = −
σ x −σ y
2τ xy
 σ x −σ y 

or 2θτ = tan −1 

 2τ
xy


(23)
13
Slide No. 26
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Development of Principal Shearing
Stress Equation
Substituting the expression for 2θτ into Eq.
13b, yields
2
 σ −σ y 
 + τ xy2
τ p = ±  x
 2 
(24)
Eq. 24 gives the maximum in-plane
shearing stress.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 27
Principal Stresses and Maximum
Shearing Stress
ENES 220 ©Assakkaf
„
Maximum In-Plane Shearing Stress
Maximum in-plane shearing stress can be
computed from
2
 σ −σ y 
 + τ xy2
τ p = ±  x
 2 
(24a)
where the subscript p refers to the plane of
maximum in-plane shearing stress τp.
14
Slide No. 28
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Location of the Plane of Maximum
Shearing Stress
σ −σ y 
1

θ τ = tan −1  x

τ
2
2
xy


(24b)
Slide No. 29
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Notes on Principal Stresses and
Maximum In-Plane Shearing Stress
Equation
1. The two angles 2θp and 2θτ differ by 900,
therefore, θp and θτ are 450 apart.
2. This means that the planes in which the
maximum in-plane shearing stress occur
are 450 from the principal planes.
15
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 30
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Notes on Principal Stresses and
Maximum In-Plane Shearing Stress
Equation
3. The direction of the maximum shearing
stress can be determined by drawing a
wedge-shaped block with two sides parallel
to the planes having the maximum and
minimum principal stresses, and with the
third side at an angle of 450. The direction of
the maximum shearing stress must oppose
the larger of the two principal stresses.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 31
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
y
σy
θp
Fig.18 Wedge-shaped Block
τ xy
σ p1 > σ P 2
σx
θp
σn
x
τ max
σ p1
σn
450
σ p1
0
45
σ p2
τ max
σ p2
16
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 32
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Useful Relationships
• The maximum value of τnt is equal to one half
the difference between the two in-plane
principal stresses, that is
τp =
σ p1 − σ p 2
(25)
2
• For plane stress, the sum of the normal
stresses on any two orthogonal planes through
a point in a body is a constant or in invariant.
σ p1 + σ p 2 = σ x + σ y
(26)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 33
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Useful Relationships
– When a state of plane exists, one of the
principal stresses is zero.
– If the values of σp1 and σp2 from Eq. 25
have the same sign, then the third principal
stress σp3 equals zero, will be either the
maximum or minimum normal stresses.
– Three possibilities:
max = (σ p1 − σ p 2 ) / 2,
max = (σ p1 − 0 ) / 2,
max = (0 − σ p 2 ) / 2
17
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 34
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 3
Normal and shearing stresses on
horizontal and vertical planes through a
point in a structural member subjected to
plane stress are shown in Figure 19.
Determine and show on a sketch the
principal and maximum shearing stresses.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 35
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 3 (cont’d)
Fig.19
4 ksi
12 ksi
The given values for use
in Eqs. 22 and 24 are:
σx = +12 ksi
σy = - 4 ksi
τxy = - 6 ksi
6 ksi
18
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 36
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 3 (cont’d)
Using Eq. 22a for the given values:
σp =
σ x +σ y
2
2
 σ x −σ y 
 + τ xy2
± 
 2 
12 + (−4)
 12 − (− 4) 
2
± 
 + (− 6) = 4 ± 10
2
2


2
=
Therefore,
σ P1 = 4 + 10 = +14 ksi = 14 ksi (T)
σ p 2 = 4 − 10 = −6 ksi = 6 ksi (C)
σ p3 = σ z = 0
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 37
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 3 (cont’d)
Since the σp1 and σp2 have opposite sign,
the maximum shearing stress is
τ max =
σ p1 − σ p 2
2
=
14 − (− 6) 20
=
= +10 ksi
2
2
The location θp of the principal stresses is
computed from Eq. 22b
1
2
 2(− 6) 
 = 18.430
 12 x − (− 4) 
y 

θ p = tan −1 
19
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 38
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
y
Example 3 (cont’d)
Sketch of principal and max
shearing stresses
4 ksi
6 ksi
σn =
12 ksi
x
6 ksi
σ x +σ y
2
=
12 − 4
= 4 ksi (T)
2
4 ksi
18.430
10 ksi
450
τ p = 10 ksi = τ max
14 ksi
6 ksi
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 39
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 4
Normal and shearing stresses on
horizontal and vertical planes through a
point in a structural member subjected to
plane stress are shown in Figure 20.
Determine and show on a sketch the
principal and maximum shearing stresses
20
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 40
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 4 (cont’d)
Fig.20
36 MPa
72 MPa
The given values for use
in Eqs. 22 and 24 are:
σx = +72 MPa
σy = +36 MPa
τxy = - 24 MPa
24 MPa
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 41
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 4 (cont’d)
Using Eq. 22a for the given values:
σp =
σ x +σ y
2
2
 σ x −σ y 
 + τ xy2
± 
 2 
72 + (+36)
 72 − (+ 36) 
2
± 
 + (− 24 ) = 54 ± 30
2
2


2
=
Therefore,
σ P1 = 54 + 30 = +84 ksi = 84 MPa (T)
σ p 2 = 54 − 30 = +24 ksi = 24 MPa (T)
σ p3 = σ z = 0
21
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 42
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 4 (cont’d)
Since the σp1 and σp2 have the same sign,
the maximum shearing stress is
τ max =
σ p1 − 0
2
=
84 − 0 84
=
= +42 MPa
2
2
The location θp of the principal stresses is
computed from Eq. 22b
 2τ xy  1
1
 = tan −1  2(−24)  = −26.57 0
θ p = tan −1 

2
 72 − 36 
 σ x −σ y  2
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 43
ENES 220 ©Assakkaf
Principal Stresses and Maximum
Shearing Stress
„
Example 4 (cont’d)
y 36 MPa
6 ksi
Sketch of principal and max
42 MPa
shearing stresses 42 MPa
450
72 MPa
x
24 MPa
τp =
τ max
2
= 42 MPa
σ p3 = 0
26.57
30 MPa
τ p ≠ τ max
σ p1 − σ p 2
84 MPa
54 MPa
0
84 − 24
=
= 30 MPa
2
450
24 MPa
84 MPa
σn =
σ x +σ y
2
=
72 + 36
= 54 MPa (T)
2
22
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 44
ENES 220 ©Assakkaf
Introduction
– Mohr’s circle is a pictorial or graphical
interpretation of the transformation
equations for plane stress.
– The process involves the construction of a
circle in such a manner that the
coordinates of each point on the circle
represent the normal and shearing
stresses on one plane through the stressed
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Maximum Shearing Stress
Slide No. 45
ENES 220 ©Assakkaf
Maximum shearing stress occurs for
σ x′ = σ ave
2
σ x −σ y 
2
 + τ xy
2


τ max = R = 
tan 2θ s = −
σ x −σ y
2τ xy
Note : defines two angles separated by 90o and
offset from θ p by 45o
σ ′ = σ ave =
σ x +σ y
2
23
Slide No. 46
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
„
Introduction
Point, and the angular position of the
radius to the point gives the orientation of
the plane.
– The proof that normal and shearing
components of stress on arbitrary plane
through a point can be represented as
points on a circle follows from Eqs. 13a
and 13b.
Slide No. 47
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
„
Plane Stresses using Mohr’s Circle
• Recall Eqs. 13a and 13b,
σn =
σ x +σ y
τ nt = −
2
+
σ x −σ y
σ x −σ y
2
cos 2θ + τ xy sin 2θ
(13a)
sin 2θ + τ xy cos 2θ
(13a)
2
• Squaring both equations, adding, and
simplifying gives
2
2
σ +σ y 

 σ −σ y 
 σ n − x
 + τ 2nt =  x
 + τ xy2
2
2




(27)
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 48
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
– The previous equation is indeed an
equation of a circle in terms of the variable
σn and τnt. The circle is centered on the the
σ axis at a distance (σx - σy)/2 from the τ
axis, and the radius of the circle is given by
2
 σ x −σ y 
 + τ xy2
R = 
 2 
(28)
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 49
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
– Normal stresses are plotted as horizontal
coordinates, with tensile stresses (positive)
plotted to the right of the origin and
compressive stresses (negative) plotted to
the left.
– Shearing stresses are plotted as vertical
coordinates, with those tending to produce
a clockwise rotation of the stress element
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
Slide No. 50
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
„
plotted above the σ-axis, and those tending
to produce counterclockwise rotation of the
stress element plotted below the σ-axis.
– Sign conventions for interpreting the
normal and shearing stresses will be
provided, and illustrated through examples.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 51
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
• Mohr’s circle for any point subjected to plane
stress can be drawn when stresses on two
mutually perpendicular planes through the point
are known.
σy
τ xy
A
θ
σx
τ yx
t
τ xy
σx
A
y
n
θ
x
τ yx
σy
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
τ
Mohr’s Circle Fig.21
σy
H (σ y ,τ yx )
σ p2
τp
τ yx
2θ p
C
σ x −σ y
2
τ nt
2θ
σx
σ p1
σ
τ xy
2
σ x +σ y
V (σ x ,τ xy )
σn
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 52
ENES 220 ©Assakkaf
Slide No. 53
ENES 220 ©Assakkaf
Mohr’s Circle
Maximum shearing stress occurs for
σ x′ = σ ave
σ x −σ y
 2
τ max = R = 
tan 2θ s = −
2

 + τ xy2

σ x −σ y
2τ xy
Note : defines two angles separated by 90o and
offset from θ p by 45o
σ ′ = σ ave =
σ x +σ y
2
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 54
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
Drawing Procedure for Mohr’s Circle
1. Choose a set of x-y coordinate axes.
2. Identify the stresses σx, σy and τxy = τyx and list
them with proper sign.
3. Draw a set of στ-coordinate axes with σ and τ
positive to the right and upward, respectively.
4. Plot the point (σx, -τxy) and label it point V
(vertical plane).
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 55
ENES 220 ©Assakkaf
Plane Stresses using Mohr’s Circle
Drawing Procedure for Mohr’s Circle (cont’d)
5. Plot the point (σy, τyx) and label it point H
(horizontal plane).
6. Draw a line between V and H. This establishes
the center and the radius R of Mohr’s circle.
7. Draw the circle.
8. An extension of the radius between C and V
can be identified as the x-axis or reference line
for the angle measurements (I.e., θ =0).
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 56
ENES 220 ©Assakkaf
Sign Conventions
– In a given face of the stressed element, the
shearing stresses that tends to rotate the
element clockwise will be plotted above the
σ-axis in the circle.
– In a given face of the stressed element, the
shearing stresses that tends to rotate the
element counterclockwise will be plotted
below the σ-axis in the circle.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Sign Conventions
σ
•
τ
The following jingle may
be helpful in remembering
this conventions:
“In the kitchen, the clock
is above, and the counter
is below.”
Beer and Johnston (1992)
τ
Slide No. 57
ENES 220 ©Assakkaf
Fig.22
σ
(a) Clockwise → Above
τ
•
σ
τ
σ
(b) Counterclockwise → Below
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Slide No. 58
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
„
Points of Interests on Mohr’s Circle
1. Point D that provides the principal stress
σp1.
2. Point E that gives the principal stress σp2.
3. Point A that provides the maximum inplane shearing stress -τp and the
accompanied normal stress σavg that acts
on the plane.
Slide No. 59
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
• With the physical significance of Mohr’s circle
for plane stress established, it may be applied
with simple geometric considerations. Critical
values are estimated graphically or calculated.
• For a known state of plane stress σ x , σ y ,τ xy
plot the points X and Y and construct the
circle centered at C.
σ ave =
σ x +σ y
2
2
σ x −σ y 
2
 + τ xy
R = 
2


• The principal stresses are obtained at A and B.
σ max,min = σ ave ± R
tan 2θ p =
2τ xy
σ x −σ y
The direction of rotation of Ox to Oa is
the same as CX to CA.
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 60
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
• With Mohr’s circle uniquely defined, the state
of stress at other axes orientations may be
depicted.
• For the state of stress at an angle θ with
respect to the xy axes, construct a new
diameter X’Y’ at an angle 2θ with respect to
XY.
• Normal and shear stresses are obtained
from the coordinates X’Y’.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 61
ENES 220 ©Assakkaf
Mohr’s Circle for Plane Stress
• Mohr’s circle for centric axial loading:
σx =
P
, σ y = τ xy = 0
A
σ x = σ y = τ xy =
P
2A
• Mohr’s circle for torsional loading:
σ x = σ y = 0 τ xy =
Tc
J
σx =σy =
Tc
τ xy = 0
J
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 62
ENES 220 ©Assakkaf
Example 5
The stresses shown in Figure 23 act at a point
on the free surface of a stressed body. Use
Mohr’s circle to determine the normal and
shearing stresses at this point on the inclined
plane AB shown in the figure.
14 ksi
5
Fig.23
12
B
20 ksi
A
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 63
ENES 220 ©Assakkaf
Example 5 (cont’d)
The given values for use in
drawing Mohr’s circle are:
τ
C
σ x = σ p1 = 20 ksi
τ nt
σ y = σ p 2 = 14 ksi
σ z = σ p3 = 0
 − 12 
0
2θ = 2 tan −1 
 = −134.76
 5 
R=3
σ
45.24
σn
R=3
134.76 0
20 + 14
= 17 ksi
2
20 − 14
τ nt = R sin (45.24 ) = 3 sin (45.24 ) = 2.13 ksi
R = radius =
=3
2
σ n = C − R cos (45.24) = 17 − 3(0.7) = 14.89 ksi
C=
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LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 64
ENES 220 ©Assakkaf
Example 6
For the state of plane stress shown,
(a) construct Mohr’s circle,
determine (b) the principal planes,
(c) the principal stresses, (d) the
maximum shearing stress and the
corresponding normal stress.
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Mohr’s Circle for Plane Stress
„
Slide No. 65
ENES 220 ©Assakkaf
Example 6 (cont’d)
SOLUTION:
• Construction of Mohr’s circle
(50) + (− 10) = 20 MPa
=
2
2
CF = 50 − 20 = 30 MPa FX = 40 MPa
σ ave =
σx +σ y
R = CX =
(30)2 + (40)2 = 50 MPa
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Slide No. 66
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
ENES 220 ©Assakkaf
Example 6 (cont’d)
• Principal planes and stresses
σ max = OA = OC + CA = 20 + 50
σ max = 70 MPa
σ max = OB = OC − BC = 20 − 50
σ max = −30 MPa
FX 40
=
CP 30
2θ p = 53.1°
tan 2θ p =
θ p = 26.6°
LECTURE 22. FAILURE CRITERIA: MOHR’S CIRCLE AND PRINCIPAL STRESSES (7.4)
Slide No. 67
ENES 220 ©Assakkaf
Example 6 (cont’d)
• Maximum shear stress
θ s = θ p + 45°
τ max = R
σ ′ = σ ave
θ s = 71.6°
τ max = 50 MPa
σ ′ = 20 MPa
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